The case of a non-compact semisimple Lie group in .NET Writer barcode 3/9 in .NET The case of a non-compact semisimple Lie group

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net 3 of 9 G is a semisimple Lie group with nite centre, and K is a maximal compact subgroup of G. In this way, we obtain Theorem A stated at the beginning of this section. Actually, in this case, a stronger result is true, namely: the G-equivariant harmonic mapping F : G/K H can be taken to be locally isometric, that is, for every x G/K, dFx (Y ) = Y , for all Y Tx (G/K).

. To see this, we rst recal l some classical facts on semisimple Lie groups. Let G be a semisimple Lie group with nite centre, and let K be a maximal compact subgroup of G. There exists a Cartan involution , that is, an involutive automorphism of G such that K is the set of - xed elements in G.

Let g be the Lie algebra of G, and let k be the subalgebra corresponding to K. Let p = {X g : d e (X ) = X }. The decomposition g = k p is the Cartan decomposition of g.

Fix a maximal abelian subspace a of p, and let A = exp a. We have the Cartan decompositions G = KAK and G = K exp p. of G (see [Helga 62, V, barcode code39 for .NET 6, III, 7 ]). Every element X p de nes a tangent vector DX Tx0 (G/K) for x0 = K, given by DX f (x0 ) = d f (exp tXx0 ).

t=0 , dt f C (G/K).. The mapping X DX is a li visual .net barcode 39 near bijection between p and Tx0 (G/K). This allows us to identify Tx0 (G/K) with p.

. Reduced cohomology Observe that, for k K, t Visual Studio .NET Code 3 of 9 he linear automorphism d (k)x0 of Tx0 (G/K) corresponds to the linear mapping Ad(k) : p p, since d d f (k exp tXx0 ). t=0 = f (k exp(tX )k 1 x0 ). t=0 dt dt d = f (exp (tAd( Code 39 for .NET k)X ) x0 ). t=0 dt for all f C (G/ K). Proposition 3.3.

17 Let G/K be a Riemannian symmetric space of the noncompact type. Assume that the action of K on the tangent space Tx0 (G/K) at x0 = K G/K is irreducible. Let be an af ne isometric action of G on a Hilbert space H, and let F : G/K H be a G-equivariant C 1 -mapping.

If F is not constant, then there exists > 0 such that F is a local isometry. Proof Let be the linear part of . We identify the tangent space Tx0 (G/K) with p as above and we identify the tangent space at a vector H with H.

Using the G-equivariance of F, we have, for every k K and X p, dFx0 (Ad(k)X ) = d F(exp (tAd(k)X ) x0 ). t=0 dt d = F(k exp tXx0 ). t=0 dt d = ( (k)0 + (k)F( exp tXx0 )) . t=0 dt d = (k) F(exp tXx0 ). t=0 dt = (k)dFx0 (X ). He nce, the symmetric bilinear form Q on Tx0 (G/K) p de ned by = Q(X , Y ) = dFx0 (X ), dFx0 (Y ) is Ad(K)-invariant. Let A be the non-negative symmetric linear operator de ned on the Euclidean space Tx0 (G/K) by Q(X , Y ) = AX , Y , for all X , Y Tx0 (G/K).

. Then A commutes with Ad(k) Code39 for .NET , for every k K. Hence, the eigenspaces of A are K-invariant.

Since, by assumption, K acts irreductibly on Tx0 (G/K), it follows that A has a unique eigenvalue, that is, A = cI for some c 0. Therefore, dFx0 (X ). =c X for all X Tx0 (G/K).. 3.3 Property (T) for Sp(n, barcode 3 of 9 for .NET 1) By G-equivariance of F, we have, for g G and x = gx0 , F = (g) F (g 1 ) and dFx = (g)dFx0 d (g 1 )x .

Since (g) is an isometry, it follows that dFx (d (g)x0 X ) = dFx0 (X ) = c X = c d (g)x0 (X ).
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