Machine-translated from Chinese.
一开始看到流形上的等距变换, 自然想到保持范数$\Leftrightarrow$保持内积,
从而保持黎曼度量. 但事实上等距(依照初始定义)保持的是流形上的距离,
而不是切空间上的距离, 因此这件事情是需要来证明的. 更重要的一点是,
前述的等价关系仅对线性变换成立,
而切映照是线性变换建立在映照可微的基础上.
事实上我们不需要事先要求等距变换可微, 这就是Myers-Steenrod定理.
定理 1. 若$\phi:M\rightarrow N$为等距满射, 则$\phi$为同构, 特别地它是光滑的.
证: 由于测地线有局部极短性, 等距变换将测地线映至测地线.
设$\phi(p)=q,$ 选取$v\in T_pM,$ $|v|$充分小.
取从$p$点出发以$v$为初始切向量的测地线$\gamma,$
那么$d(p,\gamma(1))=|v|.$ 由于$\phi$是等距变换,
$d(q,\phi\circ\gamma(1))=|v|.$ 若$\phi$可微,
那么由于$\phi\circ\gamma$也是测地线,
$|v|=|(\phi\circ\gamma)'(0)|=|d\phi(v)|.$
此时$d\phi$是保持范数的线性变换, 因此$d\phi$保持内积, 即保持黎曼度量,
故$\phi$是同构.
当$\phi$不见得可微时, 由于$\phi$保持测地线,
我们可以通过测地线的切向量来定义$d\phi:T_pM\rightarrow T_pN,$
它也是保持范数的. 但此时它不见得是线性变换, 不过我们可以来证明这一点.
假设已知如下结论:
$$
\lim_{A,B\rightarrow 0\in T_pM}\frac{d(\exp_p(A),\exp_p(B))}{|A-B|}=1.
$$
选取$A,B\in T_pM,$ 取对应的测地线$\gamma_A,\gamma_B.$ 记$A'=d\phi(A),$
$B'=d\phi(B),$ 以及对应的测地线$\gamma_{A'},\gamma_{B'},$
我们有$d(\gamma_A(t),\gamma_B(t))=d(\gamma_{A'}(t),\gamma_{B'}(t)),$
从而:
$$
1=\lim_{t\rightarrow 0}\frac{d(\exp_p(tA),\exp_p(tB))}{t|A-B|}=\lim_{t\rightarrow 0}\frac{d(\exp_q(tA'),\exp_q(tB'))}{t|A-B|}=\frac{|A'-B'|}{|A-B|},
$$
因此$|d\phi(A)-d\phi(B)|=|A-B|,$ $\,\forall\,A,B\in T_pM.$
由Mazur-Ulam定理, 结合$d\phi(0)=0,$ 便说明$d\phi$是线性变换,
从而命题得证.
引理 2. $\lim\limits_{A,B\rightarrow 0\in T_pM}\frac{d(\exp_p(A),\exp_p(B))}{|A-B|}=1.$
证: 我们首先证明如下事实, 对曲线$c:I\rightarrow T_pM,$
若$|c'|$保持有界,
$$
\lim_{c(I)\rightarrow 0}\frac{l(\exp_p(c))}{l(c)}=1.
$$
我们有$l(c)=\int |c'|,$
$l(\exp_p(c))=\int |d\exp_p(c'(t))|_{\exp_p(c(t))}.$
由于$d\exp_p$在原点处为恒等变换, 同时$g$是光滑的,
$T_{\exp_p(c(t))}M$与$T_pM$度量在$c(I)\rightarrow 0$时充分接近,
$l(\exp_p(c))=l(c)+o(l(c)),$ 由此即知事实成立.
接下来, 记$\gamma$为连接$A,B$的曲线, 使得$\exp_p(\gamma)$为测地线;
$\xi$为连接$A,B$的直线. 那么
$$
|A-B|=l(\xi)\le l(\gamma),
$$
$$
d(\exp_p(A),\exp_p(B))=l(\exp_p(\gamma))\le l(\exp_p(\xi)).
$$
因此,
$$
\frac{l(\exp_p(\gamma))}{l(\gamma)}\le \frac{d(\exp_p(A),\exp_p(B))}{|A-B|}=\frac{l(\exp_p(\gamma))}{l(\xi)}\le \frac{l(\exp_p(\xi))}{l(\xi)}.
$$
当$A,B\rightarrow 0\in T_pM$时, 两侧趋于$1,$ 从而中间项也趋于$1.$
参考: https://ncatlab.org/nlab/show/Myers-Steenrod+theorem
文章最后更新于 2022-03-24 15:24:34
When I first saw the isometric transformation on the manifold, I naturally thought of maintaining the norm $\Leftrightarrow$ and maintaining the inner product.
