In [Zhu's seminal paper](https://www.ams.org/journals/jams/1996-9-01/S0894-0347-96-00182-8/), he proves (5.3.2) that if $V$ is a vertex algebra the character of all of its modules are ***modular forms***! (This is not literally true- there are conditions).


I have always found this statement very mysterious. How on earth do modular forms/elliptic curves come into the picture?


1. Is there a proof using the moduli space of elliptic curves $\mathcal{M}\_{1,1}$ more directly? e.g. if $M$ is a vertex algebra module, it can be localised onto any elliptic curve $E$. Is Zhu's proof saying we can vary $E$ and thereby get some structure living over $\mathcal{M}\_{1,1}$, from which we can take a character (which will just be a modular form because it's a section of a line bundle on $\mathcal{M}\_{1,1}$)?


I am also very confused about the conditions ($C\_2$-finiteness, rationality).


2. Are these conditions necessary for the theorem? If they are, is there any "geometric" meaning to them, say in terms of the associated chiral algebras, or the structure in 1. living over $\mathcal{M}\_{1,1}$?


Hopefully, with all the progress on vertex/chiral algebras in the 25 years since the proof was published, there is a clean modern answer to the above.



Why are VOA characters modular forms (geometrically)?
I'm sure someone here can handwave the intuition behind these statements much better than me. This handwaving version goes like this, one is interested in computing vacuum 1-point functions on a torus, but you can cut a torus along an $S^1$ making it into a cylinder, put a module `M` on the boundaries, now you start with a vector $m \in M$, you *evolve* it with $q^{L\_0}$ and get a vector $m' \in M$. Gluing back the torus corresponds to taking the trace.


Now to actually answer the different parts of the question.



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> if MM is a vertex algebra module, it can be localised onto any elliptic curve EE. Is Zhu's proof saying we can vary EE and thereby get some structure living over M1,1M1,1, from which we can take a character
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There are a number of misinterpretations here. It is true that given a vertex algebra module (with very mild conditions) you can get a sheaf on an elliptic curve corresponding to it (a module for the corresponding chiral algebra). However, this has nothing to do with coinvariants of $V$. Coinvariants of $V$, or conformal blocks, have to do with the bundle/sheaf that you get on an elliptic curve by localizing $V$ itself, and not a module.


The statement from Zhu can be translated in a way very similar to what you say though: For each genus g curve $X$ with $n$ marked points $\{x\_i\} \subset X$, one can construct a vector space $C(X, V, \{x\_i\})$ of coinvariants. In the language of Frenkel-Ben-Zvi cited in the comments these are coinvariants of $V$ itself: you put the vacuum module $V$ at each of the marked points $x\_i$. So far no module $M$ is involved. Now as you move $X$ in the moduli space $\mathcal{M}\_{g,n}$, these spaces arrange into a twisted $\mathcal{D}$-module on this moduli space.


In the particular case of 1-marked elliptic curves, ie. $g=1$ and $n=1$, under some conditions (below we'll deal with that part of your question) these vector spaces are finite dimensional and you get an actual vector bundle with a flat connection. Notice that this is simply the statement that the twisted $D$-modules are coherent sheaves, since the space is one dimensional, this has to be a vector bundle with a flat connection. Zhu's statement is that **characters of irreducible representations of $V$ are flat sections of these bundles (rather their duals), and moreover they form a basis of the space of flat sections**.


So the role of the module $M$ is to produce a section of a geometric object constructed from $V$, not from $M$ directly.



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> Is there a proof using the moduli space of elliptic curves M1,1M1,1 more directly?
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Yes indeed, and this was already included in Emile's comments, but I'll repeat them essentially here: We have a vector bundle with a flat connection on the moduli space of elliptic curves given by the dual bundle to coinvariants of $V$. This flat connection is rather explicit (I'll describe it below), so to have a flat section is to solve an explicit differential equation. This in turn leads to prove that flat sections give rise to a solution of an explicit ODE. This ODE has a singularity in the boundary of the moduli space at $q = 0$, but it is a mild singularity in the sense that it is a regular singular point. What you do is prove that the formal character of a module $M$ gives a formal power series that solves this differential equation (there is a subtlety here between taking $\tau \in \mathbb{H}$ as the parameter or $q = e^{2 \pi i \tau}$). Once you have this, the general theory of Frobenius expansions of solutions of these ODEs tell you that these characters converge (the hard part) and therefore they form vector valued modular forms (the easy part: we already knew that they were sections of a bundle on the moduli space of elliptic curves).



