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局部黎曼几何基本定理
命题 1. 令$\omega_A$为$U\subset X$中的余标架场. 存在唯一一组$1$-形式$\omega_{AB}$满足:
$$
\omega_{AB}+\omega_{BA}=0,\quad d\omega_A=\omega_B\wedge\omega_{BA}=\omega_{AB}\wedge\omega_B.
$$
唯一性容易证明. 设有两组满足条件的$\omega_{AB},$
令$\varphi_{AB}=\omega'_{AB}-\omega_{AB},$
那么由于$\{\omega_C\}$为一组基, $\varphi_{BA}=a_{BAC}\omega_C.$ 由条件,
$a_{BAC}+a_{ABC}=0$显然, 我们还有$a_{BAC}=a_{CAB}.$ 这是因为
$$
0=d\omega_A-d\omega_A=\omega_B \wedge \varphi_{BA}=a_{BAC}\omega_B\wedge\omega_C=\sum_{B<C}(a_{BAC}-a_{CAB})\omega_B\wedge\omega_C.
$$
此时立即得到$\varphi_{BA}=0.$ 这是因为
$$
a_{ABC}=-a_{BAC}=-a_{CAB}=a_{ACB}=a_{BCA}=-a_{CBA}=-a_{ABC}.
$$
存在性直接给出计算即可. 设$\omega_{BA}=b_{BAC}\omega_C,$ 那么
$$
d\omega_A=b_{BAC}\omega_B\wedge\omega_C, \quad d\omega_A(e_B,e_C)=b_{BAC}-b_{CAB}.
$$
轮换$A,B,C,$ 由反称性, 我们有:
$$
\begin{aligned}
&d\omega_A(e_B,e_C)+d\omega_C(e_B,e_A)-d\omega_{B}(e_A,e_C)\\
=&(b_{BAC}-b_{CAB})+(b_{BCA}-b_{ACB})-(b_{ABC}-b_{CBA})\\
=&2b_{BAC}
\end{aligned}
$$
这就给出了$\omega_{BA}$的具体表示. 特别的, 系数可进一步写为
$$
\begin{aligned}
b_{BAC}&=\frac{1}{2}(d\omega_A(e_B,e_C)+d\omega_C(e_B,e_A)-d\omega_{B}(e_A,e_C))\\
&=\frac{1}{2}(\omega_B([e_A,e_C])+\omega_A([e_C,e_B])-\omega_C([e_B,e_A]))
\end{aligned}
$$
协变微分
$\omega_{AB}$称为联络形式, 可用于定义协变微分.
对于标架场$\{e_A\},$ 向量场$\xi=\xi_Ae_A,$ 定义
$$
D\xi=D\xi_A\otimes e_A,\quad D\xi_A=d\xi_A+\xi_B\omega_{BA}.
$$
我们验证它就是Levi-Civita联络. 只需验证两个性质. 在正交标架下,
$$
\left<{}D_\eta\xi,\gamma\right>=D\xi_A(\eta)\left<{}e_A,\gamma\right>=(\eta(\xi_A)+\xi_B\omega_{BA}(\eta))\gamma_A,
$$
这样由反称性,
$$
\eta\left<{}\xi,\gamma\right>=\eta(\xi_A\gamma_A)=\eta(\xi_A)\gamma_A+\xi_A\eta(\gamma_A)=\left<{}D_\eta\xi,\gamma\right>+\left<{}\xi,D_{\eta}\gamma\right>.
$$
另外,
$$
D_{\xi}\eta-D_{\eta}\xi=(D\eta_A(\xi)-D\xi_A(\eta))e_A=(\xi(\eta_A)-\eta(\xi_A)+\eta_B\omega_{BA}(\xi)-\xi_B\omega_{BA}(\eta))e_A.
$$
而由对称性,
$$
\eta_B\omega_{BA}(\xi_Ce_C)-\xi_B\omega_{BA}(\eta_Ce_C)=(\eta_B\xi_C-\xi_B\eta_C)b_{BAC}=0,
$$
因此,
$$
[\xi,\eta]=(\xi(\eta_A)-\eta(\xi_A))e_A=D_\xi\eta-D_\eta\xi.
