Graphs of groups, Lecture 5

Transcription

1 Graphs of groups, Lecture 5 Olga Kharlampovich NYC, Sep / 32

2 Amalgamated free products Graphs of groups. Definition Let A, B, C be groups and let σ : C A and τc A be injective homomorphisms. If the diagram below is a push out then we write G = A C B and we say that G is the amalgamated (free) product of A and B over C. C A B G 2 / 32

3 If G is a group, then there exists a connected CW-complex K(G, 1) (the Eilenberg-MacLane Space) such that π 1 (K(G, 1)) = G. For A, B, C, σ, τ as above, let X = K(A, 1), Y = K(B, 1), Z = K(C, 1) and realize σ and τ as maps δ + : Z X, δ : Z Y. Now let W = X (Z [ 1, 1]) Y /, where (z, ±1) δ ±1 (z). By the Seifert-Van Kampen theorem, π 1 (W ) = A C B. Suppose that A =< S 1 R 1 >, B =< S 2 R 2 >, then A C B =< S 1 S2 R 1, R 2, {σ(c) = τ(c)}, c C} >. 3 / 32

4 Example Let Σ be a connected surface and let γ be a separating, simple closed curve. Let Σ/γ = Σ Σ+. Then, π 1 (Σ) = π 1 (Σ ) <γ> π 1 (Σ + ). If γ is non-separating (but still 2-sided), then there are two natural maps δ ±1 : S 1 Σ 0 representing γ, where Σ 0 = Σ im(γ). 4 / 32

5 Amalgamated free products Graphs of groups. Let G =< A B c = φ(c), c C >. Choose a system of right coset representatives T C and T D, where D = φ(c). Definition A C-normal form is a sequence (x 0, x 1,..., x n ) such that 1) x 0 C, 2) x i T C {1} or x i T D {1}, and the consecutive terms x i and x i+1 lie in distinct systems of representatives. Similarly one can define a D-normal form. Theorem Any element g G = A C=D B can be uniquely written in the form g = x 0 x 1... x n, where (x 0, x 1,... x n ) is a C-normal form. Proof to be given 5 / 32

6 Trees and amalgamated free products This graph is called a segment Theorem Let G 1 A G 2. Then there exists a tree X, on which G acts without inversion on edges such that the factor graph G\X is a segment. Moreover this segment can be lifted to a segment in X with the property that the stabilizers in G of its vertices and edges are equal to G 1, G 2 and A respectively. 6 / 32

7 Trees and amalgamated free products Proof Let X 0 = G/G 1 G/G2 (union of left cosets) and X 1 + = G/A. Put σ(ga) = gg 1, τ(ga) = gg 2, and let T be the segment in X with the vertices G 1, G 2 and the positively oriented edge A. G acts on X by left multiplication. X is connected. Indeed, let g = g 1... g n with g i G 1 or g i G 2 depending on the parity of i. Then gg 1 is connected by a path to G 1. If g i G 1, then g 1... g i 1 G 1 = g 1... g i G 1, if g i G 2, then g 1... g i 1 G 1 and g 1... g i G 1 are connected by the edges to g 1... g i 1 G 2 = g 1... g i G 2. Now, the connectivity follows by induction on n. Suppose there is a reduced loop e 1... e n. WLOG σ(e 1 ) = G 1. Since adjacent vertices are cosets of different subgroups, n is even and there exists x i G 1 A, y i G 2 A such that σ(e 2 ) = x 1 G 2, σ(e 3 ) = x 1 y 1 G 1,..., τ(e n ) = x 1 y 1... x n/2 y n/2 G 1. Since τ(e n ) = σ(e 1 ) = G 1 this contradicts to the uniqueness of normal form. 7 / 32

8 Trees and amalgamated free products Proof Let X 0 = G/G 1 G/G2 (union of left cosets) and X 1 + = G/A. Put σ(ga) = gg 1, τ(ga) = gg 2, and let T be the segment in X with the vertices G 1, G 2 and the positively oriented edge A. G acts on X by left multiplication. X is connected. Indeed, let g = g 1... g n with g i G 1 or g i G 2 depending on the parity of i. Then gg 1 is connected by a path to G 1. If g i G 1, then g 1... g i 1 G 1 = g 1... g i G 1, if g i G 2, then g 1... g i 1 G 1 and g 1... g i G 1 are connected by the edges to g 1... g i 1 G 2 = g 1... g i G 2. Now, the connectivity follows by induction on n. Suppose there is a reduced loop e 1... e n. WLOG σ(e 1 ) = G 1. Since adjacent vertices are cosets of different subgroups, n is even and there exists x i G 1 A, y i G 2 A such that σ(e 2 ) = x 1 G 2, σ(e 3 ) = x 1 y 1 G 1,..., τ(e n ) = x 1 y 1... x n/2 y n/2 G 1. Since τ(e n ) = σ(e 1 ) = G 1 this contradicts to the uniqueness of normal form. 7 / 32

