the algebra of the box spline - cimpa · thealgebraofthebox–spline claudioprocesi....

The algebra of the box–spline

Claudio Procesi.

Hanoi, January 2011

Claudio Procesi. Topics in Hyperplane Arrangements, Polytopes and Box-Splines

BASIC INPUT

The basic input of this theory is a(real, sometimes integer) n ×m matrix A.We always think of A as a LIST of vectors in V = Rn, its columns:

A := (a1, . . . , am)

ConstrainWe assume that 0 is NOT in the convex hull of its columns.

From A we make several constructions, algebraic, combinatorial,analytic etc..

For a complete treatment see

Series: Universitext

1st Edition., 2010, XXII, 381p. 19 illus., 4 in color., Softcover

ISBN: 978-0-387-78962-0

CONVEX POLYTOPES

We begin with a system of linear equations:m∑

i=1aixi = b, or Ax = b, A := (a1, . . . , am) (1)

The columns ai , b are vectors with n coordinates(aj,i , bj , j = 1, . . . , n).

We assume that the matrix A is real and 0 is NOT in the convexhull of its columns.We are interested in dealing with twofundamental problems.

CONVEX POLYTOPES

Variable polytopes

As in Linear Programming Theory we want to study the

VARIABLE POLYTOPES:

ΠA(b) := {x |Ax = b, xi ≥ 0, ∀i}

which are convex and bounded for every b.

Identify the spaces Ax = b and Ax = 0 then think of ΠA(b) as a

variable polytope in the space Ax = 0

Variable polytopes

VARIABLE POLYTOPES:

ΠA(b) := {x |Ax = b, xi ≥ 0, ∀i}

Variable polytopes

VARIABLE POLYTOPES:

ΠA(b) := {x |Ax = b, xi ≥ 0, ∀i}

The object of study

Basic functionsSet VA(b) to be the volume of ΠA(b).If A, b have integer coordinates, consider the finite set IA(b)of points in ΠA(b) with integer coordinates.

Arithmetic caseIn this case we want compute the number PA(b) of solutions ofthe system in which the coordinates xi are non negative integers.

The object of study

Basic functionsSet VA(b) to be the volume of ΠA(b).If A, b have integer coordinates, consider the finite set IA(b)of points in ΠA(b) with integer coordinates.

Arithmetic caseIn this case we want compute the number PA(b) of solutions ofthe system in which the coordinates xi are non negative integers.

In other words the vectors a1, . . . , am define a map:

F : Rm → Rn.

F (t1, . . . tm) := t1a1 + · · ·+ tmam.

Therefore, in this language, the polytope:

ΠA(b) = F−1(b) ∩ Rm+,

where Rm+ denotes the positive quadrant.

Our goals are therefore:1 To compute the volume of ΠA(b).2 If A, b have integer elements, to compute the number PA(b)

of points with integer coordinates in ΠA(b).

F : Rm → Rn.

F (t1, . . . tm) := t1a1 + · · ·+ tmam.

ΠA(b) = F−1(b) ∩ Rm+,

F : Rm → Rn.

F (t1, . . . tm) := t1a1 + · · ·+ tmam.

ΠA(b) = F−1(b) ∩ Rm+,

When A, b have integer elements it is natural to think of anexpression like:b = t1a1 + · · ·+ tmam with ti not negative integers as a:

partition of b with the vectors ai ,in t1 + t2 + · · ·+ tm parts, hence the name partition function forthe number PA(b), thought of as a function of the vector b.

When A, b have integer elements it is natural to think of anexpression like:b = t1a1 + · · ·+ tmam with ti not negative integers as a:

partition of b with the vectors ai ,in t1 + t2 + · · ·+ tm parts, hence the name partition function forthe number PA(b), thought of as a function of the vector b.

Some combinatorial geometry

The pictures are taken from the book:C. De Boor, K. Höllig, S. Riemenschneider,

Box splines

Applied Mathematical Sciences 98 (1993).

The basic cone

Obviously, the set of vectors b such that ΠA(b) is not empty isequal, by definition, to the set:

The cone generated by A

CA := {m∑

i=1xiai | xi ≥ 0}

CA is a convex cone in Rn.

By assumption its non zero elements lie entirely in the interior of ahalf space.

CA is usually said to be a pointed cone and 0 is its vertex.

