Machine-translated from Chinese.
协变微分
张量
回忆$T^r_s(V):=\underbrace{V\otimes \cdots\otimes V}_{r}\otimes \underbrace{V^\ast \otimes \cdots\otimes V^\ast }_{s},$
$T^\ast (V):=\bigoplus\limits_{r,s} T^r_s(V).$
若有同构$\varphi:V\rightarrow W,$
我们有伴随同构$\varphi^\ast :W^\ast \rightarrow V^\ast ,$
诱导了$\widetilde{\varphi}:T^r_s(V)\rightarrow T^r_s(W),$
$$
\widetilde{\varphi}(v_1\otimes\cdots\otimes v_r\otimes \alpha^1\otimes \cdots\otimes \alpha^s)=\varphi(v_1)\otimes\cdots\otimes \varphi(v_r)\otimes (\varphi^\ast )^{-1}(\alpha^1)\otimes\cdots\otimes (\varphi^\ast )^{-1}(\alpha^s),
$$
显然利用线性性,
可以将$\widetilde\varphi$推广定义到$T^\ast (V)\rightarrow T^\ast (W)$上.
协变导数
上节我们有平移同构$P_t:T_{\gamma(0)}M\rightarrow T_{\gamma(t)}M.$
记$T_xM$为$M_x,$
我可以取$\widetilde{P_t}:T^\ast (M_{\gamma(0)})\rightarrow T^\ast (M_{\gamma(t)})$为诱导同构.
由上节注记,
定义$D_vK:=\frac{d {} }{d {}t}[\widetilde{P_t}^{-1}(K(\gamma(t)))]_{t=0},$
称为张量场$K$关于$v$的协变导数. 可以证明它与$\gamma$选取无关.
取好张量场的一组平行基后, 协变导数便化为通常的导数.
因此我们有如下性质:
若$K$为$(r,s)$型, 则$D_vK$也是$(r,s)$型;
$D_v$是作用在张量场代数上的一个导子,
即$D_v(K_1\otimes K_2)=(D_vK_1)\otimes K_2+K_1\otimes (D_vK_2).$
$D_v$和缩并可交换, 这里缩并$\mathscr{C}$指共变与反变部分相消,
使张量降两阶,
如$\mathscr{C}(X\otimes \alpha\otimes \omega)=\omega(X)\alpha,$
$X\in V,$ $\alpha,\omega\in V^\ast .$ 该交换性表示为:
$D_v (\mathscr{C}K)=\mathscr{C}(D_vK).$
$D_vK$仅依赖于$K,$ 与曲线$\gamma$选取无关.
最后一条性质的证明只需说明$D_v\eta$与$\gamma$选取无关即可,
$\eta$为$1$-形式. 这一点利用性质$2,3,$
再结合$D_vX$确实与$\gamma$选取无关即可. 由性质$4,$
我们可以对每个张量场$K,$ 向量场$X,$ 定义新的张量场$D_XK:$
$D_XK(x)=D_{X(x)}K.$
协变微分
$D_XK$关于$X$是$\mathscr{F}$线性的, 因此可以定义$(r,s+1)$型张量场$DK:$
$$
(\omega^1,\cdots,\omega^r,X_1,\cdots,X_s,X)\mapsto (D_XK)(\omega^1,\cdots,\omega^r,X_1,\cdots,X_s).
$$
$DK$称为$K$的协变微分. 容易看出, 对函数$f,$ 有$Df=df,$
它是微分的推广. 一般地, 记$D^nK=D(D^{n-1}K),$ 但需注意,
一般来说不见得有$D^2K(\cdots,X,Y)=D_YD_XK(\cdots),$ 如计算可得对函数$f,$
$$
D^2f(X,Y)=(D_Y(df))(X)=YXf-(D_YX)f.
$$
特别地, 可以看到:
$$
D^2f(X,Y)-D^2f(Y,X)=((D_XY-D_YX)-[X,Y])f=T(X,Y)f.
$$
于是若$D$是一个对称联络, 那么$T(X,Y)\equiv 0,$ $D^2f(X,Y)=D^2f(Y,X),$
$D^2f$是一个对称的二阶协变张量场, 称为$f$的Hessian.
Laplace算子
对任意二阶协变对称张量场$S$,
取$\operatorname{tr}S(x)=\sum_i S(e_i,e_i),$
这与标准正交基$\{e_i\}$选取无关. 定义$\Delta f=\operatorname{tr}D^2 f,$
称为作用在函数上的Laplace算子. 具体展开, 它有如下形式:
$$
\Delta f= g^{ij}D^2f\left(\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}\right)=\frac{1}{\sqrt{G} }\frac{\partial {} }{\partial {}x^k}\left(g^{ik}\sqrt{G}\frac{\partial {}f}{\partial {}x^i}\right).
$$
练习
习题 1. 在具联络$D$的流形$M$上, 称张量场$K$是平行的, 若$DK\equiv 0.$ 证明度量$g$作为二阶协变对称张量是平行的.
