Combinatoire Des Polytopes TD B Faces and Operations

Combinatoire des polytopes TD B Faces and operations

1 Faces and f-vectors

Consider a d-dimensional polytope P . Its f-vector is the vector (f0(P ), f1(P ), . . . , fd(P )) where fi(P ) is d the number of -dimensional faces of . Its -polynomial is the polynomial : P i. i P f f(P, x) = i=0 fi(P )x Exercice 1 (f-vectors of the simplex, the cube, and the cross-polytope). What are the f-vectors and f-polynomials of the d-dimensional simplex, cube, and cross-polytope?

Solution. For the simplex 4d, a i-face corresponds to a (i + 1)-subset of the d + 1 vertices, thus d+1 and d+1 . For the cube , a -face corresponds to the choice of fi(4d) = i+1 f(4d, x) = (x+1) −1 /x d i one of the two possible inequalities on coordinates, thus d d−i and d. d − i fi(d) = i 2 f(d, x) = (x + 2) For the cross-polytope d, a i-face corresponds to the choice of one of the two opposite vertices on i + 1 coordinates, thus d i+1 and d d . fi( d3) = i+1 2 f( d, x) = x − 1/x + (2x + 1) /x 3 3 Exercice 2 (f-vectors and polarity). Let P be a d-dimensional polytope and P 4 denote the polar polytope of P . Show that f(P 4, x) = xd − 1/x + xd−1f(P, 1/x).

Solution. For any 0 ≤ i ≤ d − 1, the i-dimensional faces of P 4 correspond to the d − 1 − i dimensional faces of P . Therefore

d d−1 d−1 4 X 4 i d X i d X d−1−j f(P , x) = fi(P )x = x + fd−1−i(P )x = x + fj(P )x i=0 i=0 j=0 d d X d−1−j d d−1 = x − 1/x + fj(P )x = x − 1/x + x f(P, 1/x). j=0

d d−1 As a sanity check, one can verify using Exercice 1 that f(4d, x) = x − 1/x + x f(4d, 1/x) and that d d−1 . f( d, x) = x − 1/x + x f(d, 1/x) 3 Exercice 3 (f-vectors of 3-polytopes). Prove that the f-vectors of 3-dimensional polytopes are precisely the integer vectors (f0, f1, f2, 1) such that

f0 − f1 + f2 = 2 f0 ≤ 2f2 − 4 and f2 ≤ 2f0 − 4.

[Hint: For the dicult direction, compute the f-vector of a pyramid over a p-gon, and study the eect on the f-vector of the two polar operations of simple vertex truncation and simplicial facet stacking, see Exercice 5.] Which polytopes satisfy the rst (resp. second) inequality?

Solution. Consider the f-vector (f0, f1, f2, 1) of a 3-dimensional polytope. It satises f0 − f1 + f2 = 2 by Euler's formula (you can even choose your favorite proof among a list of 20 dierent proofs: https:// www.ics.uci.edu/~eppstein/junkyard/euler/). Moreover, we have 2f1 ≥ 3f0 (as every edge contains 2 vertices while every vertex is contained in at least 3 edges) and 2f1 ≥ 3f2 (as every edge is contained in 2 facets while every facet contains at least 3 edges). Therefore, we obtain using Euler's relation that 4 = 2f0−2f1+2f2 ≤ −f0+2f2 and similarly 4 ≤ 2f0−f2 which proves the two inequalities. (Equivalently, we could also have proved the rst inequality and argued the second one by polarity.)

Conversely, we need to prove that any vector (f0, f1, f2, 1) satisfying these inequalities is the f-vector of a 3-dimensional polytope. For this, we observe the following four facts:

1 • The f-vector of a pyramid over a p-gon is (p + 1, 2p, p + 1, 1). • If v is a simple vertex (contained in 3 edges), then truncating v adds (2, 3, 1, 0) to the f-vector. • If f is a simplicial facet (containing 3 edges), then stacking over f adds (1, 3, 2, 0) to the f-vector. • These two operations suce to obtain all integer vectors (f0, f1, f2, 1) with f0 − f1 + f2 = 2, f0 ≤ 2f2 − 4 and f2 ≤ 2f0 − 4 from the vectors (p + 1, 2p, p + 1, 1) with p ≥ 3. Representing the f-vectors in the (f0, f2) plane, we obtain the following picture:

f2 =2f0 - 4 (simplicial 3-polytopes) f2

f0 =2f2 - 4 (simple 3-polytopes) 4

4 f0

Figure 1: The f-vectors of 3-dimensional polytopes represented in the (f0, f2)-plane. The value of f1 is clearly obtained from Euler's formula f0 − f1 + f2 = 2.

