Let $G$ be a compact subgroup of $O(n)$. Let $\rho$ be a continuous finite dimensional representation of $G$.

Question Is it true that there exists a continuous finite dimensional representation $\pi$ of $O(n)$ such that $\rho$ is a direct summand of the restriction of $\pi$ to $G$?

I am almost sure that this is true. A reference would be very helpful.

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    $\begingroup$ No, this is false. For $n\geq 3$, any nontrivial representation of $\mathrm{O}(n)$ has its kernel contained in $\{\pm I\} $. Just take for $G$ a finite group and for $\rho $ a representation with $\#\,\mathrm{Ker} \rho \geq 3$. $\endgroup$ – abx May 3 '17 at 9:51
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    $\begingroup$ @abx: I am not sure that I understood your argument. I think that it might happen that $\sharp Ker \rho \geq 3$, but $\sharp Ker \pi\leq 2$. $\endgroup$ – makt May 3 '17 at 10:10
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    $\begingroup$ Oops, you are right, sorry. I missed the "direct summand" part. $\endgroup$ – abx May 3 '17 at 10:20

I don't have a reference, but here is a short argument. Assume $G<H$ are compact groups (in our example take $H=\text{O}(n)$) and recall that unitary representations of compact groups are completely reducible. Recall also that by Peter-Weyl every irreducible unitary representation is finite dimensional.

Assume first that $\rho$ is an irreducible finite dimensional continuous representation of $G$ and endow it with a $G$-invariant inner product. Fix $\pi$ to be any finite dimensional subrepresentation of the unitary induction $\text{Ind}_G^H(\rho)$, eg an irreducible subrepresentation. Then, by Frobenius reciprocity, $\text{Hom}_G(\rho,\text{Res}_G^H \pi)\simeq \text{Hom}_H(\text{Ind}_G^H(\rho),\pi)$ and the latter is non-empty by our choice of $\pi$. It follows that $\rho$ is a direct summand of $\text{Res}_G^H \pi$ by its irreducibility.

The case of a general $\rho$, continuous finite dimensional representation, follows easily.

  • $\begingroup$ Thank you. While $Res_G^H\pi$ is infinite dimensional, but it is easy to truncate it to imbed $\rho$ to a finite dimensional representation. $\endgroup$ – makt May 3 '17 at 13:27
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    $\begingroup$ @salev, no. We choose $\pi$ to be finite dimensional and restriction does not change the underlying space (I used the notation $\text{Res}_G^H$ for the restriction of a representation from $H$ to $G$, as is common). Maybe I should add: the fact that I could choose $\pi$ to be finite dimensional follows from Peter-Weyl, eg take $\pi$ to be an irreducible $H$ rep. $\endgroup$ – Uri Bader May 3 '17 at 13:39
  • $\begingroup$ Sorry, you are right. I misunderstood. $\endgroup$ – makt May 3 '17 at 13:56

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