Polyhedral products, right-angled Coxeter groups, and hyperbolic manifolds

Transcription

1 Polyhedral products, right-angled Coxeter groups, and hyperbolic manifolds based on joint works with Victor Buchstaber, Nikolay Erokhovets, Mikiya Masuda, Seonjeong Park and Yakov Veryovkin Taras Panov Moscow State University, ITEP and IITP Ðîæäåñòâåíñêèå ìàòåìàòè åñêèå âñòðå è ÍÌÓ, Ìîñêâà, 46 ÿíâàðÿ 2017 Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

2 1. Preliminaries Polyhedral product (X, A) = {(X 1, A 1 ),..., (X m, A m )} a sequence of pairs of spaces, A i X i. K a simplicial complex on [m] = {1, 2,..., m}, K. Given I = {i 1,..., i k } [m], set (X, A) I = Y 1 Y m where Y i = { Xi if i I, A i if i / I. The K-polyhedral product of (X, A) is (X, A) K = (X, A) I = ( X i A j ). I K I K i I j / I Notation: (X, A) K = (X, A) K when all (X i, A i ) = (X, A); X K = (X, pt) K, X K = (X, pt) K. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

3 Categorical approach Category of faces cat(k). Objects: simplices I K. Morphisms: inclusions I J. top the category of topological spaces. Dene the cat(k)-diagram D K (X, A): cat(k) top, I (X, A) I, which maps the morphism I J of cat(k) to the inclusion of spaces (X, A) I (X, A) J. Then we have (X, A) K = colim D K (X, A) = colim I K (X, A)I. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

4 Example Let (X, A) = (S 1, pt), where S 1 is a circle. Then (S 1 ) K = (S 1 ) I (S 1 ) m. I K When K = {, {1},..., {m}} (m disjoint points), the polyhedral product (S 1 ) K is the wedge (S 1 ) m of m circles. When K consists of all proper subsets of [m] (the boundary m 1 of an (m 1)-dimensional simplex), (S 1 ) K is the fat wedge of m circles; it is obtained by removing the top-dimensional cell from the m-torus (S 1 ) m. For a general K on m vertices, (S 1 ) m (S 1 ) K (S 1 ) m. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

5 Example Let (X, A) = (R, Z). Then L K := (R, Z) K = (R, Z) I R m. I K When K consists of m disjoint points, L K is a grid in R m consisting of all lines parallel to one of the coordinate axis and passing though integer points. When K = m 1, the complex L K is the union of all integer hyperplanes parallel to coordinate hyperplanes. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

6 Example Let (X, A) = (RP, pt), where RP = BZ 2. Then (RP ) K = (RP ) I (RP ) m. I K Example Let (X, A) = (D 1, S 0 ), where D 1 = [ 1, 1] and S 0 = {1, 1}. The real moment-angle complex is R K := (D 1, S 0 ) K = (D 1, S 0 ) I. I K It is a cubic subcomplex in the m-cube (D 1 ) m = [ 1, 1] m. When K consists of m disjoint points, R K is the 1-dimensional skeleton of the cube [ 1, 1] m. When K = m 1, R K is the boundary of the cube [ 1, 1] m. Also, R K is a topological manifold when K is a sphere. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

7 The four polyhedral products above are related by the two homotopy brations (R, Z) K = L K (S 1 ) K (S 1 ) m, (D 1, S 0 ) K = R K (RP ) K (RP ) m. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

8 By analogy with the polyhedral product of spaces X K = colim I K X I, we may consider the following more general construction of a discrete group. Graph product G = (G 1,..., G m ) a sequence of m discrete groups, G i {1}. Given I = {i 1,..., i k } [m], set G I = { (g 1,..., g m ) m G k : g k = 1 for k / I }. k=1 Then consider the following cat(k)-diagram of groups: D K (G): cat(k) grp, I G I, which maps a morphism I J to the canonical monomorphism G I G J. The graph product of the groups G 1,..., G m is G K = colim grp D K (G) = colim grp I K G I. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