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> I am also very confused about the conditions (C2C2-finiteness, rationality).
> Are these conditions necessary for the theorem?
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Not exactly. What $C\_2$ gives you is the finite dimensionality of the space of coinvariants (in fact you need less than this, as we'll see below), what rationality gives you is the fact that you can construct **all** flat sections starting from irreducible modules. The idea here is geometric and goes back to studying the behavior of flat sections near the nodal curve limit $q=0$.


This idea roughly goes as follows: If you give me a flat section, it gives a solution to an ODE as we talked above. As such you can expand it in power series of $q$ (and in theory possibly a polynomial in $\tau = \log q$). The lowest coefficient of this series defines a symmetric function of the Zhu algebra $Z(V)$ of $V$. This is a remarkable associative algebra that has a few properties: The list of its irreducible modules up to isomorphism is in bijection with the irreducible modules V. And in addition under the finitenes conditions listed above $Z(V)$ happens to be finite dimensional and semisimple. So the conformal block being a symmetric function, is a linear combination of traces of irreducible $Z(V)$-modules. This is the direct connection: one can start from an irreducible finite dimensional module $M\_0$ for the Zhu algebra $Z(V)$ of $V$. One induces a module $M$ for $V$ having $M\_0$ as it lowest graded piece. Then by what we have talked above we know that the character of $M$ is a flat section, and its restriction to its lowest graded piece $M\_0$ is the symmetric function we started from. There are some holes here and there in the description I just gave, but it is essentially correct.


Now, if $V$ is not rational, but it is $C\_2$ cofinite, you still have finite dimensional space of coinvariants and finitely many irreducibles. Just that now you may have extensions between these modules. Miyamoto carried out Zhu's program in this scenario: instead of looking at irreducible $V$ modules induced from $Z(V)$ we need to look at modules induced from *projective modules of higher Zhu algebras*.


Finite dimensionality of coinvariants on elliptic curves.
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The $C\_2$ condition and its role in having finite dimensionality of coinvariants is a bit confusing still (at least to me). It appears in different ways that are somewhat independent. I'll describe in detail the situation for elliptic curves which is what your question is about.


The space of coinvariants of the elliptic curve $X\_q = \mathbb{C}^\* / \mathbb{Z} \simeq C / \mathbb{Z} + \mathbb{Z}\tau$ (where the action of $Z$ is given by multiplication by $q$), with coefficients in the chiral algebra associated to a vertex algebra $V$ with supports in its vacuum module ($V$ itself) in the marked point $0 \in X\_q$ is explicitly given by the cokernel of the map


$$ d: V \otimes V \otimes \Gamma(\mathcal{O}\_{X\_q}, X\_q \setminus 0) \rightarrow V, \qquad a \otimes b \otimes f(z) \mapsto \mathrm{res}\_z f(z) Y(a,z) b.$$


So these functions $f(z)$ are biperiodic functions of $z$ with possible poles at $m + n \tau$ with $m,n \in \mathbb{Z}$. It turns out that $d(V \otimes V \otimes f') \subset d(V\otimes V \otimes f)$ so it is enough to kill a basis of functions modulo total derivatives. This is the top de Rham cohomology of $X\_q \setminus 0$ and this has a basis given by the constant function $1$ and the Weierstrass $\wp(z,q)$ function. So the space of coinvariants is simply


$$ V / \langle \mathrm{res}\_z \wp(z) Y(a,z)b, \, \mathrm{res}\_z Y(a,z)b \rangle $$


I am lying slightly here in that the vertex operation that is needed is denoted by $Y[a,z]$ in Zhu's paper and has to do with the fact that we are using an isomorphism $\mathbb{C}^\*/\mathbb{Z} \simeq \mathbb{C} / \mathbb{Z}^2$, but that doesn't really change the exposition.