证: 即$\,\forall\,v,X,Y,$ $(D_vg)(X,Y)=0.$ 由导子性与和缩并可交换性,
我们有:
$$
D_v(g\otimes X\otimes Y)=D_vg \otimes X\otimes Y+g\otimes D_vX\otimes Y+g\otimes X\otimes D_vY.
$$
$$
vg(X,Y)=(D_vg)(X,Y)+g(D_vX,Y)+g(X,D_vY).
$$
回忆联络的性质(L1),
$vg(X,Y)=g(D_vX,Y)+g(X,D_vY),$ 因此的确有$(D_vg)(X,Y)=0,$
从而度量$g$是平行的张量场.
习题 2. 写出Laplace算子的展开过程.
证: 设$A^{ij}\frac{\partial {} }{\partial {}x^j}=e_i,$
则$A^{ij}=\sqrt{g}^{ij}$时, $\{e_i\}$构成标准正交基.
$$
\Delta f=\sum_i D^2f\left(\sqrt{g}^{ij}\frac{\partial {} }{\partial {}x^j},\sqrt{g}^{ik}\frac{\partial {} }{\partial {}x^k}\right)=g^{jk}D^2f\left(\frac{\partial {} }{\partial {}x^j},\frac{\partial {} }{\partial {}x^k}\right)
$$
$$
D^2f\left(\frac{\partial {} }{\partial {}x^j},\frac{\partial {} }{\partial {}x^k}\right)=\frac{\partial^2 {}f}{\partial {}x^j\partial {}x^k}-\left(D_{\frac{\partial {} }{\partial {}x^k} }\frac{\partial {} }{\partial {}x^j}\right)f
$$
$$
D_{\frac{\partial {} }{\partial {}x^k} }\frac{\partial {} }{\partial {}x^j}=\Gamma_{jk}^i\frac{\partial {} }{\partial {}x^i}=\frac{g^{li} }{2}\left(\frac{\partial {}g_{jl} }{\partial {}x^k}+\frac{\partial {}g_{kl} }{\partial {}x^j}-\frac{\partial {}g_{jk} }{\partial {}x^l}\right)\frac{\partial {} }{\partial {}x^i}
$$
于是, 我们有:
$$
\begin{aligned}
\Delta f&=g^{jk}\frac{\partial^2 {}f}{\partial {}x^j\partial {}x^k}-\frac{g^{jk}g^{li} }{2}\left(\frac{\partial {}g_{jl} }{\partial {}x^k}+\frac{\partial {}g_{kl} }{\partial {}x^j}-\frac{\partial {}g_{jk} }{\partial {}x^l}\right)\frac{\partial {}f}{\partial {}x^i}\\
&=g^{ik}\frac{\partial^2 {}f}{\partial {}x^i\partial {}x^k}-g^{il}\frac{\partial {}g_{lj} }{\partial {}x^k}g^{jk}\frac{\partial {}f}{\partial {}x^i}+\frac{g^{jk}g^{li} }{2}\frac{\partial {}g_{jk} }{\partial {}x^l}\frac{\partial {}f}{\partial {}x^i}\\
&=g^{ik}\frac{\partial^2 {}f}{\partial {}x^i\partial {}x^k}+\frac{\partial {}g^{ik} }{\partial {}x^k}\frac{\partial {}f}{\partial {}x^i}+g^{ik}\frac{g^{lj} }{2}\frac{\partial {}g_{jl} }{\partial {}x^k}\frac{\partial {}f}{\partial {}x^i}\\
&=\frac{1}{\sqrt{G} }\frac{\partial {} }{\partial {}x^k}\left(g^{ik}\sqrt{G}\frac{\partial {}f}{\partial {}x^i}\right).
\end{aligned}
$$
其中用到了如下矩阵求导的恒等式:
$$
\frac{d {}A^{-1}(t)}{d {}t}=-A^{-1}(t)A'(t)A^{-1}(t),
$$
$$
\frac{d {}\det A(t)}{d {}t}=\det(A(t))\operatorname{tr}(A^{-1}(t)A'(t)).
$$
曲率张量
曲率算子
记$\mathscr{T}^\ast $为$M$上所有张量场对应于$C^\infty$代数所形成的模.
则对任意给定向量场$X,$ 协变导数$D_X$为$\mathscr{T}^\ast $的一个导子.
现对给定向量场$X,Y,$ 定义映射:
$$
R_{XY}=D_{[X,Y]}-(D_XD_Y-D_YD_X)=D_{[X,Y]}-[D_X,D_Y].
$$
称$R_{XY}$为由$X,Y$定义的曲率算子, 具有如下性质:
(1) $R_{XY}$为$\mathscr{T}^\ast $的一个导子;
(2) $R_{XY}$保持张量场的类型;
(3) $\,\forall\,$函数$f,$ 张量场$K,$
有$R_{(fX)Y}K=R_{X(fY)}K=R_{XY}fK=fR_{XY}K;$
(4) $\,\forall\,$函数$f,$ 有$R_{XY}f=0.$
由代数中的结论, 若$D_1,D_2$为导子, 则Lie括号$[D_1,D_2]$也是导子.
由此说明了性质(1). 性质(2)由协变导数基本性质即得. 性质(3)经由常规计算得.
性质(4)由联络的基本性质即得.
