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Let M a riemannian manifold. How can I show that the hodge-laplace-operator of a function $f$ is the negative of the laplace-operator?

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2 Answers 2

up vote 4 down vote accepted

A rather short proof can be found here.

I assume you are interested in the case when $f$ is a scalar function. Otherwise the Hodge Laplacian differs from the Laplace–Beltrami operator not only by a sign due to the Ricci curvature. See the Weitzenböck identity.

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Your link to the Weitzenböck formula is broken, btw. It should be en.wikipedia.org/wiki/Weitzenböck_identity –  Willie Wong Jun 23 '10 at 12:44
    
Thank you, Willie. I stand corrected. –  Andrey Rekalo Jun 23 '10 at 12:48

For the function case, choose local coordinate $(x_1,\cdots, x_n)$, then $\nabla_i\nabla_j f=Hessf(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j})=\frac{\partial^2 f}{\partial x_i\partial x_j}-\Gamma^k_{ij}f_k$, and \begin{equation} \Delta f =g^{ij}\nabla_i\nabla_j f=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x_i}(\sqrt{g}g^{ij}\frac{\partial f}{\partial x_j}), \end{equation} where $g=\det{g_{ij}}$.

Next, calculating $\Box f=d^*d f$. First, for $\alpha\in A^1(M)$, we calculate $d^*\alpha$. Let $\alpha=\alpha_i dx^i$, $\beta\in C^{\infty}(M)$, let $d^*\alpha=A$, then, \begin{equation} \begin{split} \int_X<d^*\alpha, \beta>dV&=\int_X<\alpha, d\beta>dV\\ &=\int_X \alpha_i\partial_j\beta g^{ij}dV\\ &=-\int_X\frac{\partial}{\partial x_j}(\alpha_i g^{ij}\sqrt{g}) \sqrt g\beta dV \end{split} \end{equation} So, $d^*\alpha=-g^{-\frac{1}{2}}\frac{\partial}{\partial x_j}(\alpha_i g^{ij} g^{\frac{1}{2}})$, and \begin{equation} \begin{split} \Box f &=d^*d f=d^*(f_i dx^i)\\ &=-\frac{1}{\sqrt{g}}\frac{\partial}{\partial x_i}(\sqrt{g}g^{ij}\frac{\partial f}{\partial x_j})=-\Delta f. \end{split} \end{equation}

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