I am looking for literature that might contain the spherical representations of $GL(n, \mathbb{C})$. Here a spherical representation is an irreducible representation $\rho$ of $G$ on $\mathbb{C}$ such that $\rho_{K}$, for $K$ a maximal compact subgroup, fixes a vector in $\mathbb{C}$. I realize my question is similar to this one, however I am looking for the spherical representations that may involve $Sp_{2n}$, $U_n$ and $SO_n$ for $n$ odd and even. I apologize if I have erred with my vocabulary or if the question lacks sufficient detail for a meaningful reference; I will gladly supply more details if there is confusion.

The motivation for this query is an attempt to find out which maximal subgroups of $GL(n^2)$, for $n \in \mathbb{N}$, stabilize one-dimensional subspaces when the representation $GL(n^2) \to GL(V)$ for $V = \mathrm{Sym}^n \mathbb{C}^{n}$ is restricted to this maximal subgroup. One such subgroup that fixes a 1-dimensional subspace that has been found is $GL(n) \times GL(n)$ under the tensor product representation, which fixes $\wedge^n \mathbb{C}^n \times \wedge^n \mathbb{C}^n$ i.e., the determinant.

As a side note, another technique I have been using for examining whether certain maximal subgroups have invariant vectors is the restriction formula found in Fulton and Harris for restricting representations of $GL(n)$ to $O(n)$ and the branching rule involved with these representations.

  • $\begingroup$ What do you mean by "fix vectors in $\mathbb{C}$? If $\mathbb{C}$ denotes the trivial representation of the group then the answer is also trivial (namely, all of them). Is the group $G$ a complex Lie group (like $GL(n,\mathbb{C})$) or do you simply wish to consider complex representations (not necessarily on $\mathbb{C}$ - perhaps, you can specify whether they are supposed to be finite-dimensional). By the way, $U_n$ is not a complex Lie group, in fact, it is compact, and so it coincides with its own maximal compact subgroup $K.$ $\endgroup$ – Victor Protsak May 14 '13 at 6:00
  • $\begingroup$ I understand the OP is interested in a classification of the unitary (or smooth, admissible) representation, which are irreducible and have a invariant vector under the maximal compact subgroup, or equivalently the trivial representation is contained in the restriction to it. I think in his context, he wants to consider either the $\mathbb{R}$- or $\mathbb{C}$-points of these classical algebraic group, $U(n)$ making no sense over $\mathbb{R}$, though, and having a trivial answer over $\mathbb{C}$. Similarly, for $SO(n)$ over $\mathbb{R}$. $\endgroup$ – Marc Palm May 14 '13 at 6:50
  • $\begingroup$ @Marc @Victor thank you for your comments. I am indeed interested in what Marc managed to extract: namely, the classification of the unitary irreducible representations which have an invariant vector under the maximal compact subgroup. I am considering the $\mathbb{C}$-points of these classical algebraic groups, as you pointed out, $U(n)$ does not makes sense over $\mathbb{R}$. I have changed some language slightly and added some details to help with the reference request. Thanks again for your time. $\endgroup$ – Moderat May 14 '13 at 13:28

In the case of $GL(n, \mathbb{C})$, it is known that every unitary, irreducible, infinite-dimensional representation (the others are one-dimensional and factor through the determinant) is given as induced representation $\pi$ from a minimal parabolic associated to the Levi $M(\mathbb{C})$ (being the group of diagonal matrices). This one is spherical iff the restriction $\pi$ to $M(\mathbb{C}) \cap U(n)$ is trivial.

The Mackey Induction Restriction formula plus the Iwasawa decomposition indicates that this is the same question for inducing the restriction of $\pi$ to $M(\mathbb{C}) \cap U(n)$ up to $U(n)$.

This is the case if and only if the restriction of $\pi$ to $M(\mathbb{C}) \cap U(n)$ is trivial by Frobenius reciprocity.

This strategy works more generally for all parabolically induced representation in real reductive groups. Then look possibly at the structure of their subquotients. It is a theorem of Casselman that for a real reductive Lie group all smooth, admissible reps are found as such subquotients of such parabolically induced representations.

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  • $\begingroup$ Note, that I don't know which one of the parabolically induced ones have irreducible subquotients or are unitarizabile, though. I am only saying a classification of the former gives pretty easily a classification of the latter. $\endgroup$ – Marc Palm May 14 '13 at 16:17
  • $\begingroup$ I also would claim that the subquotients are never spherical, but I am not sure in the generality I have stated the results. $\endgroup$ – Marc Palm May 14 '13 at 16:18

There is a paper by Kramer about pairs $(G, K)$ with $G$ connected Lie group and $K$ spherical in $G$ that is for all irreducible representations of $G$, the space of vectors fixed by $K$ is at most 1-dimensional.

If I recall correctly Kramer gives some propreties of spherical pairs and provides the full classifications of spherical pairs $(G, K)$ with $G$ compact and simple.

Here: link text

Is this what you was looking for?

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  • $\begingroup$ No. What you state is simply the fact when $(G,K)$ is a Gelfand pair. The OP is search for a set of unitary representation, e.g., for $GL_2(\mathbb{C})$, it would be all unitary unramified continuous series representation and the $| \det |^s$ with $\Re s =1$. $\endgroup$ – Marc Palm May 14 '13 at 15:01
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    $\begingroup$ @David that article is indeed interesting; however, I believe that my further edits will help to clarify exactly what I seek. Thank you for finding this though, and for those who do not have MathSciNet I have found a freely available version of the article here: eudml.org/doc/89398 in case a viewer of this question is indeed looking for the above classification. $\endgroup$ – Moderat May 14 '13 at 15:16
  • $\begingroup$ I thought it could be related to the question in the other discussion (which apparently was merged with this one). It wasn't even clear to me you meant the same thing for spherical representations :) $\endgroup$ – David P May 14 '13 at 19:07

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