曲率张量
由(2),(3)知, 对向量场$X,Y,Z,$ $R_{XY}Z$也是向量场,
且关于每个变量都是$\mathscr{F}$线性的. 从而它决定了一个$(1,3)$型张量场:
$$
(\omega,X,Y,Z)\mapsto \omega(R_{XY}Z),
$$
等价地,
也可以将它视为一个$(0,4)$型张量场:
$$
R(X,Y,Z,W):=\left<{}R_{XY}Z,W\right>.
$$
称$R_{XY}Z$或$R(X,Y,Z,W)$为黎曼度量的曲率张量.
它刻画了流形”空间弯曲”的性质.
性质
曲率张量是度量的二阶不变量, 即它包含度量张量$g$的二阶导数.
记$g_{ij}=g\left(\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}\right),$
$R_{ijkl}=R\left(\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j},\frac{\partial {} }{\partial {}x^k},\frac{\partial {} }{\partial {}x^l}\right),$
计算得到
$$
R_{ijkl}=\frac{1}{2}\left(\frac{\partial^2 {}g_{il} }{\partial {}x^j\partial {}x^k}+\frac{\partial^2 {}g_{jk} }{\partial {}x^i\partial {}x^l}-\frac{\partial^2 {}g_{ik} }{\partial {}x^j\partial {}x^l}-\frac{\partial^2 {}g_{jl} }{\partial {}x^i\partial {}x^k}\right)+g_{rs}\left(\Gamma_{jk}^r\Gamma_{il}^s-\Gamma_{jl}^r\Gamma_{ik}^s\right).
$$
这意味着, 曲率张量是度量张量的非线性函数.
这将是我们理解曲率张量主要的障碍.
引理 1. 对向量场$X,Y,Z,W,$ 我们有:
(1) $R_{XY}=-R_{YX};$
(2) $R_{XY}Z+R_{YZ}X+R_{ZX}Y=0$(第一Bianchi恒等式);
(3) $R(X,Y,Z,W)=-R(X,Y,W,Z);$
(4) $R(X,Y,Z,W)=R(Z,W,X,Y).$
证: 由于曲率算子$R$关于向量场是$C^\infty$多重线性的,
只需考虑$R_{ijkl}$是否满足性质即可. 性质(1)是即得的. 对于坐标向量场,
Lie括号$[\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}]\equiv 0,$
$R_{\frac{\partial {} }{\partial {}x^i}\frac{\partial {} }{\partial {}x^j} }=-[D_{\frac{\partial {} }{\partial {}x^i} },D_{\frac{\partial {} }{\partial {}x^j} }].$
又由Levi-Civita联络性质,
$D_{\frac{\partial {} }{\partial {}x^i} }\frac{\partial {} }{\partial {}x^j}=D_{\frac{\partial {} }{\partial {}x^j} }\frac{\partial {} }{\partial {}x^i},$
结合前式即有性质(2).
性质(3)由前面的$R_{ijkl}$表达式得到, 也等价于$R_{ijkk}=0.$
性质(4)是性质(1)-(3)的代数推论, 当然也可由$R_{ijkl}$表达式得到.
现考虑$R_{XY}:M_x\rightarrow M_x$为线性变换, 由性质(3),
它相对于$M_x$上的内积来说是反对称的. 进而,
考虑定义在$M_x\oplus M_x$上的函数
$$
Q(X,Y):=R(X,Y,X,Y),
$$
称$Q$为$R$的相配二次型.
引理 2. $Q$完全确定了曲率张量. 对两个满足引理$1$条件的张量场, 若它们的相配二次型相同, 则它们本身也相同.
证: 只需证明若张量场$R$决定的$Q=0,$ 则$R=0.$ 首先由定义与性质(4),
$$
Q(X,Y),Q(X,W),Q(X,Y+W)=0\Rightarrow R(X,Y,X,W)+R(X,W,X,Y)=2R(X,Y,X,W)=0.
$$
由于对任意向量场$W$成立, $R_{XY}X=0.$ 类似地, 可以得到
$$
R_{(X+Z)Y}(X+Z)=R_{XY}Z+R_{ZY}X=0.
$$
由第一Bianchi恒等式,
$R_{XY}Z+R_{YZ}X+R_{ZX}Y=0,$ 结合曲率算子反对称性, $2R_{ZY}X=R_{ZX}Y.$
而同理我们有$R_{YZ}X+R_{XZ}Y=0,$ 代入得到$3R_{YZ}X=0.$
由向量场选取任意性, $R=0.$
引理 3 (第二Bianchi恒等式). $(D_XR)_{YZ}+(D_YR)_{ZX}+(D_ZR)_{XY}=0.$
证: 只需对坐标向量场讨论即可. 此时$D_XY=D_YX,$
$R_{XY}=-D_XD_Y+D_YD_X.$ $R$是张量场, 我们有:
$$
D_X(R_{YZ}W)=(D_XR)_{YZ}W+R_{D_XY,Z}W+R_{Y,D_XZ}W+R_{YZ}(D_XW).
$$
轮换$X,Y,Z$即可得到结论.
