p-Adic Analysis and Lie Groups

Peter Schneider

Course in Munster, winter term 2007/08

Contents

I Foundations 1

1 Ultrametric spaces 1

2 Nonarchimedean fields 6

3 Convergent series 12

4 Differentiability 15

5 Power series 22

6 Locally analytic functions 36

II Manifolds 41

7 Charts and atlases 41

8 Manifolds 43

9 The tangent space 52

10 The topological vector space Can(M,E), part 1 70

11 Locally convex K-vector spaces 75

12 The topological vector space Can(M,E), part 2 79

III Lie groups 84

13 Definitions and foundations 84

14 The universal enveloping algebra 96

15 The concept of free algebras 101

16 The Campbell-Hausdorff formula 106

17 The convergence of the Hausdorff series 119

18 Formal group laws 127

References 142

Part I

Foundations

1 Ultrametric spaces

We begin by establishing some very basic and elementary notions.

Definition. A metric space (X, d) is called ultrametric if the strict triangleinequality

d(x, z) ≤ max(d(x, y), d(y, z)) for any x, y, z ∈ X

is satisfied.

Examples will be given later on.

Remark. i. If (X, d) is ultrametric then (Y, d |Y × Y ), for any subsetY ⊆ X, is ultrametric as well.

ii. If (X1, d1), . . . , (Xm, dm) are ultrametric spaces then the cartesian prod-uct X1 × . . .×Xm is ultrametric with respect to

d((x1, . . . , xm), (y1, . . . , ym)) := max(d1(x1, y1), . . . , dm(xm, ym)) .

Let (X, d) be an ultrametric space in the following.

Lemma 1.1. For any three points x, y, z ∈ X such that d(x, y) 6= d(y, z) wehave

d(x, z) = max(d(x, y), d(y, z)) .

Proof. We may assume that d(x, y) < d(y, z). Then

d(x, y) < d(y, z) ≤ max(d(y, x), d(x, z)) = max(d(x, y), d(x, z)) .

The maximum in question therefore necessarily is equal to d(x, z) so that

d(x, y) < d(y, z) ≤ d(x, z) .

We deduce that

d(x, z) ≤ max(d(x, y), d(y, z)) ≤ d(x, z) .

1

Let a ∈ X be a point and ε > 0 be a positive real number. We call

Bε(a) := {x ∈ X : d(a, x) ≤ ε}

the closed ball and

B−ε (a) := {x ∈ X : d(a, x) < ε}

the open ball around a of radius ε. Any subset in X of one of these two kindsis simply referred to as a ball. As the following facts show this language hasto be used with some care.

Lemma 1.2. i. Every ball is open and closed in X.

ii. For b ∈ Bε(a), resp. b ∈ B−ε (a), we have Bε(b) = Bε(a), resp. B−ε (b) =B−ε (a).

Proof. Obviously B−ε (a) is open and Bε(a) is closed in X. We first considerthe equivalence relation x ∼ y on X defined by d(x, y) < ε. The corre-sponding equivalence class of b is equal to B−ε (b) and hence is open. Sinceequivalence classes are disjoint or equal this implies B−ε (b) = B−ε (a) when-ever b ∈ B−ε (a). It also shows that B−ε (a) as the complement of the otheropen equivalence classes is closed in X.

Analogously we may consider the equivalence relation x ≈ y onX definedby d(x, y) ≤ ε. Its equivalence classes are the closed balls Bε(b), and weobtain in the same way as before the assertion ii. for closed balls. It remainsto show that Bε(a) is open in X. But by what we have established alreadywith any point b ∈ Bε(a) its open neighbourhood B−ε (b) is contained inBε(b) = Bε(a).

The assertion ii. in the above lemma can be viewed as saying that anypoint of a ball can serve as its midpoint. By way of an example we will seelater on that also the notion of a radius is not well determined.

Lemma 1.3. For any two balls B and B′ in X such that B ∩ B′ 6= ∅ wehave B ⊆ B′ or B′ ⊆ B.

Proof. Pick a point a ∈ B ∩ B′. As a consequence of Lemma 1.2.ii. thefollowing four cases have to be distinguished:

1. B = B−ε (a), B′ = B−δ (a),

2. B = B−ε (a), B′ = Bδ(a),

2

3. B = Bε(a), B′ = B−δ (a),

4. B = Bε(a), B′ = Bδ(a).

Without loss of generality we may assume that ε ≤ δ. In cases 1, 2, and 4we then obviously have B ⊆ B′. In case 3 we obtain B ⊆ B′ if ε < δ andB′ ⊆ B if ε = δ.

Remark. If the ultrametric space X is connected then it is empty or consistsof one point.

Proof. Assuming that X is nonempty we pick a point a ∈ X. Lemma 1.2.i.then implies that X = Bε(a) for any ε > 0 and hence that X = {a}.

Lemma 1.4. Let U =⋃i∈I Ui be a covering of an open subset U ⊆ X by

open subsets Ui ⊆ X; moreover let ε1 > ε2 > . . . > 0 be a strictly descendingsequence of positive real numbers which converges to zero; then there is adecomposition

U =⋃j∈J

Bj

of U into pairwise disjoint balls Bj such that:

(a) Bj = Bεn(j)(aj) for appropriate aj ∈ X and n(j) ∈ N,

(b) Bj ⊆ Ui(j) for some i(j) ∈ I.

Proof. For a ∈ U we put

n(a) := min{n ∈ N : Bεn(a) ⊆ Ui for some i ∈ I} .

The family of balls J := {Bεn(a)(a) : a ∈ U} by construction has the proper-

ties (a) and (b) and covers U (observe that for any point a in the open setUi we find some sufficiently big n ∈ N such that Bεn(a) ⊆ Ui). The balls inthis family indeed are pairwise disjoint: Suppose that

Bεn(a1)(a1) ∩Bεn(a2)

6= ∅ .

By Lemma 1.3 we may assume that Bεn(a1)(a1) ⊆ Bεn(a2)

(a2). But thenLemma 1.2.ii. implies that Bεn(a2)

(a1) = Bεn(a2)(a2) and hence Bεn(a1)

(a1) ⊆Bεn(a2)

(a1). Due to the minimality of n(a1) we must have n(a1) ≤ n(a2),resp. εn(a1) ≥ εn(a2). It follows that Bεn(a1)

(a1) = Bεn(a2)(a1) = Bεn(a2)

(a2).

3

As usual the metric space X is called complete if every Cauchy sequencein X is convergent.

Lemma 1.5. A sequence (xn)n∈N in X is a Cauchy sequence if and only iflimn→∞ d(xn, xn+1) = 0.

For a subset A ⊆ X we call

d(A) := sup{d(x, y) : x, y ∈ A}

the diameter of A.

Lemma 1.6. Let B ⊆ X be a ball with ε := d(B) > 0 and pick any pointa ∈ B; we then have B = B−ε (a) or B = Bε(a).

Proof. The inclusion B ⊆ Bε(a) is obvious. By Lemma 1.2.ii. the ball B isof the form B = B−δ (a) or B = Bδ(a). The strict triangle inequality thenimplies ε = d(B) ≤ δ. If ε = δ there is nothing further to prove. If ε < δ wehave B ⊆ Bε(a) ⊆ B−δ (a) ⊆ B and hence B = Bε(a).

Let us consider a descending sequence of balls

B1 ⊇ B2 ⊇ . . . ⊇ Bn ⊇ . . .

in X. If X is complete and if limn→∞ d(Bn) = 0 then we claim that⋂n∈N

Bn 6= ∅ .

If we pick points xn ∈ Bn then (xn)n∈N is a Cauchy sequence. Put x :=limn→∞ xn. Since each Bn is closed we must have x ∈ Bn and thereforex ∈

⋂nBn.

Without the condition on the diameters the intersection⋂nBn can be

empty (compare the exercise further below). This motivates the followingdefinition.

Definition. The ultrametric space (X, d) is called spherically complete ifany descending sequence of balls B1 ⊇ B2 ⊇ . . . in X has a nonemptyintersection.

Lemma 1.7. i. If X is spherically complete then it is complete.

ii. Suppose that X is complete; if 0 is the only accumulation point of theset d(X × X) ⊆ R+ of values of the metric d then X is sphericallycomplete.

4

Proof. i. Let (xn)n∈N be any Cauchy sequence in X. We may assume thatthis sequence does not become constant after finitely many steps. Then the

εn := max{d(xm, xm+1) : m ≥ n}

are strictly positive real numbers satisfying εn ≥ εn+1 and εn ≥ d(xn, xn+1).Using Lemma 1.2.ii. we obtain Bεn(xn) = Bεn(xn+1) ⊇ Bεn+1(xn+1). Byassumption the intersection

⋂nBεn(xn) must contain a point x. We have

d(x, xn) ≤ εn for any n ∈ N. Since the sequence (εn)n converges to zero thisimplies that x = limx→∞ xn.

ii. Let B1 ⊇ B2 ⊇ . . . be any decreasing sequence of balls in X. Obviouslywe have d(B1) ≥ d(B2) ≥ . . . By our above discussion we only need toconsider the case that infn d(Bn) > 0. Our assumption on accumulationpoints implies that d(Bn) ∈ D(X × X) for any n ∈ N and then in factthat the sequence (d(Bn))n must become constant after finitely many steps.Hence there exists an m ∈ N such that 0 < ε := d(Bm) = d(Bm+1) = . . . ByLemma 1.6 we have, for any n ≥ m and any a ∈ Bn, that

Bn = B−ε (a) or Bn = Bε(a) .

Moreover, which of the two equations holds is independent of the choiceof a by Lemma 1.2.ii. Case 1: We have Bn = Bε(a) for any n ≥ m andany a ∈ Bn. It immediately follows that Bn = Bm for any n ≥ m andhence that

⋂nBn = Bm. Case 2: There is an ` ≥ m such that B` = B−ε (a)

for any a ∈ B`. For any n ≥ ` and any a ∈ Bn ⊆ B` we then obtainB−ε (a) = B` ⊇ Bn ⊇ B−ε (a) so that B` = Bn and hence

⋂nBn = B`.

Exercise. Suppose that X is complete, and let B1 ⊃ B2 ⊃ . . . be a decreasingsequence of balls in X such that d(B1) > d(B2) > . . . and infn d(Bn) > 0.Then the subspace Y := X\(

⋂nBn) is complete but not spherically complete.

Lemma 1.8. Suppose that X is spherically complete; for any family (Bi)i∈Iof closed balls in X such that Bi ∩ Bj 6= ∅ for any i, j ∈ I we then have⋂i∈I Bi 6= ∅.

Proof. We choose a sequence (in)n∈N of indices in I such that:

- d(Bi1) ≥ d(Bi2) ≥ . . . ≥ d(Bin) ≥ . . .,

- for any i ∈ I there is an n ∈ N with d(Bi) ≥ d(Bin).

The proof of Lemma 1.6 shows that Bi = Bd(Bi)(a) for any a ∈ Bi. Ourassumption on the family (Bi)i therefore implies that:

5

- Bi1 ⊇ Bi2 ⊇ . . . ⊇ Bin ⊇ . . .,

- for any i ∈ I there is an n ∈ N with Bi ⊇ Bin .

We see that ⋂i∈I

Bi =⋂n∈N

Bin 6= ∅ .

2 Nonarchimedean fields

Let K be any field.

Definition. A nonarchimedean absolute value on K is a function

| | : K −→ R

which satisfies:

(i) |a| ≥ 0,

(ii) |a| = 0 if and only if a = 0,

(iii) |ab| = |a| · |b|,

(iv) |a+ b| ≤ max(|a|, |b|).

Exercise. i. |n · 1| ≤ 1 for any n ∈ Z.

ii. | | : K× −→ R×+ is a homomorphism of groups; in particular, |1| =| − 1| = 1.

iii. K is an ultrametric space with respect to the metric d(a, b) := |b− a|;in particular, we have |a+ b| = max(|a|, |b|) whenever |a| 6= |b|.

iv. Addition and multiplication on the ultrametric space K are continuousmaps.

Definition. A nonarchimedean field (K, | |) is a field K equipped with anonarchimedean absolute value | | such that:

(i) | | is non-trivial, i. e., there is an a ∈ K with |a| 6= 0, 1,

(ii) K is complete with respect to the metric d(a, b) := |b− a|.

6

The most important class of examples is constructed as follows. We fixa prime number p. Then

|a|p := p−r if a = pr mn with r,m, n ∈ Z and p 6 |mn

is a nonarchimedean absolute value on the field Q of rational numbers. Thecorresponding completion Qp is called the field of p-adic numbers. Of course,it is nonarchimedean as well. We note that |Qp|p = pZ ∪ {0}. Hence Qp isspherically complete by Lemma 1.7.ii. On the other hand we see that in theultrametric space Qp we can have Bε(a) = Bδ(a) even if ε 6= δ. To havemore examples we state without proof (compare [Se1] Chap. II §§1-2) thefollowing fact. Let K/Qp be any finite extension of fields. Then

|a| := [K:Qp]√|NormK/Qp(a)|p

is the unique extension of | |p to a nonarchimedean absolute value on K.The corresponding ultrametric space K is complete and spherically completeand, in fact, locally compact.

In the following we fix a nonarchimedean field (K, | |). By the stricttriangle inequality the closed unit ball

oK := B1(0)

is a subring of K, called the ring of integers in K, and the open unit ball

mK := B−1 (0)

is an ideal in oK . Because of o×K = oK \mK this ideal mK is the only maximalideal of oK . The field oK/mK is called the residue class field of K.

Exercise 2.1. i. If the residue class field oK/mK has characteristic zerothen K has characteristic zero as well and we have |a| = 1 for anynonzero a ∈ Q ⊆ K.

ii. If K has characteristic zero but oK/mK has characteristic p > 0 thenwe have

|a| = |a|− log |p|

log pp for any a ∈ Q ⊆ K;

in particular, K contains Qp.

A nonarchimedean field K as in the second part of Exercise 2.1 is calleda p-adic field.

7

Lemma 2.2. If K is p-adic then we have

|n| ≥ |n!| ≥ |p|n−1p−1 for any n ∈ N.

Proof. We may obviously assume that K = Qp. Then the reader should dothis as an exercise but also may consult [B-LL] Chap. II §8.1 Lemma 1.

Now let V be any K-vector space.

Definition. A (nonarchimedean) norm on V is a function ‖ ‖ : V −→ Rsuch that for any v, w ∈ V and any a ∈ K we have:

(i) ‖av‖ = |a| · ‖v‖,

(ii) ‖v + w‖ ≤ max(‖v‖, ‖w‖),

(iii) if ‖v‖ = 0 then v = 0.

Moreover, V is called normed if it is equipped with a norm.

Exercise. i. ‖v‖ ≥ 0 for any v ∈ V and ‖0‖ = 0.

ii. V is an ultrametric space with respect to the metric d(v, w) := ‖w−v‖;in particular, we have ‖v+w‖ = max(‖v‖, ‖w‖) whenever ‖v‖ 6= ‖w‖.

iii. Addition V × V +−−→ V and scalar multiplication K × V −→ V arecontinuous.

Lemma 2.3. Let (V1, ‖ ‖1) and (V2, ‖ ‖2) let two normed K-vector spaces;a linear map f : V1 −→ V2 is continuous if and only if there is a constantc > 0 such that

‖f(v)‖2 ≤ c · ‖v‖1 for any v ∈ V1 .

Proof. We suppose first that such a constant c > 0 exists. Consider anysequence (vn)n∈N in V1 which converges to some v ∈ V1. Then (‖vn − v‖1)nand hence (‖f(vn)−f(v)‖2)n = (‖f(vn−v)‖2)n are zero sequences. It followsthat the sequence (f(vn))n converges to f(v) in V2. This means that f iscontinuous.

Now we assume vice versa that f is continuous. We find a 0 < ε < 1such that

Bε(0) ⊆ f−1(B1(0)) .

Since | | is non-trivial we may assume that ε = |a| for some a ∈ K. In otherwords

‖v‖1 ≤ |a| implies ‖f(v)‖2 ≤ 1

8

for any v ∈ V1. Let now 0 6= v ∈ V1 be an arbitrary nonzero vector. We findan m ∈ Z such that

|a|m+2 < ‖v‖1 ≤ |a|m+1 .

Setting c := |a|−2 we obtain

‖f(v)‖2 = |a|m · ‖f(a−mv)‖2 ≤ |a|m < c · ‖v‖1 .

Definition. The normed K-vector space (V, ‖ ‖) is called a K-Banach spaceif V is complete with respect to the metric d(v, w) := ‖w − v‖.

Examples. 1) Kn with the norm ‖(a1, . . . , an)‖ := max1≤i≤n |ai| is aK-Banach space.

2) Let I be a fixed but arbitrary index set. A family (ai)i∈I of elements inK is called bounded if there is a c > 0 such that |ai| ≤ c for any i ∈ I.The set

`∞(I) := set of all bounded families (ai)i∈I in K

with componentwise addition and scalar multiplication and with thenorm

‖(ai)i‖∞ := supi∈I|ai|

is a K-Banach space.

3) With I as above let

c0(I) := {(ai)i∈I ∈ `∞(I) : for any ε > 0 we have |ai| ≥ εfor at most finitely many i ∈ I}.

It is a closed vector subspace of `∞(I) and hence a K-Banach spacein its own right. Moreover, for (ai)i ∈ c0(I) we have

‖(ai)i‖∞ = maxi∈I|ai| .

For example, c0(N) is the Banach space of all zero sequences in K.

Remark. Any K-Banach space (V, ‖ ‖) over a finite extension K/Qp whichsatisfies ‖V ‖ ⊆ |K| is isometric to some K-Banach space (c0(I), ‖ ‖∞);moreover, all such I have the same cardinality.

9

Proof. Compare [NFA] Remark 10.2 and Lemma 10.3.

Let V and W be two normed K-vector spaces. From now on we denote,unless this causes confusion, all occurring norms indiscriminately by ‖ ‖. Itis clear that

L(V,W ) := {f ∈ HomK(V,W ) : f is continuous}

is a vector subspace of HomK(V,W ). By Lemma 2.3 the operator norm

‖f‖ := sup{‖f(v)‖‖v‖

: v ∈ V, v 6= 0}

= sup{‖f(v)‖‖v‖

: v ∈ V, 0 < ‖v‖ ≤ 1}

is well defined for any f ∈ L(V,W ).

Lemma 2.4. L(V,W ) with the operator norm is a normed K-vector space.

Proof. This is left to the reader as an exercise.

Proposition 2.5. If W is a K-Banach space then so, too, is L(V,W ).

Proof. Let (fn)n∈N be a Cauchy sequence in L(V,W ). Then, on the onehand, (‖fn‖)n is a Cauchy sequence in R and therefore converges, of course.On the other hand, because of

‖fn+1(v)− fn(v)‖ = ‖(fn+1 − fn)(v)‖ ≤ ‖fn+1 − fn‖ · ‖v‖

we obtain, for any v ∈ V , the Cauchy sequence (fn(v))n inW . By assumptionthe limit f(v) := limn→∞ fn(v) exists in W . Obviously we have

f(av) = af(v) for any a ∈ K .

For v, v′ ∈ V we compute

f(v) + f(v′) = limn→∞

fn(v) + limn→∞

fn(v′) = limn→∞

(fn(v) + fn(v′))

= limn→∞

fn(v + v′) = f(v + v′) .

This means that v 7−→ f(v) is a K-linear map which we denote by f . Since

‖f(v)‖ = limn→∞

‖fn(v)‖ ≤ ( limn→∞

‖fn‖) · ‖v‖

10

it follows from Lemma 2.3 that f is continuous. Finally the inequality

‖f − fn‖ = sup{‖(f − fn)(v)‖

‖v‖: v 6= 0

}= sup

{limm→∞ ‖fm(v)− fn(v)‖

‖v‖: v 6= 0

}≤ lim

m→∞‖fm − fn‖ ≤ sup

m≥n‖fm+1 − fm‖

shows that f indeed is the limit of the sequence (fn)n in L(V,W ).

In particular,V ′ := L(V,K)

always is a K-Banach space. It is called the dual space to V .

Lemma 2.6. Let I be an index set; for any j ∈ I let 1j ∈ c0(I) denote thefamily (ai)i∈I with ai = 0 for i 6= j and aj = 1; then

c0(I)′∼=−−→ `∞(I)

` 7−→ (`(1i))i∈I

is an isometric linear isomorphism.

Proof. We give the proof only in the case I = N. The general case followsthe same line but requires the technical concept of summability (cf. [NFA]end of §3). Let us denote the map in question by ι. Because of

|`(1i)| ≤ ‖`‖ · ‖1i‖∞ = ‖`‖

it is well defined and satisfies

‖ι(`)‖∞ ≤ ‖`‖ for any ` ∈ c0(N)′ .

For trivial reasons ι is a linear map. Consider now an arbitrary nonzerovector v = (ai)i ∈ c0(N). In the Banach space c0(N) we then have theconvergent series expansion

v =∑i∈N

ai · 1i .

Applying any ` ∈ c0(N)′ by continuity leads to

`(v) =∑i∈N

ai`(1i) .

11

We obtain

|`(v)|‖v‖∞

≤ supi |ai||`(1i)|supi |ai|

≤ supi|`(1i)| = ‖ι(`)‖∞ .

It follows that‖`‖ ≤ ‖ι(`)‖∞

and together with the previous inequality that ι in fact is an isometry andin particular is injective. For surjectivity let (ci)i ∈ `∞(N) be any vector andput ε := ‖(ci)i‖∞. We consider the linear form

` : c0(N −→ K

(ai)i 7−→∑i

aici

(note that the defining sum is convergent). Using Lemma 2.3 together withthe inequality

|`((ai)i)| = |∑i

aici| ≤ supi|ai||ci| ≤ sup

i|ai| · sup

i|ci| = ε · ‖(ai)i‖∞

we see that ` is continuous. It remains to observe that

ι(`) = (`(1i))i = (ci)i .

3 Convergent series

From now on throughout the book (K, | |) is a fixed nonarchimedean field.For the convenience of the reader we collect in this section the most basic

facts about convergent series in Banach spaces (some of which we have usedalready in the proof of Lemma 2.6).

Let (V, ‖ ‖) be a K-Banach space.

Lemma 3.1. Let (vn)n∈N be a sequence in V ; we then have:

i. The series∑∞

n=1 vn is convergent if and only if limn→∞ vn = 0;

ii. if the limit v := limn→∞ vn exists in V and is nonzero then ‖vn‖ = ‖v‖for all but finitely many n ∈ N;

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iii. let σ : N→ N be any bijection and suppose that the series v =∑∞

n=1 vnis convergent in V ; then the series

∑∞n=1 vσ(n) is convergent as well

with the same limit v.

Proof. i. This is immediate from Lemma 1.5. ii. If v 6= 0 then ‖v‖ 6= 0 andhence ‖vn − v‖ < ‖v‖ for any sufficiently big n ∈ N. For these n Lemma 1.1then implies that

‖vn‖ = ‖(vn − v) + v‖ = max(‖vn − v‖, ‖v‖) = ‖v‖ .

iii. We fix an ε > 0 and choose an m ∈ N such that

‖v −s∑

n=1

vn‖ < ε for any s ≥ m .

Then also

‖vs‖ = ‖(v −s−1∑n=1

vn)− (v −s∑

n=1

vn‖ ≤ max(‖v −s−1∑n=1

vn‖, ‖v −s∑

n=1

vn‖) < ε

for any s > m. Setting ` := max{σ−1(n) : n ≤ m} ≥ m we have

{σ−1(1), . . . , σ−1(m)} ⊆ {1, . . . , `}

and hence, for any s ≥ `,

{σ(1), . . . , σ(s)} = {1, . . . ,m} ∪ {n1, . . . , ns−m}

with appropriate natural numbers ni > m. We conclude that

‖v −s∑

n=1

vσ(n)‖ = ‖

(v −

m∑n=1

vn

)− vn1 − . . .− vns−m‖

≤ max

(‖v −

m∑n=1

vn‖, ‖vn1‖, . . . , ‖vns−m‖

)< ε

for any s ≥ `.

The following identities between convergent series are obvious:

–∑∞

n=1 avn = a ·∑∞

n=1 vn for any a ∈ K.

13

– (∑∞

n=1 vn) + (∑∞

n=1wn) =∑∞

n=1(vn + wn).

Lemma 3.2. Let∑∞

n=1 an and∑∞

n=1 vn be convergent series in K and V ,respectively; then the series

∑∞n=1wn with wn :=

∑`+m=n a`vm is conver-

gent, and∞∑n=1

wn =( ∞∑n=1

an

)( ∞∑n=1

vn

).

Proof. Let A := supn |an| and C := supn ‖vn‖. The other cases being trivialwe will assume that A,C > 0. For any given ε > 0 we find an N ∈ N suchthat

|an| <ε

Cand ‖vn‖ <

ε

Afor any n ≥ N .

Then

‖wn‖ ≤ max`+m=n

|a`| · ‖vm‖ ≤ max(C ·max

`≥N|a`|, A · max

m≥N‖vm‖

)< ε

for any n ≥ 2N . By Lemma 3.1.i. the series∑∞

n=1wn therefore is convergent.To establish the asserted identity we note that its left hand side is the limitof the sequence

Ws :=s∑

n=1

wn =∑

`+m≤sa`vm

whereas its right hand side is the limit of the sequence

W ′s :=( s∑n=1

an

)( s∑n=1

vn

)=∑`,m≤s

a`vm .

It therefore suffices to show that the differences Ws −W ′s converge to zero.But we have

‖Ws −W ′s‖ = ‖∑`,m≤s`+m>s

a`vm‖ ≤ max`,m≤s`+m>s

|a`| · ‖vm‖

≤ max(C ·max

`> s2

|a`|, A ·maxm> s

2

‖vm‖).

Analogous assertions hold true for series∑∞

n1,...,nr=1 vn1,...,nr indexed bymulti-indices in N× . . .× N.

14

4 Differentiability

Let V and W be two normed K-vector spaces, let U ⊆ V be an open subset,and let f : U −→W be some map.

Definition. The map f is called differentiable in the point v0 ∈ U if thereexists a continuous linear map

Dv0f : V −→W

such that for any ε > 0 there is an open neighbourhood Uε = Uε(v0) ⊆ U ofv0 with

‖f(v)− f(v0)−Dv0f(v − v0)‖ ≤ ε‖v − v0‖ for any v ∈ Uε .

We may of course assume in the above definition that the open neigh-bourhood Uε is of the form Uε = Bδ(ε)(v0) for some sufficiently small radiusδ(ε) > 0. We claim that the linear map Dv0f is uniquely determined. Fixan ε′ > 0, and choose a basis {v′j}j∈J of the vector space V . By scaling wemay assume that ‖v′j‖ ≤ δ(ε′). We put vj := v′j + v0. Then

vj ∈ Bδ(ε′)(v0) = Uε′ .

More generally, for any 0 < ε ≤ ε′ we pick a tε ∈ K× such that

|tε|δ(ε′) ≤ δ(ε) .

Thentε(vj − v0) + v0 ∈ Bδ(ε)(v0) = Uε for any j ∈ J .

It follows that

‖f(tε(vj − v0) + v0)− f(v0)−Dv0f(tε(vj − v0))‖ ≤ ε‖tε(vj − v0)‖

and hence that

‖f(tε(vj − v0) + v0)− f(v0)tε

−Dv0f(vj − v0)‖ ≤ ε‖vj − v0‖ .

By letting ε tend to zero we obtain

Dv0f(v′j) = Dv0f(vj − v0) = limt∈K×,t→0

f(t(vj − v0) + v0)− f(v0)t

for any j ∈ J . But as a linear map Dv0f is uniquely determined by its valueon the basis vectors v′j .

The continuous linear map Dv0f : V −→ W is called (if it exists) thederivative of f in the point v0 ∈ U . In case V = K we also write f ′(a0) :=Da0f(1).

15

Remark 4.1. i. If f is differentiable in v0 then it is continuous in v0.

ii. (Chain rule) Let V,W1, and W2 be normed K-vector spaces, U ⊆ Vand U1 ⊆ W1 be open subsets, and f : U −→ U1 and g : U1 −→ W2

be maps; suppose that f is differentiable in some v0 ∈ U and g isdifferentiable in f(v0); then g ◦ f is differentiable in v0 and

Dv0(g ◦ f) = Df(v0)g ◦Dv0f .

iii. A continuous linear map u : V −→ W is differentiable in any v0 ∈ Vand Dv0u = u; in particular, in the situation of ii. we have

Dv0(u ◦ f) = u ◦Dv0f .

iv. (Product rule) Let V,W1, . . . ,Wm, and W be normed K-vector spaces,let U ⊆ V be an open subset with maps fi : U −→ Wi, and letu : W1 × . . . × Wm −→ W be a continuous multilinear map; sup-pose that f1, . . . , fm all are differentiable in some point v0 ∈ U ; thenu(f1, . . . , fm) : U −→W is differentiable in v0 and

Dv0(u(f1, . . . , fm)) =m∑i=1

u(f1(v0), . . . , Dv0fi, . . . , fm(v0)) .

Proof. These are standard arguments of which we only recall the proof ofii. Let ε > 0 and choose a δ > 0 such that

δ2, δ‖Dv0f‖, δ‖Df(v0)g‖ ≤ ε

(here ‖ ‖ refers to the operator norm of course). By assumption on g wehave

(1) ‖g(w)− g(f(v0))−Df(v0)g(w − f(w0))‖ ≤ δ‖w − f(v0)‖

for any w ∈ Uδ(f(v0)). By the differentiability and hence continuity of f inv0 there exists an open neighbourhood U(v0) ⊆ U of v0 such that f(U(v0)) ⊆Uδ(f(v0)) and

(2) ‖f(v)− f(v0)−Dv0f(v − v0)‖ ≤ δ‖v − v0‖

for any v ∈ U(v0). In particular

‖f(v)− f(v0)‖ = ‖Dv0f(v − v0) + f(v)− f(v0)−Dv0f(v − v0)‖≤ max(‖Dv0f‖ · ‖v − v0‖, δ‖v − v0‖)

(3)

16

for any v ∈ U(v0). We now compute

‖g(f(v))− g(f(v0))−Df(v0)g ◦Dv0f(v − v0)‖= ‖g(f(v))− g(f(v0))−Df(v0)g(f(v)− f(v0))

+Df(v0)g(f(v)− f(v0)−Dv0f(v − v0)

)‖

(1)

≤ max(δ‖f(v)− f(v0)‖, ‖Df(v0)g‖ · ‖f(v)− f(v0)−Dv0f(v − v0)‖

)(2)

≤ max(δ‖f(v)− f(v0)‖, ‖Df(v0)g‖ · δ‖v − v0‖

)(3)

≤ max(δ‖Dv0f‖, δ2, δ‖Df(v0)g‖

)· ‖v − v0‖

≤ ε‖v − v0‖

for any v ∈ U(v0).

Suppose that the vector space V = V1 ⊕ . . . ⊕ Vm is the direct sumof finitely many vector spaces V1, . . . , Vm and that the norm on V is themaximum of its restrictions to the Vi. Write a vector v0 ∈ U as v0 = v0,1 +. . .+v0,m with v0,i ∈ Vi. For each 1 ≤ i ≤ m there is an open neighbourhoodUi ⊆ Vi of v0,i such that

U1 + . . .+ Um ⊆ U .

