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TRANSCRIPT
Math 132 - Topology II: Smooth Manifolds
Taught by Michael HopkinsNotes by Dongryul Kim
Spring 2016
This course, which is not a continuation of Math 131, was taught by MichaelHopkins. We met three times a week, on Mondays, Wednesdays, and Fridays,from 1:00 to 2:00 in Science Center 310. We used Differential Topology byGuillemin and Pollack as a textbook. There were 19 students enrolled in thiscourse. The grading was only based on the problem sets. The course assistantfor this class was Adam Al-natsheh.
Contents
1 January 25, 2016 51.1 Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 January 27, 2016 62.1 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Smooth structure . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3 January 29, 2016 83.1 Charts and Atlases . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Projective space and the Grassmannian . . . . . . . . . . . . . . 8
4 February 1, 2016 104.1 Multivariable calculus . . . . . . . . . . . . . . . . . . . . . . . . 104.2 Tangent spaces and derivative of a smooth map . . . . . . . . . . 11
5 February 3, 2016 135.1 Inverse function theorem . . . . . . . . . . . . . . . . . . . . . . . 135.2 Immersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
6 February 5, 2016 156.1 Submersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
7 February 8, 2016 177.1 Transversality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1 Last Update: August 27, 2018
8 February 10, 2016 208.1 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208.2 Stability of familiar properties . . . . . . . . . . . . . . . . . . . . 21
9 February 12, 2016 229.1 Sard’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229.2 Non-degeneracy of a critical point . . . . . . . . . . . . . . . . . 23
10 February 17, 2016 2510.1 Morse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
11 February 19, 2016 2811.1 Embedding and immersion theorems . . . . . . . . . . . . . . . . 2811.2 Tangent bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . 2911.3 Whitney embedding theorem . . . . . . . . . . . . . . . . . . . . 29
12 February 22, 2016 3012.1 Manifold with boundary . . . . . . . . . . . . . . . . . . . . . . . 3012.2 Transversality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
13 February 24, 2016 3213.1 Classification of 1-manifolds . . . . . . . . . . . . . . . . . . . . . 32
14 February 26, 2016 3514.1 Transversality in a general family . . . . . . . . . . . . . . . . . . 35
15 February 29, 2016 3715.1 Partitions of unity . . . . . . . . . . . . . . . . . . . . . . . . . . 37
16 March 2, 2016 3916.1 Normal bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3916.2 Tubular neighborhood theorem . . . . . . . . . . . . . . . . . . . 39
17 March 4, 2016 4117.1 Mod 2 intersection numbers . . . . . . . . . . . . . . . . . . . . . 41
18 March 7, 2016 4318.1 Cobordism groups . . . . . . . . . . . . . . . . . . . . . . . . . . 4318.2 Bordism homology . . . . . . . . . . . . . . . . . . . . . . . . . . 44
19 March 9, 2016 4619.1 Surgery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4619.2 The surgery cobordism . . . . . . . . . . . . . . . . . . . . . . . . 47
20 March 11, 2016 4820.1 MO1 of a torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
2
21 March 21, 2016 4921.1 Bilinear forms on MO1 . . . . . . . . . . . . . . . . . . . . . . . 4921.2 Classification of finite-dimensional non-degenerate symmetric bi-
linear forms over k = Z/2 . . . . . . . . . . . . . . . . . . . . . . 50
22 March 23, 2016 5122.1 Embeddings represent every class . . . . . . . . . . . . . . . . . . 5122.2 Separation theorems . . . . . . . . . . . . . . . . . . . . . . . . . 51
23 March 25, 2016 5323.1 Non-degeneracy of the intersection form . . . . . . . . . . . . . . 53
24 March 28, 2016 5524.1 Classification of compact connected 2-manifolds . . . . . . . . . . 55
25 March 30, 2016 5725.1 Winding numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
26 April 4, 2016 5926.1 More on Gluing . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5926.2 Isotopy of diffeomorphisms . . . . . . . . . . . . . . . . . . . . . 60
27 April 6, 2016 6227.1 Immersions of surfaces . . . . . . . . . . . . . . . . . . . . . . . . 6227.2 Linking number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
28 April 8, 2016 6428.1 Quadratic forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
29 April 11, 2016 6629.1 Quadratic forms taking values in Z/4 . . . . . . . . . . . . . . . . 6629.2 Brown-Arf invariant . . . . . . . . . . . . . . . . . . . . . . . . . 67
30 April 13, 2016 6830.1 Boy’s surface and immersions of surfaces . . . . . . . . . . . . . . 68
31 April 15, 2016 7031.1 Multilinear algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 70
32 April 20, 2016 7232.1 Integration on manifolds . . . . . . . . . . . . . . . . . . . . . . . 72
33 April 22, 2016 7433.1 Stokes theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
34 April 25, 2016 7634.1 Degree of a map . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3
35 April 27, 2016 7935.1 The Gauss-Bonnet theorem . . . . . . . . . . . . . . . . . . . . . 79
4
Math 132 Notes 5
1 January 25, 2016
1.1 Outline
We will be studying something called a smooth manifold. Objects like toruses,Klein bottles, double toruses, etc., are smooth manifolds, while cones are not,because they have these corners. A natural question is “What are all the picturesof smooth surfaces that can be drawn?” The first step is to define a smoothsurface, or more generally, a smooth manifold of dimension d. Then we needto define a ‘picture’ of a surface, which we will call an immersion f : Σ # R3.Next, we need a notion of equivalence between pictures; if one can be smoothlydeformed into another, then we call the two isotopic. Now our problem becomes:
a) Classify all smooth surfaces.
b) For each smooth surface Σ, describe the set of isotopy classes of immersionsof Σ # R3.
To approach the problem, we define the intersection form. This is a non-degenerate symmetric bilinear form on a vector space over F2. If something is aboundary of a sub manifold, we consider it to be zero. For two curves, we mayconsider the number of intersection modulo 2. This is the intersection form.It turns out that the intersection form determines the manifold, and thus theclassification of immersions into the classification of bilinear forms.
Theorem 1.1. The immersions of Σ in R3 is in one-to-one correspondence withthe quadratic q : V → Z/4Z whose underlying bilinear form is the intersectionform.
The reason for the strange Z/4Z is because a strip with 4 half-twists is thesame as a strip with no twists.
Math 132 Notes 6
2 January 27, 2016
2.1 Manifolds
Today I’m going to start by giving the official definition of a manifold. I wantto consider a set X ⊂ Rn sitting inside the Euclidean space.
Definition 2.1. Let U ⊆ Rn be a open set. A function f : U → R is calledsmooth if it there exists derivatives of all orders. A function f : X → Ris smooth if there is a neighborhood X ⊆ U ⊆ Rn and a smooth functiong : U → R such that f(x) = g(x) for all x ∈ X.
Definition 2.2. A subset X ⊂ R is a smooth manifold of dimension d if foreach x ∈ X if for each x ∈ X, there is a neighborhood x ∈ U ⊂ Rd and aninjective function f : U ∩X → Rd such that f and f−1 : f(U ∩X)→ U ∩X areboth smooth.
Example 2.3. Consider the n-sphere
Sn = (x1, . . . , xn) ∈ Rn :∑
x2i = 1.
Let n = 1 for simplicity, and consider the function f(x, y) = y on the subset ofpoints (x, y) with x > 0. Then this function and its inverse is smooth, because−1 < y < 1. Likewise, we can do this for x < 0 and y < 0 and y > 0. Thesefour open sets covers the whole space and thus S1 is a 1-manifold sitting in R2.
Example 2.4. Consider the set
X = (x, y) : y = 0, x ≥ 0 or x = 0, y ≥ 0.
We can define a function f : X → R defined by
f(x, y) =
x if x ≥ 0
−y if y ≥ 0.
Then the inverse f−1(t) is not differentiable at t = 0.
Example 2.5. Consider a disjoint union of S2 and a line sitting in R3. We donot consider this to be a smooth manifold, because it does not have an evendimension.
2.2 Smooth structure
Definition 2.6. A topological manifold of dimension d is a topological space1
X, such that each x ∈ X has a neighborhood U that is homeomorphic to anopen subset of Rd.
1Actually, we include conditions that it should be Hausdorff and have a countable basis.
Math 132 Notes 7
When I give a mathematical object, I should give some data and some condi-tions that data has to satisfy. The notion of a topological manifold only involvesa topology, but we need more than that to give the condition that it is smooth.
Definition 2.7. A chart on X on is a pair (U , f) consisting of an open subsetU ⊂ X and a homeomorphism f : U → f(U) ⊂ Rd with and open subset of Rd.
Definition 2.8. Let X ⊂ Rn be a topological manifold. A chart f : U →f(U) ⊂ Rd on X is smooth if f and f−1 are smooth. The manifold X is calledsmooth if X can be covered by smooth charts.
Definition 2.9. A smooth structure on X is a homeomorphism of X witha subspace of Rn for some n, which is a smooth manifold.
There are some fundamental problems in differential topology.
1) If X is a topological manifold, can it be given a smooth structure?
2) If so, if how many ways?
But clearly there are infinitely many smooth structure if there is one. Thismeans that we need a notion of equivalence so that we don’t consider the essen-tial same structures as different.
Definition 2.10. Let f1 : X → Z1 ⊂ Rn and f2 : X → Z2 ⊂ Rm be two smoothstructures. We consider them to be equivalent if there is a map g : Z1 → Z2
such that both g and g−1 are smooth and the following diagram commutes.
Z1
X
Z2
g
f1
f2
Now the problems make sense. If you live in 3 dimensional space, then infact, every topological manifold has a unique smooth structure. But around1956, Milnor discovered the fact that this is not true for higher dimensions.
Theorem 2.11 (Milnor, 1956). S7 has least 14 different smooth structures.
And around 1961, Kervaire and Milnor determined the number of smoothstructures on Sn where n 6= 3, 4, 2k−2, 2k−3 for k ≥ 7. Also, in the early 1960s,Kervaire gave an example of a topological manifold with no smooth structureof dimension 10.
Math 132 Notes 8
3 January 29, 2016
There will be a point-set topology review section next week.Let us recall what we did last time. A smooth manifold X ⊂ Rn of dimension
d is a manifold having a smooth atlas, i.e., locally diffeomorphic to Rd.
3.1 Charts and Atlases
So for any point x ∈ X, there is diffeomorphism between an open neighborhoodof x inside X, and an open set of Rd. This diffeomorphism is called a chart,and is sometimes called a coordinate chart.
Example 3.1. Consider X = R2\0. Then we can think of a polar coordinatesystem and the rectangular coordinate system. Both gives a chart on a smallneighborhood of X.
Suppose you have a chart Uα → Rd and Uβ → Rd. On Uα, the chart givesa coordinate system. Also we have a coordinate system on Uβ . Then in theintersection Uα ∩ Uβ , we have two coordinate systems. One thing we can seeis the the two coordinates should be smooth with respect to each other. Thisis how a smooth manifold is defined abstractly. An atlas is a covering of amanifold by charts, and a smooth atlas is an atlas such that the coordinatechanges between the charts are smooth.
Example 3.2. Let us show that Sn ⊂ Rn+1 is a smooth manifold. Given anyv ∈ Sn, we let
Uv = x ∈ Sn : x · v > 0.
Then the projection map Uv → v⊥ ∼= Rd gives a diffeomorphism from Uv toy ∈ v⊥ : |y| < 1. We could define an atlas by taking these coordinate chartsfor v = ±e1, . . . ,±en.
3.2 Projective space and the Grassmannian
We look at a more complex example. The real projective space RPn isdefined by
RPn = space of lines throughout the origin in Rn+1.
Here is another way to think about it. Because any line intersects Sn in exactlytwo antipodal points, we have
RPn = Sn/(v ∼ −v).
I want to prove that RPn be given a smooth structure. Here is the problem.This RPn is not defined as a subset of a Euclidean space. So we have to usethe more abstract definition of a smooth manifold.
Math 132 Notes 9
We introduce a more convenient way of representing a point in RPn.For anx ∈ RPn, we write it as some
[x0, . . . , xn] = [−x0, . . . ,−xn]
where x20 + · · ·+ x2
n = 1. I’m using the square brackets to indicate that this isthe equivalence class.
Cover RPn by U0, . . . ,Un, where
Ui = [x0, . . . , xn] : xi 6= 0.
Then every x = [x0, . . . , xn] ∈ Ui has a unique representation such that x0 > 0.Then we get an chart on Ui which is
[x0, . . . , xn]→ (x0
xi, x1
xi, . . . , xnxi ).
An easy exercise is to check that the change of coordinate functions are smooth.
Definition 3.3. The Grassmannian Grk(Rn) is defined as the set of k-dimensionalvector subspaces of Rn.
This is quite hard to embed into the Euclidean space, but it is relativelyeasy to describe an atlas. For a V ∈ Grk(Rn), we let
UV = W ∈ Grk(Rn) : W ∩ V ⊥ = 0.
Then the orthogonal projection Rn → V gives an isomorphism W → V . Think-ing W as the graph of the linear transformation TW : V → V ⊥, we see that UVis diffeomorphic to Hom(V, V ⊥), which is simply a k(n− k) dimensional vectorspace. This shows that
Proposition 3.4. Grk(Rk) is a smooth manifold of dimension k(n− k).
Math 132 Notes 10
4 February 1, 2016
4.1 Multivariable calculus
So today I want to talk about derivatives in the case of smooth manifolds.Suppose that I have a smooth map f : Rk → R`, and suppose that this is givenby x1
...xk
7→f1(x1, . . . , xk)
...f`(x1, . . . , xk)
.Then given a point x ∈ Rk, we can write down all the partial derivatives
Df(x) =
∂f1(x)∂x1
· · · ∂f1(x)∂xk
.... . .
...∂f`(x)∂x1
· · · ∂f`(x)∂xk
.If we have two maps
Rk R` Rm,f g
the chain rule says that
D(g f)(x) = Dg(f(x)) Df(x).
This is just a fancy way of writing down the chain rule
∂(gs f)
∂xt=∑i
∂gs∂yi
∂yi∂xt
we all know.Now suppose that we have a smooth map f : X → Y between smooth
manifolds. We want to make sense of what is the derivative of f . We wouldlike to define for f : X → Y , where dimX = k and dimY = `, a lineartransformation
Df : V →W
such that Df becomes “the matrix of derivatives” in any local coordinate sys-tem.
Let us describe it more precisely. Given a point x ∈ X and f(x) ∈ Y , thereis a local coordinate system of a neighborhood UX of x and UY of f(x).
UX UY
VX ⊂ Rk VY ⊂ R`
f
fΦΨ
Φ Ψ
Then we push down the map f to fΦΨ. Then we would want to do differentiatethings on fΦΨ instead. But we actually don’t have a vector space, and thus weneed more things.
Math 132 Notes 11
4.2 Tangent spaces and derivative of a smooth map
We have to associate to each point x ∈ X an abstract vector space of dimensiond. Intuitively we can think of it as a space that is actually tangent to themanifold at x ∈ X.
Let X ⊂ Rn be a smooth manifold of dimension d. Then by definition, thereis neighborhood x ∈ U ⊂ X, an open set U ⊂ Rd, and a smooth map Φ : U → U .Then we define
Definition 4.1. The tangent space X at x ∈ X is the image of DΦ : Rd →Rn.
Now my claim is
Proposition 4.2. The tangent space is independent of the change of local co-ordinates and has dimension d.
Proof. Suppose we have two local coordinate systems Φ and Ψ near x. Wemight as well assume they are defined for the same neighborhood U of x.
U
U ˆU
Φ
Ψ−1Φ=F
Ψ
Φ−1Ψ=G
Note that G F = IdRd and F G = IdRd . Then the chain rule tells us that
DF DG = Id, DG DF = Id
and thusDΨ−1 DG = Id .
Thus DΦ has an inverse and so kerDΦ has image of dimension d. Also,
DΦ = DΨ(DF )
implies that the image of DΦ is contained in the image of DΨ. Then reversingthe roles, we see that the image of DΦ is the same as the image of DΨ.
