Polytopes, rings and K-theory

Transcription

1 Polytopes, rings and K-theory Winfried Bruns Universität Osnabrück Joseph Gubeladze San Francisco State University Preliminary incomplete version July 30, 2007

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3 To our children and grandchildren Annika, Nils, Julia and Linus Nino, Ana, Jimsher, Tamar and Lazare

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5 Contents Part I Cones, monoids, and triangulations 1 Polytopes, cones, and complexes A Polyhedra and their faces B Finite generation of cones C Finite generation of polyhedra D Polyhedral complexes E Subdivisions and triangulations F Regular subdivisions G Rationality and integrality Exercises Notes Affine monoids and their Hilbert bases A Affine monoids B Normal affine monoids C Generating normal affine monoids D Normality and unimodular covering Exercises Notes Multiples of lattice polytopes A Knudsen-Mumford triangulations B Unimodular triangulations of multiples of polytopes C Unimodular covers of multiples of polytopes Notes

6 vi Contents Part II Affine monoid algebras 4 Monoid algebras A Graded rings B Monoid algebras C Representations of monoid algebras D Monomial prime and radical ideals E Normality F Divisorial ideals and the class group G The Picard group and seminormality Exercises Isomorphisms and Automorphisms A Linear algebraic groups B Invariants of diagonalizable groups C The isomorphism theorem D Automorphisms Exercises Homological properties and Hilbert functions A Hochster s theorem B Graded homological algebra C The canonical module D Hilbert functions E Applications to enumerative combinatorics F Divisorial linear algebra Exercises Gröbner bases, triangulations, and Koszul algebras A Gröbner bases and initial ideals B Initial ideals of toric ideals C Toric ideals and triangulations D Multiples of lattice polytopes Exercises Part III K-theory of monoid algebras 8 Projective modules over monoid rings A Projective modules B The main theorem and the plan of the proof C Projective modules over polynomial rings D Reduction to the interior E Graded Weierstraß Preparation

7 Contents vii 8.F Pyramidal descent G How to shrink a polytope H Converse results I Generalizations Exercises Bass-Whitehead groups of monoid rings A The functors K 1 and K B The nontriviality of SK 1.RŒM / C Further results: a survey Exercises Varieties A Vector bundles, coherent sheaves and Grothendieck groups B Intersection theory and Grothendieck-Riemann-Roch C Toric varieties D Intersection theory on toric varieties E The equivariant Serre Problem for abelian groups F Toric varieties with huge Grothendieck group References Notation Index

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9 Preface Not yet written.

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11 Part I Cones, monoids, and triangulations

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13 1 Polytopes, cones, and complexes The algebraic objects discussed in this book are determined by two types of structures: continuous ones related to convexity, and discrete ones determined by lattices. In this chapter we develop notions of convex geometry and of combinatorial topology related to convexity. The basic convex objects are polyhedra, polytopes and cones, and the related combinatorial constructions are polyhedral complexes, triangulations and fans. In the last section we unite convexity and lattice structures. 1.A Polyhedra and their faces We use the symbols Z, Q, R and C to denote the integral, rational, real and complex numbers respectively. The subsets of nonnegative elements in Z, Q, R will be referred to by Z C, Q C, R C. The symbol N stands for the set of positive integers. Polyhedra are subsets of finite dimensional R-vector spaced V, but to some extent only the affine structure of V is essential. We recall that an affine subspace of V (of dimension d) is a subset of the form v C U for some v 2 V and a vector subspace U (of dimension d); the empty set is also an affine subspace. Equivalently, an affine subspace is a subset closed under all affine linear combinations nx a i x i ; a i 2 R; id1 nx a i D 1 (Exercise 1.1). For a subset X of V we let aff.x/ denote the affine hull of X, i. e. the smallest affine subspace of V containing X. For x 0 ; : : : ; x n 2 V one has dim aff.x 0 ; : : : ; x n / n. If dim aff.x 0 ; : : : ; x n / D n, then x 0 ; : : : ; x n are affinely independent; in other words, if x D a 0 x 0 C C a n x n is an affine combination of x 0 ; : : : ; x n, then a 0 ; : : : ; a n are uniquely determined. They are called the barycentric coordinates of x with respect to x 0 ; : : : ; x n. Affine subspaces v C U and v 0 C U 0 are parallel if U U 0 or U 0 U. Subsets of V are parallel if their affine hulls are parallel. id1

14 4 1 Polytopes, cones, and complexes A subset X of V is convex if it contains the line segment Œx; y D fax C.1 a/y W 0 a 1g for all points x; y 2 X. (Open or halfopen line segments.x; y/ and.x; y are defined analogously.) The intersection of convex sets is obviously convex. Therefore one may define the convex hull conv.x/ of an arbitrary subset X of V to be the smallest convex subset of V containing X. If X D fx 1 ; : : : ; x n g is finite, then nx conv.x/ D a i x i W 0 a i 1; i D 1; : : : ; n; id1 nx id1 a i D 1 ; and for arbitrary X the convex hull of X is just the union of the convex hulls of the finite subsets of X. Linear combinations as in the formula for conv.x/ are called convex linear combinations. The functions considered in affine geometry are the affine forms W V! R. An affine form is given by.x/ D.x/ C a 0 ; a 0 D.0/; with a unique linear form on V. The space of linear forms on V is denoted by V. More generally, a map f from V into an R-vector space W is affine if f.v/ D g.v/ C w 0 for some linear map g W V! W and some w o 2 W. Affine forms are continuous with respect to the natural topology on V Š R d. So the open halfspaces H > D fx 2 V W.x/ > 0g; associated with the nonconstant forms are open sets. With such a form we associate also the hyperplane and the closed halfspace For flexibility of notation we set H D fx 2 V W.x/ D 0g H C D fx 2 V W.x/ 0g: H < D H > and H D H C : Every hyperplane H in V bounds exactly two open halfspaces that we denote by H > and H < (depending on the choice of with H D H ). The corresponding closed halfspaces are denoted H C and H. With respect to the standard topology we use the topological notions of the interior int.x/ and the However, in dealing with polyhedra we will usually have to consider the relative interior and the relative boundary of P that are taken within the affine subspace aff.x/. For simplicity of notation we therefore use

