用户: Solution/ 习题: do Carmo

0Differentiable Manifolds

2.

Prove that the tangent bundle of a differentiable manifold is orientable (even though $M$ may not be). Given any smooth chart $\phi :U\subset M\to {\mathbb{R}}^n$, define $\tilde{\phi }:{\pi }^{-1}(U)\subset TM\to {\mathbb{R}}^{2n}$ by $$v{|}_p\mapsto (\phi (p),v).$$For any two overlapping charts $(U,\phi )$ and $(V,\psi )$, the transition map $\tilde{\psi }\circ {\tilde{\phi }}^{-1}:\phi (U\cap V)\times {\mathbb{R}}^n\to \psi (U\cap V)\times {\mathbb{R}}^n$ is given by $$(x,v)\mapsto (\tilde{x},\frac{\partial \tilde{x}}{\partial x}v),$$where $\psi \circ {\phi }^{-1}(x):=\tilde{x}(x)=({\tilde{x}}^1(x),\cdots ,{\tilde{x}}^n(x))$. Then the determinant is $(\det \frac{\partial \tilde{x}}{\partial x}{)}^2>0$.

1Riemannian Metrics

2.	In terms of $\tilde{\pi }:{\mathbb{R}}^n\stackrel{\pi }{⟶}{T}^n\hookrightarrow {\mathbb{R}}^{2n}$, which is,$$\tilde{\pi }(x_1,\cdots ,x_n)=(\cos x_1,\sin x_1,\cdots ,\cos x_n,\sin x_n),$$we have $d\tilde{\pi }=\mathrm{diag}((-\sin x_1,\cos x_1),\cdots ,(-\sin x_n,\cos x_n))$. So the induced metric on ${\mathbb{R}}^n$ is $$(-\sin x_{j}d{u}^j)\otimes (-\sin x_{j}d{u}^j)+(\cos x_{j}d{u}^j)\otimes (\cos x_{j}d{u}^j)=d{u}^j\otimes d{u}^j.$$
4.	(a) $$⟨-,-{⟩}_{(0,1)}=⟨d(g(t)=yt+x)-,d(g(t)=yt+x)-{⟩}_{(x,y)}=y^2⟨-,-{⟩}_{(x,y)}.$$(b) Check that $-\frac{4dzd\bar{z}}{(z-\bar{z}{)}^2}$ is invariant under $z\mapsto az+b$ and $z\mapsto 1/z$.
6.	Exercise 2 provides a local isometry $\pi :{\mathbb{R}}^n\to T^n$, but there is no local isometry $\rho :T^n\to {\mathbb{R}}^n$. Indeed, write $\rho =({\rho }_1,\cdots ,{\rho }_n)$. ${\rho }_1$ mapping the compact set $T^n$ to $\mathbb{R}$ attains its maximum at some $p\in T^n$, and $d{\rho }_1(p)=0\in {\mathbb{R}}^n$. Then $d\rho (p)=(0,\cdots )\in {\mathbb{R}}^{n\times n}$ is singular.

