# Showing that the conformal Laplacian is a conformally covariant operator

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!

• Did you try to expand the term $e^{-(\frac{n}{2}+1)f}\Delta_g \, e^{(\frac{n}{2}-1)f}$? Jul 29, 2019 at 8:45
• @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. Jul 29, 2019 at 17:30
• 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. Jul 31, 2019 at 10:51
• @YuriVyatkin: Could you ellaborate on your comment? Sep 17 at 7:00
• @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. Sep 17 at 12:16

[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.

References.

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