15 1.A Polyhedra and their faces 5 int.x/ to denote the relative interior and relative boundary of X. Consequently we drop the attribute relative from now on. The closure of X is denoted by X. (There is no need for a relative closure.) Note that int.x/ D X if X D fxg consists of a single point. Every neighborhood of 0 contains a basis of the vector space V. It follows that V is the affine hull of each of its open subsets. The interior of a convex set in V is also convex: If x; y 2 int.x/, then there exists an open neighborhood U of 0 such that xcu; y CU X, and for z 2 Œx; y the neighborhood z C U is contained in X. We now introduce the main objects of this section. Definition 1.1. A subset P V is called a polyhedron if it is the intersection of finitely many closed halfspaces. The dimension of P is given by dim aff.p /. A d- polyhedron has dimension d. A polytope is a bounded polyhedron. A 2-polytope is called a polygon. A morphism of polyhedra P and Q is a map ' W P! Q that can be extended to an affine map Q' W aff.p /! aff.q/. Fig A polyhedron and a polytope Equivalently, P is a polyhedron if it is the set of solutions of a linear system of inequalities. Since halfspaces (open or closed) are convex, polyhedra are convex subsets of V, and the interior of a polyhedron is again convex (since the interior of a convex set is convex, as observed above). It will be shown in Section 1.C that a subset P of V is a polytope if and only if it is the convex hull of a finite subset of V. In Proposition 1.29 we will see that '.P / is a polyhedron if ' is a morphism of polyhedra. For simplicity we will always consider a morphism ' W P! Q to be defined on aff.p / so that we need not distinguish ' and Q'. It is convenient to always orient the halfspaces cutting out a polyhedron P in such a way that P D H C 1 \ \H C m. Often we simply write H C i for H C i, assuming that a suitable affine form i has been chosen to define the halfspace. From the representation of P as an intersection of halfspaces we can easily determine the affine hull aff.p /. Proposition 1.2. Let P D H C 1 \ \ H m C be a polyhedron. Then aff.p / is the intersection of those hyperplanes H i, i D 1; : : : ; m, that contain P. Proof. Let A be the intersection of the hyperplanes H i containing P. Then P is contained in the affine space A. Replacing V by A, we must show that aff.p / D V

16 6 1 Polytopes, cones, and complexes if P 6 H i for i D 1; : : : ; m. In this case P contains a point x i 2 H i > for i D 1; : : : ; m. It follows that 1 m.x 1 C C x m / belongs to H i > for i D 1; : : : ; m. Therefore P H 1 > \ \ H m > ; contains a nonempty open subset of V, and V D aff.p /. ut Support hyperplanes, facets and faces. If we visualize the notion of polytope in R 3, then we see a solid body whose boundary is composed of polygons. These facets (unless they are affine subspaces) are bounded by line segments, called edges, and each edge has two endpoints. All these endpoints form the vertices of P. It is our first goal to describe the face structure for polyhedra of arbitrary dimension. Definition 1.3. A hyperplane H is called a support hyperplane of the polyhedron P if P is contained in one of the two closed halfspaces bounded by H and H \P ;. The intersection F D H \P is a face of P, and H is called a support hyperplane associated with F. A facet of P is a face of dimension dim P 1. The polyhedron P itself and ; are the improper faces of P. Faces of dimension 0 are called vertices, and vert.p / is the set of vertices of P. An edge is a face of dimension 1. Below we will often use that the faces of P only depend on P in the following sense: if V 0 V is an affine subspace and P V 0, then the faces of P in V are its faces in V 0, and conversely. That a face in V is also a face in V 0 is trivial, and the converse follows immediately from the fact that a halfspace G C associated with a hyperplane G in V 0 can be extended to a halfspace H C of V such that G C D H C \ V 0, as observed above. Another easy observation: if F is a face of P, then F D aff.f / \ P. In fact, aff.f / is contained in each support hyperplane associated with F. Proposition 1.4. Let P V be a polyhedron, and H a hyperplane such that H \ P ;. Then H is a support hyperplane associated with a proper face of P if and only if H \ Proof. We may assume that V D aff.p /. Suppose first that H is a support hyperplane. If H \int.p / ;, then both inclusions P H C or P H are impossible since none of the closed halfspaces contains a neighborhood of x 2 H \ int.p /. Thus H \ As to the converse implication, if H intersects P, then one of the open halfspaces, say H >, must contain an interior point x of P since int.p / H is impossible. If H is not a support hyperplane, then the other open halfspace H < must also contain a point y 2 P. But all points on the half open line segment Œx; y/ belong to int.p /, as is easily checked. Exactly one of these points is in H so that H \ int.p / ;. ut

17 In the representation P D H C 1 \ \ H C n 1.A Polyhedra and their faces 7 the halfspace H C i can be omitted if it contains the intersection of the remaining halfspaces. Therefore P has an irredundant representation as an intersection of halfspaces. It is evident from two-dimensional polyhedra in R 3 that we can achieve the uniqueness of these halfspaces only if dim P D dim V. But this is not an essential restriction, since we can always consider P as a polyhedron in aff.p /. First a lemma: Lemma 1.5. Suppose P D H C 1 \ \ H n is contained in H i for some i. is a polyhedron. Then a convex set X Proof. On the contrary we assume that X 6 H i for all i and choose x i 2 X n H i. Then x D 1 n.x 1 C C x n / 2 X P; but x 2 H i > for all i (as one sees immediately upon choosing an affine form defining H C i ). The intersection H 1 > \ \ H n > is open and contained in P. Thus x 2 int.p /, a contradiction. ut Theorem 1.6. Let P V be a polyhedron such that d D dim P D dim V. Then the halfspaces H C 1 ; : : : ; H C n in an irredundant representation P D H C 1 \ \HC n are uniquely determined. In fact, the sets F i D P \ H i, i D 1; : : : ; n, are the facets of P. Proof. We choose an irredundant representation P D H C 1 \ \ H n C. By Proposition 1.4 and Lemma 1.5, every facet F of P is contained in one of the sets F i. Let H be the support hyperplane defining F. Then H D aff.p \ H/ D aff.f / since F is a facet. But H is contained in H i, and so H D H i and F D P \ H D P \ H i D F i. It remains to show that each F i is a facet of P. It is enough to consider F n. Let P 0 D H C 1 \ \ H C n 1 and set H D H n. Then H \ P 0 ;, but H is not a support hyperplane of P 0. By Proposition 1.4 H intersects the interior of P 0. Therefore H \ P 0 D H \ P contains a nonempty subset X that is open in H. But then aff.p \ H/ aff.x/ D H, and so dim P \ H D dim H D dim P 1. ut The face structure of a polyhedron. We note an immediate corollary of Theorem 1.6: Corollary 1.7. Let P be a polyhedron. (a) is the union of the facets of P. (b) Each proper face of P is contained in a facet. Part (b) of the corollary is very useful for inductive arguments, provided the relation F is a face of P is transitive, and this is indeed true.

