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.

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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