Anytime a one-dimensional central extension appears in the physics literature, immediately they assume that in any irreducible representation the central charge will be a multiple of the identity, implicitly (and sometimes explicitly) using Schur's Lemma (for Lie algebras). However, the version of the Schur's Lemma which tells us this is only valid for finite-dimensional modules, and physicists are usually interested in representing Hilbert spaces which in general can have infinite dimension.

Is it somehow still justified to say that the central charge is a multiple of the identity even for infinite-dimensional irreducible representations?

The two examples I have in mind are the Virasoro algebra as one dimensional central extension of the Witt algebra and the universal central extension of a loop algebra, both appearing in CFT texts, and in both cases they use Schur's Lemma as described.

  • $\begingroup$ I’m pretty sure Schur’s Lemma doesn’t assume finite dimensionality. Either way, if you’re working with a reductive group over the complex numbers then you only have finite dimensional irreducible representations $\endgroup$
    – leibnewtz
    Aug 3, 2019 at 12:15
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    $\begingroup$ Where is the question? $\endgroup$ Aug 3, 2019 at 12:47
  • $\begingroup$ @StevenLandsburg I added it explicitly. $\endgroup$
    – Soap
    Aug 3, 2019 at 14:56
  • $\begingroup$ @leibnewtz It depends where you read it. That's why I said "that version", which says that for finite dimensional irreducible modules one has that elements of the center are represented as multiples of the identity. You can see this as a corollary of the Schur's lemma you are thinking about. I am wondering if this corollary somehow extends to infinite-dimensional modules. $\endgroup$
    – Soap
    Aug 3, 2019 at 15:01
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    $\begingroup$ First, there certainly do exist infinite-dimensional irreducible unitary repns of simple Lie groups, e.g., $SL_2(\mathbb R)$. Second, at least a weak version of a spectral theorem is needed to prove the corresponding Schur's lemma. $\endgroup$ Aug 3, 2019 at 17:46

1 Answer 1


Let $\mathfrak{g}$ be a complex Lie algebra with a distinguished nonzero central element $x$, and let $V$ be an irreducible representation of $\mathfrak{g}$. The usual proof of Schur's lemma can be adapted to show that if $x$ admits an eigenvector in $V$, then $x$ acts by a scalar: If $v$ is an eigenvector with eigenvalue $\lambda$, then $v$ lies in a submodule, namely the kernel of the central transformation $x - \lambda$, so irreducibility implies this kernel is equal to $V$.

Your question becomes: What makes it safe for physicists to assume a central element has an eigenvector?

The usual answer is that the representations that appear naturally tend to have gradings with at least one finite dimensional component, so the action of the central element restricts to such finite dimensional parts where one necessarily has an eigenvector.

  • $\begingroup$ Do you know of some book or text which expands a bit on this? $\endgroup$
    – Soap
    Sep 5, 2019 at 10:18
  • $\begingroup$ @Soap I do not know of a general reference, but the idea can be extracted from the first few lectures in Kac, Raina "Bombay Lectures on highest weight representations" $\endgroup$
    – S. Carnahan
    Sep 5, 2019 at 14:48
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    $\begingroup$ Thanks, I've been looking into that book. By the way, I just found out that this is discussed and proved in the book "Lie algebras with triangular decompositions" by Moody and Pianzola, page 33, in case someone reads this in the future. $\endgroup$
    – Soap
    Sep 5, 2019 at 14:58

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