$$
于是该定义与原先的定义是一致的. 特别的, 我们有
$$
De_A=\omega_{AB}\otimes e_B.
$$
这样定义的好处就是形式简单, 如
$$
D\xi=D(\xi_Ae_A)=d\xi_A \otimes e_A+\xi_A De_A.
$$
对于高阶张量也可以定义协变导数. 设$T=T_{ABC}e_Ae_Be_C,$ 那么,
$DT:=DT_{ABC}e_Ae_Be_C,$
$$
DT_{ABC}=dT_{ABC}+T_{EBC}\omega_{EA}+T_{AEC}\omega_{EB}+T_{ABE}\omega_{EC}=T_{ABC,E}\omega_E.
$$
曲率
对命题1中的等式做外微分, 我们有
$$
0=d(d\omega_A)=d\omega_B\wedge\omega_{BA}-\omega_B\wedge d\omega_{BA}=\omega_C\wedge\omega_{CB}\wedge\omega_{BA}-\omega_C\wedge d\omega_{CA}.
$$
记$\Omega_{BA}=d\omega_{BA}-\omega_{BC}\wedge\omega_{CA},$
那么我们即有$\omega_B\wedge\Omega_{BA}=0.$ 由于$\Omega_{BA}$为$2$-形式,
其有展开 $\Omega_{BA}=-\frac{1}{2}R_{BACE}\omega_C\wedge\omega_E.$
这就引出了曲率项. 显然其满足性质
$$
R_{ABCE}=-R_{BACE}=-R_{ABEC}.
$$
我们还有第一Bianchi恒等式
$$
R_{ABCE}+R_{ACEB}+R_{AEBC}=0.
$$
这是因为
$$
\begin{aligned}
0&=\omega_{B}\wedge\Omega_{BA}=\frac{1}{2}R_{ABCE}\omega_B\wedge\omega_C\wedge\omega_E\\
&=\frac{1}{2}\sum_{B<C<E}(R_{ABCE}-R_{ABEC}-R_{ACBE}+R_{ACEB}+R_{AEBC}-R_{AECB})\omega_B\wedge\omega_C\wedge\omega_E\\
&=\sum_{B<C<E}(R_{ABCE}+R_{ACEB}+R_{AEBC})\omega_B\wedge\omega_C\wedge\omega_E
\end{aligned}
$$
结合前面的性质, 我们有$R_{ABCE}=R_{CEAB}.$ 这称为黎曼张量,
给出了黎曼度量的全部局部信息. 黎曼张量, 数量曲率的定义如下:
$$
R_{AB}=R_{BA}=R_{ACBC},\quad R=R_{AA}.
$$
Laplacian
对前面曾举例的$T,$ 我们定义它的同阶Laplacian为
$$
(\Delta T)_{ABC}=T_{ABC,E,E}.
$$
特别的, 若$u$是一个函数, 那么
$$
du=u_A \omega_A,\quad Du_A=du_A+u_B\omega_{BA}=u_{A,B}\omega_B,\quad \Delta u=u_{A,A}.
$$
我们还定义
$$
|\operatorname{grad}u|^2=u_A^2.
$$
若$\varphi(u)$为关于$u$的光滑函数, 那么
$$
\Delta\varphi(u)=\varphi'(u)\Delta u+\varphi''(u)|\operatorname{grad}u|^2.
$$
这是因为$\varphi(u)_A=\varphi'(u)u_A,$
$\varphi(u)_B\omega_{BA}=\varphi'(u)u_B\omega_{BA},$
$d(\varphi(u)_A)=\varphi''(u)u_Au_B\omega_B+\varphi'(u)du_{A},$
$$
D\varphi(u)_A=\varphi''(u)u_Au_B\omega_B+\varphi'(u)(du_A+u_B\omega_{BA})=\varphi''(u)u_Au_B\omega_B+\varphi'(u)Du_A.
