Machine-translated from Chinese.
第二基本形式
设$\widetilde{M}$为$\widetilde{n}$维黎曼流形, $M$为$n$维子流形,
$n<\widetilde{n}.$
记$\mathscr{N}(M)$为$\widetilde{M}$中$M$的法向量场全体,
$\mathscr{T}(M)$为切向量场全体. 为了考察子流形的外观,
我们取法向量场$N\in \mathscr{N}(M),$ 考虑$\widetilde{D}_XN.$
这等价于考察所有$\left<{}\widetilde D_XN,Y\right>,$
$Y\in \mathscr{T}(M).$ 这就定义了第二基本形式.
对给定$\nu\in \mathscr{N}(M),$ 定义:
$$
\Pi_\nu(X,Y)=\left<{}\widetilde{D}_X\nu,Y\right>,\quad X,Y\in \mathscr{T}(M),
$$
称为$M$对应于$\nu$的第二基本形式.
引理 1. 对每个固定的$\nu,$ $\Pi_\nu$是$M$上的二次对称协变张量场.
证: 由于$\left<{}\nu,Y\right>\equiv 0,$ $Y\in \mathscr{T}(M),$
对于任意$X,Y\in \mathscr{T}(M),$ 我们有:
$$
0=X\left<{}\nu,Y\right>=\left<{}\widetilde{D}_X\nu,Y\right>+\left<{}\nu,\widetilde{D}_XY\right>,
$$
因此我们也有Weingarten方程:
$$
\Pi_\nu(X,Y)=-\left<{}\nu,\widetilde{D}_XY\right>.
$$
接下来,
由于$[X,Y]\in \mathscr{T}(M),$
$$
\left<{}\nu,\widetilde{D}_XY\right>=\left<{}\nu,\widetilde{D}_YX\right>+\left<{}\nu,[X,Y]\right>=\left<{}\nu,\widetilde{D}_YX\right>,
$$
那么由Weingarten方程, $\Pi_\nu(X,Y)=\Pi_\nu(Y,X).$ 另外, 容易看出
$$
\Pi_\nu(fX,Y)=f\Pi_\nu(X,Y)=\Pi_\nu(X,fY),
$$
因此$\Pi_\nu$的确是$M$上的张量场. 引理得证.
可以看出, $\Pi_\nu(X,Y)(x)$也只依赖于$\nu(x),$
且关于$\nu$是$C^\infty$线性的.
记$M_x^\perp$是$M_x$在$\widetilde{M}_x$中的正交补, 可以定义二次型
$$
S:M_x\otimes M_x\rightarrow M_x^\perp,
$$
使得
$$
\left<{}S(v,w),\nu\right>:=\Pi_\nu(v,w),\quad \,\forall\,\nu\in \mathscr{N}(M).
$$
容易看出$S(v,w)$就是$-\widetilde{D}_XY(x)$在$M_x^\perp$上的投影,
$X(x)=v,Y(x)=w,$ 取值与向量场选取无关.
$S$的迹$\sum_{i=1}^nS(e_i,e_i)\in M_x^\perp,$ 记为$\eta(x),$
$\{e_i\}\subset M_x$为标准正交基. 那么$\eta\in \mathscr{N}(M),$
称为$M$的平均曲率法向量场, $|\eta|$称为$M$的平均曲率.
有时也定义对称线性变换$A_\nu:M_x\rightarrow M_x,$ 使得
$$
\left<{}A_\nu(v),w\right>=\Pi_\nu(v,w)=\left<{}v,A_\nu(w)\right>,
$$
即$A_\nu(v)$为$\widetilde{D}_v\nu$在$M_x$上的投影,
我们也称$A_\nu$为对应于$\nu$的第二基本形式. 它与平均曲率的关系是:
$$
\left<{}\eta(x),\nu\right>=\sum_{i=1}^n\left<{}S(e_i,e_i),\nu\right>=\sum_{i=1}^n\Pi_\nu(e_i,e_i)=\sum_{i=1}^n\left<{}A_\nu(e_i),e_i\right>=\operatorname{tr}A_\nu.
$$
当$\widetilde{n}=n+1$时, $M$是$\widetilde{M}$中的超曲面,
此时局部可固定单位法向量场$\nu,$ 记$A:=A_\nu.$
此时$\widetilde{D}_v\nu\in M_x,$ $A(v)=\widetilde{D}_v\nu.$
称$A$的特征值为主曲率, 由对称性它们全是实的, 记为$p_1,\cdots,p_n.$
$\det A$称为$M$的Gauss曲率. $n=2$时, 它与截面曲率一致,
这就是Gauss的Egregium(绝妙)定理.
设$\widetilde{M}=\mathbb{R}^{n+1},$ 若超曲面$M$是可定向的,
那么整体上有单位法向量场$\nu,$ 可定义$\Gamma:M\rightarrow S^n,$
$x\mapsto \nu(x),$ 称为Gauss映射. 那么,
$$
d\Gamma(X)=\widetilde{D}_X\nu=A(X),\quad \frac{\Gamma^\ast \Omega_0}{\Omega}=\det d\Gamma=\det A,
$$
其中$\Omega_0,\Omega$分别是$S^n,M$的体积元,
由此我们可以看出Gauss曲率$\det A$的几何意义.
