2
$\begingroup$

I have two vertex operator algebras and I would like to show that as graded vector spaces, they are isomorphic, rather than as algebras.

The issue is I have not found anything in the literature that has done this for a particular case. One idea I had was since there is a cohomology theory for VOAs, there are tools to compare invariants of them as algebras, and I am hoping there may be something in the literature which uses this (or some other method), to make a statement about them as graded vector spaces.

I would highly appreciate if anyone can point me to a paper that develops any of these ideas, that is, if one can 'descend' a statement of two VOAs to something about them as graded vector spaces.

$\endgroup$
3
  • $\begingroup$ The universal enveloping affine algebra associated two two different Lie algebras of the same dimension will have the same graded dimension and will be different VOAs. $\endgroup$ Feb 18, 2020 at 23:49
  • $\begingroup$ That just means that the "characters" $\operatorname{tr}(q^{L_0})$ are the same. For example, the VOA's associated with the $E_8\oplus E_8$ and the $D_{16}^+$ lattice have the same characters and thus are isomorphic as graded vector spaces. $\endgroup$ Feb 19, 2020 at 3:16
  • $\begingroup$ @MarcelBischoff So actually I have fermionic generators so you get a $(-1)^F$ that allows for cancellations to happen. I don't think in that case it's enough for characters to match. $\endgroup$
    – JamalS
    Mar 31, 2020 at 15:05

1 Answer 1

2
$\begingroup$

This was answered by Marcel Bischoff in a comment. For any vertex operator algebra, there is an invariant called the "character" or "graded dimension", that captures precisely the isomorphism type of the underlying graded vector space. In many cases, this is a modular form, and you may find yourself using tools from the theory of modular forms to identify characters of vertex algebras.

$\endgroup$
6
  • $\begingroup$ I am a bit unfamiliar with this, but quick question: does this method of finding the characters circumvent the need to know the null vectors/relations for each vector space? $\endgroup$
    – JamalS
    Mar 26, 2020 at 14:50
  • $\begingroup$ @JamalS Perhaps it would help if you explained why you find yourself confronted with a computation involving null vectors. $\endgroup$
    – S. Carnahan
    Apr 5, 2020 at 14:10
  • $\begingroup$ I think the reason I haven't found this method in the literature used is because with the VOAs I am dealing with, they have fermionic generators as well as bosonic, and so cancellations may occur in computing the characters, such that they can no longer be used to establish isomorphism on the level of graded vector spaces. $\endgroup$
    – JamalS
    Apr 6, 2020 at 8:11
  • $\begingroup$ @JamalS I see. In that case, you can resolve the problem by using a vector-valued character. One of the components is the graded trace of the parity operator, and one of the components is total dimension. $\endgroup$
    – S. Carnahan
    Apr 6, 2020 at 10:32
  • $\begingroup$ Thanks for your quick reply. Do you have a reference to a paper or textbook that provides a straightforward example of this? $\endgroup$
    – JamalS
    Apr 6, 2020 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.