# Relation between mean curvature and conformal metric

We'll consider $$(N, g)$$ a Riamannian Manifold and $$\overline{g} = e^{2f}g$$ a conformal metric. Let M be a hypersurface in N, $$\overline{H}_M$$ and $$H_M$$ the mean curvature of M with respect to the metrics $$\overline{g}$$ and g, respectively. I would like some help to prove that

$$\overline{H}_M = e^{-f}( H_M -2g( \nabla f, \eta))$$

where $$\nabla$$ is the gradient with respect to metric g and $$\eta$$ is a normal vector field in M.

• Your formula seems to be slightly off in a few ways. For example, do you really want $\eta$ to be a tangent vector field to $M$ rather than the normal vector field? Jan 27 at 22:24

Let me use the transformation $$\overline g = e^{2f}g$$ to simplify some notations (and I guess your formula also use this convention). Near a point $$p\in N,$$ let $$\{e_i\}$$ be an orthonormal frame with respect to $$g,$$ and $$\eta$$ be a normal. Then with respect to $$\overline g,$$ we have $$\overline e_i = e^{-f}e_i$$ form an orthonormal frame near $$p$$ and $$\overline\eta=e^{-f}\eta$$ being the normal. Then \begin{align} \overline h_{ij} & = \langle\overline\nabla_{\overline e_i}\overline e_j,\overline\eta \rangle_{\overline g}\\ & = e^{2f}\langle e^{-2f}\overline\nabla_{e_i}e_j, e^{-f}\eta\rangle_g\\ & = e^{-f}\langle \nabla_{e_i}e_j-\delta_{ij}\nabla f,\eta \rangle_g\\ & = e^{-f}( h_{ij} -\langle \nabla f,\eta\rangle_g\delta_{ij})\\ \end{align} where we use the transformation of the Levi-Civita connection, i.e., $$\overline\nabla_X Y = \nabla_XY + (Xf)Y + (Yf)X - \langle X,Y\rangle_g\nabla f.$$ Thus \begin{align} \overline H & = \sum_{i=1}^{n-1}\overline h_{ii}\\ & = \sum_{i=1}^{n-1}e^{-f}(h_{ii} - \langle \nabla f,\eta\rangle_g)\\ & = e^{-f} (H-(n-1) \langle \nabla f,\eta\rangle_g), \end{align} where $$n$$ is the dimension of $$N,$$ so in your case $$n-1=2.$$
• A couple of comments to get a correct formula: First, if $\bar g = \mathrm{e}^{2f} g$ and $\mathbf{e}_i$ is a $g$-orthonormal frame near $p$, then $\mathrm{e}^{-f} \mathbf{e}_i$ is a $\bar g$-orthonormal frame. (Note the minus sign.) Second, the mean curvature is the average of the principal curvatures, not the sum, so you are missing a normalization factor. For example, check your formula for the unit sphere $M=S^{n-1}\subset\mathbb{E}^n$ with $g = |dx|^2$, $\mathrm{e}^{2f} = |x|^{-2}$, and $\eta$ the outward normal. You should get $H=-1$ and $\bar H = 0$. Jan 31 at 11:25
• @Robert Bryant Hello Robert. Thanks for your comment! Just one question. If $\overline g = e^{2f}g$ and $\overline e=e^{-f}e,$ isn't it correct that $\langle \overline e,\overline e\rangle_{\overline g} = e^{-2f}\langle e^{-f} e, e^{-f} e\rangle_{g} = e^{-4f}|e|^2_g?$
• No, that is not correct. $g$ is a a tensor of type $(0,2)$ while vector fields are tensors of type $(1,0)$. Maybe you are confused by the notation $\langle X,Y \rangle_g$, which means $g(X,Y)$, so $\bar g(X,Y) = e^{2f} g(X,Y) = g(e^fX,e^fY)$. In particular, if $g(X,X)=1$, then $\bar g(e^{-f}X,e^{-f}X) = 1$. Jan 31 at 11:52
• @RobertBryant So when people write $\overline g = e^{2f} g,$ it doesn't mean that locally $\overline g_{ij} = e^{2f}g_{ij}$ (which implies $\overline g^{ij} = e^{2f}g^{ij}$), is it?
• Yes, $\bar g = e^{2f} g$ means what you say, and also $\bar g^{ij} = -e^{2f} g^{ij}$, but that's not relevant. The point is that a vector field in local coordinates is $X = X^i\,\partial_i$, with $|X|_g^2 = g_{ij}X^iX^j$. Jan 31 at 14:03