# Are vertex operator algebras ever conspiratorial?

I have a vertex operator algebra (VOA) $$V$$ with all niceness properties (unitary, rational, CFT type, etc). Its Lie algebra $$\mathfrak{g} = V_1$$ of spin-$$1$$ fields is large, and I understand how the spaces of low-spin fields decompose as $$\mathfrak{g}$$-irreps. I know that, for specific spins $$r,s,t$$ that I care about, the OPE of a spin-$$r$$ field with a spin-$$s$$ field has a generally-nonzero spin-$$t$$ component: the VOA picks out a nonzero map $$f: V_r \otimes V_s \to V_t$$.

For the particular values I care about, $$V_r$$ and $$V_s$$ are simple $$\mathfrak{g}$$-modules, but $$V_t$$ decomposes over $$\mathfrak{g}$$ as a direct sum of simple modules, all with multiplicity $$1$$. So I can choose a simple submodule $$I \subset V_t$$ and ask for the $$I$$-component of the OPE: it is a map $$f_I : V_r \otimes V_s \to I$$. I can compute that the space of $$\mathfrak{g}$$-invariant maps $$V_r \otimes V_s \to I$$ is one dimensional. So up to a normalization that I don't care about, I know $$f_I$$, provided $$f_I \neq 0$$.

Is it possible for the VOA to conspire to set $$f_I = 0$$, even though $$\mathfrak{g}$$-invariance and degree considerations (and (skew-)symmetry, when $$r=s$$) do not require this?

• One way this could in principle happen is if $\operatorname{Aut}(V)$, which is a group with Lie algebra $\mathfrak{g}$, is disconnected. Then perhaps the unique $\mathfrak{g}$-invariant map $V_r \otimes V_s \to I$ is not invariant for all of $\operatorname{Aut}(V)$. But this does not happen in my case: my $V$ has a connected group of automorphisms. So any conspiracy requires somehow protecting $f_I = 0$ without using any symmetries. – Theo Johnson-Freyd Jun 29 '19 at 17:16