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?

  • 2
    $\begingroup$ 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. $\endgroup$ – Theo Johnson-Freyd Jun 29 '19 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.