Consider a Riemannian manifold of dimension $n\geq3$. Consider the conformal Laplacian \begin{equation*} P_g=\Delta_g+\frac{n-2}{4(n-1)}R_g, \end{equation*} where $R_g$ is the scalar curvature associated to the metric $g$.

I would like to show that $P_g$ is a conformally covariant operator in the sense that under the conformal change $\tilde{g}=e^{2f}g$, the following transformation law is satisfied: \begin{align*} P_{\tilde{g}}&=e^{-(\frac{n}{2}+1)f}P_g \,e^{(\frac{n}{2}-1)f}\\ &= e^{-(\frac{n}{2}+1)f}\Delta_g \,e^{(\frac{n}{2}-1)f}+\frac{n-2}{4(n-1)}R_g\,e^{-2f} \tag{1} \end{align*} I know that under such a conformal change, the Laplace-Beltrami operator transforms like \begin{equation*} \Delta_{\tilde{g}}=e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i}. \end{equation*} For the scalar curvature, one can make the substitution $e^{2f}=\varphi^{4/(n-2)}$ (where $\varphi$ is positive) to get \begin{equation*} R_{\tilde{g}}=\varphi^{-(n+2)/(n-2)}\bigg(4\frac{n-1}{n-2}\Delta_g \varphi + R_g\varphi\bigg) \end{equation*} which is really just \begin{equation*} R_{\tilde{g}}=4\frac{n-1}{n-2}e^{-(\frac{n}{2}+1)f}\Delta_g \, e^{(\frac{n}{2}-1)f}+R_g e^{-2f}. \end{equation*} So, I get \begin{equation*} \tag{2} P_{\tilde{g}}= e^{-2f}\Delta_g-(n-2)e^{-2f}g^{ij}\frac{\partial f}{\partial x_j}\frac{\partial}{\partial x_i} + e^{-(\frac{n}{2}+1)f}\Delta_g \, e^{(\frac{n}{2}-1)f}+\frac{n-2}{4(n-1)}R_g\, e^{-2f}. \end{equation*}

The problem is that I do not see how (1) is the same as (2). Have I made a mistake somewhere? Or do the first two terms in (2) somehow cancel each other out?

Any help would be greatly appreciated!

  • $\begingroup$ Did you try to expand the term $e^{-(\frac{n}{2}+1)f}\Delta_g \, e^{(\frac{n}{2}-1)f}$? $\endgroup$ Jul 29, 2019 at 8:45
  • $\begingroup$ @YuriVyatkin I have attempted to do that, using the local coordinate form of $\Delta_g$, but all I got was $\frac{n-2}{2}\bigg[ e^{-2f}\Delta_g f - \frac{1}{\sqrt{|\det g|}}(\frac{n-2}{2})e^{-2f}\frac{\partial f}{\partial x_i}\bigg( \sqrt{|\det g|}g^{ij}\frac{\partial f}{\partial x_j} \bigg) \bigg]$. I'm not sure if continuing the calculation will somehow yield the first three terms in equation (2) of my question. $\endgroup$
    – Quantagon
    Jul 29, 2019 at 17:30
  • $\begingroup$ Oh, I am sorry, you don't need to expand this term. The calculation is straightforward, but you need to be more accurate with examining on what object the operator $P_g$ is acting. Hint: the rescaled operator $P_{\tilde{g}}$ acts on a product of functions. $\endgroup$ Jul 31, 2019 at 10:51
  • $\begingroup$ @YuriVyatkin: Could you ellaborate on your comment? $\endgroup$ Sep 17, 2022 at 7:00
  • $\begingroup$ @AniruddhaDeshmukh I believe that I did so in my answer. Could you please specify what remains unclear? You can add comments to my answer below. $\endgroup$ Sep 17, 2022 at 12:16

1 Answer 1


[An incomplete draft]

I would adjust the notation to make it more suitable for the upcoming calculation. The conformally rescaled metric will be denoted by $\widehat{g} = e^{2\varphi} g$, where $g$ is a Riemannian metric on a given manifold $M$ (the consideration is purely local, so we can be a little less specific here). It is customary then to denote the quantities corresponding to the conformally rescaled metric $\widehat{g}$ by placing the $\widehat{}$ on top of the respective symbol. Thus, the Levi-Civita connection $\nabla^{\widehat{g}}$ of the rescaled metric $\widehat{g}$ will be denoted simply by $\widehat{\nabla}$, and $\nabla$ will mean the Levi-Civita connection, corresponding to the metric $g$. Furthermore, $\Delta_{\widehat{g}}$ will be denoted by $\widehat{\Delta}$, whereas $\Delta_{g}$ will be simply denoted as $\Delta$. The same conventions apply to the scalar curvatures $\widehat{R}$ and $R$, and so on.