9 Trees and amalgamated free products Proof Let X 0 = G/G 1 G/G2 (union of left cosets) and X 1 + = G/A. Put σ(ga) = gg 1, τ(ga) = gg 2, and let T be the segment in X with the vertices G 1, G 2 and the positively oriented edge A. G acts on X by left multiplication. X is connected. Indeed, let g = g 1... g n with g i G 1 or g i G 2 depending on the parity of i. Then gg 1 is connected by a path to G 1. If g i G 1, then g 1... g i 1 G 1 = g 1... g i G 1, if g i G 2, then g 1... g i 1 G 1 and g 1... g i G 1 are connected by the edges to g 1... g i 1 G 2 = g 1... g i G 2. Now, the connectivity follows by induction on n. Suppose there is a reduced loop e 1... e n. WLOG σ(e 1 ) = G 1. Since adjacent vertices are cosets of different subgroups, n is even and there exists x i G 1 A, y i G 2 A such that σ(e 2 ) = x 1 G 2, σ(e 3 ) = x 1 y 1 G 1,..., τ(e n ) = x 1 y 1... x n/2 y n/2 G 1. Since τ(e n ) = σ(e 1 ) = G 1 this contradicts to the uniqueness of normal form. 7 / 32

10 Trees and amalgamated free products Remark In X, all edges with initial vertex gg 1 have the form gg 1 A, where g 1 runs over the set of representatives of the left cosets of A in G 1. The degree of gg 1 is G 1 : A. The stabilizer of gg 1 is gg 1 g 1. 8 / 32

11 Trees and amalgamated free products Theorem Let G act without inversions on edges on a tree X and suppose that the factor graph G\X is a segment. Let T be an arbitrary lift of this segment in X. Denote its vertices by P, Q, and the edge by e, and let G p, G q, G e be their stabilizers. Then the homomorphism φ : G P Ge G Q G which is the identity on G p and G Q is an isomorphism. 9 / 32

12 Trees and amalgamated free products Proof Write G =< G p, G q > and prove that G = G. IfG < G then the graph G T and (G G ) T are disjoint (proof?), but G T = X is connected, contradiction. Injectivity of φ. Let G = G P Ge G Q and let X be the tree constructed from G as in the proof of the previous theorem. Define a morphism ψ : X X by gg r φ(g)t, where r {P, Q, e}. It is surjective because X = G T and G =< G P, G Q >, and is locally injective morphism from a tree to a tree, therefore injective. Let g G G P. Then the vertices G P and gg P of the tree X are distinct, therefore vertices P and φ(g)p of the tree X are also distinct. Hence φ(g) / 32

13 Trees and amalgamated free products Proof Write G =< G p, G q > and prove that G = G. IfG < G then the graph G T and (G G ) T are disjoint (proof?), but G T = X is connected, contradiction. Injectivity of φ. Let G = G P Ge G Q and let X be the tree constructed from G as in the proof of the previous theorem. Define a morphism ψ : X X by gg r φ(g)t, where r {P, Q, e}. It is surjective because X = G T and G =< G P, G Q >, and is locally injective morphism from a tree to a tree, therefore injective. Let g G G P. Then the vertices G P and gg P of the tree X are distinct, therefore vertices P and φ(g)p of the tree X are also distinct. Hence φ(g) / 32

14 Action of SL 2 (Z) on the hyperbolic plane H 2 = {z C Im(z) > 0}. A hyperbolic line is an open half-circle or an open half-line (in the Euclidean sense) in H 2 such that its closure meets the real axis at right angles. A Mobius transformation of H 2 is a map z az+b cz+d, where a, b, c, d R, ad bc = 1. SL 2 (R) acts on H 2 by the rule The kernel is {±I }. ( ) a b c d : z az + b cz + d. 11 / 32