若假定$M$具正Gauss曲率, 那么Gauss映射$\Gamma$就是一个浸入.
取$S^n$上标准黎曼度量$g_0,$ 可赋予$M$新的黎曼度量$\Gamma^\ast g_0.$
当$M$紧时, $\Gamma^\ast g_0$是完备的. 由度量选取, $\Gamma$是局部等距,
因此是覆盖映射. 又由于$S^n$是单连通的, $\Gamma$是一个微分同胚.
从而我们有引理:
引理 2. 设$M$是$\mathbb{R}^{n+1}$中紧致超曲面, 具正Gauss曲率, 那么Gauss映射$\Gamma:M\rightarrow S^n$为微分同胚.
推论 3. $\mathbb{R}^{n+1}$中紧致的正曲率超曲面必是凸曲面, 即它总落在其切平面的一侧.
现定义两个正交投影:
$$
\top:\widetilde{M}_x\rightarrow M_x,\quad \perp:\widetilde{M}_x\rightarrow M_x^\perp,
$$
容易验证$\top(\widetilde{D}_XY)$满足$M$上Levi-Civita联络的性质,
因此$D_XY=\top(\widetilde{D}_XY).$
回忆前面我们说明了$S(X,Y)$是$-\widetilde{D}_XY(x)$在$M_x^\perp$上的投影,
因此我们有:
$$
S(X,Y)=\perp(-\widetilde{D}_XY)=D_XY-\widetilde{D}_XY.
$$
这恰是前面讨论全测地子流形时引入的定义.
现在我们可以说$M$是全测地的当且仅当它的第二基本形式恒为零.
因此第二基本形式刻画了与全测地的差异.
Gauss方程
回忆全测地子流形任意$2$维切平面的截面曲率与原流形相同,
对于一般情形我们有如下方程:
命题 4 (Gauss方程). $\,\forall\,X,Y\in \mathscr{T}(M),$ 我们有:
$$
\left<{}\widetilde{R}_{XY}X,Y\right>=\left<{}R_{XY}X,Y\right>-\left<{}S(X,X),S(Y,Y)\right>+|S(X,Y)|^2.
$$
证: 证明是直接的计算:
$$
\left<{}\widetilde{R}_{XY}X,Y\right>=\left<{}\widetilde{D}_{[X,Y]}X,Y\right>-\left<{}\widetilde{D}_X\widetilde{D}_YX,Y\right>+\left<{}\widetilde{D}_Y\widetilde{D}_XX,Y\right>,
$$
其中, 由于$S$的像落在$M_x^\perp$中,
$$
\left<{}\widetilde{D}_{[X,Y]}X,Y\right>=\left<{}D_{[X,Y]}X,Y\right>-\left<{}S([X,Y],X),Y\right>=\left<{}D_{[X,Y]}X,Y\right>,
$$
$$
\begin{aligned}
\left<{}\widetilde{D}_X\widetilde{D}_YX,Y\right>&=\left<{}\widetilde{D}_X(D_YX-S(X,Y)),Y\right>\\
&=\left<{}\widetilde{D}_XD_YX,Y\right>-\left<{}\widetilde{D}_XS(X,Y),Y\right>\\
&=\left<{}D_XD_YX,Y\right>-\left<{}S(X,D_YX),Y\right>-X\left<{}S(X,Y),Y\right>+\left<{}S(X,Y),\widetilde{D}_XY\right>\\
&=\left<{}D_XD_YX,Y\right>+\left<{}S(X,Y),D_XY-S(X,Y)\right>\\
&=\left<{}D_XD_YX,Y\right>-\left<{}S(X,Y),S(X,Y)\right>
\end{aligned}
$$
类似地,
$$
\left<{}\widetilde{D}_Y\widetilde{D}_XX,Y\right>=\left<{}D_YD_XX,Y\right>-\left<{}S(X,X),S(Y,Y)\right>.
$$
整合上述等式即可得到结论.
注: Gauss方程刻画了$\top\widetilde{R}_{XY}Z,$
而$\perp \widetilde{R}_{XY}Z$将由后面的Codazzi方程给出. 事实上,
利用Gauss方程证明中同样的方法, 可以证明如下等式成立:
$$
\left<{}\widetilde{R}_{XY}Z,W\right>=\left<{}R_{XY}Z,W\right>-\left<{}S(X,Z),S(Y,W)\right>+\left<{}S(X,W),S(Y,Z)\right>.
$$
推论 5. 设$M$是$\widetilde{M}$中的超曲面, $A$是关于单位法向量场$\nu$的第二基本形式. 若$P$是$M_x$中的$2$-平面, 令$A|_P:P\rightarrow P$为$A:P\rightarrow M_x$后再复合正交投影$M_x\rightarrow P,$ 则
$$
\widetilde{K}(P)=K(P)-\det(A|_P).
$$
证: $M$为超曲面时,
由于$\left<{}S(X,Y),\nu\right>=\Pi_\nu(X,Y)=\left<{}A(X),Y\right>,$
$S(X,Y)=\left<{}A(X),Y\right>\nu.$ 因此取$\{e_1,e_2\}$为$P$的标准正交基,
则
$$
\left<{}S(e_1,e_1),S(e_2,e_2)\right>-|S(e_1,e_2)|^2=\left<{}A(e_1),e_1\right>\left<{}A(e_2),e_2\right>-\left<{}A(e_1),e_2\right>^2=\det(A|_P).
$$
由于$\widetilde{K}(P)=\left<{}\widetilde{R}_{e_1e_2}e_1,e_2\right>,$
$K(P)=\left<{}R_{e_1e_2}e_1,e_2\right>,$ 由Gauss方程立即得到结论.
