Kähler Identities: from the untwisted to the twisted case

Question

For any Kähler manifold $M$, we have the well known Kähler identities \begin{align*} [L,\partial^*] = i\overline{\partial}, & & [L,\overline{\partial}^*]=-i\partial, & & [L,\partial] = 0, & & [L,\overline{\partial}] = 0, \\ [\Lambda, \partial] = i\overline{\partial}^*, & & [\Lambda,\overline{\partial}] = -i\partial^*, & & [\Lambda, \partial^*] = 0, & & [\Lambda,\overline{\partial}^*] = 0, \end{align*} Moreover, these generalise to the analogous formuls for vector-valued forms in some holomorphic vector bundle $E$ endowed with a connection. What I would like to ask is whether the vector-valued version can be "simply" derived from the untwisted case. In the literature I have found no such derivation.

Answer 1

The most clear proof (through differential operators, graded Jacobi identities and superalgebra) actually works in both twisted and untwisted cases, no need to derive one from another. See http://arxiv.org/abs/math/0112215, section 8, for example. The idea is to show that $d^*$ is a "differential operator of second order" on the de Rham algebra (in the algebraic sense), and its commutator with L is a differential operator of first order, hence satisfies the Leibnitz identity, and check that it's $d^c$ on functions.

However, if you like, you can deduce one from another. Locally, the twisted $\partial$, $\partial^*$ and so on are obtained from untwisted ones by adding connection terms, and to prove the commutator relations, you prove them for untwisted ones and for the connection terms as well (the latter ones are linear, so it's easy).