问题及解答

第二 Bianchi 恒等式

对于(1,3)型张量 $R_{XY}Z$, 有

\[
(\nabla_X R)_{YZ}+(\nabla_Y R)_{ZX}+(\nabla_Z R)_{XY}=0.
\]

Ref.

梅加强, 《流形与几何初步》, 命题 3.3.2

任给向量场 $W$, 根据定义

\[
\begin{split}
(\nabla_X R)_{YZ}W &=\nabla_X (R_{YZ}W)-R_{YZ}(\nabla_X W)-R_{\nabla_X Y,Z}W-R_{Y,\nabla_X Z}W\\
&=[\nabla_X,R_{YZ}]W-R_{\nabla_X Y,Z}W-R_{Y,\nabla_X Z}W\\
\end{split}
\]

将上式中的 $X,Y,Z$ 依次轮换, 得到

\[
\begin{split}
(\nabla_Y R)_{ZX}W &=\nabla_Y (R_{ZX}W)-R_{ZX}(\nabla_Y W)-R_{\nabla_Y Z,X}W-R_{Z,\nabla_Y X}W\\
&=[\nabla_Y,R_{ZX}]W-R_{\nabla_Y Z,X}W-R_{Z,\nabla_Y X}W\\
\end{split}
\]

\[
\begin{split}
(\nabla_Z R)_{XY}W &=\nabla_Z (R_{XY}W)-R_{XY}(\nabla_Z W)-R_{\nabla_Z X,Y}W-R_{X,\nabla_Z Y}W\\
&=[\nabla_Z,R_{XY}]W-R_{\nabla_Z X,Y}W-R_{X,\nabla_Z Y}W\\
\end{split}
\]

将这三个式子相加, 并注意到 $R_{YZ}=R(Y,Z)=\nabla_Z \nabla_Y-\nabla_Y \nabla_Z+\nabla_{[Y,Z]}=[\nabla_Z,\nabla_Y]+\nabla_{[Y,Z]}$, 我们得到

\[
\begin{split}
&(\nabla_X R)_{YZ}W+(\nabla_Y R)_{ZX}W+(\nabla_Z R)_{XY}W\\
=\quad &[\nabla_X,R_{YZ}]W-R_{\nabla_X Y,Z}W-R_{Y,\nabla_X Z}W\\
+&[\nabla_Y,R_{ZX}]W-R_{\nabla_Y Z,X}W-R_{Z,\nabla_Y X}W\\
+&[\nabla_Z,R_{XY}]W-R_{\nabla_Z X,Y}W-R_{X,\nabla_Z Y}W\\
=\quad &[\nabla_X,[\nabla_Z,\nabla_Y]]W+[\nabla_X,\nabla_{[Y,Z]}]W-([\nabla_Z,\nabla_{\nabla_X Y}]+\nabla_{[\nabla_X Y,Z]})W-([\nabla_{\nabla_X Z},\nabla_Y]+\nabla_{[Y,\nabla_X Z]})W\\
+&[\nabla_Y,[\nabla_X,\nabla_Z]]W+[\nabla_Y,\nabla_{[Z,X]}]W-([\nabla_X,\nabla_{\nabla_Y Z}]+\nabla_{[\nabla_Y Z,X]})W-([\nabla_{\nabla_Y X},\nabla_Z]+\nabla_{[Z,\nabla_Y X]})W\\
+&[\nabla_Z,[\nabla_Y,\nabla_X]]W+[\nabla_Z,\nabla_{[X,Y]}]W-([\nabla_Y,\nabla_{\nabla_Z X}]+\nabla_{[\nabla_Z X,Y]})W-([\nabla_{\nabla_Z Y},\nabla_X]+\nabla_{[X,\nabla_Z Y]})W\\
\end{split}
\]

回忆有 $[\nabla_X,[\nabla_Y,\nabla_Z]]=0$, 因此可化简为

\[
\begin{split}
=\quad \Bigl\{ &[\nabla_X,\nabla_{[Y,Z]}]-[\nabla_X,\nabla_{\nabla_Y Z}]-[\nabla_{\nabla_Z Y},\nabla_X]\\
+ & [\nabla_Y,\nabla_{[Z,X]}]-[\nabla_Y,\nabla_{\nabla_Z X}]-[\nabla_{\nabla_X Z},\nabla_Y]\\
+ & [\nabla_Z,\nabla_{[X,Y]}]-[\nabla_Z,\nabla_{\nabla_X Y}]-[\nabla_{\nabla_Y X},\nabla_Z]\Bigr\}W\\
- &\Bigl(\nabla_{[\nabla_X Y,Z]}+\nabla_{[Z,\nabla_Y X]}+\nabla_{[\nabla_Y Z,X]}+\nabla_{[X,\nabla_Z Y]}+\nabla_{[\nabla_Z X,Y]}+\nabla_{[Y,\nabla_X Z]}\Bigr)W
\end{split}
\]

注意到

\[
\begin{split}
&[\nabla_X,\nabla_{[Y,Z]}]-[\nabla_X,\nabla_{\nabla_Y Z}]-[\nabla_{\nabla_Z Y},\nabla_X]\\
=&[\nabla_X,\nabla_{[Y,Z]}]-([\nabla_X,\nabla_{\nabla_Y Z}]-[\nabla_X,\nabla_{\nabla_Z Y}])\\
=&[\nabla_X,\nabla_{[Y,Z]}]-[\nabla_X,\nabla_{{\nabla_Y Z}-{\nabla_Z Y}}]\\
=&[\nabla_X,\nabla_{[Y,Z]}]-[\nabla_X,\nabla_{[Y,Z]}]\\
=&0.
\end{split}
\]

因此化简为

\[
\begin{split}
=&-\Bigl(\nabla_{[\nabla_X Y,Z]}+\nabla_{[Z,\nabla_Y X]}+\nabla_{[\nabla_Y Z,X]}+\nabla_{[X,\nabla_Z Y]}+\nabla_{[\nabla_Z X,Y]}+\nabla_{[Y,\nabla_X Z]}\Bigr)W\\
=&-\Bigl(\nabla_{[\nabla_X Y-\nabla_X Y,Z]}+\nabla_{[\nabla_Y Z-\nabla_Y Z,X]}+\nabla_{[\nabla_Z X-\nabla_Z X,Y]}\Bigr)W\\
=&-\Bigl(\nabla_{[[X,Y],Z]}+\nabla_{[[Y,Z],X]}+\nabla_{[[Z,X],Y]}\Bigr)W\\
=&-\Bigl(\nabla_{[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]}\Bigr)W\\
=&0
\end{split}
\]

上面的计算从形式上看还是比较复杂的, 为此我们引进符号 $\mathcal{C}(f(X,Y,Z))$ 表示对"函数" $f(X,Y,Z)$ 进行轮换相加. 即

\[
\mathcal{C}(f(X,Y,Z)):=f(X,Y,Z)+f(Y,Z,X)+f(Z,X,Y).
\]

任取向量场 $W$, 根据定义

\[
\begin{split}
(\nabla_Z R)(X,Y)W &=\nabla_Z(R(X,Y)W)-R(\nabla_Z X,Y)W-R(X,\nabla_Z Y)W-R(X,Y)\nabla_Z W\\
&=[\nabla_Z,R(X,Y)]W-R(\nabla_Z X,Y)W-R(X,\nabla_Z Y)W.
\end{split}
\]

将上式关于 $X,Y,Z$ 轮换并相加, 

\[
\mathcal{C}\bigl((\nabla_X R)(Y,Z)\bigr)W=\mathcal{C}\bigl([\nabla_X,R(Y,Z)]\bigr)W-\mathcal{C}\bigl(R(\nabla_X Y,Z)\bigr)W-\mathcal{C}\bigl(R(Y,\nabla_X Z)\bigr)W.
\]

注意到 $R(Y,Z)=\nabla_Z\nabla_Y-\nabla_Y\nabla_Z+\nabla_{[Y,Z]}=[\nabla_Z,\nabla_Y]+\nabla_{[Y,Z]}$, 并且 $\mathcal{C}([\nabla_X,[\nabla_Y,\nabla_Z]])=0$, 故有

\[
\begin{split}
\mathcal{C}\bigl((\nabla_X R)(Y,Z)\bigr)W&=\mathcal{C}\bigl([\nabla_X,[\nabla_Z,\nabla_Y]]\bigr)W+\mathcal{C}\bigl([\nabla_X,\nabla_{[Y,Z]}]\bigr)W-\mathcal{C}\bigl(R(\nabla_X Y,Z)\bigr)W-\mathcal{C}\bigl(R(Y,\nabla_X Z)\bigr)W\\
&=\mathcal{C}\bigl([\nabla_X,\nabla_{[Y,Z]}]\bigr)W-\mathcal{C}\bigl(R(\nabla_X Y,Z)\bigr)W+\mathcal{C}\bigl(R(\nabla_Y X,Z)\bigr)W\\
&=\mathcal{C}\bigl([\nabla_X,\nabla_{[Y,Z]}]\bigr)W-\mathcal{C}\bigl(R([X,Y],Z)\bigr)W\\
&=\mathcal{C}\bigl([\nabla_X,\nabla_{[Y,Z]}]\bigr)W-\mathcal{C}\bigl(\nabla_Z\nabla_{[X,Y]}-\nabla_{[X,Y]}\nabla_Z+\nabla_{[[X,Y],Z]}\bigr)W\\
&=\mathcal{C}\bigl([\nabla_X,\nabla_{[Y,Z]}]\bigr)W-\mathcal{C}\bigl([\nabla_Z,\nabla_{[X,Y]}]\bigr)W-\mathcal{C}\bigl(\nabla_{[[X,Y],Z]}\bigr)W\\
&=\mathcal{C}\bigl([\nabla_X,\nabla_{[Y,Z]}]\bigr)W-\mathcal{C}\bigl([\nabla_Z,\nabla_{[X,Y]}]\bigr)W\\
&=0.
\end{split}
\]

此即

\[
(\nabla_X R)(Y,Z)+(\nabla_Y R)(Z,X)+(\nabla_Z R)(X,Y)=0.
\]