推论 6 (Gauss绝妙定理). 设$M$是$\mathbb{R}^3$中的曲面, 那么它的截面曲率等于它的Gauss曲率.
推论 7. 设$\{\nu_1,\cdots,\nu_{\widetilde{n}-n}\}$为$M_x^\perp$的幺正基, 那么$\,\forall\,X,Y\in M_x,$
$$
\left<{}\widetilde{R}_{XY}X,Y\right>=\left<{}R_{XY}X,Y\right>-\sum_i \Pi_{\nu_i}(X,X)\Pi_{\nu_i}(Y,Y)+\sum_i \Pi_{\nu_i}(X,Y).
$$
推论 8 (Synge引理). 设$M$包含一条$\widetilde{M}$的测地线$\gamma,$ 则对任意切平面$P$满足$\dot\gamma\in P\subset M_x,$ 我们有$K(P)\le \widetilde{K}(P).$
证: 设$P$由$\dot\gamma,Y$张成. 由于$\gamma$是测地线,
$S(\dot\gamma,\dot\gamma)=0.$ 那么令Gauss方程中$X=\dot\gamma,$
立即得到结论.
推论 9. $\mathbb{R}^3$中直纹面(单参数直线族)具非正Gauss曲率.
推论 10. 设$M$是$\widetilde{M}$中$2$维曲面, 包含一条$\widetilde{M}$的测地线$\gamma,$ 则$K(M_{\gamma(t)})\equiv \widetilde{K}(M_{\gamma(t)})$当且仅当$M_{\gamma(t)}$在$\widetilde{M}$中沿$\gamma$平行.
证: 设$\gamma$是正规测地线, $X$为$M$中单位向量场, 与$\gamma$垂直, 则
$$
\widetilde{K}(M_{\gamma(t)})=K(M_{\gamma(t)})+|S(\dot\gamma(t),X(t))|^2.
$$
因此等式恒成立当且仅当$\widetilde{D}_{\dot\gamma}X\equiv D_{\dot\gamma}X.$
$\dot\gamma$在$M,\widetilde{M}$中沿$\gamma$当然都是平行的, 而
$$
\left<{}D_{\dot\gamma}X,\dot\gamma\right>=\dot\gamma\left<{}X,\dot\gamma\right>-\left<{}X,D_{\dot\gamma}\dot\gamma\right>=0, \quad \left<{}D_{\dot\gamma}X,X\right>=0\:\Rightarrow\: D_{\dot\gamma}X=0,
$$
$X$在$M$中沿$\gamma$是平行的.
因此等式恒成立又当且仅当$X$在$\widetilde{M}$中沿$\gamma$也是平行的,
即$M_{\gamma(t)}$在$\widetilde{M}$中沿$\gamma$平行.
推论 11. 设$M$为黎曼流形, $x\in M.$ 取$\{x^i\}$为$x$处法坐标系, $P\subset M_x$由$\{\frac{\partial {} }{\partial {}x^1},\frac{\partial {} }{\partial {}x^2}\}$张成, $\mathscr{S}$为曲面$\{x^i=0:i\ge 3\}.$ 则$P$的截面曲率等于$\mathscr{S}$在诱导度量下的Gauss曲率.
证: 只需注意到对于法坐标系, $O$点处$S\equiv 0,$ 第二基本形式为零,
那么由Gauss方程即得.
Codazzi方程
Codazzi方程又称Codazzi-Mainardi方程,
直接进行计算可得到如下关于$\widetilde{R}_{XY}Z$的结果.
命题 12 (Codazzi方程). 若$X,Y,Z\in \mathscr{T}(M),$ $R,\widetilde{R}$为$M,\widetilde{M}$的曲率张量, 则
$$
\begin{aligned}
\perp\widetilde{R}_{XY}Z=&\left(\perp \widetilde{D}_XS(Y,Z)-S(D_XY,Z)-S(Y,D_XZ)\right)\\
&-\left(\perp \widetilde{D}_YS(X,Z)-S(D_YX,Z)-S(X,D_YZ)\right)
\end{aligned}
$$
往往把Codazzi方程与Gauss方程合称为Gauss-Codazzi方程.
Codazzi方程一般来看较为复杂, 但在特殊情形时较为简洁.
