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Let $M$ be a closed manifold in $\mathbb{R}^3$ and $\partial M$ its surface. I want to find (in general terms) the manifold that minimizes a functional of the form $$I[M]=\int_{\partial M}f\,\mathrm{d}S+\int_MF\,\mathrm{d}V,$$ for some functions $F,f$. Let $g_{ij}$ be the metric tensor of the manifold and $g$ its determinant. The volume term is easily rewritten: $$\int_MF\,\mathrm{d}V=\int F(\vec{x})\sqrt{g}\,\mathrm{d}^3x,$$ but what should I do with the surface term? The metric tensor is three-dimensional but for the surface integral I need only two dimensions. I've read at some point of induced metrics, they might be what I need. But then again, how do I vary the surface integral in terms of an induced metric with respect to the actual metric?

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You are correct that you are going to need an induced metric. Namely:

\begin{equation} \int _{\partial M} f\, dS = \int d^{2}y \, \sqrt{|h|} \, f(y^{i}) \end{equation}

where $h = \det{(h_{ij})}$, with $h_{ij}$ being the induced metric and $y^{i}$ the (two) coordinates by which you parametrize it. For a particular embedding of the surface, the induced metric is derived from the metric tensor of $M$ as:

\begin{equation} h_{ij} = g_{\mu \nu} \frac{\partial x^{\mu}}{\partial y^{i}} \frac{\partial x^{\nu}}{\partial y^{j}} \end{equation}

with summation implied for the the contracted indices as usual. Therefore the variation of the induced metric is proportional to the variation of the manifold metric, times the pullback i.e. the embedding of the submanifold $\partial M$ with coordinates $y^{i}$ in the manifold $M$, which itself is embedded in Euclidean space, with coordinates $x^{\mu}$:

\begin{equation} \delta h_{ij} = \delta g_{\mu \nu} \left (\frac{\partial x^{\mu}}{\partial y^{i}} \frac{\partial x^{\nu}}{\partial y^{j}} \right ) \end{equation}

In the case of the metric being the Euclidean space, you of course have $y^{i} = y_{i}$ and $x^{\mu} = x_{\mu}$.

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  • $\begingroup$ Is there a natural way to go from the complete metric to the one of the surface? E.g. for a sphere you can just discard the $\mathrm{d}r^2$-component, so are there ways to write this into a formula? $\endgroup$ Aug 2 at 13:32
  • $\begingroup$ The reason why I'm asking is that in order to get the metric tensor, under the hood, you already did a parametrization, taking derivatives, etc. so are there ways to strip the full metric of it's components $\endgroup$ Aug 2 at 13:35
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    $\begingroup$ The mapping I mention is the most natural way to go from the metric of $M$ to that of $\partial M$. The example you mention is very specific; you could define a submanifold on a sphere so that the coordinate that remains constant upon it is the polar or azimuthal angle instead of the radial coordinate. If you seek a greater level of abstraction in the sense of a mapping from $g_{\mu \nu}$ to $h_{ij}$ with the coordinate independence manifest, there is no particular "formula". For a concrete physical formula, you need a certain embedding for both i.e. choosing specific coordinates. $\endgroup$
    – rhomaios
    Aug 2 at 15:26
  • $\begingroup$ okay, I see. Thanks! $\endgroup$ Aug 2 at 16:42

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