几何分析中的计算

Geometric Measure Theory

Derivatives of functions and vector fields on hypersufaces can also be defined in terms of projections from $\mathbb{R}^{n+1}$ onto the tangent space of $M$. This is the framework used in geometric measure theory where coordinate systems are not available. The notions defined below can be adapted to countable $n-$ rectifiable subsets of $\mathbb{R}^{n+1}$.
$$\nabla^Mf(x)=Df(x)-(Df,\nu(x))\nu(x)$$
for $x\in M$ where $Df$ denotes the usual gradient of $f$ in $\mathbb{R}^{n+1}$ (or of its extension into $\mathbb{R}^{n+1}$).

This can also be written as
\[\nabla^Mf(x)=\sum_{i=1}^n\tau_iD_{\tau_i}f(x)\]
where $\tau_1,...,\tau_n$ form a orthogonal basis of $T_xM$.

The Laplace-Beltrami operator on $M$ of a twice differentiable function $f$ is defined by

$$\Delta_Mf=div_M(grad^M\ f),$$
$$\vec{H}=-H\vec{\nu}=-div_M(\nu)\nu,$$
$$\Delta_Mf=\Delta_{R^{n+1}}f-D^2f(\vec{\nu},\vec{\nu})-div_M(\vec{\nu})\vec{\nu}\cdot Df$$
$$=\Delta_{R^{n+1}}f-D^2f(\vec{\nu},\vec{\nu})+\vec{H}\cdot Df.$$
It follows that
$$\Delta_M x_i=\overrightarrow{H}\cdot \mathbf{e_i},$$
$$\Delta_M\mathbf{X}=\overrightarrow{H}.$$

It is also easy to check that ($X, \mathbf{e_i}$ are a column vectors)

$$\delta_if=D_if-(Df,\nu)\nu_i,\ i=1,...,n+1,$$
$$\nabla^Mf=\delta f=\delta f=(I_{n+1}-\nu\otimes\nu)Df,\ \mathbf{Orthogonal\ Projection},$$
$$D_{\tau}X=DX\tau=D_jX^i\tau_j\mathbf{e_i},\ \tau=\sum^n_{j=1}\tau_j\mathbf{e_j}\in T_pM,$$
$$div_M(X)=\sum_{i=1}^n\delta_i(X^i)=\sum_{i=1}^n\tau_i\cdot D_{\tau_i}X=Trace((I_{n+1}-\nu\otimes\nu)DX)$$
$$=div_{\mathbb{R}^{n+1}}DX-\nu^T DX\nu,\ \ where \ X=\sum_{i=1}^{n+1}X^i\mathbf{e}_i$$

$$=\nabla^{M}_i(X^i) DX\nu$$

$$H=\sum_{i=1}^n\delta_i(\nu^i),\ where\ \ \nu=\sum_{i=1}^{n+1}\nu^i\mathbf{e}_i,$$
$$\Delta_Mf=div_M(grad^M\ f)=\sum_{i=1}^n\delta_i(\delta_if),$$
$$=div_{M}(Df)+\overrightarrow{H}\cdot Df,$$
$$=\Delta_{\mathbb{R}^{n+1}}f-D^2f(\vec{\nu},\vec{\nu})+\overrightarrow{H}\cdot Df$$

It is also clear that

$$\Delta_{M} f=div_{M}(Df-<\nu,Df>\nu)$$
$$=div_{M}(Df)-<\nu,Df>\delta_i\nu^i$$
$$=div_{M}(Df)+\vec{H}\cdot Df.$$

Naturally, we can imagine that

$$\nabla^M_{\tau}X=DX\tau-<DX\tau,\nu>\nu,$$

For any $\xi,\eta\in T_pM$, let
$$B(\xi,\eta)=-\langle D_{\xi}\nu,\eta\rangle$$
$$=\xi^iD_{e_i}\nu^j\eta_j$$
$$=\xi^i(\delta_i\nu^j+\nu^i\nu^hD_hv^j)\eta_j$$

$$=\xi^i\delta_i\nu^j\eta_j$$

By using the sign distance function with $\nu=Dd$ (See E.Giusti, Chapter 8) or level set method with $\nu=\frac{DF}{|DF|}$,
it is clear that
$$D_i\nu^j=D_j\nu^i,  \delta_i\nu^j=\delta_j\nu^i,$$
then $B(\xi,\eta)$ is symmetric with respect to $\xi,\eta\in T_pM$.