命题 13. 设$\widetilde M$为常曲率空间, $M$为其中的超曲面, 那么$(D_X\Pi)(Y,Z)$关于$X,Y,Z$是对称的, 其中$X,Y,Z\in \mathscr{T}(M).$
证: 由于$\Pi(Y,Z)=\Pi(Z,Y),$ 显然关于$Y,Z$是对称的. 因此只需证
$$
(D_X\Pi)(Y,Z)=(D_Y\Pi)(X,Z).
$$
取单位法向量场$\nu,$ 那么$S(X,Z)=\Pi(X,Z)\nu.$ 因此
$$
\perp\widetilde{D}_YS(X,Z)=\perp\left((Y\Pi(X,Z))\nu+\Pi(X,Z)\widetilde{D}_Y\nu\right)=(Y\Pi(X,Z))\nu.
$$
当$\widetilde{M}$具常截面曲率$c$时,
$$
\widetilde{R}_{XY}Z=c(\left<{}X,Z\right>Y-\left<{}Y,Z\right>X).
$$
这是因为右式构成了$\widetilde{M}$上的曲率张量, 且同样具常截面曲率$c.$
此时便可发现$\perp\widetilde{R}_{XY}Z=0.$
代入Codazzi方程, 即有:
$$
Y\Pi(X,Z)-\Pi(D_YX,Z)-\Pi(X,D_YZ)=(D_Y\Pi)(X,Z)=(D_X\Pi)(Y,Z)=\cdots.
$$
Gauss-Codazzi方程的一个基本应用是说明给定第二基本形式的超曲面的存在唯一性.
证明核心是考虑由下式决定的过定一阶偏微分方程组:
$$
S(Y_i,Y_j)=\Pi(Y_i,Y_j)\nu=D_{Y_i}Y_j-\widetilde{D}_{Y_i}Y_j,\quad \widetilde{D}_{Y_i}\nu=A(Y_i).
$$
由于方程是过定的, 不见得有解.
事实上可解的充要条件是满足前面的Gauss-Codazzi方程,
这是由考虑Frobenius定理得到的. 下面我们仅对存在唯一性做定理叙述.
定理 14 (唯一性). 设$M,M'$为$\mathbb{R}^{n+1}$中的两个超曲面, 具单位法向量场$\nu,\nu'.$ 设$\varphi:M\rightarrow M'$为等距同构, 使得$\Pi(X,Y)=\Pi'(d\varphi(X),d\varphi(Y)),$ 则$M,M'$是合同的, 即存在等距同构$\Phi:\mathbb{R}^{n+1}\rightarrow \mathbb{R}^{n+1},$ 使得$\Phi_M=\varphi.$
定理 15 (存在性). 给定$n$维黎曼流形和其上的对称张量$\Pi,$ 满足Gauss-Codazzi方程, 那么$M$可局部等距地嵌入到$\mathbb{R}^{n+1}$中, 使得它相对于某一单位法向量场的第二基本形式恰是$\Pi.$ 此外这个嵌入在相差整体等距同构$\overline{\varphi}:\mathbb{R}^{n+1}\rightarrow \mathbb{R}^{n+1}$的意义下是唯一的.
我们可以将定理推广到$n+1$维黎曼流形$\widetilde{M},$
但Gauss-Codazzi方程的形式将更为复杂. 此外, 也可以推广到余维数更高的情形,
但此时要增添额外的可积条件Ricci方程,
由考虑$\perp\widetilde{R}_{XY}\nu$项得到.
Ricci方程
取$\{\nu_i\}$为$M_x^\perp$的标准正交基,
定义法基本形式$\beta_{r}^s(X_p)=\left<{}\widetilde{D}_{X_p}\nu_r,\nu_s\right>.$
记$\Pi_r:=\Pi_{\nu_r},$ 取$\{U_i\}$为$\{M_x\}$的标准正交基,
定义第二基本形式间的卷积:
$$
\Pi_r\ast \Pi_s(X,Y)=\sum_i \Pi_r(X,U_i)\Pi_s(U_i,Y),
$$
易见该等式可以退广地定义为二次对称张量场间的卷积.
计算得到如下的Ricci方程:
$$
\begin{aligned}
\left<{}\widetilde{R}_{XY}\nu_r,\nu_s\right>=&\left(\Pi_r*\Pi_s(Y,X)+(D_Y\beta_r^s)(X)+\beta_w^s(Y)\beta_r^w(X)\right)\\
&-\left(\Pi_r*\Pi_s(X,Y)+(D_X\beta_r^s)(Y)+\beta_w^s(X)\beta_r^w(Y)\right)\\
\end{aligned}
$$
文章最后更新于 2022-04-27 21:03:42
Second Fundamental Form
Let $\widetilde{M}$ be the $\widetilde{n}$-dimensional Riemannian manifold, $M$ be the $n$-dimensional submanifold,
$n<\widetilde{n}.$
Record $\mathscr{N}(M)$ as $M$ in $\widetilde{M}$ normal vector field all,
$\mathscr{T}(M)$ is tangent vector field Overall. In order to examine the appearance of the submanifold,
We take the normal vector field $N\in \mathscr{N}(M),$ and consider $\widetilde{D}_XN.$
This is equivalent to examining all $\left<{}\widetilde D_XN,Y\right>,$
$Y\in \mathscr{T}(M).$ This defines Second Fundamental Form.
