I have two reducible representations of a finite group $G$ of Lie type, $\rho, \pi$. They both have multiplicity one, and I know that they share exactly one irreducible subrepresentation.

Is there a method to explicitly obtain the character of the common subrepresentation using their characters? (so, not using character tables)

Using Mackey theory I have found an intertwining operator $M:\rho\rightarrow\pi$, so the subrepresentation is isomorphic to the image of $M$ in $\pi$. But the dimension of $\pi$ is quite large, "objectively" - not just compared to that of $\rho$, so it seems tedious to compute the image.

  • $\begingroup$ I would not be surprised if it were possible to write down two groups $G, H$, a pair of characters of $G$, and a pair of characters of $H$ such that both pairs have inner product $1$ and also have the same numerical entries (and the conjugacy classes are the same sizes) but such that the character of their common subrepresentation is different. $\endgroup$ – Qiaochu Yuan Jun 21 '11 at 21:38
  • $\begingroup$ I wouldn't either. But that is quite a narrow interpretation of my question. One should also use the group structure. For example, the character of induction can be explicitly given using the original character, but the group structure is involved. $\endgroup$ – simplequestions Jun 22 '11 at 15:31

Given a finite or compact group $G$ (without additional structure) and two finite-dim'l representations $(\rho, V_\rho)$ and $(\pi, V_\pi)$ no matter what multiplicity, you can define $G$-equivariant projections $V_\pi \rightarrow V_\pi$ by $$ v \mapsto \int\limits_{G} \overline{( tr \rho(g))} \pi(g)v d g.$$ You may consider sums, of course.

It surjects onto $V_{\pi \cap \rho}$, where $V_{\pi \cap \rho}$ is the maximal invariant subspace which is isomorphic (as $G$-reps) to a subrepresentation of a multiple of $V_\pi$.

This follows from the Schur orthogonality relations.

Given multiplicty one for both $\pi$ and $\rho$, we obtain $V_{\pi \cap \rho} \cong V_{\rho \cap \pi}$. Lets say $\pi \cap \rho = \rho \cap \pi$ is the gcm.

Given a orthonormal basis $(v)$ of $V_\rho$, we can define $$ tr \rho \cap \pi(g) = \sum_v \langle v, P \pi(g) v \rangle.$$

  • $\begingroup$ Thanks Marc! This is pretty cool. It's also nice that $\rho$ can even be taken to be a virtual character. $\endgroup$ – simplequestions Oct 9 '13 at 22:40

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