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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.

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    $\begingroup$ 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? $\endgroup$
    – Robert Bryant
    Commented Jan 27, 2021 at 22:24

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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 $n-1=2.$

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    $\begingroup$ 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 $\bar H = 0$. $\endgroup$
    – Robert Bryant
    Commented Jan 31, 2021 at 11:25
  • $\begingroup$ @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 $\langle \overline e,\overline e\rangle_{\overline g} = e^{-2f}\langle e^{-f} e, e^{-f} e\rangle_{g} = e^{-4f}|e|^2_g?$ $\endgroup$
    – User
    Commented Jan 31, 2021 at 11:37
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    $\begingroup$ 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 $\bar g(e^{-f}X,e^{-f}X) = 1$. $\endgroup$
    – Robert Bryant
    Commented Jan 31, 2021 at 11:52
  • $\begingroup$ @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? $\endgroup$
    – User
    Commented Jan 31, 2021 at 12:00
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    $\begingroup$ 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$. $\endgroup$
    – Robert Bryant
    Commented Jan 31, 2021 at 14:03

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