We can define it using the abstract definition of a smooth manifold. Supposewe have a smooth map f : X → Y . The derivative of f is the linear map
Df : TxX → Tf(x)Y
defined in the following way. (TxX is just the tangent space of X at x.) Bydefinition we can draw
U ⊂ X V ⊂ Y
U ⊂ Rk V ⊂ Rk
f
f
Φ Ψ
Math 132 Notes 12
and then we can define Df by the diagram
TxX Tf(x)Y
Rk Rk
Df
Df
DΦ DΨ
We can also state the chain rule.
Theorem 4.3. If f : X → Y and g : Y → Z are smooth maps of smoothmanifolds, then
D(g f)|x = D(g)|f(x) D(f)|x.
There is an alternative definition of a tangent space.
Definition 4.4 (Version 1). The tangent space TxX is the equivalence classesof curves γ : R1 with γ(0) = x, where the equivalence relation is γ1 ∼ γ2 if forany smooth U → R on a neighborhood U of x we have
d
dtf γ1|t=0 =
d
dtf γ2|t=0.
Definition 4.5 (Version 2). We let
TxX = derivations d : C∞(U)→ R at x,
where U is a neighborhood of x, and C∞(U) is the set of smooth functions onU , and a derivation is a map C∞(U)→ R which satisfies
d(fg) = df · g(x) = f(x) · dg
for any two smooth maps.
This last definition is very nice, because it does not make up anything new.In fact, this is what is most used in algebraic geometry.
Math 132 Notes 13
5 February 3, 2016
Now we get the idea that this is just some kind of curved linear algebra. Weassociated to each point a linear space and the derivative of smooth maps turnedout to be linear transformations on the spaces.
5.1 Inverse function theorem
Theorem 5.1 (Inverse function theorem). Suppose that M and N are smoothmanifolds of dimension d and f : M → N is a smooth map. Let x ∈ M bea point. If Df |x : TxM → Tf(x)N is an isomorphism, then f restricts to adiffeomorphism of a neighborhood of x with a neighborhood of f(x).
Sketch of Proof. We first set up a coordinate neighborhood of x and get thefollowing diagram, where x ∈ U and f(x) ∈ V.
M N
U V
U ⊂ Rd V ⊂ Rd
f
f
Φ
g
Ψ
Note that we can shrink U so that the image f(U) lies in V. For convenience,we can assume Φ(0) = x and Ψ(0) = f(x). We just have g = Ψ−1 f Φ.
Because Φ and Ψ are diffeomorphisms, we see that DΦ0 and DΨ0 are iso-morphisms. Also, we have
Dg|0 = (DΨ(0))−1 Df(x) DΦ(0).
Then Df(x) is an isomorphism if and only if Dg(0) is an isomorphism.So the theorem reduces to the inverse function theorem in Rd: “Suppose
U , V ⊂ Rd and g : U → V is smooth. If Dg(0) is an isomorphism then g is adiffeomorphism in a neighborhood of zero.”
Locally, every diffeomorphism looks like the identity. That is, we can letg = id. This is because given
M N
U V
f
g
Φ Ψ
we can consider the composite map Ψ g. Since D(Ψg)(0) is an isomorphism,
we can use the inverse function theorem to get a smallˆU of 0 on which Ψ g is
Math 132 Notes 14
a diffeomorphism. Therefore we can use the Ψ g as the coordinate chart andget the following.
M N
ˆU
f
Φ Ψg
5.2 Immersion
Definition 5.2. A map f : Mk → N `2 is an immersion at x ∈M if Df(x) isan monomorphism. The map f is an immersion if it is an immersion at eachpoint.
Example 5.3. Let k ≤ `. The map Rk → R` given by
(x1, . . . , xk) 7→ (x1, . . . , xt, 0, . . . , 0)
is an immersion. This is called the standard immersion.
Theorem 5.4. Every immersion locally looks like a standard immersion.
Example 5.5. The map S1 → R1 that simply sends the circle to the figureeight is a immersion.
Example 5.6. Define the torus as
T = R2/(x, y) ∼ (x+ 1, y) ∼ (x, y + 1).
There is an smooth structure on T given by the map T → R4 defined by
(x, y) 7→ (cos(2πx), sin(2πx), cos(2πy), sin(2πy).
Choose θ ∈ R and let fθ : R1 → T be the map given by
t 7→ (t, tθ).
If θ is a irrational number, the image is dense in T , and this is a kind of pathology.We will show later that this does not happen if the manifold is compact.
2This just means that M is of dimension k and N is of dimension `.
Math 132 Notes 15
6 February 5, 20163
Last time, we defined an immersion. The local immersion theorem states that iff : X → Y is a immersion then for any x and f(x), we can set local coordinatesystems so that df looks like the standard immersion.
We can ask: is f(X) always a submanifold of Y ? The answer is no in general,for instance, the figure 8 is a immersion of S1. But if we add a tiny conditionthe that the map is proper, it becomes a true statement.
6.1 Submersion
Definition 6.1. Let X and Y be manifolds of dimensions k and l, where k ≥ l.A smooth map f : X → Y is called a submersion if
dfx : TxX → Tf(x)Y
is surjective for any x.
Because a submersion is simply the dual notion of an immersion, we canhave an analogous theorem.
Theorem 6.2 (Local submersion theorem). Let f : X → Y be a submersionon x ∈ X. Then there is a local coordinate around x and f(x) so that f lookslike the canonical submersion.
Why do we care about this? The point of both immersions and submersionsis to create new manifolds from old ones. In the case of the immersion, undergood enough conditions, we have seen that the image f(X) is a manifold. Inthe case of the submersion, the preimage of a point is a manifold.
Example 6.3. We look at the map R4 → R2 given by
(x1, x2, x3, x4) 7→ (x1, x2).
Then the preimage of (0, 0) is the points of form (0, 0, x3, x4), and is in fact, aR2 parametrized by x3 and x4.
Definition 6.4. Let f : X → Y be a smooth map. For a point y ∈ Y , we saythat y is a regular value if for every x ∈ f−1(y), the derivative
dfx : Tx(X)→ Ty(Y )
is surjective. If it is not surjective, we call y a critical value.
Theorem 6.5 (Preimage theorem). Suppose that f : X → Y is a smooth map,with y ∈ Y being its regular value. Then f−1(y) is a submanifold of X ofdimension dimY − dimX.
3For some reason, Xiaolin(Danny) Shi, a graduate student, taught this day instead ofHopkins.
Math 132 Notes 16
Example 6.6. Consider the map f : Rk → R given by
(x1, . . . , xk) 7→ x21 + x2
2 + · · ·+ x2k.
Then the derivative is
df(a1,...,an) =(2a1 2a2 · · · 2ak
).
It is not zero whenever (a1, . . . , an) 6= (0, . . . , 0). Thus 0 is a critical value off , and f is locally a submersion at every other point. If we look at f−1(1), weimmediately see that Sn−1 is a manifold.
Let us look at one application. Let X be a smooth manifold of dimensionk, and let g1, . . . , gl : X → R be real valued smooth functions, where k ≥ l. Wewant to know whether
g−11 (0) ∩ g−1
2 (0) ∩ · · · ∩ g−1l (0)
is a smooth manifold.We can look at the product g : X → Rl given by
x 7→ (g1(x), g2(x), . . . , gl(x)).
We want to know whether g−1((0, . . . , 0)) is a smooth manifold or not. Us-ing the preimage theorem, we can say that, if for any x ∈ g−1(0) the vectorsdg1|x, . . . , dgl|x are linearly independent, then g−1(0) is a smooth manifold.
Math 132 Notes 17
7 February 8, 2016
Today I want to introduce arguable one of the most important ideas in the 20thcentury.
7.1 Transversality
Let us first look something in linear algebra. Let dimV = k, and let dimW = l,where k ≥ l. Then most linear transformations V → W are surjective. Andnow in differential topology, the analogue of that is that f : Mk → N l is asubmersion at some point.
Likewise, if k ≤ l then most linear transformations are injective. In differ-ential topology, for most cases f : Mk → N l will be a local immersion.
Let us suppose that there is a big vector space Wn, and let Uk ⊂ W andV l ⊂W be subspaces. Then for most U and V , we will have
dimU ∩ V = k + l − n.
For instance, most of the time, two planes in space intersect in a line.
Definition 7.1. For a subspace V ⊂W , we define the codimension of V as
codimV = dimW − dimV.
We see thatU ∩ V = ker(U ⊕ V →W ),
where the map is given by (x, y) 7→ i(x)−j(y) for the inclusion maps i : U →Wand j : V →W . In most cases, this map will be surjective and thus dim(U ∩V )will be dim(U ⊕V )−dimW . Mathematicians like to write this as the pullback.
U ∩ V U
V W
i
j
Let us now move the stage to differential topology. Suppose we have asmooth map f : X l → Y n be a smooth map, and Zk ⊂ Y be a submanifold.
Question. When is f−1(z) a smooth manifold? What is its expected dimen-sion?
We first look at the case when Z = y ∈ Y is a single point. As we haveseen last week, if y is a regular value of f , then f−1(y) is a smooth manifold ofdimension l − n. Furthermore, we know that
Tx(f−1(y)) = ker dfx.
Now suppose we look at the general case. Because whether something isa smooth manifold is a local property we will look at a single neighborhood.
Math 132 Notes 18
Suppose we have x ∈ X such that f(x) ∈ Z. We can introduce a coordinateneighborhood U of f(x) so that U ≡ Rn and Z ∩ U ≡ Rk.
X ⊃ f−1(U)
Y ⊃ U ≡ Rn Rkf h
π
Now f−1(Z) ∩ f−1(U) = h−1(0) is a smooth manifold if and only if 0 is aregular value of h. If we now move to tangent spaces, we se that this is true ifand only if
Tf(x)Z ⊕ TxX Tf(x)Y(i,df)
is surjective.
Definition 7.2. A map f : X → Y is transverse to Z ⊂ Y if for all (z, x) ∈Z ×X with f(x) = z the map
TzZ ⊕ TxX → Tf(x)=zY
is onto. This is denoted f t Z. If f is also an inclusion, then we write X t Z.
Theorem 7.3. If f is transverse to Z then f−1(Z) is a smooth manifold ifdimension
dimZ + dimX − dimY.
Example 7.4. The set X = (x, 0) : x ∈ R is transverse to Z = (x, x2 − 1) :x ∈ R. However, X is not transverse to Z ′ = (x, x2) : x ∈ R.
Example 7.5. The hyperboloid z = xy and the plane z = 1 are transverse.But z = xy and the plane z = 0 is not transverse.
Example 7.6. Consider two lines in R3. They are transverse if and only if theydo not intersect at all.
In the next lecture, we will give a better interpretation for “almost all,” andprove such a theorem.
Lastly I want to point out that our definition of transversality is asymmetric.One is an inclusion and the other is a smooth map.
Definition 7.7. Let f : X → Y and g : Z → Y be two smooth maps. Thetwo maps f and g are transverse if for all pairs x ∈ X and z ∈ Z such thatf(x) = g(z), the map
TxX ⊕ TzZ 7→ Tf(x)=g(z)Y
is onto. In such a case, we denote f t g.
Definition 7.8. If f t g then
W = (z, x) ∈ Z ×X : g(z) = f(x)
Math 132 Notes 19
is a smooth submanifold of dimension
dimW = dimZ + dimX − dimY.
Moreover, the tangent space is the pullback.
TwW TxX
TzZ TyY
Proof. This reduces to the previous case:
W Z ×X
Y Y × Y∆
Math 132 Notes 20
8 February 10, 2016
Today I’m going to make precise the notion of generic things.
8.1 Stability
Let f : Mk → N l, and consider a property P , such as ‘embedding’ or ‘immer-sion’ or etc.
Definition 8.1. A property P is called stable if for every smooth homotopyh : M × [0, 1] → N with h(x, 0) = f(x) there exists ε > 0 such that for any0 ≤ t < ε the map ft : M → N given by x 7→ h(x, t) has property P .
So intuitively it means that if you move it a little bit, it still has the property.
Example 8.2. Let Z ⊂ N . The property “f(X) intersects Z” is not stable.However, the property “f(X) intersects Z transversally” is stable.
These all come from facts in linear algebra. Suppose I have a linear trans-formation T : Rk → Rl of rank r, where r ≤ min(k, l). Suppose I also have ahomotopy h : Rk × [0, 1] → R` that is (1) continuous, (2) h(v, 0) = T (v), and(3) for all t, the map v 7→ h(v, t) is linear. Alternatively, we can say that wehave a continuous map
H : [0, 1]→Ml×k(R), t 7→ h(−, t).
Proposition 8.3. If T = H(0) has rank at least r, then there exists an ε > 0such that H(t) also has rank at least r for all 0 ≤ t < ε.
Proof. Since the rank of H(0) is at least r, there is an r× r submatrix of t withnonzero determinant. Suppose without loss of generality, that it is the upperleft r × r block. Now for any t, we denote the upper left r × r block by A(t).Then we have a continuous map Ml×k → det(A(t)), and composing it with H,we get another continuous map g : [0, 1]→ R.
[0, 1] Ml×k
R
H
gdet(A(t))
Then g(0) 6= 0 and thus there is a ε > 0 such that g(t) 6= 0 for any 0 ≤ t < ε.Then H(t) will have rank at least r.
In particular, the class of linear maps with trivial kernel is stable. Also, theclass of surjective linear maps is also stable.
Math 132 Notes 21
8.2 Stability of familiar properties
Theorem 8.4. Let M,N be smooth manifolds, and assume that M is compact.The following properties of f : Mk → N l are stable:
a) local diffeomorphism
b) immersion
c) submersion
d) transverse to fixed Z ⊂ Ne) embedding
f) diffeomorphism
Proof. We can do all a), b), c), and d) in one swoop. A map f is called a localdiffeomorphism if and only if for any x ∈ M the map dfx : TxM → Tf(x)Nis an isomorphism, i.e., or rank ≥ k. Likewise, f is a immersion if and onlyif the rank of dfx is ≥ k and f is a submersion if and only if the rank ofdfx is ≥ l. Similarly for d), we check that f is transverse to Z if and onlydf : Tx(M)→ Tf(x)N/Tf(x)Z is surjective.
Now suppose f satisfies P for h : M × [0, 1]→ N . Let x ∈M , and from ourproposition, there has to be a little neighborhood of x in M such that the rankof dfx is ≥ r in that neighborhood. Likewise, if we move t around a little bit,the rank will still be ≥ r. That is, we can find, for each x an open neighborhoodUx × [0, εx) ⊂M × [0, 1] of (x, 0) such that for any (y, t) in that neighborhood,dft|y has rank ≥ r.
Since M is compact, M is covered by finitely many of the U1, . . . , Uj , andletting ε = minε1, . . . , εj, we have
M × [0, ε) ⊂ (U1 × [0, ε1)) ∪ · · · ∪ (Uj × [0, εj)).
Then for all y, t ∈M × [0, ε), the derivative dft|y has rank ≥ r.Now we have to do e) and f). These are both immersions that are one-to-one.
So we need to show that throwing in the one-to-one property does not affectstability.
Suppose for every ε > 0 there are points x 6= y and t < ε such that h(x, t) =h(y, t). We can find an infinte sequence xi, yi, ti such that h(xi, ti) = h(yi, ti),and because M is compact, there is a subsequence xi → x and yi → y. Thenby continuity, we have
f(x) = h(x, 0) = limi→∞
h(xi, ti) = limi→∞
h(yi, ti) = h(y, 0) = f(y).
Since f is one-to-one, we have x = y.We now use the property that f is an immersion. Let g : M × [0, 1] →
N × [0, 1] be the map given by
(x, t) 7→ (h(x, t), t).
It is not difficult to check that dg(x,0) is injective and thus g is an immersionnear (x, 0). Then g has to be one-to-one on some neighborhood of (x, 0). Thiscontradicts the fact that there are such xi, yi, tis.