Definition for the given $\nu\in \mathscr{N}(M),$:
$$
\Pi_\nu(X,Y)=\left<{}\widetilde{D}_X\nu,Y\right>,\quad X,Y\in \mathscr{T}(M),
$$
Called $M$ corresponds to the second fundamental form of $\nu$.
Lemma 1. For each fixed $\nu,$ $\Pi_\nu$ is a quadratic symmetric covariant tensor field on $M$.
Certificate: Due to $\left<{}\nu,Y\right>\equiv 0,$ $Y\in \mathscr{T}(M),$
For any $X,Y\in \mathscr{T}(M),$ we have:
$$
0=X\left<{}\nu,Y\right>=\left<{}\widetilde{D}_X\nu,Y\right>+\left<{}\nu,\widetilde{D}_XY\right>,
$$
Therefore we also have Weingarten equation:
$$
\Pi_\nu(X,Y)=-\left<{}\nu,\widetilde{D}_XY\right>.
$$
Next,
Due to $[X,Y]\in \mathscr{T}(M),$
$$
\left<{}\nu,\widetilde{D}_XY\right>=\left<{}\nu,\widetilde{D}_YX\right>+\left<{}\nu,[X,Y]\right>=\left<{}\nu,\widetilde{D}_YX\right>,
$$
Then from the Weingarten equation, $\Pi_\nu(X,Y)=\Pi_\nu(Y,X).$ In addition, it is easy to see
$$
\Pi_\nu(fX,Y)=f\Pi_\nu(X,Y)=\Pi_\nu(X,fY),
$$
Therefore $\Pi_\nu$ is indeed the tensor field on $M$. The lemma is proved.
It can be seen that $\Pi_\nu(X,Y)(x)$ also only depends on $\nu(x),$
And it is linear with respect to $C^\infty$ with respect to $\nu$.
Note that $M_x^\perp$ is the orthogonal complement of $M_x$ in $\widetilde{M}_x$, which can be defined Quadratic type
$$
S:M_x\otimes M_x\rightarrow M_x^\perp,
$$
make
$$
\left<{}S(v,w),\nu\right>:=\Pi_\nu(v,w),\quad \,\forall\,\nu\in \mathscr{N}(M).
$$
It is easy to see that $S(v,w)$ is the projection of $-\widetilde{D}_XY(x)$ on $M_x^\perp$,
$X(x)=v,Y(x)=w,$ The value has nothing to do with the vector field selection.
The trace $\sum_{i=1}^nS(e_i,e_i)\in M_x^\perp,$ of $S$ is denoted as $\eta(x),$
$\{e_i\}\subset M_x$ is an orthonormal basis. Then $\eta\in \mathscr{N}(M),$
called $M$ Mean curvature normal vector field, $|\eta|$ is called $M$ mean curvature.
Sometimes a symmetric linear transformation $A_\nu:M_x\rightarrow M_x,$ is also defined such that
$$
\left<{}A_\nu(v),w\right>=\Pi_\nu(v,w)=\left<{}v,A_\nu(w)\right>,
$$
That is, $A_\nu(v)$ is the projection of $\widetilde{D}_v\nu$ on $M_x$,
We also call $A_\nu$ corresponding to $\nu$ Second Fundamental Form. Its relationship to the mean curvature is:
$$
\left<{}\eta(x),\nu\right>=\sum_{i=1}^n\left<{}S(e_i,e_i),\nu\right>=\sum_{i=1}^n\Pi_\nu(e_i,e_i)=\sum_{i=1}^n\left<{}A_\nu(e_i),e_i\right>=\operatorname{tr}A_\nu.
$$
When $\widetilde{n}=n+1$, $M$ is in $\widetilde{M}$ hypersurface,
At this time, the unit normal vector field $\nu,$ can be locally fixed, denoted as $A:=A_\nu.$
At this time $\widetilde{D}_v\nu\in M_x,$ $A(v)=\widetilde{D}_v\nu.$
The eigenvalues of $A$ are called principal curvatures. Due to symmetry, they are all real and are denoted as $p_1,\cdots,p_n.$.
$\det A$ is called $M$ Gauss curvature. When $n=2$, it is consistent with the cross-section curvature,
This is Gauss's Egregium (brilliant) theorem.
Let $\widetilde{M}=\mathbb{R}^{n+1},$ If the hypersurface $M$ is orientable,
Then there is a unit normal vector field $\nu,$ as a whole that can be defined $\Gamma:M\rightarrow S^n,$
$x\mapsto \nu(x),$ is called Gauss mapping. Then,
$$
d\Gamma(X)=\widetilde{D}_X\nu=A(X),\quad \frac{\Gamma^\ast \Omega_0}{\Omega}=\det d\Gamma=\det A,
$$
Where $\Omega_0,\Omega$ are the volume elements of $S^n,M$ respectively,
From this we can see the geometric meaning of Gauss curvature $\det A$.
If it is assumed that $M$ has positive Gauss curvature, then the Gauss map $\Gamma$ is an immersion.
