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It is well-known that the Weitzenböck formula for the real Laplacian is $$\frac12 Δ|∇f|2=|Hessf|2+⟨∇f,∇Δf⟩+Ricci(∇f,∇f)$$ where $Hess$ denotes the Hessian tensor of $f$. and $\nabla f$ denotes the gradient vector of $f$, $Ricci$ denotes the Ricci curvature of the manifold $M$.

If $\Delta_{\bar\partial}$ denotes the $\bar\partial$-Laplacian, it is well-known that it is half of the real Laplacian. So I am wondering is there any formula of the Weitzenböck formula in complex coordinates. (Assume the manifold is Kähler). Apparently one can devided the above formula by 2 to the one, but the expression I want should be expressed by $f_{i\bar j}$ and etc.

ps. The Comparison Geometry of Ricci Curvature, by Shunhui Zhu, 221-262 had a very nice introduction to this formula in real case. http://library.msri.org/books/Book30/contents.html

However I am not familiar with Kaehler case, for example, I dont know the such a formula can be derived in the same fashion as in Zhu's paper? Any book or paper with detailed calculation would be helpful.

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I would suggest also Schoen-Yau's book on differential geometry, where you can find a similar calculation for real case. The complex case can be done in the same manner. –  Ralph Jan 30 '13 at 21:06
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2 Answers

up vote 5 down vote accepted

You can just prove it yourself directly in local holomorphic coordinates. Indeed, the $\overline{\partial}$ Laplacian on functions is equal to $\Delta_{\overline{\partial}}f=g^{i\overline{j}}\partial_i \partial_{\overline{j}}f$. Apply this to $|\partial f|^2=g^{k\overline{\ell}}\partial_k f \partial_{\overline{\ell}}f$ (length squared of $\partial f=(df)^{(1,0)}$, which equals $1/2$ of the usual $|\nabla f|^2$), using if you want local holomorphic normal coordinates for $g$ at a point, and you will immediately get

$$\Delta_{\overline{\partial}}|\partial f|^2=|\nabla_i \nabla_j f|^2+|\nabla_i \nabla_{\overline{j}} f|^2+2\mathrm{Re}\langle \partial f, \partial\Delta_{\overline{\partial}}f\rangle+R^{i\overline{j}}\partial_i f\partial_{\overline{j}}f,$$ where $R^{i\overline{j}}$ is the Ricci curvature of $g$ with the indices raised.

If $g$ is not Kähler, and you define the complex Laplacian by the same formula $g^{i\overline{j}}\partial_i \partial_{\overline{j}}f,$ then a similar result holds, with the Ricci curvature now being one of the several Ricci curvatures of the Chern connection of $g$, and with several new terms involving the torsion of $g$ and its covariant derivative. The calculation is again completely strightforward, using local holomorphic coordinates (not normal anymore!), and using the definitions of covariant derivative and curvature of the Chern connection of $g$.

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Is $\partial f=\frac{1}{2}\nabla f - \sqrt{-1} J\nabla f$? I don't understand "gradient of the complex of $f$". –  Ralph Jan 30 '13 at 21:04
    
$\partial f$ is the $(1,0)$ part of the differential $1$-form $df$. I will correct my wrong wording now, sorry. –  YangMills Jan 31 '13 at 1:52
    
I meant to write "complex gradient of $f$", but I think it's clearer now. –  YangMills Jan 31 '13 at 1:54
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The formula is called the Bochner-Kodaira formula. It involves as all other Weitzenböck Formulars the curvature of the bundle, this time a holomorphic twisting bundle. Striking consequences are for example the Kodaira-Vanishing theorem.

You may take a look for example in Berline Getzler Vergne "Heat kernels and the dirac operator", page 135, Proposition 3.71 or in Lawson, Michelson "Spin geometry" Thm D.12

The proofs are always quite similar, you use that the symmetries of the curvature tensors and the clifford multiplication cancel each other to reduce the term which is not the connection laplace.

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