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허곡신거 [902596] · MS 2019 (수정됨) · 쪽지

2024-11-24 14:54:46
조회수 1,372
0

Mean Curvature & Minimal Surface

게시글 주소: https://h.orbi.kr/00070119579

\noindent\textbf{Theorem.} Let $M$ be a compact codimension 1 submanifold with nonempty boundary in an $(n+1)$-dimensional Riemannian manifold $(\tilde{M},\tilde{g})$. \textcolor{blue}{If $M$ is area-minimizing, then its mean curvature is identically zero.}

\bigbreak

\noindent\textit{Proof.} Let $\varphi\in C^\infty_c(U)$. For small $t$, define $M_t\subset\tilde{M}$ be the variation on $U$ via $\varphi$, i.e.

$$M_t = (M\setminus U)\cup\{z\in\tilde{U}:v(z) = t\varphi(x^1(z),\ldots,x^n(z))\},$$

where $v$ is the normal direction coordinate function of $M$. Let $\hat{g}_t = \iota^\ast_{M_t}\tilde{g}$ on $M_t$. Define a variation map $f_t:U\to\tilde{U}$ given in Fermi coordinates by

$$f_t(x^1,\ldots,x^n) = (x^1,\ldots,x^n,t\varphi(x)),$$

which naturally extends to a variation map $F_t:M\to M_t$ by

$$F_t(z) = \begin{cases}

z, & z\in M\setminus\mathrm{supp}(\varphi),\\

f_t(z), & z\in U.

\end{cases}$$

Let $g_t = F_t^\ast\hat{g}_t = F_t^\ast\tilde{g}$ on $M$. Note that when $t = 0$, then $M_0 = M$ and both $g_0$ and $\hat{g}_0$ are equal to the induced metric $g = \iota^\ast_M\tilde{g}$ on $M$. Note that the local expression of $g_t$ in Fermi coordinate is given as follows where we use $\tilde{g} = \tilde{g}_{ij}d\tilde{x}^id\tilde{x}^j$:

\begin{align*}

F^\ast_t\tilde{g} & = F^\ast_t(\tilde{g}_{ij}d\tilde{x}^id\tilde{x}^j)\\

& = \tilde{g}_{ij}\circ F_t\ d(\tilde{x}^i\circ F_t)d(\tilde{x}^j\circ F_t)\\

& = \tilde{g}_{ij}(x,t\varphi(x)) {\partial F^i_t\over\partial x^k}{\partial F^j_t\over\partial x^l}dx^kdx^l\\

& = \left(\tilde{g}_{ij}(x,t\varphi(x)) {\partial F^i_t\over\partial x^i}{\partial F^j_t\over\partial x^j} + {\partial F^{n+1}_t\over\partial x^i}{\partial F^{n+1}_t\over\partial x^j}\right)dx^idx^j\\

& = \left(\tilde{g}_{ij}(x,t\varphi(x)) + t^2{\partial \varphi\over\partial x^i}(x){\partial \varphi\over\partial x^j}(x)\right)dx^idx^j\\

& = (g_t)_{ij}dx^idx^j

\end{align*}

while on $M\setminus U$, $g_t$ is equal to $g$ and thus is independent of $t$. Note that

$$\mathrm{Area}(M_t,\hat{g}_t) = \mathrm{Area}(M,g_t) = \mathrm{Area}(M\sm U,g)+\mathrm{Area}(U,g_t).$$

Also, we have

$$\mathrm{Area}(U,g_t) = \int_U\sqrt{g_t}dx^1\wedge\cdots\wedge dx^n.$$

Therefore, we have

\begin{align*}

\left.{d\over dt}\right|_{t = 0}\mathrm{Area}(M_t,\hat{g}_t) & = \int_U\left.{d\over dt}\right|_{t = 0}\sqrt{g_t}dx^1\wedge\cdots\wedge dx^n\\

& = \int_U {1\over 2\sqrt{g}}\det(g)\tr\left(g^{-1}\left.{d g_t\over dt}\right|_{t = 0}\right) dx^1\wedge\cdots\wedge dx^n\\

& = \int_U {1\over 2\sqrt{g}}\det(g)\tr\left(g^{ik}\left.{\partial (g_t)_{ij}\over\partial t}\right|_{t = 0}\right) dx^1\wedge\cdots\wedge dx^n\\

& = \int_U {1\over 2\sqrt{g}}\det(g)\tr\left(g^{ik}{\partial g_{ij}\over\partial v}\varphi\right) dx^1\wedge\cdots\wedge dx^n\\

& = \int_U {1\over 2\sqrt{g}}\det(g)g^{ij}{\partial g_{ij}\over\partial v}\varphi dx^1\wedge\cdots\wedge dx^n\\

& = \int_U {1\over 2}\sqrt{g}g^{ij}{\partial g_{ij}\over\partial v}\varphi dx^1\wedge\cdots\wedge dx^n.

\end{align*}

We note that in Fermi coordinate, the normal component of $\tilde{\na}_{\partial_i}\partial_j$ is $\tilde{\Gamma}^{n+1}_{ij}\partial_v$. Using the formula

$$\tilde{\Gamma}^k_{ij} = {1\over 2}g^{kl}(\partial_ig_{jl}+\partial_jg_{il}-\partial_lg_{ij}),$$

we get $\tilde{\Gamma}^{n+1}_{ij} = -{1\over 2}\partial_v g_{ij}$. Using this, we compute that

$$II_{ij} = \la\tilde{\na}_{\partial_i}\partial_j,N\rangle_{\tilde{g}} = \left\la-{1\over 2}\partial_vg_{ij}\partial_v,\partial_v\right\rangle_{\tilde{g}} = -{1\over 2}\partial_vg_{ij}.$$

Since the shape operator is obtained by raising index of the second fundamental form (by definition), the local expression of the shape operator is given by

$$(B)^i_j = -{1\over 2}g^{ik}\partial_vg_{ij}.$$

Taking trace gives a local expression of the mean curvature

$$H = -{1\over 2n}g^{ij}\partial_vg_{ij}.$$

Therefore, we get the first variation formula of area functional:

$$\left.{d\over dt}\right|_{t = 0}\mathrm{Area}(M_t,\hat{g}_t) = -n\int_UH\varphi dV_g.$$

If $M$ is a minimal surface, $\mathrm{Area}$ attains its minimum at $t = 0$ so that $\int_UH\varphi dV_g\equiv 0$ for every $\varphi\in C^\infty_c(U)$ which implies $H\equiv 0$. $\square$