Exercise $3.1.7$ from Huybrechts' Complex Geometry: An Introduction

I am trying to do the following question:

Show that $L$, $d$, and $d^*$ acting on $\mathcal{A}^*(X)$ of a Kähler manifold $X$ determine the complex structure of $X$.

Here $L$ is the Lefschetz operator $L : \mathcal{A}^{p,q}(X) \to \mathcal{A}^{p+1,q+1}(X)$, $\alpha \mapsto \omega\wedge\alpha$ where $\omega$ is the Kähler form. The exterior derivative is denoted by $d$, and $d^* = -\ast d \ast$ is the adjoint of $d$, where $\ast$ denotes the Hodge dual $\mathcal{A}^{p,q}(X) \to \mathcal{A}^{n-q,n-p}(X)$. Finally $\mathcal{A}^*(X)$ denotes the exterior algebra of $X$.

I assume that the Kähler identities come in to play here, but I'm not sure how.

Added Later: The Kähler identities are

• $[L, \bar{\partial}] = 0$, $[L, \partial] = 0$, $[\Lambda, \bar{\partial}^*] = 0$, $[\Lambda, \partial^*] = 0$,
• $[L, \bar{\partial}^*] = -i\partial$, $[L, \partial^*] = i\bar{\partial}$, $[\Lambda, \bar{\partial}] = -i\partial^*$, $[\Lambda, \partial] = i\bar{\partial}^*$, and
• $\Delta_d = \Delta_{\partial} + \Delta_{\bar{\partial}}$, $\Delta_{\partial} = \Delta_{\bar{\partial}}$.

Solution 1:

Here are the ingredients of a proof:

1. The Kähler form is determined by $\omega = L(1)$.
2. The Laplace-Beltrami operator on functions is $\Delta = -d^*d$.
3. The cometric $g^*$ (i.e., the metric on $1$-forms) is the principal symbol of $\Delta$, which can be computed as follows: Given $x\in X$ and $\xi\in T_x^*X$, choose a smooth function $f$ defined on a neighborhood of $x$ satisfying $f(x)=0$ and $df_x = \xi$. Then $$ g^*(\xi,\xi) = \tfrac 1 2 \Delta(f^2) (x). $$ From this, $g^*$ is determined by polarization.
4. Once $g^*$ is known, the Riemannian metric $g$ is determined just by inverting the matrix of $g^*$ in any local coordinates.
5. Now the complex structure $J$ is determined by $g(v,w) = \omega(v, Jw)$. (In coordinates, this means $J$ is given locally by the matrix product $J = \omega^{-1} g$.)