18 8 1 Polytopes, cones, and complexes Proposition 1.8. Let F be a face of the polyhedron P and G F. Then G is a face of P if and only if it is a face of F. Proof. Evidently a face G of P is a face of every face F G. For the converse implication there is nothing to show if F D P, and this includes the case in which dim P D 0. Suppose dim P > 0. Then the proper face F is contained in a facet F 0, as we have seen in the corollary above, and, by induction, G is a face of F 0. Therefore we may assume that F itself is a facet. For simplicity we may further assume that V D aff.p /. Let H 0 be a support hyperplane in aff.f / for the face G of F, and H the (uniquely determined) support hyperplane of P through F. There exists a point x 2 H < that is contained in the interior of the polyhedron P 0 defined as the intersection of the halfspaces associated with the remaining facets of P. Set HQ D aff.h 0 [ fxg/. Then HQ is a hyperplane in V. We claim that it intersects P exactly in G. P 0 x HQ G P F Fig The construction of N H Assume that y 2 P \ QH, y G. Then y F, since H Q \ F D G. The line segment Œy; x is not contained in P, and it can leave P only through F. Since Œy; x H, Q it intersects F in a point z 2 G. On the other hand, all points on.y; x belong to the interior of P 0, since y 2 P 0 and x 2 int.p 0 /. Thus z 2 int.f / D H \ int.p 0 /. This is a contradiction. ut As we have seen, the maximal elements (with respect to inclusion) among the proper faces are the facets. Now we turn to the minimal faces: Proposition 1.9. Let P be a polyhedron and let U be the vector subspace given by the intersection of the hyperplanes through 0 parallel to the facets of P. (a) The sets x C U, x 2 P, are the maximal affine spaces contained in P. (b) Every minimal face F of P is a maximal affine subspace of P. (c) In particular, if P has a bounded face, then its minimal nonempty faces are its vertices. Proof. Clearly x CU P since the affine forms defining P are constant on x CU. On the other hand, if A is an affine subspace contained in P, then these forms must be constant on A, and so A x C U for x 2 A. This proves (a). For (b) there is nothing to prove if P has no proper face. Otherwise every minimal face of P is contained in a facet, and it follows by induction on dim P that the

19 1.A Polyhedra and their faces 9 minimal faces F are affine subspaces of V. So F x C U P for every x 2 F. If H is a support hyperplane associated with F, then U C x F, as is easily seen (for an arbitrary face F ). (c) follows immediately from (b). ut We can now prove several important properties of the set of faces of P : Theorem Let P be a polyhedron and A a maximal affine subspace of P. Then the following hold: (a) The intersection F \ G of faces F; G is a face of F and G. (b) Every face F is the intersection of facets. In particular, the number of faces is finite. (c) Let F 0 F m 1 F m D P be a strictly ascending maximal chain of nonempty faces of P. Then dim F 0 D dim A and dim F ic1 D dim F i C 1 for all i D 0; : : : ; m 1. (d) For each x 2 P there exists exactly one face F such that x 2 int.f /. It is the unique minimal element in the set of faces of P containing x. Proof. (a) If F \ G D ;, then F \ G is a face by definition. Otherwise, a support hyperplane of P associated with the face F is also a support hyperplane of the polyhedron G that intersects G exactly in F \ G. Thus F \ G is a face of G, and therefore of P. (b) If F is a facet, then it is certainly an intersection of facets. Otherwise F is strictly contained in a facet F 0, and, by induction, the intersection of facets of F 0. But the facets of F 0 arise as intersections of facets of P with F 0, as follows immediately from Theorem 1.6. (c) Follows immediately by induction from Corollary 1.7 and Proposition 1.9. (d) By (a) and (b) the intersection of all faces of P containing x is a face F of P. By Corollary 1.7 x 2 int.f /. Every other face G of P with x 2 G has F as a proper face, and so x ut It follows from Theorem 1.10 that for all faces F; G of P there are a unique maximal face contained in F \G (namely F \G itself) and a unique minimal face containing F [G. Therefore the set of faces, partially ordered by inclusion, forms a lattice, the face lattice of P. (A partially ordered set M is called a lattice if each two elements x; y 2 M have a unique infimum and a unique supremum. Most often we will use the term lattice in a completely different meaning; see Section 1.G.) In general a polyhedron need not have a bounded face, and as we see from Proposition 1.9, it has a bounded face if and only if its minimal nonempty faces are vertices. As the next proposition shows, the structure of polyhedra in general is essentially determined by those with vertices. For each affine subspace A of V we can find a complementary subspace A 0 characterized by the conditions that A \ A 0 consists of a single point x 0 and V D A C.A 0 x 0 / D.A x 0 / CA 0. The map A A 0! V,.x; x 0 / 7! x Cx 0 x 0, is then a bijection, and we may write V D A A 0.

