Ngô Quốc Anh

October 29, 2013

The mean curvature under conformal changes of Riemannian metrics: A global approach

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 22:46

In a previous post, I showed how the mean curvature changes under a conformal change using slightly local coordinates approach. Today, I want to reconsider that topic using global approach.

As usual, suppose $(M,g)$ is an $n$-dimensional Riemannian manifold with boundary $\partial M$. We also assume that $N$ is an outward unit normal vector field along the boundary $\partial M$. By an unit normal vector field we mean $g(N,N)=1$ and $g(N,X)=0$ for any tangent vectors $X$ of $\partial M$. Note that $\partial M$ is just a hypersurface of $M$ and we also use $g$ to denote the induced metric of $g$ onto $\partial M$. Then we have the so-called second fundamental form $\mathrm{I\!I}$ associated to $\partial M$ defined to be $\displaystyle \mathrm{I\!I} (X,Y)=g(X,\nabla_Y N)$

for any tangent vectors $X,Y$ of $\partial M$. Regarding to the mean curvature $H$, we shall use the following definition $\displaystyle H=\text{trace}_g (\mathrm{I\!I})=g^{ij}\mathrm{I\!I}(\partial_i, \partial_j)=g^{ij} g(\partial_i,\nabla_{\partial_j} N).$

Our aim is to calculate the mean curvature $H$ under the following conformal change $\widehat g=\phi^\kappa g$ for some smooth positive function $\phi$ and a real number $\kappa$. It is important to note that by $\widehat g=\phi^\kappa g$ we mean $\displaystyle \widehat g_{ij} = \phi^\kappa g_{ij}.$

Therefore, for the inverse metric, there holds $\displaystyle \widehat g^{ij} = \phi^{-\kappa} g^{ij}.$

Since the normal vector field $N$ depends on the metric $g$, it turns out that under the new metric $\widehat g$, $\displaystyle \widehat N = \phi^{-\kappa/2}N$

is a new normal vector field along $\partial M$, i.e. $\widehat g(\widehat N, \widehat N)=1$. To continue, we need to understand the conformal change for the Levi-Civita connection, let us recall the following formula $\displaystyle {{\widehat\nabla }_X}Y - {\nabla _X}Y = \frac{\kappa }{2}\left( {\frac{{{\nabla _X}\phi }}{\phi }Y + \frac{{{\nabla _Y}\phi }}{\phi }X - g(X,Y)\frac{{\nabla \phi }}{\phi }} \right).$

According to the note, the formula we quoted should be $\displaystyle {\widehat\nabla _X}Y = {\nabla _X}Y + X(f)Y + Y(f)X - g(X,Y) {\rm grad}f$

provided $\widehat g=e^{2f}g$. However, in view of our notation used, there holds $f = \frac{\kappa }{2}\log \phi$. Hence $\displaystyle {\rm grad}f=\frac{\kappa }{2}\frac{\nabla\phi}{\phi}, \quad X(f)=\nabla_Xf=\frac{\kappa }{2}\frac{\nabla_X\phi}{\phi},\quad Y(f)=\frac{\kappa }{2}\frac{\nabla_Y \phi}{\phi}.$

Then we calculate the new second fundamental form given by $\displaystyle \widehat{\mathrm{I\!I}} (X,Y)=\widehat g(X,\widehat\nabla_Y \widehat N)$