Math 132 Notes 22
9 February 12, 2016
Consider the projection map f : T → R that maps the torus embedded in3-space to a line. Then there are four critical points and four critical valuescorresponding to each critical point. (Recall that critical points are x ∈ Msuch that dfx is not onto, and critical values are y ∈ N such that there exists acritical point x ∈M with f(x) = y.)
T
R
f
critical values
critical points
9.1 Sard’s theorem
Theorem 9.1 (Sard). If f : M → N is smooth then almost all y ∈ N areregular values. Here, almost all means that its complement is of measure zeroin the Lebesgue sense.
Definition 9.2. A subset S ⊂ Rk has measure zero if for every ε > 0 thereis a covering of S by countably many rectangles whose total value is less thanε. A rectangle in Rk is a set of the form
(x1, . . . , xk) : ai ≤ xi ≤ bi
and it volume is defined as∏
(bi − ai).
Example 9.3. The line R1 ⊂ R2 is of measure zero. We consider the set ofrectangles
[−n, n]×[− ε
n2n+2 ,ε
n2n+2 ].
Then it clearly covers R1, and its total volume is∑ ε
2n= ε.
Thus it has measure zero. More generally, any proper subspace of a real vectorspace has measure zero.
We also need the following theorems, which we will not prove.
Theorem 9.4 (Fubini). Let S ⊂ Rk+l, and let
Sa = (x, y) ∈ S : x = a ⊂ Rl
for a ∈ Rk. If each Sa has measure zero in Rl, then S also has measure zero inRk+l.
Math 132 Notes 23
Theorem 9.5. If S ⊂ Rk has measure zero and f : Rk → Rk is a smooth map(in particular, a diffeomorphism) then f(S) has measure zero.
Theorem 9.6. A countable union of sets of measure zero has measure zero.
Proof. Let Si∞i=1 be a countable collection of measure zero sets. Given ε > 0,we can choose a cover of Sn by rectangles whose total volume is less than2−nε. then the cover of S by all these things will have total volume less than∑
2−nε = ε.
Now the notion of measure zero in a manifold can be made sense.
Definition 9.7. A subset S ⊂ Mk has measure zero if any of the followingequivalent statements hold.
(1) If for every local coordinate system Φ : U →M , the inverse image Φ−1(S)has measure zero in Rk.
(2) For some countable atlas Φi : Ui → M the sets Φ−1(S) ⊂ Ui ⊂ Rk hasmeasure zero.
9.2 Non-degeneracy of a critical point
Let f : M → R be a smooth map. From Sard’s theorem, we see that almostall values are regular values. However, if M is compact, then there has to be amaximum and a minimum, and those can’t be regular values. So we can lookat the next best thing that can happen.
Suppose that f : Rk → R is smooth and 0 ∈ Rk is a critical point and
∂f
∂x1=
∂f
∂x2= · · · = ∂f
∂xk= 0.
Then we look at the second derivative, which is described bas the Hessian off ,
Hf =
∂2f∂x2
1· · · ∂2f
∂x1∂xk...
. . ....
∂2f∂xk∂x1
· · · ∂2f∂x2k
.
Definition 9.8. The critical point 0 ∈ Rk is called non-degenerate if detHf 6=0 at 0.
If there are no degenerate critical points, then f is called a Morse function,and there are a lot of cool stuff if we use those functions.
Definition 9.9. Suppose f : M → R is smooth and x ∈ M is a critical point.The point x is a non-degenerate critical point if in all coordinate systemsΦ : U →M at x with Φ(0) = x,
det(HfΦ) 6= 0
at 0.
Math 132 Notes 24
Proposition 9.10. It suffices to check this in one coordinate system.
Proof. Suppose that we have
U U ′ RΨ g
with Ψ(0) = 0. Using the chain rule, we can check that
HgΨ = dΨT Hg dΨ.
Then since Ψ is a diffeomorphism, detHgΨ = 0 if and only if detHg = 0.
Math 132 Notes 25
10 February 17, 2016
We are getting to the theorem which is the culmination of chapter 1. Let usrecall what we did last time. We defined the notion of measure zero, and statedbut not proved Sard’s theorem.
Theorem 10.1. The set of critical values of a smooth map has measure zero.
Suppose I have a function f : Rk → R. A critical point a ∈ Rk is callednon-degenerated if detHf (a) is non-zero. We checked last time that if
Rk Rk RΨ f
with Ψ(0) = 0 and 0 ∈ Rk is a critical point, then
HfΨ = dΨT Hf dΨ
and thus critical points are independent of coordinate charts. This suggests thatthere is a definition of non-degeneracy that is independent of any coordinates.For now we are not going to get into more sophisticated because I think it ismore important to get familiar with the tools first. That said, let us defineMorse functions.
10.1 Morse functions
Definition 10.2. A Morse function f : Mk → R is a smooth function withonly non-degenerate critical points.
Example 10.3. Let us look at k = 2. It can be proved that Hf , in suitablecoordinates, looks like one of[
1 00 1
],
[1 00 −1
],
[−1 00 −1
].
x2 + y2 x2 − y2 −x2 − y2
We note that the “monkey saddle” is not a Morse function.
The miraculously cool theorem is that every manifold has a Morse function.
Theorem 10.4. Every manifold M has a Morse function.
In fact, almost every function is a Morse function; if you have any function, youcan move it a little bit and you get a Morse function.
Theorem 10.5. Let M ⊂ RN be a smooth manifold. Suppose f : M → R isany smooth function, Then for almost all a1, . . . , an ∈ RN ,
fa = f + a1x1 + · · ·+ aNxN
is a Morse function.
Math 132 Notes 26
Lemma 10.6. Suppose f : Rk → R is any smooth function. Then for almostall (a1, . . . , ak) ∈ Rk,
fa = f + a1x1 + · · ·+ akxk
is a Morse function.
This is just the case when M is the whole space Rk.
Proof. We define g : Rk → Rk as
g =( ∂f∂x1
, . . . ,∂f
∂xk
).
Then x is a critical point of fa if and only if g(x) = −a.The derivative of g
dg =( ∂2f
∂xi∂xj
)= Hf = Hfa
happens to be the Hessian of f . By Sard’s theorem, almost all −a are regularvalues of g. Let us now think what this means. We see that −a if a regularvalue if and only if for any critical point of fa, the determinant of dg = Hfa isnonzero. This simply means that fa is a Morse function.
Proof of theorem 10.5. The manifold M is already embedded in RN . We seethat for each point x ∈M , there is a neighborhood for which
xσ(1), . . . , xσ(k)
is a coordinate system, where σ(1), . . . , σ(k) ⊂ 1, . . . , N. This means thatwe can cover M by countably many charts Ui ⊂ M ⊂ RN with the coordinatecharts being (x1, . . . , xN ) 7→ (xσ(1), . . . , xσ(k)).
Note that f on Ui is a Morse function if and only if the composite of f withthe coordinate chart is a Morse function. Thus we are reduced to the followingsituation.
U RN
Rk(x1,...,xk)
Consider the function
fa = f + a1x1 + · · ·+ akxk + ak+1xk+1 + · · ·+ aNxN .
By our lemma applied to
f + ak+1xk+1 + · · ·+ aNxN ,
we see that for almost all (a1, . . . , ak) ∈ Rk the function
f + a1x1 + · · ·+ aNxN
Math 132 Notes 27
is Morse. Then by Fubini’s theorem, we see that it is Morse for almost all(a1, . . . , aN ) ∈ RN . Now because there are countably many charts, we have totake the union of countably many measure zero sets, but this again measurezero.
Math 132 Notes 28
11 February 19, 2016
Today I want to finish out chapter 1.
11.1 Embedding and immersion theorems
Theorem 11.1 (Whitney embedding theorem). Every smooth n-manifold em-beds in R2n+1 and immerses in R2n.
Actually there is a harder version, which we won’t prove.
Theorem 11.2. Every smooth n-manifold embeds in R2n.
For instance, a figure-eight 1-manifold is immersed in R2, but moving it alittle, we get an embedding in R3. This generalizes into the following problem.
Immersion/Embedding problem. Given Mn, what is the smallest m suchthat M → Rm, or M # Rm?
Theorem 11.3 (R. Cohen). Every Mn immerses in R2n−α(n), where α(n) =e0 + · · ·+ ek for n = e1 + e121 + · · ·+ ek2k.
So for example, any 2-manifold is immersed in R3. This
Problem. What is the smallest m such that RPn # Rm?
This looks like a neat problem, but the answer is not that clean; in fact,it is still open. Every decade there is a new technique and then makes someimprovement and then people lose interest and so on. There is a hall of famefor this problem.
I will first give an outline of the proof of the Whitney embedding theorem.
Proposition 11.4. There is sufficiently large N 0 such that Mn → RN .
Actually it is easy to see that it embeds into an infinite-dimensional Eu-clidean space. If we let
S = smooth f : M → R
then the smooth functions give a embedding M → RS .
Proposition 11.5. If N > 2n+ 1 there is a vector v 6= 0 ∈ RN such that
M RN
H ∼= RN−1
πv
where πv is the projection to H = v⊥.
Math 132 Notes 29
11.2 Tangent bundles
Definition 11.6. Let M ⊂ RN be a smooth k-manifold. We define the tangentbundle TM ⊂ RN × RN as
TM = (x, v) : x ∈M,v ∈ TxM.
We can check that TM is a smooth manifold of dimension 2k. This is becauseM is locally Rk, and thus TM is locally T (Rk) = R2k. There is a canonicalprojection map TM → M that sends (x, v) 7→ x. Then given a smooth mapf : M → N , the maps dfx assembles to give a smooth map df : TM → TN .
TM TN
M N
df
f
11.3 Whitney embedding theorem
Proof of proposition 11.5. We want to find a vector v such that the projectionof M inside RN is a smooth manifold. First, we need to avoid vectors that areparallel to a line connecting two points in M ; if we pick such a v, then the twopoints will be sent to the same point. Also, we need to avoid a vector that istangent to the manifold.
So we look at the map g : M ×M × R→ RN given by
(x, y, t) 7→ t(f(x)− f(y))
and the map h : TM → RN given by
(x, v) 7→ v.
We know that N > 2n + 1, and thus Sard’s theorem tells us that there is aregular value a 6= 0 ∈ RN for both g and h. This is equivalent to saying that ais not in the image of both f and g.
We now claim that the composite map M → RN → H = a⊥ is an embed-ding. This is because a is not in the image of both g and h.
It is possible to show that there is always an immersion M # R2n withat worst double points. Once we have this immersion, we can always make adouble point at a random point. Then we can move things around so that thetwo double points cancel out. This is how the hard Whitney theorem is proved.
Math 132 Notes 30
12 February 22, 2016
Now we are moving into the material in chapter 2, which is quite important.When we look at a smooth function, the inverse image of a regular value isa submanifold. But when we look at the things below that manifold, it issomething that is not a manifold but something that we want to consider.
12.1 Manifold with boundary
Let H+n ⊂ Hn ⊂ Rn be
Hn = (x1, . . . , xn) : xn ≥ 0 and H+n = (x1, . . . , xn) : xn > 0.
Definition 12.1. A subset X ⊂ RN is a (smooth) manifold with boundaryif it is locally diffeomorphic to an open subset of Hn.
On a manifold with boundary, an interior point is a point correspondingto a point in H+
n , and a boundary point that corresponding to a point in Hn
with xn = 0. Actually it is not very clear that what I said even makes sense.But before addressing that issue, let me remark that every manifold is also amanifold with boundary.
Lemma 12.2. Suppose U, V ⊂ Hn is open and f : U → V is a diffeomorphism.Then f restrictions to a diffeomorphism of U ∩ Rn−1 → V ∩ Rn−1, where
Rn−1 = (x1, . . . , xn) ∈ Hn : xn = 0.
Proof. It suffices to show that f(U ∩ Rn−1) ⊂ V ∩ Rn−1. Suppose that x ∈U ∩ Rn−1 but f(x) ∈ H+
n . By the definition of a smooth map, there is aneighborhood W of x in Rn and a smooth g : W → Rn such that f is grestricted to Hn ∩W .
Since f is a diffeomorphism, df(x) = dg(x) is an isomorphism. So g restric-tions to a diffeomorphism of a neighborhood W ′ of x in Rn with a neighborhoodof f(x). We might as well suppose g(W ′) ⊂ H+
n . This implies that W ′ ⊂ Hn,and this is a contradiction.
This shows that everything what we did so far makes sense on manifolds onboundaries.
We introduce some terminology. If M is a manifold with boundary of di-mension n, then we denote
∂M = x ∈M : x corresponds to a point a in Rn−1 ⊂ Hn.
This is a manifold of dimension (n− 1).
Example 12.3. Suppose M is a closed manifold, i.e., ∂M = ∅. If f : M → Ris a smooth function and a ∈ R is a regular value. Then f−1([a,∞)) = N is amanifold with boundary, with ∂N = f−1(a).
Math 132 Notes 31
Proof. Since (a,∞) ⊂ R is open, f−1((a,∞)) is open in M , and so each pointof f−1((a,∞)) has a neighborhood diffeomorphic to an open subset of Rn.
Suppose f(x) = a. Since a is a regular value, we can choose local coordinatesx1, . . . , xn around x with f = xn gives a. Then there is a chart Φ : U ⊂ Rn → U ,and then
(f Φ)−1([a,∞)) = U ∩Hn.
Now let me state some facts that I am sure you can even figure out by your-self. Given a manifold Mn with boundary, we can consider the tangent spaceTxM at a boundary point x ∈ ∂M . This is because in the local coordinates,smooth maps are those that can be extended to an open neighborhood. Thisgives the point at the boundary some kind of potential to “move out the man-ifold” a little bit. Because ∂M is a manifold with dimension n − 1, we have anatural inclusion
Tx∂M → TxM.
Let f : M → N be manifolds with boundary. This induces, for each x ∈M ,a map
df(x) : TxM → Tf(x)N
and moreover it satisfies the chain rule.If M is a manifold with boundary and N is manifold with empty boundary,
then the product M×N is a manifold with boundary, and ∂(M×N) = (∂M)×N . But noted that this is not true if both M and N has nonempty boundary.
12.2 Transversality
Let’s suppose I have a manifold M with boundary, and I have a Z sitting insideX both with empty boundary.
f−1(Z) M
Z X
f
codim=k
Theorem 12.4. If f t Z and ∂f = f |∂M t Z then f−1(Z) ⊂ M is a smoothmanifold with boundary of codimension k.
Proof. Because the whole question is local, we can work locally. Locally, Z lookslike g−1(0) for some map g : X → Rk. Then f−1(Z) = (gf)−1(0).
Now M look like an open neighborhood of Hn, and so we can say that gflooks locally like a map Hn → Rk. If some point z ∈ F−1(Z) is an interiorpoint of M , then by transversality, its neighborhood will look like Rn−k. Ifz ∈ F−1(Z) is a boundary point, then one can check that its neighborhoodlooks like a half space.
Math 132 Notes 32
13 February 24, 2016
Today we are going to prove the following theorem.
Theorem 13.1. If X is a compact connected 1-manifold with boundary then Xis diffeomorphic to [0, 1] or S1.
The important consequence for us will be
Theorem 13.2. If M is a compact 1-manifold with boundary, then
#∂M ≡ 0 (mod 2).
Corollary 13.3. Suppose X is an n-manifold with boundary ∂X. There doesnot exists a smooth map f : X → ∂X such that f(x) = x for all x ∈ ∂X, i.e.,∂X is not a retract of X.
Proof. Suppose f : X → ∂X is a retraction. By Sard’s theorem, there is aregular value a ∈ ∂X. We see that f−1(a) is a 1-manifold with boundary, and
∂f−1(a) = f−1(a) ∩ ∂X = x ∈ ∂X : f(x) = a = a
since f(x) = x for all x ∈ ∂X. So #∂f−1(a) = 1. This contradicts thetheorem.
Corollary 13.4 ((Smooth) Brouwer fixed point theorem). If f : Dn → Dn isa smooth map, then there is an x ∈ Dn such that f(x) = x.