Taking the standard Riemannian metric $g_0,$ on $S^n$ can give $M$ a new Riemannian metric $\Gamma^\ast g_0.$
When $M$ is compact, $\Gamma^\ast g_0$ is complete. Selected by the metric, $\Gamma$ is locally isometric,
Therefore, it is a covering map. And since $S^n$ is simply connected, $\Gamma$ is a diffeomorphism.
Thus we have the lemma:
Lemma 2. Assume $M$ is a compact hypersurface in $\mathbb{R}^{n+1}$ with positive Gauss curvature, then the Gauss map $\Gamma:M\rightarrow S^n$ is a diffeomorphism.
Corollary 3. The compact hypersurface of positive curvature in $\mathbb{R}^{n+1}$ must be a convex surface, that is, it always falls on one side of its tangent plane.
Now define two orthogonal projections:
$$
\top:\widetilde{M}_x\rightarrow M_x,\quad \perp:\widetilde{M}_x\rightarrow M_x^\perp,
$$
It is easy to verify that $\top(\widetilde{D}_XY)$ satisfies the properties of Levi-Civita contact on $M$,
Therefore $D_XY=\top(\widetilde{D}_XY).$
Recall that we explained earlier that $S(X,Y)$ is the projection of $-\widetilde{D}_XY(x)$ on $M_x^\perp$,
Therefore we have:
$$
S(X,Y)=\perp(-\widetilde{D}_XY)=D_XY-\widetilde{D}_XY.
$$
This is exactly the definition introduced when discussing fully geodesic submanifolds earlier.
Now we can say $M$ is fully geodesic if and only if its second fundamental form is always zero.
The second fundamental form thus characterizes the difference from the full geodesic.
Gauss equation
Recall that the cross-sectional curvature of any $2$ dimensional tangent plane of the full geodesic submanifold is the same as the original manifold,
For the general case we have the following equation:
Proposition 4 (Gauss equation). $\,\forall\,X,Y\in \mathscr{T}(M),$ We have:
$$
\left<{}\widetilde{R}_{XY}X,Y\right>=\left<{}R_{XY}X,Y\right>-\left<{}S(X,X),S(Y,Y)\right>+|S(X,Y)|^2.
$$
Certificate: The proof is straightforward to compute:
$$
\left<{}\widetilde{R}_{XY}X,Y\right>=\left<{}\widetilde{D}_{[X,Y]}X,Y\right>-\left<{}\widetilde{D}_X\widetilde{D}_YX,Y\right>+\left<{}\widetilde{D}_Y\widetilde{D}_XX,Y\right>,
$$
Among them, since the image of $S$ falls in $M_x^\perp$,
$$
\left<{}\widetilde{D}_{[X,Y]}X,Y\right>=\left<{}D_{[X,Y]}X,Y\right>-\left<{}S([X,Y],X),Y\right>=\left<{}D_{[X,Y]}X,Y\right>,
$$
$$
\begin{aligned}
\left<{}\widetilde{D}_X\widetilde{D}_YX,Y\right>&=\left<{}\widetilde{D}_X(D_YX-S(X,Y)),Y\right>\\
&=\left<{}\widetilde{D}_XD_YX,Y\right>-\left<{}\widetilde{D}_XS(X,Y),Y\right>\\
&=\left<{}D_XD_YX,Y\right>-\left<{}S(X,D_YX),Y\right>-X\left<{}S(X,Y),Y\right>+\left<{}S(X,Y),\widetilde{D}_XY\right>\\
&=\left<{}D_XD_YX,Y\right>+\left<{}S(X,Y),D_XY-S(X,Y)\right>\\
&=\left<{}D_XD_YX,Y\right>-\left<{}S(X,Y),S(X,Y)\right>
\end{aligned}
$$
Similarly,
$$
\left<{}\widetilde{D}_Y\widetilde{D}_XX,Y\right>=\left<{}D_YD_XX,Y\right>-\left<{}S(X,X),S(Y,Y)\right>.
$$
The conclusion can be obtained by integrating the above equations.
Note: The Gauss equation describes $\top\widetilde{R}_{XY}Z,$
And $\perp \widetilde{R}_{XY}Z$ will be replaced by the following Codazzi equation given. In fact,
Using the same method used to prove the Gauss equation, we can prove that the following equation holds:
$$
\left<{}\widetilde{R}_{XY}Z,W\right>=\left<{}R_{XY}Z,W\right>-\left<{}S(X,Z),S(Y,W)\right>+\left<{}S(X,W),S(Y,Z)\right>.
$$
Corollary 5. Let $M$ be the hypersurface in $\widetilde{M}$, $A$ be the second basic form about the unit normal vector field $\nu$. If $P$ is the $2$-plane in $M_x$, let $A|_P:P\rightarrow P$ be $A:P\rightarrow M_x$ and then compound the orthogonal projection $M_x\rightarrow P,$, then
$$
\widetilde{K}(P)=K(P)-\det(A|_P).