9 The graph product G K depends only on the 1-skeleton (graph) of K. Namely, Proposition The is an isomorphism of groups G K = m k=1 G k/ (gi g j = g j g i for g i G i, g j G j, {i, j} K), where m k=1 G k denotes the free product of the groups G k. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

10 Example Let G i = Z. Then G K is the right-angled Artin group RA K = F (g 1,..., g m ) / (g i g j = g j g i for {i, j} K), where F (g 1,..., g m ) is a free group with m generators. When K is a full simplex, we have RA K = Z m. When K is m points, we obtain a free group of rank m. Example Let G i = Z 2. Then G K is the right-angled Coxeter group RC K = F (g 1,..., g m ) / (g 2 i = 1, g i g j = g j g i for {i, j} K). Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

11 2. Classifying spaces The homotopy brations L K (S 1 ) K (S 1 ) m and R K (RP ) K (RP ) m are generalised as follows. Proposition There is a homotopy bration (E G, G) K (BG) K m BG k. k=1 A missing face (a minimal non-face) of K is a subset I [m] such that I / K, but J K for each J I. K a ag complex if each of its missing faces consists of two vertices. Equivalently, K is ag if any set of vertices of K which are pairwise connected by edges spans a simplex. Every ag complex K is determined by its 1-skeleton K 1. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

12 Theorem Let G K be a graph product group. 1 π 1 ((BG) K ) = G K. 2 Both spaces (BG) K and (E G, G) K are aspherical if and only if K is ag. Hence, B(G K ) = (BG) K whenever K is ag. 3 π i ((BG) K ) = π i ((E G, G) K ) for i 2. 4 π 1 ((E G, G) K ) is isomorphic to the kernel of the canonical projection G K m k=1 G k. Proof (1) Proceed inductively by adding simplices to K one by one and use van Kampen's Theorem. The base of the induction is K consisting of m disjoint points. Then (BG) K is the wedge BG 1 BG m, and π 1 ((BG) K ) is the free product G 1 G m. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

13 Proof (2) To see that B(G K ) = (BG) K when K is ag, consider the map colim I K BG I = (BG) K B(G K ). (1) According to [PRV], the homotopy bre of (1) is hocolim I K G K /G I, which is homeomorphic to the identication space ( B cat(k) G K)/. (2) Here B cat(k) is homeomorphic to the cone on K. The equivalence relation is dened as follows: (x, gh) (x, g) whenever h G I and x B(I cat(k)), where I cat(k) is the undercategory, and B(I cat(k)) is homeomorphic to the star of I in K. When K is a ag complex, the identication space (2) is contractible by [PRV]. Therefore, the map (1) is a homotopy equivalence, which implies that (BG) K is aspherical when K is ag. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

14 Proof Assume now that K is not ag. Choose a missing face J = {j 1,..., j k } [m] with k 3 vertices. Let K J = {I K: I J}. Then (BG) K J is the fat wedge of the spaces {BG j, j J}, and it is a retract of (BG) K. The homotopy bre of the inclusion (BG) K J j J BG j is Σ k 1 G j1 G jk, a wedge of (k 1)-dimensional spheres. Hence, π k 1 ((BG) K J ) 0 where k 3. Thus, (BG) K J and (BG) K are non-aspherical. The rest of the proof (the asphericity of (E G, G) K and statements (3) and (4)) follow from the homotopy exact sequence of the bration (E G, G) K (BG) K m k=1 BG k. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