Therefore the maps

fi : Ui −→W

vi 7−→ f(v0,1 + . . .+ vi + . . .+ v0,m)

are well defined. If it exists the continuous linear map

D(i)v0 f := Dv0,ifi : Vi −→W

is called the i-th partial derivative of f in v0. We recall that differentiabilityof f in v0 implies the existence of all partial derivatives together with theidentity

Dv0f =m∑i=1

D(i)v0 f .

Let us go back to our initial situation V ⊇ U f−→W .

17

Definition. The map f is called strictly differentiable in v0 ∈ U if thereexists a continuous linear map Dv0f : V −→ W such that for any ε > 0there is an open neighbourhood Uε ⊆ U of v0 with

‖f(v1)− f(v2)−Dv0f(v1 − v2)‖ ≤ ε‖v1 − v2‖ for any v1, v2 ∈ Uε .

Exercise. Suppose that f is strictly differentiable in every point of U . Thenthe map

U −→ L(V,W )v 7−→ Dvf

is continuous.

Our goal for the rest of this section is to discuss the local invertibilityproperties of strictly differentiable maps.

Lemma 4.2. Let Bδ(v0) be a closed ball in a K-Banach space V and letf : Bδ(v0) −→ V be a map for which there exists a 0 < ε < 1 such that

(4) ‖f(v1)− f(v2)− (v1 − v2)‖ ≤ ε‖v1 − v2‖ for any v1, v2 ∈ Bδ(v0);

then f induces a homeomorphism

Bδ(v0) '−−→ Bδ(f(v0)) .

Proof. By Lemma 1.1 we have

(5) ‖f(v1)− f(v2)‖ = ‖v1 − v2‖ for any v1, v2 ∈ Bδ(v0) .

In particular, f is a homeomorphism onto its image which satisfies f(Bδ(v0))⊆ Bδ(f(v0)). It remains to show that this latter inclusion in fact is anequality. Let w ∈ Bδ(f(v0)) be an arbitrary but fixed vector. For any v′ ∈Bδ(v0) we put

v′′ := w + v′ − f(v′) .

We compute

‖v′′ − v0‖ ≤ max(‖v′′ − v′‖, ‖v′ − v0‖)≤ max(‖w − f(v′)‖, δ)≤ max(‖w − f(v0)‖, ‖f(v0)− f(v′)‖, δ)≤ max(δ, ‖f(v0)− f(v′)‖)(5)

≤ max(δ, ‖v0 − v′‖)≤ δ

18

which means that v′′ ∈ Bδ(v0). Hence we may define inductively a sequence(vn)n≥0 in Bδ(v0) by

vn+1 := w + vn − f(vn) .

Using (4) we see that

‖vn+1 − vn‖ = ‖vn − f(vn)− (vn−1 − f(vn−1))‖= ‖f(vn−1)− f(vn)− (vn−1 − vn)‖≤ ε‖vn − vn−1‖

and therefore‖vn+1 − vn‖ ≤ εn‖v1 − v0‖ ≤ εnδ

for any n ≥ 1. It follows that (vn)n is a Cauchy sequence and, since Vis complete, is convergent. Because Bδ(v0) is closed in V the limit v :=limn→∞ vn lies in Bδ(v0). By passing to the limit in the defining equationand using the continuity of f we finally obtain that f(v) = w.

Proposition 4.3. (Local invertibility) Let V and W be K-Banach spaces,U ⊆ V be an open subset, and f : U −→ W be a map which is strictlydifferentiable in the point v0 ∈ U ; suppose that the derivative Dv0f : V

∼=−→Wis a topological isomorphism; then there are open neighbourhoods U0 ⊆ U ofv0 and U1 ⊆W of f(v0) such that:

i. f : U0'−−→ U1 is a homeomorphism;

ii. the inverse map g : U1 −→ U0 is strictly differentiable in f(v0), and

Df(v0)g = (Dv0f)−1 .

Proof. We consider the map

f1 := (Dv0f)−1 ◦ f : U −→ V .

As a consequence of the chain rule it is strictly differentiable in v0 and

Dv0f1 = (Dv0f)−1 ◦Dv0f = idV .

Hence, fixing some 0 < ε0 < 1 we find a neighbourhood Bδ0(v0) ⊆ U of v0

such that the condition (4) in Lemma 4.2 is satisfied. The lemma then saysthat

f1 : U0 := Bδ0(v0) '−−→ Bδ0(f1(v0))

19

is a homeomorphism. Since Dv0f is a homeomorphism by assumption U1 :=Dv0f(Bδ0(f1(v0))) is an open neighbourhood of f(v0) in W and f : U0

'−→ U1

is a homeomorphism.To prove ii. let ε > 0. We have ‖(Dv0f)−1‖ > 0 since (Dv0f)−1 is bijec-

tive. Hence we find a δ > 0 such that

δ‖(Dv0f)−1‖ < 1 and δ‖(Dv0f)−1‖2 ≤ ε .

By the strict differentiability of f in v0 we have

‖f(v1)− f(v2)−Dv0f(v1 − v2)‖ ≤ δ‖v1 − v2‖ for any v1, v2 ∈ Uδ .

Applying (Dv0f)−1 gives

‖(Dv0f)−1(f(v1)− f(v2))− (v1 − v2)‖ ≤ δ‖(Dv0f)−1‖ · ‖v1 − v2‖ .

By our choice of δ and Lemma 1.1 we deduce

‖(Dv0f)−1(f(v1)− f(v2))‖ = ‖v1 − v2‖ .

Combining the last two formulas we obtain

‖v1 − v2 − (Dv0f)−1(f(v1)− f(v2))‖≤ δ‖(Dv0f)−1‖ · ‖(Dv0f)−1(f(v1)− f(v2))‖≤ δ‖(Dv0f)−1‖2 · ‖f(v1)− f(v2)‖≤ ε‖f(v1)− f(v2)‖

for any v1, v2 ∈ Uδ. It follows that

‖g(w1)− g(w2)− (Dv0f)−1(w1 − w2)‖ ≤ ε‖w1 − w2‖

for any w1, w2 ∈ f(U0 ∩ Uδ). Since f(U0 ∩ Uδ) is an open neighbourhood off(v0) in U1 this establishes ii.

In regard to the assumption on the derivative in the above propositionit is useful to have in mind the open mapping theorem (cf. [NFA] Cor. 8.7)which says that any continuous linear bijection between K-Banach spacesnecessarily is a topological isomorphism. We also point out the trivial factthat any linear map between two finite dimensional K-Banach spaces iscontinuous.

20

Corollary 4.4. Let U ⊆ Kn be an open subset and f : U −→ Km be a mapwhich is strictly differentiable in v0 ∈ U ; suppose that Dv0f is injective;then there are open neighbourhoods U0 ⊆ U of v0 and U1 ⊆ Km of f(v0) aswell as a ball Bε(0) ⊆ Km−n around zero and linearly independent vectorsw1, . . . , wm−n ∈ Km such that the map

U0 ×Bε(0) '−−→ U1

(v, (a1, . . . , am−n)) 7−→ f(v) + a1w1 + . . .+ am−nwm−n

is a homeomorphism.

Proof. We choose the vectors wi in such a way that

Km = im(Dv0f)⊕Kw1 ⊕ . . .⊕Kwm−n .

Then the linear map

u : Kn ×Km−n ∼=−−→ Km

(v, (a1, . . . , am−n)) 7−→ (Dv0f)(v) + a1w1 + . . .+ am−nwm−n

is a topological isomorphism. One checks that the map

f : U ×Km−n −→ Km

(v, (a1, . . . , am−n)) 7−→ f(v) + a1w1 + . . .+ am−nwm−n

is strictly differentiable in (v0, 0) with D(v0,0)f = u. Now apply Prop. 4.3.i.

Corollary 4.5. Let U ⊆ Kn be an open subset and f : U −→ Km be a mapwhich is strictly differentiable in v0 ∈ U ; suppose that Dv0f is surjective;then there are open neighbourhoods U0 ⊆ U of v0 and U1 ⊆ Km of f(v0) aswell as a ball Bε(0) ⊆ Kn−m around zero and a linear map p : Kn −→ Kn−m

such that the map

U0'−−→ U1 ×Bε(0)

v 7−→ (f(v), p(v)− p(v0))

is a homeomorphism; in particular, the restricted map f : U0 −→ Km isopen.

21

Proof. We choose a decomposition

Kn = ker(Dv0f)⊕ C

and letp : Kn −→ ker(Dv0f) ∼= Kn−m

be the corresponding projection map. Then

u : Kn −→ Km ×Kn−m

v 7−→ ((Dv0f)(v), p(v))

is a topological isomorphism. One checks that

f : U −→ Km ×Kn−m

v 7−→ (f(v), p(v)− p(v0))

is strictly differentiable in v0 with Dv0 f = u. Now apply Prop. 4.3.i.

We finish this section with a trivial observation. A map f : X −→ Afrom some topological space X into some set A is called locally constant iff−1(a) is open (and closed) in X for any a ∈ A.

Lemma 1.4 implies that in our standard situation of two normed K-vector spaces V and W and an open subset U ⊆ W there are plenty oflocally constant maps f : U −→ W . They all are strictly differentiable inany v0 ∈ U with Dv0f = 0.

5 Power series

Let V be a K-Banach space. By a power series f(X) in r variables X =(X1, . . . , Xr) with coefficients in V we mean a formal series

f(X) =∑α∈Nr0

Xαvα with vα ∈ V .

Here and in the following we use the usual conventions for multi-indices

Xα := Xα11 · . . . ·X

αrr and |α| := α1 + . . .+ αr

if α = (α1, . . . , αr) ∈ Nr0 (with N0 := N ∪ {0}).

For any ε > 0 the power series f(X) =∑

αXαvα is called ε-convergent

iflim|α|→∞

ε|α|‖vα‖ = 0 .

22

Remark. If f(X) is ε-convergent then it also is δ-convergent for any 0 <δ ≤ ε.

The K-vector space

Fε(Kr;V ) := all ε-convergent power series f(X) =∑α∈Nr0

Xαvα

is normed by‖f‖ε := max

αε|α|‖vα‖ .

By a straightforward generalization of the argument for c0(N) it is shownthat Fε(Kr;V ) is a Banach space. By the way, in case ε = |c| for somec ∈ K× the map

c0(Nr0)

∼=−−→ F|c|(Kr;K)

(aα)α 7−→∑α

aα

c|α|Xα

is an isometric linear isomorphism.

Remark. The vector space Fε(Kr;V ) together with its topology only de-pends on the topology of V (and not on its specific norm).

Proof. Let ‖ ‖′ be a second norm on V which induces the same topology as‖ ‖. Applying Lemma 2.3 to the identity map idV we obtain two constantsc1, c2 > 0 such that

c1‖ ‖ ≤ ‖ ‖′ ≤ c2‖ ‖ .

Then obviously

lim|α|→∞

ε|α|‖vα‖′ = 0 if and only if lim|α|→∞

ε|α|‖vα‖ = 0 .

This means that using ‖ ‖′ instead of ‖ ‖ leads to the same vector spaceFε(Kr;V ) but which carries the two norms ‖ ‖ε and ‖f‖′ε := maxα ε|α|‖vα‖′.The above inequalities immediately imply the analogous inequalities

c1‖ ‖ε ≤ ‖ ‖′ε ≤ c2‖ ‖ε .

By Lemma 2.3 for the identity map on Fε(Kr;V ) this means that ‖ ‖ε and‖ ‖′ε induce the same topology.

23

Consider a convergent series

f =∞∑i=0

fi

in the Banach space Fε(Kr;V ). Suppose that

f(X) =∑α

Xαvα and fi(X) =∑α

Xαvi,α .

We have

‖f −n∑i=0

fi‖ε = ‖∑α

Xα(vα −n∑i=0

vi,α)‖ε = maxα

ε|α|‖vα −n∑i=0

vi,α‖ .

This shows that

vα =∞∑i=0

vi,α for any α ∈ Nr0

which means that limits in Fε(Kr;V ) can be computed coefficientwise.Let Bε(0) denote the closed ball around zero in Kr of radius ε. We recall

that Kr always is equipped with the norm ‖(a1, . . . , ar)‖ = max1≤i≤r |ai|.By Lemma 3.1.i. we have the K-linear map

Fε(Kr;V ) −→ K-vector space of maps Bε(0) −→ V

f(X) =∑α

Xαvα 7−→ f(x) :=∑α

xαvα .

Remark 5.1. For any x ∈ Bε(0) the linear evaluation map

Fε(Kr;V ) −→ V

f 7−→ f(x)

is continuous of operator norm ≤ 1.

Proof. We have

‖f(x)‖ = ‖∑α

xαvα‖ ≤ maxα

ε|α|‖vα‖ = ‖f‖ε .

24

Proposition 5.2. Let u : V1 × V2 −→ V be a continuous bilinear mapbetween K-Banach spaces; then

U : Fε(Kr;V1)×Fε(Kr;V2) −→ Fε(Kr;V )(∑α

Xαvα,∑α

Xαwα)7−→

∑α

Xα( ∑β+γ=α

u(vβ, wγ))

is a continuous bilinear map satisfying

U(f, g)∼(x) = u(f(x), g(x)) for any x ∈ Bε(0)

and any f ∈ Fε(Kr;V1) and g ∈ Fε(Kr;V2).

Proof. By a similar argument as for Lemma 2.3 the bilinear map u is con-tinuous if and only if there is a constant c > 0 such that

‖u(v1, v2)‖ ≤ c‖v1‖ · ‖v2‖ for any v1 ∈ V1, v2 ∈ V2 .

We therefore have

ε|α|‖∑

β+γ=α

u(vβ, wγ)‖ ≤ c maxβ+γ=α

(ε|β|‖vβ‖ · ε|γ|‖vγ‖

).

This shows (compare the proof of Lemma 3.2) that with f and g also U(f, g)is ε-convergent and that

‖U(f, g)‖ε ≤ c‖f‖ε‖g‖ε .

Hence U is well defined, bilinear, and continuous. The asserted identitybetween evaluations is an immediate generalization of Lemma 3.2.

Proposition 5.3. Fε(Kr;K) is a commutative K-algebra with respect tothe multiplication(∑

α

bαXα)(∑

α

cαXα)

:=∑α

(∑β+γ

bβcγ)Xα ;

in addition we have

(fg)∼(x) = f(x)g(x) for any x ∈ Bε(0)

as well as‖fg‖ε = ‖f‖ε‖g‖ε

for any f, g ∈ Fε(Kr;K).

25

Proof. Apart from the norm identity this is a special case of Prop. 5.2 (andits proof) for the multiplication in K as the bilinear map. It remains to showthat

‖fg‖ε ≥ ‖f‖ε‖g‖ε .

Let ≥ denote the lexicographic order on Nr0, and let µ and ν be lexicograph-

ically minimal multi-indices such that

|bµ|e|µ| = ‖f‖ε and |cν |ε|ν| = ‖g‖ε, respectively.

Put λ := µ+ ν and consider any equation of the form λ = β + γ.Claim: β ≤ µ or γ ≤ ν.Otherwise we would have β > µ and γ > ν. This means that there are

1 ≤ i, j ≤ r such that

β1 = µ1, . . . , bi−1 = µi−1, and βi > µi

as well asγ1 = ν1, . . . , γj−1 = νj−1, and γj > νj .

By symmetry we may assume that i ≤ j. We then obtain the contradiction

λi = µi + νi < βi + γi = λi .

This establishes the claim.We of course have

|bβ|ε|β| ≤ ‖f‖ε and |cγ |ε|γ| ≤ ‖g‖ε .

But in case (β, γ) 6= (µ, ν) the fact that β < µ or γ < ν together with theminimality property of µ and ν implies that

|bβ|ε|β| < ‖f‖ε or |cγ |ε|γ| < ‖g‖ε .

It follows that

|bβcγ |ε|λ| = |bβ|ε|β| · |cγ |ε|γ| < ‖f‖ε‖g‖ε

whenever β + γ = λ but (β, γ) 6= (µ, ν). We conclude that

‖fg‖ε ≥ |∑

β+γ=λ

bβcγ |ε|λ| = |bµcν |ε|λ| = ‖f‖ε‖g‖ε .

26

Proposition 5.4. Let g ∈ Fδ(Kr;Kn) such that ‖g‖δ ≤ ε; then

Fε(Kn;V ) −→ Fδ(Kr;V )

f(Y ) =∑β

Y βvβ 7−→ f ◦ g(X) :=∑β

g(X)βvβ

is a continuous linear map of operator norm ≤ 1 which satisfies

(f ◦ g)∼(x) = f(g(x)) for any x ∈ Bδ(0) ⊆ Kr .

Proof. Using the obvious identification

Fδ(Kr;Kn) =n∏i=1

Fδ(Kr;K)

g = (g1, . . . , gn)

we have maxi ‖gi‖δ = ‖g‖δ ≤ ε. The Prop. 5.3 therefore implies that g(X)β ∈Fδ(Kr;K) for each β and

‖g(X)β‖δ = ‖n∏i=1

gβii ‖δ =n∏i=1

‖gi‖βiδ ≤ ε|β| .

It follows that g(X)βvβ ∈ Fδ(Kr;V ) for each β with

‖g(X)βvβ‖δ = ‖g(X)β‖δ‖vβ‖ ≤ ε|β|‖vβ‖ .

Since the right hand side goes to zero by the ε-convergence of f the se-ries f ◦ g(X) =

∑β g(X)βvβ is convergent in the Banach space Fδ(Kr;V ).

Moreover, we have

‖f ◦ g‖δ ≤ maxβ‖g(X)βvβ‖δ ≤ max

βε|β|‖vβ‖ = ‖f‖ε .

To see the asserted identity between evaluations we first note that by Remark5.1 the map g indeed maps the ball Bδ(0) ⊆ Kr into the ball Bε(0) ⊆ Kn.Using Remark 5.1 together with Prop. 5.3 we obtain

(f ◦ g)∼(x) =∑β

(gβ)∼(x)vβ =∑β

g(x)βvβ = f(g(x)) .

27

As a consequence of the discussion before Remark 5.1 and of Prop. 5.3the power series f ◦ g can be computed by formally inserting g into f .

The reader is warned that although, for any g ∈ Fδ(Kr;Kn), we have,by Remark 5.1, the inequality

supx∈Bδ(0)

‖g(x)‖ ≤ ‖g‖δ

it is, in general, not an equality. This means that we may have g(Bδ(0)) ⊆Bε(0) even if ε < ‖g‖δ. Then, for any f ∈ Fε(Kn;V ), the composite of mapsf ◦ g exists but the composite of power series f ◦ g ∈ Fδ(Kr;V ) may not.

Exercise. An example of such a situation is

g(X) := Xp −X ∈ F1(Qp; Qp) and f(Y ) :=∞∑n=0

Y n ∈ F 1p(Qp; Qp) .

Corollary 5.5. (Point of expansion) Let f ∈ Fε(Kr;V ) and y ∈ Bε(0);then there exists an fy ∈ Fε(Kr;V ) such that ‖fy‖ε = ‖f‖ε and

f(x) = fy(x− y) for any x ∈ Bε(0) = Bε(y) .

Proof. Let e1, . . . , er denote the standard basis of Kr. Applying Prop. 5.4to the power series g(X) := X + y =

∑ri=1Xiei + y ∈ Fε(Kr;Kr) which

satisfies ‖g‖ε ≤ ε we obtain the existence of fy(X) := f(X + y) satisfying

‖fy‖ε ≤ ‖f‖ε and fy(x) = f(x+ y) for x ∈ Bε(0) .

By symmetry we also have

‖f‖ε = ‖(fy)−y‖ε ≤ ‖fy‖ε .

It will be convenient in the following to use the short notation

i := (0, . . . , 1, . . . , 0)

for the multi-index whose only nonzero entry is a 1 in the i-th place.Suppose that the power series f(X) =

∑αX

αvα is ε-convergent. Thenalso, for any 1 ≤ i ≤ r, its i-th formal partial derivative

∂f

∂Xi(X) :=

∑α

Xα−iαivα

28

is ε-convergent (since ‖αivα‖ ≤ ‖vα‖). In case our field K has characteristiczero it follows inductively that

vα = 1αi!·...·αr!

((∂

∂X1

)α1 . . .(

∂∂Xr

)αrf)∼(0) for any α .

Proposition 5.6. The map f is strictly differentiable in every point z ∈Bε(0) and satisfies

D(i)z f(1) =

(∂f

∂Xi

)∼(z) .

Proof. Case 1: We assume that z = 0, and introduce the continuous linearmap

D : Kr −→ V

(a1, . . . , ar) 7−→r∑i=1

aivi .

Let δ > 0 and choose a 0 < δ′ < ε such that

δ′‖f‖εε2≤ δ .

By induction with respect to |α| one checks that

|xα − yα| ≤ (δ′)|α|−1‖x− y‖ for any x, y ∈ Bδ′(0) .

We now compute

‖f(x)− f(y)−D(x− y)‖ = ‖∑|α|≥2

(xα − yα)vα‖

≤ max|α|≥2

|xα − yα| · ‖vα‖

≤ ‖f‖ε · max|α|≥2

|xα − yα|ε|α|

≤ ‖f‖ε · max|α|≥2

(δ′)|α|−1

ε|α|· ‖x− y‖

= ‖f‖ε ·δ′

ε2· ‖x− y‖

≤ δ‖x− y‖

for any x, y ∈ Bδ′(0). This proves that f is strictly differentiable in 0 withD0f = D and hence

D(i)0 f(1) = vi =

(∂f

∂Xi

)∼(0) .

29

Case 2: Let z ∈ Bε(0) be an arbitrary point. By Cor. 5.5 we find a powerseries fz(X) =

∑αX

αvα(z) in Fε(Kr;V ) such that

f(x) = fz(x− z) for any x ∈ Bε(0) .

Using the chain rule together with the first case we see that f is strictlydifferentiable in z with

D(i)z f(1) = D

(i)0 fz(1) = vi(z) .

Since fz(X) can be computed by formally inserting X+z into f(X) we have∑α

Xαvα(z) =∑α

(X + z)αvα

and hence

vi(z) =∑α

zα−iαivα =(∂f

∂Xi

)∼(z) .

By Prop. 5.6 the map

∂f

∂xi: Bε(0) −→ V

x 7−→ D(i)x f(1)

is well defined and satisfies

∂f

∂xi=(∂f

∂Xi

)∼.

Corollary 5.7. (Taylor expansion) If K has characteristic zero then wehave

f(X) =∑α

Xα 1α1!·...·αr!

(( ∂

∂x1

)α1 . . .( ∂

∂xr

)αr f)(0) .

Corollary 5.8. (Identity theorem for power series) If K has characteristiczero then for any nonzero f ∈ Fε(Kr;V ) there is a point x ∈ Bε(0) suchthat f(x) 6= 0.

In fact much stronger results than Cor. 5.8 hold true. In particular, theassumption on the characteristic of K is superfluous. But this requires a

30

different method of proof (cf. [BGR] 5.1.4 Cor. 5 and subsequent comment).In any case the map

Fε(Kr;V ) −→ strictly differentiable maps Bε(0) −→ V

f 7−→ f

is injective and commutes with all the usual operations as considered above.We therefore will simplify notations in the following and write very often ffor the power series as well as the corresponding map.

Proposition 5.9. (Invertibility for power series) Let f(X) ∈ Fε(Kr;Kr)such that f(0) = 0, and suppose that D0f is bijective; we fix a 0 < δ <

ε2

‖f‖ε‖(D0f)−1‖2 ; then δ < ‖f‖ε, and there is a uniquely determined g(Y ) ∈Fδ(Kr;Kr) such that

g(0) = 0, ‖g‖δ < ε, and f ◦ g(Y ) = Y ;

in particular, the diagram

Bδ(0)⊆

%%JJJJJJJJJ

g{{wwwwwwwww

Bε(0)f

// B‖f‖ε(0)

is commutative.

Proof. Case 1: We assume that D0f = idKr . Let

f = (f1, . . . , fr) and fi(X) =∑α

ai,αXα .

Since f(0) = 0 we have ai,0 = 0. Moreover, by Prop. 5.6, the matrix of D0fin the standard basis of Kr is equal to(

∂fi∂Xj

(0))i,j

= (ai,j)i,j .

But we are in the special case that this matrix is the identity matrix. Henceai,j = 0 for i 6= j and ai,i = 1. We therefore see that

(6) fi(X) = Xi +∑|α|≥2

ai,αXα .

31

It follows in particular that‖f‖ε ≥ ε

and hence

(7) δ <ε2

‖f‖ε‖(D0f)−1‖2=

ε2

‖f‖ε≤ ε ≤ ‖f‖ε .

In case 1 of the proof of Prop. 5.6 we have computed that

‖f(x)− f(y)− (x− y)‖ ≤ δ‖f‖εε2‖x− y‖ for any x, y ∈ Bδ(0) .

As δ ‖f‖εε2

< 1 this is the condition (4) in Lemma 4.2. We therefore concludethat

f : Bδ(0) '−−→ Bδ(0)

is a homeomorphism. Furthermore, for |α| ≥ 2 we have

|ai,j |δ|α| ≤ ‖f‖ε(δε

)|α| ≤ ‖f‖ε( δε)2 = δ(δ ‖f‖ε

ε2

)< δ .

Hence it follows from (6) that

‖f‖δ = δ .

In a next step we establish the existence of a formal power series g =(g1, . . . , gr) with

gi(Y ) =∑|β|≥1

bi,βYβ

such thatf(g(Y )) = Y .

First of all let us check that formally inserting any such g into f is a welldefined operation. We formally compute

fi(g(Y )) =∑|α|≥1

ai,αg1(Y )α1 · . . . · gr(Y )αr

=∑|α|≥1

ai,α( ∑|β|≥1

b1,βYβ)α1 · . . . ·

( ∑|β|≥1

br,βYβ)αr

=∑|γ|≥1

(∑...

ai,αb1,β(1) · . . . · b1,β(α1)b2,β(α1+1) · . . . · br,β(α1+···+αr))Xγ

32

where in the last expression the multi-indices in the inner sum run over allα, β(1), . . . , β(α1 + . . .+ αr) such that |α|, |β(1)|, . . . , |β(α1 + . . .+ αr)| ≥ 1and

β(1) + . . .+β(α1) +β(α1 + 1) + . . .+β(α1 +α2) + . . .+β(α1 + . . .+αr) = γ.

Because of |β(ν)| ≥ 1 this condition enforces |α| ≤ |γ| so that these innersums in fact are finite. We now set

Yi = fi(g(Y ))

and compare coefficients. For γ = i we obtain

1 =r∑`=1

ai,`b`,i = bi,i .

For γ 6= i we have

0 =∑...

ai,αb1,β(1) · . . . = bi,γ +∑

2≤|α|≤|γ|

ai,αC(α, γ)

where C(α, γ) is a (finite) sum of products of the form

b1,β(1) · . . . · br,β(α1+...+αr) with |β(ν)| ≥ 1 and∑ν

β(ν) = γ .

In particular, ∑ν

|β(ν)| = |γ| and |β(ν)| ≥ 1 .

Since the number of summands |β(ν)| is equal to |α| ≥ 2 it follows that|β(ν)| < |γ|. We see that on the right hand side of the equation

bi,γ = −∑

2≤|α|≤γ

ai,αC(α, γ)

only coefficients b`,β appear with |β| < |γ|. This means that the coefficientsβi,γ can be computed recursively from these equations. Hence g exists andis uniquely determined. In addition we check inductively that

|bi,γ | ≤(‖f‖εε2

)|γ|−1

33

holds true. For |γ| = 1 we have βi,γ = 0 or 1 and the inequality is trivial. If|γ| ≥ 2 then the induction hypothesis implies

|C(α, γ)| ≤ max...|b1,β(1)| · . . . · |br,β(α1+...+αr)|

≤ max...

(‖f‖εε2

)(|β(1)|−1)+...+(|β(α1+...+αr)|−1)

=(‖f‖εε2

)|γ|−|α|.

Hence we have

|bi,γ | ≤ max2≤|α|≤|γ|

|ai,α| · |C(α, γ)| ≤ max2≤|α|≤|γ|

‖f‖εε|α||C(α, γ)|

≤ max2≤|α|≤|γ|

‖f‖εε|α|

·(‖f‖εε2

)|γ|−|α| = max2≤|α|≤|γ|

( ε

‖f‖ε)|α|−2 ·

(‖f‖εε2

)|γ|−1

=(‖f‖εε2

)|γ|−1

(the last identity since ε ≤ ‖f‖ε). We deduce that

|bi,γ |δ|γ| ≤(δ‖f‖εε2

)|γ|−1δ .

Because of δ ‖f‖εε2

< 1 this shows that g is δ-convergent with ‖g‖δ ≤ δ. Butbi,i = 1 then implies that ‖g‖δ = δ. Altogether we have shown so far that:

– f is δ-convergent with ‖f‖δ = δ;

– there is a uniquely determined δ-convergent g with

g(0) = 0, ‖g‖δ = δ < ε, and f ◦ g(Y ) = Y .

Using Prop. 5.4 we see that f ◦ g = id so that

Bδ(0)f //

Bδ(0)g

oo

are homeomorphisms which are inverse to each other. But we also concludethat g ◦ f(X) exists as a δ-convergent power series as well and satisfies(g ◦ f)∼ = g ◦ f = id. The identity theorem Cor. 5.8 then implies that

g ◦ f(X) = X .

34

Case 2: Let D0f be arbitrary bijective. If (aij)i,j is the matrix of (D0f)−1

in the standard basis of Kr then the operator norm is given by

‖(D0f)−1‖ = maxi,j|aij | .

Viewed as a power series (D0f)−1 is ε′-convergent for any ε′ > 0 with

‖(D0f)−1‖ε′ = ε′ ·maxi,j|ai,j | = ε′‖(D0f)−1‖ .

In the following we put ε′ := ε‖(D0f)−1‖ . Then ‖(D0f)−1‖ε′ = ε, and from

Prop. 5.4 we obtain

f0 := f ◦ (D0f)−1 ∈ Fε′(Kr;Kr) and ‖f0‖ε′ ≤ ‖f‖ε .

Any δ as in the assertion then satisfies

δ <ε2

‖f‖ε‖(D0f)−1‖2=

ε′2

‖f‖ε≤ ε′2

‖f0‖ε′.

Obviously f0(0) = 0, and D0f0 = idKr by the chain rule. So we may applythe first case to f0 and obtain a uniquely determined g0 ∈ Fδ(Kr;Kr) suchthat

g0(0) = 0, ‖g0‖δ = δ, and f0 ◦ g0(Y ) = Y

as well as

δ <ε′2

‖f0‖ε′≤ ε′ = ε

‖(D0f)−1‖≤ ‖f0‖ε′ ≤ ‖f‖ε

by (7). We define

g := (D0f)−1 ◦ g0 ∈ Fδ(Kr;Kr) .

Then g(0) = 0 and

f(g(Y )) = f((D0f)−1 ◦ g0(Y )) = f ◦ (D0f)−1(g0(Y )) = f0(g0(Y )) = Y .

In addition, Prop. 5.4 implies

‖g‖δ ≤ ‖(D0f)−1‖δ = δ‖(D0f)−1‖ < ε .

The unicity of g easily follows from the unicity of g0.

The next result is more or less obvious.