The basic cone

CA := {m∑

i=1xiai | xi ≥ 0}

The basic cone

CA := {m∑

i=1xiai | xi ≥ 0}

A 2–DIMENSIONAL EXAMPLE

∣∣∣∣∣1 1 0 −11 0 1 1

∣∣∣∣∣

OO ??��

__???????

A 2–DIMENSIONAL EXAMPLE

the associated cone C(A) has three big cells

. . . . . . . . .

. . . . . . . .

. . . . . . .

. . . . . .

OO >>}}}}}}}

``AAAAAAA

}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA. . . .

It is useful to consider also the dual cone CA of CA.

In the dual space it consists of the (row) vectors that have nonnegative scalar product with all vectors ai . I.e. with all vectors ofthe cone CA.

CA is a cone with non empty interior.

DUAL CONES

. . . . . . . . .

. . . . . . . .

. . . . . . .

. . . . . .

DUAL CONES

. . . . .

. . . .

}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

DUAL CONES

. . . . . . . . .

. . . . . . . .

. . . . . . .

. . . . . .

. . . . .

. . . .

}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

Some preliminaries

In order to attack the problem of computing volumes and partitionfunctions we need to

describe some combinatorial geometry of the cone C(A)

generated by the columns of A

Some preliminaries

Some intuitive pictures

We want to decompose the cone C(A) into big cells and defineits singular and regular points.

We do everything on a transversal section, where the cone lookslike a bounded convex polytope and then project.

ExampleLet us start with some pictures where A is the list of positive rootsfor type A3.

α1, α2, α3, α1 + α2, α2 + α3, α1 + α2 + α3

Some intuitive pictures

We want to decompose the cone C(A) into big cells and defineits singular and regular points.

We do everything on a transversal section, where the cone lookslike a bounded convex polytope and then project.

ExampleLet us start with some pictures where A is the list of positive rootsfor type A3.

α1, α2, α3, α1 + α2, α2 + α3, α1 + α2 + α3

EXAMPLE Type A3 in section (big cells):

DDDDDDDDDDDDDDDDD

zzzzzzzzzzzzzzzzz

α1 + α2

��α2 + α3

6666666666666666

α1 + α2 + α3

VVVVVVVVVVVVVVVVVVVVV

hhhhhhhhhhhhhhhhhhhhh

nnnnnnnnnnnn

PPPPPPPPPPPP

α1 α3

We have 7 big cells.

EXAMPLE Type A3 in section (small cells):

DDDDDDDDDDDDDDDDD

zzzzzzzzzzzzzzzzz

α1 + α2

��α2 + α3

6666666666666666

α1 + α2 + α3

VVVVVVVVVVVVVVVVVVVVV

hhhhhhhhhhhhhhhhhhhhh

nnnnnnnnnnnn

PPPPPPPPPPPP

α1 α3

We have 8 small cells.

EXAMPLE 2 in section (big cells):

6666666666666666666666666666666

��

α1 + α2 + α3

SSSSSSSSSSSSSSS

kkkkkkkkkkkkkkk

α1 α3

We have 3 big cells.

EXAMPLE 2 in section (small cells):

6666666666666666666666666666666

��

α1 + α2 + α3

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmα3

QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ

We have 6 small cells.

Geometry of polyhedra

We start with a general remark.Let be given m points Pi in n dimensional space and X betheir convex envelop. We assume that the points are notcontained in any proper linear subspace.

If we choose in any possible way k ≤ n among these pointsthat are independent they generate a simplex of dimensionk − 1, let Y be the union of all these simplices.Y is a closed subset of X of dimension n − 1.The connected components of X − Y are called big cellsassociated to the points.The small cells, which we will not use, are obtained byremoving to X all the points laying on the subspacesgenerated by such sets of points.

PROJECTING A POLYHEDRON TO FORM A CONE

Same geometry for the cone C(A).

Let be given a list A of vectors ai in n dimensional space, thatgenerate a pointed cone C(A).

Let us take an hyperplane that intersects the half lines R+aiin the points Pi (there exists one by hypothesis).The cone C(A) is the projection from the origin of the convexenvelop X of the points Pi

Every simplex of dimension ≤ n − 2 projects to a simplicialcone of dimension ≤ n − 1.The union of such cones projects Y and the big cells of C(A)are the projections of the big cells of X minus the origin.

Regular pointsThe points of the big cells are also called regular, the otherssingular. X ,Y represent C(A) and its singular points insection.

