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Schur定理
若$R_{AB}=c\delta_{AB},$ 则称$M$为Einstein流形.
定理 1 (Schur). 若$R_{AB}=\lambda \delta_{AB},$ 那么$M$也是Einstein流形.
曲率
回忆有
$$
d\omega_{AB}=\omega_{AC}\wedge\omega_{CB}-\frac{1}{2}\widetilde R_{ABCD}\omega_C\wedge\omega_D.
$$
拉回得到
$$
d\theta_{ij}-\theta_{ik}\wedge\theta_{kj}=-\theta_{i\alpha}\wedge\theta_{j\alpha}-\frac{1}{2}\widetilde R_{ijkl}\theta_k\wedge\theta_l=-\frac{1}{2}R_{ijkl}\theta_k\wedge\theta_l,
$$
$$
R_{ijkl}=\widetilde{R}_{ijkl}+h_{i\alpha k}h_{j\alpha l}-h_{i\alpha l}h_{j\alpha k}.
$$
这就是Gauss方程. 从而
$$
R_{ik}=\widetilde R_{ik}+nH^\alpha h_{i\alpha k}-h_{i\alpha j}h_{j\alpha k},
$$
$$
R=\widetilde R+n^2|H|^2-S.
$$
对于法向联络$\{\theta_{\alpha\beta}\},$
$$
d\theta_{\alpha \beta}-\theta_{\alpha i}\wedge\theta_{i\beta}-\theta_{\alpha \gamma}\wedge\theta_{\gamma\beta}=-\frac{1}{2}\widetilde R_{\alpha\beta kl}\theta_k\wedge\theta_l
$$
定义法向曲率张量
$$
d\theta_{\alpha\beta}-\theta_{\alpha\gamma}\wedge\theta_{\gamma\beta}=-\frac{1}{2}R_{\alpha\beta kl}\theta_k\wedge\theta_l.
$$
同理, 我们有Ricci方程
$$
R_{\alpha\beta ij}=\widetilde R_{\alpha\beta ij}+h_{i\alpha k}h_{k\beta j}-h_{j\alpha k}h_{k\beta i}.
$$
这只对余维数大于等于$2$的子流形有意义.
最后, 考虑
$$
d\theta_{i\alpha}-\theta_{ik}\wedge\theta_{k\alpha}-\theta_{i\beta}\wedge\theta_{\beta\alpha}=-\frac{1}{2}\widetilde R_{i\alpha kl}\theta_k\wedge\theta_l.
$$
$$
(dh_{i\alpha l}+h_{i\alpha j}\theta_{jl}+h_{k\alpha l}\theta_{ki}+h_{i\beta l}\theta_{\beta \alpha})\wedge\theta_l=-\frac{1}{2}\widetilde{R}_{i\alpha kl}\theta_k\wedge\theta_l.
$$
$$
(h_{i\alpha l,k}+\frac{1}{2}\widetilde R_{i\alpha kl})\theta_k\wedge\theta_l=0.
$$
这就推出了Codazzi方程
$$
h_{i\alpha l,k}-h_{i\alpha k,l}=\widetilde R_{i\alpha lk}=\widetilde R_{\alpha ikl}.
$$
我们希望$kl$对称, 即$\widetilde R_{\alpha ikl}=0.$
一个例子是常截面曲率$c$的黎曼流形都如此. 若$N=R^{n+p}(c),$
$$
\widetilde R_{ABCD}=c(\delta_{AC}\delta_{BD}-\delta_{AD}\delta_{BC}).
$$
此时Codazzi方程, Ricci方程均可简化.
考虑二次协变导数, 我们有
$$
h_{i\alpha j,k,l}-h_{i\alpha j,l,k}=h_{m\alpha j}R_{mikl}+h_{i\alpha m}R_{mjkl}+h_{i\beta j} R_{\beta\alpha kl}.
$$
类似地, 也可以考虑$H^\alpha$的协变导数.
文章最后更新于 2023-03-16 15:52:33
Schur's theorem
If $R_{AB}=c\delta_{AB},$, then $M$ is called an Einstein manifold.