引理 4 (Ricci恒等式). 若$T$为一个张量场, 则
$$
D^2T(\cdots,X,Y)-D^2T(\cdots,Y,X)=(R_{XY}T)(\cdots).
$$
证:
$$
D^2T(\cdots,X,Y)=(D_Y(DT))(\cdots,X)=D_YD_XT(\cdots)-D_{D_YX}T(\cdots).
$$
交换$X,Y$后做差发现恰好得到等式.
特别地, 当$T$为函数时, 得到了前面$D^2f$对称的结论. 该等式是更一般的推广.
练习
习题 3. 等距变换保持Levi-Civita联络, 进而它保持曲率张量.
证: 设$\varphi:M\rightarrow N$为等距变换,
在$M$上定义联络$\widetilde{D}_XY=(d\varphi)^{-1}D'_{d\varphi(X)}d\varphi(Y),$
显然它确实是一个联络. 只需证明其满足Levi-Civita联络的性质,
由唯一性即得$D_XY=\widetilde{D}_XY=D'_{d\varphi(X)}d\varphi(Y).$
$$
\begin{aligned}
X\left<{}Y,Z\right>&=d\varphi(X)\left<{}d\varphi(Y),d\varphi(Z)\right>\\
&=\left<{}D'_{d\varphi(X)}d\varphi(Y),d\varphi(Z)\right>+\left<{}d\varphi(Y),D'_{d\varphi(X)}d\varphi(Z)\right>\\
&=\left<{}\widetilde{D}_XY,Z\right>+\left<{}Y,\widetilde{D}_XZ\right>
\end{aligned}
$$
$$
\begin{aligned}
\widetilde{D}_XY-\widetilde{D}_YX=&(d\varphi)^{-1}(D'_{d\varphi(X)}d\varphi(Y)-D'{D\varphi(Y)}d\varphi(X))\\
=&(d\varphi)^{-1}[d\varphi(X),d\varphi(Y)]\\
=&[X,Y]=XY-YX
\end{aligned}
$$
因此该联络确实满足Levi-Civita联络的性质. 由于联络决定了曲率张量,
等距变换也是保持曲率张量不变的.
曲率
截面曲率
一般地, 我们对$Q$做归一化处理. 设$\Pi$为$M_x$的一个二维子空间,
$\{v_1,v_2\}$为其任意一组基, 则定义$\Pi$的截面曲率为
$$
K(\Pi):=\frac{R(v_1,v_2,v_1,v_2)}{|v_1\wedge v_2|^2},
$$
容易看到,
$K$与二维子空间的基选取无关. 特别地, 若选取一组标准正交基$\{e_1,e_2\},$
$K(\Pi)=R(e_1,e_2,e_1,e_2).$ 由前面的引理,
若我们知道所有二维子空间的截面曲率, 就等价于知道了曲率张量.
截面曲率自然和曲率张量一样, 都是度量不变量, 在等距变换下保持.
对二维黎曼流形$M,$ 我们熟知有高斯曲率$K.$
此时高斯曲率与上述的截面曲率是一致的.
因此可以说截面曲率是高斯曲率的推广. 特别地, 注意到高斯曲率同样是内蕴的.
这便是高斯绝妙定理.
称$\mathbb{R}^n$上通常的度量是平坦的. 此时$\,\forall\,X,Y,Z,$
$R_{XY}Z=0,$ 曲率张量恒为零. 反之,
我们后面会证明若黎曼流形$M$曲率张量为零,
它必局部等距于配备平坦度量的$\mathbb{R}^n.$ 称这样的流形是平坦的.
对于$n$维球面$S^n,$ 由于任意两点处任意两个二维切平面间存在一个等距变换,
可以证明它是具常截面曲率的.
Ricci张量
Ricci张量为一个二阶协变张量场, 定义为:
$$
\operatorname{Ric}(X,Y):=\sum_i R(e_i,X,e_i,Y),
$$
$\{e_i\}$为$M_x$任意一组标准正交基. 由$R$的多重线性性, 它与基选取无关.
且由其性质, Ricci张量是对称的,
即$\operatorname{Ric}(X,Y)=\operatorname{Ric}(Y,X).$
在$M$的切单位球面上, 定义函数$X\mapsto \operatorname{Ric}(X,X),$
称为在向量$X$下的Ricci曲率. 若$X=e_1$在选取的基中,
则$\operatorname{Ric}(X,X)=\sum_{i=2}^n R(e_i,e_1,e_i,e_1),$
是$n-1$个截面曲率的和.
数量曲率$s$定义为$\operatorname{Ric}$的迹,
$s(x):=\sum_{i,j}R(e_i,e_j,e_i,e_j),$
是幺正基中$n$个向量所张成的各种可能的二维平面截面曲率之和.