Exercice 4 (Non-unimodality of f-vector). A sequence x0, . . . , xd is unimodal if there exists 0 ≤ i ≤ d such that x0 ≤ x1 ≤ · · · ≤ xi−1 ≤ xi ≥ xi+1 ≥ · · · ≥ xd−1 ≥ xd or the opposite. We say that i is the unimodality peak. (1) Show that the f-vectors of d-polytopes are unimodal for d ≤ 5. (2) Using Exercice 1, show that the f-vectors of simplices, cubes and cross-polytopes are unimodal and study their unimodality peak. (3) Show that there exist simplicial polytopes whose f-vector is not unimodal. [Hint: use iterative stackings on facets of the cross-polytope, see Exercice 5.] (4) ? Are the f-vectors of d-polytopes all unimodal for d = 6, d = 7, d = 8,...? Solution.

(1) The result is clear for d ≤ 4. For d = 5, we have 5f0 ≤ 2f1 and 2f3 ≥ 5f4 so that (f0, f1, f2, f3, f4) is unimodal unless f1 > f2 < f3. Using Euler's relation f0 − f1 + f2 − f3 + f4 = 2, we would obtain 10 = 5f0 − 5f1 + 5f2 − 5f3 + 5f4 ≤ −3f1 + 5f2 − 3f3 < −f2 a contradiction. (2) We compute that

f (4 ) i + 2 f ( ) 2(i + 1) f ( ) i + 2 i d = , i d = , and i d = , fi+1(4d) d − i fi+1(d) d − i fi+13( d) 2(d − i − 1) 3 which are smaller than if and only if d−2 , 2d−4 and 2d−4 respectively, so that the - 1 i ≤ 2 i ≤ 3 i ≤ 3 f vectors of the simplex, cube and cross-polytope are all unimodal with a peaks around d/2, around d/3 and around 2d/3 respectively. (For the cross-polytope, we could equivalently argue by polarity from the cube.) In fact, we can even study the f-functions ΦP (t) := ftd(P ) of the simplex, cube and cross-polytope when d tends to innity. We obtain that

d + 1 1 d d 21−t d φ (t) = ∼ , φ (t) = 2(1−t)d ∼ , 4d td + 1 tt(1 − t)1−t d td tt(1 − t)1−t d + 1 1 d and φ (t) = ∼ , 4d td + 1 tt(1 − t)1−t

2 d d d 1 21−x 2x φ4d (x) ∼ xx(1−x)1−x φd (x) ∼ xx(1−x)1−x φ d (x) ∼ xx(1−x)1−x 3 Figure 2: The -functions , , and for . f φ4d φd φ d d ∈ [10] 3 have a peak at t = 1/2, at t = 1/3, and at t = 2/3 which gets sharper when d increases. These functions are illustrated in Figure 2.

(3) We observe that the unimodality peak of the simplex is at 1/2 while that of the cross-polytope is at 2/3. Therefore, we consider the polytope obtained from a d-dimensional cross-polytope by p iterative stacking operations (see Exercice 5). If d is big enough and p is well-chosen, the f-vector of the resulting polytope will have two peaks around 1/2 and 2/3. To properly choose the values of d and p, we write a small sage program:

sage: def is_unimodal(v): ....: i = 0 ....: while i < len(v)-1 and v[i] <= v[i+1]: ....: i += 1 ....: while i < len(v)-1 and v[i] >= v[i+1]: ....: i += 1 ....: return i == len(v)-1 ....: sage: is_unimodal([1,2,4,4,5,3,2]) True sage: is_unimodal([1,2,4,3,5,3,2]) False

sage: for e in range(4,32): ....: d = 2*e ....: v = vector([binomial(d,i+1)*2^(i+1) for i in range(d)]) ....: w = vector([binomial(d,i) for i in range(d)]) - vector([0]*(d-1) + [1]) ....: p = ceil((v[e+1]-v[e])/(w[e]-w[e+1])) ....: if not is_unimodal(v + p*w): ....: print d, p ....: 60 52344913920 62 108935988690 64 226370041013 66 469746708832 68 973525920427

Figure 3 gives examples for three values of d and p.

(4) This is an open problem when d = 6 and d = 7. Using similar ideas as vertex stacking on facets, it is known that there exist 8-dimensional polytopes with non-unimodal f-vectors.

3 d = 60 d = 80 d = 100 p = 52344913920 p = 75395083047498 p = 99598837913001354

Figure 3: Three non-unimodal f-vector shapes.

2 Operations on polytopes

Exercice 5 (Truncating and stacking). Let P be a d-dimensional polytope, v be a simple vertex of P (contained in precisely d facets) and f be a simplicial facet of P (containing precisely d vertices). We consider the polytopes obtained by • truncating the vertex v of P by a hyperplane separating v from all other vertices of P , • stacking a vertex w on the facet f of P , where the vertex w is separated from P by f, but close enough to f so that it sees the vertices of f and no other vertex of P . These operations are illustrated on Figure 4.