Theorem 1 (Schur). If $R_{AB}=\lambda \delta_{AB},$ then $M$ is also an Einstein manifold.
curvature
There are memories
$$
d\omega_{AB}=\omega_{AC}\wedge\omega_{CB}-\frac{1}{2}\widetilde R_{ABCD}\omega_C\wedge\omega_D.
$$
Pull back to get
$$
d\theta_{ij}-\theta_{ik}\wedge\theta_{kj}=-\theta_{i\alpha}\wedge\theta_{j\alpha}-\frac{1}{2}\widetilde R_{ijkl}\theta_k\wedge\theta_l=-\frac{1}{2}R_{ijkl}\theta_k\wedge\theta_l,
$$
$$
R_{ijkl}=\widetilde{R}_{ijkl}+h_{i\alpha k}h_{j\alpha l}-h_{i\alpha l}h_{j\alpha k}.
$$
this is Gauss equation.thereby
$$
R_{ik}=\widetilde R_{ik}+nH^\alpha h_{i\alpha k}-h_{i\alpha j}h_{j\alpha k},
$$
$$
R=\widetilde R+n^2|H|^2-S.
$$
for Legal contact $\{\theta_{\alpha\beta}\},$
$$
d\theta_{\alpha \beta}-\theta_{\alpha i}\wedge\theta_{i\beta}-\theta_{\alpha \gamma}\wedge\theta_{\gamma\beta}=-\frac{1}{2}\widetilde R_{\alpha\beta kl}\theta_k\wedge\theta_l
$$
definition normal curvature tensor
$$
d\theta_{\alpha\beta}-\theta_{\alpha\gamma}\wedge\theta_{\gamma\beta}=-\frac{1}{2}R_{\alpha\beta kl}\theta_k\wedge\theta_l.
$$
In the same way, we have Ricci equation
$$
R_{\alpha\beta ij}=\widetilde R_{\alpha\beta ij}+h_{i\alpha k}h_{k\beta j}-h_{j\alpha k}h_{k\beta i}.
$$
This only makes sense for submanifolds with codimension greater than or equal to $2$.
Finally, consider
$$
d\theta_{i\alpha}-\theta_{ik}\wedge\theta_{k\alpha}-\theta_{i\beta}\wedge\theta_{\beta\alpha}=-\frac{1}{2}\widetilde R_{i\alpha kl}\theta_k\wedge\theta_l.
$$
$$
(dh_{i\alpha l}+h_{i\alpha j}\theta_{jl}+h_{k\alpha l}\theta_{ki}+h_{i\beta l}\theta_{\beta \alpha})\wedge\theta_l=-\frac{1}{2}\widetilde{R}_{i\alpha kl}\theta_k\wedge\theta_l.
$$
$$
(h_{i\alpha l,k}+\frac{1}{2}\widetilde R_{i\alpha kl})\theta_k\wedge\theta_l=0.
$$
This leads to the Codazzi equation
$$
h_{i\alpha l,k}-h_{i\alpha k,l}=\widetilde R_{i\alpha lk}=\widetilde R_{\alpha ikl}.
$$
We want $kl$ to be symmetrical, that is, $\widetilde R_{\alpha ikl}=0.$
An example is the Riemannian manifold with constant cross-section curvature $c$. If $N=R^{n+p}(c),$
$$
\widetilde R_{ABCD}=c(\delta_{AC}\delta_{BD}-\delta_{AD}\delta_{BC}).
$$
At this time, both the Codazzi equation and the Ricci equation can be simplified.
Considering quadratic covariant derivatives, we have
$$
h_{i\alpha j,k,l}-h_{i\alpha j,l,k}=h_{m\alpha j}R_{mikl}+h_{i\alpha m}R_{mjkl}+h_{i\beta j} R_{\beta\alpha kl}.
$$
Similarly, one can also consider covariant derivatives of $H^\alpha$.
The article was last updated on 2023-03-16 15:52:33