35

Proposition 5.10. Let u : V −→ W be a continuous linear map betweenK-Banach spaces; then

Fε(Kr;V ) −→ Fε(Kr;W )

f(X) =∑α

Xαvα 7−→ u ◦ f(X) :=∑α

Xαu(vα)

is a continuous linear map of operator norm ≤ ‖u‖ which satisfies

u ◦ f(x) = u(f(x)) for any x ∈ Bε(0) .

6 Locally analytic functions

Let U ⊆ Kr be an open subset and V be a K-Banach space.

Definition. A function f : U −→ V is called locally analytic if for anypoint x0 ∈ U there is a ball Bε(x0) ⊆ U around x0 and a power seriesF ∈ Fε(Kr;V ) such that

f(x) = F (x− x0) for any x ∈ Bε(x0) .

The set

Can(U, V ) := all locally analytic functions f : U −→ V

is a K-vector space with respect to pointwise addition and scalar multi-plication. For f1, f2 ∈ Can(U, V ) and x0 ∈ U let Fi ∈ Fεi(K

r;V ) suchthat fi(x) = Fi(x − x0) for any x ∈ Bεi(x0). Put ε := min(ε1, ε2). ThenF1 + F2 ∈ Fε(Kr;V ) and

(f1 + f2)(x) = (F1 + F2)(x− x0) for any x ∈ Bε(x0) .

The vector space Can(U, V ) carries a natural topology which we will discusslater on in a more general context.

Example. By Cor. 5.5 we have F ∈ Can(Bε(0), V ) for any F ∈ Fε(Kr;V ).

Proposition 6.1. Suppose that f : U −→ V is locally analytic; then f isstrictly differentiable in every point x0 ∈ U and the function x 7−→ Dxf islocally analytic in Can(U,L(Kr, V )).

36

Proof. Let F ∈ Fε(Kr;V ) such that

f(x) = F (x− x0) for any x ∈ Bε(x0) .

From Prop. 5.6 and the chain rule we deduce that f is strictly differentiablein every x ∈ Bε(x0) and

Dxf((a1, . . . , ar)) =r∑i=1

aiD(i)x−x0

F (1) =r∑i=1

ai( ∂F∂Xi

)∼(x− x0) .

Let∂F

∂Xi(X) =

∑α

Xαvi,α .

For any multi-index α we introduce the continuous linear map

Lα : Kr −→ V

(a1, . . . , ar) 7−→ a1v1,α + . . .+ arvr,α .

Because of ‖Lα‖ ≤ maxi ‖vi,α‖ we have

G(X) :=i∑α

XαLα ∈ Fε(Kr;L(Kr, V ))

andDxf = G(x− x0) for any x ∈ Bε(x0) .

Remark 6.2. If K has characteristic zero then, for any function f : U −→V , the following conditions are equivalent:

i. f is locally constant;

ii. f is locally analytic with Dxf = 0 for any x ∈ U .

Proof. This is an immediate consequence of the Taylor formula in Cor. 5.7.

We now give a list of more or less obvious properties of locally analyticfunctions.

1) For any open subset U ′ ⊆ U we have the linear restriction map

Can(U, V ) −→ Can(U ′, V )f 7−→ f |U ′ .

37

2) For any open and closed subset U ′ ⊆ U we have the linear map

Can(U ′, V ) −→ Can(U, V )

f 7−→ f!(x) :=

{f(x) if x ∈ U ′,0 otherwise

called extension by zero.

3) If U =⋃i∈I Ui is a covering by pairwise disjoint open subsets then

Can(U, V ) ∼=∏i∈I

Can(Ui, V )

f 7−→ (f |Ui)i .

4) For any two K-Banach spaces V and W we have

Can(U, V ⊕W ) ∼= Can(U, V )⊕ Can(U,W )f 7−→ (prV ◦f,prW ◦f) .

In particular

Can(U,Kn) ∼=n∏i=1

Can(U,K) .

5) For any continuous bilinear map u : V1×V2 −→ V between K-Banachspaces we have the bilinear map

Can(U, V1)× Can(U, V2) −→ Can(U, V )(f, g) 7−→ u(f, g)

(cf. Prop. 5.2). In particular, Can(U,K) is a K-algebra (cf. Prop. 5.3),and Can(U, V ) is a module over Can(U,K).

6) For any continuous linear map u : V −→W between K-Banach spaceswe have the linear map

Can(U, V ) −→ Can(U,W )f 7−→ u ◦ f

(cf. Prop. 5.10).

38

Lemma 6.3. Let U ′ ⊆ Kn be an open subset and let g ∈ Can(U,Kn) suchthat g(U) ⊆ U ′; then the map

Can(U ′, V ) −→ Can(U, V )f 7−→ f ◦ g

is well defined and K-linear.

Proof. Let x0 ∈ U and put y0 := g(x0) ∈ U ′. We choose a ball Bε(y0) ⊆ U ′

and a power series F ∈ Fε(Kn;V ) such that

f(y) = F (y − y0) for any y ∈ Bε(y0) .

We also choose a ball Bδ(x0) ⊆ U and a power series G ∈ Fδ(Kr;Kn) suchthat

g(x) = G(x− x0) for any x ∈ Bδ(x0) .

Observing that

‖G−G(0)‖δ′ ≤δ′

δ‖G−G(0)‖δ for any 0 < δ′ ≤ δ

we may decrease δ so that

‖G− y0‖δ = ‖G−G(0)‖δ ≤ ε

(and, in particular, g(Bδ(x0)) ⊆ Bε(y0)) holds true. It then follows fromProp. 5.4 that F ◦ (G− y0) ∈ Fδ(Kr;V ) and

(F ◦ (G− y0))∼(x− x0) = F (G(x− x0)− y0)= F (g(x)− y0)= f(g(x))

for any x ∈ Bδ(x0).

The last result can be expressed by saying that the composite of locallyanalytic functions again is locally analytic.

Proposition 6.4. (Local invertibility) Let U ⊆ Kr be an open subset andlet f ∈ Can(U,Kr); suppose that Dx0f is bijective for some x0 ∈ U ; thenthere are open neighbourhoods U0 ⊆ U of x0 and U1 ⊆ Kr of f(x0) suchthat:

i. f : U0'−−→ U1 is a homeomorphism;

39

ii. the inverse map g : U1 −→ U0 is locally analytic, i. e., g ∈ Can(U1,Kr).

Proof. According to Prop. 4.3 we find open neighbourhoods U ′0 ⊆ U of x0

and U ′1 ⊆ Kr of f(x0) such that

f : U ′0'−−→ U ′1 is a homeomorphism.

We choose a ball Bε(x0) ⊆ U ′0 and a power series F ∈ Fε(Kr;Kr) such that

f(x) = F (x− x0) for any x ∈ Bε(x0) .

The power series F1(X) := F (X)− f(x0) ∈ Fε(Kr;Kr) satisfies F1(0) = 0.Moreover, D0F1 = Dx0f is invertible. By Prop. 5.9 we therefore find, for asufficiently small 0 < δ < ε, a power series G1(Y ) ∈ Fδ(Kr;Kr) such that

G1(0) = 0, ‖G1‖δ < ε, and F1 ◦G1(Y ) = Y .

In particular, G1 : Bδ(0) −→ Bε(0) is locally analytic. Hence the composite

g : U1 := Bδ(f(x0))−f(x0)−−−−−→ Bδ(0) G1−−→ Bε(0) +x0−−−→ Bε(x0)

is locally analytic and satisfies

f ◦ g(y) = f(G1(y − f(x0)) + x0) = F (G1(y − f(x0)))= F1(G1(y − f(x0))) + f(x0) = y − f(x0) + f(x0) = y

for any y ∈ U1. By further decreasing δ we may assume that U1 ⊆ U ′1, andby setting U0 := g(U1) we obtain the commutative diagram

U ′0f

'// U ′1

Bε(x0)

⊆

OO

U0

⊆OO

f //U1

goo

⊆

OO

in which the two lower horizontal arrows both are locally analytic and areinverse to each other.

There are “locally analytic versions” of the Corollaries 4.4 and 4.5 theformulation of which we leave to the reader.

We have done so already and we will systematically continue to call amap f : U −→ U ′ between open subsets U ⊆ Kr and U ′ ⊆ Kn locallyanalytic if the composite U

f−→ U ′⊆−−→ Kn is a locally analytic function.

40

Part II

Manifolds

We continue to fix the nonarchimedean field (K, | |). But we change oursystem of notations insofar as from now on we will denote K-Banach spacesby letters like E whereas we reserve letters like U and V for open subsetsin a topological space.

7 Charts and atlases

Let M be a Hausdorff topological space.

Definition. i. A chart for M is a triple (U,ϕ,Kn) consisting of an opensubset U ⊆M and a map ϕ : U −→ Kn such that:

(a) ϕ(U) is open in Kn,

(b) ϕ : U '−−→ ϕ(U) is a homeomorphism.

ii. Two charts (U1, ϕ1,Kn1) and (U2, ϕ2,K

n2) for M are called compati-ble if both maps

ϕ1(U1 ∩ U2)ϕ2◦ϕ−1

1 //ϕ2(U1 ∩ U2)

ϕ1◦ϕ−12

oo

are locally analytic.

We note that the condition in part ii. of the above definition makessense since ϕ1(U1 ∩ U2) is open in Kni . If (U,ϕ,Kn) is a chart then theopen subset U is called its domain of definition and the integer n ≥ 0 itsdimension. Usually we omit the vector space Kn from the notation andsimply write (U,ϕ) instead of (U,ϕ,Kn). If x is a point in U then (U,ϕ) isalso called a chart around x.

Lemma 7.1. Let (Ui, ϕi,Kni) for i = 1, 2 be two compatible charts for M ;if U1 ∩ U2 6= ∅ then n1 = n2.

Proof. Let x ∈ U1∩U2 and put xi := ϕi(x). We consider the locally analyticmaps

ϕ1(U1 ∩ U2)f :=ϕ2◦ϕ−1

1 //ϕ2(U1 ∩ U2) .

g:=ϕ1◦ϕ−12

oo

41

They are differentiable and inverse to each other, and x2 = f(x1). Hence,by the chain rule, the derivatives

Kn1

Dx1f //Kn2

Dx2goo

are linear maps inverse to each other. It follows that n1 = n2.

Definition. i. An atlas for M is a set A = {(Ui, ϕi,Kni)}i∈I of chartsfor M any two of which are compatible and which cover M in the sensethat M =

⋃i∈I Ui.

ii. Two atlases A and B for M are called equivalent if A ∪ B also is anatlas for M .

iii. An atlas A for M is called maximal if any equivalent atlas B for Msatisfies B ⊆ A.

Remark 7.2. i. The equivalence of atlases indeed is an equivalence re-lation.

ii. In each equivalence class of atlases there is exactly one maximal atlas.

Proof. i. Let A,B, and C be three atlases such that A is equivalent to B andB is equivalent to C. Then A is equivalent to C if we show that any chart(U1, ϕ1) in A is compatible with any chart (U2, ϕ2) in C. By symmetry itsuffices to show that the map ϕ2 ◦ ϕ−1

1 : ϕ1(U1 ∩ U2) −→ ϕ2(U1 ∩ U2) islocally analytic in a sufficiently small open neighbourhood of ϕ1(x) for anypoint x ∈ U1 ∩ U2. Since B covers M we find a chart (V, ψ) around x in B.By assumption (V, ψ) is compatible with both (U1, ϕ1) and (U2, ϕ2). Thenϕ1(U1∩V ∩U2) is an open neighbourhood of ϕ1(x) in ϕ1(U1∩U2) on whichthe map ϕ2 ◦ϕ−1

1 is the composite of the two locally analytic maps ϕ2 ◦ψ−1

and ψ ◦ ϕ−11 . Hence it is locally analytic by Lemma 6.3.

ii. If the given equivalence class consists of the atlases Aj for j ∈ J thenA :=

⋃j∈J Aj is the unique maximal atlas in this class.

Lemma 7.3. If A is a maximal atlas for M the domains of definition of allthe charts in A form a basis of the topology of M .

Proof. Let U ⊆M be an open subset. We have to show that U is the unionof the domains of definition of the charts in some subset of A, or equivalentlythat for any point x ∈ U we find a chart (Ux, ϕx) around x in A such that

42

Ux ⊆ U . Since A covers M we at least find a chart (U ′x, ϕ′x) around x in A.

We put Ux := U ′x ∩ U and ϕx := ϕ′x|Ux. Clearly (Ux, ϕx) is a chart aroundx for M such that Ux ⊆ U . We claim that (Ux, ϕx) is compatible with anychart (V, ψ) in A. But we do have the locally analytic maps

ϕ′x(U ′x ∩ V )ψ◦ϕ′−1

x //ψ(U ′x ∩ V )

ϕ′x◦ψ−1oo

which restrict to the locally analytic maps

ϕx(Ux ∩ V )ψ◦ϕ−1

x //ψ(Ux ∩ V ) .

ϕx◦ψoo

Hence B := A ∪ {(Ux, ϕx)} is an atlas equivalent to A. The maximality ofA then implies that B ⊆ A and a fortiori (Ux, ϕx) ∈ A.

Definition. An atlas A for M is called n-dimensional if all the charts inA with nonempty domain of definition have dimension n.

Remark 7.4. Let A be an n-dimensional atlas for M ; then any atlas Bequivalent to A is n-dimensional as well.

Proof. Let (V, ψ) be any chart in B and choose a point x ∈ V . We find achart (U,ϕ) in A around x. Since A and B are equivalent these two chartshave to be compatible. It then follows from Lemma 7.1 that both have thesame dimension u.

8 Manifolds

Definition. A (locally analytic) manifold (M,A) (over K) is a Hausdorfftopological space M equipped with a maximal atlas A. The manifold is calledn-dimensional (we write dimM = n) if the atlas A is n-dimensional.

By abuse of language we usually speak of a manifold M while consideringA as given implicitly. A chart for M will always mean a chart in A.

Example. Kn will always denote the n-dimensional manifold whose maxi-mal atlas is equivalent to the atlas {(U,⊆,Kn) : U ⊆ Kn open}.

Remark 8.1. Let (U,ϕ,Kn) be a chart for the manifold M ; if V ⊆ U is anopen subset then (V, ϕ|V,Kn) also is a chart for M .

43

Proof. This was shown in the course of the proof of Lemma 7.3.

Let (M,A) be a manifold and U ⊆M be an open subset. Then

AU := {(V, ψ,Kn) ∈ A : V ⊆ U} ,

by Lemma 7.3, is an atlas for U . We claim that AU is maximal. Let (V0, ψ0)be a chart for U which is compatible with any chart in AU . To see that(V0, ψ0) ∈ AU it suffices, by the maximality of A, to show that (V0, ψ0)is compatible with any chart (V, ψ) in A. The Remark 8.1 implies that(V ∩U,ψ|V ∩U) is a chart in A and hence in AU . By assumption (V0, ψ0) iscompatible with (V ∩U,ψ|V ∩U). Since V0∩V ⊆ V ∩U the compatibility of(V0, ψ0) with (V, ψ) follows trivially. The manifold (U,AU ) is called an opensubmanifold of (M,A).

As a nontrivial example of a manifold we discuss the d-dimensional pro-jective space Pd(K) over K. We recall that Pd(K) = (Kd+1 \ {0})/ ∼ is theset of equivalence classes in Kd+1 \ {0} for the equivalence relation

(a1, . . . , ad+1) ∼ (ca1, . . . , cad+1) for any c ∈ K× .

As usual we write [a1 : . . . : ad+1] for the equivalence class of (a1, . . . , ad+1).With respect to the quotient topology from Kd+1 \ {0} the projective spacePd(K) is a Hausdorff topological space. For any 1 ≤ j ≤ d + 1 we have theopen subset

Uj := {[a1 : . . . : ad+1] ∈ Pd(K) : |ai| ≤ |aj | for any 1 ≤ i ≤ d+ 1}

together with the homeomorphism

ϕj : Uj'−−→ B1(0) ⊆ Kd

[a1 : . . . : ad+1] 7−→(a1aj, . . . ,

aj−1

aj,aj+1

aj, . . . ,

ad+1

aj

).

The (Uj , ϕj ,Kd) are charts for Pd(K) such that⋃j Uj = Pd(K). We claim

that they are pairwise compatible. For 1 ≤ j < k ≤ d+ 1 the composite

f : V := {x ∈ B1(0) : |xk−1| = 1}ϕ−1j−−−→ Uj ∩ Uk

ϕk−−→ {y ∈ B1(0) : |yj | = 1}

is given by

f(x1, . . . , xd) =(

x1xk−1

, . . . ,xj−1

xk−1, 1xk−1

,xjxk−1

, . . . ,xk−2

xk−1, xkxk−1

, . . . , xdxk−1

).

44

Let a ∈ V be a fixed but arbitrary point and choose a 0 < ε < 1. ThenBε(a) ⊆ V . We consider the power series

Fj(X) := 1ak−1

∑n≥0

(− 1ak−1

)nXnk−1

and

Fi(X) := Fj(X) ·

{(Xi + ai) if 1 ≤ i < j or k ≤ i ≤ d,(Xi−1 + ai−1) if j < i < k .

Because of |ak−1| = 1 we have F := (F1, . . . , Fd) ∈ Fε(Kd;Kd). For x ∈Bε(a) we compute

Fj(x− a) = 1ak−1

∑n≥0

(− xk−1−ak−1

ak−1

)n = 1ak−1

· 1

1+xk−1−ak−1

ak−1

= 1xk−1

and thenf(x) = F (x− a) .

Hence f ist locally analytic. In case j > k the argument is analogous. Theabove charts therefore form a d-dimensional atlas for Pd(K).

Exercise. Let (M,A) and (N,B) be two manifolds. Then

A× B := {(U × V ), ϕ× ψ,Km+n) : (U,ϕ,Km) ∈ A, (V, ψ,Kn) ∈ B}

is an atlas for M × N with the product topology. We call M × N equippedwith the equivalent maximal atlas the product manifold of M and N .

Let M be a manifold and E be a K-Banach space.

Definition. A function f : M −→ E is called locally analytic if f ◦ ϕ−1 ∈Can(ϕ(U), E) for any chart (U,ϕ) for M .

Remark 8.2. i. Every locally analytic function f : M −→ E is contin-uous.

ii. Let B be any atlas consisting of charts for M ; a function f : M −→E is locally analytic if and only if f ◦ ϕ−1 ∈ Can(ϕ(U), E) for any(U,ϕ) ∈ B.

45

The set

Can(M,E) := all locally analytic functions f : M −→ E

is a K-vector space with respect to pointwise addition and scalar multipli-cation. It is easy to see that a list of properties 1) - 6) completely analogousto the one given in section 6 holds true. In a later section we will come backto a more detailed study of this vector space.

Let now M and N be two manifolds. The following result is immediate.

Lemma 8.3. For a map g : M −→ N the following assertions are equiva-lent:

i. g is continuous and ψ ◦ g ∈ Can(g−1(V ),Kn) for any chart (V, ψ,Kn)for N ;

ii. for any point x ∈ M there exist a chart (U,ϕ,Km) for M around xand a chart (V, ψ,Kn) for N around g(x) such that g(U) ⊆ V andψ ◦ g ◦ ϕ−1 ∈ Can(ϕ(U),Kn).

Definition. A map g : M −→ N is called locally analytic if the equivalentconditions in Lemma 8.3 are satisfied.

Lemma 8.4. i. If g : M −→ N is a locally analytic map and E is aK-Banach space then

Can(N,E) −→ Can(M,E)f 7−→ f ◦ g

is a well defined K-linear map.

ii. With Lf−→M

g−→ N also g ◦ f : L −→ N is a locally analytic map ofmanifolds.

Proof. This follows from Lemma 6.3.

Examples 8.5. 1) For any open submanifold U of M the inclusion map

U⊆−−→M is locally analytic.

2) Let g : M −→ N be a locally analytic map; for any open submanifoldV ⊆ N the induced map g−1(V )

g−→ V is locally analytic.

46

3) The two projection maps

pr1 : M ×N −→M and pr2 : M ×N −→ N

are locally analytic.

4) For any pair of locally analytic maps g : L −→ M and f : L −→ Nthe map

(g, f) : L −→M ×Nx 7−→ (g(x), f(x))

is locally analytic.

For the remainder of this section we will discuss a certain technical butuseful topological property of manifolds. First let X be an arbitrary Haus-dorff topological space. We recall:

- Let X =⋃i∈I Ui and X =

⋃j∈J Vj be two open coverings of X. The

second one is called a refinement of the first if for any j ∈ J there isan i ∈ I such that Vj ⊆ Ui.

- An open covering X =⋃i∈I Ui of X is called locally finite if every

point x ∈ X has an open neighbourhood Ux such that the set {i ∈ I :Ux ∩ Ui 6= ∅} is finite.

- The space X is called paracompact, resp. strictly paracompact, if anyopen covering of X can be refined into an open covering which is locallyfinite, resp. which consists of pairwise disjoint open subsets.

Remark 8.6. i. Any ultrametric space X is strictly paracompact.

ii. Any compact space X is paracompact.

Proof. i. This follows from Lemma 1.4. ii. This is trivial.

Proposition 8.7. For a manifold M the following conditions are equivalent:

i. M is paracompact;

ii. M is strictly paracompact;

iii. the topology of M can be defined by a metric which satisfies the stricttriangle inequality.

47

Proof. The implication iii. =⇒ ii. is Lemma 1.4, and the implication ii. =⇒i. is trivial.

i. =⇒ ii. We suppose that M is paracompact. From general topology werecall the following property of paracompact Hausdorff spaces (cf. [B-GT]Chap. IX §4.4 Cor. 2). Let A ⊆ U ⊆ M be subsets with A closed and Uopen. Then there is another open subset V ⊆M such that

A ⊆ V ⊆ V ⊆ U .

Step 1: We show that the open and closed subsets of M form a basis ofthe topology. Given a point x in an open subset U ⊆M we have to find anopen and closed subset W ⊆ M such that x ∈ W ⊆ U . By Lemma 7.3 wemay assume that U is the domain of definition of a chart (U,ϕ,Kn) for M .As recalled above there is an open neighbourhood V ⊆ M of x such thatV ⊆ U . We then have the vertical homeomorphisms

V⊆ //

'��

V⊆ //

'��

U⊆ //

'��

M

ϕ(V )⊆ // ϕ(V )

⊆ // ϕ(U)⊆ // Kn .

Since ϕ(V ) is open in Kn there is a ball B := Bε(ϕ(x)) ⊆ ϕ(V ) aroundϕ(x). We put W := ϕ−1(B) ⊆ V . Clearly x ∈ W ⊆ U . The ball B is openand hence B is open in V and M . But the ball B also is closed in Kn. HenceW is closed in V and therefore in M . This finishes step 1.

Let now M =⋃i∈I Ui be a fixed but arbitrary open covering. By Lemma

7.3 we may assume, after refinement, that any Ui is the domain of defini-tion of some chart for M . By the first step and Remark 8.1 we may evenassume, after a further refinement, that each Ui is open and closed in M andis the domain of definition of some chart for M . In particular, each Ui hasthe topology of an ultrametric space. By assumption we may pick a locallyfinite refinement (Vj)j∈J of (Ui)i∈I . So we have the locally finite open cover-ing M =

⋃j∈J Vj , and for each j ∈ J there is an i(j) ∈ I such that Vj ⊆ Ui(j).

Step 2: We construct a covering M =⋃j∈JWj by open and closed

subsets Wj ⊆ M such that Wj ⊆ Vj for any j ∈ J . For this purpose weequip J with a well-order ≤ (recall that this is a total order on J with theproperty that each nonempty subset of J has a minimal element - by theaxiom of choice such a well-order always exists). We now use transfiniteinduction to find open and closed subsets Wj ⊆M such that

48

(a) Wj ⊆ Vj for any j ∈ J , and

(b) M =(⋃

j≤kWj

)∪(⋃

j>k Vj)

for any k ∈ J .

We fix a k ∈ J and suppose that the Wj for j < k are constructed already.Claim: M =

(⋃j<kWj

)∪(⋃

j≥k Vj).

Let x ∈M . Since the covering (Vj)j is locally finite the set

{j ∈ J : x ∈ Vj} = {j1 < . . . < jr}

is finite. If jr ≥ k then x ∈ Vjr ⊆⋃j≥k Vj . If jr < k then x 6∈ Vj for

any j > jr and the induction hypothesis (property (b) for jr) implies x ∈⋃j≤jr Wj ⊆

⋃j<kWj . This establishes the claim.

We see that the closed subset

W := M \(( ⋃

j<k

Wj

)∪( ⋃j>k

Vj))

of M satisfiesW ⊆ Vk ⊆ Ui(k) .

Claim: Let (X, d) be an ultrametric space; for any subsets A ⊆ U ⊆ Xwith A closed and U open there exists an open and closed subset V ⊆ Xsuch that

A ⊆ V ⊆ U .

For any subset D ⊆ X and any x ∈ X we put

d(x,D) := infy∈D

d(x, y) .

The strict triangle inequality implies that the function d(., D) on X is con-tinuous and that

D(ε) := {x ∈ X : d(x,D) = ε} ,

for any ε > 0, is open in X. Moreover, D(0) = D. The closed subsets Aand B := X \ U of X satisfy A ∩B = ∅. By the continuity of the functionsd(., A) and d(., B) the subset

V := {x ∈ X : d(x,A) < d(x,B)}

therefore is open in X and satisfies A ⊆ V ⊆ U . Similarly V ′ := {x ∈ X :d(x,A) > d(x,B)} is open in X. It follows that V as the complement in Xof the open subset V ′ ∪

(⋃ε>0(A(ε) ∩B(ε))

)is closed. This establishes the

claim.

49

We apply this claim to W ⊆ Vk ⊆ Ui(k) and obtain an open and closedsubset Wk ⊆ Ui(k) such that W ⊆ Wk ⊆ Vk. With Ui(k) also Wk is openand closed in M . As W ⊆ Wk the index k has the property (b). It remainsto show that the Wj for j ∈ J actually cover M . Let x ∈ M . As arguedbefore the set {j ∈ J : x ∈ Vj} = {j1 < . . . < jr} is finite. Then x 6∈ Vjfor any j > jr. The property (b) for the index jr therefore implies thatx ∈

⋃j≤jr Wj . This finishes step 2.

Step 3: At this point we have constructed a locally finite refinement(Wj)j∈J of our initial covering which consists of open and closed subsetsWj ⊆M .

Claim: W ′L :=⋃j∈LWj , for any subset L ⊆ J , is open and closed in M .

Obviously W ′L is open. To see that its complement M \W ′L is open aswell let x ∈ M \W ′L be any point. In particular, x 6∈ Wj for any j ∈ L.Since the covering (Wj)j is locally finite we find an open neighbourhoodUx ⊆M of x such that the set {j ∈ L : Ux∩Wj 6= ∅} = {j1, . . . , js} is finite.Then Ux \ (Wj1 ∪ . . .∪Wjs) is an open neighbourhood of x in M \W ′L. Thisestablishes the claim.

We finally define a new index set P by

P := all nonempty finite subsets of J ,

and for any L ∈ P we put

WL :=( ⋂j∈L

Wj

)\( ⋃j∈J\L

Wj

)=( ⋂j∈L

Wj

)\W ′J\L .

Clearly any WL is contained in some Wj . By the above claim each WL isopen and closed in M . To check that

M =⋃L∈P

WL

holds true let x ∈ M be any point. Then x ∈ WL for the finite set L :={j ∈ J : x ∈ Wj}. Moreover, the WL are pairwise disjoint: Let L1 6= L2

be two different indices in P . By symmetry we may assume that there is aj ∈ L1 \ L2. Then WL1 ⊆ Wj and WL2 ⊆ M \Wj . It follows that (WL)L∈Pis a refinement of our initial covering by pairwise disjoint open subsets. Thisproves that M is strictly paracompact.

ii. =⇒ iii. We start with an open covering of M by domains of definitionof charts for M . By assumption we may refine it into a covering M =

⋃i∈I Ui

50

by pairwise disjoint open subsets. According to Remark 8.1 each Ui also isthe domain of definition of some chart for M . In particular, the topology ofUi can be defined by a metric d′i which satisfies the strict triangle inequality.We put

di(x, y) := d′i(x,y)1+d′i(x,y)

for any x, y ∈ Ui .

Obviously we have di(x, y) = di(y, x) and di(x, y) = 0 if and only if x = y.To see that di satisfies the strict triangle inequality we compute

di(x, z) = d′i(x,z)1+d′i(x,z)

≤ max(d′i(x,y),d′i(y,z))1+max(d′i(x,y),d′i(y,z))

= max( d′i(x,y)

1+d′i(x,y),

d′i(y,z)1+d′i(y,z)

)= max(di(x, y), di(y, z)) .

Here we have used the simple fact that t ≥ s ≥ 0 implies t(1 + s) = t+ ts ≥s+ st = s(1 + t) and hence t

1+t ≥s

1+s . For trivial reasons we have di ≤ d′i.On the other hand

d′i = di1−di

and hence, for 0 < ε ≤ 1,

d′i(x, y) ≤ ε if di(x, y) ≤ ε2 .

This shows that the metrics d′i and di define the same topology on Ui. Wenote that

di(x, y) < 1 for any x, y ∈ Ui .We now define

d : M ×M −→ R≥0

(x, y) 7−→

{di(x, y) if x, y ∈ Ui for some i ∈ I,1 otherwise .

This is a metric on d. The strict triangle equality

d(x, z) ≤ max(d(x, y), d(y, z))

only needs justification if not all three points lie in the same subset Ui. Butthen the right hand side is ≥ 1 whereas the left hand side is ≤ 1. We claimthat this metric d defines the topology of M . First consider any ball Bε(x)with respect to d in M . If ε ≥ 1 then Bε(x) = M , and if ε < 1 then Bε(x) isopen in some Ui. Hence Bε(x) is open in M . Vice versa let V ⊆ M be anyopen subset and let x ∈ V . We choose an i ∈ I such that x ∈ Ui. Then V ∩Uiis an open neighbourhood of x in Ui. Hence, for some 0 < ε < 1, the ballBε(x) with respect to d (or equivalently di) is contained in V ∩Ui ⊆ V .

51

Corollary 8.8. Open submanifolds and product manifolds of paracompactmanifolds are paracompact.

9 The tangent space

Let M be a manifold, and fix a point a ∈M . We consider pairs (c, v) where

– c = (U,ϕ,Km) is a chart for M around a and

– v ∈ Km.

Two such pairs (c, v) and (c′, v′) are called equivalent if we have

Dϕ(a)(ϕ′ ◦ ϕ−1)(v) = v′ .

It follows from the chain rule that this indeed defines an equivalence relation.

Definition. A tangent vector of M at the point a is an equivalence class[c, v] of pairs (c, v) as above.

We define

Ta(M) := set of all tangent vectors of M at a .

Lemma 9.1. Let c = (U,ϕ,Km) and c′ = (U ′, ϕ′,Km) be two charts for Maround a; we then have:

i. The map

θc : Km ∼−−→ Ta(M)v 7−→ [c, v]

is bijective.

ii. θ−1c′ ◦ θc : Km

∼=−−→ Km is a K-linear isomorphism.

Proof. (We recall from Lemma 7.1 that the dimensions of two charts aroundthe same point necessarily coincide.) i. Surjectivity follows from

[c′′, v′′] = [c,Dϕ′′(a)(ϕ ◦ ϕ′′−1)(v′′)] .