Proof. Suppose not. We draw a line joining each x and f(x), and let g(x) bethe intersection of ∂Dn and the line on the f(x) side. Then we can see that gis a retraction of Dn to ∂Dn.
13.1 Classification of 1-manifolds
A 1-manifold is something like gluing intervals together, and what can happen?You can either glue the intervals back to the starting point and get a circle, ordon’t and get a interval. But the problem we have to solve is that somethinglike “not getting to the end” does not happen. Morse functions give a way tokind of cut the manifold into a finite number of parts.
Let M be a compact connected 1-manifold, and let us choose a Morse func-tion f : M → R. Let S be the union of the set of critical points and ∂M .Let
M \ S = L1 q L2 q · · · q Lkwhere each Li are connected. We see that for each L = Li, the image f(L) ⊂ Ris connected and bounded. Thus f(L) = (a, b) for some a < b.
Proposition 13.5. The restriction f : L→ (a, b) is a diffeomorphism.
Math 132 Notes 33
Sketch of proof. Since f has no critical points, it is a local diffeomorphism. Thismeans that it is a covering space of (a, b), but I will explain more.
Choose a t ∈ (a, b) and x ∈ L with f(x) = t. Consider an interval [c, d]containing a neighborhood of t and g such that g(t) = x and f g(s) = s forany s ∈ [c, d].
L
(a, b) (c, d)
fg
We first note that if g exists on [c, d] then such a g is unique. This is becausethe set
s ∈ [c, d] : g0(x) = gx(2)
is both open and closed.Now we consider
supd : [c, d] such that g exists = d′.
If d′ < a, then we may look at the local diffeomorphism around d′ and thenextend g a bit further. So d′ = a and gluing such gs together, we get the desiredmap.
We also need do some induction, so we need to understand what happens atthe boundary of the Lis, that is, what happens at the critical points.
Lemma 13.6 (Morse lemma). If f : M1 → R is a Morse function and x ∈M isa critical pint with f(x) = a then there is a coordinate system Φ : U ⊂ R1 → Usuch that f Φ(t) = ±t2.
Sketch of proof of theorem 13.1. We know that f : Li → (ai, bi) is a diffeomor-phism. Using this, we get that each Li is diffeomorphic to [0, 1], and thus Li \Lihas two points. Now we need a way to order the Li so that we can glue thedifferent segments.
Definition 13.7. A chain is a sequence
Li1 , Li2 , Li3 , . . .
such that ∂Lij ∩ ∂Lij−1is nonempty and ij 6= ij−1.
Proposition 13.8. Any maximal chain contains all the Li.
Proof. We show that a maximal chain is both open and closed in M .
So suppose L1, . . . , Lk is a maximal chain. We can rearrange the Morsefunction so that f on Li is replaced by (−1)i−1f + ci so that they agree onthe common boundary points. Let this function be F . This F is continuous,monotone, and smooth except at the pi. We can now apply this lemma.
Math 132 Notes 34
Lemma 13.9 (Smoothing lemma). We can change F a little to be smooth atthe pi and monotone.
Now if p0 6= pk, then F is diffeomorphic with a closed interval. When p0 = pkthen e2πicF is diffeomorphic with S for some c.
Math 132 Notes 35
14 February 26, 2016
Suppose I have a manifoldM possibly with boundary, and it maps to f : M → Xwhere X is closed. Also let Z ⊂ X be a closed submanifold. Our goal is to showthat f can always be moved a little bit so that f and ∂f are transverse to Z.
14.1 Transversality in a general family
We introduce the notion of a family. If there is a map S ×M → X, then wecan consider it as a set of maps fs collected for s ∈ S.
S ×M s ×M
X
Ffs
Definition 14.1. A “general” family F : S →M → X is a map which is either1) a submersion or 2) transverse to Z with ∂F also transverse to Z.
Theorem 14.2. If F : S ×M → X has the properties F t Z and ∂F t Z,then for almost all s ∈ S, both fs and ∂fx is transverse to Z.
Proof. Let us look at the pullback:
S ×M X
W Z
Then we can look at the projection map
W S ×M
S
g
We claim that if s is a regular value of g then fs t Z.Suppose that s is a regular value of g, and let x ∈ M . We must show that
if z ∈ Z with fs(x) = F (s, x) = z then the direct sum of the two maps
TxM TzX
TzZ
dfx
is onto, or equivalently that
TxM → TzX/TzZ := V
Math 132 Notes 36
is onto.Our assumption that s is a regular value gives that
TsX ⊕ TxM = T(s,x)S ×M VdF
is onto. Since T(s,x)W is just the kernel, we see that if s is regular value, then
T(s,x)W TsS ⊕ TxM
TsS
After this, some linear algebra will show that fs is transverse to M .4
Theorem 14.3 (Existence of general families). If f : M → X is a smoothmap, then there is a submersion F : Rn ×M → X with F (0, x) = f(x). (Forconvenience, we assume that both M and X are compact.)
Proof. We may assume that X ⊂ Rn. By the “tubular neighborhood theorem,”which we will prove next time, X is a retract of open set U ⊂ Rn containing X,i.e., a map r : U → X such that r(x) = x for any x ∈ X.
Now let us construct a map G : M × Rn → Rn given by
(x, v) 7→ f(x) + v.
We look at the inverse image G−1(U) and because this contains M , there is anε > 0 such that G−1(U) contains M ×Bε. Then we can construct a new map
M ×Bε U
X
G
Fr
This F is clearly a submersion, because both G and r are submersions.
4The full proof was given in the lecture, but I did not understand it then. In any case,some uninteresting linear algebra does finish the proof.
Math 132 Notes 37
15 February 29, 2016
15.1 Partitions of unity
Definition 15.1. Let X be a topological space, and let f : X → R be acontinuous function. We define the support of f as
supp f = closure of x : f(x) 6= 0
andsupp+ f = x : f(x) > 0.
Let U = Uα be an open covering of X.
Definition 15.2. A partition of unity subordinate to U is a sequence θ1, θ2, . . . , of functions θi : X → [0, 1] ⊂ R such that
1) Each x ∈ X has a neighborhood on which only finitely many θi are non-zero.
2) For each i there exists an α such that supp θi ⊂ Uα.
3) For all x ∈ X, the sum of all θi is∑θi(x) = 1.
We note that we can replace the third condition by
3′) For all x there exists an i such that θi(x) 6= 0.
If we have this, then ni = θi/(∑θi) form a partition of unity.
Theorem 15.3. If X ⊂ M is a subspace of a manifold, then partitions ofunity exist for any open cover of X. Moreover the θi can be constructed as therestriction of smooth functions on M .
There is a half-page proof in the book, but I want to break it into digestibleterms. We are not going to assume that M is embedded in the Euclidean space.
By definition of a manifold, M is Hausdorff and has a countable basis forthe topology. So is, if Vα is an open cover, then some countable number ofthe Vα actually cover M .
Lemma 15.4. Every x ∈M has a neighborhood with compact closure.
Proof. Take the open ball in any coordinate neighborhood.
Lemma 15.5. Every compact K ⊂ M has neighborhood K ⊂ V with compactclosure.
Proof. For each x ∈ K choose a neighborhood Bx with Bx compact. Thena finite number of B1, . . . , Bl cover K. Take V = B1 ∪ · · · ∪ Bl. Then V =B1 ∪ · · · ∪Bl is compact.
Lemma 15.6. M is a countable union of compact subspaces.
Math 132 Notes 38
Proof. For each x choose a neighborhood Bx with compact closure. Now Bxcovers M and so some countable subset B1, B2, . . . covers M . Then we haveM =
⋃iBi.
Lemma 15.7. There is a sequence
K1 ⊂ V1 ⊂ K2 ⊂ V2 ⊂ K3 ⊂ V3 ⊂ · · ·
where 1)⋃iKi = M , 2) each Ki is compact and Vi is open, 3) Vi ⊂ Ki+1.
Proof. Pick a family of compact sets B1, B2, . . . such that M =⋃iBi. We
start with K1 = B1 and find an K1 ⊂ V1 with compact closure. Then takeK2 = V1 ∪B2 and so on.
Definition 15.8. A subspace X ⊂ M is a disconnected union of compactsets if
X =
∞⋃i=1
Ki
where each Ki is compact, and Ki ∩ Kj = ∅ there are open set Ki ⊂ Vi suchthat Vi ∩ Vj = ∅.
Theorem 15.9. M can be written as M = I1 ∪ I2 ∪ I3 where each Ii is adisconnected union of compact sets.
Proof. If we letLn = Kn − Vn−2, Wn = Vn −Kn−2,
we see that Ln ⊂Wn and M = ∪Ln. Also, Wn ∩Wn−3 = ∅. Thus we can let
Ii =⋃n
L3n+i
and get a cover of M by three disconnected unions of compact sets.
Proof of theorem 15.3. Let us first look at the special case when X ⊂ M iscompact. We can choose for each x ∈ X a smooth θx : X → [0, 1] ⊂ R suchthat there exists an α with supp(θx) ⊂ Uα. This can be done by using thebump function. Then the open sets supp−1 θx cover X, and since X is compact,finitely many θ1, . . . , θk do. This shows that θ1, . . . , θk satisfies 1), 2), and 3′).
Now suppose that X is a disconnected union of compact sets Ki, whereKi ⊂ Vi and Vi ∩ Vj = ∅. For each i, there is a partition of union on Ki
subordinate to Uα ∩ Vi. Let this be θi1, . . . , θij. Then the set θij is apartition of unity.
Next we look at the case X = M . Since M is the union of three disconnectedunion of compact sets, we can take the union of all θs for the three unions. ThusX = M has a partition of unity.
Finally, let us look at the general case X ⊂ M . Since Uα give a open coverof X, we see that there is an open Uα ⊂ M such that Uα ∩X = Uα. If we letM ′ =
⋃Uα, it must be a manifold. Then we can look at the partition of unity
on M ′ and its restriction to X will give a partition of unity on X.
Math 132 Notes 39
16 March 2, 2016
Theorem 16.1 (Tubular neighborhood theorem). X is a retract of a neighbor-hood U of X.
In the last two lectures, we have used this without proving it. Now let usprove this.
16.1 Normal bundle
Let X ⊂ Rm be a smooth manifold if dimension d. The tangent space TxXhas dimension d, and does not depend on the embedding. Let the normalspace NxX be the orthogonal complement of TxX. Now this depends on theembedding, and will have dimension m− d.
We have defined the tangent bundle as
TX = (x, v) ∈ Rm × Rm : v ∈ TxX
and showed that TX is a manifold of dimension 2d. Likewise, we define thenormal bundle as
NX = (x, v) ∈ X × Rm : v ∈ NxX.
The is actually bad notation, because NX depends on the information of theembedding, which is not present in the notation.
Proposition 16.2. NX is a manifold if dimension d+ (m− d) = m. In fact,NX → X is a smooth fiber bundle.
Let us assume the proposition for a moment.
16.2 Tubular neighborhood theorem
Proof of the Tubular neighborhood theorem. We construct a map g : N → Rmas g(x, v) 7→ x + v. The derivative of this map at (x, 0) looks like dg(x,0) :T(x,0)N → Rm, and we see that T(x,0)N = TxX⊕NxX ∼= Rm. Therefore dg(x,0)
is an isomorphism.By the inverse function theorem, g is an diffeomorphism in a neighborhood
of (x, 0) in N . Since X is compact, there is an ε such that g|Bε(N) : Bε(N)→ Rmis a diffeomorphism with its image, where
Bε(N) = (x, v) ∈ N : |v| < ε.
This is a local diffeomorphism, by definition. Why is it one-to-one? This, wecan choose ε small enough so that the images of g don’t overlap, because X iscompact.
Then Bε(N) is diffeomorphic to an open set of Rm containing X, and wecan set Bε(N)→ X to be our retraction.
Math 132 Notes 40
More generally, let Xd ⊂ Y l ⊂ Rm be manifolds. We can define
NYX = (x, v) : x ∈ X, v ∈ TxY, x ⊥ TxX.
A similar argument gives an embedding BεNYX → Y as an open set. Actually
there is a small technical problem. When we map Bε(NYX ) → Rm, the image
might not lie in Y . So what we do is consider the tubular neighborhood U ofY , and consider a small enough ε so that Bε(N
YX ) lives in the U .
Y
Bε(NYX ) U ⊂ Rm
π
Then we can project it down to Y .Now let me justify why the normal bundle is a smooth manifold.
Proof. Locally, we can find a function F : Rm → Rm−d so that X = F−1(0),that is 0 is a regular value. Then Tx is the kernel of the map dFx : Rm → Rm−d.Since Nx ∩ Tx = 0, we see that the composite
Nx Rm Rm−ddF
is an isomorphism.Now let us show that N → X is a smooth fiber bundle around each x ∈ X.
If F is defined on a neighborhood W of x, then since dF : Nx → Rm−d is anisomorphism, the map N(W ) → W × Rm−d is a local diffeomorphism. Thususing the inverse function theorem, we can choose a neighborhood so that N(V )and V × Rm−d are diffeomorphic.
Math 132 Notes 41
17 March 4, 2016
So far we have done transversality, partitions of unity and all this stuff, andtoday we are going to start applying these machinery.
Let f : M → X, and let Z ⊂ X.
Question 1. Can f be homotopic to a map that misses Z?
Also, let Zd−1 ⊂ Xd.
Question 2. Can one find a smooth f : X → R for which 0 is a regular valueand f−1(0) = Z?
These two seemingly unrelated questions are actually related very much. Wecan answer the second question completely, and the first question is not quitethe right question to ask.
17.1 Mod 2 intersection numbers
For the first question, we can make f transverse to Z and then study W =f−1(Z) ⊂M , which is a smooth manifold.
W Mk
Zl X
f
If dimZ + dimM = dimX, then dimW = 0. In this case, we can simplycount the number of points.
Definition 17.1. We define the intersection number I(f, Z) as
#f−1(Z) mod 2
where f is transverse to Z and is homotopic to f .
Lemma 17.2. The intersection number I(f, Z) is independent of f .
Proof. Suppose we have two f and f ′. Then we can concatenate two homotopiesbetween f and f , and f and f . That gives us a homotopy
H : M × [−1, 1]→ X
where H(•,−1) = f and H(•, 1) = f ′. Actually this might not be smoothat t = 0, so we have to modify this a little bit, using a bump function, sothat it slows down and almost become constant at t = 0. Then H becomes asmooth homotopy. Next, we can move H a little bit without changing H in aneighborhood of M × −1, 1 so that H is transverse to Z.
Y M × [−1, 1]
Z X
H
Math 132 Notes 42
Now Y is a 1-manifold with
∂Y = ∂−1Y q ∂1Y = M × −1, 1 ∩ Y,
where ∂−1Y = f ′−1(Z) and ∂1Y = f−1(Z). By the classification of 1-manifoldswe have proved, we have #∂Y = 0 (mod 2) and thus #∂−1Y = #∂1Y (mod 2).
Now this gives you a well-defined intersection number modulo 2. This is aninvariant, so if you want some thing to not happen then the easiest thing todo is check that the invariants are different. If you want something to happen,then it requires something much more sophisticated constructive methods.
There is an obvious variation.
Theorem 17.3. If Z = ∂L for some submanifold L of X, then I(f, Z) = 0.
W M
Z = ∂L X
f
Let us recall the question 2 I started with: when is Z ⊂ X of codimension 1some f−1(0)? If Z = f−1(0), then clearly Z is the boundary of f−1(x : x ≥ 0).Then for any g : S1 → X, the intersection number is I(g, Z) = 0. This gives anecessary condition. But in fact, this is also a sufficient condition.
Theorem 17.4. Let Zn−1 ⊂ Xn be closed manifolds. Then Z = f−1(0) forsome f : X → R where 0 is a regular value of f , if and only if I(g, Z) = 0 forany g : S1 → X.
In the problem set, we showed that the close Mobius band is not diffeo-morphic to the closed cylinder. This was done by looking at the connectedcomponents of the boundary. But this does not work if the manifolds are open.First consider the open Mobius band.
