TITLE: Global Algebraic Proof of the Kahler Identities?
QUESTION [11 upvotes]: I'm looking at Kahler geometry at the moment and admiring how it manages to do so much with clean global algebraic arguments. One of the big exceptions to all this, however, is the proof of the Kahler identities
$$
[\Lambda,\overline{\partial}]=-i \partial^\ast, ~~~~~~ 
[\Lambda,\partial]=-i \overline{\partial}^\ast.
$$
In the two standard references, Voisin, and Griff + Harris, the identities are proved using arguments that are local and somewhat analaytic. Does there exists anywhere a nice global algebraic proof?

REPLY [7 votes]: Actually, there is an argument that doesn't use neither Weil identities nor flat up to a second order coordinates, instead, it uses the symplectic Hodge star, invented by Brylinski. When you have a 2n-dimensional manifold $M$ with the non-degenerate 2-form $\omega$, you can define an operator $*_s:\Lambda^k(M) \mapsto \Lambda^{2n-k}(M)$ by the standard identity $\alpha \wedge *_s\beta = \omega(\alpha,\beta)\omega^n$. Then, for a closed $\omega$ a really simple computation (you can do it only for the case of two variables, and then use induction on dimension) in Darboux coordinates shows that, up to some sign ($(-1)^{k+1}$, i guess), $*_sd*_s=[\Lambda, d].$ It is written in the Brylinski's article "A differential complex for Poisson manifolds". And then you need to observe that for a Kaehler $M$ symplectic and Riemannian Hodge stars differ by an action of $I$. The Kaehler identity $d^*=[\Lambda, d^c]$ follows from that.