2Affine and Riemannian Connections

1.	Denote $P$ by $P_t$ instead. Since $$\frac{d}{dt}⟨P_{t}v,P_{t}w⟩=⟨\frac{D}{dt}(P_{t}v),P_{t}w⟩+⟨P_{t}v,\frac{D}{dt}(P_{t}w)⟩=0,$$$P$ is an isometry. Then $\det P_t\in \{1,-1\}$. Because $P_t$ is continuously dependent on $t$ and $P_{t_0}=\mathrm{id}$, we have $\det P_t$ is always $1$, and in particular positive.
2.	Proved in [Chen] Chapter 2 Theorem 7.2 (pp.142-3).
4.	a) According to Exercise 3, ${\nabla }_{\dot{c}}V$ is the tangential component of ${\overline{\nabla }}_{\dot{c}}V$, i.e. the orthonormal projection from $T_{c(t)}{\mathbb{R}}^3$ onto $T_{c(t)}M$, where $\overline{\nabla }$ is the Riemannian connection of ${\mathbb{R}}^3$. ${\overline{\nabla }}_{\dot{c}}V$ is simply the usual derivative $\frac{dV}{dt}$, so ${\nabla }_{\dot{c}}V=0$ iff the projection of $\frac{dV}{dt}$ to $T_{c(t)}M$ vanishes. b) The acceleration of the unit velocity field along a great circle, namely $\frac{dV}{dt}$, points to the center of the sphere, perpendicular to the tangent space embedded in ${\mathbb{R}}^3$ (or ${\mathbb{R}}^{n+1}$).
7.	Method I. Consider the cone $C$ tangent to $S^2$ along $c$, then the parallel transport of $V_0$ along $c$ is the same, whether taken relative to $S^2$ or to $C$, as the tangent spaces along $c$ whether taken relative to $S^2$ or to $C$ are the same. Flatten the cone (locally, or at least minus a line) to a sector on the plane (in other words, $C$ is locally isometric to a sector in ${\mathbb{R}}^2$ with $c$ being its arc). The parallel transport of $V_0$ on the plane is merely translation in ${\mathbb{R}}^2$. From the perspective of one walking along $c$, $V_0$ appears to be rotating at constant speed. For details, refer to [dC] 4.4 Example 1 (p.246). 運命のルーレット廻して、ずっと君を見ていた. Method II. Consider the cone $C$ tangent to $S^2$ along $c$, parametrized as $$\{(av\cos u,av\sin u,bv)\in {\mathbb{R}}^3:u,v\in \mathbb{R}\},$$where $a^2+b^2=1$. The parallel transport of $V_0$ along $c$ is the same, whether taken relative to $S^2$ or to $C$, as the tangent spaces along $c$ whether taken relative to $S^2$ or to $C$ are the same. Now we carry out a bit of calculations to see the effect of parallel–translating $V_0$ along $c$, written $U(t){\partial }_u+V(t){\partial }_v$. Besides, let $u(t)=t,v(t)=v_0$ be the local expression for $c(t)$. First, $$g_{uu}=a^2{v}^2,g_{uv}=0,g_{vv}=1,$$and then $${\Gamma }_{uu}^u=0,{\Gamma }_{vu}^u=\frac{1}{v},{\Gamma }_{uu}^v=-a^{2}v,{\Gamma }_{vu}^v=0.$$Substituting into $$\{\begin{array}{r}\dot{U}+({\Gamma }_{uu}^u\dot{u}+{\Gamma }_{uv}^u\dot{v})U+({\Gamma }_{vu}^u\dot{u}+{\Gamma }_{vv}^u\dot{v})V=0,\\ \dot{V}+({\Gamma }_{uu}^v\dot{u}+{\Gamma }_{uv}^v\dot{v})U+({\Gamma }_{vu}^v\dot{u}+{\Gamma }_{vv}^v\dot{v})V=0,\end{array}$$where $\dot{u}=1,\dot{v}=0$, we get $$\{\begin{array}{r}\dot{U}+\frac{1}{v}V=0,\\ \dot{V}-a^{2}vU=0.\end{array}$$Hence, $$\{\begin{array}{rl}& U=c\cos (at+d),\\ & V=a{v}_0\sin (at+d)\end{array}$$for two constants $c,d$. We conclude that the vector rotates at angular velocity $a$, if the angular velocity of $c$ along the parallel is taken to be $1$.