文章最后更新于 2022-02-14 15:31:56
covariant differential
Tensor
Memories$T^r_s(V):=\underbrace{V\otimes \cdots\otimes V}_{r}\otimes \underbrace{V^\ast \otimes \cdots\otimes V^\ast }_{s},$
$T^\ast (V):=\bigoplus\limits_{r,s} T^r_s(V).$
If there is isomorphism $\varphi:V\rightarrow W,$
We have adjoint isomorphism $\varphi^\ast :W^\ast \rightarrow V^\ast ,$
Induced $\widetilde{\varphi}:T^r_s(V)\rightarrow T^r_s(W),$
$$
\widetilde{\varphi}(v_1\otimes\cdots\otimes v_r\otimes \alpha^1\otimes \cdots\otimes \alpha^s)=\varphi(v_1)\otimes\cdots\otimes \varphi(v_r)\otimes (\varphi^\ast )^{-1}(\alpha^1)\otimes\cdots\otimes (\varphi^\ast )^{-1}(\alpha^s),
$$
Obviously using linearity,
$\widetilde\varphi$ can be generalized and defined to $T^\ast (V)\rightarrow T^\ast (W)$.
covariant derivative
In the previous section we had translational isomorphism $P_t:T_{\gamma(0)}M\rightarrow T_{\gamma(t)}M.$
Record $T_xM$ as $M_x,$
I can take $\widetilde{P_t}:T^\ast (M_{\gamma(0)})\rightarrow T^\ast (M_{\gamma(t)})$ to be an induced isomorphism.
From the notes in the previous section,
Definition$D_vK:=\frac{d {} }{d {}t}[\widetilde{P_t}^{-1}(K(\gamma(t)))]_{t=0},$
is called the tensor field $K$ with respect to $v$ covariant derivative. It can be proved that it has nothing to do with $\gamma$ selection.
After taking a set of parallel bases of the tensor field, the covariant derivatives become ordinary derivatives.
Therefore we have the following properties:
If $K$ is of type $(r,s)$, then $D_vK$ is also of type $(r,s)$;
$D_v$ is a derivative acting on the tensor field algebra,
That is $D_v(K_1\otimes K_2)=(D_vK_1)\otimes K_2+K_1\otimes (D_vK_2).$
$D_v$ and contraction are interchangeable, here contraction $\mathscr{C}$ means that the covariation and contravariance parts cancel,
Reduce the tensor by two orders,
Such as $\mathscr{C}(X\otimes \alpha\otimes \omega)=\omega(X)\alpha,$
$X\in V,$ $\alpha,\omega\in V^\ast .$ The commutativity is expressed as:
$D_v (\mathscr{C}K)=\mathscr{C}(D_vK).$
$D_vK$ only depends on $K,$ and has nothing to do with the selection of curve $\gamma$.
The proof of the last property only needs to show that $D_v\eta$ has nothing to do with the selection of $\gamma$.
$\eta$ is of the $1$-form. This makes use of the property $2,3,$
Combined with $D_vX$, it really has nothing to do with the selection of $\gamma$. According to the property $4,$
We can define a new tensor field $D_XK:$ for each tensor field $K,$ vector field $X,$
$D_XK(x)=D_{X(x)}K.$
covariant differential
$D_XK$ is $\mathscr{F}$ linear with respect to $X$, so the $(r,s+1)$ type tensor field $DK:$ can be defined
$$
(\omega^1,\cdots,\omega^r,X_1,\cdots,X_s,X)\mapsto (D_XK)(\omega^1,\cdots,\omega^r,X_1,\cdots,X_s).
$$
$DK$ is called $K$ covariant differential. It is easy to see that for function $f,$ there is $Df=df,$
It is a generalization of differential. Generally, it is recorded as $D^nK=D(D^{n-1}K),$ but it should be noted that
Generally speaking, there may not be $D^2K(\cdots,X,Y)=D_YD_XK(\cdots),$. For example, the function $f,$ can be obtained by calculation.
$$
D^2f(X,Y)=(D_Y(df))(X)=YXf-(D_YX)f.
$$
In particular, one can see:
$$
D^2f(X,Y)-D^2f(Y,X)=((D_XY-D_YX)-[X,Y])f=T(X,Y)f.
$$
So if $D$ is a symmetric connection, then $T(X,Y)\equiv 0,$ $D^2f(X,Y)=D^2f(Y,X),$
$D^2f$ is a symmetric second-order covariant tensor field, called the Hessian of $f$.
Laplace operator
For any second-order covariant symmetric tensor field $S$,
Take $\operatorname{tr}S(x)=\sum_i S(e_i,e_i),$
This has nothing to do with the selection of the orthonormal basis $\{e_i\}$. Definition $\Delta f=\operatorname{tr}D^2 f,$
called acting on a function Laplace operator. Specifically expanded, it has the following form:
$$
\Delta f= g^{ij}D^2f\left(\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}\right)=\frac{1}{\sqrt{G} }\frac{\partial {} }{\partial {}x^k}\left(g^{ik}\sqrt{G}\frac{\partial {}f}{\partial {}x^i}\right).
$$
Practice
Exercise 1. On the manifold $M$ with connection $D$, the tensor field $K$ is said to be parallel , if $DK\equiv 0.$ proves that the metric $g$ is parallel as a second-order covariant symmetric tensor.
Certificate: That is, $\,\forall\,v,X,Y,$ $(D_vg)(X,Y)=0.$ is commutative by derivation and contraction,
We have:
$$
D_v(g\otimes X\otimes Y)=D_vg \otimes X\otimes Y+g\otimes D_vX\otimes Y+g\otimes X\otimes D_vY.
$$
$$
vg(X,Y)=(D_vg)(X,Y)+g(D_vX,Y)+g(X,D_vY).
$$
Recall the nature of contact (L1),
$vg(X,Y)=g(D_vX,Y)+g(X,D_vY),$ So there is $(D_vg)(X,Y)=0,$
Thus the metric $g$ is a parallel tensor field.