If [c, v] = [c, v′] then v′ = Dϕ(a)(ϕ ◦ϕ−1)(v) = v. This proves the injectivity.ii. From [c, v] = [c′, Dϕ(a)(ϕ′ ◦ ϕ−1)(v)] we deduce that

θ−1c′ ◦ θc = Dϕ(a)(ϕ

′ ◦ ϕ−1) .

52

The set Ta(M), by Lemma 9.1.i., has precisely one structure of a topo-logical K-vector space such that the map θc is a K-linear homeomorphism.Because of Lemma 9.1.ii. this structure is independent of the choice of thechart c around a.

Definition. The K-vector space Ta(M) is called the tangent space of M atthe point a.

Remark. The manifold M has dimension m if and only if dimK Ta(M) = mfor any a ∈M .

Let g : M −→ N be a locally analytic map of manifolds. By Lemma8.3.ii. we find charts c = (U,ϕ,Km) for M around a and c = (V, ψ,Kn) forN around g(a) such that g(U) ⊆ V . The composite

Ta(g) : Ta(M) θ−1c−−−→ Km Dϕ(a)(ψ◦g◦ϕ−1)

−−−−−−−−−−−→ Kn θc−−→ Tg(a)(N)

is a continuous K-linear map. We claim that Ta(g) does not depend on theparticular choice of charts. Let c′ = (U ′, ϕ′) and c′ = (V ′, ψ′) be other chartsaround a and g(a), respectively. Using the identity in the proof of Lemma9.1.ii. as well as the chain rule we compute

θc ◦Dϕ(a)(ψ ◦ g ◦ ϕ−1) ◦ θ−1c

= θc′ ◦Dψ(g(a))(ψ′ ◦ ψ−1) ◦Dϕ(a)(ψ ◦ g ◦ ϕ−1) ◦Dϕ(a)(ϕ

′ ◦ ϕ−1)−1 ◦ θ−1c′

= θc′ ◦Dϕ′(a)(ψ′ ◦ g ◦ ϕ′−1) ◦ θ−1

c′ .

Definition. Ta(g) is called the tangent map of g at the point a.

Remark. Ta(idM ) = idTa(M) .

Lemma 9.2. For any locally analytic maps of manifolds Lf−→ M

g−→ Nwe have

Ta(g ◦ f) = Tf(a)(g) ◦ Ta(f) for any a ∈ L .

Proof. This is an easy consequence of the chain rule.

Proposition 9.3. (Local invertibility) Let g : M −→ N be a locally analyticmap of manifolds, and suppose that Ta(g) : Ta(M)

∼=−−→ Tg(a)(N) is bijectivefor some a ∈ M ; then there are open neighbourhoods U ⊆ M of a andV ⊆ N of g(a) such that g restricts to a locally analytic isomorphism

g : U '−−→ V

of open submanifolds.

53

Proof. By Lemma 8.3.ii. we find charts c = (U ′, ϕ,Km) for M around a andc = (V ′, ψ,Kn) for N around g(a) such that g(U ′) ⊆ V ′. We consider thelocally analytic function

ϕ(U ′)ϕ−1

−−−→ U ′g−→ V ′

ψ−−→ ψ(V ′) ⊆ Kn .

By assumption the derivative

Dϕ(a)(ψ ◦ g ◦ ϕ−1) = θ−1c ◦ Ta(g) ◦ θc

is bijective. Prop. 6.4 therefore implies the existence of open neighbourhoodsW0 ⊆ ϕ(U ′) of ϕ(a) and W1 ⊆ ψ(V ′) of ψ(g(a)) such that

ψ ◦ g ◦ ϕ−1 : W0'−−→W1

is a locally analytic isomorphism. Hence

g : U := ϕ−1(W0) '−−→ V := ψ−1(W1)

is a locally analytic isomorphism as well (observe the subsequent exercise).

Exercise. Let (U,ϕ,Km) be a chart for the manifold M ; then ϕ : U '−−→ϕ(U) is a locally analytic isomorphism between the open submanifolds U ofM and ϕ(U) of Km.

Let M be a manifold, E be a K-Banach space, f ∈ Can(M,E), and a ∈M . If c = (U,ϕ,Km) is a chart for M around a then f ◦ϕ−1 ∈ Can(ϕ(U), E).Hence

daf : Ta(M) θ−1c−−→ Km Dϕ(a)(f◦ϕ−1)

−−−−−−−−−→ E

[c, v] p−−−−−−−−−−−→ Dϕ(a)(f ◦ ϕ−1)(v)

is a continuous K-linear map. If c′ = (U ′, ϕ′,Km) is another chart arounda then

Dϕ(a)(f ◦ ϕ−1) ◦ θ−1c = Dϕ(a)(f ◦ ϕ−1) ◦Dϕ(a)(ϕ

′ ◦ ϕ−1)−1 ◦ θ−1c′

= Dϕ′(a)(f ◦ ϕ′−1) ◦ θ−1c′ .

This shows that daf does not depend on the choice of the chart c.

Definition. daf is called the derivative of f in the point a.

54

Remark 9.4. For E = Kr viewed as a manifold and for the chart c0 =(Kr, id, E) for E we have

Ta(f) = θc0 ◦ daf .

Obviously the map

Can(M,E) −→ L(Ta(M), E)f 7−→ daf

is K-linear.

Lemma 9.5. (Product rule)

i. Let u : E1×E2 −→ E be a continuous bilinear map between K-Banachspaces; if fi ∈ Can(M,Ei) for i = 1, 2 then u(f1, f2) ∈ Can(M,E) and

da(u(f1, f2)) = u(f1(a), daf2) + u(daf1, f2(a)) for any a ∈M .

ii. For g ∈ Can(M,K) and f ∈ Can(M,E) we have

da(gf) = g(a) · daf + dag · f(a) for any a ∈M .

Proof. i. It is a straightforward consequence of Prop. 5.2 that the functionu(f1, f2) is locally analytic (compare the property 5) of the list in section 6).Let c = (U,ϕ) be a chart of M around a. Using the product rule in Remark4.1.iv. we compute

da(u(f1, f2))([c, v]) = Dϕ(a)(u(f1, f2) ◦ ϕ−1)(v)

= Dϕ(a)(u(f1 ◦ ϕ−1, f2 ◦ ϕ−1))(v)

= u(f1 ◦ ϕ−1(ϕ(a)), Dϕ(a)(f2 ◦ ϕ−1)(v))

+ u(Dϕ(a)(f1 ◦ ϕ−1)(v), f2 ◦ ϕ−1(ϕ(a)))

= u(f1(a), daf2([c, v]))+u(daf1([c, v]), f2(a)) .

ii. This is a special case of the first assertion.

Let c = (U,ϕ,Km) be a chart for M . On the one hand, by definition, wehave daϕ = θ−1

c for any a ∈ U ; in particular

daϕ : Ta(M)∼=−−→ Km

55

is a K-linear isomorphism. On the other hand viewing ϕ = (ϕ1, . . . , ϕm) asa tuple of locally analytic functions ϕi : U −→ K we have

daϕ = (daϕ1, . . . , daϕm) .

This means that {daϕi}1≤i≤m is a K-basis of the dual vector space Ta(M)′.Let {(

∂∂ϕi

)(a)}

1≤i≤m

denote the corresponding dual basis of Ta(M), i. e.,

daϕi((

∂∂ϕj

)(a))

= δij for any a ∈ U

where δij is the Kronecker symbol. For any f ∈ Can(M,E) we define thefunctions

∂f∂ϕi

: U −→ E

a 7−→ daf((

∂∂ϕi

)(a)).

Lemma 9.6. ∂f∂ϕi∈ Can(U,E) for any 1 ≤ i ≤ m, and

daf =m∑i=1

daϕi · ∂f∂ϕi (a) for any a ∈ U .

Proof. We have

∂f∂ϕi

(a) = Dϕ(a)(f ◦ ϕ−1) ◦ θ−1c

((∂∂ϕi

)(a))

= Dϕ(a)(f ◦ ϕ−1)(ei)

where e1, . . . , em denotes the standard basis of Km. Hence ∂f∂ϕi

is the com-posite

Uϕ−→ ϕ(U)

x 7→Dx(f◦ϕ−1)−−−−−−−−−−→ L(Km, E)D 7→D(ei)−−−−−−−→ E .

The function in the middle is locally analytic by Prop. 6.1. Clearly, D 7−→D(ei) is a continuous K-linear map. Hence the composite of the right twomaps is locally analytic by the property 6) in section 6. That the full com-posite ∂f

∂ϕiis locally analytic now follows from Lemma 8.4.i. Let

t =m∑i=1

ci(∂∂ϕi

)(a) ∈ Ta(M)

56

be an arbitrary vector. By the definition of the dual basis we have ci =daϕi(t). We now compute

daf(t) =m∑i=1

ci · daf((

∂∂ϕi

)(a))

=m∑i=1

daϕi(t) · ∂f∂ϕi (a) .

In a next step we want to show that the disjoint union

T (M) :=⋃a∈M

Ta(M)

in a natural way is a manifold again. We introduce the projection map

pM : T (M) −→M

t 7−→ a if t ∈ Ta(M) .

Hence Ta(M) = p−1M (a).

Consider any chart c = (U,ϕ,Km) for M . By Lemma 9.1.i. the map

τc : U ×Km ∼−−→ p−1M (U)

(a, v) 7−→ [c, v] viewed in Ta(M)

is bijective. Hence the composite

ϕc : p−1M (U) τ−1

c−−−→ U ×Km ϕ×id−−−−→ Km ×Km = K2m

is a bijection onto an open subset in K2m. The idea is that

cT :=(p−1M (U), ϕc,K2m

)should be a chart for the manifold T (M) yet to be constructed. Clearly wehave

T (M) =⋃

c=(U,ϕ)

p−1M (U) .

Let c = (V, ψ,Km) be another chart for M such that U ∩ V 6= ∅. Thecomposed map

ϕ(U ∩ V )×Km ϕ−1c−−−→ p−1

M (U ∩ V ) = p−1M (U)∩ p−1

M (V )ψc−−→ ψ(U ∩ V )×Km

57

is given by

(8) (x, v) 7−→ (ψ ◦ ϕ−1(x), Dx(ψ ◦ ϕ−1)(v)) .

The first component ψ◦ϕ−1 of this map is locally analytic on ϕ(U ∩V ) sinceM is a manifold. The second component can be viewed as the composite

ϕ(U ∩ V )×Km −→ L(Km,Km)×Km −→ Km

(x, v) 7−→ (Dx(ψ ◦ ϕ−1), v)(u, v) 7−→ u(v) .

The left function is locally analytic by Prop. 6.1. The right bilinear map iscontinuous. Hence the composite is locally analytic by Lemma 9.5.i. Thisshows that, once cT and cT are recognized as charts for T (M) with respectto a topology yet to be defined, they in fact are compatible, and hence thatthe set {cT : c a chart for M} is an atlas for T (M).

We have shown in particular that the composed map

(9) (U ∩ V )×Km τc−→ p−1M (U ∩ V )

τ−1c−−−→ (U ∩ V )×Km

is a homeomorphism.

Definition. A subset X ⊆ T (M) is called open if τ−1c (X ∩ p−1

M (U)) is openin U ×Km for any chart c = (U,ϕ,Km) for M .

This defines the finest topology on T (M) which makes all composedmaps

U ×Km τc−−→ p−1M (U) ⊆−−→ T (M)

continuous.

Lemma 9.7. i. The map τc : U ×Km '−−→ p−1M (U) is a homeomorphism

with respect to the subspace topology induced by T (M) on p−1M (U).

ii. The map pM is continuous.

iii. The topological space T (M) is Hausdorff.

Proof. i. The continuity of τc holds by construction. Let Y ⊆ U × Km

be an open subset. We will show that τc(Y ) is open in T (M), i. e., thatτ−1c (τc(Y )∩ p−1

M (V )) is open in V ×Km for any chart c = (V, ψ,Kn) for M .

58

We may of course assume that U ∩V 6= ∅ so that n = m. Clearly the subsetY ∩ ((U ∩ V )×Km) is open in (U ∩ V )×Km. By (9) the subset

τ−1c (τc(Y ∩ ((U ∩ V )×Km))) = τ−1

c (τc(Y ) ∩ p−1M (U ∩ V ))

= τ−1c (τc(Y ) ∩ p−1

M (U) ∩ p−1M (V ))

= τ−1c (τc(Y ) ∩ p−1

M (V ))

is open in (U ∩ V )×Km and therefore in V ×Km.ii. The above reasoning for Y = U ×Km shows that τc(Y ) = p−1

M (U) isopen in T (M) where U is the domain of definition of any chart for M . Itthen follows from Lemma 7.3 that pM is continuous.

iii. Let s and t be two different points in T (M). Case 1: We have pM (s) 6=pM (t). Since M is Hausdorff we find open neighbourhoods U ⊆M of pM (s)and V ⊆M of pM (t) such that U∩V = ∅. By ii. then p−1

M (U) and p−1M (V ) are

open neighbourhoods of s and t, respectively, such that p−1M (U) ∩ p−1

M (V ) =p−1M (U ∩ V ) = ∅. Case 2: We have a := pM (s) = pM (t). We choose a chartc = (U,ϕ,Km) for M around a. The two points s and t lie in the open (byii.) subset p−1

M (U) of T (M). But by i. the subspace p−1M (U) is homeomorphic,

via the map τc, to the Hausdorff space U ×Km. Hence p−1M (U) is Hausdorff

and s and t can be separated by open neighbourhoods in p−1M (U) and a

fortiori in T (M).

The Lemma 9.7 in particular says that cT indeed is a chart for T (M).Altogether we now have established that {cT : c a chart for M} is an atlasfor T (M). We always view T (M) as a manifold with respect to the equivalentmaximal atlas.

Definition. The manifold T (M) is called the tangent bundle of M .

Remark. If M is m-dimensional then T (M) is 2m-dimensional.

Lemma 9.8. The map pM : T (M) −→M is locally analytic.

Proof. Let c = (U,ϕ,Km) be a chart for M . It suffices to contemplate thecommutative diagram

T (M)

pM

��

p−1M (U)

⊇oo

��

// ϕc(p−1M (U)) = ϕ(U)×Km

pr1��

⊆ // K2m

M U⊇oo ϕ // ϕ(U)

⊆ // Km .

59

Let g : M −→ N be a locally analytic map of manifolds. We define themap

T (g) : T (M) −→ T (N)

byT (g)|Ta(M) := Ta(g) for any a ∈M .

In particular, the diagram

T (M)T (g) //

pM

��

T (N)

pN

��M

g // N

is commutative.

Proposition 9.9. i. The map T (g) is locally analytic.

ii. For any locally analytic maps of manifolds Lf−→M

g−→ N we have

T (g ◦ f) = T (g) ◦ T (f) .

Proof. i. We choose charts c = (U,ϕ,Km) for M and c = (V, ψ,Kn) for Nsuch that g(U) ⊆ V . The composite

ϕ(U)×Km ϕ−1c−−−→ p−1

M (U)T (g)−−−→ p−1

N (V )ψc−−→ ψ(V )×Kn

is given by

(x, v) 7−→ (ψ ◦ g ◦ ϕ−1(x), Dx(ψ ◦ g ◦ ϕ−1)(v)) .

It is locally analytic by the same argument as for (8).ii. This is a restatement of Lemma 9.2.

The following is left to the reader as an exercise.

Remark 9.10. i. If U ⊆ M is an open submanifold then T (⊆) inducesan isomorphism between T (U) and the open submanifold p−1

M (U).

ii. For any two manifolds M and N the map

T (pr1)× T (pr2) : T (M ×N) '−−→ T (M)× T (N)

is an isomorphism of manifolds.

60

Now let M be a manifold and E be a K-Banach space. For any f ∈Can(M,E) we define

df : T (M) −→ E

t 7−→ dpM (t)f(t) .

Lemma 9.11. We have df ∈ Can(T (M), E).

Proof. Let c = (U,ϕ,Km) be a chart for M . The composed map

ϕ(U)×Km ϕ−1c−−−→ p−1

M (U)df−−→ E

is given by(x, v) 7−→ Dx(f ◦ ϕ−1)(v)

and hence is locally analytic by the same argument as for (8).

Lemma 9.12. Let g : M −→ N be a locally analytic map of manifolds; forany f ∈ Can(N,E) we have

d(f ◦ g) = df ◦ T (g) .

Proof. This is a consequence of the chain rule.

Exercise. The map

d : Can(M,E) −→ Can(T (M), E)f 7−→ df

is K-linear.

Remark 9.13. If K has characteristic zero then a function f ∈ Can(M,E)is locally constant if and only if df = 0.

Proof. Let c = (U,ϕ) be any chart for M . As can be seen from the proof ofLemma 9.11 we have df |p−1

M (U) = 0 if and only if Dx(f ◦ ϕ−1) = 0 for anyx ∈ ϕ(U). By Remark 6.2 the latter is equivalent to f ◦ ϕ−1 being locallyconstant on ϕ(U) which, of course, is the same as f being locally constanton U .

Definition. Let U ⊆M be an open subset; a vector field ξ on U is a locallyanalytic map ξ : U −→ T (M) which satisfies pM ◦ ξ = idU .

61

We define

Γ(U, T (M)) := set of all vector fields on U .

It follows from Remark 9.10.i. that

Γ(U, T (M)) = Γ(U, T (U)) .

Suppose that U is the domain of definition of some chart c = (U,ϕ,Km) forM . Because of the commutative diagram

p−1M (U) '

ϕc//

pM""FFFFFFFFF

ϕ(U)×Km

(x,v)7→ϕ−1(x)yyttttttttttt

U

the map

Can(U,Km) ∼−−→ Γ(U, T (M))

f 7−→ ξf (a) := ϕ−1c ((ϕ(a), f(a))) = τc(a, f(a))

is bijective. The left hand side is a K-vector space. On the right hand sidethis vector space structure corresponds to the pointwise addition and scalarmultiplication of maps which makes sense since each Ta(M) is a K-vectorspace. The latter we can define on any open subset U ⊆M . For any c ∈ Kand ξ, η ∈ Γ(U, T (M)) we define

(c · ξ)(a) := c · ξ(a) and (ξ + η)(a) := ξ(a) + η(a) .

Obviously the result are again maps ? : U −→ T (M) satisfying pM◦ ? = idU .But since U can be covered by domains of definition of charts for M theabove discussion actually implies that these maps are locally analytic again.We see that

Γ(U, T (M)) is a K-vector space.

We have the bilinear map

Γ(M,T (M))× Can(M,E) −→ Can(M,E)(ξ, f) 7−→ Dξ(f) := df ◦ ξ .

Lemma 9.14. Let u : E1×E2 −→ E be a continuous bilinear map betweenK-Banach spaces; for any ξ ∈ Γ(M,T (M)) and fi ∈ Can(M,Ei) we have

Dξ(u(f1, f2)) = u(Dξ(f1), f2) + u(f1, Dξ(f2)) .

62

Proof. This follows from the product rule in Lemma 9.5.i.

Corollary 9.15. For any vector field ξ ∈ Γ(M,T (M)) the map

Dξ : Can(M,K) −→ Can(M,K)

is a derivation, i. e.:

(a) Dξ is K-linear,

(b) Dξ(fg) = Dξ(f)g + fDξ(g) for any f, g ∈ Can(M,K).

Proposition 9.16. Suppose that M is paracompact; then for any derivationD on Can(M,K) there is a unique vector field ξ on M such that D = Dξ.

The proof requires some preparation. In the following we always assumeM to be paracompact. At first we fix a point a ∈ M . A K-linear map∆ : Can(M,K) −→ K will be called an a-derivation if

∆(fg) = ∆(f)g(a) + f(a)∆(g) for any f, g ∈ Can(M,K) .

The a-derivations form a K-vector subspace

Dera(M,K)

of the dual vector space Can(M,K)∗.

Lemma 9.17. Suppose that M is paracompact, and let ∆ be an a-derivation;if f ∈ Can(M,K) is constant in a neighbourhood of the point a then ∆(f) =0.

Proof. Case 1: We assume that f vanishes in the neighbourhood U ⊆M ofa. By Prop. 8.7 we may assume that U is open and closed in M . Then thefunction

g(x) :=

{1 if x 6∈ U ,

0 if x ∈ U

lies in Can(M,K) and satisfies gf = f . It follows that

∆(f) = ∆(gf) = ∆(g)f(a) + g(a)∆(f) = 0 .

Case 2: We assume that f is constant on M with value c. Let 1M denotethe constant function with value one on M . Then f = c1M and hence

∆(f) = c∆(1M ) = c∆(1M1M ) = c∆(1M ) + c∆(1M ) = 2c∆(1M ) = 2∆(f)

63

which means ∆(f) = 0.Case 3: In general we write

f = f(a)1M + (f − f(a)1M )

and use the K-linearity of ∆ together with the first two cases.

As a consequence of the product rule Lemma 9.5.ii. we have the K-linearmap

Ta(M) −→ Dera(M,K)t 7−→ ∆t(f) := daf(t) .

(10)

Proposition 9.18. If M is paracompact then (10) is an isomorphism.

Proof. We fix a chart c = (U,ϕ,Km) for M around a point a and writeϕ = (ϕ1, . . . , ϕm). Since M is paracompact we may assume by Prop. 8.7that U is open and closed in M . Then each ϕi extends by zero to a functionϕi! ∈ Can(M,K). In the discussion before Lemma 9.6 we had seen that

daϕ = (daϕ1, . . . , daϕm) : Ta(M)∼=−−→ Km

is an isomorphism, and we had introduced theK-basis ti := ∂∂ϕi

(a) of Ta(M).Injectivity: Let t ∈ Ta(M) such that ∆t(f) = 0 for any f ∈ Can(M,K).

In particular

0 = ∆t(ϕi!) = daϕi!(t) = daϕi(t) for any 1 ≤ i ≤ m .

This means daϕ(t) = 0 and hence t = 0.Surjectivity: From the injectivity which we just have established we de-

duce that the ∆ti are linearly independent in Dera(M,K). It therefore suf-fices to write an arbitrarily given ∆ ∈ Dera(M,K) as a linear combinationof the ∆ti . In fact, we claim that

∆ =m∑i=1

∆(ϕi!) ·∆ti

holds true. Let f ∈ Can(M,K). We find an open and closed neighbour-hood V ⊆ U of a such that ϕ(V ) = Bε(ϕ(a)) and a power series F (X) =∑

α cαXα ∈ Fε(Km;K) such that

f(x) = F (ϕ(x)− ϕ(a)) for any x ∈ V .

64

We may write

f(x) =∑α

cα(ϕ(x)− ϕ(a))α = f(a) +m∑i=1

(ϕi(x)− ϕi(a))gi(x)

for any x ∈ V where the gi ∈ Can(V,K) are appropriate functions satisfyinggi(a) = ci (recall that i = (0, . . . , 1, . . . , 0)). From the proof of Prop. 6.1 weknow that

Dϕ(a)(f ◦ ϕ−1)(ei) =∂F

∂Xi(0) = ci

where e1, . . . , em denotes, as usual, the standard basis of Km. We thereforeobtain

gi(a) = ci = Dϕ(a)(f ◦ ϕ−1)(ei) = daf(θc(ei)) .

By the construction of the ti we have θc(ei) = ti. It follows that

gi(a) = daf(ti) .

On the other hand we extend each gi by zero to a function gi! ∈ Can(M,K).The function

f −m∑i=1

(ϕi! − ϕi(a))gi!

is constant (with value f(a)) in a neighbourhood of a. Using Lemma 9.17we compute

∆(f) = ∆( m∑i=1

(ϕi! − ϕi(a))gi!)

=m∑i=1

∆(ϕi! − ϕi(a)

)gi(a)

=m∑i=1

∆(ϕi!) · daf(ti)

=m∑i=1

∆(ϕi!) ·∆ti(f) .

Since f was arbitrary this establishes our claim.

Proof of Proposition 9.16: First of all we note that the relation be-tween derivations and a-derivations on Can(M,K) is given by the formula

(11) Dξ(f)(a) = df(ξ(a)) = daf(ξ(a)) = ∆ξ(a)(f) .

65

Therefore, if Dξ = 0 then ∆ξ(a) = 0 for any a ∈ M . The Prop. 9.18 thenimplies that ξ(a) = 0 for any a ∈ M , i. e., that ξ = 0. This shows that theξ in our assertion is unique if it exists. For the existence we first fix a pointa ∈M and consider the a-derivation ∆(f) := D(f)(a). By Prop. 9.18 thereis a tangent vector ξ(a) ∈ Ta(M) such that ∆ = ∆ξ(a). For varying a ∈ Mthis gives a map ξ : M −→ T (M) which satisfies pM ◦ξ = idM . It remains toshow that ξ is locally analytic, since D = Dξ then is a formal consequenceof (11). So let c = (U,ϕ,Km) be a chart for M . In the proof of Prop. 9.18we have seen that

D(f)(a) =m∑i=1

D(ϕi!)(a) ·∆θc(ei)(f) .

It follows that

ξ(a) =m∑i=1

D(ϕi!)(a) · θc(ei) = θc((D(ϕ1!)(a), . . . , D(ϕm!)(a))

).

Using the commutative diagram

Kmv 7−→(a,v) //

θc ∼=��

U ×Km

τc'��

Ta(M)⊆ // p−1

M (U)

we rewrite this as

ξ(a) = τc(a, (D(ϕ1!)(a), . . . , D(ϕm!)(a))) .

This means that under the identification discussed after the definition ofvector fields we have

ξ|U = ξf with f := (D(ϕ1!), . . . , D(ϕm!)) ∈ Can(U,Km) .

Hence ξ is locally analytic. �

Lemma 9.19. For any derivations B,C,D : Can(M,K) −→ Can(M,K) wehave:

i. [B,C] := B ◦ C − C ◦B again is a derivation;

ii. [ , ] is K-bilinear;

66

iii. [B,B] = 0 and [B,C] = −[C,B];

iv. (Jacobi identity) [[B,C], D] + [[C,D], B] + [[D,B], C] = 0.

Proof. These are straightforward completely formal computations.

Definition. A K-vector space g together with a K-bilinear map

[ , ] : g× g −→ g

which is antisymmetric (i. e., [z, z] = 0 for any z ∈ g) and satisfies theJacobi identity is called a Lie algebra over K.

If M is paracompact then, using Prop. 9.16 and Lemma 9.19, we maydefine the Lie product [ξ, η] of two vector fields ξ, η ∈ Γ(M,T (M)) by therequirement that

D[ξ,η] = Dξ ◦Dη −Dη ◦Dξ

holds true. This makes Γ(M,T (M)) into a Lie algebra over K.

Proposition 9.20. Suppose that M is paracompact, and let E be a K-Banach space and ξ, η ∈ Γ(M,T (M)) be two vector fields; on Can(M,E) wethen have

D[ξ,η] = Dξ ◦Dη −Dη ◦Dξ .

Proof. Let f ∈ Can(M,E). We have to show equality of the functions

D[ξ,η](f) = df ◦ [ξ, η]

and

(Dξ ◦Dη −Dη ◦Dξ)(f) = d(Dη(f)) ◦ ξ − d(Dξ(f)) ◦ η= d(df ◦ η) ◦ ξ − d(df ◦ ξ) ◦ η .

This, of course, can be done after restriction to the domain of definition Uof any chart c = (U,ϕ,Km) for M . Since M is paracompact we furthermoreneed only to consider charts for which U is open and closed in M . Letϕ = (ϕ1, . . . , ϕm) and denote, as before, by ϕi! ∈ Can(M,K) the extensionby zero of ϕi. We now make use of the following identifications. If ? denotesthe restriction to U of any of the vector fields ξ, η, and [ξ, η] then, as discussedafter the definition of vector fields, we have a commutative diagram

p−1M (U) '

ϕc// ϕ(U)×Km

U

?

OO

'ϕ

// ϕ(U)

x 7→(x,g?(x))

OO

67

with g? ∈ Can(ϕ(U),Km). On the other hand, as noticed already in the proofof Lemma 9.11, we have, for any function ? ∈ Can(U,E), the commutativediagram

E

p−1M (U)

d?

<<xxxxxxxxx'ϕc

// ϕ(U)×Km .

(x,v)7→Dx(?◦ϕ−1)(v)

eeKKKKKKKKKKK

These identifications reduce us to proving the equality of the following twofunctions of x ∈ ϕ(U) given by

(12) Dx(f ◦ ϕ−1)(g[ξ,η](x))

and

Dx(df ◦ η ◦ ϕ−1)(gξ(x))−Dx(df ◦ ξ ◦ ϕ−1)(gη(x))

= Dx

(D.(f ◦ ϕ−1)(gη(.))

)(gξ(x))−Dx

(D.(f ◦ ϕ−1)(gξ(.))

)(gη(x)) ,

(13)

respectively. By viewing D.(f ◦ϕ−1), resp. gη(.) and gξ(.), as functions fromϕ(U) into L(Km, E), resp. into Km, we may apply the product rule Remark4.1.iv. for the continuous bilinear map

L(Km, E)×Km −→ E

(u, v) 7−→ u(v)

to both summands in the last expression for (13) and rewrite it as

= [Dx(D.(f ◦ ϕ−1))(gξ(x))](gη(x)) +Dx(f ◦ ϕ−1)[Dxgη(gξ(x))]

− [Dx(D.(f ◦ ϕ−1)(gη(x))](gξ(x))−Dx(f ◦ ϕ−1)[Dxgξ(gη(x))] .(14)

To simplify this further we establish the following generalClaim: For any open subset V ⊆ Km, any point x ∈ V , any vectors v =

(v1, . . . , vm) and w = (w1, . . . , wm) in Km, and any function h ∈ Can(V,E)we have

Dx(D.h(v))(w) = Dx(D.h(w))(v) .

(Note that the function D.h(v) is the composite

Vy 7→Dyh−−−−−→ L(Km, E)

u7→u(v)−−−−−→ E .)

We expand h around the point x into a power series

h(y) = H(y − x) .

68

By the proof of Prop. 6.1 we then have

Dyh(v) =m∑i=1

vi∂H

∂Yi(y − x)

and

Dx(D.h(v))(w) =m∑i=1

viDx

(∂H∂Yi

(.− x))(w)

=m∑i=1

vi

m∑j=1

wj( ∂

∂Yj

∂

∂YiH)(0)

=m∑j=1

wj

m∑i=1

vi( ∂

∂Yi

∂

∂YjH)(0)

...= Dx(D.h(w))(v) .

Applying this claim to (14) we see that the expression for the function(13) simplifies to

Dx(f ◦ ϕ−1)[Dxgη(gξ(x))−Dxgξ(gη(x))] .

Comparing this with (12) we are reduced to showing that the identity

(15) g[ξ,η](x) = Dxgη(gξ(x))−Dxgξ(gη(x))

holds true in Can(ϕ(U),Km). But in case E = K our assertion and thewhole computation above holds by construction. In particular we have

Dx(ϕi! ◦ ϕ−1)(g[ξ,η](x)) = Dx(ϕi! ◦ ϕ−1)[Dxgη(gξ(x))−Dxgξ(gη(x))]

for any 1 ≤ i ≤ m. Since

Dx(ϕi! ◦ ϕ−1) : Km −→ K

(v1, . . . , vm) 7−→ vi

the identity (15) follows immediately.

