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Invariant Tensor Product | OMICS International
ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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Invariant Tensor Product

He H*

Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803, USA

*Corresponding Author:
He H
Department of Mathematics, Louisiana State University
Baton Rouge, LA 70803, USA
Tel: 1 225-578-1665
E-mail: [email protected]

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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Abstract

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(V1V2). Suppose that dimV1<dimV2. Let π be a discrete series representation of G(V1V2). We prove that the functor πG(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.

Keywords

Representation theory; Tomography; Topology; Algebra

Invariant Tensor Products

Various forms of invariant tensor products appeared in the literature implicitly, for example, in Schur’s orthogonality for finite groups [1]. In many cases, they are employed to study the space HomG1, 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].

Definition 1

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

equation

as follows, ∀u,v∈V, u1,v1∈V1,

equation (1)

equation (2)

Suppose that equation is well-defined. The image of equation will be called the invariant tensor product. It will be denoted by equation. Whenever we use the notation equation, we assume VGV1 is well-defined, that is, the integral (1) converges for all equation. Denote equation. Define

equation

For any unitary representation equation of G, let equation be the same unitary representation of G equipped with the conjugate linear multiplication. If V is a subspace of equation, let Vc be the corresponding subspace of equation.

Lemma 1.1

Let G be a unimodular group. Suppose that equation is well-defined. Then the form (,)G is a well-defined Hermitian form on equation.

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 equation be the space of smooth vectors in Hπ. Then equation is well-defined. Suppose that equation . Then (,)G(n) is positive definite. Furthermore, equation completes to a unitary representation of G(m).

Example: π1 Irreducible

Example I

Let G be a compact group. Let (π1,Hπ1) and equation be two unitary representations of G. Then equation is always well-defined. Suppose that π1 is irreducible. Then the dimension of equation is the the multiplicty of equation occuring in Hπ.

Example II

Let G be a real reductive group. Let π and π1 be two discrete series representations. Then equation is always well-defined. It is one dimensional if and only if equation. 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 equation is well-defined. Then equation induces an injection from equation to equation , the space of G-equivariant homomorphisms from Vc to the Hermitian dual equation .

Proof: For each equation define equation as follows:

equation

We have for every equation

equation

equation

equation

equation

We see that equation is in the Hermitian dual of V1. In addition, equation is G-equivariant:

equation (3)

equation (4)

equation (5)

equation (6)

equation (7)

equation (8)

Here dh is a left invariant measure if G is not unimodular. Now it is easy to see that equation for every u if and only if equation. So

equation

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 equation induces an injection from equation into equation.

Proof: When V is K-finite, equation will land in the K-finite subspaces of V h

1 which is isomorphic to V1.

 

A Geometric Realization

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 equation be a unitary representation of G. For any equation, define

equation

Then equation is a -valued function on X. We shall see that it realizes equation 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

equation

Let Cc(X)be the set of continuous functions with compact support. Let equation be a representation of G. Then equation where equation is the set of continuous compactly supported sections of the vector bundle

equation

Furthermore,

equation

and for every equation and equation

equation

Proof: Let equation and equation. It is easy to see that equation is compactly supported in X/G. In addition

equation

So equation. Observe that

equation (9)

equation (10)

equation (11)

equation (12)

equation (13)

equation (14)

equation (15)

Absolute convergence are guaranteed since f(g) is compactly supported. Notice that

equation

We have

equation

Invariant Tensor Product and Representation Theory

Definition 2

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 equation 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 equation diagonally. We say that G1×G2 is diagonally embedded in G.

Definition 3

Let (G1,G2) be diagonally embedded in G. Let equation be a unitary representation of G and equationbe a unitary representation of G1. Let V be a subspace of equation that is invariant under G2. Let V1 be a subspace of equation such that equation is well-defined. Define a linear G2- representation equation as follows:

equation

Since the Lie group action of G2 commutes with the integration over G1, the action of G2 on equation is well-defined.

The linear representation equation is not neccessarily continuous because no topology has been defined on equation

Lemma 4.1

The form equation on equation is G2-invariant.

Proof: Let equation and equation Write equation . Then

equation (16)

equation (17)

equation (18)

equation (19)

equation (20)

equation (21)

Hence equation is G2-invariant.

ITP Associated with Discrete Series Representations

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 equation be the Frechet space of smooth vectors.

Theorem 5.1: Suppose that equation is a discrete series representation of G(m+n). Suppose that m>n and (π1,H1) is a unitary representation of G(n). Then equation is well-defined. Suppose that equation . Then (,)G(n) is positive definite. Furthermore, equation completes to a unitary representation of G(m).

The key of the proof is to realize equation as a subspace of the L2-sections of the Hilbert bundle

equation

Proof: Write G= G(m+n). Fix a maximal compact subgroup K of G such that

equation

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

equation

Let A be the analytic group generated by a. The function log:Aa is well-defined. Let equation for each equation

Since equation is a discrete series representation, without loss of generality, realize equation on a right K-finite subspace of L2(G). So equation.

Let equation be Harish-Chandra’s basic spherical function. Let equation be the space of Harish-Chandra’s Schwartz space. It is well-known that every equation satisfies equation for some equation see for example Ch. 12.4 [?]). For every equation for a constant Ch. Observe that equation is G(m)-invariant.

Fix a positive root system in equation. Let A+ be the corresponding closed Weyl Chamber. Let be the half sum of positive roots. Let equation. Then equation for a positive constant Cu,v [5,6]. Notice that for equation,

equation

for some q0 and C>0. Let ρ(n) be the half sum of positive roots of the restricted root system equation. Let equation be the positive Weyl chamber of equation with respect to the root system equation. Since

equation.

equation. It follows that equation for every equation. Notice that equation is always bounded for equation. We see that

equation always converges. So equation is well-defined. Now suppose that equation

Notice that

equation is bounded by a multiple of equation. So equation. For each equation and equation, define equation to be the equation -valued function on G:

equation in the strong sense. Notice that for gG,h1∈G(n),

equation (22)

equation (23)

equation (24)

equation (25)

equation (26)

So equation can be regarded as a section of the Hilbert bundle

equation

In addition, we have

equation (27)

equation (28)

equation (29)

equation (30)

equation (31)

equation (32)

equation (33)

equation (34)

equation (35)

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

equation

To see this, recall that equation. In particular, for any N>0 and a∈A+, k1,k2∈K, there exists Cu,N>0 such that:

equation.

Write equation for g= k1ak2. Then there also exists Cv,N>0 such that

equation

Fix an N such that equation. In particular, equation Observe that the function

equation

is bounded by a multiple of (L(h)WN(g),WN(g)), which, by a Theorem of Cowling-Haagerup-Howe [5], is bounded by a multiple of equation. Hence

equation

equation

Eqn. (29) converges absolutely. Therefore Eqn. (31) holds.

Now we have

equation

It follows that equation. Realize equation as equation, which is a subspace of L2-sections of the Hilbert bundle:

equation

Clearly (,)G(n) is positive definite. Let equation be the completion of equation.

Since G(m) acts on equation and it commutes with G(n), G(m) acts on equation and it preserves (,)G(n). So the action of each g2∈G(m) can be extended into a unitary operator on equation. The group structure is kept in this completion essentially due to the fact that each extension is unique. Therefore equation completes to a unitary representation of G(m).

Definition 4

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 equation 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).

Conclusion

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 [7].

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