complexes vs. Graph powers Michał Adamaszek Warwick Institute and DIMAP [email protected]

Combinatorial algebraic topology

Graphs and their powers Clique complexes Bringing the two together

r Let G be a finite connected graph. The r-th distance power Various simplicial constructions encode topologically the The space Cl(G ) is the Vietoris-Rips complex of subsets of of G is a graph Gr with the same vertex set in which two global combinatorial structure of G, for example: diameter at most r in G. G r vertices are adjacent if their distance in is at most . • The clique complex Cl(G) is a simplicial complex whose • Problem 1. What are the interesting features of the r We have a sequence of graph inclusions faces are the complete subgraphs of G. spaces Cl(G ) for higher r? 2 3 • Problem 2. What are the interesting features of the in- -G ֒→ G ֒→ G ֒→ • The Ind(G) is a simplicial com plex whose faces are the independent sets of G. duced inclusions which stabilize at the . C2 2 3 ? →֒ ( Example. The clique complex Cl( 6) is the union Cl(G) ֒→ Cl(G ) ֒→ Cl(G C C2 Example. For the 6-cycle the inclusion 6 ֒→ 6 is Warm-up exercise. Prove that the induced map of funda- mental groups S →֒ r π1(Cl(G)) → π1(Cl(G )) of two disks glued along the boundary, that is S2. is surjective.

Stability Universality

i G G2 When is : Cl( ) ֒→ Cl( ) a homotopy equivalence? For every r ≥ 1 and every finite simplicial complex K there exists a graph G with a The most general condition we provide is homotopy equivalence Gr K 2 2 Cl( ) ≃ . Every maximal clique in G (i.e. every maximal face of Cl(G )) has the form NG[v] (it is spanned by a vertex v of G together with all its neighbours in G). • A folklore result when r = 1, even with a homeomorphism K ≡ Cl(G) for G = (bdK)(1), the 2 Sketch of proof. The simplices NG[v] form a covering of Cl(G ) and all their nonempty in- 1-skeleton of the barycentric subdivision of K. −1 tersections Xv v = NG[v ] ∩ ∩ NG[vk] are contractible. One shows that i (Xv v ) are s (1) 1,..., k 1 1,..., k • For r ≥ 2 we use G = (bd K) , the 1-skeleton of a large iterated i also contractible (in fact cones) and a Mayer-Vietoris type argument shows is a weak subdivision (roughly s = O(logr)). equivalence. r r • Cover Cl(G ) with subcomplexes Cl((Gv) ) where Gv consists of i,i ,i C2 The condition is violated for example by the cliques { + 2 + 4} in 6 above. the vertices which belong to the open star of v in K (see fig.). r Some special cases of this result include: • Dochtermann proved Gv is dismantlable. It follows also for (Gv) , G r G has girth at least 7. so the covering complexes Cl(( v) ) are contractible. An analy- sis of the intersections of Gv and the nerve lemma do the rest. More generally: for any r, if G has girth at least 3r +1 then Cl(Gr) collapses to Cl(G) which, since G is triangle-free, is just G. Another representation, which is off by just some 2-cells, comes from an equivalence G is (r−1)t(G) • t(G) – number of triangles in G – free r r ( −1)G G S2 r , , , Cl((sd ) ) ≃ Cl( ) ∨ • sd( −1)G – an edge-subdivision of G by (it does not have either of the four graphs as an .) _ r − 1 new vertices on each edge

The first example: cycles

We want to know the homotopy types of For example, this process for T14,3 is The answer r Cl(Cn) n n r For ≤ r < the clique complexes Cl(Cn) of cycle powers With the parameter k = n − 2r − 1 consider the complements 3 2 r satisfy Tn,k = Cn known as the circular complete graphs. → → 0 r n r 0 13 1 r 2 3r−n 2 −1( −2 ) Cl(Cn) ≃ Σ Cl(C r n) = Σ Cl(C ). 8 1 4 − n−2(n−2r) 12 2

11 3 r 7 2 C T14,3 → S14,3 → T6,3 It means that all Cl( n) are generated by the double sus- 2 10 4 pension operator Σ , acting along lines of slope (2,1), as

6 3 9 5 In general: shown by the arrows below.