20 10 1 Polytopes, cones, and complexes Proposition Let P V be a nonempty polyhedron, A a maximal affine subspace of P, A 0 a complementary subspace, and P 0 D A 0 \ P. Then P D A P 0 under the identification V D A A 0, and P 0 has vertices. Proof. Let x 0 be the intersection point of A and A 0. We can assume that x 0 D 0. For a point y D x C x 0 2 P, x 2 A, x 0 2 A 0, one has y x 2 P, since y C A P for all y 2 P. But y x D x 0 2 P 0. Conversely, if x 0 2 P 0 P, then x 0 C x 2 P for all x 2 A. This shows P D A P 0. Since A B P for each affine subspace B of P 0, the largest affine subspaces contained in P 0 are points. ut There meet exactly two edges in a vertex of a polyhedron in R 2 and two facets in an edge of a 3-dimensional polyhedron in R 3. This fact can be generalized to all dimensions: Proposition Let F be a nonempty face of the polyhedron P. Then dim F D dim P 2 if and only if there exist exactly two facets F 0 and G 0 containing F. In this case one has F D F 0 \ G 0. Proof. If there exist exactly two facets F 0 and G 0 containing F, then dim F D dim P 2 as follows immediately from Theorem 1.10, and clearly F D F 0 \ G 0. For the converse implication we can assume that V D aff.p /. Let dim F D dim P 2, and let F 0 be a facet containing F. Then F is a facet of F 0. By Theorem 1.10 the facets of F 0 are of the form F 0 \ G 0 where G 0 is another facet of P. It is enough to show that G 0 is uniquely determined. Suppose that F D F 0 \ G 00 for a third facet G 00 of P. We choose affine-linear forms ; ˇ; with nonnegative values on P such that F 0 D P \ H, G 0 D P \ Hˇ and G 00 D P \ H. The affine linear forms vanishing on F, and therefore on aff.f /, form a two-dimensional vector subspace of the space of all affine-linear forms. Thus there is a linear relation a C bˇ C c D 0: Since ; ˇ; define pairwise different facets, none of a; b; c can be zero. We can assume that two of a; b; c are positive, say a and b. If c < 0, then H C contains H C \ H Cˇ impossible, since the facets define an irreducible representation of P. But c > 0 is also impossible, since ; ˇ; are simultaneously positive on an interior point of P. ut Finally we want to give another, internal characterization of faces. We call a subset X of P extreme, if it is convex and if it contains every line segment Œx; y for which x; y 2 P and.x; y/ \ X ;. Proposition Let P be a polyhedron. Then the faces of P are exactly its extreme subsets. Proof. Every face is an extreme subset for trivial reasons. Conversely, let X be an extreme subset. If X \ int.p / D ;, then X is contained in a facet by Lemma 1.5,

21 1.B Finite generation of cones 11 and we are done by induction on dim P. Otherwise X contains an interior point x of P. Let y 2 P, x y. Then the line through x and y contains a point z 2 P such that x is strictly between y and z. Hence Œy; z X, and in particular y 2 X which shows X D P. ut The notion of face and extreme subset make sense for arbitrary convex sets. However, while every face of a convex set is an extreme subset, the converse is false in general (the reader should find an example). Convex and strictly convex functions. Let X V be a convex set. (As above, V is a finite dimensional R-vector space.) Then a function f W X! R is convex if f.txc.1 t/y/ tf.x/c.1 t/f.y/ for all x; y 2 X and t 2 Œ0; 1. In other words: f is convex if the graph of f jœx; y is below the line segment.x; f.x//;.y; f.y// V R. (We use j to denote the restriction of functions or mappings to subsets.) One can immediately prove Jensen s inequality: f.t 1 x 1 C C t n x n / t 1 f.x 1 / C C t n f.x n / for all convex combinations of points in X. The function f is strictly convex if f.tx C.1 t/y/ < tf.x/ C.1 t/f.y/ for all x; y 2 X, x y, and t 2.0; 1/. Concave and strictly concave functions are defined in the same manner, the inequalities being replaced by the opposite ones. 1.B Finite generation of cones We have introduced polyhedra as the set of solutions of finite linear systems of inequalities. In this subsection we study the special case in which the inequalities are homogeneous, or, equivalently, the affine forms are linear forms on the finitedimensional vector space V (over the field R, just as in Section 1.A). The main advantage of the linear case is the duality between a vector space V and the space V D Hom R.V; R/, for which there is no affine analogue. Definition A cone in V is the intersection of finitely many linear closed halfspaces. (We say that a halfspace is linear if it is defined by a linear form on V.) A morphism of cones C and C 0 is a map ' W C! C 0 that extends to a linear map Q' W RC! RC 0. A subset X of V is conical if it is closed under nonnegative linear combinations: a 1 x 1 C C a n x n 2 C for all x 1 ; : : : ; x n 2 C and all a 1 ; : : : ; a n 2 R C, n 2 Z C. A cone C is evidently conical. Often this property is used to define the notion of cone. However, all our conical sets are polyhedra, and therefore the term cone will always include the attribute polyhedral. The faces of a cone are themselves cones since each support hyperplane H of a cone C must contain 0: the ray R C x, x 2 C \ H, x 0, cannot leave C, and 0 2 H > is impossible. The intersection of conical sets is certainly conical. Therefore we may define the conical hull R C X of a subset X of V as the smallest conical set containing X. In analogy with the convex hull, one has