$$
这就给出了$\Delta\varphi(u).$
文章最后更新于 2023-03-02 19:47:28
Fundamental Theorem of Local Riemannian Geometry
Proposition 1. Let $\omega_A$ be the co-standard field in $U\subset X$. There is a unique set of $1$-form $\omega_{AB}$ that satisfies:
$$
\omega_{AB}+\omega_{BA}=0,\quad d\omega_A=\omega_B\wedge\omega_{BA}=\omega_{AB}\wedge\omega_B.
$$
Uniqueness is easy to prove. There are two sets of $\omega_{AB},$ that satisfy the conditions
Order$\varphi_{AB}=\omega'_{AB}-\omega_{AB},$
Then since $\{\omega_C\}$ is a set of basis, $\varphi_{BA}=a_{BAC}\omega_C.$ is determined by the condition,
$a_{BAC}+a_{ABC}=0$Obviously, we still have $a_{BAC}=a_{CAB}.$ because
$$
0=d\omega_A-d\omega_A=\omega_B \wedge \varphi_{BA}=a_{BAC}\omega_B\wedge\omega_C=\sum_{B<C}(a_{BAC}-a_{CAB})\omega_B\wedge\omega_C.
$$
At this time, $\varphi_{BA}=0.$ is obtained immediately. This is because
$$
a_{ABC}=-a_{BAC}=-a_{CAB}=a_{ACB}=a_{BCA}=-a_{CBA}=-a_{ABC}.
$$
The existence can be calculated directly. Assume $\omega_{BA}=b_{BAC}\omega_C,$ then
$$
d\omega_A=b_{BAC}\omega_B\wedge\omega_C, \quad d\omega_A(e_B,e_C)=b_{BAC}-b_{CAB}.
$$
Rotation $A,B,C,$ By inverse symmetry, we have:
$$
\begin{aligned}
&d\omega_A(e_B,e_C)+d\omega_C(e_B,e_A)-d\omega_{B}(e_A,e_C)\\
=&(b_{BAC}-b_{CAB})+(b_{BCA}-b_{ACB})-(b_{ABC}-b_{CBA})\\
=&2b_{BAC}
\end{aligned}
$$
This gives a specific representation of $\omega_{BA}$. In particular, the coefficient can be further written as
$$
\begin{aligned}
b_{BAC}&=\frac{1}{2}(d\omega_A(e_B,e_C)+d\omega_C(e_B,e_A)-d\omega_{B}(e_A,e_C))\\
&=\frac{1}{2}(\omega_B([e_A,e_C])+\omega_A([e_C,e_B])-\omega_C([e_B,e_A]))
\end{aligned}
$$
covariant differential
$\omega_{AB}$ is called Contact form, can be used to define covariant differential.
For the frame field $\{e_A\},$ the vector field $\xi=\xi_Ae_A,$ is defined
$$
D\xi=D\xi_A\otimes e_A,\quad D\xi_A=d\xi_A+\xi_B\omega_{BA}.
$$
We verify that it is the Levi-Civita connection. We only need to verify two properties. Under the orthogonal frame,
$$
\left<{}D_\eta\xi,\gamma\right>=D\xi_A(\eta)\left<{}e_A,\gamma\right>=(\eta(\xi_A)+\xi_B\omega_{BA}(\eta))\gamma_A,
$$
In this way, by antisymmetry,
$$
\eta\left<{}\xi,\gamma\right>=\eta(\xi_A\gamma_A)=\eta(\xi_A)\gamma_A+\xi_A\eta(\gamma_A)=\left<{}D_\eta\xi,\gamma\right>+\left<{}\xi,D_{\eta}\gamma\right>.
$$
In addition,
$$
D_{\xi}\eta-D_{\eta}\xi=(D\eta_A(\xi)-D\xi_A(\eta))e_A=(\xi(\eta_A)-\eta(\xi_A)+\eta_B\omega_{BA}(\xi)-\xi_B\omega_{BA}(\eta))e_A.
$$
And due to symmetry,
$$
\eta_B\omega_{BA}(\xi_Ce_C)-\xi_B\omega_{BA}(\eta_Ce_C)=(\eta_B\xi_C-\xi_B\eta_C)b_{BAC}=0,
$$
Therefore,
$$
[\xi,\eta]=(\xi(\eta_A)-\eta(\xi_A))e_A=D_\xi\eta-D_\eta\xi.