By the GMT definition, we can also define the corresponding operators for codimension bigger than one case.

In the mean curvature flow case, if $h(p,t)=f(x,t)$ for $x=F(p,t)$. The partial derivative of $h$ with respect to time, equals the total time derivative of $f$ along ($M_t$), that is,
$$\frac{\partial h}{\partial t}(p,t)=\frac{d f}{d t}(x,t)=\frac{d f}{d t}(F(p,t),t)$$

$$=\frac{\partial f}{\partial t}(x,t)+Df(x,t)\cdot \frac{\partial F}{\partial t}(p,t)$$

for $x=F(p,t)$ or in short
$$\frac{\partial h}{\partial t}=\frac{d f}{d t}=\frac{\partial f}{\partial t}+Df \cdot \vec{H}.$$

Notice that

$$\Delta_{M_t}f=\Delta_{R^{n+1}}f-D^2f(\vec{\nu},\vec{\nu})+\vec{H}\cdot Df$$


It follows that

$$(\frac{d }{dt}-\Delta_{M_t})f=\frac{\partial f }{\partial t}-div_{M_t}Df$$
$$=(\frac{\partial}{\partial t}-\Delta_{R^{n+1}})f+D^2f(\vec{\nu},\vec{\nu})$$

for the difference between the ambient heat operator and the heat operator on $M_t$.

Recall that

$$(\frac{\partial}{\partial t}-\Delta_{M_t})h=(\frac{d }{dt}-\Delta_{M_t})f.$$

As a example, consider
\[f(x,t)=|x-x_0|^2+2n(t-t_0).\]
Since
\[\frac{\partial f}{\partial t}(x,t)=2n,\]
\[Df(x,t)=2(x-x_0)\]
and
\[div_{M_t}(x-x_0)=n,\]
we obtain
$$(\frac{d }{dt}-\Delta_{M_t})(|x-x_0|^2+2n(t-t_0))=0.$$

As for the Monotonicity formula for functions on minimal surface $M$, we need the following simple fact:
$$\Delta_{M}(fg)=\Delta_{M}(f)g+f\Delta_{M}(g)+2\langle\nabla_M f,\nabla_M g \rangle$$

$$\Delta_{M} X=0,$$

$$\Delta_{M} |X|^2=2\langle\nabla_M X,\nabla_M X \rangle=2n.$$

Second fundamental form for level set

Let $\nu=\frac{DF}{|DF|}$, note that $grad F=DF$ in $R^{n+1}$. It follows that

$$II_{\nu}(X,Y)=<\nu,D_XY>=\frac{1}{|DF|}<grad F,D_XY>$$
$$=\frac{1}{|DF|}\nabla_XYF$$
$$=\frac{1}{|DF|}\{-YXF+\nabla_XYF\}$$
$$=-\frac{1}{|DF|}D^2F(X,Y),$$

Hence,
$$II_\nu=-\frac{D^2F}{|DF|}.$$

Gradient, Divergence, Laplace , Ricci Identity

Let $G=det(g_{ij})$. We have
$$\frac{1}{\sqrt{G}}\cdot\frac{\partial \sqrt{G}}{\partial x_k}=\Gamma_{ki}^i.$$

By the definition of Christoffol symbols, we have
$$\Gamma_{ki}^i=\frac{1}{2}g^{il}\{\frac{\partial g_{il}}{\partial x_k}+\frac{\partial g_{kl}}{\partial x_i}-\frac{\partial g_{ik}}{\partial x_l}\}=\frac{1}{2}g^{il}\frac{\partial g_{il}}{\partial x_k}.$$

By using the differentiate w.r.t. determinant, we have
$$\frac{1}{\sqrt{G}}\cdot\frac{\partial \sqrt{G}}{\partial x_k}=\frac{1}{2G}\cdot\frac{\partial G}{\partial x_k}=\frac{g^{*}_{ij}}{2G}\cdot\frac{\partial g_{ij}}{\partial x_k}=\frac{g^{ij}}{2}\cdot\frac{\partial g_{ij}}{\partial x_k}.$$
This finish the proof.

