The first cohomology, derivations, and the reductivity of a (meromorphic open-string) vertex algebra

Abstract:

We give a criterion for the complete reducibility of modules satisfying a composability condition for a meromorphic open-string vertex algebra $V$ using the first cohomology of the algebra. For a $V$-bimodule $M$, let $\hat {H}^{1}_{\infty }(V, M)$ be the first cohomology of $V$ with the coefficients in $M$, which is canonically isomorphic to the quotient space of the space of derivations from $V$ to $W$ by the subspace of inner derivations. Let $\hat {Z}^{1}_{\infty }(V, M)$ be the subspace of $\hat {H}^{1}_{\infty }(V, M)$ canonically isomorphic to the space of derivations obtained from the zero mode of the right vertex operators of weight $1$ elements such that the difference between the skew-symmetric opposite action of the left action and the right action on these elements are Laurent polynomials in the variable. If $\hat {H}^{1}_{\infty }(V, M)= \hat {Z}^{1}_{\infty }(V, M)$ for every $\mathbb {Z}$-graded $V$-bimodule $M$, then every left $V$-module satisfying a composability condition is completely reducible. In particular, since a lower-bounded $\mathbb {Z}$-graded vertex algebra $V$ is a special meromorphic open-string vertex algebra and left $V$-modules are in fact what has been called generalized $V$-modules with lower-bounded weights (or lower-bounded generalized $V$-modules), this result provides a cohomological criterion for the complete reducibility of lower-bounded generalized modules for such a vertex algebra. We conjecture that the converse of the main theorem above is also true. We also prove that when a grading-restricted vertex algebra $V$ contains a subalgebra satisfying some familiar conditions, the composability condition for grading-restricted generalized $V$-modules always holds and we need $\hat {H}^{1}_{\infty }(V, M)= \hat {Z}^{1}_{\infty }(V, M)$ only for every $\mathbb {Z}$-graded $V$-bimodule $M$ generated by a grading-restricted subspace in our complete reducibility theorem.