Remarks 9.21. i. The identity (15) shows that Can(V,Km), for anyopen subset V ⊆ Km, is a Lie algebra with respect to

[f, g](x) := Dxf(g(x))−Dxg(f(x)) .

69

ii. The identity (15) can be made into a definition of which one then canshow that it is compatible with any change of charts for M . In thisway a Lie product [ξ, η] can be obtained and Prop. 9.20 can be provedeven for manifolds which are not paracompact.

10 The topological vector space Can(M, E), part 1

Throughout this section M is a paracompact manifold and E is a K-Banachspace. Following [Fea] we will show that Can(M,E) in a natural way is atopological vector space.

To motivate the later construction we first consider a fixed functionf ∈ Can(M,E). Since, by Prop. 8.7, M is strictly paracompact we find afamily of charts (Uj , ϕj ,Kmj ), for j ∈ J , for M such that the Uj are pairwisedisjoint and M =

⋃j∈J Uj . According to Remark 8.2.ii. the function f is

locally analytic if and only if all f ◦ ϕ−1j : ϕj(Uj) −→ E, for j ∈ J , are

locally analytic. For each ϕj(Uj) we find balls Bεj,ν (xj,ν) ⊆ Kmj and powerseries Fj,ν ∈ Fεj,ν (Kmj ;E) such that

(16) ϕj(Uj) =⋃ν

Bεj,ν (xj,ν)

andf ◦ ϕ−1

j (x) = Fj,ν(x− xj,ν) for any x ∈ Bεj,ν (xj,ν) .

By Lemma 1.4 the covering (16) can be refined into a covering by pair-wise disjoint balls Bδj,α(yj,α). Consider a fixed α. We find a ν such thatBδj,α(yj,α) ⊆ Bεj,ν (xj,ν). In fact we then have

Bmin(δj,α,εj,ν)(yj,α) = Bδj,α(yj,α) ⊆ Bεj,ν (xj,ν) = Bεj,ν (yj,α) .

Hence we may assume that δj,α ≤ εj,ν . Using Cor. 5.5 we may change Fj,νinto a power series Fj,α ∈ Fδj,α(Kmj ;E) such that

f ◦ ϕ−1j (x) = Fj,α(x− yj,α) for any x ∈ Bδj,α(yj,α) .

We put Uj,α := ϕ−1j (Bδj,α(yj,α)). The (Uj,α, ϕj ,Kmj ) again are charts for M

such that the Uj,α cover M and are pairwise disjoint.

Resume: Given f ∈ Can(M,E) there is a family of charts (Ui, ϕi,Kmi),for i ∈ I, for M together with real numbers εi > 0 such that:

(a) M =⋃i∈I Ui, and the Ui are pairwise disjoint;

70

(b) ϕi(Ui) = Bεi(xi) for one (or any) xi ∈ ϕi(Ui);

(c) there is a power series Fi ∈ Fεi(Kmi ;E) with

f ◦ ϕ−1i (x) = Fi(x− xi) for any x ∈ ϕi(Ui) .

We note that by Cor. 5.5 the existence of Fi as well as its norm ‖Fi‖εiis independent of the choice of the point xi.

Let (c, ε) be a pair consisting of a chart c = (U,ϕ,Km) for M and areal number ε > 0 such that ϕ(U) = Bε(a) for one (or any) a ∈ ϕ(U). As aconsequence of the identity theorem for power series Cor. 5.8 the K-linearmap

Fε(Km;E) −→ Can(U,E)F 7−→ F (ϕ(.)− a)

is injective. Let F(c,ε)(E) denote its image. It is a K-Banach space withrespect to the norm

‖f‖ = ‖F‖ε if f(.) = F (ϕ(.)− a) .

By Cor. 5.5 the pair (F(c,ε)(E), ‖ ‖) is independent of the choice of the pointa.

Definition. An index for M is a family I = {(ci, εi)}i∈I of charts ci =(Ui, ϕi,Kmi) for M and real numbers εi > 0 such that the above conditions(a) and (b) are satisfied.

For any index I for M we have

FI(E) :=∏i∈IF(ci,εi)(E) ⊆

∏i∈I

Can(Ui, E) = Can(M,E) .

Our above resume says that

Can(M,E) =⋃IFI(E)

where I runs over all indices for M . Hence Can(M,E) is a union of directproducts of Banach spaces. This is the starting point for the construction ofa topology on Can(M,E).

But first we have to discuss the inclusion relations between the subspacesFI(E) for varying I. Let I = {(ci = (Ui, ϕi,Kmi), εi)}i∈I and J = {(dj =(Vj , ψj ,Knj ), δj)}j∈J be two indices for M .

71

Definition. The index I is called finer than the index J if for any i ∈ Ithere is a j ∈ J such that:

(i) Ui ⊆ Vj ,

ii) there is an a ∈ ϕi(Ui) and a power series Fi,j ∈ Fεi(Kmi ;Knj ) with‖Fi,j − Fi,j(0)‖εi ≤ δj and

ψj ◦ ϕ−1i (x) = Fi,j(x− a) for any x ∈ ϕi(Ui) .

We observe that if the condition (ii) holds for one point a ∈ ϕi(Ui) thenit holds for any other point b ∈ ϕi(Ui) as well. This follows from Cor. 5.5which implies that Gi,j(X) := Fi,j(X + b− a) ∈ Fεi(Kmi ;Knj ) with

ψj ◦ ϕ−1i (x) = Gi,j(x− b) for any x ∈ ϕi(Ui)

and

‖Gi,j −Gi,j(0)‖εi = ‖(Fi,j − Fi,j(0))(X + b− a) + Fi,j(0)−Gi,j(0)‖εi≤ max(‖(Fi,j − Fi,j(0))(X + b− a)‖εi , δj)= max(‖Fi,j − Fi,j(0)‖εi , δj)= δj .

Lemma 10.1. If I is finer than J then we have FJ (E) ⊆ FI(E).

Proof. Let f ∈ FJ (E). We have to show that f |Ui ∈ F(ci,εi)(E) for any i ∈ I.In the following we fix an i ∈ I. We have ϕi(Ui) = Bεi(a). By assumptionwe find a j ∈ J and an Fi,j ∈ Fεi(Kmi ;Knj ) such that

- Ui ⊆ Vj ,

- ‖Fi,j − Fi,j(0)‖εi ≤ δj , and

- ψj ◦ ϕ−1i (x) = Fi,j(x− a) for any x ∈ ϕi(Ui) .

We putb := ψj ◦ ϕ−1

i (a) = Fi,j(0) ∈ ψj(Vj) .

Since f ∈ FJ (E) we also find a Gj ∈ Fδj (Knj ;E) such that

f ◦ ψ−1j (y) = Gj(y − b) for any y ∈ ψj(Vj) = Bδj (b) .

As a consequence of Prop. 5.4 then the power series

Fi := Gj ◦ (Fi,j − Fi,j(0)) ∈ Fεi(Kmi ;E)

72

exists and satisfies

Fi(x− a) = Gj(Fi,j(x− a)− b) = f ◦ ψ−1j (ψj ◦ ϕ−1

i (x)) = f ◦ ϕ−1i (x)

for any x ∈ ϕi(Ui).

The relation of being finer only is a preorder. If the index I is finer thanthe index J and J is finer than I one cannot conclude that I = J . But itdoes follow that FI(E) = FJ (E) which is sufficient for our purposes.

Lemma 10.2. For any two indices J1 and J2 for M there is a third indexI for M which is finer than J1 and J2.

Proof. Let J1 = {((Ui, ϕi,Kni), εi)}i∈I and J2 = {((Vj , ψj ,Kmj ), δj)}j∈J .We have the covering

M =⋃i,j

Ui ∩ Vj

by pairwise disjoint open subsets. For any pair (i, j) ∈ I × J the function

ψj ◦ ϕ−1i : ϕi(Ui ∩ Vj) −→ Kmj

is locally analytic. Hence we may cover ϕi(Ui ∩ Vj) by a family of ballsBi,j,k = Bβi,j,k(ai,j,k) such that

- βi,j,k ≤ min(εi, δj), and

- there is a power series Fi,j,k ∈ Fβi,j,k(Kni ;Kmj ) with

ψj ◦ ϕ−1i (x) = Fi,j,k(x− ai,j,k) for any x ∈ Bi,j,k .

Using the fact that

‖Fi,j,k − Fi,j,k(0)‖α ≤ αβi,j,k‖Fi,j,k − Fi,j,k(0)‖βi,j,k for any 0 < α ≤ βi,j,k

together with Cor. 5.5 we may, after possibly decreasing the βi,j,k, assumein addition that

- ‖Fi,j,k − Fi,j,k(0)‖βi,j,k ≤ δj .

After a possible further refinement based on Lemma 1.4 (compare the argu-ment for the resume at the beginning of this section) we finally achieve thatthe Bi,j,k are pairwise disjoint. We put

Wi,j,k := ϕ−1i (Bi,j,k)

73

and obtain the index I := {((Wi,j,k, ϕi,Kni), βi,j,k)}i,j,k for M . By construc-

tion I is finer than J2. Moreover, observing that ϕi◦ϕ−1i : ϕi(Wi,j,k)

⊆−−→ Kni

is the inclusion map and that βi,j,k ≤ εi we see that I is finer than J1 fortrivial reasons.

Given any index I for M we consider FI(E) =∏i∈I F(ci,εi)(E) from

now on as a topological K-vector space with respect to the product topol-ogy of the Banach space topologies on the F(ci,εi)(E). Obviously FI(E) isHausdorff. But it is not a Banach space if I is infinite. Suppose that thetopology of FI(E) can be defined by a norm. The corresponding unit ballB1(0) is open. By the definition of the product topology there exist finitelymany indices i1, . . . , ir ∈ I such that∏

i 6=i1,...,ir

F(ci,εi)(E)× {0} × . . .× {0} ⊆ B1(0) .

As a vector subspace the left hand side then necessarily is contained in anyball Bε(0) for ε > 0. The intersection of the latter being equal to {0} itfollows that I is finite.

Lemma 10.3. If I is finer than J then the inclusion map FJ (E) ⊆−−→FI(E) is continuous.

Proof. For any i ∈ I there exists, by assumption, a j(i) ∈ J such that theconditions (i) and (ii) in the definition of ”finer” are satisfied. The inclusionmap in question can be viewed as the map∏

j

F(dj ,δj)(E) −→∏i

F(ci,εi)(E)

(fj)j 7−→ (fj(i)|Ui)i .

Hence it suffices to show that each individual restriction map

F(dj(i),δj(i))(E) −→ F(ci,εi)(E)

is continuous. But we even know from Prop. 5.4 that the operator norm ofthis map is ≤ 1.

We point out that, for I finer than J , the topology of FJ (E) in generalis strictly finer than the subspace topology induced by FI(E).

In the present situation there is a certain universal procedure to constructfrom the topologies on all the FI(E) a topology on their union Can(M,E) =⋃I FI(E). Since this construction takes place within the class of locally

convex topologies we first need to review this concept in the next section.

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11 Locally convex K-vector spaces

This section serves only as a brief introduction to the subject. The readerwho is interested in more details is referred to [NFA]. Let E be any K-vectorspace.

Definition. A (nonarchimedean) seminorm on E is a function q : E −→ Rsuch that for any v, w ∈ E and any a ∈ K we have:

(i) q(av) = |a| · q(v),

(ii) q(v + w) ≤ max(q(v), q(w)).

It follows immediately that a seminorm q also satisfies:

(iii) q(0) = |0| · q(0) = 0;

(iv) q(v) = max(q(v), q(−v)) ≥ q(v − v) = q(0) = 0 for any v ∈ E;

(v) q(v + w) = max(q(v), q(w)) for any v, w ∈ E such that q(v) 6= q(w)(compare the proof of Lemma 1.1);

(vi) −q(v − w) ≤ q(v)− q(w) ≤ q(v − w) for any v, w ∈ E.

Let (qi)i∈I be a family of seminorms on E. We consider the coarsesttopology on E such that:

(1) All maps qi : E −→ R, for i ∈ I, are continuous,

(2) all translation maps v + . : E −→ E, for v ∈ E, are continuous.

It is called the topology defined by (qi)i∈I . For any finitely many qi1 , . . . , qirand any w ∈ E and ε > 0 we define

Bε(qi1 , . . . , qir ;w) := {v ∈ E : qir(v − w), . . . , qir(v − w) ≤ ε} .

The following properties are obvious:

(a) Bε(qi1 , . . . , qir ;w) = Bε(qi1 ;w) ∩ . . . ∩Bε(qir ;w);

(b) Bε1(qi1 ;w1) ∩ Bε2(qi2 ;w2) =⋃w Bmin(ε1,ε2)(qi1 , qi2 ;w) where w runs

over all points in the left hand side;

(c) Bε(qi1 , . . . , qir ;w) = w +Bε(qi1 , . . . , qir ; 0);

(d) a ·Bε(qi1 , . . . , qir ;w) = B|a|ε(qi1 , . . . , qir ; aw) for any a ∈ K×.

75

Lemma 11.1. The subsets Bε(qi1 , . . . , qir ;w) form a basis for the topologyon E defined by (qi)i∈I .

Proof. The Bε(qi1 , . . . , qir ;w), by (a) and (b), do form a basis for a (unique)topology T ′ on E. On the other hand let T denote the topology defined by(qi)i∈I . We first show that T ′ ⊆ T . By (a), (c), and (2) it suffices to checkthat Bε(qi; 0) ∈ T for any i ∈ I and ε > 0. As a consequence of (1) and (2)we certainly have that

B−δ (qi;w) := {v ∈ E : qi(v − w) < δ} ∈ T

for any w ∈ E and δ > 0. But using (ii) we see that

Bε(qi; 0) = B−ε (qi; 0) ∪⋃

qi(w)=ε

B−ε (qi;w) .

To conclude that actually T ′ = T holds true it now suffices to show that T ′satisfies (1) and (2). The continuity property (2) follows immediately from(c). To establish (1) for T ′ we have to show that q−1

i ((α, β)) ∈ T ′ for anyi ∈ I and any open interval (α, β) ⊆ R. Because of (iv) we may assume thatβ > 0. Let w ∈ q−1

i ((α, β)) be any point. Case 1: We have qi(w) > 0. Chooseany 0 < ε < qi(w). It then follows from (v) that Bε(qi;w) ⊆ q−1

i (qi(w)) ⊆q−1i ((α, β)). Case 2: We have qi(w) = 0. Choose any 0 < ε < β. We obtainBε(qi;w) ⊆ q−1

i ([0, ε]) ⊆ q−1i ((α, β)) since necessarily α < 0 in this case.

Lemma 11.2. E is a topological K-vector space, i. e., addition and scalarmultiplication are continuous, with respect to the topology defined by (qi)i∈I .

Proof. Using Lemma 11.1 this easily follows from the following inclusions:

– Bε(qi1 , . . . , qir ;w1) +Bε(qi1 , . . . , qir ;w2) ⊆ Bε(qi1 , . . . , qir ;w1 + w2);

– Bδ(a) · B|a|−1ε(qi1 , . . . , qir ;w) ⊆ Bε(qi1 , . . . , qir ; aw) provided δ ≤ |a|and δ ·max(qi1(w), . . . , qir(w)) ≤ ε;

– Bδ(0) · Bε(qi1 , . . . , qir ;w) ⊆ Bε(qi1 , . . . , qir ; 0) provided δ ≤ 1 and δ ·max(qi1(w), . . . , qir(w)) ≤ ε.

The details are left to the reader as an exercise.

Exercise. The topology on E defined by (qi)i∈I is Hausdorff if and only iffor any vector 0 6= v ∈ E there is an index i ∈ I such that qi(v) 6= 0.

76

Definition. A topology on a K-vector space E is called locally convex if itcan be defined by a family of seminorms. A locally convex K-vector space isa K-vector space equipped with a locally convex topology.

Obviously any normed K-vector space and in particular any K-Banachspace is locally convex.

Remark 11.3. Let {Ej}j∈J be a family of locally convex K-vector spaces;then the product topology on E :=

∏j∈J Ej is locally convex.

Proof. Let (qj,i)i be a family of seminorms which defines the locally convextopology on Ej . Moreover, let prj : E −→ Ej denote the projection maps.Using Lemma 11.1 one checks that the family of seminorms (qj,i ◦ prj)i,jdefines the product topology on E.

Exercise 11.4. Let {Ej}j∈J be a family of locally convex K-vector spacesand let E :=

∏j∈J Ej with the product topology; for any continuous semi-

norm q on E there is a unique minimal finite subset Jq ⊆ J such that

q( ∏j∈J\Jq

Ej × {0} × . . .× {0})

= {0} .

For our purposes the following construction is of particular relevance.Let E be a any K-vector space, and suppose that there is given a family{Ej}j∈J of vector subspaces Ej ⊆ E each of which is equipped with a locallyconvex topology.

Lemma 11.5. There is a unique finest locally convex topology T on E suchthat all the inclusion maps Ej

⊆−−→ E, for j ∈ J , are continuous.

Proof. Let Q be the set of all seminorms q on E such that q|Ej is continuousfor any j ∈ J , and let T be the topology on E defined by Q. It followsimmediately from Lemma 11.1 that all the inclusion maps Ej

⊆−−→ (E, T )are continuous. On the other hand, let T ′ be any topology on E definedby a family of seminorms (qi)i∈I such that Ej

⊆−−→ (E, T ′) is continuous forany j ∈ J . Obviously we then have (qi)i∈I ⊆ Q. This implies, using againLemma 11.1, that T ′ ⊆ T .

The topology T on E in the above Lemma is called the locally convexfinal topology with respect to the family {Ej}j∈J . Suppose that the family{Ej}j∈J has the additional properties:

- E =⋃j∈J Ej ;

77

- the set J is partially ordered by ≤ such that for any two j1, j2 ∈ Jthere is a j ∈ J such that j1 ≤ j and j2 ≤ j;

- whenever j1 ≤ j2 we have Ej1 ⊆ Ej2 and the inclusion map Ej1⊆−−→ Ej2

is continuous.

In this case the locally convex K-vector space (E, T ) is called the locallyconvex inductive limit of the family {Ej}j∈J .

Lemma 11.6. A K-linear map f : E −→ E into any locally convex K-vector space E is continuous (with respect to T ) if and only if the restrictionsf |Ej, for any j ∈ J , are continuous.

Proof. It is trivial that with f all restrictions f |Ej are continuous. Let ustherefore assume vice versa that all f |Ej are continuous. Let (qi)i∈I be afamily of seminorms which defines the topology of E. Then all seminormsqi := qi◦f , for i ∈ I, lie in the set of seminorms Q which defines the topologyT of E. It follows that

f−1(Bε(qi1 , . . . , qir ; f(w)) = Bε(qi1 , . . . , qir ;w)

is open in E. Because of Lemma 11.1 this means that f is continuous.

Lemma 11.7. Let {Ej}j∈J be a family of locally convex K-vector spacesand let E :=

∏j∈J Ej with the product topology; suppose that each Ej has

the locally convex final topology with respect to a family of locally convexK-vector spaces {Ej,k}k∈Ij and that Ej =

⋃k Ej,k; for any k = (kj)j ∈ I :=∏

j∈J Ij we put Ek :=∏j∈J Ej,kj with the product topology; then the topology

of E is the locally convex final topology with respect to the family {Ek}k∈I .

Proof. By the proof of Lemma 11.5 the locally convex topology of Ej isdefined by the set Qj of all seminorms q such that q|Ej,k is continuous for anyk ∈ Ij . Let prj : E −→ Ej denote the projection maps. By Remark 11.3 thetopology of E is defined by the set of seminorms Q :=

⋃j∈J{q◦prj : q ∈ Qj}.

For any q ∈ Qj and any k ∈ I we have the commutative diagram

Ek⊆ //

prj��

E

prj

��

q◦prj

&&MMMMMMMMMMMMM

Ej,kj⊆ // Ej

q // R .

Hence the restriction of any seminorm in Q to any Ek is continuous. Thismeans that the locally convex final topology on E with respect to the family

78

{Ek}k is finer than the product topology. Vice versa, let q be any seminormon E such that q|Ek, for any k ∈ I, is continuous. We have to show that q iscontinuous. By Exercise 11.4 we find, for any k ∈ I, a unique minimal finitesubset Jq,k ⊆ J such that the restriction q|Ek factorizes into

Ekpr //

q��????????

∏j∈Jq,k Ej,kj

yyssssssssss

R .

In particular, q|Ej,kj 6= 0 for any j ∈ Jq,k. We claim that the set

Jq :=⋃k∈I

Jq,k

is finite. We define ` = (`j)j ∈ I in the following way. If j ∈ Jq we choosea k ∈ I such that j ∈ Jq,k and we put `j := kj ; in particular, q|Ej,`j =q|Ej,kj 6= 0. For j ∈ J \ Jq we pick any `j ∈ Ij . By construction we haveJq ⊆ Jq,` so that Jq necessarily is finite. This means that the seminorm q onE factorizes into

Epr //

q��========

∏j∈Jq Ej

{{vvvvvvvvv

R .

It follows that

q(v) ≤ maxj∈Jq

(q|Ej) ◦ prj(v) for any v ∈ E .

Since each q|Ej is continuous by assumption we conclude that q is continu-ous.

12 The topological vector space Can(M, E), part 2

As in section 10 we let M be a paracompact manifold and E be a K-Banachspace. We have seen that

Can(M,E) =⋃IFI(E)

where I runs over all indices for M . Each FI(E) by Remark 11.3 is locallyconvex as a product of Banach spaces. By lemmas 10.1 - 10.3 we may and

79

always will view Can(M,E) as the locally convex inductive limit of the family{FI(E)}I (where I ≤ J if J is finer than I). All our earlier constructionsinvolving Can(M,E) are compatible with this topology. In the following webriefly discuss the most important ones.

Proposition 12.1. For any a ∈M the evaluation map

δa : Can(M,E) −→ E

f 7−→ f(a)

is continuous.

Proof. It suffices, by Lemma 11.6, to show that the restriction δa|FI(E) iscontinuous for any index I for M . Let I = {(ci = (Ui, ϕi,Kmi), εi)}i∈I .There is a unique i ∈ I such that a ∈ Ui. Then ϕi(Ui) = Bεi(ϕi(a)), and wehave the commutative diagram

FI(E)δa //

pr

��

E

F(ci,εi)(E) Fεi(Kmi ;E) .∼=

F (ϕi(.)−ϕi(a))←[Foo

F 7→F (0)

OO

The left vertical projection map clearly is continuous. The lower horizontalmap is a topological isomorphism by construction. By Remark 5.1 the rightvertical evaluation map is continuous of operator norm ≤ 1.

Corollary 12.2. The locally convex vector space Can(M,E) is Hausdorff.

Proof. Let f 6= g be two different functions in Can(M,E). We find a pointa ∈ M such that f(a) 6= g(a). Since E is Hausdorff there are open neigh-bourhoods Vf of f(a) and Vg of g(a) in E such that Vf ∩Vg = ∅. Using Prop.12.1 we see that Uf := δ−1

a (Vf ) and Ug := δ−1a (Vg) are open neighbourhoods

of f and g, respectively, in Can(M,E) such that Uf ∩ Ug = ∅.

Remark 12.3. With M also its tangent bundle T (M) is paracompact.

Proof. Since M is strictly paracompact by Prop. 8.7 we find a family ofcharts {ci = (Ui, ϕi,Kmi)}i∈I for M such that the Ui are pairwise disjointand M =

⋃i Ui. Then the ci,T = (p−1

M (Ui), ϕi,ci ,K2mi) form a family of

charts for T (M) such that T (M) is the disjoint union of the open subsetsp−1M (Ui). Each p−1

M (Ui) being homeomorphic to an open subset in K2mi car-ries the topology of an ultrametric space. The construction in the proof of

80

implication ii. =⇒ iii. in Prop. 8.7 then shows that the topology of T (M) canbe defined by a metric which satisfies the strict triangle inequality. HenceT (M) is paracompact by Lemma 1.4.

Proposition 12.4. i. The map d : Can(M,E) −→ Can(T (M), E) iscontinuous.

ii. For any locally analytic map of paracompact manifolds g : M −→ Nthe map

Can(N,E) −→ Can(M,E)f 7−→ f ◦ g

is continuous.

iii. For any vector field ξ on M the map Dξ : Can(M,E) −→ Can(M,E)is continuous.

Proof. i. By Lemma 11.6 we have to show that d|FI(E) is continuous forany index I = {(ci = (Ui, ϕi,Kmi), εi)}i∈I for M . Let f ∈ FI(E). We havethe commutative diagrams

E

p−1M (Ui)

df

55kkkkkkkkkkkkkkkkkk ϕi,ci //

pM

��

ϕi(Ui)×Kmi

pr

��

(x,v)7→Dx(f◦ϕ−1i )(v)

OO

Uiϕi // ϕi(Ui)

(cf. the proof of Lemma 9.11). We also have power series Fi ∈ Fεi(Kmi ;E)such that

f ◦ ϕ−1i (x) = Fi(x− ai) for any x ∈ ϕi(Ui) = Bεi(ai) .

From the proof of Prop. 6.1 we recall the formula

Dx(f ◦ ϕ−1i )(v) =

mi∑j=1

vj∂Fi∂Xj

(x− ai)

for any x ∈ ϕi(Ui) and any v = (v1, . . . , vmi) ∈ Kmi . We now cover Kmi bypairwise disjoint balls Bεi(w

(i)k ) where w(i)

k = (w(i)k,1, . . . , w

(i)k,mi

) runs over an

81

appropriate family of vectors in Kmi , and we put

Gi,k(X1, . . . , Xmi , Y1, . . . , Ymi) :=mi∑j=1

(Yj + w(i)k,j)

∂Fi∂Xj

∈ Fεi(K2mi ;E) .

Thendf ◦ ϕ−1

i,ci(x, v) = Dx(f ◦ ϕ−1

i )(v) = Gi,k(x− ai, v − w(i)k )

for any (x, v) ∈ ϕi(Ui) × Bεi(w(i)k ) = Bεi(ai, w

(i)k ). This means that df ∈

FJ (E) ⊆ Can(T (M), E) for the index

J := {((ϕ−1i,ci

(Bεi(ai, w(i)k )), ϕi,ci ,K

2mi), εi)}i,k .

In other words we have the commutative diagram

Can(M,E) d // Can(T (M), E)

FI(E)

⊆

OO

// FJ (E) .

⊆

OO

Since the vertical inclusion maps are continuous by construction this reducesus to showing the continuity of the lower horizontal map FI(E) −→ FJ (E).But this easily follows from the inequalities

‖Gi,k‖εi ≤ max(1,|w(i)k,1|εi

, . . . ,|w(i)k,mi|

εi) · ‖Fi‖εi .

ii. We only sketch the argument and leave the details to the reader. LetI = {((Ui, ϕi,Kni), εi)}i∈I be an index for N . We refine the covering M =⋃i g−1(Ui) into a covering M =

⋃j∈J Vj which underlies an appropriate

index J = {((Vj , ψj ,Kmj ), δj)}j∈J and such that, for any i ∈ I and j ∈J with Vj ⊆ g−1(Ui), there is a power series Gi,j ∈ Fδj (Kmj ;Kni) with‖Gi,j −Gi,j(0)‖δj ≤ εi and

ϕi ◦ g ◦ ψ−1j (x) = Gi,j(x− aj) for any x ∈ ψj(Vj) = Bδj (aj) .

In this situation we have the commutative diagram

Can(N,E)f 7→f◦g // Can(M,E)

FI(E) //

⊆

OO

FJ (E)

⊆

OO

82

where the lower horizontal arrow in terms of power series is given by themaps

Fεi(Kni ;E) −→ Fδj (Kmj ;E)

F 7−→ F ◦ (Gi,j −Gi,j(0))

whose continuity was established in Prop. 5.4.iii. follows from i. and ii.

Proposition 12.5. For any covering M =⋃i∈I Ui by pairwise disjoint open

subsets Ui we haveCan(M,E) =

∏i∈I

Can(Ui, E)

as topological vector spaces.

Proof. Using Lemma 10.2 one checks that in the construction of Can(M,E)as a locally convex inductive limit it suffices to consider indices for M whoseunderlying covering of M refines the given covering M =

⋃i Ui. Then the

assertion is a formal consequence of Lemma 11.7.

83

Part III

Lie groups

As before we fix the nonarchimedean field (K, | |).

13 Definitions and foundations

Definition. A Lie group G (over K) is a manifold (over K) which alsocarries the structure of a group such that the multiplication map

m = mG : G×G −→ G

(g, h) 7−→ gh

is locally analytic.

In the following let G be a Lie group, and let e ∈ G denote the unitelement.

Lemma 13.1. For any h ∈ G the maps

`h : G '−−→ G and rh : G '−−→ G

g 7−→ hg g 7−→ gh

are locally analytic isomorphisms (of manifolds).

Proof. By symmetry we only need to consider the case of the left multipli-cation `h. This map can be viewed as the composite

G −→ G×G m−−→ G

g 7−→ (h, g) .

The left arrow is locally analytic by Example 8.5.4) and the right arrowby assumption. Hence the map `h is locally analytic by Lemma 8.4.ii. Weobviously have `h ◦ `h−1 = `hh−1 = `e = idG and then also `h−1 ◦ `h = idG.It follows that `−1

h = `h−1 is locally analytic as well.

Corollary 13.2. For any two elements g, h ∈ G the map

Tg(`hg−1) : Tg(G)∼=−−→ Th(G)

is a K-linear isomorphism; in particular,

Te(`g) : Te(G)∼=−−→ Tg(G)

is an isomorphism for any g ∈ G.

84

Proof. Use Lemma 9.2.

Corollary 13.3. Every Lie group is n-dimensional for some n ≥ 0.

Proof. We have dimG = dimK Te(G) = dimK Tg(G) for any g ∈ G.

Examples. 1) Kn and more generally any ball Bε(0) (as an open sub-manifold of Kn) with the addition is a Lie group.

2) K× and more generally B−1 (1) and Bε(1) for any 0 < ε < 1 (as opensubmanifolds of K) with the multiplication (observe that ab − 1 =(a− 1)(b− 1) + (a− 1) + (b− 1)) are Lie groups.

3) GLn(K) viewed as the open submanifold in Kn2defined by ”det 6= 0”

with the matrix multiplication is a Lie group.

Let g, h ∈ G. We know from Remark 9.10.ii. that the map

T (pr1)× T (pr2) : T(g,h)(G×G)∼=−−→ Tg(G)× Th(G)

is a K-linear isomorphism. In order to describe its inverse we introduce themaps

ih : G −→ G×G and jg : G −→ G×Gx 7−→ (x, h) x 7−→ (g, x)

which are locally analytic by Example 8.5.4). We have

pr1 ◦ ih = idG and pr2 ◦ ih = constant map with value h

and henceT (pr1) ◦ T (ih) = T (idG) = idT (G)

andT (pr2) ◦ T (ih) = T (constant map) = 0 .

This means that the composed map

Tg(G)T (ih)−−−−→ T(g,h)(G×G)

T (pr1)×T (pr2)−−−−−−−−−−→ Tg(G)× Th(G)

sends t to (t, 0). Analogously the composed map

(17) Th(G)T (jg)−−−−→ T(g,h)(G×G)

T (pr1)×T (pr2)−−−−−−−−−−→ Tg(G)× Th(G)

85

sends t to (0, t). We conclude that

Tg(ih) + Th(jg) : Tg(G)× Th(G) −→ T(g,h)(G×G)

(t1, t2) 7−→ Tg(ih)(t1) + Th(jg)(t2)

is the inverse of T(g,h)(pr1)× T(g,h)(pr2).

