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If $G$ is a finite group, the characters of its irreps satisfy $$ \langle \chi_1,\chi_2\rangle := \frac{1}{|G|}\sum_{g\in G} \chi_1(g)\; \overline{\chi_2(g)} = \delta_{\chi_1,\chi_2}. $$

Alexei Latyntsev asked me the following question.

Instead of a group $G$, let us consider a vertex algebra $V$ (let's say a rational unitary VOA).

Is there an analog of the above orthogonality relations which applies to the characters of irreducible $V$-modules? (Or maybe genus one 1-point functions?)

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  • $\begingroup$ It is reminiscent of the statement that if you take the twice-punctured P^1 and insert two different modules, you will get zero.. $\endgroup$
    – Theo Johnson-Freyd
    Apr 2, 2019 at 14:26
  • $\begingroup$ @Theo Johnson-Freyd. Absolutely correct! And if you take your twice-punctured P^1, and label the points by the same module (or maybe the module and its dual), you get a one-dimensional space of conformal blocks. Also, in the above situation, you can "sew the two points to each other" and get a genus one curve... So it all looks very suggestive. But does this translate into something like orthogonality relations??? $\endgroup$
    – André Henriques
    Apr 3, 2019 at 12:24
  • $\begingroup$ I don't see it that way cause that's just the fact of not having pairings between non-isomorphic irreps which is akin to Schur's Lemma more than the asked orthogonality. On the other hand. What you ask seems more about Hermitian structures on bundles on $P^1$ when viewed as the moduli space of elliptic curves. Let $V$ be a rational $C_2$ cofinite vertex algebra. Then characters are flat sections of a local system on $P^1$ with a singularity at infinity. It seems to me that you're asking if the corresponding vector bundle has a natural hermitian metric making the sections an orthonormal basis. $\endgroup$
    – Reimundo Heluani
    Apr 3, 2019 at 13:55
  • $\begingroup$ ... Using that any such bundle decomposes as a sum of $\mathcal{O}(n)$ you may try to prove that there exists a unique such hermitian metric. $\endgroup$
    – Reimundo Heluani
    Apr 3, 2019 at 13:56
  • $\begingroup$ I guess this is how the inner product is defined :b Basically one needs to define an inner product on the vector spaces of genus 1 conformal blocks so that the mapping class groups are acting unitarily. For such purpose the characters have to be orthogonal, and their lengths are chosen in a particular way depending on their quantum dimensions. $\endgroup$
    – Bin Gui
    Aug 21, 2019 at 22:39

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