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More precisely, I often read or listen that Lorentz group has not (non trivial) unitary finite dimensional irreducible representations because it is not compact. Now, I know that the "converse" part of this theorem is part of the statement of Peter-Weyl theorem (i.e. if $G$ is a compact group on a Hilbert space $V$ and $\phi$ is a unitary representation of $G$, then $V$ is the orthogonal sum of finite dimensional irreducible invariant subspaces), but I don't know which is the theorem that states the preceding claim.

I really appreciate proofs, references or suggestions for readings! (keep in mind that at the present day I never studied represention theory, but I need some deeper notion because I'm a physicist with the aim to be a mathematical physicist and the current duty to study QFT!)

Thank you in advance.

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    $\begingroup$ The Lorentz group is non-compact, because one of its continuous paramters is defined on closed interval $0\le v<c$, physically the velocity of a boost must be less than and not equal to the speed of light. $\endgroup$ – innisfree Oct 1 '13 at 13:46
  • $\begingroup$ Yeah, sorry, it was a typo. I have fixed it! $\endgroup$ – Federico Oct 1 '13 at 13:49
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    $\begingroup$ You might also want to add the requirement that the sought for representation be irreducible. Otherwise $G$ has many fin. dim. reps.: sums of trivial one. $\endgroup$ – Marek Oct 1 '13 at 14:16
  • $\begingroup$ Sure, you're right. Added, thank you. $\endgroup$ – Federico Oct 1 '13 at 14:19
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TL; DR There is no faithful unitary representation of a non-compact Lie group that realizes it as a closed Lie subgroup of some $U(n)$.

First of all, it's not true that a non-compact group can't have unitary representations. Because what is a unitary representation? It's simply a homomorphism into some $U(n)$ that then acts linearly on an $n$-dimensional vector space (in other words, a unitary representation factors through a unitary group). So we want to find a smooth homomorphisms into $U(n)$ from non-compact groups. But they clearly exist! For example $\rho : (\mathbb R, +) \to U(1), x \mapsto \exp(2\pi ix)$ is such.

The correct statement you are after (I think) is that it's not possible to find a faithful unitary representation. That is, a homomorphism that doesn't map any element of $G$ to identity of $U(n)$ (note that $\rho$ above sends $\mathbb Z$ to $1$). Altogether, we want to smoothly embed $G$ as a Lie subgroup of $U(n)$ (there are caveats here regarding words such as immersion, embedding and smooth structure that I don't want to get into).

Let $\psi : G \to U(n)$ be a continuous homomorphism. Suppose $\psi(G)$ is closed in $U(n)$. Then it is compact (since $U(n)$ is) and since this is to be homeomorphic to $G$, $G$ must be compact as well.

Note that not every subgroup of a Lie group needs to be closed. For example the line of an irational slope in the $U(1) \times U(1)$ (a torus) is a subgroup, isomorphic and homeomorphic to $\mathbb R$ whose closure is all of $U(1) \times U(1)$. So there remains a possibility of finding a homomorphism with a non-closed image. This, for a general $G$ other than $\mathbb R$, seems like a very hard task, but I don't see any way of excluding this possibility and am actually pretty sure these examples do exist. But I'm quite of my depth here, so that's it from me.

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  • $\begingroup$ Maybe you want to fix the initial typo. (I think it should be "Theorem", not "TL; DR".) $\endgroup$ – Federico Oct 2 '13 at 20:34
  • $\begingroup$ Well, it certainly is a theorem but more importantly it's a summary of the rest of the answer for those who don't want to read it all of it; hence tldr. Anyway, thanks for accepting. $\endgroup$ – Marek Oct 2 '13 at 20:38

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