15 Specialising to the cases G k = Z and G k = Z 2 respectively we obtain: Corollary Let RA K be a right-angled Artin group. 1 π 1 ((S 1 ) K ) = RA K. 2 Both (S 1 ) K and L K = (R, Z) K are aspherical i K is ag. 3 π i ((S 1 ) K ) = π i (L K ) for i 2. 4 π 1 (L K ) is isomorphic to the commutator subgroup RA K. Corollary Let RC K be a right-angled Coxeter group. 1 π 1 ((RP ) K ) = RC K. 2 Both (RP ) K and R K = (D 1, S 0 ) K are aspherical i K is ag. 3 π i ((RP ) K ) = π i (R K ) for i 2. 4 π 1 (R K ) is isomorphic to the commutator subgroup RC K. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

16 Example Let K be an m-cycle (the boundary of an m-gon). A simple argument with Euler characteristic shows that R K is homeomorphic to a closed orientable surface of genus (m 4)2 m (This observation goes back to a 1938 work of Coxeter.) Therefore, the commutator subgroup of the corresponding right-angled Coxeter group RC K is a surface group. Similarly, when K = S 2 (which is equivalent to K being the boundary of a 3-dimensional simplicial polytope), R K is a 3-dimensional manifold. Therefore, the commutator subgroup of the corresponding RC K is a 3-manifold group. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

17 3. The structure of the commutator subgroups We have ( Ker G K m ) G k = π 1 ((E G, G) K ). k=1 In the case of right-angled Artin or Coxeter groups (or when each G k is abelian), the group above is the commutator subgroup (G K ). The next goal is to study the group π 1 ((E G, G) K ), identify the class of K for which this group is free, and describe a generator set. A graph Γ is called chordal (in other terminology, triangulated) if each of its cycles with 4 vertices has a chord. By a result of FulkersonGross, a graph is chordal if and only if its vertices can be ordered in such a way that, for each vertex i, the lesser neighbours of i form a complete subgraph. (A perfect elimination order.) Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

18 Theorem (PVeryovkin) The following conditions are equivalent: 1 Ker(G K m k=1 G k) is a free group; 2 (E G, G) K is homotopy equivalent to a wedge of circles; 3 K 1 is a chordal graph. Proof ( (2) (1) Because Ker G K ) m k=1 G k = π 1 ((E G, G) K ). (3) (2) Use induction and perfect elimination order. (1) (3) Assume that K 1 is not chordal. Then, for each chordless cycle of length 4, one can nd a subgroup in Ker(G K m k=1 G k) which is a surface group. Hence, Ker(G K m k=1 G k) is not a free group. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

19 Corollary Let RA K and RC K be the right-angled Artin and Coxeter groups corresponding to a simplicial complex K. (a) The commutator subgroup RA K is free if and only if K1 is a chordal graph. (b) The commutator subgroup RC K is free if and only if K1 is a chordal graph. Part (a) is the result of Servatius, Droms and Servatius. The dierence between (a) and (b) is that the commutator subgroup RA K is innitely generated, unless RA K = Z m, while the commutator subgroup is nitely generated. We elaborate on this in the next theorem. RC K Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

20 Let (g, h) = g 1 h 1 gh denote the group commutator of g, h. Theorem (PVeryovkin) The commutator subgroup RC K has a nite minimal generator set consisting of J [m] rank H 0 (K J ) iterated commutators (g j, g i ), (g k1, (g j, g i )),..., (g k1, (g k2, (g km 2, (g j, g i )) )), where k 1 < k 2 < < k l 2 < j > i, k s i for any s, and i is the smallest vertex in a connected component not containing j of the subcomplex K {k1,...,k l 2,j,i}. Idea of proof First consider the case K = m points. Then R K is the 1-skeleton of an m-cube and RC K = π 1(R K ) is a free group of rank m l=2 (l 1)( ) m l. It agrees with the total number of nested commutators in the list. Then eliminate the extra nested commutators using the commutation relations (g i, g j ) = 1 for {i, j} K. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

21 Idea of proof To see that the given generating set is minimal, argue as follows. The rst homology group H 1 (R K ) is RC K /RC K. On the other hand, H 1 (R K ) = H 0 (K J ). J [m] Hence, the number of generators in the abelian group H 1 (R K ) = RC K /RC K is J [m] rank H 0 (K J ), and the latter number agrees with the number of iterated commutators in the in generator set for RC K constructed above. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

