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If a group $ G $ admits a faithful irreducible finite dimensional complex representation $ \pi: G \to GL_n(\mathbb{C}) $ then the algebra generated by $ \pi(G) $ is all of $ M_n(\mathbb{C}) $. So in particular the center of $ G $ must be contained in the center of the full matrix algebra, which is the diagonal copy of $ \mathbb{C}^* $ , $ \mathbb{C}^* I $. See

If $(\pi,V)$ is irreducible, then $\pi(G)$ spans $\operatorname{End}(V)$

If, moreover, $ G $ admits a faithful irreducible finite dimensional unitary representation (which is equivalent to the original hypothesis together with $ G $ being compact) then the center of $ G $ must be contained in the diagonal copy of $ U_1 $.

What happens if we drop the hypothesis of finite dimensionality? If a Lie group $ G $ admits a faithful unitary irrep can we still conclude that the center of $ G $ is isomorphic to a closed subgroup of $ U_1 $?

Ian Gershon Teixeira
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  • Consider nilpotent Lie groups. – markvs Feb 10 '22 at 03:32
  • @markvs could you perhaps expand on that thought? – Ian Gershon Teixeira Feb 11 '22 at 20:21
  • Compactness is assumed? – markvs Feb 11 '22 at 20:26
  • No I didn't assume compactness ( although I'm always especially interested in the compact case). It's just what the question says. "If a Lie group $ G $ admits a faithful unitary irrep can we still conclude that the center of $ G $ is isomorphic to a closed subgroup of $ U_1 $?" – Ian Gershon Teixeira Feb 11 '22 at 20:31
  • Then take the direct product of two Heisenberg groups. – markvs Feb 11 '22 at 20:33
  • ok that sounds like the kind of counterexample I was expecting. Could you say a little more about that? I see an obvious way to make a faithful unitary representation and an obvious way to make a unitary irreducible representation but I'm not sure how to make one that is faithful unitary and irreducible. – Ian Gershon Teixeira Feb 11 '22 at 20:37
  • See https://en.wikipedia.org/wiki/Heisenberg_group#Representation_theory – markvs Feb 11 '22 at 20:42
  • I'm aware of the schrodinger representation but the obvious thing to do is take a direct sum of two schrodinger representations but that's not irreducible. Are you suggesting something like take both schrodinger representations on the same hilbert space but then scale one by some other function so the rep is still faithful? It's not clear to me how to get a representation of the product that is simultaneously faithful and irreducible – Ian Gershon Teixeira Feb 11 '22 at 20:55
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    I would take the tensor product instead of direct sum. But I have not thought about it. – markvs Feb 11 '22 at 20:57

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