SPLINES TA(x) =√detAAt−1VA(x)

The volume of the variable polytope VA(x) equals, up to theconstant

√detAAt to the

multivariate splinethat is the function TA(x) characterized by the formula:∫

Rnf (x)TA(x)dx =

f (m∑

i=1tiai )dt,

where f (x) is any continuous function with compact support.

√detAAt to the

Rnf (x)TA(x)dx =

f (m∑

i=1tiai )dt,

√detAAt to the

Rnf (x)TA(x)dx =

f (m∑

i=1tiai )dt,

WARNING

We have given a weak definition for TA(x)

In generalTA(x) is a tempered distribution, supported on the cone C(A)!

Only when A has maximal rank TA(x) is a function.

WARNING

Assume A spans Rn.

Let h be the minimum number of columns that one can removefrom A so that the remaining columns do not span Rn.

The basic function TA(x) is a spline1 TA(x) has support on the cone C(A)

2 TA(x) is continuous if h ≥ 23 In general TA(x) is of class h − 24 TA(x) coincides with a homogeneous polynomial of degree

m − n on each big cell.

Assume A spans Rn.

SIMPLE EXAMPLE

∣∣∣∣∣1 0 10 1 1

∣∣∣∣∣

OO ??~~~~~~~

The three vectors

C(A) is the

first quadrant

The function TA(x , y) is

of class C0

the associated cone C(A) has two big cells

OO ??��

��

TA is 0 outside the first

quadrant and on the two cells:0

OO ??��

��

TA is 0 outside the first quadrant and on the two cells:0

OO ??��

��

SUMMARIZING

In order to compute TA(x) we need to1 Determine the decomposition of C(A) into cells2 Compute on each big cell the homogeneous polynomial of

degree m − n coinciding with TA(x) .

SUMMARIZING

BOX SPLINES

While the function TA(x) is the basic object, the more interestingobject for numerical analysis is the

box splinethat is the function BA(x) characterized by the formula:∫

Rnf (x)BA(x)dx =

∫[0,1]m

f (m∑

i=1tiai )dt,

where f (x) is any continuous function.

BOX SPLINES

Rnf (x)BA(x)dx =

∫[0,1]m

f (m∑

i=1tiai )dt,

BOX SPLINES

Rnf (x)BA(x)dx =

∫[0,1]m

f (m∑

i=1tiai )dt,

THE BOX

The box spline BA(x) is supported in the compact polytope:

The Box B(A)

that is the compact convex polytope

B(A) := {m∑

i=1tiai}, 0 ≤ ti ≤ 1, ∀i .

THE BOX

The box spline BA(x) is supported in the compact polytope:

The Box B(A)

that is the compact convex polytope

B(A) := {m∑

i=1tiai}, 0 ≤ ti ≤ 1, ∀i .

The box B(A) has a nice combinatorial structure, and can bepaved by a set of parallelepipeds indexed by:all the bases which one can extract from A!

ExampleIn the next example

∣∣∣∣∣1 0 1 −1 2 10 1 1 1 1 2

∣∣∣∣∣we have 15 bases and 15 parallelograms.

The box B(A) has a nice combinatorial structure, and can bepaved by a set of parallelepipeds indexed by:all the bases which one can extract from A!

ExampleIn the next example

∣∣∣∣∣1 0 1 −1 2 10 1 1 1 1 2

∣∣∣∣∣we have 15 bases and 15 parallelograms.

EXAMPLE paving the box

∣∣∣∣∣1 0 1 −1 2 10 1 1 1 1 2

∣∣∣∣∣

∣∣∣∣∣1 0 1 −1 2 10 1 1 1 1 2

∣∣∣∣∣START WITH

∣∣∣∣∣1 00 1

∣∣∣∣∣

∣∣∣∣∣1 0 10 1 1

∣∣∣∣∣

��

∣∣∣∣∣1 0 1 −10 1 1 1

∣∣∣∣∣

��

???????

??????????????

��

��???????

��

∣∣∣∣∣1 0 1 −1 20 1 1 1 1

∣∣∣∣∣

???????ooooooooooooo

ooooooooooooo��

???????

ooooooooooooo???????

��

ooooooooooooo

��???????

