Gauss-Codazzi-Ricci方程

Theorem 1. 假设$M^m,N^n$是两个个黎曼流形, $R$, $\bar R$分别是它们的曲率张量, 它们的黎曼联络分别记为$\nabla$, $\overline\nabla$。 又设$f:M\to N$是一个浸入(即$f_*:TM\to TN$ 是单射), 则有如下的Gauss-Codazzi-Ricci方程成立
\begin{equation}
\begin{split}
\bar R(X,Y,Z,W)&=R(X,Y,Z,W)+\left\langle A(X,Z),A(Y,W)\right\rangle-\left\langle A(X,W),A(Y,Z)\right\rangle,\\
\bar R(X,Y,Z,U)&=\Big\langle (\widetilde\nabla_XA)(Y,Z)-(\widetilde\nabla_YA)(X,Z),U\Big\rangle,\\
\bar R(X,Y,U,V)&=\Big\langle R^\perp(X,Y)U-\left\langle [P_U,P_V]X,Y \right\rangle,V\Big\rangle.
\end{split}
\end{equation}
其中$X,Y, Z,W$是$M$的切向量场, 而$U,V$是$M$的法向量场(即法丛$T^\perp M\subset TN$的截面); $A$是第二基本型而$P$是形状算子, $\widetilde\nabla$是$\nabla$与法联络$\nabla^\perp$诱导的联络.

Proof . 按照第二基本型与形状算子(shape operator)的定义:
\begin{align*}
\overline\nabla_XY&=\nabla_XY+A(X,Y)\in TM\oplus T^\perp M\\
\overline\nabla_XU&=\nabla^\perp_XU-P(U;X)\in T^\perp M\oplus TM.
\end{align*}
可以证明$\nabla^\perp$是法丛$T^\perp M$上一个保持度量(由$N$上的度量诱导)的联络(我们不能谈无挠性)。易见$A(X,Y)=A(Y,X)$.

注意到
\begin{align*}
\overline\nabla_X\overline\nabla_YZ&=\overline\nabla_X(\nabla_YZ+A(Y,Z))\\
&=\nabla_X\nabla_YZ+A(X,\nabla_YZ)+\nabla^\perp_XA(Y,Z)-P(A(Y,Z);X).
\end{align*}
因此, 按照黎曼曲率张量的定义:
\begin{align*}
\bar R(X,Y,Z,W)&=\left\langle \bar R(X,Y)Z,W\right\rangle=\left\langle\overline\nabla_X\overline\nabla_Y Z-\overline\nabla_Y\overline\nabla_XZ-\overline\nabla_{[X,Y]}Z,W\right\rangle\\
&=\Big\langle\nabla_X\nabla_YZ+A(X,\nabla_YZ)+\nabla^\perp_XA(Y,Z)-P(A(Y,Z);X)\\
&\qquad-\nabla_Y\nabla_XZ-A(Y,\nabla_XZ)-\nabla^\perp_YA(X,Z)+P(A(X,Z);Y)\\
&\qquad-\nabla_{[X,Y]}Z-A([X,Y],Z),W\Big\rangle\\
&=\Big\langle R(X,Y)Z,P(A(X,Z);Y)-P(A(Y,Z);X)\\
&\qquad +\nabla^\perp_XA(Y,Z)-A(Y,\nabla_XZ)-A(\nabla_XY,Z)\\
&\qquad -\left(\nabla_Y^\perp A(X,Z)-A(X,\nabla_YZ)-A(\nabla_YX,Z)\right),W\Big\rangle.
\end{align*}
注意到$A$可视为$TM\otimes TM\to T^\perp M$的张量, 故利用$\nabla$与法联络$\nabla^\perp$, 可以定义张量丛$T^*M\otimes T^*M\otimes T^\perp M$上的联络:
$$
(\widetilde\nabla_X A)(Y,Z):=\nabla^\perp_XA(Y,Z)-A(\nabla_XY,Z)-A(Y,\nabla_XZ).
$$
因此,
\begin{align*}
\bar R(X,Y,Z,W)&=\Big\langle R(X,Y)Z,P(A(X,Z);Y)-P(A(Y,Z);X)\\
&\qquad (\widetilde\nabla_XA)(Y,Z)-(\widetilde\nabla_YA)(X,Z),W\Big\rangle.
\end{align*}
注意到, 我们有如下的Weingarten方程:
\begin{equation}
\left\langle P(U;X),Y\right\rangle
=-\left\langle\overline\nabla_XU,Y\right\rangle
=\left\langle\overline\nabla_XY,U\right\rangle
=\left\langle A(X,Y),U\right\rangle.
\end{equation}
因此, 我们得到Gauss方程:
\begin{equation}
\bar R(X,Y,Z,W)=R(X,Y,Z,W)+\left\langle A(X,Z),A(Y,W)\right\rangle-\left\langle A(X,W),A(Y,Z)\right\rangle.
\end{equation}

