Proving that $L_{-1}v=0$ implies $v\in V_0$ in a vertex operator algebra

Question

The discussion of the actual problem is labelled in bold after the notoriously long definition of vertex operator algebra is given for the sake of completeness (Note: "fields" and "locality" are not relevant for this exercise, as far as I can tell). (The definition of vertex algebra is taken from Vertex Algebras and Algebraic Curves by Frankel and Ben-Zvi, and the conditions for a vertex operator algebra as well as the exercise are from the notes "Vertex operators and modular forms" by Mason and Tuite.)

Definition of vertex algebra. A vertex algebra consists of:

$\bullet$ (space of states) a vector space $V$;

$\bullet$ (vacuum vector) a vector $|0\rangle\in V$;

$\bullet$ (translation operator) a linear operator $T:V\to V$;

$\bullet$ (vertex operators) a linear operation $$Y(\cdot,v):V\to \text{End}V[[z^{\pm 1}]];$$ taking each $A\in V$ to a field $$Y(A,z)=\sum_{n\in \mathbb{Z}} A_{(n)}z^{-n-1}$$ acting on $V$.

These data are subject to the axioms

$\bullet$ (vacuum axiom) $Y(|0\rangle,z)=\text{Id}V$. Furthermore, for any $A\in V$ we have $$Y(A,z)|0\rangle\in V[[z]],$$ so that $Y(A,z)|0\rangle$ has a well-defined value at $z=0$, and $$Y(A,z)|0\rangle|_{z=0}=A.$$

$\bullet$ (translation axiom) For any $A\in V$, $$[T,Y(A,z)]=\partial_z Y(A,z)$$ and $T|0\rangle=0$.

$\bullet$ (locality axiom) All fields $Y(A,z)$ are local with respect to each other.

A vertex algebra is called a vertex operator algebra if $V=\bigoplus_{n\in \mathbb{Z}} V_n$, and $V$ has a distinguished vector $\omega$ called the conformal vector such that $Y(\omega,z)=\sum_{n\in\mathbb{Z}} L_n z^{-n-2}$, where the $L_n$ satisfy the Virasoro relation

$$[L_m,L_n]=(m-n)L_{m+n}+\frac{m^3-m}{12}c\text{Id}_V.$$

We also have that $V_n=\{v\in V:L_0v=nv\}$, dim$V_n=0$ for $n<<0$, and $Y(L_{-1}v,z)=\partial Y(v,z)$, i.e. $T=L_{-1}$.

Facts from previous exercises. We know that $Y(|0\rangle,z)=\text{Id}$, $|0\rangle\in V_0$, $\omega\in V_2$, $L_n |0\rangle =0$ for $n\geq -1$, and $(L_{-1}v)_n=-n v_{n-1}$.

The problem.

To show that $L_{-1}v=0$ implies that $v\in V_0$, I started by thinking of how one can use this data to generate expressions that relate $L_0v$ and $L_{-1}v$, such as

$$[L_1,L_{-1}]v=2L_0v=L_1(L_{-1}v)-L_{-1}(L_1v)=-L_{-1}(L_1v)$$ and $$[L_0,L_{-1}]v=L_{-1}v=0=L_0(L_{-1}v)-L_{-1}(L_0v)=-L_{-1}(L_0v).$$ The first expression doesn't seem to help since I can't see how to find/use the value of $L_1v$, and I don't see how the second expression could imply that $L_0v=0$, since if $v\in V_1$ for example we still have $L_{-1}(L_0v)=L_{-1}v=0$. Further, I can't see how any of the other data might apply.

Answer 1

Here I've adapted the proof from Lepowsky and Li's Introduction to Vertex Operator Algebras and Their Representations. They prove a more general result, and we need not refer to the concepts of 'centralizer' or 'vertex operator subalgebra' to establish our result, namely that $L_{-1}v=0$ implies $v\in V_0$, i.e. $L_0v=0$.

The proof uses the commutator identity $$[u_m,v_n]=\sum_{i\geq 0} {m\choose i} (u_iv)_{m+n-i},$$ which I didn't include in the original post since (1) I didn't think it was relevant and (2) there are several other identities (the Jacobi identity and the associator and skew-symmetry identities), and if I included all of these my post would have been a giant wall of definitions. Lesson learned that I should have been more skeptical of my intuition.