��

ooooooooooooo

• •

@@@@@@@

oooooooooooooo •

~~~~~~~•

��•

@@@@@@@

��•

��

oooooooooooooo •

��

~~~~~~~•

@@@@@@@•

@@@@@@@

oooooooooooooo •

��

@@@@@@@

~~~~~~~•

oooooooooooooo •

~~~~~~~

@@@@@@@@•

~~~~~~~

oooooooooooooo

∣∣∣∣∣1 0 1 −1 2 10 1 1 1 1 2

∣∣∣∣∣Claudio Procesi. Topics in Hyperplane Arrangements, Polytopes and Box-Splines

two box splines of class C 0, h = 2 and C 1, h = 3

∣∣∣∣∣1 0 10 1 1

∣∣∣∣∣ Hat functionor Courant element

∣∣∣∣∣1 0 1 −10 1 1 1

∣∣∣∣∣Zwart-Powell element

EXAMPLE OF A two dimensional box spline

Non continuous, A =

∣∣∣∣∣1 2 1 1 10 0 0 0 2

∣∣∣∣∣ (h = 1)

3 reasons WHY the BOX SPLINE ?

BA(x)dx = 1

recursive definition

B[A,v ](x) =

0BA(x − tv)dt

in the case of integral vectors, we have

PARTITION OF UNITYThe translates BA(x − λ), λ runs over the integral vectors form a

partition of 1.

BA(x)dx = 1

B[A,v ](x) =

0BA(x − tv)dt

partition of 1.

BA(x)dx = 1

B[A,v ](x) =

0BA(x − tv)dt

partition of 1.

BA(x)dx = 1

B[A,v ](x) =

0BA(x − tv)dt

partition of 1.

TRIVIAL EXAMPLE

A = {1, 1} we have for the box spline

???????��

Now let us add to it its translates!

???????��

???????

???????��

��

???????��

???????

???????��

��

???????��

???????

???????��

��

???????��

???????

???????��

��

???????��

???????

???????��

��

???????��

???????

???????��

��

From TA to BA

For a given subset S of A define aS :=∑

a∈S a

the basic formula is:

BA(x) =∑S⊂A

(−1)|S|TA(x − aS).

So TA is the fundamental object.

Notice that the local pieces of BA are no more homogeneouspolynomials.

From TA to BA

a∈S a

BA(x) =∑S⊂A

(−1)|S|TA(x − aS).

From TA to BA

a∈S a

BA(x) =∑S⊂A

(−1)|S|TA(x − aS).

EXAMPLE OF THE HAT FUNCTION

∣∣∣∣∣1 0 10 1 1

∣∣∣∣∣From the function TA

��

EXAMPLE OF THE HAT FUNCTIONwe get the box spline hat function summing over the 6translates of TA

Here comes the algebra

How to compute TA? or the partition function PA?We use the Laplace transform which will change the analyticproblem to one in

algebra

LAPLACE TRANSFORM from Rs = V to U = V ∗.

Lf (u) :=

∫Ve−〈u | v〉f (v)dv .

basic propertiesp ∈ U,w ∈ V , write p, Dw for the linear function 〈p | v〉 and thedirectional derivative on V

L(Dw f )(u) = wLf (u), L(pf )(u) = −DpLf (u),

L(epf )(u) = Lf (u − p), L(f (v + w))(u) = ewLf (u).

Lf (u) :=

Basic transformation rules

Consider any a ∈ V as a LINEAR FUNCTION ON U we have:

An easy computation gives the Laplace transforms:

LBA =∏a∈A

1− e−a

aand LTA =

∏a∈A

LBA =∏a∈A

1− e−a

aand LTA =

∏a∈A

LBA =∏a∈A

1− e−a

aand LTA =

∏a∈A

BASIC NON COMMUTATIVE ALGEBRAS

Algebraic Fourier transform

Weyl algebrasSet W (V ),W (U) be the two Weyl algebras of differentialoperators with polynomial coefficients on V and U.

Fourier transformThere is an algebraic Fourier isomorphism between them, so anyW (V ) module M becomes a W (U) module M

D−modules in Fourier duality:

Two modules Fourier isomorphic

1. The D−module DA := W (V )TA generated, in the space oftempered distributions, by TA under the action of the algebraW (V ) of differential operators on V with polynomial coefficients.

2. The algebra RA := S[V ][∏

a∈A a−1] obtained from thepolynomials on U by inverting the element dA :=

∏a∈A a

RA = W (U)d−1A

1. The D−module DA := W (V )TA generated, in the space oftempered distributions, by TA under the action of the algebraW (V ) of differential operators on V with polynomial coefficients.

2. The algebra RA := S[V ][∏

a∈A a−1] obtained from thepolynomials on U by inverting the element dA :=

∏a∈A a

RA = W (U)d−1A

TheoremRA = W (U)d−1

AUnder Laplace transform, L(TA) = d−1

A andDA is mapped isomorphically onto RA as W (U)-modules.

