Assume that $V$ is a vertex operator algebra, and the VOA $V'$ is a vertex subalgebra of $V$. The notion that $V'\subset V$ is a conformal inclusion has different meanings in different literatures. Physicists often mean that $V'$ and $V$ have the same central charge, while mathematicians often mean that $V'$ and $V$ have the same conformal vector. Clearly the second meaning implies the first one. Is the converse also true? i.e., if $V'$ and $V$ have the same central charge, is the conformal vector of $V'$ also a conformal vector of $V$? (Feel free to add conditions on $V$ and $V'$, e.g., CFT type, rational, etc.)

| cite | improve this question | | | | |

The answer is no: it is possible to have a vertex algebra map between vertex operator algebras of equal central charge that does not take preserve the distinguished conformal vectors.

Given a vertex algebra $V$, the set of conformal vectors of central charge $c$ in $V$ is in natural bijection with the set of vertex algebra homomorphisms from the Virasoro vertex algebra at central charge $c$ to $V$. It is quite easy to find vertex algebras with more than one conformal vector of central charge $c$. For example, any framed vertex operator algebra with central charge at least $1$ has at least 2 Ising vectors, giving different conformal structures at central charge 1/2.

In this case, it is easy to set $V'$ to be $V$ as a vertex algebra, but choose different Ising vectors to be the conformal vectors. Then the identity map is a conformal embedding in the first sense, but not the second.

| cite | improve this answer | | | | |
  • $\begingroup$ I find that one can prove the inverse statement if the Virasoro vector \nu' of V' satisfies L_0\nu'=2\nu' and L_1\nu'=0 (L_0,L_1 come from the Virasoro vector \nu of V). This is because one can do the coset construction under these conditions. $\endgroup$ – Bin Gui Oct 4 '17 at 5:36
  • $\begingroup$ I'm not sure what definition of conformal vector you are using, but typically you need to include $L_{-1} = T$, and this does not follow from having a morphism $Vir^c \rightarrow V$. For example looking at $Vir^c \rightarrow Vir^c \otimes Vir^0$ is a morphism between vertex algebras of the same central charge that does not send the conformal vector to a conformal vector. $\endgroup$ – Reimundo Heluani Oct 1 '18 at 16:58

Your Answer

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

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