P f w

Figure 4: Truncating a vertex v (middle) and stacking over a facet f (right).

Observe that truncating and stacking are dual operations: the polytope obtained by stacking a vertex on the facet v of P 4 is the polar of the polytope obtained by truncating vertex v of P . Describe the f-vector of the resulting polytopes in terms of the f-vector of P . What can you say in the case v is not simple or f is not simplicial? Solution. The -vector of the truncation is the -vector of plus d . The - f f P i+1 i∈[d] − (1, 0,..., 0) f vector of the stacking is the -vector of plus d . When is not simple, d f P i i∈[d] −(0,..., 0, 1) v i+1 i∈[d] is replaced by the -vector of the vertex gure of . Dually, when is not simplicial, d is replaced f v f i i∈[d] by the f-vector of the facet f.

Exercice 6 (Cartesian product of polygons). (1) Describe the i-dimensional faces of the Cartesian product P × Q in terms of the faces of the two polytopes P and Q. (2) Describe the f-vector of the product P × Q in terms of f-vector of the two polytopes P and Q. (3) What is the f-vector of a product of q copies of a p-gon? Do you recognize something when p = 4? Solution. (1) The faces of the Cartesian product P ×Q are the Cartesian products of the faces of P by the faces of Q. p More precisely, let F be a f-dimensional face of P maximizing a linear functional φ : R → R and G q be a g-dimensional face of Q maximizing a linear functional ψ : R → R. Then F × G is a (f + g)- p+q dimensional face of P × Q maximizing the functional θ : R → R dened by θ(x, y) = φ(x) + ψ(y). p+q Conversely, the face of P × Q maximizing a linear functional θ : R → R is the product of the face p q p p+q of P maximizing θ ◦ ι1 : R → R by the face of Q maximizing θ ◦ ι2 : R → R (where ι1 : R → R q p+q and ι2 : R → R are the two natural inclusions).

4 (2) We obtain that P . Therefore, . fi(P × Q) = j fj(P ) · fi−j(Q) f(P × Q, x) = f(P, x) · f(Q, x) (3) According to (2), the f-polynomial of a product R of q copies of a p-gon is f(R, x) = (p + px + x2)q so that the -vector given by P q j+k P q q−` q−`. When , we f fi(R) = j+k+`=q j,k,` p = `≤i/2 ` i−2` p p = 4 k+2`=i recognize the f-polynomial of the 2q-dimensional cube, although it is not immediate on the f-vector. Exercice 7 (Normal fans and polytope operations). Describe the normal fans of the Cartesian product P × Q and the Minkowski sum P + Q in terms of the normal fans of the polytopes P and Q. Solution. Using the description of the faces in Exercice 6, we observe that the normal fan of the Cartesian product P × Q is the Cartesian product of the normal fan of P by the normal fan of Q, meaning that the cones of N (P × Q) are the cones generated by the unions of the rays of a cone in N (P ) with the rays of a cone in N (Q). The normal fan of the Minkowski sum P + Q is the common renement of the normal fan of P and the normal fan of Q, meaning that the maximal cones of N (P + Q) are the intersections of a maximal cone of N (P ) with a maximal cone of N (Q).

Exercice 8 (Hanner polytopes). A Hanner polytope is either the interval I := [−1, 1] or a product or a direct sum of two Hanner polytopes. (1) What are the Hanner polytopes of dimension 1, 2, 3, 4? (2) Show that each Hanner polytope has 3d + 1 faces. (3) Is it true that each Hanner polytope is a prism or a bipyramid?

Solution. (1) The small dimensional Hanner polytopes are: • the interval I in dimension 1, • the square I ⊕ I ∼ I × I in dimension 2, • the cube I×3 := I × I × I and the cross-polytope I⊕3 := I ⊕ I ⊕ I in dimension 3, • the cube I×4, the cross-polytope I⊕4, the prism over an octahedron I⊕3 × I and the bipyramid over a cube I×3 ⊕ I in dimension 4.

(2) We prove (2) by induction on the dimension. Consider two Hanner polytopes, P of dimension p with 3p + 1 faces and Q of dimension q with 3q + 1 faces. The non-empty faces of P × Q are all products of a non-empty faces of P by non-empty faces of Q. Therefore, P × Q has 3p · 3q = 3p+q non-empty faces, thus 3p+q +1 faces. We argue the direct sum by polarity since P ⊕Q = (P 4 ×Q4)4.

(3) Consider the Hanner polytope P := (I × I × I) ⊕ (I × I × I). It has 16 vertices each of degree 11, so that it cannot be a bipyramid (would require two vertices of degree 14). It has 36 facets each of degree 8, so it cannot be a prism (would require two facets of degree 34).