DA is the space of tempered distributions which are linearcombinations of polynomial functions on the cones C(B), B ⊂ A alinearly independent subset and their distributional derivatives.

RA and it is the coordinate ring of the open set AAcomplement of the union of the hyperplanes of Uof equations a = 0, a ∈ A.

It is a cyclic module under W (U) generated by d−1A .

PROJECTIVE PICTURE OF THE ARRANGEMENT A3

We have drawn in theprojective plane of 4−tuples ofreal numbers with sum 0, the6 lines

xi − xj = 0, 1 ≤ i < j ≤ 4

the 7 intersection points are also subspaces of the arrangement!

PROJECTIVE PICTURE OF THE ARRANGEMENT A3

We have drawn in theprojective plane of 4−tuples ofreal numbers with sum 0, the6 lines

xi − xj = 0, 1 ≤ i < j ≤ 4

the 7 intersection points are also subspaces of the arrangement!

The module RA

LocalizationIt is well known that once we invert an element in a polynomialalgebra we get a holonomic module over the algebra of differentialoperators.

In particular it has a finitecomposition series and it is cyclic

We want to describe a composition series of RA.

The module RA

The building blocks

For each subspace W of U we have an irreducible module NW(generated by the δ function of W ).

The NW as W runs over the subspaces of the hyperplanearrangement given by the equations ai = 0, ai ∈ A are thecomposition factors of RA.

Take coordinates x1, . . . , xn

W = {x1 = x2 = . . . = xk = 0}

NW is generated by an element δ satisfying:

xiδ = 0, i ≤ k, ∂

∂xiδ, i > k.

It is free generated by δ over:

C[x1, x2, . . . , xk ,∂

∂xk+1, . . . ,

∂xn].

Take coordinates x1, . . . , xn

W = {x1 = x2 = . . . = xk = 0}

NW is generated by an element δ satisfying:

xiδ = 0, i ≤ k, ∂

∂xiδ, i > k.

It is free generated by δ over:

C[x1, x2, . . . , xk ,∂

∂xk+1, . . . ,

∂xn].

The filtration of RA by polar order

DefinitionIntroduce a filtration by W (U)−submodules:

filtration degree ≤ kRA,k is the span of all the fractions f

∏a∈A a−ha , ha ≥ 0

for which the set of vectors a, with ha > 0, spans a space ofdimension ≤ k.

We have RA,s = RA.

DefinitionIntroduce a filtration by W (U)−submodules:

filtration degree ≤ kRA,k is the span of all the fractions f

∏a∈A a−ha , ha ≥ 0

for which the set of vectors a, with ha > 0, spans a space ofdimension ≤ k.

We have RA,s = RA.

RA,k/RA,k−1 is semisimple.Its isotypic components are of type NW as W runs over thesubspaces of the arrangement of codimension k.In particular RA,s/RA,s−1 is a direct sum of modules N0.The annihilator of the cyclic generator of N0 is generated byall the variables, i.e. by all the elements ai ∈ A.N0 is a free rank 1 module over the commutative polynomialring S[U] of differential operators with constant coefficients.

A basis for RA,s/RA,s−1

By the previous theorem RA,s/RA,s−1 is a free module over

S[U] = C[∂

∂x1, . . . ,

∂xs]

(in coordinates).It is important to choose a basis.

No broken circuits

A basis comes from the theory of matroids.Let c := ai1 , . . . , aik ∈ A, i1 < i2 · · · < ik , be a sublist of linearlyindependent elements.

DefinitionWe say that ai breaks c if there is an index 1 ≤ e ≤ k suchthat:

i ≤ ie .ai is linearly dependent on aie , . . . , aik .

No broken circuits

A basis comes from the theory of matroids.Let c := ai1 , . . . , aik ∈ A, i1 < i2 · · · < ik , be a sublist of linearlyindependent elements.

DefinitionWe say that ai breaks c if there is an index 1 ≤ e ≤ k suchthat:

i ≤ ie .ai is linearly dependent on aie , . . . , aik .

The no broken bases extracted from A

no broken basesFor a basis b := ai1 , . . . , ais extracted from A set:B(b) := {a ∈ A | a breaks b} and n(b) = |B(b)| the cardinalityof B(b).

DefinitionWe say that b is no broken if B(b) = b or n(b) = s.

EXAMPLE of the elements braking a basis

Example A3 ordered as: