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I know it's true that every irreducible representation of a compact Lie group is finite-dimensional, and I've seen it mentioned that this result is a corollary of the Peter-Weyl Theorem. However, I don't really see how that's possible. The Peter Weyl theorem gives us the following isomorphism of $G \times G$-representations for a compact Lie group $G$:

$$\widehat{\bigoplus_{\rho \in \widehat{G}}}\,\, \rho^* \boxtimes \rho \simeq L^2(G),$$

where $\widehat{G}$ denotes the space of finite-dimensional irreducible representations of $G$. Is there any way to use this statement to deduce that every irreducible representation is finite-dimensional?

Here is what I have thus far. The Schur orthogonality relations, which state that the images of two different irreducible representations under the Peter-Weyl map are orthogonal, can be proven in the case where only one of the two representations is finite-dimensional. So for an infinite-dimensional $\rho$, the image of $\rho^* \boxtimes \rho$ in $L^2(G)$ is orthogonal to the images of all the finite-dimensional representations. If the Peter-Weyl map embeds $\rho^* \boxtimes \rho$ as a subrepresentation of $L^2(G)$, then we would have a contradiction because the image of the direct sum of all the finite-dimensional representations is dense in $L^2(G)$. But I don't know how to show that the Peter-Weyl map is injective without first proving that it's an isometry, and this uses finite-dimensionality.

Clarification: I am not putting any restrictions on what sort of infinite-dimensional vector space we're representing $G$ on. I just want to use the Peter-Weyl Theorem to show that every infinite-dimensional representation of a compact Lie group has a proper invariant subspace, with no additional assumptions on the vector space.

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    $\begingroup$ You are understating the P-W theorem, see Part II in en.wikipedia.org/wiki/Peter%E2%80%93Weyl_theorem $\endgroup$ – Moishe Kohan Nov 25 '17 at 0:03
  • $\begingroup$ @MoisheCohen Good point! But in part II of the theorem, it is assumed that the representation under consideration is unitary, and the averaging trick only works in the finite-dimensional case. Is it true that any infinite-dimensional representation is unitarizable? $\endgroup$ – Ashvin Swaminathan Nov 25 '17 at 0:10
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    $\begingroup$ To average you do not need finite dimensionality. Even more so: Every continuous representation of an amenable group is unitarizable. $\endgroup$ – Moishe Kohan Nov 25 '17 at 0:13
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    $\begingroup$ Since you are interested in "exotic" vector spaces you should spell out the assumptions on the vector spaces you are willing to make. For instance, are you OK with normable vector spaces? Etc. $\endgroup$ – Moishe Kohan Nov 25 '17 at 0:29
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    $\begingroup$ OK, then you should say what do you mean by a "proper invariant subspace": Do you require a subspace to be closed? (This is usually required.) Furthermore, without any restrictions on the vector space, you cannot derive the desired conclusion form P-W. $\endgroup$ – Moishe Kohan Nov 25 '17 at 12:48
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First of all, the Peter-Weyl Theorem is about unitary representations, it will be of no use if you want to study representations on topological vector spaces which are not isomorphic to Hilbert spaces (or pre-Hilbert spaces). Thus, formally speaking, the answer to your question (at the end of the 1st paragraph) is negative. Nevertheless, one has:

Theorem. Let $V$ be a Hausdorff locally convex quasicomplete topological vector space, $G$ a compact (and Hausdorff) topological group and $\rho: G\to Aut(V)$ a continuous irreducible representation (meaning that $V$ contains no proper closed invariant subspaces). Then $V$ is finite-dimensional.

You can find a proof in

R.A.Johnson, Representations of compact groups on topological vectors spaces: some remarks, Proceedings of AMS, Vol. 61, 1976.

The point of considering this class of topological vector spaces (which includes, for instance, all Banach spaces) is that one has a satisfactory theory of integration for maps to such spaces. As for more general vector spaces, I have no idea, I suppose that the finite-dimensionality claim is simply false. For instance, if you drop the assumption that $V$ is Hausdorff (which one usually assumes) and take a vector space with trivial topology, you will have irreducible representations of any group on such a vector space.

One last thing: In the context of Hilbert spaces, there is a very short and direct (avoiding PWT) proof of finite dimensionality, given in this mathoverflow post.

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  • $\begingroup$ I have a question about representations in Banach spaces. For Hilbert spaces, you have a correspondence between all classes of finite-dimensional irreps of the Hausdorff compact group G and all irreducible components of the regular representation of $L^2(G)$. If we consider $C(G)$ instead of $L^2$ with the uniform norm, do we have the same correspondence? $\endgroup$ – Ángel Valencia Jun 30 '18 at 15:35
  • $\begingroup$ @ÁngelValencia: I suggest, you ask a separate question and not at MSE but Mathoverflow. $\endgroup$ – Moishe Kohan Jul 6 at 4:49

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