$$
Certificate: When $M$ is a hypersurface,
Due to $\left<{}S(X,Y),\nu\right>=\Pi_\nu(X,Y)=\left<{}A(X),Y\right>,$
$S(X,Y)=\left<{}A(X),Y\right>\nu.$ Therefore, take $\{e_1,e_2\}$ as the orthonormal basis of $P$,
rule
$$
\left<{}S(e_1,e_1),S(e_2,e_2)\right>-|S(e_1,e_2)|^2=\left<{}A(e_1),e_1\right>\left<{}A(e_2),e_2\right>-\left<{}A(e_1),e_2\right>^2=\det(A|_P).
$$
Due to $\widetilde{K}(P)=\left<{}\widetilde{R}_{e_1e_2}e_1,e_2\right>,$
$K(P)=\left<{}R_{e_1e_2}e_1,e_2\right>,$ Get the conclusion immediately from the Gauss equation.
Corollary 6 (Gauss's brilliant theorem). Assume $M$ is the surface in $\mathbb{R}^3$, then its cross-sectional curvature is equal to its Gauss curvature.
Corollary 7. Let $\{\nu_1,\cdots,\nu_{\widetilde{n}-n}\}$ be the unitary basis of $M_x^\perp$, then $\,\forall\,X,Y\in M_x,$
$$
\left<{}\widetilde{R}_{XY}X,Y\right>=\left<{}R_{XY}X,Y\right>-\sum_i \Pi_{\nu_i}(X,X)\Pi_{\nu_i}(Y,Y)+\sum_i \Pi_{\nu_i}(X,Y).
$$
Corollary 8 (Synge’s Lemma). Assume $M$ contains a geodesic $\gamma,$ of $\widetilde{M}$, then for any tangent plane $P$ satisfying $\dot\gamma\in P\subset M_x,$ we have $K(P)\le \widetilde{K}(P).$
Certificate: Let $P$ be formed by $\dot\gamma,Y$. Since $\gamma$ is a geodesic,
$S(\dot\gamma,\dot\gamma)=0.$ Then let $X=\dot\gamma,$ in the Gauss equation
Get to the conclusion immediately.
Corollary 9. $\mathbb{R}^3$ Medium Ruled surface (Single parameter straight line family) with non-positive Gauss curvature.
Corollary 10. Assume $M$ is a $2$-dimensional surface in $\widetilde{M}$, including a geodesic $\gamma,$ of $\widetilde{M}$, then $K(M_{\gamma(t)})\equiv \widetilde{K}(M_{\gamma(t)})$ if and only if $M_{\gamma(t)}$ is parallel to $\gamma$ in $\widetilde{M}$.
Certificate: Assume $\gamma$ is a regular geodesic, $X$ is the unit vector field in $M$, and is perpendicular to $\gamma$, then
$$
\widetilde{K}(M_{\gamma(t)})=K(M_{\gamma(t)})+|S(\dot\gamma(t),X(t))|^2.
$$
Therefore the equation remains true if and only if $\widetilde{D}_{\dot\gamma}X\equiv D_{\dot\gamma}X.$
$\dot\gamma$ is of course parallel to $M,\widetilde{M}$ along $\gamma$, and
$$
\left<{}D_{\dot\gamma}X,\dot\gamma\right>=\dot\gamma\left<{}X,\dot\gamma\right>-\left<{}X,D_{\dot\gamma}\dot\gamma\right>=0, \quad \left<{}D_{\dot\gamma}X,X\right>=0\:\Rightarrow\: D_{\dot\gamma}X=0,
$$
$X$ is parallel to $M$ along $\gamma$.
Therefore, the equation is always true if and only if $X$ is parallel to $\gamma$ in $\widetilde{M}$,
That is, $M_{\gamma(t)}$ is parallel to $\gamma$ in $\widetilde{M}$.
Corollary 11. Assume $M$ is the Riemannian manifold, $x\in M.$ takes $\{x^i\}$ as the normal coordinate system of $x$, $P\subset M_x$ is spanned by $\{\frac{\partial {} }{\partial {}x^1},\frac{\partial {} }{\partial {}x^2}\}$, and $\mathscr{S}$ is the surface $\{x^i=0:i\ge 3\}.$. Then the cross-sectional curvature of $P$ is equal to the Gauss curvature of $\mathscr{S}$ under the induced metric.
Certificate: Just note that for the normal coordinate system, the second fundamental form of $S\equiv 0,$ at point $O$ is zero,
Then it is obtained from the Gauss equation.
Codazzi equation
Codazzi equation is also called Codazzi-Mainardi equation,
By directly calculating, the following results about $\widetilde{R}_{XY}Z$ can be obtained.
Proposition 12 (Codazzi equation). If $X,Y,Z\in \mathscr{T}(M),$ $R,\widetilde{R}$ is the curvature tensor of $M,\widetilde{M}$, then
$$
\begin{aligned}
\perp\widetilde{R}_{XY}Z=&\left(\perp \widetilde{D}_XS(Y,Z)-S(D_XY,Z)-S(Y,D_XZ)\right)\\
&-\left(\perp \widetilde{D}_YS(X,Z)-S(D_YX,Z)-S(X,D_YZ)\right)
\end{aligned}
$$
The Codazzi equation and the Gauss equation are often collectively referred to as Gauss-Codazzi equation.
