5
$\begingroup$

Edit 2: Please note the new, more specific version of the question: Does there always exist an irreducible representation occurring with multiplicity one when inducing from $M=Z_K(A)$ to $K$?

Let $K$ be a compact connected Lie group and $M\subset K$ a closed (hence compact Lie) subgroup, not necessarily connected. Let $\tau$ be an irreducible (hence finite-dimensional) complex $M$-representation. Does there exist an irreducible $K$-representation $\sigma$ such that $\tau$ occurs with multiplicity one in $\sigma|_M$?

By Frobenius reciprocity, this is equivalent to the question:

Does the $K$-representation $\mathrm{Ind}^K_M(\tau)$ contain an irreducible subrepresentation that occurs in it only once?


Edit: As a motivation, I am reading a thesis at the moment which says that for $(K,M)=(\mathrm{SO}(n+1),\mathrm{SO}(n))$ or $(K,M)=(\mathrm{SU}(n+1),\mathrm{S}(\mathrm{U}(1)\times \mathrm{U}(n)))$, one has even the stronger statement that $$ [\sigma|_M:\tau]\leq 1 $$ for any pair of irreducible representations $(\sigma,\tau)$ of $K$ and $M$. In particular, for such pairs of groups, the answer to my question is positive according to the thesis. However, it does not cite references where I could directly check this, or where I could find an answer to my question.

$\endgroup$

2 Answers 2

7
$\begingroup$

Let $K=SU(2)$, $M=Z(SU(2)) = \{\pm I\}$, and $\tau$ the sign representation of $M$. Then given an irreducible representation $\sigma$ of $SU(2)$, $\sigma|_M$ is necessarily a character with multiplicity $dim(V)$. Moreover, the sign character appears exactly for the even dimensional representations (i.e. those of odd highest weight). Thus $\tau$ always appears with even multiplicity in any irreducible representation.

Regarding your motivation, those particular pairs of groups are special - I think the key word here is Gelfand pair.

$\endgroup$
-1
$\begingroup$

In Vogan's green book, on page 400 it is shown that given an irreducible $(\mathfrak g, K)-$module, then, each lowest $K-$type has multiplicity one. Perhaps this gives a partial answer to your question. best regards

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.