Lemma 13.4. The diagram

T(g,h)(G×G)T (m) //

∼=T (pr1)×T (pr2) ((QQQQQQQQQQQQQ

Tgh(G)

Tg(G)× Th(G)Tg(rh)+Th(lg)

77ooooooooooo

is commutative for any g, h ∈ G.

Proof. We compute

T(g,h)(m) ◦ (T(g,h)(pr1)× T(g,h)(pr2))−1 = T(g,h)(m) ◦ (Tg(ih) + Th(jg))

= Tg(m ◦ ih) + Th(m ◦ jg)= Tg(rh) + Th(lg) .

Corollary 13.5. The diagram

T(e,e)(G×G)T (m) //

∼=T (pr1)×T (pr2) ((QQQQQQQQQQQQQ

Te(G)

Te(G)× Te(G)

+

77ppppppppppp

is commutative.

Proposition 13.6. The map

ι = ιG : G −→ G

g 7−→ g−1

is a locally analytic isomorphism (of manifolds).

86

Proof. Because of ι2 = idG it suffices to show that the map ι is locallyanalytic. To do so we use the bijective locally analytic map

µ : G×G −→ G×G(x, y) 7−→ (xy, y) .

We claim that the tangent map T(g,h)(µ), for any g, h ∈ G, is bijective. ByLemma 13.4 the diagram

T(g,h)(G×G)T(g,h)(µ)

//

∼=T (pr1)×T (pr2)

��

T(gh,h)(G×G)

∼= T (pr1)×T (pr2)

��Tg(G)× Th(G) // Tgh(G)× Th(G)

in which the lower horizontal arrow is given by

(t1, t2) 7−→ (Tg(rh)(t1) + Th(lg)(t2), t2)

is commutative. Suppose that (t1, t2) lies in the kernel of this latter map.Then t2 = 0 and hence 0 = Tg(rh)(t1) + Th(lg)(t2) = Tg(rh)(t1). The analogof Cor. 13.2 for the right multiplication implies that t1 = 0. We see thatthis lower horizontal map and therefore T(g,h)(µ) are injective. But all vec-tor spaces in the diagram have the same finite dimension. Our claim thatT(g,h)(µ) is bijective follows. We now may apply the criterion for local invert-ibility in Prop. 9.3 and we conclude that the inverse µ−1 is locally analyticas well. It remains to note that ι is the composite

Gje−−→ G×G µ−1

−−−→ G×G pr1−−−→ G .

Corollary 13.7. For any g ∈ G the diagram

Te(G)Te(lg) //

−1

��

Tg(G)

Tg(ι)��

Te(G)Te(rg−1 )

// Tg−1(G)

is commutative; in particular, the map Te(ι) coincides with the multiplicationby −1.

87

Proof. The composed map

Glg−−→ G

ι−→ Grg−−→ G

ι−→ G

is the identity. Hence the diagram

Te(G)Te(lg) //

Te(ι)

��

Tg(G)

Tg(ι)��

Te(G)Te(rg−1 )

// Tg−1(G)

is commutative. This reduces us to showing the special case in our assertion.We consider the diagram

Te(G)t7→(0,t)

''NNNNNNNNNNNT (je)

xxqqqqqqqqqq

T(e,e)(G×G)

T(e,e)(µ−1)=T(e,e)(µ)−1

��

T (pr1)×T (pr2) // Te(G)× Te(G)

(t1,t2)7→(t1−t2,t2)

��T(e,e)(G×G)

T (pr1)×T (pr2) //

T (pr1) &&MMMMMMMMMMTe(G)× Te(G)

pr1wwppppppppppp

Te(G) .

In the proof of Prop. 13.6 we have seen that the map µ−1(x, y) = (xy−1, y)is locally analytic and that the central square in the above diagram is com-mutative. The top triangle is commutative by (17). The commutativity ofthe bottom triangle is trivial. It remains to observe that passing from topto bottom along the left, resp. right, hand side is equal to Te(ι), resp. to themultiplication by −1.

Corollary 13.8. For every n ∈ Z the map

fn : G −→ G

g 7−→ gn

is locally analytic, and Te(fn) coincides with the multiplication by n.

88

Proof. Case 1: For n = 0 the map f0 is the constant map with value e andTe(f0) = 0.

Case 2: Let n ≥ 1. We may view fn as the composite

Gdiag−−−→ G× . . .×G mult−−−→ G

g 7−→ (g, . . . , g)(g1, . . . , gn) 7−→ g1 . . . gn .

Both maps are locally analytic, the left diagonal map by Example 8.5.4) andthe right multiplication map by assumption. Hence in the diagram

Te(G× . . .×G)

∏i T (pri)

��

Te(mult)

((QQQQQQQQQQQQQ

Te(G)

Te(diag)66mmmmmmmmmmmmm

diag((QQQQQQQQQQQQQ Te(G)

Te(G)× . . .× Te(G)

(t1,...,tn)7→t1+...+tn

66mmmmmmmmmmmmm

the top, resp. bottom, composed map is equal to Te(fn), resp. the multipli-cation by n. But this diagram is commutative, the left triangle for trivialreasons and the right triangle as a consequence of Cor. 13.5.

Case 3: Let n ≤ −1. Since fn = f−n ◦ ι we obtain, using the previouscase and Cor. 13.7, that

Te(fn) = Te(f−n) ◦ Te(ι) = (−n · id) ◦ (− id) = n · id .

Cor. 13.2 already indicates that the tangent space Te(G) in the unitelement of G plays a distinguished role. We want to investigate this in greaterdetail.

Proposition 13.9. The maps

rT : Te(G)×G '−−→ T (G) and lT : G× Te(G) '−−→ T (G)(t, g) 7−→ Te(rg)(t) (g, t) 7−→ Te(lg)(t)

are locally analytic isomorphisms (of manifolds); the diagram

Te(G)×G rT //

pr2))RRRRRRRRRRRRRRRT (G)

pG

��

G× Te(G)lToo

pr1uulllllllllllllll

G

is commutative.

89

Proof. By symmetry it suffices to discuss the map rT . We choose a chartc = (U,ϕ,Kn) for G around e. By Lemma 9.1.i. the map

θc : Kn ∼=−−→ Te(G)v 7−→ [c, v]

is a K-linear isomorphism. We equip Te(G) with the unique structure of amanifold such that θc becomes a locally analytic isomorphism of manifolds.By Lemma 9.1.ii. this structure does not depend on the choice of the chartc. Of course, we then view Te(G)×G as the product manifold of Te(G) and

G. The inclusion map Te(G) ⊆−−→ T (G) is locally analytic since it can beviewed as the composite of the locally analytic maps

Te(G) θ−1c−−−→ Kn v 7→(e,v)−−−−−−→ U ×Kn τc−−→ p−1

G (U) ⊆−−→ T (G) .

We recall that τc((g, v)) = [c, v] ∈ Tg(G) is locally analytic by the construc-tion of T (G) as a manifold. Let

ξ0 : G −→ T (G)g 7−→ 0 ∈ Tg(G)

denote the ”zero vector field”, i. e., the zero vector in the vector spaceΓ(G,T (G)). Using Lemma 13.4 we see that the composed locally analyticmap

Te(G)×G ⊆× id−−−−→ T (G)×G id×ξ0−−−−→ T (G)× T (G)(T (pr1)×T (pr2))−1

−−−−−−−−−−−−−→ T (G×G)T (m)−−−−→ T (G)

sends (t, g) to Te(rg)(t) +Tg(le)(0) = Te(rg)(t) and hence coincides with rT .This shows that the map rT is locally analytic. It is easy to check that themap

T (G) −→ Te(G)×Gt 7−→ (TpG(t)(rpG(t)−1)(t), pG(t))

is inverse to rT . Its second component pG is locally analytic by Lemma 9.8.It therefore remains to prove that the map

f : T (G) −→ Te(G)t 7−→ TpG(t)(rpG(t)−1)(t)

90

is locally analytic. Using again Lemma 13.4 we compute that the composedlocally analytic map

T (G)id×pG−−−−→ T (G)×G id×ι−−−−→ T (G)×G id×ξ0−−−−→ T (G)× T (G)

(T (pr1)×T (pr2))−1

−−−−−−−−−−−−−→ T (G×G)T (m)−−−−→ T (G)

sends t to TpG(t)(rpG(t)−1)(t). It follows that the left vertical composite inthe commutative diagram

T (G)

f��

T (G)

f��

=oo

f

((QQQQQQQQQQQQQQ

Te(G)

⊆��

Te(G)=oo

t7→(e,θ−1c (t))

��

= // Te(G)

T (G) U ×Knτcoo (g,v)7→θc(v)

??

is locally analytic. Since τc is an open embedding we conclude that the rightvertical composite is locally analytic. With the lower oblique arrow thereforealso the upper oblique arrow f is locally analytic.

Corollary 13.10. The maps

Γ(G,T (G))∼=←−− Can(G,Te(G))

∼=−−→ Γ(G,T (G))ξlf (g) := lT ((g, f(g))) ←− [ f 7−→ ξrf (g) := rT ((f(g), g))

are isomorphisms of K-vector spaces.

Proof. The maps ξ 7−→ pr2 ◦(lT )−1◦ξ and ξ 7−→ pr1 ◦(rT )−1◦ξ, respectively,are inverses.

In Can(G,Te(G)) we have, for any t ∈ Te(G), the constant map

constt(g) := t .

We put

ξlt(g) := ξlconstt(g) = Te(lg)(t) and ξrt (g) := ξrconstt(g) = Te(rg)(t) .

Definition. A vector field ξ ∈ Γ(G,T (G)) is called left invariant, resp. rightinvariant, if ξ(g) = Te(lg)(ξ(e)), resp. ξ(g) = Te(rg)(ξ(e)), holds true for anyg ∈ G.

91

Corollary 13.11. The maps

Te(G)∼=−−→ {ξ ∈ Γ(G,T (G)) : ξ is left invariant}

t 7−→ ξlt

and

Te(G)∼=−−→ {ξ ∈ Γ(G,T (G)) : ξ is right invariant}

t 7−→ ξrt

are K-linear isomorphisms.

Proof. The map ξ 7−→ ξ(e) is the inverse in both cases.

Let E be a K-Banach space. With any vector field ξ on G we had asso-ciated the K-linear map

Dξ : Can(G,E) −→ Can(G,E)f 7−→ Dξ(f) = df ◦ ξ .

If ξ is left or right invariant what consequence does this have for the mapDξ? We observe that as a consequence of Lemma 13.1 we have a left K-linearaction by ”left translation”

G× Can(G,E) −→ Can(G,E)

(h, f) 7−→ hf(g) := f(h−1g)

of the group G on the vector space Can(G,E) as well as a right K-linearaction by ”right translation”

Can(G,E)×G −→ Can(G,E)

(f, h) 7−→ fh(g) := f(gh−1) .

Lemma 13.12. If ξ ∈ Γ(G,T (G)) is right, resp. left, invariant then we have

Dξ(fh) = Dξ(f)h, resp. Dξ(hf) = hDξ(f) ,

for any f ∈ Can(G,E) and h ∈ G. In the case E = K the converse holdstrue as well.

92

Proof. By symmetry we only consider the ”right” case. First we supposethat ξ is right invariant, i. e., ξ(g) = Te(rg)(ξ(e)). It follows that

T (rh−1) ◦ ξ(g) = Tg(rh−1) ◦ Te(rg)(ξ(e)) = Te(rgh−1)(ξ(e)) = ξ(gh−1) .

We now compute

Dξ(fh)(g) = dfh ◦ ξ(g) = d(f ◦ rh−1) ◦ ξ(g)= df ◦ T (rh−1) ◦ ξ(g)

= df ◦ ξ(gh−1) = Dξ(f)(gh−1)

= Dξ(f)h(g)

where for the second line we have used Lemma 9.12. If vice versa Dξ satisfiesthe asserted identity (for some E) then we have

df ◦ T (rh−1) ◦ ξ(g) = Dξ(fh)(g) = Dξ(f)h(g) = Dξ(f)(gh−1) = df ◦ ξ(gh−1)

for any f and any g, h. We rewrite this as

df ◦ T (rh−1) ◦ ξ(gh) = df ◦ ξ(g) .

With ξ alsoξh(g) := T (rh) ◦ ξ(gh−1)

is a vector field on G. Hence we obtain the identity

Dξh = Dξ for any h ∈ G .

Later on (Cor. 18.8) we will see that G is paracompact. In the case E = Kthe map ξ 7→ Dξ therefore is injective by Prop. 9.16. It follows that ξh = ξ,i. e., that

T (rh) ◦ ξ(gh−1) = ξ(g)

holds true for any g, h ∈ G. In particular, for g = h we obtain

T (rg)(ξ(e)) = ξ(g) for any g ∈ G

which means that ξ is right invariant.

The Lie product of vector fields is characterized by the identity

Dξ ◦Dη −Dη ◦Dξ = D[ξ,η]

(which even holds for general E by Prop. 9.20 and Cor. 18.8).

93

Corollary 13.13. If the vector fields ξ and η on G both are left or rightinvariant then so, too, is the vector field [ξ, η].

Proof. With Dξ and Dη obviously also D[ξ,η] satisfies the identity assertedin Lemma 13.12.

Using Lemma 13.11 and Cor. 13.13 we see that for any s, t ∈ Te(G) thereare uniquely determined tangent vectors [s, t]l and [s, t]r in Te(G) such that

ξl[s,t]l = [ξls, ξlt] and ξr[s,t]r = [ξrs , ξ

rt ] .

Then(Te(G), [ , ]l)

t7→ξlt−−−−→ (Γ(G,T (G)), [ , ])

and(Te(G), [ , ]r)

t7→ξrt−−−−→ (Γ(G,T (G)), [ , ])

are injective maps of Lie algebras. Is there a relation between the two Lieproducts [ , ]l and [ , ]r on Te(G)? For any ξ ∈ Γ(G,T (G)) also

ιξ(g) := Tg−1(ι) ◦ ξ(g−1)

is a vector field on G. This provides us with an involutory K-linear auto-morphism

ι. : Γ(G,T (G)) −→ Γ(G,T (G)) .

Remark 13.14. The diagram

Te(G)ξl. //

−1��

Γ(G,T (G))

ι.��

Te(G)ξr.

// Γ(G,T (G))

is commutative.

Proof. Using Cor. 13.7 we compute

ι(ξlt)(g) = T (ι) ◦ ξlt(g−1) = T (ι) ◦ T (lg−1)(t) = −T (rg)(t) = −ξrt (g) .

Lemma 13.15. Any vector fields ξ and η on G satisfy

[ιξ, ιη] = ι[ξ, η] .

94

Proof. Using Lemma 9.12 we compute

Dιξ(f)(g) = df ◦ ιξ(g) = df ◦T (ι) ·ξ(g−1) = d(f ◦ι)◦ξ(g−1) = Dξ(f ◦ι)(g−1).

This amounts to the identity

Dιξ(f) ◦ ι = Dξ(f ◦ ι) .

We continue computing

(Dιξ ◦Dιη)(f)(g) = Dξ(Dιη(f) ◦ ι)(g−1)

= Dξ(Dη(f ◦ ι))(g−1)

= (Dξ ◦Dη)(f ◦ ι)(g−1)

and consequently

D[ιξ,ιη](f)(g) = [Dιξ, Dιη](f)(g)

= [Dξ, Dη](f ◦ ι)(g−1)

= D[ξ,η](f ◦ ι)(g−1)

= Dι[ξ,η](f)(g) .

Corollary 13.16. We have

[s, t]r = −[s, t]l for any s, t ∈ Te(G) .

Proof. Using Remark 13.14 and Lemma 13.15 we compute

ξr[s,t]r = [ξrs , ξrt ] = [−ξrs ,−ξrt ] = [ξr−s, ξ

r−t]

= [ιξls,ιξlt] = ι[ξls, ξ

lt] = ιξl[s,t]l

= ξr−[s,t]l.

From now on we simplify the notation by setting

[s, t] := [s, t]r and Dt := Dξrt

for any s, t ∈ Te(G). We then have the identity

D[s,t] = Ds ◦Dt −Dt ◦Ds

on Can(G,E).

95

Definition. Lie(G) := (Te(G), [ , ]) is called the Lie algebra of G.

We obviously have

dimK Lie(G) = dimG .

The guiding question for the rest of this book is how much informationthe Lie algebra Lie(G) retains about the Lie group G. The answer requiresseveral purely algebraic concepts which we discuss in the next few sections.

Definition. Let G1 and G2 be two Lie groups over K; a homomorphism ofLie groups f : G1 −→ G2 is a locally analytic map which also is a grouphomomorphism.

Definition. If (g1, [ , ]1) and (g2, [ , ]2) are two Lie algebras over K thena homomorphism (of Lie algebras) σ : g1 −→ g2 is a K-linear map whichsatisfies

[σ(x), σ(y)]2 = σ([x, y]1) for any x, y ∈ g1 .

We write HomK((g1, [ , ]1), (g2, [ , ]2)) for the set of all homomorphismsof Lie algebras σ : g1 −→ g2.

Exercise. For any homomorphism of Lie groups f : G1 −→ G2 the mapLie(f) := Te(f) : Lie(G1) −→ Lie(G2) is a homomorphism of Lie algebras.

14 The universal enveloping algebra

In this section K is allowed to be a completely arbitrary field.

Exercise. i. Let A be an associative K-algebra with unit. Then (A, [ , ]A)with

[x, y]A := xy − yx

is a Lie algebra over K. In the case of a matrix algebra A = Mn×n(K)the corresponding Lie algebra is denoted by gln(K).

ii. If the field K is nonarchimedean then we have gln(K) = Lie(GLn(K)).

How general are the Lie algebras in this exercise? Obviously (A, [ , ]A)may have Lie subalgebras which do not correspond to associative subalge-bras. We want to show that any Lie algebra in fact arises as a subalgebraof an associative algebra. A K-linear map σ : g −→ A from a Lie algebra g

96

into a associative algebra A, of course, will be called a homomorphism if itsatisfies

σ([x, y]) = σ(x)σ(y)− σ(y)σ(x) for any x, y ∈ g .

At this point we need to recall the following general construction from mul-tilinear algebra. Let E be any K-vector space. Then

T (E) := ⊕n≥0E⊗n where E⊗n := E ⊗K . . .⊗K E (n factors)

is an associative K-algebra with unit (note that E⊗0 = K). The multiplica-tion is given by the linear extension of the rule

(v1 ⊗ . . .⊗ vn)(w1 ⊗ . . .⊗ wm) := v1 ⊗ . . .⊗ vn ⊗ w1 ⊗ . . .⊗ wm .

This algebra T (E) is called the tensor algebra of the vector space E. It hasthe following universal property.

Any K-linear map σ : E −→ A into any associative K-algebra with unitA extends in a unique way to a homomorphism of K-algebras with unitσ : T (E) −→ A. In fact, this extension satisfies

σ(v1 ⊗ . . .⊗ vn) = σ(v1) · . . . · σ(vn) .

Let g be a Lie algebra over K. Viewed as a K-vector space we mayform the tensor algebra T (g). In T (g) we consider the two sided ideal J(g)generated by all elements of the form

x⊗ y − y ⊗ x− [x, y] for x, y ∈ g .

Note that x⊗ y − y ⊗ x ∈ g⊗2 whereas [x, y] ∈ g⊗1. Then

U(g) := T (g)/J(g)

is an associative K-algebra with unit and

ε : g −→ U(g)x 7−→ x+ J(g)

is a homomorphism.

Definition. U(g) is called the universal enveloping algebra of the Lie algebrag.

97

This construction has the following universal property. Let σ : g −→ A beany homomorphism into any associative K-algebra with unit A. It extendsuniquely to a homomorphism σ : T (g) −→ A of K-algebras with unit.Because of

σ(x⊗ y − y ⊗ x− [x, y]) = σ(x)σ(y)− σ(y)σ(x)− σ([x, y]) = 0

we haveJ(g) ⊆ ker(σ) .

Hence there is a uniquely determined homomorphism of K-algebras withunit

σ : U(g) −→ A with σ ◦ ε = σ ,

i. e., the diagram

g

ε

��==================σ //

⊆

''NNNNNNNNNNNNN A

T (g)

σ

77ppppppppppppp

pr

��U(g)

σ

@@������������������

is commutative.The tensor algebra T (E) has the increasing filtration

T0(E) ⊆ T1(E) ⊆ . . . ⊆ Tm(E) ⊆ . . .

defined byTm(E) := ⊕0≤n≤mE

⊗n .

The Tm(E) do not form ideals in T (E). But they satisfy

Tl(E) · Tm(E) ⊆ Tl+m(E) for any l,m ≥ 0 .

Correspondingly we obtain an increasing filtration

U0(g) ⊆ U1(g) ⊆ . . . ⊆ Um(g) ⊆ . . .

in U(g) defined byUm(g) := Tm(g) + J(g)/J(g)

and which satisfies

(18) Ul(g) · Um(g) ⊆ Ul+m(g) for any l,m ≥ 0 .

98

For example, we have U0(g) = K · 1 and U1(g) = K · 1 + ε(g). We define

gr• U(g) := ⊕m≥0 grm U(g) with grm U(g) := Um(g)/Um−1(g)

(and the convention that U−1(g) := {0}). Because of (18) the K-bilinearmaps

grl U(g)× grm U(g) −→ grl+m U(g)(y + Ul−1(g), z + Um−1(g)) 7−→ yz + Ul+m−1(g)

are well defined. Together they make gr• U(g) into an associative K-algebrawith unit.

Theorem 14.1. (Poincare-Birkhoff-Witt) The algebra gr• U(g) is isomor-phic to a polynomial ring over K in possibly infinitely many variables Xi

and, in particular, is commutative. More precisely, let {xi}i∈I be a K-basisof g; then

K[{Xi}i∈I ]∼=−−→ gr• U(g)

Xi 7−→ ε(xi) + U0(g) ∈ gr1 U(g)

is an isomorphism of K-algebras with unit.

Proof. Compare [B-LL] Chap. I §2.7 or [Hum] §17.3.

Corollary 14.2. The map ε : g −→ U(g) is injective.

Because of this fact the map ε usually is viewed as an inclusion and isomitted from the notation. We see that g indeed is a Lie subalgebra of anassociative algebra.

Corollary 14.3. Let d := dimK g < ∞; if x1, . . . , xd is an (ordered) K-basis of g then {xi1 · . . . · xim : m ≥ 0, 1 ≤ i1 ≤ . . . ≤ im ≤ d} is a K-basisof U(g).

Proof. The Theorem 14.1 of Poincare-Birkhoff-Witt implies that, for anym ≥ 0, the set

{xi1 · . . . · xim + Um−1(g) : 1 ≤ i1 ≤ . . . ≤ im ≤ d}

is a K-basis of Um(g)/Um−1(g) (recall the convention that the empty pro-duct, in the case m = 0, is equal to the unit element).

99

This last corollary obviously remains true, by choosing a total orderingof a K-basis of g, even if g is not finite dimensional.

Let τ : g1 −→ g2 be a homomorphism of Lie algebras. Applying theuniversal property gives a homomorphism of K-algebras with unit

U(τ) : U(g1) −→ U(g2)

such that the diagram

g1τ //

⊆��

g2

⊆��

U(g1)U(τ)

// U(g2)

is commutative. We want to apply this in two specific situations. First letg1 and g2 two Lie algebras. Obviously, g1 × g2 again is a Lie algebra withrespect to the componentwise Lie product. There are the correspondingmonomorphisms of Lie algebras

i1 : g1 −→ g1 × g2 and i2 : g2 −→ g1 × g2

x 7−→ (x, 0) y 7−→ (0, y) .

Lemma 14.4. The map

U(g1)⊗K U(g2)∼=−−→ U(g1 × g2)

a⊗ b 7−→ U(i1)(a) · U(i2)(b)

is an isomorphism of K-algebras with unit.

Proof. Obviously, we have the K-bilinear map

U(g1)× U(g2) −→ U(g1 × g2)(a, b) 7−→ U(i1)(a) · U(i2)(b) .

By the universal property of the tensor product it induces the map in theassertion as a K-linear map. The latter is bijective by a straightforwardapplication of Cor. 14.3. Since we have

[i1(x), i2(y)] = [(x, 0), (0, y)] = ([x, 0], [0, y]) = (0, 0)

for any x ∈ g1 and any y ∈ g2 it follows easily that U(i1)(a) and U(i2)(b), forany a ∈ U(g1) and any b ∈ U(g2), commute with one another. This impliesthat the asserted map is a homomorphism and hence an isomorphism ofK-algebras with unit.

100

We point out that under the isomorphism in the above lemma the ele-ments

x⊗ 1 + 1⊗ y ←→ (x, y)

correspond to each other.Secondly, for any Lie algebra g the diagonal map

∆ : g −→ g× g

x 7−→ (x, x)

is a homomorphism of Lie algebras. We obtain the commutative diagram

g ∆ //

⊆��

g× g

⊆��

U(g)U(∆)

// U(g× g)∼= // U(g)⊗K U(g) .

Definition. The composed map U(g) −→ U(g) ⊗K U(g) in the lower lineof the above diagram is denoted (by abuse of notation) again by ∆ and iscalled the diagonal (or comultiplication) of the algebra U(g).

We note that for x ∈ g ⊆ U(g) we have

∆(x) = x⊗ 1 + 1⊗ x .

Proposition 14.5. If the field K has characteristic zero then we have

g = {a ∈ U(g) : ∆(a) = a⊗ 1 + 1⊗ a} .

Proof. Compare [B-LL] Chap. II §1.5 Cor.

15 The concept of free algebras

In this section K again is an arbitrary field. We will discuss the followingproblem. Let A be a specific class (or category) of K-algebras. We have inmind the following list of examples:

– ComK := all commutative and associative K-algebras with unit;

– AssK := all associative K-algebras with unit;

– LieK := all Lie algebras over K;

101

– AlgK := all K-algebras, i. e., all K-vector spaces A equipped with aK-bilinear ”multiplication” map A×A −→ A.

We suppose given a finite set X = {X1, . . . , Xd}, and we ask for an algebraAX in the class A together with a map X −→ AX which have the followinguniversal property: For any map X −→ A from the set X into any algebraA in the class A there is a unique homomorphism AX −→ A of algebras inA such that the diagram

X //

!!BBBBBBBB A

AX

>>||||||||

is commutative. If it exists AX is called the free A-algebra on X.The case ComK : The polynomial ring AX := K[X1, . . . , Xd] over K in

the variables X1, . . . , Xd has the requested universal property.The case AssK : As we have recalled in section 14 the tensor algebra

AX := AsX := T (Kd)

of the standard K-vector space Kd together with the map

X −→ Kd ⊆ T (Kd)Xi 7−→ i-th standard basis vector ei

satisfies the requested universal property. It sometimes is useful to view AsXas the ring of all ”noncommutative” polynomials

P (X1, . . . , Xd) =∑

(i1,...,im)

a(i1,...,im)Xi1 · . . . ·Xim

with coefficients a(i1,...,im) ∈ K where the sum runs over finitely many tu-ples (i1, . . . , im) with entries from the set {1, . . . , d} (including possibly theempty tuple). The multiplication is determined by the rule that the variablescommute with the coefficients but not with each other. The algebra AsX ina natural way is graded by As

(n)X := Kd ⊗K . . . ⊗K Kd (n factors) which

means that

AsX = ⊕n≥0As(n)X with As

(l)X ·As

(m)X ⊆ As(l+m)

X for any l,m ≥ 0 .

102

The case AlgK : Here we have to preserve the information about theorder in which the multiplications in a ”monomial” Xi1 · . . . ·Xim are per-formed (and we have to omit the unit element). This can be done in thefollowing way. We inductively define sets X(n) for n ≥ 1 by X(1) := X and

X(n) := disjoint union of all X(p)×X(q) for p+ q = n ,

and we putMX := disjoint union of all X(n) .

The obvious inclusion maps X(m) × X(n) ⊆−−→ X(m + n) combine into a”multiplication” map

µX : MX ×MX −→MX .

We now form the K-algebra

AX := the K-vector space on the basis MX

in which the multiplication is given by the linear extension of the map µX .There are the obvious inclusions X ⊆MX ⊆ AX .

Let γ : X −→ A be any map into any K-algebra A. We inductivelyextend γ to a map γ : MX −→ A by

γ : X(n) ⊇ X(p)×X(q) −→ A

(x, y) 7−→ γ(x)γ(y) .

This extension by construction is multiplicative in the sense that the diagram

MX ×MXµX //

γ×γ��

MX

γ

��AX ×AX

· // AX

is commutative. Hence it further extends by linearity to a homomorphismof K-algebras

γ : AX −→ A .

We stress that the algebra AX is graded by

A(n)X := the K -vector space on the basis X(n) ,

i. e., we have

AX = ⊕n∈NA(n)X with A

(l)X ·A

(m)X ⊆ A(l+m)

X for any l,m ≥ 1 .

103

The case LieK : In AX we consider the two sided ideal JX which isgenerated by all expressions of the form

aa and (ab)c+ (bc)a+ (ca)b for a, b, c ∈ AX .

ThenLX := AX/JX with [a+ JX , b+ JX ] := ab+ JX

is a Lie algebra over K.Let γ : X −→ g be any map into any Lie algebra g over K. As discussed

above it extends to a homomorphism of K-algebras γ : AX −→ g. Weobviously have

JX ⊆ ker(γ) .

Hence there is a uniquely determined homomorphism of Lie algebras γ :LX −→ g such that the diagram

X

��==================γ //

⊆

''NNNNNNNNNNNNN g

AX

γ

88ppppppppppppp

pr

��LX

γ

@@������������������

is commutative.

Exercise. i. We have JX = ⊕n∈NJX ∩A(n)X and hence

LX = ⊕n∈NL(n)X with [L(l)

X , L(m)X ] ⊆ L(l+m)

X for any l,m ≥ 1

if we define L(n)X := A

(n)X /JX∩A(n)

X (i. e., the Lie algebra LX is graded).

ii. The set X is (more precisely, maps bijectively onto a) K-basis of L(1)X .

iii. The set {[Xi, Xj ] : i < j} is a K-basis of L(2)X .

The inclusion map X ⊆−−→ AsX extends uniquely to a homomorphism ofLie algebras

φ : LX −→ (AsX , [ , ]AsX ) .

By the universal property of the universal enveloping algebra this map φfurther extends uniquely to a homomorphism of associative K-algebras withunit

Φ : U(LX) −→ AsX .

104

Proposition 15.1. The map Φ is an isomorphism of algebras.

Proof. The composed map

X −→ LX −→ U(LX)

extends uniquely to a homomorphism of associative K-algebras with unit

Ψ : AsX −→ U(LX) .

We have the commutative diagram

X

⊆

}}||||||||||||||||||||

�� ⊆

!!BBBBBBBBBBBBBBBBBBBB

LX

�� φ ((PPPPPPPPPPPPPP

AsX Ψ// U(LX)

Φ// AsX .

It therefore follows from the unicity in the universal property of AsX thatΦ ◦Ψ = id. We also have the commutative diagram

X

||xxxxxxxx

⊆

��

""FFFFFFFF

LX

{{wwwwwwwww

φ""EEEEEEEE LX

##GGGGGGGGG

U(LX)Φ

// AsX Ψ// U(LX) .

