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.


1 Answer 1


A proof of this property can be found in Proposition 2.2 of Huang's paper Vertex operator algebras and the Verlinde conjecture. 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) 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. 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 $\mathcal Y_k=0$.


1 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 Lemma 14.5 and section 15.4.

  • $\begingroup$ Thank you Bin Gui, this is an excellent answer. Sorry for having taken so long to process it (and to accept it). $\endgroup$ Apr 8, 2019 at 21:14
  • 2
    $\begingroup$ I was aware of Proposition 2.2 of Huang's paper when I asked the question, but was having difficulty understanding the logical step between equations (2.17) and (2.18) in the proof. With the help of Huang, I ended up figuring out how to bridge the small gap. But Bin Gui's proof above follows a different strategy, and is very elegant $\endgroup$ Apr 8, 2019 at 21:32

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.