Z
For the circle Z on the Mobius band, we see that I(Z,Z) = 1, because whenwe move one end up, the other goes down. On the other hand, we see thatany loop has intersection number 0 with itself, because we can move them upand down so that they don’t meet. This proves that the Mobius band and thecylinder are not isomorphic.
Math 132 Notes 43
18 March 7, 2016
Today I want to introduce something that is not usually taught this level ofcourse.
18.1 Cobordism groups
Definition 18.1. Suppose M1 and M2 are closed n-manifolds. A cobordism ofM1 with M2 is an (n+1)-manifold N with boundary such that ∂N = M1qM2.If there exists a cobordism between M1 and M2 we say that they are cobordant.
Figure 1: Cobordism between M and N .
Theorem 18.2. “Cobordant” is an equivalence relation.
Proof. It is obvious that it is symmetric and reflexive. For transitivity, suppose∂N12 = M1 qM2 and ∂N23 = M2 qM3. Then we can glue these two manifoldstogether along M2 and get a new manifold
N12 ∪M2
N23 = (N12 qN23)/(x ∈M2 ⊂ ∂N12 ∼ x ∈M2 ⊂ N23)
with boundaryM1qM3. Actually this requires a bit of machinery in the problemset.
Theorem 18.3 (Collar neighborhood theorem). If M = ∂N then there existsa neighborhood U of M and a diffeomorphism N ∼= M × [0,∞).
Once we have this we can glue N12 and N13 along the collar neighborhood, andwe no longer have any problems at the boundary.
Definition 18.4. The n-th cobordism group MOn is the quotient
closed n-manifolds/cobordant.5
For addition, we make MOn into a commutative monoid by letting [M1]+[M2] =[M1 qM2].
Since [M ] + [M ] = 0, we immediately see that MOn is a Z/2-vector space.
5Actually there is a set-theoretic issue here but don’t worry about it.
Math 132 Notes 44
Example 18.5. The 0-th cobordism group MO0 = Z/2 because the cobordismclasses are determined by the number of points. The next one is MO1 = 0because the circle is a boundary of the disk. Then MO2 = Z/2 generated byRP 2, and this is not obvious.
The amazing thing is that we exactly know what MOn is.
Theorem 18.6 (Thom).⊕n≥0
MOn = Z/2[xj | j ≥ 1, n 6= 2j − 1].
18.2 Bordism homology
Suppose X is a topological space. A manifold over X is a manifold M togetherwith a continuous map f : M → X. We let ∂f = f |∂M : ∂M → X.
Definition 18.7. Suppose f1 : M1 → X and f2 : M2 → X are two closedn-manifolds over X. A cobordism between them is an (n + 1)-manifold witha map h : N → X and an isomorphism ∂h = f1 q f2.
Proposition 18.8. Cobordism of maps is an equivalence relation.
Definition 18.9. The n-th bordism homology group MOn(X) of X is the setof all closed n-manifolds over X modulo cobordism.
Again, MOn(X) is always a Z/2-vector space.
Example 18.10. We have MO1(R2) = 0. This is because any maps f : S1 →R2 extends to a map g : D2 → R2 given by
g(r, θ) = rf(θ).
There is actually a technical problem because this g might not be smoothat the origin. The following proposition deals with this issue.
Proposition 18.11. If f1 : M1 → X and f2 : m2 → X are smooth andh : N → X is a cobordism between them, then there exists a smooth cobordismh : N → X between them.
Example 18.12. We have MO1(S2) = 0. This is because for any f : M1 →S2, by Sard’s theorem there is a regular value x ∈ S2. Then we can use thestereographic projection S2 \ x → R2 the make a map
f : M1 S2 \ x
g : N2 R2
that shows that f cobordant to the constant map.
Math 132 Notes 45
Example 18.13. We have MO1(T 2) 6= 0. To prove this, we will produce a non-zero homomorphism MO1(T 2)→ Z/2. Consider a fixed loop Z around a torus,and for any f : M1 → T 2, let us look at the intersection number I2(f, Z) ∈ Z/2.From the last lecture, we see that I(f, Z) depends only on the cobordism classesof f . Thus I(f, Z) defines a map
MO1(T 2)→ Z/2.
This is not the zero map, and hence MO1(T 2) 6= 0. In fact, we can take twoloops and construct a map
MO1(T 2)→ Z/2⊕ Z/2.
We shall prove later that this is an isomorphism.
Math 132 Notes 46
19 March 9, 2016
Last time we talked about this bordism homology. There is a really useful toolthat enables us to do a lot of interesting stuff.
19.1 Surgery
Suppose we have two 1-manifolds. Pick two points on the manifold and cutout holes (or intervals in this case) and connect them by smooth curves. Orsuppose we have two points on a surface. We can cut out neighborhoods of thepoints and attach a cylinder connecting the two holes. Then we kind of attacha handle to it.
Definition 19.1. Let Ak ⊂Mn. A framing of A in M is an open A ⊂ U ⊂Mwith a diffeomorphism
U A× Rn−k
A A× 0
≈
We remark that a framing on A exists if and only if there is a smoothf : M → Rn−k with 0 as a regular value and A = f−1(0).
Definition 19.2. A bounding manifold of A is a manifold B with A = ∂B.
Let U ′ be the open subset of U so that
U A× Rn−k
U ′ A×Bn−k
Φ
⊃ ⊃
where Bn−k is the open unit ball. Now the boundary of M \ U ′ is
∂(M \ U ′) = ∂M q (A× Sn−k−1).
SinceA× Sn−k−1 = ∂(B × Sn−k−1),
we can now glue M \U ′ with B×Sn−k−1 along A×Sn−k−1. There is always aproblem when gluing; we need to somehow smooth the boundaries so that theyare smooth manifolds, but this can be done. It is too technical to go over it inlecture.
We note that a surgery needs
• a submanifold A ⊂M ,
• a framing U ≈ A× Rn−k of A,
• and a bounding manifold A = ∂B,
Math 132 Notes 47
and collar neighborhoods. It is possible to check that a surgery is isomorphicup to diffeomorphism. Usually we will set A = Sn−k and B = Dn−k+1.
Example 19.3. Suppose that M1 and M2 are n-manifolds. Choose pointsx ∈ M1 and y ∈ M2 with coordinate neighborhoods Φx : Ux → Rn and Φy :Uy → Rn. For A = S0 = −1, 1 →M1qM2 that sends −1 7→ x and 1 7→ y, wecan perform a surgery. Then we new manifold that is like attaching a cylinderbetween M1 and M2 at x and y. This is called the connected sum of M1 andM2, and is denoted by M1#M2.
Now what does surgery have to do with cobordism? When you perform asurgery, you are not changing the cobordism.
19.2 The surgery cobordism
Suppose M is closed and compact, and also suppose that B is compact. Let
N = M × [0, 1]qB ×Dn−k
where Dn−k = x ∈ Rn−k : |x| ≤ 1 is the closed ball. Then we have
∂N = M × 0 q M
where M is the result of the surgery.This looks hard, but it is actually just something like this. Suppose you
have a 2-sphere and want to attach a handle. Then you first thicken the ball,and then attach a solid handle to the outer part. Then you get a 3-manifoldwith boundary the sphere and the handled sphere. This shows that the sphereis cobordant with the handled sphere, which is the torus.
We can also perform surgeries on maps. Suppose that we have a map f :M → X, and A = ∂B ⊂ M and a framing Φ : U → A × Rn−k. Suppose thatwe have an extension
g : B ×Dn−k → X
extending A×Dn−k → X. Then we can map N → X by first mapping
M × [0, 1] M Xf
and then mapping
B ×Dn−k X.g
We see that these agree on A×Dn−k and thus gives a well-defined map. Actuallywe can canonically give the map g, but leaving it this way, we can make ourown choices.
Math 132 Notes 48
20 March 11, 2016
20.1 MO1 of a torus
Let Z ⊂ Σ be a connected and closed 1-manifold lying inside a 2-manifold.
Proposition 20.1. Every f : M → Σ is cobordant to a map f which is trans-verse to Z and with 0 ≤ #f−1(Z) < 2.
Proof. Moving f by a homotopy if necessary, we may assume f t Z. Letz1, . . . , zk ∈ Z be f(M) ∩ Z. We choose z1 and z2 to be neighboring pointson Z, and close an embedding [0, 1] → Z so that g(0) = z1, g(1) = z2, andg(t) /∈ Z ∩ f(M) for 0 < t < 1.
Choose m1,m2 ∈M such that f(m1) = z1, f(m2) = z2. Now we do surgerywith A = m1,m2 and B = [0, 1]. Then we get a map f that does not meetM at z1 and z2. Then the number of intersections decrease by 2.
There is a variation of this fact.
Proposition 20.2. Let Z1, . . . , Zn ⊂ Σ be connected closed 1-manifolds trans-verse to each other. Then any f : M → Z is cobordant to a map transverse toall Zi with #f−1(Zi) < 2.
To prove this, you can simply make f transverse to all Zi and Zi∩Zj . Thenf−1(Zi ∩ Zj) = ∅, and do the argument on each Zi.
Example 20.3. Let Σ be the torus and consider two loops α and β that meetseach other once. We can look at the homomorphism MO1(Σ)→ (Z/2)× (Z/2)given by
f 7→ (I(f, α), I(f, β)).
Since α 7→ (0, 1) and β 7→ (1, 0), we see that this map is onto.
Proposition 20.4. The kernel of this map is zero.
Proof. Suppose f : M → Σ satisfies I(f, α) = I(f, β) = 0. Then f is cobordantto a map f : M → Σ\ (α∪β) ∼= R2. Because we already know that MO1(R2) =0, we see that f is zero.
So we conclude that MO1(Σ) ∼= Z/2×Z/2 and moreover α and β form a basis.
Math 132 Notes 49
21 March 21, 2016
Let Σ be a compact closed 2-manifold. We have computed the first bordismhomology MO1(Σ) for Σ = S2 and Σ = T 2. We remark that MO1(Σ) comesequipped with a symmetric bilinear form
MO1(Σ)×MO1(Σ)→ Z/2
that maps (f1, f2) 7→ I(f1, f2).
S M2
M1 Σ
f2
f1
The pair (MO1(Σ), I) is called the (mod 2) intersection form of Σ.
21.1 Bilinear forms on MO1
Suppose V is a vector space over a field k.
Definition 21.1. A bilinear form on V is a map B : V × V → k satisfy-ing B(v, λ1w1 + λ2w2) = λ1B(v, w1) + λ2B(v, w2) and B(λ1v1 + λ2v2, w) =λ1B(v1, w) + λ2B(v2, w). A bilinear form is symmetric if B(x, y) = B(y, x).A symmetric bilinear form is non-degenerate if for every v ∈ V there is a wsuch that B(v, w) 6= 0.
Suppose V is finite dimensional, with basis e1, . . . , en. Then a vectorcorresponds to a column, and a bilinear form B corresponds to a matrix B,given by B(x, y) = xT By.
Example 21.2. Let Σ = T 2. Then the bilinear form on MO1(Σ) is
B =
(0 11 0
)with a certain basis. Also, if Σ = T 2#T 2 then the bilinear form is
B =
0 1 0 01 0 0 00 0 0 10 0 1 0
.
Note that B is non-degenerate in both cases.
Theorem 21.3. If Σ is a compact closed 2-manifold then MO1(Σ) is finitedimensional.
Theorem 21.4. The form MO1(Σ) is non-degenerate.
Math 132 Notes 50
We will prove this later and use it only.
Definition 21.5. If (V1, B1) and (V2, B2) are symmetric bilinear forms then(V1, B1)⊕ (V2, B2) is defined as
B((v1, w1), (v2, w2)) = B1(v1, v2) +B2(w1, w2).
If B1 and B2 are non-degenerate, so is B1 ⊕B2.
Example 21.6. We let
R =
V = Z/2B = (1)
and H =
V = Z/2⊕ Z/2
B =
(0 1
1 0
).
Lemma 21.7. Suppose (V,B) is a non-degenerate symmetric bilinear form andW ⊂ V . Defined BW : W ×W → k as BW (x, y) = B(x, y). If BW is non-degenerate, then (V,B) = (W,BW )⊕ (W⊥, BW⊥), where W⊥ is the orthogonalcomplement with respect to the bilinear form.
21.2 Classification of finite-dimensional non-degenerate sym-metric bilinear forms over k = Z/2
First suppose there is a v ∈ V with B(v, v) = 1. Then W = (v) with BW is justR. Then
(V,B) = R⊕ (V ′, B′).
Repeating this, we find
(V,B) = R⊕a ⊕ (V ′, B′).
Here, (V ′, B′) is even, i.e., B(x, x) = 0 for any x.Now choose 0 6= x ∈ V . Since B is non-degenerate, there is y with B(x, y) =
1. Also, B(y, y) = 0. Then if we set W = (x, y) then (W,BW ) = H. This showsthat (V,B) = H ⊕ (V ′, B′). Hence
(V,B) = R⊕a ⊕H⊕b.
But this is not a full classification, because some of they might be the same.In fact, R+R+R = R+H. Summing up, we get the following proposition.
Proposition 21.8. Every non-degenerate symmetric bilinear form over Z/2 isisometric to one of
1) H⊕a,
2) R⊕H⊕a,
3) R⊕R⊕H⊕a.
In fact, they are all different because 1) is even and the dimension is even, 2)is odd and the dimension is also odd, and 3) is odd and the dimension is even.
Math 132 Notes 51
22 March 23, 2016
Last class I talked about symmetric bilinear forms, and now we are going toprove the non-degeneracy assertion and how surgery directly corresponds tooperations on symmetric bilinear forms.
22.1 Embeddings represent every class
Theorem 22.1. Let X be a smooth 2-manifold. Every class in MO1(X) isrepresented by an embedding f : M → X.
We will assume the following theorem of Whitney.
Theorem 22.2 (Whitney). Every map Mk → X2k is homotopic to a regularimmersion with at most double points. (Here a regular immersion is a map fsuch that for any x 6= y ∈ M with f(x) = f(y), the map df : TxM + TyM →Tf(x)X is onto.)
Proof of theorem 22.1. Let us look at the special case M = Rq R, and look atthe map M → X = R2 that maps one line to the x-axis an the other line to they-axis.
Now let us do surgery so that this map becomes an embedding. Then thesurgery cobordism will guarantee that they are in the same cobordism class.We set A = 01, 02 ⊂ M and consider the interval B = [0, 1] so that A→ ∂Bis given by 01 → 0 and 02 → 1. We already have the framing, and we need athickening B× [−1, 1]→ X that extends A→ X. We can do this by connectingthe relevant points by segment. Then we can do a surgery and make it into anembedding.
Now let us go back to the general case. Using Whitney’s theorem, we canassume that f is a regular immersion with at most double points. Then locally(in X) f : M → X either looks like an embedding or there are two points pand q mapping to x such that TpM +TqM = TxX. By the standard immersiontheorem, they both look like R1 → R2 locally (in M) and therefore we can usesurgery to remove the double point.
In fact, we can connect things by surgery, and even show that every bordismclass is represented by a embedding of a circle.
22.2 Separation theorems
Definition 22.3. A submanifold Z ⊂ X of codimension k is cut out by ϕ :X → Rk if 0 is a regular value of f and f−1(0) = Z.
Question. How can we tell when Z ⊂ X of codimension 2 is cut out by afunction X → R?
First note that Z is cut out by f : X → R and ∂X = ∅ then Z = ∂f−1[0,∞).Then I(Z, f) = 0 for all f : M1 → X. But the converse is also true.
Math 132 Notes 52
Theorem 22.4. Z is out out by a function ϕ if and only if I(Z, f) = 0 for allf ∈MO1(X).
Proof. Suppose X is connected, and choose a point p ∈ X−Z. Given x ∈ X−Z,choose a path γ : [−1, 1] → X such that γ(−1) = p and γ(1) = x and istransverse to Z. Then set S(x) = I(γ, Z). Then this actually does not dependof γ, because given any two paths γ and γ, we can glue them to get a mapf : S1 → X and
0 = I(f, Z) = I(γ, Z) + I(γ, Z).
This defines a function S : X − Z → 0, 1.This S is locally constant, because given any x ∈ X − Z we can let V be a
connected open neighborhood of x and for every x′ ∈ V the concatenation of γand the path joining x and x′ gives a path with the same intersection number.This shows that S is constant on V .
Math 132 Notes 53
23 March 25, 2016
Last time I proved the theorem that if X is a smooth 2-manifold, every α ∈MO1(X) is represented by an embedded 1-manifold Z ⊂ X. In fact you canshow in addition that if X is connected then Z can be taken to be diffeomorphicto S1.
Suppose Zk−1 ⊂ Xk be and embedded manifold.
Definition 23.1. Z is cut out by a function f : X → R if 0 is a regular valueof f and f−1(0) = Z.
Theorem 23.2. The manifold Z is cut out by a function f if and only ifIZ(α) = 0 for every α ∈MO1(X).
Proof. Choose a point p ∈ X − Z and define a map S : X − Z → 0, 1 bychoosing a γ : [−1, 1] → X with γ(−1) = p and γ(1) = x and γ t Z. ThenS(x) = #γ−1(Z) is well-defined since I(α,Z) = 0 for any α ∈ MO1X. Also,this function is locally constant.
Find a countable number of diffeomorphisms Φi : Ui → R2 covering Z suchthat Z ∩ Ui corresponds to the x-axis. Moreover, we can assume that S(x) = 0corresponds to the upper half plane y > 0 in each Φi.
For x ∈ Z, define T+x ⊂ TxX to be the set of vectors v ∈ Tx such that
dΦi(v) ∈ TΦi(x)R2 = R2 is in the upper half space. Note that this is independentof the choice of Ui. We can then choose a partition of unity θi subordinateto Ui and set f =
∑θiϕi, where
ϕi : Ui R2 R.Φi y−coord
This f works on a neighborhood of Z, because f(x) = 0 for x ∈ Z, f(x) > 0for S(x) = 0 and f(x) < 0 for S(x) = 1. Also, df(v) > 0 for v ∈ T+
x , so 0 is aregular value. To make in be defined on X, we throw in the open set X − Z inthe cover and do the same thing.
Now this implies that if Z is diffeomorphic to S1, then f can be regarded asa connected sum of two manifolds. I will come back to this.
23.1 Non-degeneracy of the intersection form
Theorem 23.3. If X is a compact closed 2-manifold, the intersection form
(MO1X, I(−,−))
is non-degenerate.
Proof. We must show that if I(α, β) = 0 for every β, then α = 0. By “specialposition” we may assume α is represented by an embedded Z ⊂ X. Then bythe separation theorem, Z ⊂ X is cut out by a smooth f : X → R. The inverseimage f−1([0,∞)) is a manifold with boundary over X with boundary equal toZ. Therefore Z = 0 inside MO1(X).
Math 132 Notes 54
Theorem 23.4 (Neighborhood theorem). Suppose g : S1 → X2 is an embed-ding such that g(S1) = Z ⊂ intX. If I(g, g) = 0 then Z has a neighborhood Udiffeomorphic to a cylinder Z ⊂ U . If I(g, g) = 1, then Z has a neighborhood Udiffeomorphic to the Mobius band.
Proof. Let N be the normal bundle of Z inside X. By the ε-neighborhoodtheorem, there is a neighborhood U of Z diffeomorphic to N . We must showthat either N is diffeomorphic to the cylinder or the Mobius band.
Now letC = (x, v) ∈ NZ : |v| = 1
and look at the diagram.
0 C
R S1 Z
(p,v)
eit
γ
g
Because C is a covering space of Z, by the path lifting property, we have aunique such γ. Then γ(1) = (p,±v). In each cases, we see that the normalbundle is the cylinder and the Mobius band.
Example 23.5. Suppose α ∈ MO1(X) with I(α, α) = 1. Represent α be anembedded circle, take the Mobius band neighborhood Z of α. If we take thethe boundary W of Z, then this is diffeomorphic to a circle, and by surgery, wecan make it into a connected sum of an RP2 and something. We will exploitthis property more.
Math 132 Notes 55
24 March 28, 2016
We have the following facts so far.
1. If X a connected surface, every α ∈MO1(X) is represented by an embed-ded circle.
2. An embedded circle Z ⊂ X can be cut out by a smooth f : X → R if andonly if I(Z,α) = 0 for all α ∈MO1(X).
3. An embedded circle Z ⊂ X has a neighborhood U , diffeomorphic to S1×Rif I(Z,Z) = 0, and diffeomorphic to the Mobius band if I(Z,Z) = 1.
24.1 Classification of compact connected 2-manifolds
Let us recall the classification of non-degenerate symmetric bilinear forms (B, V )over Z/2. If there is a v ∈ V with B(v, v) = 1, then we can decompose V intoR⊕ v⊥, and if there isn’t, we can choose v, w such that B(v, w) and decomposeV = H ⊕ (v, w)⊥.
We do the analogous thing for manifolds. We ask the question “Is there av ∈MO1(X) with I(v, v) = 1?” If yes, then represent v by an embedded circleZ ⊂ X. Then there is a neighborhood U of Z such that U = R × R/(x, y) ∼(x+ 1,−y). Low let Z ⊂M ⊂ X be the set defined as
M = R× [−1, 1]/(x, y) ∼ (x+ 1,−y).
Then W = ∂X has intersection number zero with any other α, and hence thereis a function f that cut out W . One side of W is the Mobius band, and theother side is some X ′.
This is actually the connected sum. Let M1 and M2 be two manifolds. Wecan choose two points each on M1 and M2 and take the neighborhood disks ofthe points and do surgery. The resulting manifold do not depend on the choiceof points, because there is a diffeomorphisms of a manifold that sends any pointto any point. It might depend on the choice of the framing; essentially theorientation might be different. In fact, when you do surgery on two points onS2, the resulting manifold might become a torus of a Klein bottle dependingon the orientation. But if M1 and M2 are not connected, then orientation doesnot affect the surgery. Therefore, the connected sum M1#M2 is well-definedas the result of the surgery.
Given a M = M1#M2, we can even disconnect them using surgery. Choosea circle on the cylinder we have attached, choose a neighborhood that looks likethe cylinder, remove a cylinder inside the cylinder, and attach disks. Then weget M1 and M2 back.
Now given a function f that cut outW , we can do surgery onW . The Mobiusband gets attached a disk, and it becomes RP 2. That is, given a v ∈MO1(X)with I(v, v) = 1, we can write X = RP 2#X ′. Moreover, it is possible to prove
MO1(X1#X2) = MO1(X1)⊕MO1(X2).
Math 132 Notes 56
Then we getMO1(X) = MO1(RP 2)⊕MO1(X ′).
Continuing this process, we get
X = RP 2# · · ·#RP 2#X ′
such that I(α, α) = 0 for any α ∈MO1(X ′). Choose x, y ∈MO1(X ′) such thatI(x, y) = 1, and we can represent x, y by embedded circles Z1, Z2 which meetonly once. Because I(Z1, Z1) = I(Z2, Z2) = 0, both Z1 and Z2 thickens up totwo bands intersecting at a single part. Then the union of the two bands is apunctured torus, and thus using the same argument, we can represent X ′ as aconnected sum of the torus T and some other manifold. That is, continuing theprocess, we can write
X = RP 2# · · ·#RP 2#T# · · ·#T#X ′,
where MO1(X ′) = 0.
Theorem 24.1. If X is a compact closed 2-manifold with MO1(X) = 0 thenX = S2.
This theorem, whose proof uses Morse theory, shows that X is the connectedsum of RP 2s and T s. In fact, we have proven something stronger.
Theorem 24.2. Suppose X is a compact connected closed 2-manifold and wehave chosen an isometry
R⊕ · · · ⊕R⊕H ⊕ · · · ⊕H → (MO1(X), I).
Then X = RP 2# · · ·#RP 2#T# · · ·#T wit he the standard basis going to thestand basis.
Theorem 24.3. If X1 and X2 are compact connected closed 2-manifolds and
T : MO1(X1)→MO1(X2)
is an isometry then there is a diffeomorphism X1 → X2 inducing T by MO1(−).
Math 132 Notes 57
25 March 30, 2016
For the next few lectures, we are going to do winding numbers and the Jordan-Brouwer separation theorem. Classically, it states
Theorem 25.1 (Jordan curve theorem). Every simple closed curve in R2 di-vides the plane into two pieces.
25.1 Winding numbers
Let X be a contact, connected (n−1)-manifold and let f : X → Rn be a smoothmap. The winding number “studies how f wraps X round Rn.”
z
X
Pick a point z not on X and consider the function u(x) = (x − z)/|x − z|.Then this is a map u : X → Sn−1, and because they are both of dimension(n − 1), mod 2 intersection theory says that the intersection number I(u, p) iswell-defined for a point p. This is independent of p and is called the degree ofu and we define the winding number as
W2(X,Z) = deg2(u) = I(u, p).
There are several things to verify.
Proposition 25.2. Let U be a connected component of Rn −X. If p0, p2 ∈ U ,then W2(X, p0) = W2(X, p1).
Proof. First assume that p0 and p1 can be joined by a segment. Let pt =(1− t)p0 + tp1 and look at the homotopy F : X × I → Sn−1 given by
(x, t) 7→ x− pt|x− pt|
.
Then we see that the winding number for t = 0 and t = 1 are the same.So it suffices to show that any two points can be connected by a finite number
of segments. Let
A = the points that can be joined by a sequence of line segments
and B = U \A. Then both B and A are open, and because U is not connected,it follows that B = ∅.
Math 132 Notes 58
Proposition 25.3. Let X ⊂ Rn be an (n− 1)-dimensional submanifold. Thenset Rn −X has at most 2 connected components.
Proof. We first show that for any q ∈ X, there exists a Bε(q) such that Bε(q)−Xhas two connected components. Let f : X → Rn be the inclusion map. Becauseclearly f is an immersion at q, there exist charts such that the diagram commutes
U1 U2
V1 V2
f
ϕ1 ϕ2
and V1 → V2 is the inclusion map. Then the hyperplane divides the space intotwo parts, and therefore Bε(q)−X will have two components.
Now for any p ∈ Rn − X and q ∈ X with a neighborhood q ∈ U , there isa path joining p and U . This is because the set of points q ∈ X such that thestatement is true is a nonempty open and closed subset of X. Then if we letU = Bε(q) it follows that Rn −X has at most two components.
Math 132 Notes 59
26 April 4, 2016
We used surgery to do the classification of closed compact 2-manifolds.
Theorem 26.1. If M is a compact closed connected 2-manifold then MO1(M)is finite dimensional.
Theorem 26.2. If (V,B) is a nondegenerate symmetric bilinear form over Z/2,then there exists an M2 with (MO1(M), I) ∼= (V,B).
Theorem 26.3. If M1,M2 are compact closed connected 2-manifolds and thereexists an isometry MO1(M1) → MO1(M2), then there is a diffeomorphismf : M1 →M2 inducing T on MO1.
The first one is not proved, the second is proved, and the third is almostproved with a few loose ends. We need the following theorem.
Theorem 26.4. If MO1(M) = 0 then M is diffeomorphic to S2.
Example 26.5. Becasue the intersection form of RP 2#RP 2#RP 2 is isometricto the intersection form of RP 2#T , we see that they are diffeomorphic.
Example 26.6. The bordism homology of the Klein bottle is (Z/2)〈e1, e2〉 withB(e1, e1) = B(e1, e2) = 1, B(e2, e2) = 0. This is isometric to MO1(RP 2) ⊕MO1(RP 2) and hence the Klein bottle is diffeomorphic to RP 2#RP 2.
There are a few more things I have to prove: the connected sum is well-defined, and sowing a disk on the boundary of a Mobius band always gives anRP 2 no matter how you glue it.
26.1 More on Gluing
Consider
M Nf0
f1
diffeomorphisms with distinct boundary components. So for instance, M mapsto two boundary circles of N . Let us now glue the manifolds along the boundaryas
N ′ = N/(f0(x) = f1(x)).
The new manifold N ′ is a topological manifold, but it does not have a smoothstructure.
Theorem 26.7. The space N ′ can be given the structure of a smooth manifoldfor which the map N → N ′ is smooth. Moreover anay two smooth structures arediffeomorphic by a diffeomorphism which is the identity on M and the identityoutside a specified neighborhood of M .
Theorem 26.8. Any diffeomorphism of S1 extends to a diffeomorphism of D2.
Math 132 Notes 60
This is something special to dimension 2 and somthing strange happens oncewe get to dimenison 6.
These two theormes imply that any two manifold obtained from the Mobiusband by attaching a disk is diffeomorphic to RP 2. Suppose there are two bound-ary maps f, f ′ → ∂M that are related by a diffeomorphism g = f−1 f ′. Thenwe can extend g to a diffeomorphism of the disk:
∂M S1 S1
D2 D2
f g
g
Then it follows that the result is independent of f .
26.2 Isotopy of diffeomorphisms
Let
M Mf0
f1
be two diffeomorphisms.
Definition 26.9. An isotopy of f0 and f1 is a smooth map
h : M × [0, 1]→M
such that h(x, 0) = f0(x), h(x, 1) = f1(x), and ht : M →M is a diffeomorphismfor each 0 ≤ t ≤ 1.
Clearly isotopy is reflexive and symmetric. The only thing that might requirea proof is transitivity. This can be attained by gluing the two isotopies andslowing down using a bump function at the midpoint.
Theorem 26.10. Every diffeomorphism of S1 is isotopic to either (x, y) 7→(x, y) or (x, y) 7→ (x,−y).
Proof. Consider a diffeomorphism φ : S1 → S1. There is a covering map
0 R
R1 S1 S1
eitf
eit φ
and by the lifting property, there exists a unique f . This funciton f satisfiesf(0) = 0 and f ′(x) 6= 0 for every x.
If f ′(0) > 0, because f(0) = 0 the intermediate value theorem and somemore arguments imply that f(2π) = 2π and f(x+ 2πk) = f(x) + 2πk. We canthen simply set
h(x, t) = (1− t)f(x) + tx
Math 132 Notes 61
and h(cos θ, sin θ, t) = (cos h(θ, t), sin h(θ, t)) gives an isotopy between φ and theidentity map.
Likewise in the f ′(0) < 0 case, we can do the same thing and get a isotopywith (x, y) 7→ (x,−y).
Proof of theorem 26.8. We can look at the isotopy S1 × [0, 1] → S1 between φand one of the two standard maps and map it into a neighborhood of S1 in D2.Then we can fill in the inner circle as the identity map.
Math 132 Notes 62
27 April 6, 2016
27.1 Immersions of surfaces
Let X be a closed compact connnected 2-manifold. Our goal is to classify allimmersions f : X # R3, up to isotopy and diffeomorphism.
Definition 27.1. An isotopy of immersions X → Y is a map h : X×[0, 1]→Y such that for any t the map ht : X → Y is an immersion.
Example 27.2. There are several immersions of the torus. One way is to mapit to an ordinary torus. Another way is to first twist one direction into a figureeight and then connect the two ends. This will look like some kind of two toriattached together. When we connect the two ends, we can even twist it into afigure eight form.
Given an α ∈ MO1(X), choose an embedded circle z on X representingα. Assume that I(z, z) = 0. We know that there is a neighborhood of zdiffeomorphic to S1 × [−1, 1]. Then f : X # R3 restrics to an immersion ofS1 × [−1, 1] # R3.
By moving things around, we can show that every such immersion is isotpicto an embedding that looks like a some kind of Mobius band with k full-twists.This can be given a structure of a group, by defining addition as the result ofsurgery. Moreover, the group is generated by “1 twist” and actually “1 twist”is isotopic to “−1 twist.”
∼= ∼=
But we don’t really know whether the “0 twist” and the “1 twist” are distinct.We need an invariant to check this.