$$
This definition is then consistent with the original definition. In particular, we have
$$
De_A=\omega_{AB}\otimes e_B.
$$
The advantage of this definition is that the form is simple, such as
$$
D\xi=D(\xi_Ae_A)=d\xi_A \otimes e_A+\xi_A De_A.
$$
Covariant derivatives can also be defined for higher-order tensors. Let $T=T_{ABC}e_Ae_Be_C,$ then,
$DT:=DT_{ABC}e_Ae_Be_C,$
$$
DT_{ABC}=dT_{ABC}+T_{EBC}\omega_{EA}+T_{AEC}\omega_{EB}+T_{ABE}\omega_{EC}=T_{ABC,E}\omega_E.
$$
curvature
Taking the exterior derivative of the equation in Proposition 1, we have
$$
0=d(d\omega_A)=d\omega_B\wedge\omega_{BA}-\omega_B\wedge d\omega_{BA}=\omega_C\wedge\omega_{CB}\wedge\omega_{BA}-\omega_C\wedge d\omega_{CA}.
$$
Remember$\Omega_{BA}=d\omega_{BA}-\omega_{BC}\wedge\omega_{CA},$
Then we have $\omega_B\wedge\Omega_{BA}=0.$ Since $\Omega_{BA}$ is of the form $2$,
It has expansion $\Omega_{BA}=-\frac{1}{2}R_{BACE}\omega_C\wedge\omega_E.$
This leads to the curvature term. Obviously it satisfies the property
$$
R_{ABCE}=-R_{BACE}=-R_{ABEC}.
$$
We also have the first Bianchi identity
$$
R_{ABCE}+R_{ACEB}+R_{AEBC}=0.
$$
This is because
$$
\begin{aligned}
0&=\omega_{B}\wedge\Omega_{BA}=\frac{1}{2}R_{ABCE}\omega_B\wedge\omega_C\wedge\omega_E\\
&=\frac{1}{2}\sum_{B<C<E}(R_{ABCE}-R_{ABEC}-R_{ACBE}+R_{ACEB}+R_{AEBC}-R_{AECB})\omega_B\wedge\omega_C\wedge\omega_E\\
&=\sum_{B<C<E}(R_{ABCE}+R_{ACEB}+R_{AEBC})\omega_B\wedge\omega_C\wedge\omega_E
\end{aligned}
$$
Combining the previous properties, we have $R_{ABCE}=R_{CEAB}.$ which is called Riemann tensor,
All local information of the Riemannian metric is given. The Riemannian tensor, quantity curvature is defined as follows:
$$
R_{AB}=R_{BA}=R_{ACBC},\quad R=R_{AA}.
$$
Laplacian
For the $T,$ given in the previous example, we define its Laplacian of the same order as
$$
(\Delta T)_{ABC}=T_{ABC,E,E}.
$$
In particular, if $u$ is a function, then
$$
du=u_A \omega_A,\quad Du_A=du_A+u_B\omega_{BA}=u_{A,B}\omega_B,\quad \Delta u=u_{A,A}.
$$
We also define
$$
|\operatorname{grad}u|^2=u_A^2.
$$
If $\varphi(u)$ is a smooth function about $u$, then
$$
\Delta\varphi(u)=\varphi'(u)\Delta u+\varphi''(u)|\operatorname{grad}u|^2.
$$
This is because $\varphi(u)_A=\varphi'(u)u_A,$
$\varphi(u)_B\omega_{BA}=\varphi'(u)u_B\omega_{BA},$
$d(\varphi(u)_A)=\varphi''(u)u_Au_B\omega_B+\varphi'(u)du_{A},$
$$
D\varphi(u)_A=\varphi''(u)u_Au_B\omega_B+\varphi'(u)(du_A+u_B\omega_{BA})=\varphi''(u)u_Au_B\omega_B+\varphi'(u)Du_A.
$$
This gives $\Delta\varphi(u).$
The article was last updated on 2023-03-02 19:47:28