22 12 1 Polytopes, cones, and complexes R C X D fa 1 x 1 C C a n x n W n 2 Z C ; x 1 ; : : : ; x n 2 X; a i 2 R C ; i D 1; : : : ; ng if X D fx 1 ; : : : ; x n g is a finite set, and if X is infinite, then R C X is the union of the sets R C X 0, where X 0 X is finite. We say that a conical set is finitely generated if it is the conical hull of a finite set. The central theorem in this subsection shows that finitely generated conical sets are just cones, and conversely. In addition to the external representation of cones as intersections of halfspaces one has an internal description by finitely many elements generating C as a conical set. Theorem Let C V be a conical set. Then the following are equivalent: (a) C is finitely generated; (b) C is a cone. Proof. Suppose that C D R C x 1 C C R C x m. For (a) H) (b) we have to find linear forms 1 ; : : : ; n such that C D H C 1 \ \ H C n. We use induction on m, starting with the trivial case m D 0 for which C D f0g is certainly polyhedral. Suppose that C 0 D R C x 1 C C R C x m 1 D H C ~ 1 \ \ H C ~ r for linear forms ~ 1 ; : : : ; ~ r. The process by which we now construct 1 ; : : : ; n is known as (the dual version of) Fourier-Motzkin elimination. (Instead of solving a system of inequalities we construct such a system for a given set of solutions.) Set x D x m. We may assume that 8 ˆ< D 0 i D 1; : : : ; p; ~ i.x/ > 0 i D p C 1; : : : ; q; ˆ: < 0 i D q C 1; : : : ; r: Set ij D ~ i.x/~ j ~ j.x/~ i ; i D p C 1; : : : ; q; j D q C 1; : : : ; r: Then ij.x u / 0 for u D 1; : : : ; m 1 and ij.x/ D 0. Let f 1 ; : : : ; n g D f~ 1 ; : : : ; ~ q g [ f ij W i D p C 1; : : : ; q; j D q C 1; : : : ; rg: Evidently C D D H C 1 \ \ H C n : In fact, x 1 ; : : : ; x m are contained in D, and so this holds for each of their nonnegative linear combinations. It remains to prove the converse inclusion D C. Choose y 2 D. We must find t 2 R C and z 2 C 0 for which y D tx C z. So we consider the ray y tx, t 2 R C, and have to show that it meets C 0. We have to find the values of t 2 R for which y tx 2 C 0. Equivalently, we have to find out when y tx 2 H~ C i for i D 1; : : : ; r. We distinguish three cases:

23 1.B Finite generation of cones 13 z y x Fig Construction of z (i) i p. Then ~ i.x/ D 0, ~ i 2 f 1 ; : : : ; n g, and so ~ i.y/ 0. Clearly y tx 2 H~ C i for all t 2 R. (ii) p C 1 i q. Then ~ i.x/ > 0, ~ i 2 f 1 ; : : : ; n g, and ~ i.y/ 0. Evidently y tx 2 H~ C i if and only if t ~ i.y/=~ i.x/. (iii)qc1 i r. Then ~ i.x/ < 0, and y tx 2 H~ C i if and only if t ~ i.y/=~ i.x/. Thus the range of values of t 2 R for which y tx 2 C 0 is an intersection of intervals of type. 1; a i, i D p C 1; : : : ; q, and Œb j ; 1/, j D q C 1; : : : ; r. The intersection of all these intervals and R C is nonempty if and only if a i 0 and a i b j for all values of i and j. But a i D ~ i.y/=~ i.x/ 0 since ~ i.y/ 0 and ~ i.x/ > 0, and a i b j, as follows immediately from the inequality ij.y/ 0 (observe that ~ i.x/~ j.x/ < 0). The implication (a) H) (b) has been proved, and we will see below that (b) H) (a) follows from it (and is in fact equivalent to it). ut We are now justified in speaking of the cone C generated by x 1 ; : : : ; x m. Clearly, each face F of C is generated by the intersection F \ fx 1 ; : : : ; x m g. For the efficiency of Fourier Motzkin elimination one should always produce a shortest possible list 1 ; : : : ; n of linear forms in each step. Duality. The concept by which we will complete the proof of Theorem 1.15 is that of duality of cones. For a subset X of V we set X D f 2 V W.x/ 0 for all x 2 Xg; and call (the evidently conical set) X the dual conical set of X. Since dim V < 1, the natural map h W V! V, defined by.h.v//./ D.v/; is an isomorphism of vector spaces. This shows that X is not only conical, but also an intersection of linear halfspaces. (The symbol has two different meanings; applied to V it always denotes the space of linear forms.) Furthermore, the identification V Š V allows us to consider the bidual conical set X as a subset of V, and, by definition, X X : Since X is always conical and an intersection of halfspaces, the inclusion X X is strict in general. Using the implication (a) H) (b) of Theorem 1.15 we now derive the duality theorem for cones:

24 14 1 Polytopes, cones, and complexes Theorem Let C V. Then the following hold: (a) C is the intersection of (possibly infinitely many) halfspaces if and only if C D C. (b) In particular, C D C if C is a cone. (c) Suppose C is a cone, and let 1 ; : : : ; n 2 V. Then C D H C 1 \ \ H C n if and only if C D R C 1 C C R C n. Proof. (a) By definition, C is the intersection of linear halfspaces, namely of the halfspaces H C, 2 C. On the other hand, if C D T 2 H C for a subset of V, then C, and C D \ \ H C D C: 2C H C 2 Together with the inclusion C C, this shows that C D C. (b) follows immediately from (a). (c) Suppose that C D H C 1 \ \H C n V and set C 0 D R C 1 C CR C n. Then C 0 is a finitely generated conical set, C 0 C, and C D.C 0 /. By Theorem 1.15(a) H) (b) C 0 is a cone, and so (b) implies C 0 D.C 0 / D..C 0 / / D C : Conversely, if C D R C 1 C C R C n, then.c / D f 1 ; : : : ; n g, and so C D C D f 1 ; : : : ; n g, as desired. ut As a first application we complete the proof of Theorem Let C be a cone. We have to show that C is finitely generated as a conical set. By Theorem 1.16(c) the dual conical set C is finitely generated. So we can apply Theorem 1.15(a) H) (b) and obtain that C is a cone. It follows that C is finitely generated, and since C D C, we are done. Let F be a face of the cone C. Then F C D f 2 C W.F / D 0g is a conical set. Moreover, if a C b, ; 2 C, a; b > 0, vanishes on F, then ; 2 FC. So F C is an extreme subset of C, and by Proposition 1.13 it is a face of C, the face dual to F (the subscript C is necessary since FC depends on C in an essential way). Theorem Let C be a cone in the vector space V. Then the assignment F 7! F C defines an order reversing bijection of the face lattices of C and C. Moreover, one has dim F C dim F C D dim V. Proof. Let us first show that the assignment is injective. Choose a support hyperplane H of C associated with F and a linear form such that H D H and C H C. Then is contained in G C for a face G of C if and only if G F. Thus FC D G C implies F D G.

