In Huang & Lepowsky's series of papers A theory of tensor products for module categories for a vertex operator algebra, they defined for a rational vertex algebra $V$ the $P(z)$ tensor product of two modules $W_1$ and $W_2$ to be $$ W_1\boxtimes_{P(z)} W_2=\coprod_k{(\mathcal{M}[P(z)]^{M_k}_{W_1~W_2})}^*\otimes M_k $$ where $\mathcal{M}[P(z)]^{M_k}_{W_1~W_2}$ is the (finite dimensional) vector space of conformal blocks for modules $M'_k$, $W_1$ and $W_2$ at points $\infty$, $z$ and $0$ on the compact Riemann surface $\mathbb{P}^1$. Here $M'_k$ is the module contragedient to $M_k$. They showed that under certain conditions this tensor product gives a vertex tensor category. Now if we define the tensor product by using conformal blocks on an arbitary compact Riemann surface, can we get a vertex tensor category (and hence a braided tensor category) under certain reasonable conditions? If this is true, can we show furthermore the rigidity and the modularity for this category?


1 Answer 1


In general, you won't get a vertex tensor category, because you don't get well-defined unit behavior when you use conformal blocks on higher genus surfaces.

Huang-Lepowsky assume the vertex operator algebra is rational and $C_2$-cofinite, and this is conjecturally strong enough to obtain strong factorization properties for conformal blocks in higher genus. In particular, we expect the higher genus conformal blocks to be derived from genus zero blocks by a decomposition into pants. You can define a product structure by fixing a Riemann surface, and moving marked points in a small neighborhood to get a braiding, but all of the spaces of conformal blocks will be too large to describe a tensor category with unit if the original representation category has non-identity simple modules.

On the other hand, if your vertex operator algebra is holomorphic and $C_2$-cofinite, then the tensor category derived from genus zero conformal blocks is equivalent to the category of finite dimensional vector spaces, and you should get essentially the same tensor category if you look at higher genus blocks.


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.