First, we prove the useful fact from Remark 3.2.4 (more or less exactly as in the book) that for $u,v\in V$, $[Y(u,z_1),Y(v,z_2)]=0$ if and only if $u_nv=0$ for $n\geq 0$.

If $[Y(u,z_1),Y(v,z_2)]=0$, then $$\sum_{m,n\in \mathbb{Z}}[u_m,v_n]z_1^{-m-1}z_2^{-n-1}=0,$$ therefore $[u_m,v_n]=0\Rightarrow u_mv_n=v_nu_m$ for all $m,n\in \mathbb{Z}$. By the creativity axiom, $v=v_{-1}|0\rangle$ and $u_n|0\rangle=0$ for all $n\geq 0$, so for all $n\geq 0$ we have $$u_nv=u_nv_{-1}|0\rangle=v_{-1}u_n|0\rangle=0.$$ On the other hand, if we assume $u_nv=0$ for all $n\geq 0$, it follows immediately from the commutator identity that $[u_m,v_n]=0$ for all $n,m\in \mathbb{Z}$. This completes the proof.

Now to the original problem. I showed in my other answer that $L_{-1}v=0$ implies that $Y(v,z)=v_{-1}$. Hence $v_n=0$ for $n\geq 0$, so for any $u\in V$ we have $v_nu=0$ for all $n\geq 0$. Remark 3.2.4 implies $Y(v,z)$ commutes with all other fields, i.e. for all $u\in V$ we have $[Y(v,z_1),Y(u,z_2)]=0$. This clearly implies that $[Y(u,z_2),Y(v,z_1)]=0$, hence $u_nv=0$ for all $n\geq 0$. This is true in particular for $u=\omega$; since $\omega_n=L_{n-1}$, for all $n\geq 0$ we have $$\omega_nv=L_{n-1}v=0,$$ so it follows that $L_{-1}v=L_0v=0$. Thus, $v\in V_0$.

Answer 2

No solution yet, but a seemingly more promising approach than guesswork with the Virasoro bracket.

Another way of stating the problem is that $\ker L_{-1}\subseteq \ker L_0$, so perhaps one has to consider how these operators act on all of $V$ first.

We already know how $L_0$ acts on $V$: it acts as a map $V_k\to V_k$ on the homogeneous spaces, sending $v\mapsto kv$. $\ker L_0=V_0$ by definition.

We don't immediately know how $L_{-1}$ acts on vectors besides the vacuum, but we do know that it acts as a partial derivative on fields: $Y(L_{-1}v,z)=\partial_z Y(v,z)$. By the vacuum axiom, $$Y(L_{-1}v,z)|0\rangle|_{z=0}=L_{-1}v=\partial_zY(v,z)|0\rangle|_{z=0}=v_{-2}|0\rangle.$$ So for all $v\in V$, $L_{-1}v=v_{-2}|0\rangle.$ So $\ker L_{-1}=\{v\in V: v_{-2}|0\rangle=0\}.$

An important fact I forgot to mention (since I failed to see that it is relevant) is the injectivity of the state-field correspondence $v\mapsto Y(v,z)$. It follows from this that if $L_{-1}v=0$, then $Y(L_{-1}v,z)=\sum_{n\in \mathbb{Z}} 0_n z^{-n-1}$, where $0_n$ is the zero endomorphism for all $n$. Hence $0_n=(L_{-1}v)_n=-nv_{n-1}$ for all $n$, so in particular $v_n|0\rangle=0$ for all $n\leq -2$. It follows that $Y(v,z)|0\rangle=v.$ An arbitrary $u\in V$ satisfies $Y(u,z)|0\rangle=u+O(z)$, so perhaps this is strong enough to show that $v\in V_0$ already; I still don't quite see it yet.

Edit. I consulted Lepowski and Li's Introduction to Vertex Operator Algebras and Their Representations, and Proposition 3.11.2 gives a proof that $\ker L_{-1}\subset \ker L_0$ by defining the centralizer $C_V(U)$ of a vertex operator subalgebra $U$ of $V$ and showing that $C_V(U)$ is independent of $U$ and that $$C_V(U)=\ker L_{-1} \subset V_0=\ker L_0.$$ I am still curious if there is a simpler proof of this without referencing centralizers or vertex operator subalgebras, since this exercise came from the notes http://library.msri.org/books/Book57/files/40mason.pdf which did not define or use these concepts (at least not prior to the exercise, Exercise 2.28). I will see if I can adapt the argument from the book of Lepowski and Li in the meantime.