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For fixed closed smooth manifolds $M^n$ and $N^{n+1}$, two $C^{k,\alpha}$ embeddings $f, f' : M \to N$ are said to be equivalent if there exists $\varphi \in \operatorname{Diff}(M)$ such that $f' = f \circ \varphi$. Let $\mathcal{M}$ be the set of pairs $([f], g)$, where $[f]$ is an equivalence class of CMC embeddings into the Riemannian manifold $(N,g)$ and $g$ belongs to a fixed open set $\Gamma$ of $C^q$ metrics. Brian White showed that $\mathcal{M}$ is a $C^{q-j}$ Banach manifold modelled on $\Gamma$.

Now let $\mathcal{A} : \mathcal{M} \to \mathbb{R}$ be the area functional. I am trying to find its critical points. To this end, let $([f],g)$ be given and consider a curve $([f_t], g_t)$ in $\mathcal{M}$ such that $([f_0],g_0) = ([f],g)$ and $d/dt ([f_t], g_t) = (X, h)$ at $t = 0$, where $X$ is a vector field along $f$ and $h$ is a symmetric 2-tensor. Then, by the usual variation formulae for the area and volume elements, we arrive at

$$\left. \frac{\mathrm{d}}{\mathrm{d}t} \right\vert_{t=0} \mathcal{A}([f_t], g_t) = - \int_M H_{f} g(X, \nu) \, \mathrm{d}A_{f^\ast g} + \int_M \operatorname{tr}_{f^\ast g}(f^\ast h) \, \mathrm{d}A_{f^\ast g},$$

where $H_f$ is the mean curvature of $f : M \to N$, $\nu$ is a unit normal for $M$ along $f$ and $\operatorname{tr}_{f^\ast g}(f^\ast h) $ is just the trace of $f^\ast h$ with respect to the metric $f^\ast g$. How do I conclude? What are the critical points of $\mathcal{A}$?

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    $\begingroup$ You can always shrink the metric $g$, e.g. taking an appropriate multiple of $h=-g$, so that the area decreases. Thus there are no critical points. $\endgroup$
    – Sebastian
    Commented Jul 23, 2021 at 7:37

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