25 1.B Finite generation of cones 15 The assignment is evidently order reversing, and if we apply it to C and C D C, then the composition of the two maps is an injective order preserving map of the face lattice of C to itself. But the face lattice is a finite set, and so the composition must be the identity. Since we can exchange the roles of C and C, the first statement of the theorem has been proved. By Theorem 1.10, which describes the chain structure of the face lattice, we must have dim FC D dim G C C 1 if dim G D dim F C 1. Therefore it is enough for the second statement to show dim C C dim CC D dim V. This however is evident: all linear forms vanishing on C belong to C, and a linear form vanishes on C if and only if it vanishes on the vector subspace generated by C. ut Pointed cones. A cone C is a pointed if 0 is a face of C. Equivalently we can require that x 2 C; x 2 C H) x D 0; and every cone decomposes into the direct sum of a pointed one and a vector subspace. This follows immediately from Proposition Let C be a cone and C 0 D fx 2 C W x 2 C g. (a) C 0 is the maximal vector subspace of C and the unique minimal face of C. (b) Let U be a vector subspace complement of C 0. Then C D C 0.C \ U /, and C \ U is a pointed cone. The proposition is just a special case of Proposition By C D C 0 C 0 we indicate that each element of C has a unique decomposition x D x 0 C x 0 with x 0 2 C 0 and x 0 2 C 0. Pointed cones can be characterized in terms of their dual cones: Proposition A cone C in the vector space V is pointed if and only if dim C D dim V. Proof. A vector subspace U of V is contained in C if and only if C is contained in the dual vector space U. ut The proposition shows that the class of full-dimensional pointed cones is closed under duality. Very often we will use that a pointed cone has an essentially unique minimal system of generators: Proposition Let C be a pointed cone, and choose an element x i 0 from each 1-dimensional face R 1 ; : : : ; R n of C. Then x 1 ; : : : ; x n is, up to order and positive scalar factors, the unique minimal system of generators of C. Proof. By Proposition 1.19 one has dim C D dim V D dim V. By Theorem 1.6 C has a unique description as an irredundant intersection of halfspaces H C 1 ; : : : ; H n C. So Theorem 1.16 implies that C has a unique minimal system of generators, up to positive scalar factors, since the elements x i of V D V with Hx C i D H C i are uniquely determined up to positive scalar factors. Moreover, the faces of C that are dual to the facets of C are exactly the 1-dimensional faces of C, and the face dual to C \ H i contains x i. ut

26 16 1 Polytopes, cones, and complexes The 1-dimensional faces of a pointed cone C are called its extreme rays. As a consequence of Proposition 1.20 we obtain that pointed cones have polytopes as cross-sections: Proposition Let C be a pointed cone, 2 int.c /, and a 2 R, a > 0. Then the hyperplane H D fx W.x/ D ag intersects C in a polytope P, and one has R C P D C. Proof. Since 2 int.c /, the hyperplane H 0 D fx W.x/ D 0g intersects C only in 0, as follows from Theorem 1.17 and Proposition Therefore a.x/ 1 x 2 P for all x 2 C, x 0, and C D R C P. It remains to be verified that P is bounded. Let R be an extreme ray of C. It suffices to show that R intersects H in exactly one point. Then Proposition 1.20 implies that P is the convex hull of the finite set of the intersection points of H with the extreme rays of C. Let G be the facet of C dual to R, and U the vector subspace generated by G. Then the linear forms vanishing on R are exactly the elements of U (note that dim U D dim C 1). Since U, the hyperplane H intersects R in exactly one point. ut In the situation of Proposition 1.21 we will say that H defines a cross-section of C. 1.C Finite generation of polyhedra We have introduced polyhedra as the set of solutions of linear systems of inequalities. It is an elementary fact from linear algebra that the set S of solutions of a linear system of equations is an affine subspace of the form S D x C S 0 where S 0 is the vector space of solutions of the associated homogeneous system of equations. As we will see, polyhedra have a similar description that in fact characterizes them. Recession cone and projectivization. In the same way as one associates a homogeneous system of equations with an inhomogeneous one, we associate a cone with a polyhedron: Definition Let P D H C 1 \ \H n C of P is rec.p / D HQ C 1 be a polyhedron. Then the recession cone \ \ Q H C n where HQ C i is the vector halfspace parallel to the affine halfspace H C i, i D 1; : : : ; n (so HQ C i D H C i x for some x 2 H C i ). (Halfspaces are parallel if their bounding hyperplanes are parallel.) It is easy to see and left to the reader that the recession cone can also be described as the set of infinite directions in P :

27 1.C Finite generation of polyhedra 17 x x C rec.p / 0 rec.p / Fig A polyhedron and its recession cone Proposition Let P be a polyhedron, x 2 P, and v 2 V. Then v 2 rec.p / if and only if x C av 2 P for all a 2 R C. We have constructed the recession cone by passing from an inhomogeneous system of linear equations to the associated homogeneous system. There is a second construction leading from an inhomogeneous system to a homogeneous one, namely projectivization (or homogenization): we introduce a new variable and view the constant as its coefficient. In other words, with an affine form D C ˇ; ˇ D.0/; we associate the linear form W V R! R;.v; h/ D.v/ C ˇh: The affine hyperplane and the affine halfspaces associated with are then extended to the linear hyperplane and the linear halfspaces associated with. We indicate the extension by. We have chosen the letter h for the auxiliary variable since we interpret h as the height of the point.v; h/ over v. In the following we have to embed subsets of V into V D V R at a specific height. Thus we set for X V..X; b/ D X fbg V Definition Let P D H C 1 \ \ H n C the cone over P by V be a polyhedron. Then we define C.P / D H C 1 \ \ H C n \ H C where W V! R is given by.v; h/ D h. Clearly.P; 1/ D f.v; h/ 2 C.P / W h D 1g. Thus C.P / is just the cone with cross-section P at height 1, a statement made precise by Proposition Let C be the conical hull of.p; 1/ in V. (a) Then C.P / \.V; 0/ D.rec.P /; 0/. (b) C.P / \.V; a/ D C \.V; a/ for a 0.