22 Example Let K = Then the commutator subgroup RC K is free with the following basis: (g 3, g 1 ), (g 4, g 1 ), (g 4, g 2 ), (g 4, g 3 ), (g 2, (g 4, g 1 )), (g 3, (g 4, g 1 )), (g 1, (g 4, g 3 )), (g 3, (g 4, g 2 )), (g 2, (g 3, (g 4, g 1 ))). Example Let K be an m-cycle with m 4 vertices. Then K 1 is not a chordal graph, so the group RC K is not free. In fact, R K is an orientable surface of genus (m 4)2 m 3 + 1, so RC K = π 1 (R K ) is a one-relator group. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

23 The are similar results of Grbic, P., Theriault and Wu describing the commutator subalgebra of the graded Lie algebra given by L K = FL u 1,..., u m /( [u i, u i ] = 0, [u i, u j ] = 0 for {i, j} K ), where FL u 1,..., u m is the free graded Lie algebra on generators u i of degree one, and [a, b] = ( 1) a b [b, a] denotes the graded Lie bracket. The commutator subalgebra is the kernel of the Lie algebra homomorphism L K CL u 1,..., u m to the commutative (trivial) Lie algebra. The graded Lie algebra L K is a graph product similar to the right-angled Coxeter group RC K. It has a similar colimit decomposition, with each G i = Z 2 replaced by the trivial Lie algebra CL u = FL u /([u, u] = 0) and the colimit taken in the category of graded Lie algebras. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

24 4. Right-angled polytopes and hyperbolic manifolds Let P be a polytope in n-dimensional Lobachevsky space L n with right angles between adjacent facets (a right-angled n-polytope). Denote by G(P) the group generated by reections in the facets of P. It is a right-angled Coxeter group given by the presentation G(P) = g 1,..., g m g 2 i where g i denotes the reection in the facet F i. = 1, g i g j = g j g i if F i F j, The group G(P) acts on L n discretely with nite isotropy subgroups and with fundamental domain P. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

25 Lemma Consider an epimorphism ϕ: G(P) Z k 2. The subgroup Ker ϕ G(P) does not contain elements of nite order if and only if the images of the reections in any k facets of P that have a common vertex are linearly independent in Z k 2. In this case the group Ker ϕ acts freely on L n. The quotient N = L n / Ker ϕ is a hyperbolic n-manifold. It is composed of Z k 2 = 2k copies of P and has a Riemannian metric of constant negative curvature. Furthermore, the manifold N is aspherical (the EilenbergMac Lane space K(Ker ϕ, 1)), as its universal cover L n is contractible. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

26 Which combinatorial n-polytopes have right-angled realisations in L n? In dim 3, there is a nice criterion going back to Pogorelov's work of 1967: Theorem (Pogorelov, Andreev) A combinatorial 3-polytope P 3 can be realised as a right-angled polytope in L 3 if and only if it is simple, and does not have 3- and 4-belts of facets. Furthermore, such a realisation is unique up to isometry. We refer to the above class of 3-polytopes as the Pogorelov class P. A polytope from the class P does not have triangular or quadrangular facets. The Pogorelov class contains all fullerenes (simple 3-polytopes with only pentagonal and hexagonal facets). There is no classication of right-angled polytopes in L 4. For n 5, right-angled polytopes in L n do not exist [Vinberg]. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

27 Given a right-angled polytope P, how to nd an epimorphism ϕ: G(P) Z k 2 with Ker ϕ acting freely on Ln? One can consider the abelianisation: G(P) ab Z m 2, with Ker ab = G (P), the commutator subgroup. The corresponding n-manifold L n /G (P) is the real moment-angle manifold R P, described as an intersection of quadrics in the beginning of this talk. Corollary If P is a right-angled polytope in L n, then the real moment-angle manifold R P admits a hyperbolic structure as L n /G (P), where G (P) is the commutator subgroup of the corresponding right-angled Coxeter group. The manifold R P is composed of 2 m copies of P. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

