Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA
Received Date: January 19, 2017; Accepted Date: January 25, 2017; Published Date: February 03, 2017
Citation: He H (2017) Invariant Tensor Product. J Generalized Lie Theory Appl 11: 252. doi:10.4172/1736-4337.1000252
Copyright: © 2017 He H. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
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In this paper, we define invariant tensor product and study invariant tensor products associated with discrete series representations. Let G(V1)×G(V2) be a pair of classical groups diagonally embedded in G(V1⊕V2). Suppose that dimV1<dimV2. Let π be a discrete series representation of G(V1⊕V2). We prove that the functor π ⊗G(V1) *maps unitary representations of G(V1) to unitary representations of G(V2). Here we enlarge the definition of unitary representations by including the zero dimensional representation.
Various forms of invariant tensor products appeared in the literature implicitly, for example, in Schur’s orthogonality for finite groups . In many cases, they are employed to study the space HomG(π1, π2) where one of the representations π1 and π2 is irreducible. In this paper, we formulate the concept of invariant tensor product uniformly. We also study the invariant tensor functor associated with discrete series representations for classical groups. For motivations and applications [2-4].
Let G be a locally compact tomography group and dg be a left invariant Haar measure. Let (π,Hπ) and (π1,Hπ1) be two unitary representations of G. Let V and V1 be two dense subspaces of Hπ and Hπ1 . Formally, define the averaging operator
as follows, ∀u,v∈V, u1,v1∈V1,
Suppose that is well-defined. The image of will be called the invariant tensor product. It will be denoted by . Whenever we use the notation , we assume VGV1 is well-defined, that is, the integral (1) converges for all . Denote . Define
For any unitary representation of G, let be the same unitary representation of G equipped with the conjugate linear multiplication. If V is a subspace of , let Vc be the corresponding subspace of .
Let G be a unimodular group. Suppose that is well-defined. Then the form (,)G is a well-defined Hermitian form on .
The main result proved in this paper is as follows.
Theorem 1.1: Let G(m+n) be a classical group of type I with m>n. Let (G(n),G(m)) be diagonally embedded in G (see Def. 2). Suppose that (π,Hπ) is a discrete series representation of G(m+n) and (1,H1) is a unitary representation of G(n). Let be the space of smooth vectors in Hπ. Then is well-defined. Suppose that . Then (,)G(n) is positive definite. Furthermore, completes to a unitary representation of G(m).
Let G be a compact group. Let (π1,Hπ1) and be two unitary representations of G. Then is always well-defined. Suppose that π1 is irreducible. Then the dimension of is the the multiplicty of occuring in Hπ.
Let G be a real reductive group. Let π and π1 be two discrete series representations. Then is always well-defined. It is one dimensional if and only if . Otherwise, it is zero dimensional.
Theorem 2.1: Let 1 be an irreducible unitary representation of G. Suppose that V1 and V are both closed under the action of G. Suppose that is well-defined. Then induces an injection from to , the space of G-equivariant homomorphisms from Vc to the Hermitian dual .
Proof: For each define as follows:
We have for every
We see that is in the Hermitian dual of V1. In addition, is G-equivariant:
Here dh is a left invariant measure if G is not unimodular. Now it is easy to see that for every u if and only if . So
is an injection.
Corollary 2.1: Under the same assumption as in Theorem 2.1, let G be a real reductive group and K a maximal compact Lie group of G. Suppose that V and V1 are both smooth and K-finite. Then induces an injection from into .
Proof: When V is K-finite, will land in the K-finite subspaces of V
1 which is isomorphic to V1.
Let G be a Lie algebra group and dg a left invariant Haar measure. Let X be a manifold with a continuous free (right) G action. Suppose that X/G is a smooth manifold. Let be a unitary representation of G. For any , define
Then is a -valued function on X. We shall see that it realizes in the following sense.
Theorem 3.1: Let G be a Lie group and dg a left invariant Haar measure. Let X be a manifold with a continuous free (right) G action such that the topological quotient X/G is a smooth manifold. Suppose there exist measures (X,μ) and (X/G,d[x]) such that
Let Cc(X)be the set of continuous functions with compact support. Let be a representation of G. Then where is the set of continuous compactly supported sections of the vector bundle
and for every and
Proof: Let and . It is easy to see that is compactly supported in X/G. In addition
So . Observe that
Absolute convergence are guaranteed since f(g) is compactly supported. Notice that
Let G be a classical group that preserves a nondegenerate sesquilinear form Ω. Write G=G(V,Ω) or simply G(V), where V is a vector field over equipped with the nondegenerate sesquilinear form Ω. Let V=V1V2 such that Ω(V1,V2)=0. Let G1=G(V1, Ω|V1) and G2=G(V2, Ω|V2). For each g1G1,g2G2, let (g1,g2) acts on diagonally. We say that G1×G2 is diagonally embedded in G.