Thus maintaining the Riemannian metric. But in fact isometric (according to the initial definition) maintains the distance on the manifold,
Rather than the distance in tangent space, this matter needs to be proved. The more important point is,
The aforementioned equivalence relationship only holds true for linear transformations.
The tangent mapping is a linear transformation based on the differentiability of the mapping.
In fact, we do not need to require the isometric transformation to be differentiable in advance, this is the Myers-Steenrod theorem.
Theorem 1. If $\phi:M\rightarrow N$ is an isometric surjection, then $\phi$ is isomorphic, especially it is smooth.
Certificate: Since geodesics have local extreme shortness, isometric transformation maps geodesics to geodesics.
Let $\phi(p)=q,$ select $v\in T_pM,$ and $|v|$ be sufficiently small.
Take the geodesic $\gamma,$ starting from point $p$ and taking $v$ as the initial tangent vector.
Then $d(p,\gamma(1))=|v|.$ since $\phi$ is an isometric transformation,
$d(q,\phi\circ\gamma(1))=|v|.$ If $\phi$ is differentiable,
Then since $\phi\circ\gamma$ is also a geodesic,
$|v|=|(\phi\circ\gamma)'(0)|=|d\phi(v)|.$
At this time $d\phi$ is a linear transformation that maintains the norm, so $d\phi$ maintains the inner product, that is, maintains the Riemannian metric,
Therefore $\phi$ is isomorphism.
When $\phi$ is not necessarily differentiable, since $\phi$ maintains the geodesic,
We can define $d\phi:T_pM\rightarrow T_pN,$ by the tangent vector of the geodesic
It also maintains the norm. But it is not necessarily a linear transformation at this time, but we can prove this.
Assume that the following conclusion is known:
$$
\lim_{A,B\rightarrow 0\in T_pM}\frac{d(\exp_p(A),\exp_p(B))}{|A-B|}=1.
$$
Select $A,B\in T_pM,$ and take the corresponding geodesic $\gamma_A,\gamma_B.$ and record $A'=d\phi(A),$
$B'=d\phi(B),$ and the corresponding geodesic $\gamma_{A'},\gamma_{B'},$
We have $d(\gamma_A(t),\gamma_B(t))=d(\gamma_{A'}(t),\gamma_{B'}(t)),$
Thus:
$$
1=\lim_{t\rightarrow 0}\frac{d(\exp_p(tA),\exp_p(tB))}{t|A-B|}=\lim_{t\rightarrow 0}\frac{d(\exp_q(tA'),\exp_q(tB'))}{t|A-B|}=\frac{|A'-B'|}{|A-B|},
$$
Therefore $|d\phi(A)-d\phi(B)|=|A-B|,$ $\,\forall\,A,B\in T_pM.$
According to the Mazur-Ulam theorem, combined with $d\phi(0)=0,$, it shows that $d\phi$ is a linear transformation,
Thus the proposition is proved.
Lemma 2. $\lim\limits_{A,B\rightarrow 0\in T_pM}\frac{d(\exp_p(A),\exp_p(B))}{|A-B|}=1.$
Certificate: We first prove the following fact, for the curve $c:I\rightarrow T_pM,$
If $|c'|$ remains bounded,
$$
\lim_{c(I)\rightarrow 0}\frac{l(\exp_p(c))}{l(c)}=1.
$$
We have $l(c)=\int |c'|,$
$l(\exp_p(c))=\int |d\exp_p(c'(t))|_{\exp_p(c(t))}.$
Since $d\exp_p$ is an identity transformation at the origin, and $g$ is smooth,
$T_{\exp_p(c(t))}M$ and $T_pM$ metrics are sufficiently close at $c(I)\rightarrow 0$,
$l(\exp_p(c))=l(c)+o(l(c)),$ From this we know that the facts are established.
Next, let $\gamma$ be the curve connecting $A,B$, making $\exp_p(\gamma)$ a geodesic;
$\xi$ is a straight line connecting $A,B$. Then
$$
|A-B|=l(\xi)\le l(\gamma),
$$
$$
d(\exp_p(A),\exp_p(B))=l(\exp_p(\gamma))\le l(\exp_p(\xi)).
$$
Therefore,
$$
\frac{l(\exp_p(\gamma))}{l(\gamma)}\le \frac{d(\exp_p(A),\exp_p(B))}{|A-B|}=\frac{l(\exp_p(\gamma))}{l(\xi)}\le \frac{l(\exp_p(\xi))}{l(\xi)}.
$$
When $A,B\rightarrow 0\in T_pM$, both sides tend to $1,$ and the middle term also tends to $1.$
Reference: https://ncatlab.org/nlab/show/Myers-Steenrod+theorem
The article was last updated on 2022-03-24 15:24:34
