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the set of vertices is V , (x, y) V V is an edge if and only if y = sx for some s S. using barcode generator for none control to generate, create none image in none applications.barcode service report c# Observe tha none for none t, for every vertex x V , we have deg(x) k, where k = #S. We assume further that the graph G(V , S) is connected; this is the case if and only if acts transitively on V . Remark 6.

1.6 If H is a subgroup of , we consider the natural action of on /H . The corresponding Schreier G( /H , S) is connected.

Every connected Schreier graph G(V , S) is of the form G( /H , S) for a subgroup H of G. Indeed, if we choose a base point v0 V , the -space V can be identi ed with /H , where H is the stabiliser of v0 in ..

Microsoft SQL Server Some applic ations of Property (T). With the pr none none evious notation, let V be the quasi-regular representation of on 2 (V ), thus V ( ) (x) = ( 1 x), . (V ), x V , .. From now on none none , we assume that V is nite. Then the constant functions on V belong to 2 (V ) and the subspace. 2 0 (V ). = . (V ) :. (x) = 0 = {1V } . 0 is -invar iant. The corresponding representation V of on 2 (V ) has no 0 non-zero invariant vectors. The following crucial lemma establishes a link between the expanding con0 stant of the graph G(V , S) and the Kazhdan constant ( , S, V ) associated to 0 (see Remark 1.

1.4). Recall that S and V 0 ( , S, V ) = inf max V (s) s S.

: . 2 0 (V ),. =1 .. Lemma 6.1.7 With the previous notation, we have h(G(V , S)) .

0 ( , S, V )2 . 4. Proof Let A be a proper non-empty subset of V . We have to show that there exists a unit vector 2 (V ) such that 0 1 # A max V (s) min{#A, #(V \ A)} 4 s S. Set B = V \ none none A, n = #V , a = #A, and b = #B = n a. Let f : V C be de ned by b if x A f (x) = a if x B. Then f .

2 (V ) 0. and f = ab2 + ba2 = nab. Fix s S. We have b + a if s 1 x A and x B f (s 1 x) f (x) = a b if s 1 x B and x A 0 otherwise.. 6.1 Expander graphs It follows that V (s)f f where = (a + b)2 #Es = n2 #Es ,. Es = (B s A) (A sB) = (B sA) s(B s 1 A).. It is clear none for none that #Es 2# A. Hence, with 1 = f, nab we have = 1 and # A 1 #Es = 2 V (s)f f 2 2 2n nab ab = 2 V (s) 2 = V (s) 2n 2n. As min{#A, #(V \ A)} = min{a, b} we have 2ab 2ab = , a+b n 1 # A max V (s) min{#A, #(V \ A)} 4 s S and this ends the proof. Theorem 6.1 none for none .8 Let be a group generated by a nite set S with S 1 = S.

Let (Hn )n be a sequence of subgroups of of nite index with lim #( /Hn ) = .. Assume that there exists > 0 such that, for every n, there is no (S, )-invariant vector in 2 ( /Hn ). 0 Then the family of the Schreier graphs G( /Hn , S) is a family of (k, 2 /4)expanders, where k = #S. Proof Let n N.

Write n for /Hn . Since there is no (S, )-invariant vector in 2 ( /Hn ), we have 0 max n (s) . for all lemma. 2( 0. /Hn ) with = 1. Hence, h(Gn ) 2 /4, by the previous Some applic none for none ations of Property (T). Let now be a group with Property (T), and let S be a nite generating subset of with S 1 = S. Recall that there exists > 0 such that (S, ) is a Kazhdan pair (Proposition 1.3.

2.i). Lemma 6.

1.7 shows that, for any subgroup of nite index H of , we have h(G( /H , S)) 2 /4 for the corresponding Schreier graph. Corollary 6.

1.9 Let be a residually nite, in nite group with Property (T). Let S be a nite generating subset of with S 1 = S, and let > 0 be a Kazhdan constant for S.

For every decreasing sequence (Hn )n of nite index subgroups of with n Hn = {1}, the family of Schreier graphs (G( /Hn , S))n , is a family of (k, 2 /4)-expanders, where k = #S. Remark 6.1.

10 Let be a group with Property (T); let (S, ) be as in the previous corollary. Let H be a subgroup of nite index in . In view of Remark 6.

1.5, we obtain a uniform lower bound for the smallest non-zero eigenvalue 1 of the Laplace operator of the Schreier graph G( /H , S) of H . In fact, a direct proof yields the better bound 2 1 .

2 2 ( /H ) be an eigenfunction of Indeed, let f 0 with f = 1. Denoting by V and E the set of vertices and the set of edges of G( /H , S), we have by Proposition 5.2.

2.iv, 1 = = 1 2 1 2 . f ( y) f (x). 2.
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