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Could anyone tell me where I'm wrong with the following elementary calculation? Given a smooth Riemannian manifold $(M, g)$, I'd prove that if $\tilde{g}$ is conformally equivalent to $g$ (that is, $\tilde{g} = e^{2w}g$ for some smooth function $w$), then $\Delta_{\tilde{g}} = e^{-2w}\Delta_g$. Now, recalling that the Laplace-Beltrami operator $\Delta_g$ is defined as (or better, this is a possible definition)

$$\Delta_g:=g^{ij}\left(\frac{\partial^2}{\partial x_i \partial x_j} - \Gamma^k_{i j}\frac{\partial }{\partial x_k}\right).$$

With elementary calculations, I obtained (and I'm pretty sure that, at least this, is right...)

$$\tilde{\Gamma}^k_{ij} = \Gamma^k_{ij} - \left(\delta_{ik}\frac{\partial w}{\partial x_j} + \delta_{kj}\frac{\partial w}{\partial x_i} - g^{k\ell}g_{ij}\frac{\partial w}{\partial x_{\ell}}\right),$$

where $\tilde{\Gamma}^k_{ij}$ are the Christoffel's symbols of the Levi-Civita's connection associated to the metric $\tilde{g}$ and $\delta_{ij}$ is the Kronecker's $\delta$. At this point, I have

$$ \Delta_{\tilde{g}} = e^{-2w}\Delta_g + e^{-2w}g^{ij}\left(\delta_{ik}\frac{\partial w}{\partial x_j} + \delta_{kj}\frac{\partial w}{\partial x_i} - g^{k\ell}g_{ij}\frac{\partial w}{\partial x_{\ell}}\right)\frac{\partial}{\partial x_k} $$

But

$$ g^{ij}\delta_{ki}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_k} = g^{ij}\frac{\partial w}{x_j}\frac{\partial }{\partial x_i}, $$

$$ g^{ij}\delta_{kj}\frac{\partial w}{\partial x_i} \frac{\partial }{\partial x_k} = g^{ij}\frac{\partial w}{\partial x_i}\frac{\partial}{\partial x_j} \stackrel{i \leftrightarrow j}{=} g^{ji}\frac{\partial w}{\partial x_j}\frac{\partial}{\partial x_i} \stackrel{g^{ji} = g^{ij}}{=} g^{ij}\frac{\partial w}{\partial x_j}\frac{\partial}{\partial x_i}, $$

$$ \underbrace{g^{ij}g_{ij}}_{= g^{ij}g_{ji} = 1} g^{k\ell} \frac{\partial w}{\partial x_{\ell}}\frac{\partial }{\partial x_k} = g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i}. $$

That is, one the last terms seems survive. Where am I wrong?

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2 Answers 2

The proposed relationship $$\Delta_{\tilde{g}} = e^{-2w} \Delta_g$$ only holds for surfaces. In your final calculation, assuming $M$ is a surface, you should have $g^{ij} g_{ij} = 2$, so that the final term is \begin{align*} e^{-2w}g^{ij}\left(\delta_{ik}\frac{\partial w}{\partial x_j} + \delta_{kj}\frac{\partial w}{\partial x_i} - g^{k\ell}g_{ij}\frac{\partial w}{\partial x_{\ell}}\right)\frac{\partial}{\partial x_k} & \\ = g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i} + g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i} - 2g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i} & = 0. \end{align*}

An easier approach which avoids using Christoffel symbols or messing around with indices is to use the formula $$\Delta_g = - \frac{1}{\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \right).$$ With this formula, you can compute \begin{align*} \Delta_{\tilde{g}} & = - \frac{1}{e^{dw}\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( e^{dw} \sqrt{|\det g|} e^{-2w} g^{ij} \frac{\partial}{\partial x^i} \right) \\ & = - \frac{e^{-dw}}{\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( e^{(d-2)w} \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \right) \\ & = - \frac{(d-2)e^{-2w}}{\sqrt{|\det g|}} \frac{\partial w}{\partial x_j} \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} - \frac{e^{-2w}}{\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \right) \\ & = \Delta_g - (d-2)e^{-2w} g^{ij} \frac{\partial w}{\partial x_j} \frac{\partial}{\partial x^i}, \end{align*} so $\Delta_{\tilde{g}} = e^{-2w} \Delta_g$ only holds when $d = 2$. (You'll get the same formula when you correctly state $g^{ij}g_{ij} = d$ in your work above).

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For Riemannian manifolds of dimension $d$, $g^{ij}g_{ji} = d$. This means your formula works, but only in dimension two.

There is an alternative expression for the Laplace-Beltrami operator involving $\sqrt{\det g}$ which gives you the factor $d-2$ directly.

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Is there a coordinate-free way to arrive at this expression for $\Delta$? –  William Jan 25 '13 at 23:43

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