Exercise 2. Write down the expansion process of the Laplace operator.
Certificate: Let $A^{ij}\frac{\partial {} }{\partial {}x^j}=e_i,$
Then when $A^{ij}=\sqrt{g}^{ij}$, $\{e_i\}$ forms an orthonormal basis.
$$
\Delta f=\sum_i D^2f\left(\sqrt{g}^{ij}\frac{\partial {} }{\partial {}x^j},\sqrt{g}^{ik}\frac{\partial {} }{\partial {}x^k}\right)=g^{jk}D^2f\left(\frac{\partial {} }{\partial {}x^j},\frac{\partial {} }{\partial {}x^k}\right)
$$
$$
D^2f\left(\frac{\partial {} }{\partial {}x^j},\frac{\partial {} }{\partial {}x^k}\right)=\frac{\partial^2 {}f}{\partial {}x^j\partial {}x^k}-\left(D_{\frac{\partial {} }{\partial {}x^k} }\frac{\partial {} }{\partial {}x^j}\right)f
$$
$$
D_{\frac{\partial {} }{\partial {}x^k} }\frac{\partial {} }{\partial {}x^j}=\Gamma_{jk}^i\frac{\partial {} }{\partial {}x^i}=\frac{g^{li} }{2}\left(\frac{\partial {}g_{jl} }{\partial {}x^k}+\frac{\partial {}g_{kl} }{\partial {}x^j}-\frac{\partial {}g_{jk} }{\partial {}x^l}\right)\frac{\partial {} }{\partial {}x^i}
$$
So, we have:
$$
\begin{aligned}
\Delta f&=g^{jk}\frac{\partial^2 {}f}{\partial {}x^j\partial {}x^k}-\frac{g^{jk}g^{li} }{2}\left(\frac{\partial {}g_{jl} }{\partial {}x^k}+\frac{\partial {}g_{kl} }{\partial {}x^j}-\frac{\partial {}g_{jk} }{\partial {}x^l}\right)\frac{\partial {}f}{\partial {}x^i}\\
&=g^{ik}\frac{\partial^2 {}f}{\partial {}x^i\partial {}x^k}-g^{il}\frac{\partial {}g_{lj} }{\partial {}x^k}g^{jk}\frac{\partial {}f}{\partial {}x^i}+\frac{g^{jk}g^{li} }{2}\frac{\partial {}g_{jk} }{\partial {}x^l}\frac{\partial {}f}{\partial {}x^i}\\
&=g^{ik}\frac{\partial^2 {}f}{\partial {}x^i\partial {}x^k}+\frac{\partial {}g^{ik} }{\partial {}x^k}\frac{\partial {}f}{\partial {}x^i}+g^{ik}\frac{g^{lj} }{2}\frac{\partial {}g_{jl} }{\partial {}x^k}\frac{\partial {}f}{\partial {}x^i}\\
&=\frac{1}{\sqrt{G} }\frac{\partial {} }{\partial {}x^k}\left(g^{ik}\sqrt{G}\frac{\partial {}f}{\partial {}x^i}\right).
\end{aligned}
$$
The following identity for matrix derivation is used:
$$
\frac{d {}A^{-1}(t)}{d {}t}=-A^{-1}(t)A'(t)A^{-1}(t),
$$
$$
\frac{d {}\det A(t)}{d {}t}=\det(A(t))\operatorname{tr}(A^{-1}(t)A'(t)).
$$
curvature tensor
curvature operator
Let $\mathscr{T}^\ast $ be the module formed by all tensor fields on $M$ corresponding to the algebra of $C^\infty$.
Then for any given vector field $X,$, the covariant derivative $D_X$ is a derivative of $\mathscr{T}^\ast $.
Now define the mapping for the given vector field $X,Y,$:
$$
R_{XY}=D_{[X,Y]}-(D_XD_Y-D_YD_X)=D_{[X,Y]}-[D_X,D_Y].
$$
Call $R_{XY}$ defined by $X,Y$ curvature operator, has the following properties:
(1) $R_{XY}$ is a derivative of $\mathscr{T}^\ast $;
(2) $R_{XY}$ maintains the type of tensor field;
(3) $\,\forall\,$ function $f,$ tensor field $K,$
Yes $R_{(fX)Y}K=R_{X(fY)}K=R_{XY}fK=fR_{XY}K;$
(4) $\,\forall\,$ function $f,$ has $R_{XY}f=0.$
According to the conclusion in algebra, if $D_1,D_2$ is a derivative, then the Lie bracket $[D_1,D_2]$ is also a derivative.
This illustrates property (1). Property (2) is obtained from the basic properties of covariant derivatives. Property (3) is obtained by conventional calculations.
