6
$\begingroup$

I have some conceptual doubts to clear up, in terms of piecing together what we learn of a vertex operator algebra (VOA) in conformal field theory, and how it is defined by a mathematician, say from Kac's book. In particular:

  • Because of the state-field correspondence, can we equally think of $V$ as a space of fields, rather than space of states?
  • If we have $a,b \in V$, and we wish to find say, $a_{-1}b$, in physicist's notation what would this be precisely equivalent to?
  • I presume a null state $v \in V$ is such that for a suitable norm $||v|| = 0$ however, $V$ is not taken to be a normed space in the axioms of a VOA, so how is a null state defined in this context?
Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
4
$\begingroup$

  • Yes.

  • In the case of the Virasoro algebra, we have the mode decomposition $T(y)=\sum_{n\in\mathbb{Z}} \frac{L_n}{(y-z)^{n+2}}$, so $(L_{-1}T)(z) = \frac{1}{2\pi i} \oint_z dy\ T(y)T(z)$.

  • No need to have a norm for defining null states. In the case of the Virasoro algebra, a null state is a state that is killed by the annihilation modes $L_{n>0}$, while also being a descendant state.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Regarding null states, if you have something like a W-algebra, with some additional mode like $J_n$, do you also require that $J_n v = 0$ for $n >0$? I recall you can check if a state is null if its OPE with all operators, contains no term with an operator of dimension higher than itself, so it seems like it must depend on its annihilation with respect to all operators? $\endgroup$
    – GRNS
    Commented Aug 17, 2020 at 19:07
  • 1
    $\begingroup$ Yes, in the presence of a larger algebra, you would define null states to also be killed by $J_{n>0}$. You could still consider "Virasoro-null" states that are only killed by $L_{n>0}$, but these are less relevant. (Typically, the Virasoro-null states do not vanish.) $\endgroup$
    – Sylvain Ribault
    Commented Aug 17, 2020 at 19:16
  • $\begingroup$ With your answer, I think I have answered the $a_{-1}b$ question since your answer is not general. Correct me if I'm wrong: I take the fields associated to $a,b$, then I compute $\mathrm{Res} \, (z-w) a(z)b(w)$ which gives me something that is a field. Then I take the state associated to that field, which is $a_{-1}b$. Do I have it right? $\endgroup$
    – GRNS
    Commented Aug 17, 2020 at 19:17
  • 1
    $\begingroup$ In physics the usual convention is $a(y) = \sum_{n\in\mathbb{Z}} \frac{a_n}{(y-z)^{n+h}}$ where $h$ is the dimension of $a$, so $h=2$ in my example with the Virasoro algebra. Your formula seems to hold if $h=1$ only. But maybe mathematicians' conventions are different from physicists'. $\endgroup$
    – Sylvain Ribault
    Commented Aug 17, 2020 at 19:29
  • $\begingroup$ The more general formula I propose is $\mathrm{Res}\, (z-w)^n a(z)b(w)$ if you want to know $a_{-n}b$, though indices are sometimes shifted because it is convention to use $z^{-n-1}$ in the expansion regardless of $h$. For example a mathematician will write $T = \sum \omega_n z^{-n-1}$ so that $L_n = \omega_{n+1}$. This sometimes leads to some confusion. $\endgroup$
    – GRNS
    Commented Aug 17, 2020 at 19:34

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.