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