用户: Solution/ 习题: Riemann几何

1.	Let $F:M^n\to {\mathbb{R}}^N$ be an immersion. $F$ induces a metric $F^{\ast }(d{s}^2)$ on $M$, as well as a connection $${\nabla }_Y^{M}X:=\sum _{i=1}^n⟨{\nabla }_{Y}X,e_i⟩e_i$$on $M$, where $e_1,\cdots ,e_n$ is a unitary frame on $M$. Verify that ${\nabla }^M$ is the Levi-Civita connection on $M$.
Pf.	The problem is a special case of Exercise 3 of Chapter 2, of which we use the notation and provide a proof here. That is mainly to check (i) let $f\in C^{\infty }(M)$ be extended arbitrarily to $\overline{M}$, using $\overline{X}f=Xf$, then $${\nabla }_X^{\top }(fY)=({\overline{\nabla }}_{\overline{X}}(f\overline{Y}){)}^{\top }=(f{\overline{\nabla }}_{\overline{X}}\overline{Y}{)}^{\top }+(\overline{X}f)\overline{Y}=f{\nabla }_X^{\top }Y+(Xf)Y.$$(ii) ${\nabla }^{\top }$ is also symmetric, i.e. (the last equality is since $[X,Y]$ is tangent to $M$ as long as $X,Y$ are) $${\nabla }_X^{\top }Y-{\nabla }_Y^{\top }X=({\overline{\nabla }}_{\overline{X}}\overline{Y}-{\overline{\nabla }}_{\overline{Y}}\overline{X}{)}^{\top }=[\overline{X},\overline{Y}{]}^{\top }=[X,Y].$$(iii) ${\nabla }^{\top }$ is compatible with the induced metric, i.e. $$\begin{array}{rl}X⟨Y,Z{⟩}_M& =\overline{X}⟨\overline{Y},\overline{Z}{⟩}_{\overline{M}}=⟨{\overline{\nabla }}_{\overline{X}}\overline{Y},\overline{Z}{⟩}_{\overline{M}}+⟨\overline{Y},{\overline{\nabla }}_{\overline{X}}\overline{Z}{⟩}_{\overline{M}}\\ & =⟨({\overline{\nabla }}_{\overline{X}}\overline{Y}{)}^{\top },Z{⟩}_M+⟨Y,({\overline{\nabla }}_{\overline{X}}\overline{Z}{)}^{\top }{⟩}_M\\ & =⟨{\nabla }_X^{\top }Y,Z{⟩}_M+⟨Y,{\nabla }_X^{\top }Z{⟩}_M.\end{array}$$
2.	Show that $R(X,Y)(fZ)=fR(X,Y)Z$.
Pf.	We have $$\begin{array}{rl}{\nabla }_Y{\nabla }_X(fZ)& ={\nabla }_Y(f{\nabla }_{X}Z+(Xf)Z)\\ & =f{\nabla }_Y{\nabla }_{X}Z+(Yf){\nabla }_{X}Z+(Xf){\nabla }_{Y}Z+(Y(Xf))Z.\end{array}$$Therefore, $${\nabla }_Y{\nabla }_X(fZ)-{\nabla }_X{\nabla }_Y(fZ)=f({\nabla }_Y{\nabla }_X-{\nabla }_X{\nabla }_Y)Z+((YX-XY)f)Z,$$hence $$\begin{array}{rl}R(X,Y)(fZ)& =f({\nabla }_Y{\nabla }_X-{\nabla }_X{\nabla }_Y)Z+([Y,X]f)Z\\ & +([X,Y]f)Z+f{\nabla }_{[X,Y]}Z=fR(X,Y)Z.\end{array}$$
3.	证明 [Xin] 第 7 章最后的附注 (P45).
Pf.	见 [Jo] Theorem 5.5.2 (P231-3).
4.	验证 $B^3$ 在双曲度量下的截面曲率是 $-1$.
Pf.	见 [Xin] 命题 10.2 (P53-4), 最后一步用到了 [dC] Lemma 3.4 (P96).
5.	在双曲空间的上半空间模型下, 计算它的截面曲率, 并确定它的测地线.
Pf.	见 [dC] Chapter 8 section 3 (P160-2).
6.	对于 [Xin] 第 11 章定义的 Laplace 算子 (P65), 验证其局部表达式为 (记 $g=\det (g_{ij})$) $$△f=\frac{1}{\sqrt{g}}{\partial }_i(\sqrt{g}{g}^{ij}{\partial }_{j}f).$$
Pf.	幺正坐标系下, [Xin] 与上式均有 $△f=\sum {\partial }_{ii}f$, 故说明上式不取决于坐标系的选取即可. 这只需注意到, 对任何局部坐标区域 $U$, 对于测试函数 $\varphi \in C_c^{\infty }(U)$, $${\int }_U△f\varphi \sqrt{g}dx={\int }_U{\partial }_i(\sqrt{g}{g}^{ij}{\partial }_{j}f)\varphi dx=-{\int }_U{g}^{ij}{\partial }_{j}f{\partial }_i\varphi \sqrt{g}dx,$$ $g^{ij}{\partial }_i\otimes {\partial }_j$ 是 $g_{ij}d{x}^i\otimes d{x}^j$ 的对偶, 所以在不同坐标系下是相同的; $\sqrt{g}dx$ 亦同坐标系无关. 再用 Riesz 表示定理, 得 $△f$ 在 $U$ 上无关乎坐标系.
7.	设 $M$ 是完备单连通的 Riemann 流形, 其截面曲率 $\le \kappa$, $\kappa >0$. 设 $o\in M$, 距离函数 $\rho (x)=d(o,x)$, 那么 ${\rho }^2(x)$ 在 $B_{\frac{\pi }{\sqrt{\kappa }}}(o)$ 上是光滑凸的.
Pf.	基本是 [Xin] 定理 11.3 的证明, 除了改成$$L^{\prime \prime }(0)\ge {\int }_0^r|{\dot{U}}^{\perp }{|}^2-\kappa |U^{\perp }{|}^{2}dt.$$上式对 $r\le \frac{\pi }{\sqrt{\kappa }}$ 非负, 是用 $U^{\perp }(r)=0$ 与 Wirtinger 不等式, 后者见 [dC] Exercise 11.2 (P250).
8.	设 Riemann 流形 $M$ 的截面曲率 $K$ 满足 $0<L\le K\le H$, 其中 $H$ 和 $L$ 是常数. 设 $\gamma$ 是 $M$ 中的测地线, 那么, $\gamma$ 上相邻共轭点沿着 $\gamma$ 的距离 $d$ 满足$$\frac{\pi }{\sqrt{H}}\le d\le \frac{\pi }{\sqrt{L}}.$$
Pf.	见 [dC] Chapter 10 Proposition 2.4 (P218).

Reference

[Xin]	忻元龙. 黎曼几何讲义. 研究生教学用书. 复旦大学出版社, 2010.
[dC]	M.P. do Carmo. Riemannian geometry. Mathematics: Theory & Applications. Boston: Birkhäuser, 1992.
[Jo]	J. Jost. Riemannian Geometry and Geometric Analysis. Universitext. Springer, 2011.