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Let $(M^3,g)$ be a complete Riemannian manifold. Fix a two-sided immersion $\varphi : \Sigma^2 \to M$ from a closed surface into $M$, with unit normal $N$. Given $f \in C^\infty(\Sigma)$ and $\alpha : (-\varepsilon, \varepsilon) \to \mathbb{R}$ of sufficiently small $C^0$-norm, define $\varphi_t : \Sigma \to M$, for each $t \in (-\varepsilon, \varepsilon)$, by

$$ \varphi_t(x) = \exp_{\varphi(x)}(\alpha(t) f(x) N(x)), \quad x \in \Sigma, t \in (-\varepsilon, \varepsilon), $$

where $\exp$ is the exponential map of $(M,g)$.

Is there a reasonable formula for the area of the Riemannian surface $(\Sigma, \varphi_t^\ast g)$? For fixed $f$ satisfying $\int_{\Sigma} f H_{\Sigma} \, \mathrm{d}A = 0$ (here $H_\Sigma$ is the mean curvature of $\varphi$), is it possible to choose $\alpha$ of the form $\alpha(t) = t \beta(t)$ so that this area is constant with respect to $t$?

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    $\begingroup$ For the second question: Given your statement, infinitesimally you need $\int_\Sigma f(x) h(x) = 0$ where $h(x)$ is the mean curvature of $\Sigma$ in $M$. So for a given immersion and a given function $f$, certainly it is not always possible for the area to be constant (unless you admit the case $\alpha(t) = \text{const.}$. Example: $M = \mathbb{R}^3$ and $\Sigma$ the standard unit sphere. If you choose $f\equiv 1$ then any non-trivial $\alpha$ will give a change of area. $\endgroup$ Commented May 3, 2022 at 17:38
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    $\begingroup$ For the first question: there is no simple formula using only quantities local to $\Sigma$. (Smooth Riemannian manifolds are not very rigid.) But in principle you can compute the Taylor expansion as a function of $t$, if $\alpha$ is suitably differentiable. $\endgroup$ Commented May 3, 2022 at 17:41
  • $\begingroup$ Indeed, $f$ needs to be orthogonal to the mean curvature. I’ll correct that. $\endgroup$ Commented May 4, 2022 at 12:36

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The answer to the second question is still negative.

Let $M^3$ be the cylinder $\mathbb{S}^2 \times \mathbb{R}$. Then $\Sigma = \mathbb{S}^2\times \{0\}$ is totally geodesic in $M$, and so for any function $f$ the condition $\int f H_\Sigma = 0$ holds.

In this case the area of the $\varphi_t(\Sigma)$ can be explicitly computed (good calculus exercise). But what's important is that unless $f$ is constant, you have that the area of $\varphi_t(\Sigma)$ is strictly greater than that of $\Sigma$ for any value of $t$ such that $\alpha(t)\neq 0$.


More generally: if $\Sigma$ is minimal in $M$ (minimality here is just so I can use standard formulae for the second variation of area), you have that the second variation of area of $\Sigma$ looks something like

$$ \delta^2\text{Area} = \int_\Sigma |\nabla_\Sigma f|^2 + V f^2 ~dvol_\Sigma $$

Here the precise value of the function $V$ is unimportant: it is related to the ambient geometry of $M$ and the extrinsic geometry of $\Sigma$ in $M$. What's important is that the scalar function $V$ is smooth and bounded under our hypotheses, and so there exists functions $f$ for which the second variation is strictly positive, and again you have functions for which it is impossible to obtain constant area.

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  • $\begingroup$ The second variation in the present case depends on $\alpha$ too, right? $\endgroup$ Commented May 4, 2022 at 15:32
  • $\begingroup$ $\alpha$ just gives an overall scaling factor (it is constant independent of position). $\endgroup$ Commented May 4, 2022 at 16:57
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    $\begingroup$ @EduardoLonga incidentally: in your original question, $\alpha$ also plays almost no role, since it is constant along $\Sigma$. Changing $\alpha$ is just changing how you parametrize the level sets, and not their geometry at all. $\endgroup$ Commented May 4, 2022 at 17:00

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