The universal property of LX then implies that even the diagram

LX

vvnnnnnnnnnnnnn

((PPPPPPPPPPPPP

U(LX)Φ

// AsX Ψ// U(LX)

is commutative. Using the universal property of U(.) we conclude that Ψ ◦Φ = id as well.

Corollary 15.2. The map φ : LX −→ AsX is injective.

105

Proof. Combine Cor. 14.2 and Prop. 15.1.

Exercise. i. If we view AsX as the ring of noncommutative polynomi-als P (X1, . . . , Xd) over K in the variables X1, . . . , Xd then φ(LX) isthe smallest K-vector subspace of AsX which contains the variablesX1, . . . , Xd and is closed under the operation (P,Q) 7−→ P ·Q−Q ·P .

ii. φ(L(n)X ) ⊆ As(n)

X for any n ∈ N, i. e., the homomorphism φ is graded .

16 The Campbell-Hausdorff formula

Again K is an arbitrary field and X = {X1, . . . , Xd} is a fixed finite set. Werecall that the free associative K-algebra with unit AsX on X is graded:

AsX = ⊕n≥0As(n)X and As

(l)X ·As

(m)X ⊆ As(l+m)

X for any l,m ≥ 0 .

ThereforeAsX :=

∏n≥0

As(n)X

with the multiplication

(an)n · (bn)n :=( n∑i=0

aibn−i)n

also is an associative K-algebra with unit (containing AsX as a subalgebra).It is called the Magnus algebra on X. Similarly as for AsX it is useful toview AsX as the ring of all ”noncommutative” formal power series over Kin the variables X1, . . . , Xd. In AsX we have the two sided maximal ideal

mX := {(an)n ∈ AsX : a0 = 0} .

Lemma 16.1. i. As×X = {(an)n ∈ AsX : a0 6= 0}.

ii. 1 + mX is a subgroup of As×X .

Proof. i. The map

AsX −→ K

(an)n 7−→ a0

is a homomorphism of K-algebras with unit. The group of multiplicativeunits As

×X therefore must be contained in the complement of the kernel of

106

this map. Vice versa let a = (an)n ∈ AsX be an element such that a0 6= 0.We have

a = a0 · (1− u) where u := (0,−a−10 · a1, . . . ,−a−1

0 · an+1, . . .) .

Since um ∈ {0} × . . .× {0} ×∏n≥mAs

(n)X the sum

∑m≥0 u

m is well definedin AsX . For b := a−1

0 ·(∑

m≥0 um)

we then obtain ab = ba = 1.ii. This is obvious from i. Note that 1 + mX is the kernel of the homo-

morphism of groups

As×X −→ K×

(an)n 7−→ a0 .

In the last proof we have used a special case of the following generalprinciple. For each m ≥ 0 let u(m) ∈ {0}× . . .×{0}×

∏n≥mAs

(n)X be some

element. Then the sum∑

m≥0 u(m) ∈ AsX is well defined. In particular, forany u ∈ mX we have the well defined homomorphism of K-algebras withunit

εu : K[[T ]] −→ AsX

F (T ) 7−→ F (u) .

Proposition 16.2. If the field K has characteristic zero then the maps

exp : mX −→ 1 + mX and log : 1 + mX −→ mX

u 7−→∑n≥0

un

n! 1 + u 7−→∑n≥1

(−1)n+1 un

n

are well defined and inverse to each other.

Proof. Because of

exp(u) = εu(exp(T )) and log(1 + u) = εu(log(1 + T ))

the maps in the assertion are well defined. Applying εu to the identities

exp(log(1 + T )) = 1 + T and log(exp(T )) = T

in the ring Q[[T ]] shows that they are inverse to each other.

107

Exercise. If a, b ∈ mX commute with each other (multiplicatively) then wehave

exp(a+ b) = exp(a) · exp(b) .

Using Cor. 15.2 and the subsequent Exercise we may view LX as the Liesubalgebra of AsX ”generated” by the elements X1, . . . , Xd. Moreover, wehave

LX = L(1)X ⊕ L

(2)X ⊕ . . .

∩ ∩ ∩AsX = K⊕ As

(1)X ⊕ As

(2)X ⊕ . . .

We now defineLX :=

∏n≥1

L(n)X ⊆ mX ⊆ AsX .

Lemma 16.3. LX is a Lie subalgebra of AsX .

Proof. Let a = (an)n and b = (bn)n be any two elements of LX . We have toshow that ab− ba ∈ LX holds true. For any m ≥ 1 we put

a(m) := (0, a1, . . . , am, 0, . . .), v(m) := (0, . . . , 0, am+1, am+2, . . .),

b(m) := (0, b1, . . . , bm, 0, . . .), u(m) := (0, . . . , 0, bm+1, bm+2, . . .).

Then a(m), b(m) ∈ LX and hence a(m)b(m) − b(m)a(m) ∈ LX . Moreover

ab− ba = (a(m) + v(m))(b(m) + u(m))− (b(m) + u(m))(a(m) + v(m))

= a(m)b(m) − b(m)a(m) + (0, . . . , 0, cm+1, . . .) .

It follows that for n ≤ m we have

n-th component of ab−ba = n-th component of a(m)b(m)−b(m)a(m) ∈ L(n)X .

Since m was arbitrary we conclude that ab− ba ∈ LX .

Since U(LX) = AsX by Prop. 15.1 we may view the comultiplication ∆of U(LX) as a homomorphism of K-algebras with unit

∆ : AsX −→ AsX ⊗K AsK .

It satisfies ∆(Xi) = Xi⊗ 1 + 1⊗Xi for any 1 ≤ i ≤ d. Since the X1, . . . , Xd

form a K-basis of As(1)X it follows that

∆(As(1)X ) ⊆ As(1)

X ⊗K As(0)X +As

(0)X ⊗K As

(1)X

108

and then inductively that

∆(As(n)X ) ⊆

⊕l+m=n

[As

(l)X ⊗K As

(m)X

]for any n ≥ 0. This makes it possible to extend ∆ to the homomorphism ofK-algebras with unit

∆ : AsX =∏n≥0

As(n)X −→ AsX⊗KAsX :=

∏l,m≥0

[As

(l)X ⊗K As

(m)X

]=∏n≥0

⊕l+m=n

[As

(l)X ⊗K As

(m)X

](an)n 7−→

(∆(an)

)n.

Lemma 16.4. If the field K has characteristic zero then we have

LX = {a ∈ AsX : ∆(a) = a⊗ 1 + 1⊗ a} .

Proof. Let a = (an)n ∈ AsX be any element. We have ∆(a) = a⊗ 1 + 1⊗ aif and only if ∆(an) = an⊗1+1⊗an for any n ≥ 0. By Prop. 14.5 the latteris equivalent to an ∈ LX ∩ As(n)

X = L(n)X for any n ≥ 0 which exactly is the

condition that a ∈ LX .

Theorem 16.5. (Campbell-Hausdorff) Suppose that K has characteristiczero; then the map

exp : LX∼−−→ {b ∈ 1 + mX : ∆(b) = b⊗ b}

is a well defined bijection; moreover, the right hand side is a subgroup of1 + mX .

Proof. The second part of the assertion follows immediately from ∆ beinga ring homomorphism. Since LX ⊆ mX the map exp is defined on LX andis injective by Prop. 16.2. For the subsequent computations we observe thatthe componentwise construction of the ring homomorphism ∆ implies that∆ commutes with the maps exp and log. First let a ∈ LX . Then ∆(a) =a⊗ 1 + 1⊗ a, and we compute

∆(exp(a)) = exp(∆(a)) = exp(a⊗ 1 + 1⊗ a)= exp(a⊗ 1) · exp(1⊗ a)= (exp(a)⊗ 1) · (1⊗ exp(a))= exp(a)⊗ exp(a) .

109

This shows that exp(a) indeed lies in the target of the asserted map. Viceversa let b ∈ 1 + mX such that ∆(b) = b ⊗ b. By Prop. 16.2 we may definea := log(b) ∈ mX so that b = exp(a). We compute

∆(a) = ∆(log(b)) = log(∆(b))= log(b⊗ b) = log((b⊗ 1) · (1⊗ b))= log(b⊗ 1) + log(1⊗ b)= log(b)⊗ 1 + 1⊗ log(b)= a⊗ 1 + 1⊗ a .

Hence Lemma 16.4 implies that a ∈ LX . We see that the asserted map issurjective.

Corollary 16.6. Suppose that the field K has characteristic zero; then LXequipped with the multiplication

a� b := log(exp(a) · exp(b))

is a group whose neutral element is the zero vector 0 and such that −a isthe inverse of a.

Proof. Since exp(0) = 1 the neutral element for � must be the zero vector0. Furthermore, since a and −a commute with respect to the usual multi-plication in AsX we have exp(a) · exp(−a) = exp(−a) · exp(a) = exp(0) = 1.Hence a� (−a) = (−a)� a = log(1) = 0.

Definition. For the field K = Q and the two-element set {Y,Z} we call

H(Y, Z) := Y � Z ∈ L{Y,Z} ⊆ As{Y,Z}

the Hausdorff series (in Y and Z).

As alluded to earlier we should viewH(Y,Z) as a noncommutative formalpower series in the variables Y,Z with coefficients in the field Q. We have

exp(Y ) · exp(Z) = 1 +W with W =∑r+s≥1

Y r

r! ·Zs

s!

110

and hence

H(Y,Z) =∑m≥1

(−1)m+1Wm

m

=∑m≥1

(−1)m+1

m

( ∑r+s≥1

Y r

r! ·Zs

s!

)m=∑n≥1

∑r+s=n

n∑m=1

(−1)m+1

m

∑r1+...+rm=rs1+...+sm=s

r1+s1≥1,...,rm+sm≥1

m∏i=1

Y riri!· Zsisi! .

Here and in the following the product sign∏mi=1 always has to be under-

stood in such a way that the corresponding multiplications are carried outin the order of the enumeration i = 1, . . . ,m. It is convenient to use theabbreviations

Hr,s :=r+s∑m=1

(−1)m+1

m

∑r1+...+rm=rs1+...+sm=s

r1+s1≥1,...,rm+sm≥1

m∏i=1

Y riri!· Zsisi!

andHn :=

∑r+s=n

Hr,s .

We note that Hr,s is a sum of noncommutative monomials of degree r in Yand s in Z. As a sum of noncommutative monomials of total degree n theelement Hn lies in As

(n){Y,Z}. We have

H =∑n≥1

Hn or, more formally, H = (Hn)n .

From the theory we know that Hn ∈ L(n){Y,Z} for each n ≥ 1 but this is not

visible from the above explicit formula.

Examples. 1) H1,0 = Y , H0,1 = Z, and H1 = Y + Z.

2) Hr,0 = H0,r = 0 for any r ≥ 2 (observe, for example, that Hr,0 is theterm of degree r in log(exp(Y )) = Y ).

3) H2 = H2,0 +H1,1 +H0,2 = H1,1 = Y Z − 12(Y Z + ZY ) = 1

2 [Y,Z].

111

If g is any Lie algebra over (any) K then the K-linear map

ad(z) : g −→ g

x 7−→ [z, x] ,

for any z ∈ g, is a derivation in the sense that

ad(z)([x, y]) = [ad(z)(x), y] + [x, ad(z)(y)] for any x, y ∈ g

holds true. This is just a reformulation of the Jacobi identity in g.

Proposition 16.7. (Dynkin’s formula) For r + s ≥ 1 we have

Hr,s = 1r+s(H

′r,s +H ′′r,s)

with H ′r,s defined as

∑m≥1

(−1)m−1

m

∑r1+...+rm=r

s1+...+sm−1=s−1r1+s1≥1,...,rm−1+sm−1≥1

((m−1∏i=1

ad(Y )ri

ri!◦ ad(Z)si

si!

)◦ ad(Y )rm

rm!

)(Z)

and

H ′′r,s :=∑m≥1

(−1)m−1

m

∑r1+...+rm−1=r−1s1+...+sm−1=s

r1+s1≥1,...,rm−1+sm−1≥1

(m−1∏i=1

ad(Y )ri

ri!◦ ad(Z)si

si!

)(Y ) .

Proof. Compare [B-LL] Chap. II §6.4. We note that the above defining sumsare finite and make it visible that Hn lies in L

(n){Y,Z}.

Remark 16.8. Suppose that K has characteristic zero; then we have

a� b = H(a, b) for any a, b ∈ LX .

Proof. The above explicit computations including Dynkin’s formula werecompletely formal and therefore are valid for any a, b (instead of Y, Z). Theexpression H(a, b), of course, has to be calculated componentwise in AsXusing the observation before Prop. 16.2.

The exploitation of these ”universal” considerations is based upon thefollowing technique. For any finite dimensional K-vector space V let

Map(V × V ;V ) := K-vector space of all maps f : V × V −→ V .

We pick a K-basis e1, . . . , ed of V .

112

Definition. A map f : V ×V −→ V is called polynomial (of degree (r, s)) ifthere are (homogeneous) polynomials Pi(X1, . . . , Xd, Y1, . . . , Yd) over K (ofdegree r in X1, . . . , Xd and degree s in Y1, . . . , Yd), for 1 ≤ i ≤ d, such that

f( d∑i=1

aiei,d∑i=1

biei)

=d∑i=1

Pi(a1, . . . , ad, b1, . . . , bd)ei for any ai, bi ∈ K.

In Map (V × V ;V ) we have the vector subspace Pol(V × V ;V ) of allpolynomial maps. It decomposes into

Pol(V × V ;V ) = ⊕n≥0 Poln(V × V ;V ) = ⊕n≥0 ⊕r+s=n Polr,s(V × V ;V )

where Polr,s(V ×V ;V ) denotes the subspace of all polynomial maps of degree(r, s) and

Poln(V × V ;V ) := ⊕r+s=n Polr,s(V × V ;V )

is the subspace of all polynomial maps of total degree n.

Lemma 16.9. Given any f ∈ Polr,s(V × V ;V ) and gi ∈ Polli,mi(V × V ;V )for i = 1, 2 the map

(v, w) 7−→ f(g1(v, w), g2(v, w))

lies in Polrl1+sl2,rm1+sm2(V × V ;V ).

Proof. Straightforward.

Corollary 16.10. The property of a map f : V × V −→ V of being poly-nomial (of a certain degree) does not depend on the choice of the K-basis ofV .

Proof. View the change of bases as a polynomial map and apply Lemma16.9.

Suppose now that the vector space V is a Lie algebra g of finite dimensiond := dimK g. Then also the vector space Map(g× g; g) is a Lie algebra withrespect to the Lie product

[f, g](x, y) := [f(x, y), g(x, y)] .

Corollary 16.11. Pol(g × g; g) is a Lie subalgebra of Map(g × g; g); moreprecisely, we have

[Polr,s(g× g; g),Polr′,s′(g× g; g)] ⊆ Polr+r′,s+s′(g× g; g) .

113

Proof. Apply Lemma 16.9 to the map f := [ , ] lying in Pol1,1(g× g; g).

We identify the two-element set {Y,Z} with the subset of Pol(g × g; g)consisting of the two projection maps pri : g× g −→ g by sending Y to pr1

and Z to pr2. By the universal property of free Lie algebras this extendsuniquely to a homomorphism of graded Lie algebras

θ : L{Y,Z} −→ Pol(g× g; g) .

It satisfies

(19) θ([Y, a])(y, z) = [y, θ(a)(y, z)] and θ([Z, a])(y, z) = [z, θ(a)(y, z)]

for any a ∈ L{Y,Z}.We define

Pow(g× g; g) :=∏n≥0

Poln(g× g; g)

as a K-vector space. The elements of Pow(g × g; g) can be viewed (if K isinfinite, and after the choice of a K-basis of g) as d-tuples of usual formalpower series in the variables Y1, . . . , Yd, Z1, . . . , Zd with coefficients in K. Asa consequence of Cor. 16.11 the Lie product on Pol(g× g; g) extends by

[(fn)n, (gn)n] :=( ∑l+m=n

[fl, gm])n

to a Lie product on Pow(g × g; g). Being graded θ extends to the K-linearmap

θ : L{Y,Z} −→ Pow(g× g; g)

(fn)n 7−→ (θ(fn))n .

Using the trick in the proof of Lemma 16.3 we obtain for any m ∈ N, withthe notations in this proof, that

θ([a, b]) ≡ θ([a(m), b(m)]) ≡ [θ(a(m)), θ(b(m))]

≡ [θ(a), θ(b)] mod∏n>m

Poln(g× g; g)

for any a, b ∈ L{Y,Z}. Since m is arbitrary this means that θ also is a homo-morphism of Lie algebras.

114

From now on we assume for the rest of this section that the field K hascharacteristic zero. We put

H := Hg := θ(H) ∈ Pow(g× g; g) .

More precisely, we have

H =∑r+s≥1

Hr,s with Hr,s := θ(Hr,s) ∈ Polr,s(g× g; g) .

Using (19) Dynkin’s formula in Prop. 16.7 implies that

Hr,s = 1r+s(H

′r,s + H ′′r,s)

with

H ′r,s : g× g −→ g

(y, z) 7−→∑m≥1

(−1)m−1

m

∑...

((m−1∏i=1

ad(y)ri

ri!◦ ad(z)si

si!

)◦ ad(y)rm

rm!

)(z)

and

H ′′r,s : g× g −→ g

(y, z) 7−→∑m≥1

(−1)m−1

m

∑...

(m−1∏i=1

ad(y)ri

ri!◦ ad(z)si

si!

)(y) .

Examples 16.12. H1,0 = pr1, H0,1 = pr2, and

H1,1 : g× g −→ g

(y, z) 7−→ 12 [y, z] .

What else do we know about H? First of all we recall from Cor. 16.6and Remark 16.8 that we have

(20)H(a,H(b, c)) = H(H(a, b), c),

H(a, 0) = H(0, a) = a, and H(a,−a) = 0

for any a, b, c ∈ L{Y,Z}. In order to use this we reinterpret the evaluation ofH in a and b in the following way.

Let a, b ∈ L{Y,Z} be any two elements. By the universal property of freeLie algebras there is a unique homomorphism of Lie algebras

εa,b : L{Y,Z} −→ L{Y,Z}

115

mapping Y to a and Z to b. By construction it satisfies

εa,b(L(m){Y,Z}) ⊆ {0} × . . .× {0} ×

∏n≥m

L(n){Y,Z}

for any m ≥ 1 and therefore extends, by the observation before Prop. 16.2,to the K-linear map

εa,b : L{Y,Z} −→ L{Y,Z}

(cn)n 7−→∑n≥1

εa,b(cn) .

The same reasoning as for θ shows that εa,b in fact is a homomorphismof Lie algebras. On the other hand of course, εa,b is the restriction of acorresponding unique homomorphism of associative K-algebras with unit

εa,b : As{Y,Z} −→ As{Y,Z} .

Viewing an element in As{Y,Z} as a noncommutative polynomial G(Y,Z) itis clear that

εa,b(G) = G(a, b)

holds true. It follows that

εa,b(H) =∑n≥1

εa,b(Hn) =∑n≥1

Hn(a, b) = H(a, b) .

There is an analogous construction for the Lie algebra Pow(g × g; g).Quite generally, given any g1, g2 ∈ Map(V × V ;V ) there is the homomor-phism (of Lie algebras in case V = g)

Map(V × V ;V ) −→ Map(V × V ;V )f 7−→ f(g1, g2)(v, w) := f(g1(v, w), g2(v, w)) .

If g1, g2 ∈ Pol(V × V ;V ) then Lemma 16.9 says that it restricts to

Pol(V × V ;V ) −→ Pol(V × V ;V )

and satisfies

f(g1, g2) ∈ Polrn1+sn2(V × V ;V )if f ∈ Polr,s(V × V ;V ) and gi ∈ Polni(V × V ;V ) .

116

Hence for g1, g2 in

Pow0(g× g; g) := {0} ×∏n≥1

Poln(g× g; g)

we obtain, by the usual componentwise procedure, a homomorphism of Liealgebras

Pow(g× g; g) −→ Pow(g× g; g)f −→ f(g1, g2) .

Indeed, this is just a reformulation of the fact that a formal power serieswithout constant term can be inserted into any formal power series. Wenote that

pri(g1, g2) = gi .

As before let now a, b ∈ L{Y,Z} be any two elements. Then θ(a), θ(b) liein Pow0(g× g; g), and we have the commutative diagram

(21)

L{Y,Z}εa,b //

θY,Z 7→pr1,pr2��

Y,Z 7→θ(a),θ(b)

++VVVVVVVVVVVVVVVVVVVVVVV L{Y,Z}

�

Pow(g× g; g)f 7→f(θ(a),θ(b))

// Pow(g× g; g) .

Lemma 16.13. Suppose that K has characteristic zero; for any y ∈ g andr + s ≥ 1 we have:

i. Hr,s(y,−y) = 0;

ii. Hr,s(y, 0) =

{y if r = 1, s = 0,0 otherwise ;

iii. Hr,s(0, y) =

{y if r = 0, s = 1,0 otherwise .

Proof. We apply the commutative diagram (21) to the Hausdorff series H ∈L{Y,Z} and various choices for the elements a and b. For a := Y and b := −Ywe have θ(a) = pr1 and θ(b) = −pr1 and we obtain from (21) that

H� //

_

��

H(Y,−Y )_

��

H� // H(pr1,−pr1) .

117

Since H(Y,−Y ) = 0 by (20) the assertion i. follows. For a := Y and b := 0we similarly obtain

H� //

_

��

H(Y, 0)_

��

H� // H(pr1, 0) .

Again by (20) we have H(Y, 0) = Y and hence H(pr1, 0) = pr1 which isthe assertion ii. The last assertion iii. comes symmetrically from the choicea := 0 and b := Y .

The discussion leading to the commutative diagram (21) can easily begeneralized to the three-element set {U, Y, Z} and the Lie algebra

Pow(g× g× g; g)

of d-tuples of formal power series over K in 3d variables. We leave the detailsto the reader. This leads to the homomorphism of Lie algebras

θ : L{U,Y,Z} −→ Pow(g× g× g; g)

which sends U, Y, and Z to pr1, pr2, and pr3, respectively. For any choiceof elements a, b ∈ L{U,Y,Z} we obtain, analogously to (21), the commutativediagram

(22)

L{Y,Z}Y,Z 7→a,b //

�

L{U,Y,Z}

�

Pow(g× g; g)f 7→f(θ(a),θ(b))

// Pow(g× g× g; g) .

Lemma 16.14. Suppose that K has characteristic zero; we then have

H(pr1, H(pr2, pr3)) = H(H(pr1, pr2),pr3) .

Proof. Apply (22) to the Hausdorff series H ∈ L{Y,Z} and the two choicesa := U, b := H(Y, Z) and a := H(U, Y ), b := Z, respectively, and use thefirst part of (20).

118

17 The convergence of the Hausdorff series

We fix a Lie algebra g of finite dimension d over a field K of characteristiczero. We also pick a K-basis e1, . . . , ed of g.

Definition. The elements γkij ∈ K, for 1 ≤ i, j, k ≤ d, defined by theequations

[ei, ej ] =d∑

k=1

γkijek

are called the structure constants of g with respect to the basis {ei}1≤i≤d.

If we define the Lie product [ , ]′ on Kd by

(23) [(v1, . . . , vd), (w1, . . . , wd)]′ :=(∑i,j

γ1ijviwj , . . . ,

∑i,j

γdijviwj)

then the isomorphism g ∼= Kd becomes an isomorphism of Lie algebras.Using this same isomorphism we also may view the element

H = Hg ∈ Pow0(g× g; g)

as a d-tuple

H(Y , Z) := Hg(Y , Z) = (H(1)(Y , Z), . . . ,H(d)(Y , Z))

of formal power series H(i)(Y , Z) over K in the variables Y = (Y1, . . . , Yd)and Z = (Z1, . . . , Zd). The Examples 16.12 tell us that

H(i)(Y , Z) = Yi + Zi + 12

∑j,k

γijkYjZk + . . . .

Lemma 17.1. i. H(Y , 0) = Y , H(0, Z) = Z.

ii. H(Y ,−Y ) = 0.

iii. H(U,H(Y , Z)) = H(H(U, Y ), Z).

Proof. This is a restatement of Lemma 16.13 and Lemma 16.14.

From now on let (K, | |) be a nonarchimedean field of characteristic zero.Via the linear isomorphism g ∼= Kd we may view g as a manifold over K(but which structure does not depend on the choice of the basis).

119

Let us suppose at this point that there is an ε > 0 such that

(24) H(Y , Z) ∈ Fε(Kd ×Kd;Kd) and ‖H‖ε ≤ ε

(where Kd is equipped with the usual maximum norm). We then considerthe open submanifold

Gε := Bε(0) ⊆ Kd ∼= g .

Obviously

Gε ×Gε −→ Gε

(g, h) 7−→ gh := H(g, h)

is a well defined locally analytic map. Prop. 5.4 and Lemma 17.1 togetherimply that we have

g10 = 0g1 = g1, g1(−g1) = 0, and g1(g2g3) = (g1g2)g3

for any g1, g2, g3 ∈ Gε. This proves the following fact provided (24) holdstrue.

Proposition 17.2. Gε is a d-dimensional Lie group over K whose neutralelement is the zero vector 0 and such that −g is the inverse of g ∈ Gε.

If two ε ≥ ε′ > 0 satisfy (24) then Gε′ of course is an open subgroup Gε.

Definition. {Gε}ε is called the Campbell-Hausdorff Lie group germ of theLie algebra g.

What is the Lie algebra of Gε (still assuming (24))? We have the ”global”chart c := (Gε,⊆,Kd) for the manifold Gε and correspondingly the locallyanalytic isomorphism

τc : Gε ×Kd '−−→ T (Gε)(g, v) 7−→ [c, v] ∈ Tg(Gε)

as well as the linear isomorphism

Can(Gε,Kd)∼=−−→ Γ(Gε, T (Gε))

f 7−→ ξf (g) = τc(g, f(g)) = [c, f(g)] ∈ Tg(Gε) .

120

From Remark 9.21.i. we know that the Lie product of vector fields corre-sponds on the left hand side to the Lie product

[f1, f2](g) = Dgf1(f2(g))−Dgf2(f1(g)) .

On the other hand the Lie product on Lie(Gε) = T0(Gε) is induced via theinclusion

T0(Gε) −→ Γ(Gε, T (Gε))t 7−→ ξt(g) = T0(rg)(t)

by the Lie product of vector fields. By the construction of the tangent mapT0(rg) in section 9 we have the commutative diagram

KdD0rg //

v 7→[c,v] ∼=��

Kd

v 7→[c,v]∼=��

T0(Gε)T0(rg)

// Tg(Gε) .

We define fv(g) := D0(rg)(v) and compute

ξ[c,v](g) = T0(rg)([c, v]) = [c,D0rg(v)] = ξfv(g)

which means that the diagram

Kdv 7→fv //

v 7→[c,v] ∼=��

Can(Gε,Kd)

f 7→ξf∼=��

T0(Gε)t7→ξt

// Γ(Gε, T (Gε))

is commutative. Let [ , ]′′ denote the Lie product on Kd which under theleft perpendicular arrow corresponds to the Lie product of the Lie algebraLie(Gε) = T0(Gε). The commutativity of the diagram then says that wehave

f[v,w]′′ = [fv, fw] for any v, w ∈ Kd .

Observing thatfv(0) = D0 idGε(v) = v

we deduce that

[v, w]′′ = [fv, fw](0) = D0fv(fw(0))−D0fw(fv(0))= D0fv(w)−D0fw(v)

(25)

121

for any v, w ∈ Kd.We summarize that we have identified both Lie algebras, g and Lie(Gε),

with Kd thereby obtaining the two Lie products [ , ]′ and [ , ]′′ on Kd.

Proposition 17.3. Lie(Gε) ∼= g as Lie algebras.

Proof. By the above discussion it suffice to show that

[ , ]′ = [ , ]′′

holds true. To further compute the Lie product [ , ]′′ we start from theidentity

rg(h) = H(h, g) .

Since, by Lemma 17.1.i., H does not contain monomials of degree (0, s) in(Y , Z) with s ≥ 2 we may write

H(i)(Y , Z) = Zi +d∑j=1

P(i,j)(Z)Yj + terms of degree ≥ 2 in Y .

Using Prop. 5.6 we deduce that

D0rg =(∂H(i)(Y , g)

∂Yj∣∣Y=0

)i,j

=(P(i,j)(g)

)i,j

and hence that

fv(g) = D0(rg)(v) =( d∑j=1

vjP(1,j)(g), . . . ,d∑j=1

vjP(d,j)(g))

for any v = (v1, . . . , vd) ∈ Kd. To derive the function fv in 0 we must derivethe P(i,j)(Z) in Z and subsequently set Z = 0. By the Examples 16.12 wemay write

P(i,j)(Z) = δij + 12

d∑k=1

γijkZk + terms of degree ≥ 2 in Z

where δij denotes the Kronecker symbol. It follows that

∂P(i,j)(Z)∂Zk

∣∣Z=0= 1

2γijk

122

and hence that

D0fv =(

12

d∑j=1

γijkvj)i,k

and

D0fv(w) =(

12

d∑j,k=1

γ1jkvjwk, . . . ,

12

d∑j,k=1

γdjkvjwj)

= 12 [v, w]′

for any v = (v1, . . . , vd), w = (w1, . . . , wd) ∈ Kd. We conclude that

[v, w]′′ = D0fv(w)−D0fw(v) = 12 [v, w]′ − 1

2 [w, v]′ = [v, w]′ .

Having seen the interesting consequences of a possible convergence ofthe Hausdorff series we now must address the main question of this sectionwhether any ε satisfying (24) exists.

Using the isomorphism g ∼= Kd any element f ∈ Pol(g × g; g) can beviewed as a d-tuple f of polynomials in the variables Y and Z and hence,in particular, as an element f ∈ Fε(Kd ×Kd;Kd) for any ε > 0. Since thepolynomials in Hr,s := Hr,s are homogeneous of total degree r + s we have

‖Hr,s‖ε = ‖Hr,s‖1εr+s .

Suppose that there is a 0 < ε0 ≤ 1 such that

(26) ‖Hr,s‖1 ≤ ε−(r+s−1)0 for any r + s ≥ 1 .

It follows that for any 0 < ε < ε0 we have

‖Hr,s‖ε = ‖Hr,s‖1εr+s ≤ ‖Hr,s‖1εr+s−10 ε ≤ ε for any r + s ≥ 1

and

limr+s→∞

‖Hr,s‖ε ≤ ε · limr+s→∞

‖Hr,s‖1εr+s−1

= ε · limr+s→∞

‖Hr,s‖1εr+s−10

(εε0

)r+s−1

≤ ε · limr+s→∞

(εε0

)r+s−1

= 0 .

123

AsH =

∑r+s≥1

Hr,s

we conclude that (26) implies (24) for any 0 < ε < ε0.The coefficients of the Hausdorff series H are explicitly known and their

absolute values therefore can easily be estimated. But in order to translatethis knowledge into an estimate for the norms ‖Hr,s‖1 we need a particularlywell behaved basis of the K-vector space L{Y,Z}.