27.2 Linking number
Suppose that M1,M2 → R3 be embeddings of closed 1-manifolds that are dis-joint. We can always write M1 = ∂N . We have shown last time that it isalwasys possible to extend the map so that
M1 N
R3
f
Math 132 Notes 63
with f t M2. We define the linking number of M1 and M2 as L(M1,M2) =I(f,M2).
Proposition 27.3. The intersection number I(f,M2) is independent of thechoice of N .
Proof. Suppose we have two N ′ and N ′′. Then N ′ ∪N ′′ is a closed 2-manifold.Let f, f ′, f ′′ be the immersions N,N ′, N ′′. Then
I(f ′,M2) + I(f ′′,M2) = I(f,M2) = 0
because M2 is a boundary of something.
It can also be proved that the linking number is invariant under isotopy.Now given an embedding f : S1 × [−1, 1] → R3, we can look at the
L(f |S1×0, f |S1×1). In the “0 twist” embedding, this number is 0 and inthe “1 twist” embedding, the linking number is 1. This shows that the two arenot isotopic.
In a general setting, consider a 2-manifold X that has even intersection form(MO1(X), I). Given an immersion f : X # R3, it induces a function
φ : MO1(X)→ Z/2,
where φ(α) is the isotopy class of the immersed cylinder around an embeddedcircle representing α.
A famous mistake of Pontryagin is assuming that φ is linear. But actuallyit is not.
Example 27.4. Consider the two circles z1 and z2 on the torus. Then clearlyφ(z1) = φ(z2) = 0. But it can be checked that φ(z1 + z2) = 1.
In fact, φ is a quadratic form; it satisfies the identity
φ(x+ y)− φ(x)− φ(y) = I(x, y).
Math 132 Notes 64
28 April 8, 2016
Last time I drew three different immersions of the torus. You can just map itto the ordinary torus, or you can make a figure-eight torus, or a figure-eight-figure-eight torus.
LetX be a connected compact closed 2-manifold, and assume that (MO1(X), I)is even. We showed that given an immersion X # R3, it incudes a quadraicfunction q : MO1(X)→ Z/2 giving a bilinear form
q(x+ y)− q(x)− q(y) = I(x, y).
If we look at the quadratic forms of the torus, figure-eight torus, figure-eight-figure-eight torus, we have
q
(0 e1
e2 e1 + e2
)=
(0 00 1
) (0 10 0
) (0 11 1
)torus f-e torus f-e-f-e torus.
The theroem then tells us that the torus and the figure-eight torus are isotopic,where as the figure-eight-figure-eight torus is not.
28.1 Quadratic forms
Let V be a vector space over a field k.
Definition 28.1. A quadratic function on V is a function q : V → k suchthat
B(x, y) = q(x+ y)− q(x)− q(y)
is bilinear and q(λx) = λ2q(x) for λ ∈ k.
Example 28.2. Let V = kn. The map
q
x1
...xn
=∑i≤j
aijxixj
is a quadratic form.
Definition 28.3. A quadratic function is non-degenerate if B(−,−) is non-degenerate.
We call that q is a quadratic refinement of B if B is the bilinear formassociated to q.
Proposition 28.4. If q1 and q2 are refinemenets of B, then q1 − q2 is linear.
Proof. It follows from that (q1(x+y)−q1(x)−q1(y))−(q2(x+y)−q2(x)−q2(y)) =B(x, y)−B(x, y) = 0.
Math 132 Notes 65
Corollary 28.5. The set of quadratic refinements of B is an affine space forHom(V, k). So if k = Z/2 and dimV = n then there are 2n quadratic re-finemenets of any fixed B.
Definition 28.6. A quadratic form (over k) is a pair (V, q) consisting of avector space V and a quadratic function q : V → k.
Definition 28.7. The orthogonal sum (V1, q1) ⊕ (V2, q2) is the vector spaceV1 ⊕ V2 equipped with (q1 ⊕ q2)(x, y) = q1(x) + q2(y).
Definition 28.8. An isometry (V1, q1) → (V2, q2) is a linear transformationT : V1 → V2 such htat q2(Tx) = q1(x) for all x ∈ V1.
Now we want to classify quadratic forms over k up to isometry. Over R,every q is isometic to
q(x1, . . . , xn) = x21 + · · ·+ x2
j − x2j+1 − · · · − x2
n.
But what happens to k = Z/2? First, because q is nondegnerate and
B(x, x) = 2q(x) = 0,
we see that (V,B) = H⊕m for some m, and in particular dimV = 2m.
Lemma 28.9. Suppose (V, q) is a non-degenerate quadratic form over Z/2, withdimV > 2. Then there is a 0 6= v ∈ V such that q(v) = 0.
Proof. Because dimV is even, it is at least 4. Let e1, . . . , e4 be linearly inde-pendent vectors. If q(v) = 1 for all v 6= 0, then B(e1, e2) = B(e3, e4) = 1 andthen
q(e1 + · · ·+ e4) = q(e1 + e2) + q(e3 + e4) + 0 = 1 + 1 + 0 + 1 + 1 + 0 = 0.
Consider a 0 6= v such that q(v) = 0. Since B is nondegenerate, there is a vsuch that B(v, w) = 0. Then q(v+w) = q(w)+1 and hence we can assume thatone is 1, and we might as well suppose that it is w. Let V ′ = 0, v, w, v +w⊥,and let
H+ =
(0 10 0
), H− =
(0 11 1
)be the standard 2-dimensional quadratic forms. We still have to show that theyare not isometric.
Proposition 28.10. The quadratic form (V, q) can be decomposed into (V, q) =(V ′, q′)⊕H+.
It folows that (V, q) is either of the form H⊕m+ or H− ⊕H⊕(m−1)+ .
Definition 28.11. Given a nondegenerate quadratic form (V, q), we define itsArf invariant
Arf(q) =1√|V |
∑v∈V
(−1)q(v).
Then Arf(q1 ⊕ q2) = Arf(q1) Arf(q2). It follows that Arf(H⊕m+ ) = 1 and
Arf(H− ⊕H⊕(m−1)+ ) = −1.
Math 132 Notes 66
29 April 11, 2016
Last time, we showed that if M is an oritentable closed compact connected2-manifold, then there is a bijection
immersions M # R3
modulo isotopy
←→
quadratic refinementq : MO1(M)→ Z/2
of the intersection form
and there are 22m of these, where 2m is the dimension of MO1(M). Whenwe look at immersions up to diffeomorphism, we have to take the quotient byisometry on the right hand side.
immersions M # R3
modulo isotopy
/diff. ←→
quadratic refinementq : MO1(M)→ Z/2
of the intersection form
/
isom.
Then there are 2 of them, characterized by the Arf invariant. Now let us lookat the case when M is not orientable.
29.1 Quadratic forms taking values in Z/4If M is not orientable, then the surface M contains a Mobius band. There aretwo kinds of Mobius bands: the one with the right twist, and the one withthe left twist. These two are different, because if you add the same Mobius bysurgery then you get the 1-full-twist band, where as if you add the differentMobius bands, then you get the cylinder. So, each loop in M thickens up to anelement of Z/4.
Hence each immersion M # R3 gives a quadratic q : MO1(M)→ Z/4 thatsatisfies
q(x+ y)− q(x)− q(y) = 2I(x, y).
Then we have a similar theorem.
Theorem 29.1. The set of immmersions M # R3 modulo isotopy is in bijec-tion with the set of quadratic q : MO1(M)→ Z/4 refining I.
Definition 29.2. A quadratic form is pair (V, q) where V is a finite dimen-sional vector space over Z/2 and q : V → Z/4 such that
q(x+ y)− q(x)− q(y) = B(x, y)
is bilinear and q(λx) = λ2q(x) for λ ∈ Z.
Example 29.3. There are two 1-dimensional nondegenerate quadratic forms.Let R = 0, e and B(e, e) = 1. Then there can be two
R+ = (F2, q), q(e) = 1,
R− = (F2, q), q(e) = −1.
Math 132 Notes 67
Example 29.4. Let us look at the Klein bottle R ⊕ R with bases e1 and e2.Then there are four possibilities comming from R± ⊕R±.
Just like before, any two quadratic refinemenets of a fixed symmetric bilinearform differ by a linear function. Hence there are 2dimV quadratic refinementsof a fixed symmetric bilinear form.
29.2 Brown-Arf invariant
Looking at the quadratic forms up to isometry, we see that R+⊕R− and R−⊕R+
are isometric (clearly), but no other are isometric. We can define an invariant.
Definition 29.5. Given a quadratic form (V, q), we define the Brown-Arfinvariant
Arf(q) =1√|V |
∑v∈V
iq(v).
Again, it is not hard to check that Arf(q1 ⊕ q2) = Arf(q1) Arf(q2).
Example 29.6. For R+, its Arf invariant is
1√2
(i0 + i1) =1 + i√
2.
Any symmetric bilinear form looks like either R ⊕ · · · ⊕ R or H ⊕ · · · ⊕H.It can be proved that the possible non-degenerate quadratic forms are
R⊕k+ ⊕R⊕l− or H⊕n+ ⊕H⊕m− .
(You can do the same thing, but you have to work a bit more.) Let me juststate the theorem.
Theorem 29.7. Two quadratic forms (V1, q1) and (V2, q2) are isometric if andonly if (V1, B2) and (V2, B2) are isometric and the Arf invariants agree.
We can easily calculate
Arf(R+) = ζ, Arf(R−) = ζ−1, Arf(H+) = 1, Arf(H−) = −1,
where ζ = e2πi/8.Next time we will do talk some more and draw some pictures, and after that
we will go back to our book.
Math 132 Notes 68
30 April 13, 2016
We have seen last time that a quadratic form is determined up to isometry bythe Arf invariant. Each surface is diffeomorphic to one of H⊕n or R ⊕H⊕n orR⊕R⊕H⊕n. For each case, the possible quadratic forms are
Quadratic form ArfH⊕n+ +1
H− ⊕H⊕(n−1)+ −1
Quadratic form ArfR+ ⊕H⊕n+ ζ1
R− ⊕H⊕n+ ζ−1
R+ ⊕H− ⊕H⊕(n−1)+ ζ−3
R− ⊕H− ⊕H⊕(n−1)+ ζ3
Quadratic form ArfR+ ⊕R+ ⊕H⊕n+ ζ2
R− ⊕R+ ⊕H⊕n+ ζ0
R− ⊕R− ⊕H⊕n+ ζ−2
R+ ⊕R+ ⊕H− ⊕H⊕(n−1 ζ−2
R− ⊕R+ ⊕H− ⊕H⊕(n−1)+ ζ4
R− ⊕R− ⊕H− ⊕H⊕(n−1)+ ζ2
So, there are in total 2 or 4 different immersions up to isotopy and diffeo-morphism.
30.1 Boy’s surface and immersions of surfaces
In the case of S2, the bordism homology has dimension 0. So it should havea unique immersion up to isotopy. This is actually pretty hard to prove, andthe key part of this statement is that the sphere can be eversed. How do we dothis? We can push one side of the sphere and take it back to the other side,and this will give us a cylider with some kind of carloop in the middle of eachstrip. This can be shown to be isotopic with a Dehn twist on the middle. Thenwe can untwist it to make a inverted sphere.
So, there is a unique immersion of the sphere up to isotopy. Let us now lookat the dimension 1 case. The only such surface is RP 2. The immmersion ofRP 2 in R3 is called Boy’s surface. Hilbert assgined proving that RP 2 cannot beimmersed in 3-space as an assignment, but it was in fact false and Boy foundan immersion.
The surface can be visualized as in figure 2.6 Note that this image is chiral.This means that that there is a mirror reflection of this image, and the twoimmersions Boy+ and Boy− will be the immersions of RP 2.
For the torus, we know there are 4 different immersions. The other immer-sions will be described as connected sums of these up to isotopy and diffeomor-phism.
6Hopkins actually gave a much more detailed description of the surface, but it is impossibleto draw it.
Math 132 Notes 69
Figure 2: Boy’s surface from the top view. You can see the 3-fold symmetry.
We can try to do surgery on the surface. For instance, the ordinary Kleinbottle, after doing surgery on the handle part, retracts to a ball. This gives away to easily see that the Arf invariatn of the Klein bottle is +1.
Theorem 30.1. Let (V, q) be a nondegenerate Z/2-valued quadratic form, andlet α ∈ V be such that q(α) = 0. There is a natural quadratic form (V ′, q′)where V ′ = α⊥/α. Then Arf q = Arf q′.
This theorem implies that there are 8 surfaces immersed in R3 up to surgery.I have listed above and those will be the 8 surfaces.
Math 132 Notes 70
31 April 15, 2016
So I want to make it easy on you by going back to be book and doing integrationon manifolds.
Green’s theorem says that for a region X and a loop γ as a boudary of X,we have ∮
γ
pdx+ qdy =
∫∫X
∂q
∂y− ∂p
∂xdxdy.
This is in the form of∫∂M
something =
∫M
something else(=some kind of derivative).
Note that integration involves
(1) an orientation,
(2) the thing you integrate, which is a differential form.
Let us think about what a differential form must look like. One observationis that it should not be a function. If we have a fuction g(x, y) on R2 anda diffeomorphism (f1, f2), the interation of the pullback is different from theoriginal integration.∫∫
g(x, y)dxdy 6=∫∫
g(f1(s, t), f2(s, t))dsdt.
Instead, we have the formula∫∫g(x, y)dxdy =
∫∫g(f1(s, t), f2(s, t))|dF |dsdt.
31.1 Multilinear algebra
Let V be a vector space, and let
T : V × · · · × V → R
be a multilinear form, i.e., a function linear in each factor. You can add mul-tilinear forms, and therefore multilinear forms form a vector space. Moreover,the dimension of the vector space is (dimV )n. In our book, this is called a rankn tensor. In the literature, tensors are multilinear maps V n × V ∗m → R.
An alternating tensor is a mutilinear map T : V n → R such that
T (v1, . . . , vi, . . . , vj , . . . , vn) = −T (v1, . . . , vj , . . . , vi, . . . , vn).
More generally, for any permuation σ,
T (vσ1, . . . , vσn) = (−1)σ(v1, . . . , vn).
Math 132 Notes 71
Proposition 31.1. Suppose V has dimension n and T : V n → R is alternating.Then
T (Sv1, . . . , Svn) = det(S)T (v1, . . . , vn)
for every linear S : V → V .
There is another piece of algebra I need to teach you.Let S : V k → R and T : V ` → R be tensors of rank k, l. Then you can
construct a bilinear map S ⊗ T : V k × V ` → R given by
S ⊗ T (v, w) = S(v)T (w).
We can specifically focus on alternating tensors. Let us define the exteriorpower
Λk(V ∗) = alternating T : V k → R.
Likewise, we have a map Λk(V ∗) × Λl(V ∗) → Λk+l(V ∗). But the naıve guessdoesn’t work here. We need to somehow skew symmetrize the tensor.
Suppose T : V n → R is a rank n tensor. We will define
Alt(T )(v1, . . . , vn) =1
n!
∑σ∈Sn
(−1)σT (vσ1, . . . , vσ1
).
(The factor 1/n! arranges things so that if T were already alternating thenAltT = T .) Given this, we can define
S ∧ T = Alt(S ⊗ T ).
Example 31.2. Let V be a vector space with basis e1, . . . , en. Then Λ1(V ∗) =V ∗ has basis e∗1, . . . , e
∗n. Next, Λ2(V ∗) has basis e∗1 ∧ e∗2, . . . , e∗n−1 ∧ e∗n.
Proposition 31.3. The space Λk(V ∗) has basis e∗i1 ∧ . . . e∗ik for 1 ≤ i1 <
· · · < ik ≤ n, and thus has dimension dim Λk(V ∗) =(nk
).
Math 132 Notes 72
32 April 20, 2016
To define an integral, which is a number, we need a n-manifold M , an n-formω ∈ Ωn(M), and an orientation of M . Then you can define
∫Mω as a number.
