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
  • $\Delta_d = \Delta_{\partial} + \Delta_{\bar{\partial}}$, $\Delta_{\partial} = \Delta_{\bar{\partial}}$.
  • $\begingroup$ This question is obviously intended to a technical audience, but as a favor to us nonspecialists: Could you include the statement of the Kähler identities? $\endgroup$ – Semiclassical Jul 24 '14 at 4:13

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 $J = \omega^{-1} g$.)
  • 2
    $\begingroup$ Could you expand on 3.? I am not familiar with this $\endgroup$ – klirk Feb 4 '18 at 19:48
  • $\begingroup$ @klirk: You don't really need to know anything about symbols or differential operators to do this: Just check the computation in any local coordinates. It has nothing to do with the complex structure either; it's purely a Riemannian computation. $\endgroup$ – Jack Lee Feb 4 '18 at 19:50
  • $\begingroup$ Dear @JackLee,so if we know the almost complex structure,since it is a complex manifold,we can know the unique complex structure which induces the almost complex structure? $\endgroup$ – Steve Oct 1 at 13:36
  • $\begingroup$ @Steve: Exactly so. $\endgroup$ – Jack Lee Oct 1 at 21:45

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