Given a locally compact group $G$ with a compact subgroup $K$.

Assume we are given two irreducible, infinite dimensional, admissible representations $\pi$ and $\pi'$ of $G$.

What are examples, where $Res_K \pi$ and $Res_K \pi'$ are not isomorphic $K$ representation except for a finite dimensional representation?

Admissible in this context means that $Res_K \pi$ decomposes with finite multiplicity.

I would be mostly interested in examples of reductive groups over local fields. Also what will happen, if we replace irreducible by indecomposable?

If $G$ is connected, let $K$ be its maximal compact subgroup. If $G$ is totally disconnected, let $K$ be open.

Are there examples, where nuclear representations with linearly independant characters become isomorphic after restriction to $K$?

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    $\begingroup$ What does that mean, "isomorphic up to a finite dimensional representation". Does it mean that there is a morphism from one representation to the other with finite dimensional kernel and cockerel? That can't be right, because that there would be many trivial answers to your question, like $\pi$ any infinite dimensional representation (irreducible admissible) and $\pi'$ the trivial representation. $\endgroup$ – Joël Feb 28 '12 at 13:56
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    $\begingroup$ pm, Joël, perhaps "up to a finite dimensional representation" means the multiplicities of the $K$-types are different for only finitely many $K$-types. Of course, there are going to be trivial counter-examples when the representations are not infinite-dimensional. $\endgroup$ – B R Feb 28 '12 at 15:09
  • $\begingroup$ Yes, I am interested in infinite dimensional representations only. $\endgroup$ – Marc Palm Feb 28 '12 at 15:12

If Archimedean local fields are ok, then the simplest example probably occurs with $G=GL(2, \mathbb R)$ and $K=SO(2, \mathbb R).$ The irreducible representations of $K$ are in bijection with the integers. One can construct representations of $GL(2, \mathbb R)$ such that the set of $K$-types is the set of odd integers, the set of even integers, the set of odd integers with absolute value at least (or at most) $2n+1,$ or the set of even integers with absolute value at least (or at most) $2n.$ In particular, one can construct two representations such that the restrictions to $K$ differ by deleting a single representation of $K.$

In a bit more detail, take $s=(s_1, s_2)$ a pair of complex numbers and $\omega=(\omega_1, \omega_2)$ where for $i=1,2,$ $\omega_i$ is either trivial or the sign character. Consider smooth functions $GL(2, \mathbb R) \to \mathbb C$ such that $$ f\left( \begin{pmatrix} a& b \\ 0& d \end{pmatrix} g \right) = |a|^{s_1} \omega_1(a) |d|^{s_2} \omega_2(d) f(g). $$ The $K$-type corresponding to an integer $n$ will appear in this representation if and only if $(-1)^n = \omega_1(-1)\omega_2(-1).$ (And all have multiplicity one.) You get reducibility when $s_1-s_2$ is an integer $m$ with this parity. If $m$ is positive there is a subrepresentation which lives on the $K$-types $\ge m$ in absolute value. If $m$ is nonpositive, there is a subrepresentation which lives on the $K$-types $\le -m$ in absolute value.

For obvious reasons, the reference I know best is the book Goldfeld and I wrote. We discuss reducibility of $(\mathfrak{g}, K)$-modules on p. 267. On p. 332, we give a fairly elementary description of each invariant subspace looks like in the representation of $GL(2, \mathbb R)$ on smooth functions. I see another reference in the comments as well.

Adding: in the $p$-adic case a similar phenomenon occurs and may be easier to see. Take $F$ a nonarchimedean local field with ring of integers $\mathfrak{o}.$ Take $G=GL(2, F)$ and $K=GL(2, \mathfrak{o}).$ Consider the representation $Ind_B^G \chi$ parabolically induced from a character of the Borel subgroup $B$: $$ \chi\begin{pmatrix} a& b \\ 0& d \end{pmatrix} = |a|^{s_1}|d|^{s_2}. $$ As a $K$-module $Ind_B^G \chi$ is isomorphic to $Ind_{B\cap K}^K 1$ for all $s_1, s_2.$ (Here "$1$" means the one dimensional trivial representation of $B \cap K$.) As a $G$-module $Ind_B^G \chi$ is irreducible for most values of $s_1, s_2,$ but if $s_1=s_2$ then the subspace of $Ind_{B\cap K}^K 1$ corresponding to the trivial representation of $K$ is actually a $G$-invariant subspace spanned by the function $g \mapsto |\det g|^{s_1}.$

  • $\begingroup$ This example is okay: Can you please write down the representation in terms of induced representation, so that I can check this myself. I guess via restriction by steps, this is remains true for restriction to $O(2)$. This example you mention, does it stem from the fact $GL(2) = GL(2)^+ \rtimes \{ \pm 1 \}$? $\endgroup$ – Marc Palm Feb 28 '12 at 17:03
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    $\begingroup$ You can find this series of examples worked out completely in Howe and Tan's book on non-abelian harmonic analysis. $\endgroup$ – Victor Protsak Feb 28 '12 at 19:22
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    $\begingroup$ pm, it works for $SL(2,\mathbb R)$. It stems from the fact that the maximal torus of $SL(2,\mathbb R)$ is $\mathbb R^\times\simeq \mathbb R_{>0}\times\lbrace\pm 1\rbrace$. $\endgroup$ – B R Feb 29 '12 at 0:09

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