# Ngô Quốc Anh

## December 31, 2014

### Conformal change of the Laplace-Beltrami operator

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 23:55

Happy New Year 2015!

In the last entry in 2014, I talk about conformal change of the Laplace-Beltrami operator. Given $(M,g)$ a Riemannian manifold of dimension $n \geqslant 2$. We denote $\widetilde g = e^{2\varphi} g$ a conformal metric of $g$ where the function $\varphi$ is smooth.

Recall the following formula for the Laplace-Beltrami operator $\Delta_g$ calculated with respect to the metric $g$:

$\displaystyle \Delta_g = \frac{1}{\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \Big( \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \Big).$

where $\det g$ is the determinant of $g$. Then, it is natural to consider the relation between $\Delta_g$ and $\Delta_{\widetilde g}$ in terms of $\varphi$. Recall that by $\widetilde g = e^{2\varphi} g$ we mean, in local coordinates, the following

$\displaystyle \widetilde g_{ij} = e^{2\varphi} g_{ij},$

hence by taking the inverse, we obtain

$\displaystyle \widetilde g^{ij} = e^{-2\varphi} g^{ij}.$

Clearly,

$\displaystyle\det {\widetilde g} = e^{2n \varphi}\det g,$

hence

$\displaystyle\sqrt{| \det {\widetilde g} |} = e^{n \varphi} \sqrt{ |\det g| }.$

With all these ingredient, we easily obtain

$\begin{array}{lcl} \Delta_{\widetilde g} &=& \displaystyle\frac{1}{\sqrt{|\det \widetilde g|}} \frac{\partial}{\partial x_j} \Big( \sqrt{|\det \widetilde g|} \widetilde g^{ij} \frac{\partial}{\partial x^i} \Big)\\&=&\displaystyle\frac{1}{e^{n \varphi}\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \Big( e^{n \varphi} \sqrt{|\det g|} e^{-2\varphi} g^{ij} \frac{\partial}{\partial x^i} \Big)\\&=&\displaystyle\frac{1}{e^{n \varphi}\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \Big( e^{(n-2) \varphi} \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \Big)\\&=&\displaystyle\frac{1}{e^{2\varphi}\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \Big( \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \Big)+\frac{1}{e^{n \varphi}\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \big( e^{(n-2) \varphi} \big)\sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \\&=&\displaystyle e^{-2\varphi}\Delta_g + (n-2)e^{-2\varphi}g^{ij} \frac{\partial \varphi}{\partial x_j} \frac{\partial }{\partial x_i}. \end{array}$

Thus, conformal change of the Laplace-Beltrami operator follows the following rule

$\displaystyle \Delta_{\widetilde g}=e^{-2\varphi}\Delta_g + (n-2)e^{-2\varphi}g^{ij} \frac{\partial \varphi}{\partial x_j} \frac{\partial }{\partial x_i}.$

When $n=2$, we simple have

$\displaystyle \Delta_{\widetilde g}=e^{-2\varphi}\Delta_g.$

Keep in mind that $\Delta_g$ can be calculated using $\Delta_g = \text{div}_g (\text{grad}_g \cdot)$; hence one can use conformal changes for $\text{grad}_g$ and $\text{div}_g$ to derive the above formula.

In local coordinates,

$\displaystyle\mbox{div}_g X = \frac{1}{\sqrt{|\det g|}} \partial_i \left(\sqrt {|\det g|} X^i\right)$

and

$\displaystyle\left(\mbox{grad}_g f\right)^i = \partial^i f = g^{ij} \partial_j f.$

1. Hi,
I am not a professional Mathematician but interested in Mathematics.

http://www.math.harvard.edu/~canzani/math253/Laplacian.pdf

I found difficulty in the exercises in P.45, about conformal transformations.

http://math.stackexchange.com/questions/1071506/conformal-transformation-of-the-divergence

I have the same calculation as yours but this is mismatch with the exercise.

Can you spare some time to help me find out what’s wrong?
Thanks a lot.

Comment by CW CHU — March 3, 2015 @ 8:08

• Hi, thanks for your interest in my blog and the comment.

In “4.3 The Laplacian under conformal deformations (Exercise)”, they use the conformal transformation $\widetilde g = e^f g$; hence to derive the 3rd identity, we let $2\varphi = f$. An other remark is that they define $\Delta_g = -\text{div}_g (\text{grad}_g \cdot)$.

3. Hence, from my formula

$\displaystyle \Delta_{\widetilde g}=e^{-2\varphi}\Delta_g - (n-2)e^{-2\varphi}g^{ij} \frac{\partial \varphi}{\partial x_j} \frac{\partial }{\partial x_i}$

and thanks to

$\displaystyle \frac{{\partial \varphi }}{{\partial {x_j}}} = \frac 12\frac{{\partial f}}{{\partial {x_j}}}$

we obtain

$\displaystyle \Delta_{\widetilde g}=e^{-f}\Delta_g - (\frac n2 -1)e^{-f}g^{ij} \frac{\partial f}{\partial x_j} \frac{\partial }{\partial x_i}.$

Finally, using their notion, we can write

$\displaystyle g^{ij} \frac{\partial f}{\partial x_j} \frac{\partial }{\partial x_i} =\nabla_g f;$

hence

$\displaystyle \Delta_{\widetilde g}=e^{-f}\Delta_g +(1-\frac n2 )e^{-f}\nabla_g f$

as claimed. Note that, there is a typo in their note, you may check with Wiki.

Comment by Ngô Quốc Anh — March 3, 2015 @ 11:09

2. Thank you very much!

“typo”, oh, it cost me a lot of time!

I find that your blog is useful for me (my interest is PDE and geometry). I will bookmark your blog and check it frequently.

Comment by CW CHU — March 3, 2015 @ 13:10