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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)

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2 Answers 2

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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.

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  • $\begingroup$ Welcome to MO professor McRae, good to have you! $\endgroup$
    – Alec Rhea
    Commented Dec 9, 2017 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
    Commented Dec 9, 2017 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
    Commented Dec 9, 2017 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
    Commented Dec 9, 2017 at 14:46
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    $\begingroup$ 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
    Commented Dec 9, 2017 at 20:00
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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 $n < d$.

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  • $\begingroup$ 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
    Commented Dec 8, 2017 at 17:18

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