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Gelfand pairs using barcode maker for none control to generate, create none image in none applications. Java Reporting Library-Jasper Reports De nition 3.3.1 none for none Let G be a locally compact group, and let K be a compact subgroup of G.

The pair (G, K) is a Gelfand pair if the convolution algebra. Reduced cohomology Cc (G//K) of con none for none tinuous K-bi-invariant functions on G with compact support is commutative. Example 3.3.

2 Let G be a real semisimple Lie group with nite centre, and let K be a maximal compact subgroup of G. By a classical result of Gelfand, the pair (G, K) is a Gelfand pair (see, e.g.

, [Helga 84, IV, Theorem 3.1]). Let (G, K) be a Gelfand pair, and let ( , H) be an irreducible unitary representation of G.

The subspace HK of (K)-invariant vectors in H is invariant under the operators (f ) for f Cc (G//K). (For the de nition of (f ) L(H), see Section F.4.

) If HK = 0, then f (f ). HK is an irreduc ible -representation of the commutative -algebra Cc (G//K) (see [Helga 84, IV, Lemma 3.6]). By Schur s Lemma (Theorem A.

2.2), it follows that dimC HK = 1. Summarizing, we see that, for G, we have either dimC HK = 0 or dimC HK = 1.

De nition 3.3.3 Let (G, K) be a Gelfand pair.

An irreducible unitary representation ( , H) of G is called a spherical representation if dimC HK = 1. Lemma 3.3.

4 Let G be a topological group, and let K be a compact subgroup of G. Let ( , H) be a unitary representation of G which does not contain the unit representation. Assume that HK is nite dimensional.

Then does not almost have invariant vectors. In particular, if (G, K) is a Gelfand pair and is a non-trivial spherical representation of G, then does not almost have invariant vectors. Proof Assume, by contradiction, that almost has invariant vectors, that is, there exists a net ( )i of unit vectors such that lim (g) i i = 0.

uniformly on com none for none pact subsets of G. For i large enough, we have and we set i = 1 (k) i dk K (k) i dk,. (k) i dk = 0 3.3 Property (T) for Sp(n, 1). where dk is a Haar measure on K. Then i is a unit vector in HK such that ( ) lim (g) i i = 0 uniformly on com none none pact subsets of G. Since HK is nite dimensional, its unit sphere is compact. Hence, a subnet of ( i )i converges to some unit vector HK .

It follows from ( ) that is invariant under (G). This is a contradiction, since does not contain 1G . The following result is due to P.

Delorme [Delor 77, Proposition V.3]. The proof we give was found by Y.

Shalom (unpublished). Proposition 3.3.

5 Let (G, K) be a Gelfand pair, and let ( , H) be a spherical representation of G distinct from the unit representation; assume that G is compactly generated. Then H 1 (G, ) = 0. Proof Let be an af ne isometric action of G on H, with linear part .

We must show that has a globally xed point. Let b Z 1 (G, ) be de ned by (g) = (g) + b(g), H..

For every compac tly supported regular Borel measure on G, we can de ne an af ne transformation ( ) on H by the H-valued integral ( ) =. (g) d (g) = ( ) + b(g)d (g),. H.. Let 0 be a symm etric, K-bi-invariant probability measure on G which is absolutely continuous with respect to a left Haar measure on G and has compact support. Assume, in addition, that the subgroup generated by the support of 0 is dense in G. First claim: the operator ( 0 ) has a unique xed point 0 H.

Indeed, by the previous lemma, does not almost have invariant vectors. Hence, ( 0 ) < 1 by Proposition G.4.

2. Now, for , H ( 0 ) ( 0 ) = ( 0 ) ( 0 ) ( 0 ) . It follows that ( 0 ) is a strict contraction of the af ne Hilbert space H.

Hence, ( 0 ) has a unique xed point 0 . Second claim: the af ne subspace of the (K)- xed points has positive dimension. Indeed, observe rst that 0 is xed under (K) since (k) 0 = (k) ( 0 ) 0 = ( 0 ) 0 = 0 , k K.

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