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17) in case v is homogeneous. 10), the case v = u. 3 is not assumed symmetric. However, it will typically turn out to be symmetric. For example, we have: 54 IGOR B. 6. Suppose that the contragredient V of the adjoint module V is equivalent to V", and let (•, •) be the nondegenerate bilinear form on V corresponding to a given isomorphism V —• V . Then (•, •) is symmetric. Proof. Let vi,v2 € V. ,z)l, v2) = (v2, Y{vu z)l). 20) for v,- G V, and we shall want to refer to this computation later. 1). ) = (Y{e-'-lLW(-z-2)LWv2, = (Y(e'-lLWvo, -z)e*-lLWv0,e'L(1\-z-2)LWvl) z)e-*"li<1>(-z2)L <°>i> 2, ( - z - 2 ) L ( ( % i ) = (e-'~lLl1\-z2)Ll%2, Y(e*LW(-z-2)LW.

H) where L(—1) is the operator on V. This completes the definition. We may denote the module just defined by (W,Y). 1) is only a convenience. 3. 5 and cf. [FLM]). 4- A vertex operator algebra V is clearly a module for itself. As such, it is called the adjoint module. We are using the same notation Y(-,z) as for algebras, but no confusion should arise provided V is viewed as the adjoint module. 2. 2) z). 4) as an operator on W. 5) VERTEX OPERATOR ALGEBRAS AND MODULES 35 when v is a homogeneous vector has weight — n : wt xv(n) - -n.

0 • • • 0 vp, z) = Y(vly z) • • • (8) Y(vp, z) (vi G Vi). ) T h e tensor product module is in particular a tensor product of Virasoro algebra modules. 1: P R O P O S I T I O N 4 . 6 . 1 . 1), is a module for the product algebra. vertex tensor Also, given modules X L , . . ,p) , their tensor product is a h o m o m o r p h i s m f r om Wi®*~®Wp to X i • • • (Si Xp. %. It is worth emphasizing t h at we are considering tensor product modules not for a single algebra, b u t rather for a tensor product of algebras.

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