The reason why I make all these preparation is that we need to recall a few formulas for the conformal rescaling of the Levi-Civita connection. We are going to use this formula for the case of $1$-forms: $$ \widehat{\nabla}_i \omega_j = \nabla_i \omega_j - \omega_i \varphi_j - \varphi_i \omega_j + \varphi^k \omega_k g_{i j} $$ where $\varphi_i := \nabla_i \varphi$.

Using this formula, it is straightforward to show that $$ \widehat{\Delta} f = e^{-2 \varphi} \big( \Delta f - (n - 2) \varphi^k \nabla_k f \big) $$ and, with a little bit more work, that the scalar curvature rescales as (see. e.g. here) $$ \widehat{R} = e^{-2 \varphi} \big( R + 2(n - 1) \Delta \varphi - (n - 2)(n - 1) \varphi^k \varphi_k \big) $$ It is not too difficult to convince yourself that there is no linear combination of operators $\Delta$ and $R$ (regarded as acting by multiplication) such that all the extra terms in their transformation would cancel out.

The trick is to consider the action of $\Delta$ and $R$ on weighted functions, that is functions of the form $e^{a \varphi} f$.

One can verify that $$ \widehat{\nabla}_i (e^{a \varphi} f) = e^{a \varphi} \big( \nabla_i f + a \varphi_i f \big) $$ and, using this, that $$ \widehat{\Delta} (e^{a \varphi} f) = e^{(a - 2) \varphi} \bigg( \Delta f + (n + 2 a -2) \varphi^k \nabla_k f + a \big( \nabla^k \varphi_k + (n + a -2) \varphi^k \varphi_k \big) f \bigg) $$

For $a = 1 - n/2$ we get $$ \widehat{\Delta} (e^{ (1 - n/2) \varphi} f) = e^{(- 1 - n/2) \varphi} \bigg( \Delta f - \frac{n-2}{4} \big( 2 \Delta \varphi + (n - 2) \varphi^k \varphi_k f \bigg) $$

At the same time $$ \widehat{R}(e^{ (1 - n/2) \varphi} f) = e^{(- 1 - n/2) \varphi} \bigg( R - (n - 1) \big( 2 \Delta \varphi + (n - 2) \varphi^k \varphi_k \big) f \bigg) $$

At this point it should be easy to see that the operator $P_g$, as defined in the question, exhibits the necessary covariance property.

TODO: check the signs.


  1. Jan Slovak, Natural Operators on Conformal Manifolds, http://www.math.muni.cz/~slovak/Papers/habil.pdf

  2. M.Eastwood, Conformally invariant differential operators, https://maths-people.anu.edu.au/~eastwood/fayetteville2.pdf

  3. List of formulas in Riemannian geometry, https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry

  • $\begingroup$ Hi, this is really interesting. One question, what is the definition of the operator $\Delta$ in this context? I know that in Euclidean space the laplacian is the divergence of the gradient but I was wondering what is the difference here. Thanks $\endgroup$ May 4, 2023 at 7:42
  • 1
    $\begingroup$ @l4teLearner We are talking here about the Laplacian on a Riemannian manifold. The Euclidean space is a particular example, and the Laplacian becomes what you know about it. It's hard to give you a complete exposition in a comment. Perhaps, you can refer to the beginning of this answer to get a taste of the definitions. Simply put, the Laplacian here is the contraction of the second covariant derivative. To become comfortable in this context, you should read at least one textbook on Riemannian geometry. $\endgroup$ May 5, 2023 at 21:56
  • $\begingroup$ thanks for your reply. the reason why I am interested in the Laplacian is that I am trying to match this answer: math.stackexchange.com/questions/3126835/… with the wikipedia page en.wikipedia.org/wiki/… . I am facing the same exercise from Tu and I have found the same formula in terms of moving frames as in the other answer but I cannot correlate it with the wikipedia formula. note that in the exercise n=2. so the wiki formula has just two terms. thanks $\endgroup$ May 5, 2023 at 22:36

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