15 Action of SL 2 (Z) on the hyperbolic plane Let M = {z 1 < z, 1/2 < Re(z) 1/2} {e iα π/3 α π/2}. Theorem The set M is the fundamental domain for the action of PSL 2 (Z) on H / 32

16 Action of SL 2 (Z) on a tree Graphs of groups. Theorem The union of the images of the arc T = {e iα π/3 α π/2} under the action of the group SL 2 (Z) is a tree. SL 2 (Z) acts on this tree without inversion on edges and so that distinct points of the arc are inequivalent. ( The stabilizers ) of( endpoints ) are generated by the matrices A = and B = of orders 4 and ( ) 1 0 The stabilizer of the arc is generated by I = of order In particular SL 2 (Z) = Z 4 Z2 Z / 32

17 Action of SL 2 (Z) on a tree Graphs of groups. ( ) 0 1 HWP8 Let C =. Prove that 1 0 < A, C > = D 4, < B, C > = D 6. Deduce that GL 2 (Z) = D 4 D2 D 6. Theorem [Serre] For n 3 the groups SL n (Z) and GL n (Z) cannot be represented as nontrivial amalgamated products. 14 / 32

18 Trees and HNN extensions Graphs of groups. Let G =< H, t t 1 at = φ(a), a A, φ(a) = B >. Definition A normal form is a sequence (g 0, t ɛ 1, g 1,..., t ɛn, g n ) such that 1) g 0 is an arbitrary element of H, 2) if ɛ i = 1, then g i T A (right coset representative), 3)if ɛ i = 1, then g i T B, 4) there is no consecutive subsequence t ɛ, 1, t ɛ. Theorem HW9 [Britton s lemma] 1) Every element x G has a unique representation x = g 0 t ɛ 1 g 1... t ɛn g n, where (g 0, t ɛ 1, g 1,..., t ɛn, g n ) is a normal form. 2) H is embedded into G by the map h h. If w = g 0 t ɛ 1 g 1... t ɛn g n, and this expression does not contain subwords t 1 g i t with g i A or tg i t 1 with g i B, then w 1 in G. 15 / 32

19 Trees and HNN extensions Graphs of groups. Theorem Let G =< H, t t 1 at = φ(a), a A, φ(a) = B >. Then there exists a tree X on which G acts without inversion of edges such that the factor graph G\X is a loop. Moreover, there is a segment Ỹ in X such that the stabilizers of its vertices and edges in the group G are equal to H, tht 1 and A respectively. Proof Set X 0 = G/H, X 1 + = G/A (all cosets are left), σ(ga) = gh, τ(ga) = gth, and let Ỹ be the segment in X with vertices H, th. G acts on X by left multiplication. 16 / 32

20 Trees and HNN extensions Graphs of groups. Theorem Let G act without inversions on edges on a tree X and suppose that the factor graph Y = G\X is a loop. Let Ỹ be an arbitrary segment in X. Denote its vertices by P, Q, and the edge by e, and let G p, G q, G e = Gē be their stabilizers. Let x be an arbitrary element such that Q = xp. Put G e = x 1 G e x and let φ : G e G e be the isomorphism induced by conjugation by x. Then G e G P and the homomorphism < G P, t t 1 at = φ(a), a G e > G which is the identity on G P and sends t to x is an isomorphism. 17 / 32

21 Definition A graph of groups Γ(G, X ) consists of 1) a connected graph X ; 2) a function G which for every vertex v V (X ) assigns a group G v, and for each edge e E(X ) assigns a group G e such that Gē = G e. 3) For each edge e E(X ) there exists a monomorphism σ : G e G σe. Let Γ(G, X ) be a graph of groups. Since Gē = G e then there exists a monomorphism G e G σē = G τe which we denote by τ : G e G τe. 18 / 32

22 Fundamental groups of graphs of groups Let Γ = Γ(G,, X ) be a graph of groups, and let T be a maximal subtree of X. Suppose the groups G v are given by presentations G v = X v,, R v, v V (X ). We define a fundamental group π(γ) of the graph of groups Γ by generators and relations : Generators of π(γ): Relations of π(γ): v V (X ) v V (X ) X v {t e e E(X )} R v {t 1 e σgt e = τg g G e, e E(X )} {tē = t 1 e e E(X )} {t e = 1 e T }. We assume here that σ(g) and τ(g) are words in generators X σe and X τe, correspondingly. 19 / 32

