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For a vertex operator algebra $(V,Y,\left|0\right>)$ and a $\mathbb{Z}$-graded module $(M,Y_M)$, one can form the contragradient module $(M^{\lor},Y_{M^{\lor}})$ with underlying $\mathbb{Z}$-graded vector space given by the restricted dual $$M^{\lor} = \bigoplus_{n \in \mathbb{Z}}M_n^*$$ There's a canonical pairing $\left<-,-\right>:M \otimes M^{\lor} \to \mathbb{C}$, and we define the vertex operator by the formula $$\left<Y_{M^{\lor}}(A,z)\cdot \varphi,v\right> = \left<\varphi,Y(e^{zL_1}(-z^{-2})^{L_0}\cdot A,z^{-1})\right>$$

In the case where $V=V_{\mathfrak{g},\kappa}$ is the level-$\kappa$ Kac-Moody VOA, there's an equivalence $V_{\mathfrak{g},\kappa}\operatorname{-mod} \simeq \hat{\mathfrak{g}}\operatorname{-mod}_{\kappa}$ between vertex algebra modules and smooth level-$\kappa$ representations of the corresponding Kac-Moody affine Lie algebra.

Is there a simple description of the above contragradient module in terms of Lie algebra representations? I'm assuming it should be something like the dual representation plus twisting the action by $t \mapsto -t^{-1}$, but I'm not able to see that from the formula.

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  • $\begingroup$ What does $(-z^{-2})^{L_0}$ mean? $\endgroup$ Commented Nov 25 at 20:17
  • $\begingroup$ @KentaSuzuki it means the operator acting on $A$ as multiplication by $-z^{-2}$ to the power of $L_0(A)$. There are more details in Frenkel Ben Zvi's book section 10.4.5 $\endgroup$
    – E. KOW
    Commented Nov 26 at 0:27

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