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.

See also:

  1. Conformal Changes of the Green function for the conformal Laplacian.
  2. Conformal Changes of Riemannian Metrics
  3. Why should we call ” f=\varphi g ” conformal change?

3 Comments »

  1. Hi,
    I am not a professional Mathematician but interested in Mathematics.
    When I read the following notes about Laplacian:

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

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

    So , I google and found your useful page and this link:

    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


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