Property (4) is obtained from the basic properties of contact.
curvature tensor
From (2) and (3), we know that the vector field $X,Y,Z,$ $R_{XY}Z$ is also a vector field,
And is $\mathscr{F}$ linear with respect to each variable. Thus it determines a $(1,3)$ type tensor field:
$$
(\omega,X,Y,Z)\mapsto \omega(R_{XY}Z),
$$
Equivalently,
It can also be regarded as a $(0,4)$ type tensor field:
$$
R(X,Y,Z,W):=\left<{}R_{XY}Z,W\right>.
$$
Call $R_{XY}Z$ or $R(X,Y,Z,W)$ the Riemannian metric curvature tensor.
It depicts the "space curvature" nature of the manifold.
nature
The curvature tensor is a second-order invariant of the metric, that is, it contains the second-order derivative of the metric tensor $g$.
Note$g_{ij}=g\left(\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}\right),$
$R_{ijkl}=R\left(\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j},\frac{\partial {} }{\partial {}x^k},\frac{\partial {} }{\partial {}x^l}\right),$
Calculated
$$
R_{ijkl}=\frac{1}{2}\left(\frac{\partial^2 {}g_{il} }{\partial {}x^j\partial {}x^k}+\frac{\partial^2 {}g_{jk} }{\partial {}x^i\partial {}x^l}-\frac{\partial^2 {}g_{ik} }{\partial {}x^j\partial {}x^l}-\frac{\partial^2 {}g_{jl} }{\partial {}x^i\partial {}x^k}\right)+g_{rs}\left(\Gamma_{jk}^r\Gamma_{il}^s-\Gamma_{jl}^r\Gamma_{ik}^s\right).
$$
This means, A curvature tensor is a nonlinear function of a metric tensor.
This will be the main obstacle to our understanding of the curvature tensor.
Lemma 1. For the vector field $X,Y,Z,W,$ we have:
(1) $R_{XY}=-R_{YX};$
(2) $R_{XY}Z+R_{YZ}X+R_{ZX}Y=0$ (First Bianchi identity);
(3) $R(X,Y,Z,W)=-R(X,Y,W,Z);$
(4) $R(X,Y,Z,W)=R(Z,W,X,Y).$
Certificate: Since the curvature operator $R$ is $C^\infty$ multilinear with respect to the vector field,
Just consider whether $R_{ijkl}$ satisfies the property. Property (1) is obtained immediately. For the coordinate vector field,
Lie brackets$[\frac{\partial {} }{\partial {}x^i},\frac{\partial {} }{\partial {}x^j}]\equiv 0,$
$R_{\frac{\partial {} }{\partial {}x^i}\frac{\partial {} }{\partial {}x^j} }=-[D_{\frac{\partial {} }{\partial {}x^i} },D_{\frac{\partial {} }{\partial {}x^j} }].$
Contacted by Levi-Civita,
$D_{\frac{\partial {} }{\partial {}x^i} }\frac{\partial {} }{\partial {}x^j}=D_{\frac{\partial {} }{\partial {}x^j} }\frac{\partial {} }{\partial {}x^i},$
Combined with the previous formula, we have property (2).
Property (3) is obtained from the previous $R_{ijkl}$ expression, and is also equivalent to $R_{ijkk}=0.$
Property (4) is the algebraic corollary of properties (1)-(3), and of course it can also be obtained from the $R_{ijkl}$ expression.
Now consider $R_{XY}:M_x\rightarrow M_x$ as a linear transformation. According to property (3),
It is antisymmetric relative to the inner product on $M_x$. Furthermore,
Consider a function defined on $M_x\oplus M_x$
$$
Q(X,Y):=R(X,Y,X,Y),
$$
Call $Q$ $R$ Matches the quadratic form.
Lemma 2. $Q$ completely determines the curvature tensor. For two tensor fields that satisfy the conditions of Lemma $1$, if their matching quadratic forms are the same, then they themselves are also the same.
Certificate: It only needs to be proved that if $Q=0,$ is determined by the tensor field $R$, then $R=0.$ is first determined by the definition and property (4),
$$
Q(X,Y),Q(X,W),Q(X,Y+W)=0\Rightarrow R(X,Y,X,W)+R(X,W,X,Y)=2R(X,Y,X,W)=0.
$$
Since it holds for any vector field $W$, $R_{XY}X=0.$ similarly, we can get
$$
R_{(X+Z)Y}(X+Z)=R_{XY}Z+R_{ZY}X=0.
$$
From the first Bianchi identity,
$R_{XY}Z+R_{YZ}X+R_{ZX}Y=0,$ Combined with curvature operator antisymmetry, $2R_{ZY}X=R_{ZX}Y.$
In the same way, we have $R_{YZ}X+R_{XZ}Y=0,$ and substitute it to get $3R_{YZ}X=0.$
Selection arbitrariness by vector fields, $R=0.$
Lemma 3 (Second Bianchi identity). $(D_XR)_{YZ}+(D_YR)_{ZX}+(D_ZR)_{XY}=0.$
Certificate: Just discuss the coordinate vector field. At this time $D_XY=D_YX,$
$R_{XY}=-D_XD_Y+D_YD_X.$ $R$ is the tensor field, we have:
$$
D_X(R_{YZ}W)=(D_XR)_{YZ}W+R_{D_XY,Z}W+R_{Y,D_XZ}W+R_{YZ}(D_XW).
$$
Rotate $X,Y,Z$ to get the conclusion.