28 A more econimical way to obtain a hyperbolic manifold is to consider ϕ: G(P) Z n ab 2. Such an epimorphism factors as G(P) Z m Λ 2 Z n 2, where Λ is a linear map. The subgroup Ker ϕ acts freely on L n if and only the Λ-images of any n facets of P that meet at a vertex form a basis of Z n 2. Such Λ is called a Z 2 -characteristic function. Proposition Any simple 3-polytope admits a characteristic function. Proof. Given a 4-colouring of the facets of P, we assign to a facet of ith colour the ith basis vector ei Z 3 for i = 1, 2, 3 and the vector e 1 + e 2 + e 3 for i = 4. The resulting map Λ: Z m 2 Z3 2 satises the required condition, as any three of the four vectors e 1, e 2, e 3, e 1 + e 2 + e 3 form a basis of Z 3. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

29 Manifolds N(P, Λ) = L 3 / Ker ϕ obtained from right-angled 3-polytopes P P and characteristic functions Λ: Z m 2 Z3 2 are called hyperbolic 3-manifolds of L obell type. They were introduced and studied by A. Vesnin in Each N(P, Λ) is composed of Z 3 2 = 8 copies of P. In particular, one obtains a hyperbolic 3-manifold from any 4-colouring of a right-angled 3-polytope P. L obell was rst to consider a hyperbolic 3-manifold coming from a (unique) 4-colouring of the dodecahedron. γ α δ β δ β γ α γ β δ β δ γ γ β δ β δ α γ α γ α δ β δ γ α β γ δ β δ β α γ α γ β α δ α β γ δ β δ β α γ α δ β α γ Figure: Four non-equivalent 4-colouring of the `barell' fullerene with 14 facets. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

30 Pairs (P, Λ) and (P, Λ ) are equivalent if P and P are combinatorially equivalent, and Λ, Λ : Z m 2 Zn 2 dier by an automorphism of Zn 2. Theorem (BuchstaberErokhovetsMasudaPPark) Let N = N(P, Λ) and N = N(P, Λ ) be two hyperbolic 3-manifolds of L obell type corresponding to right-angled 3-polytopes P and P. Then the following conditions are equivalent: (a) there is a cohomology ring isomorphism ϕ: H (N; Z 2 ) = H (N ; Z 2 ); (b) there is a dieomorphism N = N ; (c) there is an equivalence of Z 2 -characteristic pairs (P, Λ) (P, Λ ). In particular, hyperbolic 3-manifolds corresponding to non-equivalent 4-colourings of P are not dieomorphic. The dicult implication is (a) (c). Its proof builds upon the wealth of cohomological techniques of toric topology. Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31

31 References [1] T. Panov, N. Ray and R. Vogt. Colimits, Stanley-Reiner algebras, and loop spaces, in: "Categorical Decomposition Techniques in Algebraic Topology" (G. Arone et al eds.), Progress in Mathematics, vol. 215, Birkhauser, Basel, 2004, pp [2] J. Grbic, T. Panov, S. Theriault and J. Wu. The homotopy types of moment-angle complexes for ag complexes. Trans. of the Amer. Math. Soc. 368 (2016), no. 9, [3] T. Panov and Ya. Veryovkin. Polyhedral products and commutator subgroups of right-angled Artin and Coxeter groups. Sbornik Math. 207 (2016), no. 11; arxiv: [4] V. Buchstaber, N. Erokhovets, M. Masuda, T. Panov and S. Park. Cohomological rigidity of manifolds dened by right-angled 3-dimensional polytopes. Russian Math. Surveys 72 (2017), to appear; arxiv: Taras Panov (MSU) Polyhedral products and hyperbolic mds IUM 4 Jan / 31