Let $X\in\chi(M)$, we define
$$div(X):=C^1_1(DX).$$

In local coordinate, we have $X=X^i\frac{\partial}{\partial x_i}$. Then
$$DX=X^i_{,j}\frac{\partial}{\partial x_i}\otimes dx^j,$$
and
$$div(X)=\sum\limits_{i=1}^n X^i_{,i},$$

Notice that

$$\quad DX(dx^i,\partial_j)$$
$$=D_{\partial_j}{X}(dx^i)=\partial_j(X(dx^i))-X(D_{\partial_j}dx^i)$$
$$=\frac{\partial X_i}{\partial x_j}+\Gamma_{jk}^iX^k$$

Therefore, we have
\[div(X)=\frac{\partial X_i}{\partial x_i}+\Gamma_{ik}^iX^k\]

By using the previous argument, we have
\[div(X)=\frac{\partial X_i}{\partial x_i}+\frac{1}{\sqrt{G}}\frac{\partial\sqrt{G}}{\partial x_k}X^k=\frac{1}{\sqrt{G}}\frac{\partial}{\partial x_i}(\sqrt{G}X^i).\]

With the help of Riemmanian connection $g$ and $df$, we can define $grad f$, i.e., for any $X\in\chi(M)$,
\[g(grad f, X)=df(X)=X(f).\]
Obviously, $grad f$ is a tangent vector field of $M$.
In local coordinate, we have $df=\frac{\partial f}{\partial x_i}dx^i$. If $grad f= a^i\frac{\partial}{\partial x_i}$, we have
\[a^i\langle \frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\rangle=df(\frac{\partial}{\partial x_j})=\frac{\partial}{\partial x_j}(f).\]

Then we have
\[a^i g_{ij}=\frac{\partial f}{\partial x_j}.\]
It follows that
\[a^i =\frac{\partial f}{\partial x_j} g^{ji}.\]
Thus,
\[grad~f=\frac{\partial f}{\partial x_j}g^{ij}\frac{\partial}{\partial x_i}.\]

Now we can define
\[\Delta f:=div(grad~f).\]

In local coordinate, we have (linear elliptic operator)
\[\Delta f=div(grad~f)=\frac{1}{\sqrt{G}}\frac{\partial}{\partial x_i}(\sqrt{G}g^{ij}\frac{\partial f}{\partial x_j}).\]

We can also define $Hess(f)$, i.e.,
\[Hess(f)=D(df).\]

In local coordinate, we have
$$Hess(f)(X,Y)=D(df)(X,Y)$$
$$=D_Y(df)(X)=Y(df(X))-(df)(D_YX)$$
$$=Y(X(f))-(D_YX)f$$
$$=X(Y(f))-(D_XY)f$$
$$=Hess(f)(Y,X).$$

Hessian of f is a symmetric (0,2) type tensor.

Let $S$ be a (0,2) symmetric covariant tensor. We can define the trace of $S$ as
\[TrS=\sum\limits_{i=1}^nS(e_i,e_i),\]
where ${e_i}$ are standard orthogonal basis.

Let $e_i=b_i^j\frac{\partial}{\partial x_j}$. Then
\[\delta_{kl}=\langle e_k,e_l\rangle=b_k^ib_l^j\langle \frac{\partial}{\partial x_k},\frac{\partial}{\partial x_l}\rangle=b_k^ib_l^jg_{ij}.\]

Let $C=(b^j_i)$. Then we have $CgC^T=I_n$. It is easy to see that $C^T$ is the inverse matrix of $Cg$. Hence $C^TCg=I_n$ and $g^{-1}=C^TC$. This is exactly the identity \[g^{ij}=b_k^ib^j_k.\]
This leads to
\[TrS=S(e_k,e_k)=S(b_{k}^i\frac{\partial}{\partial x_i},b_{k}^j\frac{\partial}{\partial x_j})=b_{k}^ib_{k}^jS(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j})=g^{ij}S(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}).\]

Recall P119 in Jiaqiang Mei, we can also define\[dix(x):=\langle D_{e_i}X, e_i\rangle,\]
where ${e_i}$ is a normal orthogonal basis.

In the following, we can check that the two definitions of $\Delta$ in previous paragraph are coincident.

Firstly, we have

$$\Delta f=g^{ij}Hess(f)(\partial x_i,\partial x_j)$$
$$=g^{ij}\frac{\partial}{\partial x_j}(\frac{\partial f}{\partial x_i})-g^{ij}\Gamma_{ij}^k\frac{\partial f}{\partial x_k}$$

 

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