The key point is that **every vertex algebra** is filtered, the relevant filtration is known as the Li filtration. And the above two-term complex is compatible with this filtration. So by looking at the associated graded we get an immediate bound for the homology of that two term complex in degree 0. It turns out that the operation
$$ a \otimes b \mapsto \mathrm{res}\_z \wp(z;q) Y(a,z)b$$ in this associated graded is simply the operation $a \otimes b \mapsto a\_{(-2)}b = \mathrm{res}\_z z^{-2} Y(a,z)b$, that is, in this filtration, we only care about the singular part of $f(z)$, not $f(z)$ itself.


There you go, just killing $d(V \otimes V \otimes \wp(z))$ we get, in the associated graded, the $C\_2$ quotient of $V$. So if $V$ is $C\_2$ cofinite, we know that the space of coinvariants is finite dimensional.


But there is more. We didn't kill the constant functions $f(z)= 1$. Zhu didn't really care since he needed rationality and finite dimensionality of $Z(V)$, but in his paper is clear that what actually is enough to prove finite dimensionality of coinvariants is not $R\_V := V/C\_2(V)$ being finite dimensional, but rather the quotient of $R\_V$ by the image of $a \otimes b \mapsto \mathrm{res}\_z Y(a,z)b = a\_{(0)}b$.


Now $R\_V$ is a Poisson algebra, with Poisson bracket given by $\{a,b\} = a\_{(0)}b$ so **the actual sufficient condition to have finite dimensionality of coinvariants is that the zeroth Poisson homology of $R\_V$, $R\_V / \{ R\_V, R\_V\}$ is finite dimensional**.


This condition fits nicely in a much wider context: it was proved that the finite dimensionality of the first Poisson homology of $R\_V$ also plays a role in the finite dimensionality of the first chiral homology of the elliptic curve with coefficients in $V$ (there is a complex whose degree $0$ homology is coinvariants, the general homology is known as *chiral homology*). We expect this to be the case for arbitrary chiral homology in fact, not only degree $0$ and $1$.


Now that I've set up the above notation, I can quickly tell you about the flat connection that I promised above it was explicit. What is a class in the dual to the coinvariants: it's a functional $\varphi: V \rightarrow \mathbb{C}$ that satisfies


$$ \varphi ( \mathrm{res}\_z f(z) Y(a,z)b ) = 0$$


For every biperiodic $f$ with possible poles at $z=0$ and every $a,b \in V$. Now there is the Weierstrass $\zeta$ function, which is not really bi-periodic, but it almost is. It's derivative is the Weierstrass $\wp$ function. The flat sections of coinvariants are $\varphi$ that satisfy the equation above and in addition satisfy:


$$ \frac{d}{d\tau } \varphi(a) = \varphi ( \mathrm{res}\_z \zeta(z) Y(\omega, z)a ) $$
where $\omega \in V$ is the conformal vector. Notice that I never talked about $V$ being conformal up to this point and indeed the conformal vector only appears in the geometric picture just to give us the flat connection.


$C\_2$ appears in a different way as a sufficient condition to prove finite dimensionality of coinvariants in arbitrary genus. I'll point you to recent work of Damiolini, Gibney and Tarasca. They use this condition to check that you indeed have a vector bundle in the moduli space. But I have a strong suspicion that this can be relaxed as well to finite dimensionality of the Poisson homology.


A final remark in an already long answer. When you ask:



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> say in terms of the associated chiral algebras,
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Not as far as I know actually. The key point was the Li filtration. And the Li filtration is a filtration of Vertex algebras, but it is not compatible with the chiral algebra structure. There is no literature that I am aware about this filtration in relation to chiral algebras. And up to not long ago at least one of the leading experts in the field of chiral algebras was not aware of the existence of this filtration. It is picking up pace and there have been many exciting results in the last couple of years, so I expect to see some new idea popping up in the arxiv soon regarding this.