The free K-algebra A{Y,Z} by construction has the K-basis M{Y,Z} =⋃n≥1{Y, Z}(n). For any x ∈M{Y,Z} we let ex denote its image in the factor

algebra L{Y,Z}. These ex obviously generate L{Y,Z} as a K-vector space.Hence there exist subsets B ⊆ M{Y,Z} such that {ex}x∈B is a K-basis ofL{Y,Z}. In the following we have to make a particularly clever choice ofsuch a subset B. But first we note that also the free associative K-algebrawith unit As{Y,Z} has an obvious K-basis which is the set Mon{Y,Z} ofall noncommutative monomials in Y and Z. All of this is valid over anarbitrary field K. Since our K is nonarchimedean we may introduce theoK-submodules

As≤1{Y,Z} :=

∑µ∈Mon{Y,Z}

oKµ

of As{Y,Z} andL≤1{Y,Z} := L{Y,Z} ∩As≤1

{Y,Z}

of L{Y,Z}.

Proposition 17.4. i. (K arbitrary) There is a subset B ⊆M{Y,Z} suchthat we have

– {ex}x∈B is a K-basis of L{Y,Z},

– {Y, Z} ⊆ B, and

– for any x ∈ B \ {Y,Z} there are x′, x′′ ∈ B with x = x′x′′ and, inparticular, ex = [ex′ , ex′′ ].

ii. (K nonarchimedean) There is a subset B ⊆ M{Y,Z} as in i. and suchthat

L≤1{Y,Z} =

∑x∈B

oKex .

Proof. Compare [B-LL] Chap. II §2.10 Prop. 11 and §2.11 Thm. 1 over Kand oK , respectively.

124

We now define the constant ε0 by

ε0 :=

|p|1p−1 ε−1

1 if K is p-adic for some p,ε−1

1 otherwise

whereε1 := max(1,max

i,j,k|γkij |) .

We note that 0 < ε0 ≤ 1. The constant ε1 has the property that

‖[f, g]‖1 ≤ ε1‖f‖1‖g‖1 for any f, g ∈ Pol(g× g; g) .

Lemma 17.5. Let {ex}x∈B be any K-basis of L{Y,Z} as in Prop. 17.4.i.; wethen have

‖θ(ex)‖1 ≤ εn−11 for any x ∈ B(n) := B ∩ {Y,Z}(n) .

Proof. We proceed by induction with respect to n. For x = Y we haveθ(eY ) = pr1 and hence θ(eY ) = (Y1, . . . , Yd) so that ‖θ(eY )‖1 = 1 = ε0

1. Thecase x = Z is analogous. Any x ∈ B(n) with n ≥ 2 can be written as

x = x′x′′ with x′ ∈ B(l), x′′ ∈ B(m), and l +m = n .

Since l,m < n we may apply the induction hypothesis to x′ and x′′ andobtain

‖θ(ex)‖1 = ‖θ([ex′ , ex′′ ])‖1 = ‖[θ(ex′), θ(ex′′)]‖1≤ ε1‖θ(ex′)‖1‖θ(ex′′)‖1≤ ε1ε

l−11 εm−1

1 = εn−11 .

Proposition 17.6. For any 0 < ε < ε0 we have H ∈ Fε(Kd×Kd;Kd) and‖H‖ε ≤ ε.

Proof. As discussed after (26) it suffices to show that

‖Hr,s‖1 ≤ ε−(r+s−1)0 for any r + s ≥ 1 .

We fix n := r + s ≥ 1. We also pick a basis {ex}x∈B as in Prop. 17.4.i. andwrite

Hr,s =∑x∈B

cxex .

125

Since Hr,s ∈ L(n){Y,Z} we in fact have

Hr,s =∑

x∈B(n)

cxex where B(n) = B ∩ {Y,Z}(n) .

Lemma 17.5 then implies that

‖Hr,s‖1 ≤ maxx∈B(n)

|cx|‖θ(ex)‖1 ≤ εn−11 max

x∈B(n)|cx| .

In order to estimate the |cx| we have to distinguish cases. But we emphasizethat this is a question solely about the Hausdorff series (and not the Liealgebra g) and therefore, in principle, can be treated over the field Q.

Case 1: K is not p-adic for any p. Since Q ⊆ K we may choose the basisalready over the field Q. Then all coefficients cx lie in Q. By Exercise 2.1.i.we have |cx| = 0 or 1 and hence ‖Hr,s‖1 ≤ εn−1

1 = ε−(n−1)0 .

Case 2: K is p-adic for some p. By Exercise 2.1.ii. we have Qp ⊆ K and

|a| = |a|− log |p|

log pp for any a ∈ Qp .

Hence we may assume without loss of generality that (K, | |) = (Qp, | |p),and we choose B as in Prop. 17.4.ii. We want to show that

maxx∈B(n)

|cx|p ≤ pn−1p−1 .

Since the left hand side is an integral power of p this amounts to showingthat

plcx ∈ Zp := oQp for any x ∈ B(n)

where l is the unique integer such that

l ≤ n−1p−1 < l + 1 .

By our particular choice of the set B this is equivalent to

plHr,s ∈ L≤1{Y,Z} and hence to plHr,s ∈ As≤1

{Y,Z} .

The explicit form of the coefficients of Hr,s then reduces us to showing that

| pl

m

m∏i=1

1ri!si!|p ≤ 1, or equivalently, |m

m∏i=1

ri!si!|p ≥ p−l

126

whenever 1 ≤ m ≤ n, r1 + . . .+ rm = r, s1 + . . .+ sm = s, and ri + si ≥ 1.But Lemma 2.2 implies

|mm∏i=1

ri!si!|p ≥ p− 1p−1 ((m−1)+(r1+s1−1)+...+(rm+sm−1))

= p−n−1p−1 .

Since the left hand side is an integral power of p it indeed must be ≥ p−l.

18 Formal group laws

Let K be any field of characteristic zero. We fix a natural number d, and letR := K[[Y1, . . . , Yd, Z1, . . . , Zd]] denote the ring of formal power series overK in the variables Y = (Y1, . . . , Yd) and Z = (Z1, . . . , Zd).

Definition. A formal group law (of dimension d over K) is a d-tuple F =(F1, . . . , Fd) of power series Fi ∈ R such that we have:

(i) F (Y , 0) = Y and F (0, Z) = Z,

(ii) F (U,F (Y , Z)) = F (F (U, Y ), Z).

We observe that the condition (i) implies that

Fi(Y , Z) = Yi + Zi + terms of degree ≥ 1 both in Y and Z .

Hence the two sides in the condition (ii) are well defined.

Examples. 1) Fi(Y , Z) = Yi + Zi.

2) F (Y, Z) = Y + Z + Y Z (for d = 1).

3) (Lemma 17.1) F := Hg for a finite dimensional Lie algebra g over K(and some choice of K-basis of g).

The last example has a converse. Let F be any formal group law. Wehave

Fi(Y , Z) = Yi + Zi +∑j,k

cijkYjZk + terms of degree ≥ 3 .

We define a bilinear map bF : Kd ×Kd −→ Kd by

bF ((v1, . . . , vd), (w1, . . . , wd)) :=(∑j,k

c1jkvjwk, . . . ,

∑j,k

cdjkvjwk),

127

and we put

[v, w]F := bF (v, w)− bF (w, v) for v, w ∈ Kd .

Lemma 18.1. [ , ]F satisfies the Jacobi identity.

Proof. Compare [Se2] Part II, Chap. IV §7 formula 6).

We see that [ , ]F is a Lie product on Kd. In the case of the formal grouplaw Hg it follows from the Examples 16.12 that [ , ]Hg coincides (up to theisomorphism g ∼= Kd) with the Lie product on g.

Next we discuss a seemingly very different construction of a formal grouplaw from a finite dimensional Lie algebra g over K by using the universalenveloping algebra U(g). We have the following list of K-linear maps:

• (multiplication) m = mg : U(g)⊗K U(g) −→ U(g),

• (unit) e = eg : K −→ U(g) sending a to a · 1,

• (comultiplication) ∆ = ∆g : U(g)U(∆)−−−−→ U(g× g) ∼= U(g)⊗K U(g),

• (counit) c = cg : U(g) = T (g)/J(g)pr−−→ g⊗0 = K.

Of course, the maps m and e satisfy the axioms for a (noncommutative) asso-ciative K-algebra with unit, and ∆ and c are homomorphisms of K-algebraswith unit. In addition, the maps ∆ and c have the following properties:

(27) (counit property) (c⊗ id) ◦∆ = id = (id⊗ c) ◦∆ ;

(28) (coassociativity) (id⊗∆) ◦∆ = (∆⊗ id) ◦∆ ;

(29) (cocommutativity) the diagram

U(g)∆

&&NNNNNNNNNNN∆

xxqqqqqqqqqqq

U(g)⊗K U(g)x⊗y 7→y⊗x

// U(g)⊗K U(g)

is commutative .

They easily follow, by applying the universal property of U(g), from thecorresponding properties of the diagonal map ∆ : g −→ g × g. We nowconsider the K-linear dual

U(g)∗ := HomK(U(g),K)

128

together with the K-linear map

µ : U(g)∗ ⊗K U(g)∗ can−−−→ [U(g)⊗K U(g)]∗ ∆∗−−−→ U(g)∗ .l1 ⊗ l2 7−→ [x⊗ y 7→ l1(x) · l2(y)]

Proposition 18.2. (U(g)∗, µ, c) is a commutative and associative K-algebrawith unit.

Proof. The required axioms are exactly dual to the properties (27) - (29).

In order to determine the algebra U(g)∗ explicitly we pick an (ordered)K-basis e1, . . . , ed of g. We know from Cor. 14.3 of the Poincare-Birkhoff-Witt theorem that the

eα := eα11α1! · . . . ·

eαddαd! for α = (α1, . . . , αd) ∈ Nd

0

form a K-basis of U(g).

Proposition 18.3. The map

U(g)∗∼=−−→ K[[U1, . . . , Ud]]

` 7−→ F`(U) :=∑α∈Nd0

`(eα)Uα

is an isomorphism of K-algebras with unit onto the ring of formal powerseries over K in the variables U = {U1, . . . , Ud}.

Proof. The fact that {eα}α is a K-basis of U(g) immediately implies thatthe asserted map is a K-linear isomorphism. The unit element c of U(g)∗

is the projection map onto Ke0∼= K which is mapped to Fc = 1. For the

multiplicativity we first recall that

∆(emk ) = ∆(ek)m = (ek ⊗ 1 + 1⊗ ek)m

=m∑i=0

(mi

)(eik ⊗ 1)(1⊗ em−ik )

=m∑i=0

(mi

)eik ⊗ em−ik

for any 1 ≤ k ≤ d and any m ≥ 0. By induction one deduces that

(30) ∆(eα) =∑

β+γ=α

eβ ⊗ eγ for any α ∈ Nd0

129

holds true. We now compute

Fµ(`1,`2)(U) =∑α

µ(`1, `2)(eα)Uα =∑α

(`1 ⊗ `2)(∆(eα))Uα

=∑α

∑β+γ=α

`1(eβ)`2(eγ)Uβ+γ

=(∑

β

`1(eβ)Uβ)(∑

γ

`2(eγ)Uγ)

= F`1(U)F`2(U) .

By dualizing the multiplication map

U(g× g) ∼= U(g)⊗K U(g) m−−→ U(g)

we obtain a K-linear map

U(g)∗ m∗−−−→ U(g× g)∗ .

Applying Prop. 18.3 to both sides (with (e1, 0), . . . , (ed, 0), (0, e1), . . . , (0, ed)as an ordered K-basis for g× g) we may view the latter as a K-linear map

K[[U1, . . . , Ud]]m∗−−−→ K[[Y1, . . . , Yd, Z1, . . . , Zd]] = R .

We define

F(i) := m∗(Ui) ∈ R and F g := (F(1), . . . , F(d)) .

At this point we have to recall a few basic facts about formal powerseries rings. First of all, the formal power series ring K[[U1, . . . , Ur]] has aunique maximal ideal mU which is the ideal generated by U1, . . . , Ur. This isan immediate consequence of the fact that any formal power series F overK with F (0) 6= 0 is invertible.

Definition. i. A commutative ring with unit is called local if it has aunique maximal ideal.

ii. A homomorphism of local rings is called local if it maps the maximalideal into the maximal ideal.

130

Consider two formal power series ringsK[[U1, . . . , Ur]] andK[[V1, . . . , Vs]].For any F = (F1, . . . , Fr) ∈ mV × . . .×mV the map

εF : K[[U1, . . . , Ur]] −→ K[[V1, . . . , Vs]]G 7−→ G(F ) := G(F1, . . . , Fr)

is a well defined local homomorphism of local rings. We have εF (Ui) = Fi.

Lemma 18.4. Let ε : K[[U1, . . . , Ur]] −→ K[[V1, . . . , Vs]] be any homomor-phism of K-algebras with unit which is local; we then have

ε = εF with Fi := ε(Ui) .

Proof. Since ε is local we have Fi ∈ mV so that εF is well defined. Both εand εF are homomorphisms of K-algebras with unit. Hence the identitiesε(Ui) = εF (Ui) imply that

ε(G) = εF (G) for any polynomial G ∈ K[U1, . . . , Ur] .

We now write an arbitrary formal power series G ∈ K[[U1, . . . , Ur]] as

G =∑n≥0

Gn

where Gn is a homogeneous polynomial of degree n. In particular, Gn liesin mn

U . Since ε is local we obtain ε(Gn) ∈ mnV . Therefore the element∑

n≥0

ε(Gn) ∈ K[[V1, . . . , Vs]]

is well defined. We have

ε(G)−∑n≥0

ε(Gn) = ε(G0) + . . .+ ε(Gk) + ε(∑n>k

Gn)−∑n≥0

ε(Gn)

= ε(∑n>k

Gn)−∑n>k

ε(Gn)

∈ mk+1V

for any k ≥ 0. Since⋂k≥0 mk+1

V = {0} we conclude that

ε(G) =∑n≥0

ε(Gn) .

131

The same reasoning, of course, applies to εF . Hence

ε(G) =∑n≥0

ε(Gn) =∑n≥0

εF (Gn) = εF (G) .

Proposition 18.5. i. m∗ is a local homomorphism.

ii. m∗ = εFg.

iii. F g is a formal group law.

iv. [ , ]Fg coincides (up to the isomorphism g ∼= Kd given by the basise1, . . . , ed) with the Lie product on g.

Proof. i. The multiplicativity of m∗ amounts to the commutativity of theouter square in the diagram

U(g)∗ ⊗K U(g)∗ m∗⊗m∗ //

can

��

U(g× g)∗ ⊗K U(g× g)∗

can

��[U(g)⊗K U(g)]∗

(m⊗m)∗ //

∆∗g��

[U(g× g)⊗K U(g× g)]∗

∆∗g×g

��U(g)∗ m∗ // U(g× g)∗ .

The upper square is commutative for trivial reasons. The lower square is thedual of the diagram

U(g× g)∆g×g //

∼=��

U(g× g)⊗K U(g× g)

∼=⊗∼=��

U(g)⊗K U(g)

m

��

[U(g)⊗K U(g)]⊗K [U(g)⊗K U(g)]

m⊗m��

U(g)∆g // U(g)⊗K U(g) .

We check its commutativity on the K-basis

eα,β := (e1,0)α1

α1! · . . . · (ed,0)αd

αd! · (0,e1)β1

β1! · . . . · (0,ed)βd

βd! ,

132

for α = (α1, . . . , αd), β = (β1, . . . , βd) ∈ Nd0, of U(g × g). Under the algebra

isomorphism U(g×g) ∼= U(g)⊗K U(g) an element (x, y) ∈ g×g correspondsto x⊗1+1⊗y. It follows that eα,β corresponds to eα⊗eβ. Using the formula(30) we obtain

eα,β � ∆g×g //_

��

∑(γ,λ)+(δ,µ)=(α,β)

eγ,λ ⊗ eδ,µ_

��

eα ⊗ eβ_

m

��

∑γ+δ=αλ+µ=β

(eγ ⊗ eλ)⊗ (eδ ⊗ eµ)

_

m⊗m��

eαeβ � ∆g //

( ∑γ+δ=α

eγ ⊗ eδ)( ∑

λ+µ=β

eλ ⊗ eµ)

=∑

γ+δ=αλ+µ=β

eγeλ ⊗ eδeµ.

This establishes the multiplicativity of m∗. The unit element in U(g)∗, resp.in U(g × g)∗, is the linear form cg, resp. cg×g, which is the projection maponto Ke0

∼= K, resp. onto Ke0,0∼= K. Since cg is a homomorphism of

algebras we have

m∗(cg)(eα,β) = cg(eαeβ) = cg(eα)cg(eβ) =

{1 if (α, β) = (0, 0),0 otherwise .

It follows that m∗(cg) = cg×g.The dual of the map e : K −→ U(g) is the linear form ` 7−→ `(e0) on

U(g)∗, resp. the linear form F 7−→ F (0) on K[[U1, . . . , Ud]]. We thereforesee that under the isomorphism K[[U1, . . . , Ud]] ∼= U(g)∗ in Prop. 18.3 themaximal ideal mU is the orthogonal complement of the subspace im(e) =K · 1 ⊆ U(g). But the multiplication map m respects unit elements, i. e.,m◦eg×g = eg. Hence m∗ respects the corresponding orthogonal complementswhich means that m∗ is local.

ii. This is immediate from i. and Lemma 18.4.iii. The associativity of the multiplication in U(g) can be expressed as

the identitym ◦ (id⊗m) = m ◦ (m⊗ id) .

Dually we obtain

(id⊗m∗) ◦m∗ = (m∗ ⊗ id) ◦m∗ .

133

By ii. the evaluation on U gives the identity

F g(U,F g(Y , Z)) = F g(F g(U, Y ), Z) .

The commutative diagram

U(g) e⊗id //

=

++WWWWWWWWWWWWWWWWWWWWWWWWWWWW U(g)⊗K U(g) ∼= U(g× g)

m

��U(g)

gives rise on the dual side to the commutative diagram

K[[Y1, . . . , Yd, Z1, . . . , Zd]]Yi 7→0,Zi 7→Ui // K[[U1, . . . , Ud]]

K[[U1, . . . , Ud]] .

m∗

OO=

33fffffffffffffffffffffff

This means that F g(0, Z) = Z. Analogously we obtain F g(Y , 0) = Y .iv. We write

m∗(Ui) = F(i)(Y , Z) = Yi + Zi +∑j,k

cijkYjZk + terms of degree ≥ 3 .

The linear form

`i : U(g) −→ K∑α∈Nd0

cαeα 7−→ ci

(where as before i = (. . . , 0, 1, 0, . . .) with 1 in the i-th place) satisfies F`i =Ui. Hence Fm∗(`i) = m∗F`i = m∗(Ui) = F(i) and therefore

cijk = m∗(`i)(ej,0e0,k) = `i(ejek) .

It follows that

[ej , ek] =d∑i=1

`i([ej , ek])ei =d∑i=1

`i(ejek − ekej)ei

=d∑i=1

(cijk − cikj)ei .

Hence the structure constants cijk − cijk of the Lie product [ , ]Fg coincidewith the structure constants of g.

134

As a last construction we will associate a formal group law with any Liegroup. Let K be a nonarchimedean field of characteristic zero, and let G bea d-dimensional Lie group over K with Lie algebra g := Lie(G). We pick achart c = (U,ϕ,Kd) for G around the unit element e ∈ G such that ϕ(e) = 0.(The latter, of course, can always be achieved by translating ϕ by ϕ(e).)Since the multiplication mG is continuous we find an open neighbourhoodV ⊆ U of e such that V · V ⊆ U . Then also (V, ϕ|V,Kd) is a chart of Garound e, by Remark 8.1, and the map

ϕ(V )× ϕ(V )ϕ◦mG◦(ϕ−1×ϕ−1)−−−−−−−−−−−−→ ϕ(U)

is locally analytic. Hence there exists, for sufficiently small ε > 0, a d-tupleof power series FG,c ∈ Fε(Kd ×Kd;Kd) such that

FG,c(v, w) = ϕ(ϕ−1(v)ϕ−1(w)) for any v, w ∈ Bε(0) .

By the identity theorem Cor. 5.8 the d-tuple FG,c(Y , Z) of formal powerseries does not depend on the choice of ε.

Proposition 18.6. i. FG,c is a formal group law.

ii. [ , ]FG,c coincides, modulo the isomorphism θ−1c : g = Te(G)

∼=−−→ Kd,with the Lie product on g.

Proof. i. Because of ϕ(e) = 0 we have

FG,c(v, 0) = v and FG,c(0, w) = w for any v, w ∈ Bε(0) .

Again using the identity theorem Cor. 5.8 this translates into the identitiesof formal power series

FG,c(Y , 0) = Y and FG,c(0, Z) = Z .

In particular, the formation of

FG,c(U,FG,c(Y , Z)) and FG,c(FG,c(U, Y ), Z)

is well defined. For sufficiently small δ > 0 these formations commute, byProp. 5.4, with the evaluation in any points u, v, w ∈ Bδ(0). But by theassociativity of the multiplication in G we have

FG,c(u, FG,c(v, w)) = FG,c(FG,c(u, v), w) for any u, v, w ∈ Bδ(0) .

135

By a third application of Cor. 5.8 this translates into the identity of formalpower series

FG,c(U,FG,c(Y , Z)) = FG,c(FG,c(U, Y ), Z) .

ii. This is exactly the same computation as in the proof of Prop. 17.3.

Lemma 18.7. The Lie group G has a family {H|a|}|a| of open subgroupsindexed by the sufficiently big |a| ∈ |K| which forms a fundamental systemof open neighbourhoods of e ∈ G and such that each H|a| is isomorphic, viaϕ, to (B 1

|a|(0), FG,c).

Proof. With ε > 0 as above we put ε0 := ‖FG,c‖ε, and we choose any|a| ≥ max(1

ε ,ε0ε2

) (so that, in particular, 1|a| ≤ ε). We claim that

‖FG,c‖ 1|a|≤ 1|a|

holds true. Let

FG,c(Y , Z) = Y + Z +∑

|α|,|β|≥1

vα,βYαZβ with vα,β ∈ Kd .

We have ‖vα,β‖ ≤ ε0ε−|α|−|β| and hence

‖FG,c‖ 1|a|

= max(

1|a| , max|α|,|β|≥1

‖vα,β‖ 1|a||α|+|β|

)≤ max

(1|a| , max|α|,|β|≥1

ε0

(1|a|ε)|α|+|β|)

≤ max(

1|a| , ε0

(1|a|ε)2)

≤ 1|a| .

By possibly enlarging the lower bound for |a| we can make exactly the sameargument for the power series expansion of the map g 7−→ g−1 on G in asufficiently small neighbourhood of ϕ(e) = 0 (cf. Prop. 13.6). The familyH|a| := ϕ−1(B 1

|a|(0)) then has the required properties.

Corollary 18.8. Every Lie group is paracompact.

Proof. By Lemma 18.7 we find an open subgroupH ⊆ G which as a manifoldis isomorphic to a ball Bε(0). Any coset gH, for g ∈ G, then is isomorphic,as a manifold, to Bε(0) as well. By Lemma 1.4 the ultrametric space Bε(0)and therefore any coset gH is strictly paracompact. As a disjoint union ofcosets gH the Lie group G also is strictly paracompact.

136

Remark 18.9. If Gε is the Campbell-Hausdorff Lie group germ of a Liealgebra g then we have

Hg = FGε,c

for the chart c := (Gε,⊆,Kd).

In the present situation of a Lie group G and with the choice of theK-basis of g which corresponds to the standard basis of Kd under the iso-morphism θ−1

c : g∼=−−→ Kd we now have the three formal group laws

Hg, F g, and FG,c

whose Lie products[ , ]Hg = [ , ]Fg = [ , ]FG,c

coincide and coincide with the Lie product on g (Example 16.12, Lemma17.1, Prop. 18.5, Prop. 18.6).

In order to compare formal group laws we need the following concept.

Definition. Let F and F ′ be formal group laws over K of dimension dand d′, respectively. A formal homomorphism Φ : F −→ F ′ is a d′-tupleΦ = (Φ1, . . . ,Φd′) of formal power series Φi ∈ K[[U1, . . . , Ud]] such thatΦi(0) = 0 and

Φ(F (Y , Z)) = F ′(Φ(Y ),Φ(Z)) .

The formal group laws F and F ′ are called isomorphic if d = d′ and if thereare formal homomorphisms Φ : F −→ F ′ and Φ′ : F ′ −→ F such thatΦ(Φ′(U)) = U = Φ′(Φ(U)).

We write Hom(F , F ′) for the set of all formal homomorphisms Φ : F −→F ′, and we consider the linear map

Hom(F , F ′) −→ HomK(Kd,Kd′)

Φ 7−→ σΦ :=(∂Φi(U)

∂Uj∣∣U=0

)i,j.

Theorem 18.10. The map

Hom(F , F ′)∼=−−→ HomK((Kd, [ , ]F ), (Kd′ , [ , ]F ′))

Φ 7−→ σΦ

is well defined and bijective; in particular, the formal group laws F and F ′

are isomorphic if and only if the corresponding Lie products [ , ]F and [ , ]F ′are isomorphic.

137

Proof. ???

Corollary 18.11. The three formal group laws Hg, F g, and FG,c are mu-tually isomorphic.

Proposition 18.12. Let G1 and G2 be two Lie groups over K and let ci =(Ui, ϕi,Kdi), for i = 1, 2, be a chart for Gi around the unit element ei ∈ Gisuch that ϕi(ei) = 0; for any formal homomorphism Φ : FG1,c1 −→ FG2,c2

there is an ε > 0 such that Φ ∈ Fε(Kd1 ;Kd2).

Proof. In a first step we consider the special case that G1 = (K,+) is theadditive group of the field K and the chart is c1 = (K, id,K). The FG1,c1 =Y +Z. We abbreviate d := d2 and F := FG2,c2 . The formal homomorphismΦ is a d-tuple of formal power series in one variable U which satisfies

Φ(0) = 0 and Φ(Y + Z) = F (Φ(Y ),Φ(Z)) .

Deriving the last identity with respect to Z and then setting Z equal to zeroleads to

Φ′(U) =∂F

∂Z(Φ(U), 0) · Φ′(0) .

We defineG(Y ) :=

∂F

∂Z(Y , 0) · Φ′(0)

and obtain the system of differential equations

Φ′(U) = G(Φ(U)) with Φ(0) = 0 .

We write

G(Y ) =∑α∈Nd0

Y α(Mα · Φ′(0)) with Mα ∈Md×d(K)

andΦ(U) =

∑n≥1

Unwnn!

with wn = (wn,1, . . . , wn,d) ∈ Kd .

Our system of differential equations now reads∑n≥0

Un wn+1

n! =∑α∈Nd0

(∑m≥1

Umwm,1m!

)α1 · . . . ·(∑m≥1

Umwm,dm!

)αd(Mα · Φ′(0)) .

138

By comparing coefficients we obtain the equations

wn+1 =∑α∈Nd0

( ∑m1,1+...+md,αd=n

∏i,j

wmi,j ,i·n!

mi,j !

)(Mα · Φ′(0))

for n ≥ 0, where the second summation runs over all |α|-tuples

(m1,1, . . . ,m1,α1 ,m2,1, . . . ,m2,α2 , . . . ,md,1, . . . ,md,αd)

of integers ≥ 1 whose sum is equal to n. Since each n!m1,1!·...·md,αd ! is an integer

it follows that

‖wn+1‖ ≤ max{(∏i,j

|wmi,j ,i|)·‖Mα ·Φ′(0)‖ : α ∈ Nd0,m1,1+. . .+md,αd = n}

≤ max{(∏i,j

‖wmi,j‖)·‖Mα‖·‖Φ′(0)‖ : α ∈ Nd0,m1,1+. . .+md,αd = n}.

In the proof of Lemma 18.7 we have seen that

‖F‖ 1|a|≤ 1|a|

holds true for any sufficiently big |a| ∈ |K|. This implies the existence ofsome |a| ≥ 1 such that

‖Mα‖ ≤ |a||α| for any α ∈ Nd0 .

We claim that

‖wn+1‖ ≤ |a|n · ‖Φ′(0)‖n+1 for any n ≥ 0 .

The case n = 0 is obvious form w1 = Φ′(0). We now proceed by inductionwith respect to n. Since 1 ≤ mi,j < n the induction hypothesis gives

‖wmi,j‖ ≤ |a|mi,j−1 · ‖Φ′(0)‖mi,j .

We deduce ∏i,j

‖wmi,j‖ ≤ |a|n−|α| · ‖Φ′(0)‖n

and therefore

‖wn+1‖ ≤ maxα|a|n−|α| · ‖Φ′(0)‖n · |a||α| · ‖Φ′(0)‖ = |a|n · ‖Φ′(0)‖n+1 .

139

Using Exercise 2.1.i. and Lemma 2.2 we conclude that there are appropriateε0, ε1 > 0 such that

‖wnn!‖ ≤ ε0ε

n1 for any n ≥ 1 .

It follows that Φ ∈ Fε(K;Kd) for any 0 < ε < ε−11 .

We now consider the general case, and we fix a K-basis x1, . . . , xd1 ofg1 (where gi := Lie(Gi)). For any x ∈ g1 we may apply Thm. 18.10 to thehomomorphism of Lie algebras

K −→ g1

a 7−→ ax

and obtain a unique formal homomorphism

Φx : F(K,+),id −→ FG1,c1

such thatΦ′x(0) = σΦx

(1) = θ−1c1 (x) .

We introduce the homomorphism of Lie algebras

σ := θc2 ◦ σΦ ◦ θ−1c1 : g1 −→ g2 .

The unicity part of Thm. 18.10 implies in addition that we must have

(31) Φσ(x)(U) = Φ(Φx(U)) .

By the special case which we have treated already we find an ε > 0 suchthat

Φxi ∈ Fε(K;Kd1) and Φσ(xi) ∈ Fε(K;Kd2) for any 1 ≤ i ≤ d1 .

Hence, for sufficiently small ε > 0, the maps

G1

Kd1 ⊇ Bε(0)

f1((a1,...,ad1 )):=ϕ−11 (Φx1

(a1))·...·ϕ−11 (Φxd1

(ad1 ))

OO

f2((a1,...,ad1 )):=ϕ−12 (Φσ(x1)(a1))·...·ϕ−1

2 (Φσ(xd1)(ad1 ))

��G2

140

are well defined and locally analytic. Using Cor. 13.5 we see that the tangentmap at 0 of the upper map is equal to (a1, . . . , ad1) 7−→ a1x1 + . . . + ad1xd1which is a bijection. Hence by Prop. 9.3 the upper map can be inverted asa locally analytic map in a sufficiently small open neighbourhood V1 ⊆ U1

of e1 ∈ G1. Because of (31) the resulting composed locally analytic mapf2 ◦ f−1

1 : V1 −→ Bε(0) −→ G2 has Φ as its power series expansion (withrespect to the charts ϕ1|V1 and ϕ2) around ϕ1(e1) = 0.

Proposition 18.13. Let G1 and G2 be Lie groups over K, and let σ :Lie(G1) −→ Lie(G2) be a homomorphism of Lie algebras; we then have:

i. There exist open subgroups Hi ⊆ Gi as well as a homomorphism ofLie groups f : H1 −→ H2 such that Lie(f) = σ;

ii. if (H ′1, H′2, f′) is another triple as in i. then f |H = f ′|H on some open

subgroup H ⊆ H1 ∩H ′1.

Proof.

141

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