8 6 5 4 7 For n ≥ 3k + 3 the independence complexes of circular For any n 3 and 0 r < n T C3 T C5 ≥ ≤ 2 9,2 = 9 14,3 = 14 complete graphs satisfy n−2r−1 S2l if r l n So the equivalent question is to identify Cr = 2l+1 l Σ2 Cl( n) ≃ 2l+1 l l+1 for some ≥ 0. Ind(Tn,k) ≃ Ind(Tn−2(k+1),k). WS n r n  if 2l+1 < < 2l+3 Ind(Tn,k)

r = 0 1 2 3 4 5 6 7 8 9 10 11 Since Tn,k are triangle-free we can exhibit their independence C 7 S0 S1 S1 S3 T 8 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ complexes Ind( n,k) as suspensions: 8 0 1 1 2 2 C9 W S S S S ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ C 9 S0 S1 S1 S1 S4 1) Every maximal independent set contains 0 or its neighbour: 10 W W ∗ ∗ ∗ ∗ ∗ ∗ ∗ All these 10 0 1 1 1 3 C11 W S S S S S ∗ ∗ ∗ ∗ ∗ ∗ ∗ entries are T w 11 0 1 1 1 3 2 5 Ind( n,k) = st(0) ∪ st( ). C12 W S S S S S S ∗ ∗ ∗ ∗ ∗ ∗ ∗ 12 0 1 1 1 1 3 w∈[N(0) C13 W S S S S WS S ∗ ∗ ∗ ∗ ∗ ∗ (contractible) 13 0 1 1 1 1 3 6 C14 W S S S S S S S ∗ ∗ ∗ ∗ ∗ because C 14 S0 S1 S1 S1 S1 4 S2 2 S4 Cr 2) Since the vertex 0 is not in any triangle, N(0) is a in 15 W ∗ ∗ ∗ ∗ ∗ n is C W15 S0 S1 S1 S1 S1 WS1 WS3 S7 complete. Ind(Tn,k) and both summands are contractible (by a result of 16 ∗ ∗ ∗ ∗ C 16 S0 S1 S1 S1 S1 S1 S3 S5 Barmak). Therefore 17 W ∗ ∗ ∗ ∗ All these 17 0 1 1 1 1 1 5 2 3 8 C18 W S S S S S S S S S ∗ ∗ ∗ entries are 18 0 1 1 1 1 1 1 3 5 1 C19 W S S S S S S WS S S ∗ ∗ ∗ Ind(Tn,k) ≃ ΣK,where K = st(0) ∩ st(w) S C W19 S0 S1 S1 S1 S1 S1 S1 S3 3 S4 S9 ∗ ∗ w [N because of 20 ∈ (0) C 20 S0 S1 S1 S1 S1 S1 S1 6 S2 S3 2 S6 stability. 21 W W ∗ ∗ 21 0 1 1 1 1 1 1 1 3 5 10 C22 W S S S S S S S WS S WS S ∗ 3) The constraints defining K can be encoded in the indepen- 22 0 1 1 1 1 1 1 1 3 3 7 C23 W S S S S S S S S S S S ∗ dence complex of some graph Sn k. 23 0 1 1 1 1 1 1 1 7 2 3 5 11 , C24 W S S S S S S S S S S S S C 24 S0 S1 S1 S1 S1 S1 S1 S1 S1 S3 4 S4 S7 4) Steps 1), 2), 3) can be repeated for the new graph Sn,k. The 25 W W W W outcome can be identified with Tn k k. n −2( +1), r n r = = 3 2

• J.A.Barmak, Star clusters in independence complexes of graphs, arxiv/1007.0418 • D.Kozlov, Combinatorial algebraic topology, Springer 2008 • A.Dochtermann, The universality of Hom complexes of graphs, Combinatorica 29 (4) (2009) 433-448 • M.Adamaszek, Clique complexes vs. graph powers, arxiv/1104.0433