28 18 1 Polytopes, cones, and complexes h Fig Cross-sections of C.P / (c) C.P / is the closure of C in V. Proof. (a) is trivial and, moreover, C.P / D C [.rec.p /; 0/, as is easily checked. This proves (b). For (c) it is therefore enough to show that.rec.p /; 0/ is contained in the closure of C. Pick v 2 rec.p / and choose y 2 P. Then y C tv 2 P for all t 2 R C, and W D fu.y C tv; 1/ W t; u 2 R C g C. But W contains a point from every neighborhood of.v; 0/ in V. Therefore.v; 0/ is contained in the closure of W, and, a fortiori, in the closure of C. ut Figure 1.5 illustrates the construction of C.P /, showing the recession cone at h D 0,.P; 1/, and.2p; 2/. The theorems of Minkowski and Motzkin. According to Theorem 1.15 C.P / is finitely generated, say C.P / D R C y 1 C CR C y m. If y i has height h > 0, then we can replace it by h 1 y i. After this operation fy 1 ; : : : ; y m g decomposes into a subset Y 0 of elements of height 0 and a subset Y 1 of elements of height 1. Let x i D.y i / where is the projection V R! V. Choose x 2 P. Then z D.x; 1/ can be written as a linear combination z D X X X a y y C a y y; a y 2 R C ; y 2 Y 0 [ Y 1 ; a y D 1: (1.1) y2y 0 y2y 1 y2y 1 A comparison of the last components shows that indeed P y2y 1 a y D 1. Furthermore.y/ 2 P for y 2 Y 1 and.y/ 2 rec.p / for y 2 Y 0. A first consequence of these observations is Minkowski s theorem: Theorem Let P V. Then the following are equivalent: (a) P is a polytope; (b) P is a polyhedron and P D conv.vert.p //; (c) P is the convex hull of a finite subset of V. Proof. For the implication (a) H) (b) one notes that rec.p / D 0 if P is a polytope. So Proposition 1.25 implies that C.P / is a pointed cone with C.P / \.V; 0/ D 0, and thus we can apply Proposition 1.20: the elements of height 1 in the extreme

29 1.C Finite generation of polyhedra 19 rays of C.P / form a minimal system of generators of C.P /, and they are evidently the vertices of P. We have Y 0 D ; in equation (1.1), and it follows immediately that P D conv.vert.p //. The implication (b) H) (c) is trivial since every polyhedron has only finitely many vertices. As to (c) H) (a), if P is the convex hull of a finite set, then it is certainly bounded, and the conical hull C of.p; 1/ is finitely generated. So Theorem 1.15 implies that C is a cone. But then C \.V; 1/ is a polyhedron that can be identified with P. To sum up: P is a polytope, and the proof of Minkowski s theorem is complete. ut If we give up the restriction rec.p / D 0, then we obtain Motzkin s theorem: Theorem Let P V. Then the following are equivalent: (a) P is a polyhedron; (b) there exist a polytope Q and a cone C such that P D Q C C. Proof. For the implication (a) H) (b) we choose Q D conv..y 1 // and let C be the cone generated by.y 0 /. The notation is as for equation (1.1), and this equation proves the claim. For the converse we choose a finite system X of generators of C. Then the conical hull of.x; 0/ [ vert.q; 1/ is a cone. Its cross-section at height 1 is not only equal to.q C C; 1/, but also a polyhedron. ut Since a polytope is the convex hull of its finite vertex set, and since a cone is finitely generated, Theorem 1.27 allows us to say that a polyhedron is finitely generated. If P has vertices, then we call the union of the bounded faces the bottom of P. (In Figure 1.4 the bottom is indicated by a thick line.)with this notion we can give a more precise statement about the generation of polyhedra. Proposition Let P be a polyhedron, A a maximal affine subspace contained in P, A 0 a complementary affine subspace and P 0 D P \A 0. Furthermore let W V! A 0 denote the parallel projection along A, and B 0 the bottom of P 0. Then P D B 0 C rec.p /. Moreover, if P D B C C for a bounded set B and a cone C, then B 0.B/ and C D rec.p /. Proof. Since P D P 0 C rec.a/ by Proposition 1.11, the equation P D B 0 C rec.p / follows from the analogous equation for P 0. In proving the first statement we may therefore assume that P D P 0 or, in other words, that P has vertices. Let x 2 P and choose y 2 rec.p /. Then x C ty 2 P for all t 2 R C, but the line x C Ry cannot be contained in P ; see Proposition 1.9. So there exists t 0 for which x 0 D x ty belongs to a facet of P. But F has vertices, too, and we can apply induction since the proposition is obviously true in dimension 0: there exists x 00 in the bottom of F and y 0 2 rec.f / such that x 0 D x 00 C y 0. But x 00 belongs to the bottom of P and y 0 2 rec.p /. So x D x 00 C.y 0 C ty/ with y 0 C ty 2 rec.p /. We leave the proof of the second assertion to the reader. ut

30 20 1 Polytopes, cones, and complexes First of all, note that the proof of the theorem contains a new demonstration of the implications (a) H) (b) of Minkowski s and Motzkin s theorems. Second, the proposition has the following interpretation: B 0 is a minimal system of generators of P as a rec.p /-module, and a subset B of P is a minimal system of generators if and only if it differs from B 0 by units of rec.p /, namely elements of the maximal subgroup contained in rec.p /. In particular, if P has vertices, then the bottom of P is the unique minimal system of generators of P over rec.b/. Exercise 1.4 compares the face lattices of P, rec.p /, and C.P /. Minkowski sums and joins. Minkowski s and Motzkin s theorems help us to understand the behavior of polyhedra under affine maps. Proposition Let V; V 0 be vector spaces over R, ' W V! V 0 an affine map, P a polyhedron in V, and P 0 a polyhedron in V 0. (a) Then '.P / and ' 1.P 0 / are polyhedra. (b) '.P / is a polytope if P is a polytope. (c) If ' is linear and P; P 0 are cones, then '.P / and ' 1.P 0 / are cones. Proof. Since ' D C v 0 for a linear map W V! V 0 and some v 0 2 V 0, we may assume that ' is linear. For the preimage we start from a description P D H C 1 \ \H n C. The preimage of a (linear) halfspace under a linear map is a (linear) halfspace, and the preimage of the intersection is the intersection of the preimages. The image '.C / of a finitely generated conical set C V is certainly conical and finitely generated. Therefore, if P is a cone, then so is '.P /. For (b) we argue similarly, using Minkowski s theorem and convex sets. In order to show that the image of a polyhedron is a polyhedron, we write it in the form QCC where Q is a polytope and C is a cone. Then '.P / D '.Q/C'.C / is a polyhedron, as follows from (b), (c) and Motzkin s theorem. ut As a consequence of Proposition 1.29 the Minkowski sum P C P 0 D fx C x 0 W x 2 P; x 0 2 P 0 g of polyhedra (polytopes) is a polyhedron (polytope): Theorem Let P; P 0 be polyhedra in V. (a) Then P C P 0 is a polyhedron. (b) P C P 0 is a polytope if (and only if) P and P 0 are polytopes. (c) If P and P 0 are cones, then P C P 0 is a cone. Proof. Consider P P 0 in V V. Together with P and P 0 it is a polyhedron, polytope or cone, respectively. Now we can apply Proposition 1.29 to the linear map V V 0! V,.v; v 0 / 7! v C v 0. ut

