Let $V$ be a vertex operator algebra with all the good finiteness properties that people usually assume (positively graded, $C\_2$-cofinite, $V\cong V'$, etc.)


Let $W$ be a module for $V$, not necessarily irreducible.


How does one prove, or where can I find a proof of the following fact:



> 
> **Claim:**
>  The map
>  $$
> \bigoplus\_{\tilde W\in\text{irrep}(V)} \operatorname{Hom}\_{\mathrm{Rep}(V)}(W\boxtimes\tilde W,\tilde W)\to \operatorname{Hom}(W,\{\text{functions of $q$ and $z$}\})
> $$
>  which sends an intertwining operator $\mathcal Y$ type $\binom {\tilde W} {W \tilde W}$ to the function that sends $w\in W$ to
>  $$
> (q,z)\mapsto \operatorname{Tr}\_{\tilde W}(\mathcal Y(w,z)q^{L(0)})
> $$
>  is injective.
> 
> 
> 


Thank you for your help.



Linear independence of genus-one correlation functions
A proof of this property can be found in Proposition 2.2 of Huang's paper *[Vertex operator algebras and the Verlinde conjecture](https://arxiv.org/abs/math/0406291)*. This proof actually uses Zhu's algebras to simplify the discussions (similar to Zhu's proof of the linear independence of characters in *[Modular invariance of characters of vertex operator algebras](https://www.jstor.org/stable/2152847?casa_token=XsJ4R-viIU8AAAAA:6zTrdHonkCHlTwxegISPO6MQUnffLUJdz_msaTTAWKV1dqrIDw6D6q9QqCXLfXmlILm1z8WX6_aKpNO-055v9x3Jpuoq8pjP76UQeTu8KkigAeVF-g&seq=1#metadata_info_tab_contents)*) although I think this is not necessary. So let me give an elementary argument without using Zhu's algebras. Indeed I think this argument should also work for proving the injectivity of the sewings of (the vector spaces of) general conformal blocks.


Let me change your notations a little bit, and restate your claim as follows:


**Proposition**. Let $M\_1,\dots,M\_n$ be a list of inequivalent irreducible $V$-modules. For any $k=1,\dots,n$ we choose any intertwining operator $\mathcal Y\_k$ of type $M\_k\choose WM\_k$. Assume that $\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)q^{L\_0})=0$ for any $w,z,q$. Then $\mathcal Y\_k=0$ for all $k$.


*Proof*. It suffices to show that there exits a $\mathcal Y\_k$ equaling $0$. Then, by induction, all $\mathcal Y\_k$ are $0$.


For each $k$ we have a natural injective linear map $A\_k:M\_k\otimes\overline M\_k\rightarrow End(M\_k)$, where $\overline M\_k$ is the contragredient (conjugate) $V$-module of $M\_k$. Set $\mathfrak M=\bigoplus\_kM\_k\otimes\overline M\_k$, and set $A=\bigoplus\_k A\_k$.
We then have $$A:\mathfrak M\hookrightarrow\bigoplus\_k End(M\_k).$$


Now let $\mathfrak N$ be the subspace of all $\xi\in\mathfrak M$ satifying that $\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)A(\xi))=0$ for any $w,z$. (Here we regard the action of $\mathcal Y\_{k\_1}$ on $M\_{k\_2}$ as zero if $k\_1\neq k\_2$.)


$$\mathfrak N := \left\{\xi\in\mathfrak M:\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)A(\xi))=0\right\}$$
We claim that $\mathfrak N\neq0$. Indeed, since $\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)q^{L\_0})=0$, if we regard it as a series of $q$: $\sum\_{\Delta}\sum\_k Tr\_{M\_k(\Delta)}(\mathcal Y\_k(w,z))q^\Delta$, then any coefficient $\sum\_k Tr\_{M\_k(\Delta)}(\mathcal Y\_k(w,z))$ must be $0$ [1](https://arxiv.org/abs/math/0406291). Choose a conformal weight $\Delta$ such that there exists $k$ such that $M\_k(\Delta)$ is non-zero. Then we immediately have the non-zeroness of $\mathfrak N$.


Note that $\mathfrak M$ is indeed a $V\otimes V$-module. We now show that $\mathfrak N$ is a $V\otimes V$-submodule. Choose any $\xi\in\mathfrak N$. Then, for any $w,z$, since $\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)A(\xi))=0$, we have $\sum\_k Tr\_{M\_k}(\mathcal Y\_k(Y(u,z\_0-z)w,z)A(\xi))=0$ for any $u$ and any $z\_0$ such that $0<|z\_0-z|<|z|$. If moreover $0<|z\_0-z|<|z\_0|<|z|$, then $\mathcal Y\_k(Y(u,z\_0-z)w,z)=\mathcal Y\_k(w,z)Y(u,z\_0)$. Thus $$\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)Y(u,z\_0)A(\xi))=0$$ for all $z\_0$ satisfying $0<|z\_0-z|<|z\_0|<|z|$. By uniqueness of analytic continuations, we have $\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)Y(u,z\_0)A(\xi))=0$ for all $z\_0$ satisfying only $0<|z\_0|<|z|$. Take contour integrals of $z\_0$. Then we obtain $\sum\_k Tr\_{M\_k}(\mathcal Y\_k(w,z)Y(u)\_mA(\xi))=0$ for any $m\in\mathbb Z$, where we write $Y(u,z\_0)=\sum\_{m}Y(u)\_mz\_0^{-m-1}$. Since $Y(u)\_mA(\xi)=A((Y(u)\_m\otimes 1)\xi)$, we've actually shown that $\mathfrak N$ is invariant under the action of $V\otimes \Omega$ (where $\Omega$ is the vacuum vector of $V$). Similarly $\mathfrak N$ is $\Omega\otimes V$-invariant. So $\mathfrak N$ is $V\otimes V$-invariant, i.e., a $V\otimes V$ submodule of $\mathfrak M$.


Since $\mathfrak M$ has irreducible decomposition $\mathfrak M=\bigoplus\_k M\_k\otimes\overline M\_k$, and since $\mathfrak N$ is a non-trivial submodule of $\mathfrak M$, there exists $k$ such that $M\_k\otimes\overline M\_k\subset \mathfrak N$. Therefore, for any $w\_1\in M\_k,w'\_2\in \overline M\_k$, we have $\langle\mathcal Y\_k(w,z)w\_1,w'\_2\rangle=Tr\_{W\_k}(\mathcal Y\_k(w,z)A(w\_1\otimes w\_2'))=0$ for any $w,z$. Thus $\mathcal Y\_k=0$. 


*Q.E.D.*




---


[1](https://arxiv.org/abs/math/0406291) We know that the coefficients $c\_n$ of a convergent power series $f(z)=\sum\_{n\in\mathbb N}c\_nz^n$ are determined by the values of $f$. This is also true if $f$ can be written as a finite sum $f(z)=\sum\_m z^{\mu\_m}g\_m(z)$ where each $g\_m(z)$ is a convergent power series of $z$ and $\mu\_m\in\mathbb C$. This is an easy exercise. See for example *[A theory of tensor products for module categories for a vertex operator algebra, IV](https://arxiv.org/abs/q-alg/9505019)* Lemma 14.5 and section 15.4.