23 Examples. i) Let Γ be G u G e G v then free product with amalgamation. π(γ) = G u Ge G v 20 / 32

24 ii) If Γ is G e G v then - HNN extension of G v. π(γ) = G v Ge 21 / 32

25 iii) If Γ is then π(γ) is called a tree product. 22 / 32

26 iv) If Γ is G e1 G e2 G e3 G v then π(γ) is a generalized HNN extension. 23 / 32

27 Let Γ = (G, X ) be a graph of groups and Y X be a connected subgraph. Then one can define a subgraph of groups Γ Y = (G Y, Y ), where G Y = G Y is the restriction of G on Y, i.e., every vertex and every edge from Y has the same associated groups as in Γ. Every maximal subtree S of Y can be extended to a maximal subtree T of X with S T. The identical map { gν G ν g ν G ν, (ν Y ) t e t e, (e Y ) gives rise to a homomorphism of the free product φ 0 : v V (Y ) G v F (E(Y )) π(γ, T ). where F (E(Y )) is a free group with basis E(Y ). 24 / 32

28 Clearly, φ 0 sends all defining relations of π(γ Y, S) into identity. Hence it induces a homomorphism φ Y : π(γ Y, S) π(γ, T ). We call φ Y the canonical homomorphism. 25 / 32

29 Theorem The canonical homomorphism is a monomorphism. φ Y : π(γ Y, S) π(γ, T ) 26 / 32

30 Proof. Case 1. Let X be a finite tree, so S T = X. If S = T then there is nothing to prove. If S T then there exists an edge e T and a subtree T 1 of T such that T = T 1 {e} and S T 1. By induction on V (T ) the canonical homomorphism π(γ S, S) φ 1 π(γ T1, T 1 ). is a monomorphism. Observe that also by induction we have canonical monomorphisms G σ(e) φ σ(e) π(γ S, S) φ π(γ T1, T 1 ). 27 / 32

31 In particular, G e σ Gσ(e) φ σ(e) π(γ T1, T 1 ), G e τ Gτ(e) are monomorphisms. This shows that the representation of π(γ T, T ) via generators and relations is a presentation of a free product with amalgamation: Hence is a monomorphism as well as as required. π(γ T, T ) = π(γ T1, T 1 ) Ge G τ(e). π(γ T1, T 1 ) φ T 1 π(γ T, T ) π(γ S, S) φ π(γ T1, T 1 ) φ T 1 π(γ T, T ), 28 / 32

32 Case 2. Suppose now that T is an infinite tree and S is finite. Then there exists an increasing chain of finite trees S = T 0 T 1... T i... such that T = i T i. Then the canonical monomorphisms π(γ T0, T 0 ) π(γ T1, T 1 )... provide an increasing chain of groups. Clearly, π(γ, T ) = lim π(γ Ti, T i ) 1 and for each i there exists an embedding π(γ Ti, T i ) π(γ, T ). In particular, π(γ S, S) π(γ, T ). 1 Notice that direct limit here is just a union of the increasing chain of groups. 29 / 32

33 Case 3. Let S T be infinite trees. Then π(γ S, S) = lim π(γ Si, S i ) for some infinite chain of finite subtrees such that S = i S i. By Case 2 S 1 S 2... S n... φ Si : π(γ Si, S i ) π(γ, T ) is a monomorphism for each i. Therefore, the canonical homomorphism is a monomorphism. π(γ S, S) = lim π(γ Si, S i ) π(γ, T ) 30 / 32

34 Case 4. Let X be an arbitrary graph and X T be finite. Then π(γ, T ) is an HNN-extension of π(γ T, T ) (see Example IV above). Case 3 implies that π(γ S, S) φ S π(γ T, T ) φ T π(γ, T ) is a monomorphism. It follows from the properties of HNN extensions that the canonical map is a monomorphism. π(γ T, T ) φ T π(γ T (Y S), T ) 31 / 32

35 Case 5. Let now X be an arbitrary graph. Then X = i X i such that T X i and X i T is finite for every i. Then and by Case 4 is monic, as well as π(γ, T ) = lim π(γ Xi, T ) π(γ Y Xi, S) φ Y X i π(γ Xi, T ) π(γ Y, S) = lim π(γ Y Xi, S) lim π(γ Xi, T ) = π(γ, T ). 32 / 32