For instance, let M = R1 and ω = f(x)dx = e−x2
dx. You don’t know what∫R e−x2
dx is, not be cause you don’t know how to integrate, but because haven’ttold you whether you’re integrating from −∞ to ∞ or from ∞ to −∞. In mostcases, we give the orientation 1 ∈ TxR1 = R.
Definition 32.1. Suppose ω ∈ Ωn(M) is an n-form. The support of ω is theclosure of the set of all x ∈M for which ωx 6= 0 ∈ Λn(T ∗xM).
If the support is finite, then∫R ω =
∫ baω <∞.
32.1 Integration on manifolds
Now our goals is defining the integral. Consider a diffeomorphism f : U → V ⊂Rn and we have learned in our nth semester of calculus that∫
V
g(x1, . . . , xn)dx1 · · · dxn =
∫U
g(f(t1, . . . , tn))|det df |dt1 · · · dtn.
Now there is a quote by Grothendieck that goes like “Let mathematics be whatit wants to be.” We try to define things as naturally as possible. If we let
ω = g(x1, . . . , xn)dx1 ∧ · · · ∧ dxn
and U and V are oriented with f : U → V preserving orientation, then we candefine the pull-back
f∗ω = g(f(t1, . . . , tn)) det(df)dt1 ∧ · · · ∧ dtn.
Since f preserves orientation, det(df) = |df |. Then we can write the change ofvariables law as ∫
V
ω =
∫U
f∗ω.
Here, in the notation, the orientation of both U and V are implicitly contained.This is not the best possible mathematical practice, but this is what people do.
We realize our goal whenM is diffeomorphic to an open subset of Rn, becauseif Φ : M → V ⊂ Rn is the diffeomorphism, we can simply define∫
M
ω =
∫V
(Φ−1)∗ω.
Moreover, even if ω ∈ Ωn and suppω is in a coordinate neighborhood U , thenwe can also define the integral because we can restrict the manifold to suppω.
Math 132 Notes 73
What if M is arbitrary? We choose a countable covering Ui∞i=1 of M bycoordinate neighborhoods Φi : Ui → Rn. Then construct a partition of unityθi subordinate to the covering, and define∫
M
ω =
∫M
∑i
θiω =∑i
∫M
θiω.
Each∫Mθiω has been defined because supp(θiω) ⊂ Ui.
We need to show that the integral is well-defined.
Proposition 32.2.∫Mω is independent of the choice of θi.
Proof. Suppose θi is another partition of unity. Then we have∑i
∫M
θiω =∑i,j
∫M
θiθjω =∑j
∫M
θjω.
Example 32.3. Consider a curve γ : [0, 1] → R2 and consider a 1-form pdx +qdy ∈ Ω1(R2). We can integrate the 1-form over the curve by∮
γ
pdx+ qdy −∫ 1
0
γ∗(pdx+ qdy).
Example 32.4. Let us try integrate a 2-form ω = f1dy∧dz+f2dz∧dx+f3dx∧dyon a surface given as the image of
X = (X1, X2, X3) : U → R3.
The pull back of for instance dx ∧ dy is
X∗(dx ∧ dy) = (X∗dx) ∧ (X∗dy) = (X1udu+X1
vdv) ∧ (X2udu+X2
vdv)
= (X1uX
2v −X1
vX2u)du ∧ dv.
If you compute everything, you get something like
X∗ω = ~f · (Xu ×Xv).
This is why you were taught to integrate vector fields in you calculus course.This theory of forms encompasses all the area component, right hand rule, etc.in simply pulling back forms.
Math 132 Notes 74
33 April 22, 2016
Today I want to introduce Stokes’ theorem.
33.1 Stokes theorem
One version is the fundamental theorem of calculus∫ b
a
f ′(x)dx = f(b)− f(a).
Another version is ∮∂R
pdx+ qdy =
∫∫R
(py − qx)dxdy.
Stokes theorem just says for a manifold Mn and ω ∈ Ωn−1(M),∫∂M
dω =
∫∂M
ω.
I have to explain what d is, and I have to explain what the orientation is.Suppose M has an orientation, and let’s see what happens at the boundary.
We see that Tx∂M ⊕R = TxM . We orient M by giving the same orientation ofnout, e1, . . . , en−1 to e1, . . . , en−1, where e1, . . . , en−1 form a basis for Tx∂Mand nout is the normal vector pointing outwards.
Now that we know the orientation for ∂M , we need a operator for takingthe derivative of differential forms.
Theorem 33.1. There is a unique linear operator d : Ωk−1(M) → Ωk(M)satisfying
(1) d(f) = df , for any 0-form, i.e., function, f ,
(2) d(ω ∧ η) = (dω) ∧ η + (−1)|ω|ω ∧ dη, (Leibniz rule)
(3) d2 = 0,
(4) any g : M → N induces a g∗ : Ωk(N) → Ωk(M) such that dg∗ω = g∗dω(naturality).
Ωk(M) Ωk(N)
Ωk−1(M) Ωk−1(N)
g∗
d d
g∗
Proof. Since d is linear, we may suppose that suppω is contained in a coordinateneighborhood. Using naturality, it suffices to defined d when M = Rn. Thenwe can look at each component and define it explicitly.
Math 132 Notes 75
Example 33.2. Let us consider M = R3. For f ∈ Ω0, we have
df = fxdx+ fydy + fzdz = grad(f) = ∇f.
If we go from Ω1 to Ω2, then
d(fdx+ gdy + hdz) = (gx − fy)dx ∧ dy + (hx − fz)dx ∧ dz + (hy − gz)dy ∧ dz
and with suitable conventions, it is the curl. Lastly, for Ω2 → Ω3, it correspondsto the divergence.
Now I want to describe the steps that go into the proof of Stokes theorem.The cool thing is that once you have the formulation, the proof is easy.
Proof of Stokes theorem. I want to prove that∫Mdω =
∫∂M
ω. Both sides arelinear in ω, so by writing ω =
∑θiωi, we can suppose that suppω is compact
and is contained in a coordinate neighborhood U . This neighborhood can beeither in the interior, or contain a part of the boundary.
Suppose first that U ∼= Rn. We might as well suppose M = Rn. Then wecan write
ω = f1dx2 ∧ · · · ∧ dxn + · · ·+ fndx1 ∧ · · · ∧ dxn−1,
and again using linearity, we may suppose that ω = fdx2 ∧ · · · ∧ dxn. Thendω = (∂f/∂x1)dx1 ∧ · · · ∧ dxn. If we choose sufficiently large a such thatsuppω ⊂ [−a, a]× Rn−1 then∫
Rndω =
∫Rn−1
∫ ∞−∞
∂f
∂x1dx1 ∧ · · · ∧ dxn
=
∫Rn−1
∫ a
−a
∂f
∂x1dx1dx2 ∧ · · · ∧ dxn
=
∫Rn−1
f(a)− f(−a)dx2 ∧ · · · ∧ dxn = 0.
We will do the other case next time.
Math 132 Notes 76
34 April 25, 2016
Remember that the Stokes theorem says∫M
dω =
∫∂M
ω.
Because the expression is linear and is diffeomorphism-invariant, we can checkit only when M = Rn and M = Hn. We showed it in the case of Rn last timeusing the fundamental theorem of calculus, and now let’s do it for M = Hn.Because we use the outward normal vector, I will unconventionally let Hn =(x1, . . . , xn) ∈ Rn : x1 ≤ 0.
Let our differential form ω ∈ Ωn−1(Hn) be
ω =∑i
fi dx1 ∧ · · · ∧ dxi ∧ · · · ∧ dxn.
Then its derivative is
dω =∑ ∂fi
∂xidxi∧dx1∧· · ·∧ dxi∧· · ·∧dxn =
(∑(−1)i−1 ∂fi
∂xi
)dx1∧· · ·∧dxn.
So ∫Hn
dω =
∫Hn
∑ ∂fi∂xi
(−1)i−1dx1 ∧ · · · ∧ dxn
=∑i
∫ ∞−∞· · ·∫ ∞−∞
∫ 0
−∞
∂fi∂xi
(−1)i−1dx1 · · · dxn.
The first term (i = 1) is∫ ∞−∞· · ·∫ ∞−∞
∫ 0
−∞
∂f1
∂x1dx1 · · · dxn =
∫Rn−1
f1(0, x2, . . . , xn)dx2 · · · dxn =
∫∂Hn
ω.
The other terms are all 0, because the ith derivative goes to zero.This abstract apparatus reducing things to simple local problems is a theme
you encounter everywhere in math.
34.1 Degree of a map
Let f : Mn → Nn be a map between compact manifolds. We have defineddeg2 f to be the number of points in f−1(y), where y is a regular value. Thisis well-defined modulo 2, because the difference of two #f−1(y) and #f−1(y′)will be given a boundary of a 1-manifold.
If both M and N are oriented, we can do better than that. If dfx : TxM →Tf(x)N is an isomorphism, we define
sign df =
+1 if df preserves orientation,
−1 if df reverses orientation.
Math 132 Notes 77
We the define
deg f =∑
x∈f−1(y)
sign dfx,
where y is a regular value. This is then a well-defined value in Z, and if f = ∂hfor some h, then deg f = 0. I won’t build too much on this, but there is also anotion of bordism homology MSO∗ of oriented manifolds.
Theorem 34.1 (Degree formula). Let f : Mn → Nn be a smooth map ofcompact oriented n-manifolds, and let ω ∈ Ωn(N). Then∫
M
f∗ω = deg(f)
∫N
ω.
This is the starting point of the relation between topological invariants andanalytic stuff. You cannot continuously deform the degree, but you can do itfor ω.
Proof. First suppose that y ∈ N is a regular value and find a neighborhoody ∈ U ⊂ N be a small neighborhood such that f−1(U) = U × S where U is afinite set. Consider the simple case of suppω ⊂ U .
f−1(U) U × S
U
f
g
Then ∫M
f∗ω =
∫f−1(U)
f∗ω =
∫U×S
g∗f∗ω = (deg f)
∫N
ω.
Now the problem is how to reduce to the case we did above.
Proposition 34.2. If f : M → N is the boundary of g : W → N , then∫Mf∗ω = 0.
Proof. By Stokes theorem,∫M
f∗ω =
∫W
dg∗ω =
∫W
g∗dω =
∫W
g∗0 = 0.
There is also the following fact.
Proposition 34.3. If N is a connected manifold, and x, y ∈ N , then there isa diffeomorphism g : N → N which is isotopic to the identity, and satisfiesg(x) = y.
Idea of proof. You can always move the point a little bit, so the set of pointsyou can reach is open, and the set of points you can’t reach is open. So you canreach every point.
Math 132 Notes 78
Going back to the original theorem, we know that the theorem is true forsuppω ⊂ U where U is a neighborhood around a regular value y. Using theproposition, for any y′ ∈ N we can choose a diffeomorphism g and an isotopyh : N × [0, 1] → N such that g(y′) = y and h(z, 0) = g(z), h(z, 1) = z. LetU ′ = g−1(U). If suppω ⊂ U ′ then∫M
f∗ω =
∫N
ω =
∫N
(g−1)∗ω =
∫M
f∗(g−1)∗ω = (deg f)
∫N
(g−1)∗ω = (deg f)
∫N
ω.
Then we can cover N by finitely many U ′s, which we will denote by U1, . . . , Uk.Using partitions of unity, we can decompose any ω into ω = ω1 + · · ·+ ωk suchthat suppωi ⊂ Ui, and then∫
M
f∗ω =∑∫
M
f∗ωi =∑
(deg f)
∫N
ωi = (deg f)
∫N
ω.
Math 132 Notes 79
35 April 27, 2016
Let Mn ⊂ Rn+1 be a compact closed manifold. Then M is oriented, and forany x ∈ M there is a unit normal vector nx pointing outward. Then there is amap g : Mn → Sn that sends
g : x 7→ nx.
For any x ∈ M , the derivative maps dgx : TxM → Tg(x)Sn. We define the
Gaussian curvature K(x) at x of M as
K(x) = deg dgx.
Example 35.1. Let M be the sphere or radius r. Then g(x) = x/r andK(x) = r−n. If M is a saddle point, then the curvature is negative.
So this curvature measures the shape of the surface. But this depends on theway the manifold is embedded in the Euclidean space.
Given a M ⊂ Rn+1, there is a thing called the ωvol ∈ Ωn(M) such that atx ∈M , if e1, . . . , en is an oriented orthonormal basis of TxM then
ωvol,x = e∗1 ∧ · · · ∧ e∗n.
We are taught in calculus that∫M
ωvol = Volume of M.
35.1 The Gauss-Bonnet theorem
Theorem 35.2 (Gauss-Bonnet). For any M ⊂ Rn+1 and n even,∫M
g∗ωvol = Vol(Sn) · 1
2χ(M).
Here, χ is something called the Euler characteristic ofM , and it only dependson M as a topological space. I haven’t given a definition in this course, but I’llgive a definition for dimM = 2 around 1:42 pm.
Actually, we already know by the degree theorem,∫M
g∗ω = deg(g) ·∫Snωvol = deg(g) ·Vol(Sn).
The Gauss-Bonnet theorem is further saying that deg(g) = 12χ(M). Further,
by the classification of 2-manifolds, we know that if M1 and M2 are connectedoriented surfaces and χ(M1) = χ(M2) then M1 is diffeomorphic to M2. So forinstance, if we sail around the world and integrate the curvature, we can knowhow the earth looks like.
So let me now tell you how the Euler characteristic is defined. The firstdefinition involves triangulating M . If we choose a triangulation and define
χ(M) = #Faces−#Edges + #Vertices,
Math 132 Notes 80
then this does not depend on the triangulation. Alternatively, we can define itas
χ(M) = dimMO0(M)− dimMO1(M) + dimMO2(M).
We know by the degree formula that the integral∫M
g∗(ωvol)
depends only on the embedding M → Rn=1 up to isotopy. So we may assumethat, for instance, if M = S2 then change M into a icosahedron. The Gaussmap at each vertex sweeps out a spherical polygon, and so∫
M
g∗(ωvol) =∑
vertices
area of some spherical polygon.
Now there is another cool theorem about the area of a spherical polygon:
Theorem 35.3. The area of a spherical n-gon is the excess angle∑angles− (n− 2)π.
Proof. You can play around with the fact that a “lune” of angle α has area2α. First prove for triangles and then add them up to get the general n-goncase.
Proof of the Gauss-Bonnet theorem. Let M be triangulated and at a vertex v,nv triangles come together. Using the cool theorem above, we get∫M
g∗(ωvol) =∑
verticesv
(∑
θ − (nv − 2)π) =∑
verticesv
(∑(π − α)− (nv − 2)π
)=
∑vertices
(2π −∑
α) = 2πV − πF.
But because each face has exactly 3 edges and each edge has exactly 2 faces,2E = 3F . So we finally get∫
M
g∗ωvol = 2πV − πF = 2π(V − E + F ) = 2πχ(M) =1
2(4π)χ(M).
There is this notion of local stuff accumulating to produce global stuff.
Index
Arf invariant, 65atlas, 8
bilinear form, 49bounding manifold, 46Brown-Arf invariant, 67
chart, 7, 8cobordism, 43, 44codimension, 17connected sum, 47critical value, 15
degree, 57, 77degree formula, 77derivative, 10, 12
Euler characteristic, 79exterior power, 71
framing, 46
Gauss-Bonnet theorem, 79Grassmannian, 9
Hessian matrix, 23
immersion, 14regular, 51
integration, 73intersection number, 41inverse function theorem, 13isotopy, 60
of immersions, 62
linking number, 63
manifold with boundary, 30measure zero set, 22, 23Morse function, 25
non-degenerate critical point, 23normal bundle, 39normal space, 39
partition of unity, 37preimage theorem, 15
quadratic form, 65quadratic function, 64
real projective space, 8regular value, 15
Sard’s theorem, 22smooth atlas, 8smooth function, 6smooth manifold, 6, 7smooth structure, 7stability, 20support, 37, 72
tangent bundle, 29tangent space, 11tensor, 70
alternating, 70topological manifold, 6transverse, 18
winding number, 57
81