类似地,
\begin{align*}
\bar R(X,Y,Z,U)&=\Big\langle R(X,Y)Z,P(A(X,Z);Y)-P(A(Y,Z);X)\\
&\qquad (\widetilde\nabla_XA)(Y,Z)-(\widetilde\nabla_YA)(X,Z),U\Big\rangle\\
&=\Big\langle (\widetilde\nabla_XA)(Y,Z)-(\widetilde\nabla_YA)(X,Z),U\Big\rangle.
\end{align*}
由此得到Codazzi方程:
\begin{equation}
\bar R(X,Y,Z,U)=\Big\langle (\widetilde\nabla_XA)(Y,Z)-(\widetilde\nabla_YA)(X,Z),U\Big\rangle.
\end{equation}

完全类似地, 注意到
\begin{align*}
\overline\nabla_X\overline\nabla_YU&=\overline\nabla_X(\nabla^\perp_YU-P(U;Y))\\
&=\nabla_X^\perp\nabla_Y^\perp U-P(\nabla^\perp_YU;X)-\nabla_XP(U;Y)-A(X,P(U;Y)).
\end{align*}
若定义
$$
R^\perp(X,Y,U,V):=\left\langle R^\perp(X,Y)U,V\right\rangle=\nabla_X^\perp\nabla_Y^\perp U-\nabla_Y^\perp\nabla_X^\perp U-\nabla^\perp_{[X,Y]}U,
$$

\begin{align*}
\bar R(X,Y,U,V)&=\left\langle \bar R(X,Y)U,V\right\rangle=\left\langle\overline\nabla_X\overline\nabla_Y U-\overline\nabla_Y\overline\nabla_XU-\overline\nabla_{[X,Y]}U,V\right\rangle\\
&=\Big\langle\nabla_X^\perp\nabla_Y^\perp U-P(\nabla^\perp_YU;X)-\nabla_XP(U;Y)-A(X,P(U;Y))\\
&\qquad-\nabla_Y^\perp\nabla_X^\perp U+P(\nabla^\perp_XU;Y)+\nabla_YP(U;X)+A(Y,P(U;X))\\
&\qquad-\nabla^\perp_{[X,Y]}U+P(U;[X,Y]),V \Big\rangle\\
&=\Big\langle R^\perp(X,Y)U-A(X,P(U;Y))+A(Y,P(U;X)),V\Big\rangle.
\end{align*}
为进一步简化记号, 记$P_U(Y):=P(U;Y)$, 并定义
$$
[P_U,P_V]X=P_UP_V(X)-P_VP_U(X).
$$
则有Weingarten方程知
\begin{align*}
\left\langle P_UX,Y\right\rangle&=\left\langle P(X;U),Y\right\rangle=\left\langle A(X,Y),U\right\rangle=\left\langle P_UY,X\right\rangle,\\
\left\langle P_VP_UX,Y\right\rangle&=\left\langle A(P_UX,Y),V \right\rangle=\left\langle X,P_UP_VY\right\rangle.
\end{align*}
因此
\begin{align*}
\left\langle A(X,P(U;Y))-A(Y,P(U;X)),V \right\rangle&=\left\langle A(X,P_UY)-A(Y,P_UX),V\right\rangle\\
&=\left\langle P_VP_UY,X\right\rangle-\left\langle X,P_UP_VY\right\rangle\\
&=-\left\langle [P_U,P_V]Y,X \right\rangle\\
&=\left\langle P_UP_VX,Y\right\rangle-\left\langle Y,P_VP_UX\right\rangle\\
&=\left\langle [P_U,P_V]X,Y \right\rangle.
\end{align*}
故, 我们得到Ricci方程:
\begin{equation}
\bar R(X,Y,U,V)=\Big\langle R^\perp(X,Y)U-\left\langle [P_U,P_V]X,Y \right\rangle,V\Big\rangle.
\end{equation}

Leave a Reply

Your email address will not be published. Required fields are marked *