Let $(M^3,g)$ be a Riemannian manifold and let $\varphi : \Sigma \to M$ be a two-sided embedding of a closed surface into $M$, with a unit normal $N$. Suppose that $\varphi$ is a regular point of the area functional (i.e. it is not a minimal embedding). Is it true that given any analytic function $f : \Sigma \to \mathbb{R}$ satisfying $\int_\Sigma f H_\Sigma \, \mathrm{d}A = 0$ there exists an analytic variation $\varphi_t : \Sigma \to M$, $t \in (-\varepsilon, \varepsilon)$, such that

• $\operatorname{Area}(\Sigma, \varphi_t^\ast g)$ is constant with respect to $t$;
• the corresponding variational vector field is equal to $fN$, i.e., $\frac{\mathrm{d}}{\mathrm{d}t} \big\vert_{t = 0} \varphi_t = fN$ ?

I suspect that the answer is affirmative in the smooth case: since $\varphi$ is a regular point of the area functional, the set of embeddings of $\Sigma$ into $M$ which are close to $\varphi$ and have the same area is locally modeled on an open set (containing the origin) of smooth functions $h$ on $\Sigma$ such that $\int_\Sigma h H_\Sigma \, \mathrm{d}A = 0$ (via the exponential map of $(M, g)$). Thus, for any $f$ as before, there should be a small curve $\varphi_t$ in the space of embeddings with the same area of $\varphi$ with tangent vector equal to $fN$. This would give the desired variation. Is this reasoning correct? What about the analytic case?

Let $(M^3,g)$ be a Riemannian manifold and let $\varphi : \Sigma \to M$ be a two-sided embedding of a closed surface into $M$, with a unit normal $N$. Suppose that $\varphi$ is a regular point of the area functional (i.e. it is not a minimal embedding). Is it true that given any analytic function $f : \Sigma \to \mathbb{R}$ satisfying $\int_\Sigma f H_\Sigma \, \mathrm{d}A = 0$ (here, $H_\Sigma$ is the mean curvature of $\varphi$) there exists an analytic variation $\varphi_t : \Sigma \to M$, $t \in (-\varepsilon, \varepsilon)$, such that

• $\operatorname{Area}(\Sigma, \varphi_t^\ast g)$ is constant with respect to $t$;
• the corresponding variational vector field is equal to $fN$, i.e., $\frac{\mathrm{d}}{\mathrm{d}t} \big\vert_{t = 0} \varphi_t = fN$ ?

I suspect that the answer is affirmative in the smooth case: since $\varphi$ is a regular point of the area functional, the set of embeddings of $\Sigma$ into $M$ which are close to $\varphi$ and have the same area is locally modeled on an open set (containing the origin) of smooth functions $h$ on $\Sigma$ such that $\int_\Sigma h H_\Sigma \, \mathrm{d}A = 0$ (via the exponential map of $(M, g)$). Thus, for any $f$ as before, there should be a small curve $\varphi_t$ in the space of embeddings with the same area of $\varphi$ with tangent vector equal to $fN$. This would give the desired variation. Is this reasoning correct? What about the analytic case?

Let $(M^3,g)$ be a Riemannian manifold and let $\varphi : \Sigma \to M$ be a two-sided embedding of a closed surface into $M$, with a unit normal $N$. Suppose that $\varphi$ is a regular point of the area functional (i.e. it is not a minimal embedding). Is it true that given any analytic function $f : \Sigma \to \mathbb{R}$ satisfying $\int_\Sigma f H_\Sigma \, \mathrm{d}A = 0$ there exists an analytic variation $\varphi_t : \Sigma \to M$, $t \in (-\varepsilon, \varepsilon)$, such that

• $\operatorname{Area}(\Sigma, \varphi_t^\ast g)$ is constant with respect to $t$;
• the corresponding variational vector field is equal to $fN$, i.e., $\frac{\mathrm{d}}{\mathrm{d}t} \big\vert_{t = 0} \varphi_t = fN$ ?

I suspect that the answer is affirmative in the smooth case: since $\varphi$ is a regular point of the area functional, the set of embeddings of $\Sigma$ into $M$ which are close to $\varphi$ and have the same area is locally modeled on an open set (containing the origin) of smooth functions $h$ on $\Sigma$ such that $\int_\Sigma h H_\Sigma \, \mathrm{d}A = 0$ (via the exponential map of $(M, g)$). Thus, for any $f$ as before, there should be a small curve $\varphi_t$ in the space of embeddings with the same area of $\varphi$ with tangent vector equal to $fN$. This would give the desired variation. Is this reasoning correct? What about the analytic case?

Let $(M^3,g)$ be a Riemannian manifold and let $\varphi : \Sigma \to M$ be a two-sided embedding of a closed surface into $M$, with a unit normal $N$. Suppose that $\varphi$ is a regular point of the area functional (i.e. it is not a minimal embedding). Is it true that given any analytic function $f : \Sigma \to \mathbb{R}$ satisfying $\int_\Sigma f H_\Sigma \, \mathrm{d}A = 0$ there exists an analytic variation $\varphi_t : \Sigma \to M$, $t \in (-\varepsilon, \varepsilon)$, such that

• $\operatorname{Area}(\Sigma, \varphi_t^\ast g)$ is constant with respect to $t$;
• the corresponding variational vector field is equal to $fN$, i.e., $\frac{\mathrm{d}}{\mathrm{d}t} \big\vert_{t = 0} \varphi_t = fN$ ?

I suspect that the answer is affirmative in the smooth case: since $\varphi$ is a regular point of the area functional, the set of embeddings of $\Sigma$ into $M$ which are close to $\varphi$ and have the same area is locally modeled on an open set (containing the origin) of smooth functions $h$ on $\Sigma$ such that $\int_\Sigma h H_\Sigma \, \mathrm{d}A = 0$ (via the exponential map of $(M, g)$). Thus, for any $f$ as before, there should be a small curve $\varphi_t$ in the space of embeddings with the same area of $\varphi$ with tangent vector equal to $fN$. This would give the desired variation. Is this reasoning correct? What about the analytic case?