Clearly, we have $\begin{array}{lcl} \widehat{\mathrm{I\!I}} (X,Y) &=& \displaystyle\widehat g(X,{{\widehat\nabla }_Y}\widehat N) \hfill \\ &=& \displaystyle {\phi ^\kappa }g(X,{{\widehat\nabla }_Y}({\phi ^{ - \kappa /2}}N)) \hfill \\ &=& \displaystyle {\phi ^\kappa }g \Big( X,{\phi ^{ - \kappa /2}}{{\widehat\nabla }_Y}N - \frac{\kappa }{2}{\phi ^{ - \kappa /2 - 1}}N{{\widehat\nabla }_Y}\phi \Big) \hfill \\ &=& \displaystyle {\phi ^{\kappa /2}}g \bigg( X,{{\widehat\nabla }_Y}N - \underbrace {\frac{\kappa }{2}N\frac{{{{\widehat\nabla }_Y}\phi }}{\phi }}_{\-0} \bigg) \hfill \\ &=& \displaystyle {\phi ^{\kappa /2}}g \Bigg( X,\underbrace {{\nabla _Y}N}_{\-\mathrm{I\!I}} + \frac{\kappa }{2}\bigg( {\underbrace {\frac{{{\nabla _Y}\phi }}{\phi }N}_{\-0} + \frac{{{\nabla _N}\phi }}{\phi }Y - \underbrace {g(Y,N)}_{ = 0}\frac{{\nabla \phi }}{\phi }} \bigg) \Bigg) \hfill \\ &=& \displaystyle {\phi ^{\kappa /2}}\mathrm{I\!I} (X,Y) + {\phi ^{\kappa /2}}g \Big( X,\frac{\kappa }{2}\frac{{{\nabla _N}\phi }}{\phi }Y \Big). \end{array}$

Thus, we have shown that $\displaystyle \widehat{\mathrm{I\!I}} (X,Y)= {\phi ^{\kappa /2}}\left( {\mathrm{I\!I} (X,Y) + \frac{\kappa }{2}\frac{{{\nabla _N}\phi }}{\phi }g(X,Y)} \right).$

Then $\begin{array}{lcl} {\widehat H_{\widehat g}} &=& \displaystyle {{\text{trace}}_{\widehat g}}(\widehat {\mathrm{I\!I}}) \hfill \\ &=&\displaystyle {{\widehat g}^{ij}}\widehat {\mathrm{I\!I}}({\partial _i},{\partial _j}) \hfill \\ &=&\displaystyle {\phi ^{ - \kappa }}{g^{ij}}\left( {{\phi ^{\kappa /2}}\mathrm{I\!I} ({\partial _i},{\partial _j}) + {\phi ^{\kappa /2}}g \Big( {\partial _i},\frac{\kappa }{2}\frac{{{\nabla _N}\phi }}{\phi }{\partial _j} \Big) } \right) \hfill \\ &=&\displaystyle {\phi ^{ - \kappa /2}}\bigg( {{H_g} + \frac{\kappa }{2}\frac{{{\nabla _N}\phi }}{\phi }\underbrace {{g^{ij}}{g_{ij}}}_{n - 1}} \bigg). \end{array}$

It is important to remind that the metric $g$ appearing in the above calculation is just the induced metric on $\partial M$ which is of dimension $n-1$, therefore $g^{ij}g_{ij}=n-1$ is clear. In particular, when $\kappa=4/(n-2)$, we obtain $\displaystyle {\widehat H_{\widehat g}} = {\phi ^{ - 2/(n - 2)}} \bigg( {H_g} + \frac{{2(n - 1)}}{{n - 2}}\frac{{{\nabla _N}\phi }}{\phi } \bigg).$

Therefore, if we normalize the mean curvature by $\displaystyle H_g =\frac{1}{n-1}\text{trace}_g (\mathrm{I\!I})$

we then have the following rule $\displaystyle {\widehat H_{\widehat g}} = {\phi ^{ - 2/(n - 2)}} \bigg( {H_g} + \frac{2}{{n - 2}}\frac{{{\nabla _N}\phi }}{\phi } \bigg).$

We now consider the case when the mean curvature $H_g$ is defined to be $\displaystyle H_g=g^{ij} g(\partial_i,\nabla_{\partial_j} (-N)).$

Clearly, $H_g=-\text{trace}_g(\mathrm{I\!I})$. Therefore, we can write $\displaystyle \widehat H_{\widehat g}=-\text{trace}_{\widehat g} (\widehat{\mathrm{I\!I}}) = -{\phi ^{ - 2/(n - 2)}} \bigg( -H_g + \frac{{2(n - 1)}}{{n - 2}}\frac{{{\nabla _N}\phi }}{\phi } \bigg).$

Thus, we have shown that $\displaystyle \nabla_N \phi - \frac{n-2}{2(n-1)}H_g \phi= -\widehat H_{\widehat g} \phi^{n/(n-2)}.$