If one has a finite dimension simple Lie algebra, one can easily calculate that taking the centralizer of a torus (or toral subalgebra), that is, summing the weight spaces that lie in some proper subspace of the dual Cartan, always gives a finite dimensional reductive Lie algebra; actually almost semi-simple, except that there is some central toral subalgebra (maybe bigger than the original torus).

For an affine Lie algebra, the picture is the same if you pick a subspace on which $(-,-)$ is positive definite, you get again something finite dimensional and reductive. If you pick a subspace containing $\delta$, however, things are a bit messier. Now you have an affine Lie algebra plus a central piece in every single one of the imaginary weight spaces; thus you have an infinite dimensional center. I would kind of like to think of this as a Borcherds algebra, where I add infinitely many rows and columns of zeros to the affine Cartan matrix.

For a hyperbolic Kac-Moody algebra, the picture is even worse; I still get a hyperbolic KM algebra attached to the root subsystem living in the subspace, but the "extra stuff" in imaginary weight spaces is much more complicated and not central anymore. This seems very complicated, but I hope to get some kind of handle on it:

Is it true that the centralizer of torus in a hyperbolic Kac-Moody algebra is a Borcherds algebra (necessarily infinite rank in most interesting cases)? Is there some nice description of its Cartan matrix?

  • $\begingroup$ Any reason you want to take the centralizer of a connected torus as opposed to a more general subgroup of the torus? (Assuming I'm dualizing your question correctly) $\endgroup$ Commented Apr 18, 2012 at 16:15
  • $\begingroup$ I'm interested in an application (well, more of a tossed-off comment) where I believe only te connected case is relevant. However I'd be happy to hear about the generalization as well. $\endgroup$
    – Ben Webster
    Commented Apr 18, 2012 at 17:19
  • $\begingroup$ I am puzzled. I thought that Borcherds algebras are something "very very very huge", how can it sit inside Kac-Moody, which is not so huge ? I mean even their Dynkin diagram typically is something like infinite lattice - very different from the typical Kac-Moody which has finite set as Dynkin diagram. $\endgroup$ Commented Apr 20, 2012 at 6:00
  • $\begingroup$ Your intuition about sizes is failing you; Kac-Moody algebras are "big" in a certain sense. Think about my example with affine Lie algebras. It's easy to cook up an infinitely generated centralizer.... $\endgroup$
    – Ben Webster
    Commented Apr 20, 2012 at 11:55
  • $\begingroup$ Another way to say this is that actually all infinite dimensional vectors spaces with some kind of countable basis are actually the same size, and you should never be surprised when you can fit one inside another no matter how "big" or "small" they looked. $\endgroup$
    – Ben Webster
    Commented Apr 20, 2012 at 19:35

1 Answer 1


The answer to your first question is "yes", and it follows from Theorem 1 in Borcherds's paper Central extensions of generalized Kac-Moody algebras, which is available online as number 11 on his papers page. The inner product and involution can be chosen as restrictions from your hyperbolic Kac-Moody algebra.

The general philosophy for recognizing these objects is that "anything that looks like a Borcherds algebra is a Borcherds algebra" (cf. Wikipedia), and it holds here. As long as your Lie algebra is long and skinny (i.e., admits a $\mathbb{Z}$-grading with finite dimensional pieces), and symmetrically shaped (i.e., admits an involution and a bilinear form), you're all set.

I don't know a nice description of the Cartan matrix, but the proof of the theorem gives a general construction. You may also want to look at Borcherds's later paper A characterization of generalized Kac-Moody algebras (number 15 on the same page). As a rule, you should try to avoid actually using the Cartan matrix of a Borcherds algebra if you possibly can.

  • $\begingroup$ Exactly what I was looking for. Thanks very much (and congrats on your recent nuptials). $\endgroup$
    – Ben Webster
    Commented Apr 20, 2012 at 11:53

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