The Codazzi equation is generally complex, but is simpler in special cases.
Proposition 13. Assume $\widetilde M$ is a constant curvature space, $M$ is the hypersurface in it, then $(D_X\Pi)(Y,Z)$ is symmetrical about $X,Y,Z$, where $X,Y,Z\in \mathscr{T}(M).$
Certificate: Since $\Pi(Y,Z)=\Pi(Z,Y),$ is obviously symmetric about $Y,Z$, we only need to prove
$$
(D_X\Pi)(Y,Z)=(D_Y\Pi)(X,Z).
$$
Taking the unit normal vector field $\nu,$ then $S(X,Z)=\Pi(X,Z)\nu.$ therefore
$$
\perp\widetilde{D}_YS(X,Z)=\perp\left((Y\Pi(X,Z))\nu+\Pi(X,Z)\widetilde{D}_Y\nu\right)=(Y\Pi(X,Z))\nu.
$$
When $\widetilde{M}$ has constant cross-section curvature $c$,
$$
\widetilde{R}_{XY}Z=c(\left<{}X,Z\right>Y-\left<{}Y,Z\right>X).
$$
This is because the right formula constitutes the curvature tensor on $\widetilde{M}$, and also has constant cross-section curvature $c.$
You can now find $\perp\widetilde{R}_{XY}Z=0.$
Substituting into the Codazzi equation, we have:
$$
Y\Pi(X,Z)-\Pi(D_YX,Z)-\Pi(X,D_YZ)=(D_Y\Pi)(X,Z)=(D_X\Pi)(Y,Z)=\cdots.
$$
A fundamental application of the Gauss-Codazzi equation is to illustrate the existence and uniqueness of a hypersurface given its second fundamental form.
The core of the proof is to consider a system of overdetermined first-order partial differential equations determined by:
$$
S(Y_i,Y_j)=\Pi(Y_i,Y_j)\nu=D_{Y_i}Y_j-\widetilde{D}_{Y_i}Y_j,\quad \widetilde{D}_{Y_i}\nu=A(Y_i).
$$
Since the equation is overdetermined, it may not have a solution.
In fact, the necessary and sufficient condition for solvability is to satisfy the previous Gauss-Codazzi equation,
This is done by considering Frobenius theorem Obtained. Below we only describe the theorem of existence and uniqueness.
Theorem 14 (uniqueness). Let $M,M'$ be two hypersurfaces in $\mathbb{R}^{n+1}$, with unit normal vector field $\nu,\nu'.$ Let $\varphi:M\rightarrow M'$ be an isometry isomorphism, so that $\Pi(X,Y)=\Pi'(d\varphi(X),d\varphi(Y)),$ then $M,M'$ is contract , that is, there is an isometric isomorphism $\Phi:\mathbb{R}^{n+1}\rightarrow \mathbb{R}^{n+1},$ such that $\Phi_M=\varphi.$
Theorem 15 (existence). Given that the $n$ Viriemannian manifold and the symmetric tensor $\Pi,$ on it satisfy the Gauss-Codazzi equation, then $M$ can be locally isometrically embedded in $\mathbb{R}^{n+1}$, so that its second basic form relative to a certain unit normal vector field is exactly $\Pi.$. In addition, this embedding is unique in the sense of the phase-difference global isometric isomorphism $\overline{\varphi}:\mathbb{R}^{n+1}\rightarrow \mathbb{R}^{n+1}$.
We can generalize the theorem to the $n+1$ dimensional Riemannian manifold $\widetilde{M},$
But the form of the Gauss-Codazzi equation will be more complex. In addition, it can also be extended to cases with higher codimension,
But at this time, additional integrable conditions need to be added Ricci equation,
Obtained by considering the $\perp\widetilde{R}_{XY}\nu$ term.
Ricci equation
Take $\{\nu_i\}$ as the orthonormal basis of $M_x^\perp$,
definition basic form of law $\beta_{r}^s(X_p)=\left<{}\widetilde{D}_{X_p}\nu_r,\nu_s\right>.$
Denote $\Pi_r:=\Pi_{\nu_r},$ and take $\{U_i\}$ as the orthonormal basis of $\{M_x\}$,
Define the second basic form between Convolution:
$$
\Pi_r\ast \Pi_s(X,Y)=\sum_i \Pi_r(X,U_i)\Pi_s(U_i,Y),
$$
It is easy to see that this equation can be defined reductively as a convolution between quadratic symmetric tensor fields.
The calculation is as follows Ricci equation:
$$
\begin{aligned}
\left<{}\widetilde{R}_{XY}\nu_r,\nu_s\right>=&\left(\Pi_r*\Pi_s(Y,X)+(D_Y\beta_r^s)(X)+\beta_w^s(Y)\beta_r^w(X)\right)\\
&-\left(\Pi_r*\Pi_s(X,Y)+(D_X\beta_r^s)(Y)+\beta_w^s(X)\beta_r^w(Y)\right)\\
\end{aligned}
$$
The article was last updated on 2022-04-27 21:03:42
