9
$\begingroup$

Let $V=\bigoplus_{d\in\mathbb N}V(d)$ be a Möbius-covariant vertex algebra with $V(0)=\mathbb C$.
Recall that a vector $v\in V$ is called quasi-primary if $L_1v=0$.

For $v\in V(d)$, we write $Y(v,z)=\sum_{n\in\mathbb Z} z^{-n-d}v_{(n)}$.
With that convention, $v_{(n)}$ is an operator $V(k)\to V(k-n)$.

Let $d>n$.
Is it true that for any quasi-primary $v\in V(d)$ and any vector $w\in V(n)$, we have $v_{(n)}w=0$?

If the above relation does not always hold, are there reasonable extra assumptions that one can impose on $V$ that imply it?

Do the above relations hold when $v$ is required to be primary instead of quasi-primary?
(Add the assumption that $V$ is a VOA so that the notion of a primary vector make sense)

$\endgroup$
4
$\begingroup$

The answer seems to be yes for quasi-primary $v$ if $V$ has a suitable invariant bilinear form. Then one can identify $v_{(n)} w$ with its pairing with the vacuum, and obtains it as the appropriate coefficient of $$(\mathbf{1}, Y(v, x)w) = (-x^{-2})^d (Y(v, x^{-1})\mathbf{1}, w) = (-x^{-2})^d (e^{x^{-1} L(-1)} v, w) = 0$$ since the weight of $v$ is greater than that of $w$.

Of course, this argument does not work if $v$ is not quasi-primary since then $v$ must be replaced with $e^{x L(1)} v$ after the first equality in the calculation.

Note that it isn’t necessary to assume the bilinear form is nondegenerate, just nondegenerate on the one-dimensional vacuum space.

$\endgroup$
  • $\begingroup$ Welcome to MO professor McRae, good to have you! $\endgroup$ – Alec Rhea Dec 9 '17 at 5:53
  • $\begingroup$ Thank you Robert. This answer is great, and I will accept it. Do you have any expectations in the absence of an invariant bilinear form? $\endgroup$ – André Henriques Dec 9 '17 at 13:49
  • $\begingroup$ Assuming $V$ is a CFT-type VOA as in your setting, $V$ will fail to have a non-zero invariant bilinear form exactly when $\mathbf{1}=L(1)w$ for some $w\in V(1)$. So then taking $v$ to be the (quasi-primary) conformal vector would give a counterexample. $\endgroup$ – Robert McRae Dec 9 '17 at 14:43
  • $\begingroup$ I am not sure what one can say in the more general Mobius vertex algebra setting when there is no conformal vector. $\endgroup$ – Robert McRae Dec 9 '17 at 14:46
  • 1
    $\begingroup$ The general result was obtained by Haisheng Li in his paper, “Symmetric invariant bilinear forms on vertex operator algebras.” It states that the space of invariant bilinear forms on a vertex operator algebra is isomorphic to $V(0)/L(1)V(1),$ so when the dimension of $V(0)$ is one, the existence of a non-zero form is equivalent to $L(1)V(1)=0$. $\endgroup$ – Robert McRae Dec 9 '17 at 20:00
6
$\begingroup$

Suppose there exists $v,w \in V(d)$ such that $v_{(d)}w \neq 0$. And now consider $(Tv)_{(d)}w = -2d \,v_{(d)}w \neq 0$. Notice also that by skew-symmetry your condition being true for $d>n$ implies the same condition for $n < d$.

$\endgroup$
  • $\begingroup$ Great! This answers the question which I had asked. But really, it points to the fact that I hadn't found the correct formulation of my question. So, if you don't mind, I'll modify my question and add the condition that $v$ is (quasi-)primary. $\endgroup$ – André Henriques Dec 8 '17 at 17:18

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.