31 1.C Finite generation of polyhedra 21 Another natural construction associated with polytopes P and P 0 is the convex hull Q D conv.p [ P 0 / of their union. Evidently Q is the convex hull of the finite set vert.p / [ vert.p 0 /, and thus it is a polytope by Minkowski s theorem. It is likewise easy to see that D D conv.c [ C 0 / D C C C 0 is a cone if C and C 0 are cones. However, conv.p [ P 0 / need not be a polyhedron if P and P 0 are just polyhedra. For example if P is a line in R 2 and P 0 consists of a single point x not in P, then the smallest polyhedron containing P and P 0 is the strip bounded by P and its parallel through x. Of all the points on the parallel, only x is in conv.p [ P 0 /. But conv.p [ P 0 / may fail to be a polyhedron only because it need not be closed. For this reason we introduce the closed convex hull as an additional construction. conv.x/ D conv.x/ Proposition Let P and P 0 be polyhedra. Then conv.p [ P 0 / is a polyhedron. Proof. We write P D Q C rec.p / and P 0 D Q 0 C rec.q/ with polytopes Q and Q 0. Then R D conv.q [ Q 0 / C rec.p / C rec.q/ is a polyhedron by Motzkin s theorem and it contains conv.p [ P 0 / as is easily verified. On the other hand, rec.t / rec.p / C rec.p 0 / for every polyhedron T P [ P 0, as follows immediately from Proposition Therefore R is the smallest polyhedron containing P [ P 0. Clearly R conv.p [ P 0 /. It remains to show that R conv.p [ P 0 /. Pick s 2 R, s D ax C.1 a/y C w C z; x 2 Q; y 2 Q 0 ; a 2 Œ0; 1 ; w 2 rec.p /; z 2 rec.p 0 /: If a 0; 1, then s 2 conv.p [ P 0 /, and if a D 0, then each neighborhood of s contains a point s 0 D a 0 x C.1 a 0 /y C w C z, a 0 > 0, with s 0 2 conv.p [ P 0 /. For a D 1 one argues similarly. ut We leave it to the reader to find the exact conditions under which conv.p [ P 0 / D conv.p [ P 0 /. As a last construction principle for polyhedra we introduce the join. Roughly speaking, it is the free convex hull that we obtain by considering polyhedra in positions independent of each other. Let P V and Q W be polyhedra. Then we set V 0 D V W R, P 0 D f.x; 0; 0/ W x 2 P g; Q 0 D f.0; y; 1/ W y 2 Qg; and join.p; Q/ D conv.p 0 [ Q 0 /:

32 22 1 Polytopes, cones, and complexes If P and Q are polyhedra in R d such that aff.p / \ aff.q/ D ;, then conv.p [ Q/ will also be called the join of P and Q. (Prove that this polyhedron is isomorphic to conv.p 0 [ Q 0 /.) By Proposition 1.31 the join of P and Q is a polyhedron. If P and Q are polytopes, then join.p; Q/ is a polytope. Separation of polyhedra. A characteristic feature of convexity are separation theorems: convex sets that are disjoint or just touch each other can be separated by a hyperplane. Theorem Let P and Q be polyhedra in the vector space V. Let F be the smallest face of P containing P \Q, and define the face G of Q analogously (F D G D ; if P \ Q D ;). Then there exists a hyperplane H such that P n F H > and Q n G H <. Proof. Replacing P and Q by C D C.P / and D D C.Q/ reduces the theorem to the case of cones P and Q. Assume first that P C. Q/ D V, and choose v 2 int.p /. Then v D v 0 w for some v 0 2 P, w 2 Q. Hence w D v C v 0 2 int.p /, and so F D P. But P C. Q/ D V implies P C Q D V as well, and we conclude that G D Q. In this case there is nothing to prove. It remains the case in which P C. Q/ V. It is a cone by Theorem 1.15, and has a support hyperplane H through its minimal face, namely the vector space of its invertible elements. Suppose that v 2 P \ H. Then v 2 H \.P C. Q//, v D v 0 w 0 with v 0 2 P, w 0 2 Q. It follows that v C v 0 D w 0 2 P \ Q. Hence v C v 0 2 F, and so v 2 F : H intersects P in F, and analogously it intersects Q in G. It follows immediately that H is the desired hyperplane. ut Remark Separation theorems like 1.32 often appear as Farkas lemma (see [187]). A far reaching generalization is the Hahn-Banach separation theorem. The most general version of Minkowski s theorem is the theorem of Krein-Milman (see [108]). 1.D Polyhedral complexes It is evident (in the true sense of the word) that a polygon, i. e. a 2-dimensional polytope can be triangulated, as illustrated by Figure 1.6(a): In higher dimensions triangles are generalized by simplices: Definition A d-polytope with exactly d C 1 vertices is called a d-simplex. In other words, a polytope is a simplex if its vertex set is affinely independent. The conical analogue is a simplicial cone, generated by a linearly independent set of vectors. In the triangulation in Figure 1.6(a) the intersection of two triangles P 1 and P 2 and, more generally, of faces F 1 P 1, F 2 P 2, is a face of P 1 and of P 2. This