12
$\begingroup$

In physics textbooks one frequently sees the name (affine) Kac Moody algebra used to describe the universal (one dimensional) central extension of the loop algebra of a semisimple algebra. But this is not an affine Kac Moody algebra: one would still need a derivation extension to obtain a true (untwisted) affine Kac Moody algebra.

Do the properties of the object that physicists call affine Kac Moody algebra resemble the properties of the true Kac Moody algebra so closely that this misuse of the terms is rendered harmless? Meaning, are physicists somehow justified to assume that properties proved for affine Kac Moody algebras hold for "their" affine Kac Moody algebras?

$\endgroup$
5
  • 6
    $\begingroup$ It is my impression that, as seems to often happen in physics-y contexts, names have a bit of free-association, context-dependence, and heuristic suggestive-ness. So, well, I think the general attitude would be that "we" are "possibly justified" in assuming this-or-that, based on some prior experiences. This may often be weaker than the "justification" insisted upon in mathematics textbooks, for example, but that doesn't mean it's not more productive in the long run. $\endgroup$ Commented Aug 2, 2019 at 21:26
  • 1
    $\begingroup$ @paulgarrett: I think that's exactly right. $\endgroup$
    – Nik Weaver
    Commented Aug 3, 2019 at 1:38
  • $\begingroup$ @paulgarrett So in this case the assumption would be that the derivation extension does not affect the main properties, it seems. That seems to be about right: in Kerf's book (a mathematical physicist from what I gather) he says that the derivation extension is a "technicality". $\endgroup$
    – soap
    Commented Aug 3, 2019 at 10:47
  • 1
    $\begingroup$ @Soap, yes, I think your reading of that is what is intended. $\endgroup$ Commented Aug 3, 2019 at 16:32
  • $\begingroup$ The accepted answer to this post and the included references might be relevant to your question. $\endgroup$ Commented Aug 9, 2019 at 10:11

1 Answer 1

9
$\begingroup$

I can't address all uses by all physicists, but in many contexts, they consider only representations at a fixed level that admit a well-behaved energy grading. That is, sometimes an energy grading is implicit in the environment. Such a grading can be given by the eigenvalues of some semisimple operator, and adjoining such an operator to the centrally extended loop algebra yields the full affine Kac-Moody Lie algebra. In this case, the representations in question naturally extend to representations of the larger algebra, but when energy is implicit, a physicist may feel that references to an explicit operator are unnecessary.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .