问题及解答

设 $M$ 为 $n$ 维光滑黎曼流形, $A(M)$ 是 $M$ 上微分形式集合. $\omega\in A(M)$. $X,Y,Y_i$ 均是 $M$ 上的向量场.
 

(1) $L_X\circ i(Y)-i(Y)\circ L_X=i([X,Y])$

(2) $L_X\circ L_Y-L_Y\circ L_X=L_{[X,Y]}$

(3) $d\circ i(X)+i(X)\circ d=L_X$

(4) $d\circ L_X=L_X\circ d$

(5) $i(X)L_X=L_X i(X)$

(6) $L_{fX}\omega=fL_X \omega+df\wedge i(X)\omega$, 其中 $f\in A^0(M)$, $\omega\in A(M)$.

(7) $L_X i(Y)-L_Y i(X)-i([X,Y])=[d,i(X)\circ i(Y)]$

第一题要用到下面的结论

(0) 设 $\omega\in A^r(M)$, 则

\[
(L_X\omega)(Y_1,\ldots,Y_r)=X\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}\omega(Y_1,\ldots,L_X Y_i,\ldots,Y_r)
\]

这是基础.

以上一些结论可以推广到一般的 $(0,p)$-型张量.

(1) 设 $\omega\in A^r(M)$, $Y_1,\ldots,Y_{r-1}$ 为 $M$ 上的光滑向量场.

$i(Y)$ 是内乘运算, 有时也记为 $i_Y$. 不妨记 $\omega_Y=i(Y)\omega=i_Y\omega$. 于是要证

\[
\bigl(L_X\circ i(Y)\omega\bigr)(Y_1,\ldots,Y_{r-1})-\bigl(i(Y)\circ L_X\omega\bigr)(Y_1,\ldots,Y_{r-1})=\bigl(i([X,Y])\omega\bigr)(Y_1,\ldots,Y_{r-1})
\]

这等价于

\[
\bigl(L_X\omega_Y\bigr)(Y_1,\ldots,Y_{r-1})-(L_X\omega)_Y(Y_1,\ldots,Y_{r-1})=\omega\bigl([X,Y],Y_1,\ldots,Y_{r-1}\bigr)
\]

左边第一项

\[
\begin{split}
&\bigl(L_X\omega_Y\bigr)(Y_1,\ldots,Y_{r-1})\\
=&X\bigl(\omega_Y(Y_1,\ldots,Y_{r-1})\bigr)-\sum_{i=1}^{r-1}\omega_Y(Y_1,\ldots,L_X Y_i,\ldots,Y_{r-1})\\
=&X\bigl(\omega(Y,Y_1,\ldots,Y_{r-1})\bigr)-\sum_{i=1}^{r-1}\omega(Y,Y_1,\ldots,L_X Y_i,\ldots,Y_{r-1})
\end{split}
\]

左边第二项

\[
\begin{split}
&(L_X\omega)_Y(Y_1,\ldots,Y_{r-1})\\
=&(L_X\omega)(Y,Y_1,\ldots,Y_{r-1})\\
=&X\bigl(\omega(Y,Y_1,\ldots,Y_{r-1})\bigr)-\omega(L_X Y,Y_1,\ldots,Y_{r-1})-\sum_{i=1}^{r-1}\omega(Y,Y_1,\ldots,L_X Y_i,\ldots,Y_{r-1})
\end{split}
\]

这两式相减, 即等于右边. 注意 $L_X Y=[X,Y]$.

(2) 若作用在向量场 $Z$ 上, 则

\[
\begin{split}
& L_X\circ L_Y Z-L_Y\circ L_X Z=L_{[X,Y]}Z\\
=& \bigl[X,[Y,Z]\bigr]-\bigl[Y,[X,Z]\bigr]=\bigl[[X,Y],Z\bigr]\\
=& \bigl[X,[Y,Z]\bigr]+\bigl[Y,[Z,X]\bigr]+\bigl[Z,[X,Y]\bigr]\\
=&0
\end{split}
\]

最后等于零, 是因为这就是 Jacobi 恒等式.

现在考虑作用在微分形式 $\omega\in A^r(M)$ 上, 设 $Y_1,\ldots,Y_r$ 是 $M$ 上的光滑向量场.

\[
\begin{split}
&(L_X\circ L_Y\omega)(Y_1,\ldots,Y_r)\\
=&X\bigl((L_Y\omega)(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}(L_Y\omega)(Y_1,\ldots,L_X Y_i,\ldots,Y_r)\\
=&X\Bigl[Y\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}\omega(Y_1,\ldots,L_Y Y_i,\ldots,Y_r)\Bigr]\\
&\quad -\sum_{i=1}^{r}\Bigl[Y\bigl(\omega(Y_1,\ldots,L_X Y_i,\ldots,Y_r)\bigr)-\sum_{j=1,j\neq i}^{r}\omega(Y_1,\ldots,L_X Y_j,\ldots,L_X Y_i,\ldots,Y_r)-\omega(Y_1,\ldots,L_Y L_X Y_i,\ldots,Y_r)\Bigl]
\end{split}
\] 

\[
\begin{split}
&(L_Y\circ L_X\omega)(Y_1,\ldots,Y_r)\\
=&Y\bigl((L_X\omega)(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}(L_X\omega)(Y_1,\ldots,L_Y Y_i,\ldots,Y_r)\\
=&Y\Bigl[X\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}\omega(Y_1,\ldots,L_X Y_i,\ldots,Y_r)\Bigr]\\
&\quad -\sum_{i=1}^{r}\Bigl[X\bigl(\omega(Y_1,\ldots,L_Y Y_i,\ldots,Y_r)\bigr)-\sum_{j=1,j\neq i}^{r}\omega(Y_1,\ldots,L_X Y_j,\ldots,L_Y Y_i,\ldots,Y_r)-\omega(Y_1,\ldots,L_X L_Y Y_i,\ldots,Y_r)\Bigl]
\end{split}
\]

或写成下面的形式更便于观察

\[
\begin{split}
&(L_X\circ L_Y\omega)(Y_1,\ldots,Y_r)\\
=&XY\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}X\bigl(\omega(Y_1,\ldots,L_Y Y_i,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}Y\bigl(\omega(Y_1,\ldots,L_X Y_i,\ldots,Y_r)\bigr)\\
&\quad +\sum_{i,j=1;i\neq j}^{r}\omega(Y_1,\ldots,L_Y Y_j,\ldots,L_X Y_i,\ldots,Y_r)+\sum_{i=1}^{r}\omega(Y_1,\ldots,L_Y L_X Y_i,\ldots,Y_r)
\end{split}
\]

\[
\begin{split}
&(L_Y\circ L_X\omega)(Y_1,\ldots,Y_r)\\
=&YX\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}Y\bigl(\omega(Y_1,\ldots,L_X Y_i,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}X\bigl(\omega(Y_1,\ldots,L_Y Y_i,\ldots,Y_r)\bigr)\\
&\quad +\sum_{i,j=1;j\neq i}^{r}\omega(Y_1,\ldots,L_X Y_j,\ldots,L_Y Y_i,\ldots,Y_r)+\sum_{i=1}^{r}\omega(Y_1,\ldots,L_X L_Y Y_i,\ldots,Y_r)
\end{split}
\]

其中上式 $\sum_{i,j=1; j\neq i}^{r}$ 这一部分是再经互换 $i,j$ 这两个指标得

\[
\sum_{j,i=1;i\neq j}^{r}\omega(Y_1,\ldots,L_X Y_i,\ldots,L_Y Y_j,\ldots,Y_r)
\]

从而两式相减, 得

\[
\bigl((L_X\circ L_Y-L_Y\circ L_X)\omega\bigr)(Y_1,\ldots,Y_r)=(XY-YX)\bigl(\omega(Y_1,\ldots,Y_r)\bigr)+\sum_{i=1}^{r}\omega(Y_1,\ldots,L_Y L_X Y_i-L_X L_Y Y_i,\ldots,Y_r)
\]

而

\[
\Bigl(L_{[X,Y]}\omega\Bigr)(Y_1,\ldots,Y_r)=[X,Y]\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}\omega(Y_1,\ldots,L_{[X,Y]}Y_i,\ldots,Y_r)
\]

利用一开始的观察, 也就是 $[L_X, L_Y]Z=L_{[X,Y]}Z$, 可得

\[
L_Y L_X Y_i-L_X L_Y Y_i=-L_{[X,Y]}Y_i
\]

成立. 从而

\[
(L_X\circ L_Y-L_Y\circ L_X)\omega=L_{[X,Y]}\omega
\]

故命题成立. 有时我们简写为

\[
[L_X, L_Y]=L_{[X,Y]}
\]

(3) 设 $\omega\in A^r(M)$, 即要证

\[
d\circ i(X)\omega+i(X)\circ d\omega=L_X\omega
\]

又设 $X_2,\ldots,X_{r+1}$ 是 $M$ 上的光滑向量场, $X_1=X$.

\[
\begin{split}
&(d\circ i(X)\omega)(X_2,\ldots,X_{r+1})=(d\omega_X)(X_2,\ldots,X_{r+1})\\
=&\sum_{i=2}^{r+1}(-1)^i X_i\bigl(\langle X_2\wedge\cdots\wedge\hat{X}_i\wedge\cdots\wedge X_{r+1},\omega_X\rangle\bigr)\\
&\qquad+\sum_{2\leq i < j\leq r+1}(-1)^{i+j}\langle[X_i,X_j]\wedge\cdots\wedge\hat{X}_i\wedge\cdots\wedge\hat{X}_j\wedge\cdots\wedge X_{r+1},\omega_X\rangle\\
=&\sum_{i=2}^{r+1}(-1)^i X_i\bigl(\omega(X,X_2,\ldots,\hat{X}_i,\ldots,X_{r+1})\bigr)\\
&\qquad+\sum_{2\leq i < j\leq r+1}(-1)^{i+j}\omega\bigl(X,[X_i,X_j],\ldots,\hat{X}_i,\ldots,\hat{X}_j,\ldots,X_{r+1}\bigr)
\end{split}
\]

另一方面,

\[
\begin{split}
&(i(X)\circ d\omega)(X_2,\ldots,X_{r+1})=d\omega(X,X_2,\ldots,X_{r+1})\\
=&X\bigl(\omega(X_2,\ldots,X_{r+1})\bigr)+\sum_{i=2}^{r+1}(-1)^{i+1} X_i\bigl(\omega(X,X_2,\ldots,\hat{X}_i,\ldots,X_{r+1})\bigr)\\
&\qquad+\sum_{2\leq j\leq r+1}(-1)^{1+j}\omega\bigl([X,X_j],X_2,\ldots,\hat{X}_j,\ldots,X_{r+1}\bigr)\\
&\qquad+\sum_{2\leq i < j\leq r+1}(-1)^{i+j}\omega\bigl([X_i,X_j],X,X_2,\ldots,\hat{X}_i,\ldots,\hat{X}_j,\ldots,X_{r+1}\bigr)
\end{split}
\]

两式相加, 注意有些项是可以消掉的, 得

\[
\begin{split}
&\bigl(d\circ i(X)\omega+i(X)\circ d\omega\bigr)(X_2,\ldots,X_{r+1})\\
=&X\bigl(\omega(X_2,\ldots,X_{r+1})\bigr)+\sum_{2\leq j\leq r+1}(-1)^{1+j}\omega\bigl([X,X_j],X_2,\ldots,\hat{X}_j,\ldots,X_{r+1}\bigr)
\end{split}
\]

而

\[
\begin{split}
&(L_X\omega)(X_2,\ldots,X_{r+1})\\
&=X\bigl(\omega(X_2,\ldots,X_{r+1})\bigr)-\sum_{j=2}^{r+1}\omega(X_2,\ldots,L_X X_j,\ldots,X_{r+1})\\
&=X\bigl(\omega(X_2,\ldots,X_{r+1})\bigr)-\sum_{j=2}^{r+1}(-1)^{j-2}\omega(L_X X_j,X_2,\ldots,\hat{X}_j,\ldots,X_{r+1})\\
&=X\bigl(\omega(X_2,\ldots,X_{r+1})\bigr)+\sum_{j=2}^{r+1}(-1)^{1+j}\omega\bigl([X,X_j],X_2,\ldots,\hat{X}_j,\ldots,X_{r+1})
\end{split}
\]

故

\[
d\circ i(X)+i(X)\circ d=L_X
\]

(4)

这是 (3) 的推论. 具体的, 将 $L_X=d\circ i(X)+i(X)\circ d$ 代人 (4) 的两边, 并注意到 $d\circ d=0$ 即得.

(5) 这是 (1) 的特殊情形. 在 (1) 中令 $Y=X$ 即得.

或者, 根据 (3), $L_X=d\circ i(X)+i(X)\circ d$, 则

\[
i(X)L_X=i(X)\circ d\circ i(X)+i(X)i(X)\circ d
\]

\[
L_X i(X)=d\circ i(X)i(X)+i(X)\circ d\circ i(X)
\]

由于 $i(X)i(X)\omega=0$ (根据 $\omega$ 的反对称性), 因此 $i(X)L_X=L_X i(X)$.

(6) 设 $f\in A^0(M)$, $\omega\in A(M)$

根据 (3),

\[
\begin{split}
L_{fX}&=d\circ i(fX)+i(fX)\circ d\\
&=d\circ \bigl(fi(X)\bigr)+fi(X)\circ d\\
&=df\wedge i(X)+fd\circ i(X)+fi(X)\circ d\\
&=f(d\circ i(X)+i(X)\circ d)+df\wedge i(X)\\
&=fL_X+df\wedge i(X)
\end{split}
\]

(7)

根据 (1), $i([X,Y])=L_X\circ i(Y)-i(Y)\circ L_X$, 代入 (7) 的左边, 得

\[
\begin{split}
LHS&=L_X i(Y)-L_Y i(X)-L_X i(Y)+i(Y)\circ L_X\\
&=i(Y)\circ L_X-L_Y i(X)\\
&\stackrel{by (3)}{=}i(Y)\bigl(d\circ i(X)+i(X)\circ d\bigr)-\bigl(d\circ i(Y)+i(Y)\circ d\bigr)i(X)\\
&=i(Y)\circ d\circ i(X)+i(Y)i(X)\circ d-d\circ i(Y)i(X)-i(Y)\circ d\circ i(X)\\
&=i(Y)i(X)\circ d-d\circ i(Y)i(X)\\
&=[i(Y)i(X),d]\\
&=[d,i(X)i(Y)]=RHS
\end{split}
\]

我们证明最基本的性质 (0). 要证明

\[
(L_X\omega)(Y_1,\ldots,Y_r)=X\bigl(\omega(Y_1,\ldots,Y_r)\bigr)-\sum_{i=1}^{r}\omega(Y_1,\ldots,L_X Y_i,\ldots,Y_r)\qquad(*)
\]

其中 $X\bigl(\omega(Y_1,\ldots,Y_r)\bigr)$ 是对行列式 $\omega(Y_1,\ldots,Y_r)$ 求导, 故

\[
X\bigl(\omega(Y_1,\ldots,Y_r)\bigr)=\sum_{i=1}^{r}\omega(Y_1,\ldots,XY_i,\ldots,Y_r)
\]

从而要证式子(*)的右边为

\[
\begin{split}
RHS&=\sum_{i=1}^{r}\omega(Y_1,\ldots,XY_i,\ldots,Y_r)-\sum_{i=1}^{r}\omega(Y_1,\ldots,[X,Y_i],\ldots,Y_r)\\
&=\sum_{i=1}^{r}\omega(Y_1,\ldots,Y_i X,\ldots,Y_r)
\end{split}
\]

另一方面, 回忆 $L_X\omega$ 的定义, 设 $\omega\in A^r(M)$,

\[
L_X\omega:=\lim_{t\rightarrow 0}\frac{\varphi^*\omega-\omega}{t}
\]

由于 (*) 式两边对于 $\omega$ 是线性的, 因此不妨设 $\omega=gdf^1\wedge\cdots df^r$. 而

\[
\varphi_t^*\omega=\widetilde{\omega}_t=gdf_t^1\wedge\cdots\wedge df_t^r
\]

这里下标带有 $t$ 表示与 $t$ 有关. 因此

\[
\begin{split}
\frac{\partial}{\partial t}(\varphi_t^*\omega)\biggl|_{t=0}&=\lim_{t\rightarrow 0}\frac{\varphi^*\omega-\omega}{t}\\
&=\lim_{t\rightarrow 0}\frac{1}{t}\Bigl[gdf_t^1\wedge\cdots\wedge df_t^r-gdf^1\wedge\cdots\wedge df^r\Bigr]\\
&=\lim_{t\rightarrow 0}\frac{1}{t}\Bigl[gdf_t^1\wedge\cdots\wedge df_t^r-gdf^1\wedge df_t^2\wedge\cdots\wedge df_t^r\\
&\qquad+gdf^1\wedge df_t^2\cdots\wedge df_t^r-gdf^1\wedge df^2\wedge df_t^3\wedge\cdots\wedge df_t^r\\
&\qquad+\cdots\\
&\qquad+gdf^1\wedge\cdots\wedge df^{r-1}\wedge df_t^r-gdf^1\wedge\cdots\wedge df^r\Bigr]\\
&=\sum_{i=1}^{r}gdf^1\wedge\cdots\wedge\frac{\partial}{\partial t}df_t^i\biggl|_{t=0}\wedge\cdots\wedge df^r
\end{split}
\]

因此 (*) 式的左边为

\[
\begin{split}
(L_X\omega)(Y_1,\ldots,Y_r)&=\sum_{i=1}^{r}\Bigl(gdf^1\wedge\cdots\wedge\frac{\partial}{\partial t}df_t^i\biggl|_{t=0}\wedge\cdots\wedge df^r\Bigr)(Y_1,\ldots,Y_r)\\
&=\sum_{i=1}^{r}\omega(Y_1,\ldots,Y_i X,\ldots,Y_r).
\end{split}
\]

这里要注意

\[
df^i(Y_i X)=Y_iXf^i=Y_i\frac{\partial}{\partial t}f_t^i\biggl|_{t=0}=d(\frac{\partial}{\partial t}f_t^i\biggl|_{t=0})(Y_i)=\frac{\partial}{\partial t}df_t^i\biggl|_{t=0}(Y_i).
\]

故 (*) 式成立.