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Consider a unitary real representation of a Lie group $G$ over a real Hilbert space $\mathcal{H}_\mathbb{R}$ \begin{equation} \rho:G\rightarrow U(\mathcal{H}_{\mathbb{R}}) \end{equation}

Taking the complexification $\mathcal{H}_{\mathbb{C}}:=\mathbb{C}\otimes_{\mathbb{R}},\mathcal{H}_{\mathbb{R}}$ it's straightforward how to extend $\rho$ to a complex unitary representation of $G$ over $\mathcal{H}_{\mathbb{C}}$.

Under which hypotheses is the irreducibility of $\rho$ preserved moving to the complexification?

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  • $\begingroup$ For finite dimensional representations of compact groups this is quite well documented in the book by Broecker and tom Dieck. For infinite dimensional representations of non-compact groups (inf. dim. reps of compact groups are never irreducible) I don't know the answer. I'm looking forward to reading the answers to this question. $\endgroup$ – Vincent Apr 23 '15 at 7:22
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There are situations where irreducibility is preserved after complexification, however by saying unitary I understand you to mean that the original representation already has an invariant complex structure $J$. In that case the answer is never. The easy way to see this is that the ring of endomorphisms $\text{End}_G(\mathcal{H}_\mathbb{C})$ which commutes with $G$ is not the complex scalars, because it contains independent commuting elements old $J$ (acting on the right factor) and new $i$ (acting on the left) which both square to $-1$. Hence by Schur's lemma, the representation is not irreducible.

If you meant something else by unitary (orthogonal?), then a sufficient condition is $$\text{End}_G(\mathcal{H}_\mathbb{R})=\mathbb{R}$$ I am completely sure about this second part for finite dimensional reps, but it should still hold for Hilbert spaces (I think).

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    $\begingroup$ I think "unitary real representation" just means it's a representation on a real Hilbert space preserving the real inner product (so it means "orthogonal representation"). $\endgroup$ – Qiaochu Yuan Apr 23 '15 at 3:59
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    $\begingroup$ Yes, I should have been more precise in my question. With unitary real/complex representation I mean exactly what @QiaochuYuan wrote in the comment: $<\rho(g)u|\rho(g)v>=<u,v>$. $\endgroup$ – moppio89 Apr 23 '15 at 8:55
  • $\begingroup$ Addendum: I'm actually working with the Heisenberg group which is not compact. $\endgroup$ – moppio89 Apr 23 '15 at 8:57

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