Let (G1,G2) be diagonally embedded in G. Let be a unitary representation of G and be a unitary representation of G1. Let V be a subspace of that is invariant under G2. Let V1 be a subspace of such that is well-defined. Define a linear G2- representation as follows:
Since the Lie group action of G2 commutes with the integration over G1, the action of G2 on is well-defined.
The linear representation is not neccessarily continuous because no topology has been defined on
The form on is G2-invariant.
Proof: Let and Write . Then
Hence is G2-invariant.
Let G(m+n) be a classical group preserving a nondegenerate sesquilinear form. Let (G(n), G(m)) be diagonally embedded in G. For any irreducible unitary representation of G(m+n), let be the Frechet space of smooth vectors.
Theorem 5.1: Suppose that is a discrete series representation of G(m+n). Suppose that m>n and (π1,H1) is a unitary representation of G(n). Then is well-defined. Suppose that . Then (,)G(n) is positive definite. Furthermore, completes to a unitary representation of G(m).
The key of the proof is to realize as a subspace of the L2-sections of the Hilbert bundle
Proof: Write G= G(m+n). Fix a maximal compact subgroup K of G such that
are maximal compact subgroups of G(m) and G(n) respectively. Let a be a maximal Abelian subalgebra in the orthogonal complement of k with respect to the Killing form (,)K, such that
Let A be the analytic group generated by a. The function log:Aa is well-defined. Let for each
Since is a discrete series representation, without loss of generality, realize on a right K-finite subspace of L2(G). So .
Let be Harish-Chandra’s basic spherical function. Let be the space of Harish-Chandra’s Schwartz space. It is well-known that every satisfies for some see for example Ch. 12.4 [?]). For every for a constant Ch. Observe that is G(m)-invariant.
for some q0 and C>0. Let ρ(n) be the half sum of positive roots of the restricted root system . Let be the positive Weyl chamber of with respect to the root system . Since
. It follows that for every . Notice that is always bounded for . We see that
always converges. So is well-defined. Now suppose that
is bounded by a multiple of . So . For each and , define to be the -valued function on G:
in the strong sense. Notice that for gG,h1∈G(n),
So can be regarded as a section of the Hilbert bundle
In addition, we have
where G(n)\G is equipped with a right G invariant measure. Eqn. (31) is valid because the integrative Eqn. (29) converges absolutely. In fact, we have
To see this, recall that . In particular, for any N>0 and a∈A+, k1,k2∈K, there exists Cu,N>0 such that:
Write for g= k1ak2. Then there also exists Cv,N>0 such that
Fix an N such that . In particular, Observe that the function
is bounded by a multiple of (L(h)WN(g),WN(g)), which, by a Theorem of Cowling-Haagerup-Howe , is bounded by a multiple of . Hence
Eqn. (29) converges absolutely. Therefore Eqn. (31) holds.
Now we have
It follows that . Realize as , which is a subspace of L2-sections of the Hilbert bundle:
Clearly (,)G(n) is positive definite. Let be the completion of .
Since G(m) acts on and it commutes with G(n), G(m) acts on and it preserves (,)G(n). So the action of each g2∈G(m) can be extended into a unitary operator on . The group structure is kept in this completion essentially due to the fact that each extension is unique. Therefore completes to a unitary representation of G(m).
Let Πu(G) be the unitary dual of G. Suppose that m>n. Let π be a discrete series representation of G(m+n). We denote the functor from π1 to the completion of by ITπ. If ITπ(1)0, ITπ(π1) is a unitary representation of G(m). Regarding the zero dimensional representation as a unitary representation, ITπ defines a functor from unitary representations of G(n) to unitary representations of G(m).
One natural question arises. That is, if π1 is irreducible, is ITπ(1) irreducible? This is beyond the scope of this paper. In fact, this problem is quite difficult. In general, ITπ(π1) is not irreducible. However, for a certain holomorphic discrete series representation π, ITπ(π1) will indeed be irreducible. For the time being, it is not clear which discrete series representation π has such a property. This question may be intrinsically related to the cohomology induction .
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