Lemma 4 (Ricci identity). If $T$ is a tensor field, then
$$
D^2T(\cdots,X,Y)-D^2T(\cdots,Y,X)=(R_{XY}T)(\cdots).
$$
Certificate:
$$
D^2T(\cdots,X,Y)=(D_Y(DT))(\cdots,X)=D_YD_XT(\cdots)-D_{D_YX}T(\cdots).
$$
After exchanging $X,Y$ and doing the difference, we find that we get exactly the equation.
In particular, when $T$ is a function, the previous symmetric conclusion of $D^2f$ is obtained. This equation is a more general generalization.
Practice
Exercise 3. The isometric transformation preserves the Levi-Civita connection, which in turn preserves the curvature tensor.
Certificate: Let $\varphi:M\rightarrow N$ be the isometric transformation,
Define contact $\widetilde{D}_XY=(d\varphi)^{-1}D'_{d\varphi(X)}d\varphi(Y),$ on $M$
Obviously it is indeed a contact. Just prove that it satisfies the properties of Levi-Civita contact,
Obtained from uniqueness$D_XY=\widetilde{D}_XY=D'_{d\varphi(X)}d\varphi(Y).$
$$
\begin{aligned}
X\left<{}Y,Z\right>&=d\varphi(X)\left<{}d\varphi(Y),d\varphi(Z)\right>\\
&=\left<{}D'_{d\varphi(X)}d\varphi(Y),d\varphi(Z)\right>+\left<{}d\varphi(Y),D'_{d\varphi(X)}d\varphi(Z)\right>\\
&=\left<{}\widetilde{D}_XY,Z\right>+\left<{}Y,\widetilde{D}_XZ\right>
\end{aligned}
$$
$$
\begin{aligned}
\widetilde{D}_XY-\widetilde{D}_YX=&(d\varphi)^{-1}(D'_{d\varphi(X)}d\varphi(Y)-D'{D\varphi(Y)}d\varphi(X))\\
=&(d\varphi)^{-1}[d\varphi(X),d\varphi(Y)]\\
=&[X,Y]=XY-YX
\end{aligned}
$$
Therefore, this connection does satisfy the properties of Levi-Civita connection. Since the connection determines the curvature tensor,
Isometric transformation also keeps the curvature tensor unchanged.
curvature
Section curvature
Generally, we normalize $Q$. Let $\Pi$ be a two-dimensional subspace of $M_x$,
$\{v_1,v_2\}$ is any set of its bases, then define $\Pi$ Section curvature for
$$
K(\Pi):=\frac{R(v_1,v_2,v_1,v_2)}{|v_1\wedge v_2|^2},
$$
easy to see,
$K$ has nothing to do with the selection of the basis of the two-dimensional subspace. In particular, if a set of orthonormal basis $\{e_1,e_2\},$ is selected
$K(\Pi)=R(e_1,e_2,e_1,e_2).$ From the previous lemma,
If We know the cross-sectional curvature of all two-dimensional subspaces, which is equivalent to knowing the curvature tensor.
Naturally, the cross-section curvature, like the curvature tensor, is a metric invariant and is maintained under isometric transformation.
For the two-dimensional Riemannian manifold $M,$ we know that there is Gaussian curvature $K.$
At this time, the Gaussian curvature is consistent with the above-mentioned cross-sectional curvature.
Therefore, it can be said that section curvature is a generalization of Gaussian curvature. In particular, note that Gaussian curvature is also intrinsic.
This is Gauss's brilliant theorem.
The usual metric on $\mathbb{R}^n$ is flat.At this time $\,\forall\,X,Y,Z,$
$R_{XY}Z=0,$ The curvature tensor is always zero. On the contrary,
We will prove later that if the curvature tensor of the Riemannian manifold $M$ is zero,
It must be locally isometric to $\mathbb{R}^n.$ equipped with a flat metric. Such a manifold is said to be flat.
For the $n$-dimensional sphere $S^n,$, since there is an isometric transformation between any two two-dimensional tangent planes at any two points,
It can be proved that it has constant cross-sectional curvature.
Ricci tensor
Ricci tensor is a second-order covariant tensor field, defined as:
$$
\operatorname{Ric}(X,Y):=\sum_i R(e_i,X,e_i,Y),
$$
$\{e_i\}$ is any set of orthonormal basis of $M_x$. Due to the multilinearity of $R$, it has nothing to do with the selection of the basis.
And by its properties, the Ricci tensor is symmetric,
That is $\operatorname{Ric}(X,Y)=\operatorname{Ric}(Y,X).$
On the tangent unit sphere of $M$, define the function $X\mapsto \operatorname{Ric}(X,X),$
is called under vector $X$ Ricci curvature. If $X=e_1$ is in the selected basis,
Then $\operatorname{Ric}(X,X)=\sum_{i=2}^n R(e_i,e_1,e_i,e_1),$
is the sum of $n-1$ section curvatures.
quantity curvature $s$ is defined as the trace of $\operatorname{Ric}$,
$s(x):=\sum_{i,j}R(e_i,e_j,e_i,e_j),$
is the sum of various possible two-dimensional plane section curvatures spanned by $n$ vectors in the unitary basis.
The article was last updated on 2022-02-14 15:31:56