3Geodesics; Convex Neighborhoods

1.	a) The induced metric is $$(d(f(v)\cos u){)}^2+(d(f(v)\sin u){)}^2+(d(g(v)){)}^2=f^{2}d{u}^2+(f^{\prime 2}+g^{\prime 2})d{v}^2.$$b) The inverse is $$g^{11}=\frac{1}{f^2},g^{12}=0,g^{22}=\frac{1}{f^{\prime 2}+g^{\prime 2}}.$$Then the coefficients in these differential equations are $$\begin{array}{rl}{\Gamma }_{12}^1+{\Gamma }_{21}^1& =2{\Gamma }_{12}^1=g^{k1}({\partial }_1{g}_{2k}+{\partial }_2{g}_{1k}-{\partial }_k{g}_{12})=g^{11}{\partial }_2{g}_{11}=\frac{2f{f}^{\prime }}{f^2},\\ {\Gamma }_{11}^2& =\frac{1}{2}{g}^{k2}({\partial }_1{g}_{1k}+{\partial }_1{g}_{1k}-{\partial }_k{g}_{11})=-\frac{1}{2}{g}^{22}{\partial }_2{g}_{11}=-\frac{f{f}^{\prime }}{f^{\prime 2}+g^{\prime 2}},\\ {\Gamma }_{22}^2& =\frac{1}{2}{g}^{k2}({\partial }_2{g}_{2k}+{\partial }_2{g}_{2k}-{\partial }_k{g}_{22})=\frac{1}{2}{g}^{22}{\partial }_2{g}_{22}=\frac{f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime }}{f^{\prime 2}+g^{\prime 2}}.\end{array}$$c) Let $$\gamma (t)=(f(v(t))\cos u(t),f(v(t))\sin u(t),g(v(t))),$$then $$|{\gamma }^{\prime }(t){|}^2=f^2{u}^{\prime 2}+(f^{\prime 2}+g^{\prime 2})v^{\prime 2}.$$Thus by the first (which gives $f^2{u}^{\prime \prime }=-2f{f}^{\prime }{u}^{\prime }{v}^{\prime }$) and the second equation subsequently, $$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}|{\gamma }^{\prime }(t){|}^2& =f^2{u}^{\prime }{u}^{\prime \prime }+f{f}^{\prime }{v}^{\prime }{u}^{\prime 2}+(f^{\prime 2}+g^{\prime 2})v^{\prime }{v}^{\prime \prime }+(f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime })v^{\prime 3}\\ & =-2f{f}^{\prime }{u}^{\prime 2}{v}^{\prime }+f{f}^{\prime }{v}^{\prime }{u}^{\prime 2}+(f^{\prime 2}+g^{\prime 2})v^{\prime }{v}^{\prime \prime }+(f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime })v^{\prime 3}\\ & =-f{f}^{\prime }{v}^{\prime }{u}^{\prime 2}+(f^{\prime 2}+g^{\prime 2})v^{\prime }{v}^{\prime \prime }+(f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime })v^{\prime 3}\\ & =v^{\prime }(f^{\prime 2}+g^{\prime 2})(-\frac{f{f}^{\prime }}{f^{\prime 2}+g^{\prime 2}}{u}^{\prime 2}+v^{\prime \prime }+\frac{f^{\prime }{f}^{\prime \prime }+g^{\prime }{g}^{\prime \prime }}{f^{\prime 2}+g^{\prime 2}}{v}^{\prime 2})=0.\end{array}$$We have $$r=\text{radius of }\{(f(v)\cos u,f(v)\sin u,g(v)):u\}=f(v),$$and ${\gamma }^{\prime }=u^{\prime }{\partial }_1+v^{\prime }{\partial }_2$, $$r\cos \beta =r\frac{⟨{\gamma }^{\prime },{\partial }_1⟩}{|{\gamma }^{\prime }||{\partial }_1|}=f\frac{u^{\prime }{f}^2}{|{\gamma }^{\prime }|f}=\frac{u^{\prime }{f}^2}{|{\gamma }^{\prime }|}.$$That $|{\gamma }^{\prime }|$ is constant has been verified, and $$(u^{\prime }{f}^2{)}^{\prime }=u^{\prime \prime }{f}^2+u^{\prime }2f{f}^{\prime }{v}^{\prime }=0$$shows that $u^{\prime }{f}^2$ is constant as well. d) Click. Or cf. [Chen] pp.465-6.
2.	a) The inner product being well–defined is because $$d\pi (V)=p^{\prime }(0),\frac{Dv}{dt}(0)=v^{\prime }(0)+p_i^{\prime }(0)v^{j\prime }(0){\nabla }_{{\partial }_i}{\partial }_j$$depends only on $V=(p^{\prime }(0),v^{\prime }(0))=(p_i^{\prime }(0),v^{j\prime }(0))$, not essentially on $\alpha$. To show that it is a Riemannian metric, properties like symmetry, bilinearity, non-negativity and smooth dependence are all obvious, and $⟨V,V⟩=0$ implies $p^{\prime }(0)=\frac{Dv}{dt}(0)=0$, which further implies $v^{\prime }(0)=0$ by the expression of $\frac{Dv}{dt}(0)$ in local coordinates above. b) We have the decomposition $$T_{(p,v)}TM\simeq T_v(T_{p}M)\oplus {\pi }^{-1}(p)\simeq T_{p}M\oplus T_{p}M,$$and $(p^{\prime }(t),v^{\prime }(t))\perp {\pi }^{-1}(p)$ iff $\frac{Dv}{dt}=0$, i.e. $v$ is parallel along $p$. c) Simply because ${\gamma }^{\prime }$ is parallel along $\gamma$. d) The hint is a solution indeed. e) With respect to the decomposition given in b), $W=d\pi (W)+0$ if $W$ is horizontal, and $W=0+W=0+\frac{Dw}{dt}(0)$ if $W$ is vertical.
4.	a) Write $A$ for the convex neighborhood and $a$ for any of its points. By definition, for any $b\in A$, there is a unique geodesic ${\gamma }_b:[0,1]\to A$ joining $a$ and $b$. Then $h:[0,1]\times A\to A,(t,b)\mapsto {\exp }_a(t{\gamma }_b^{\prime })$ is a homotopy between $k_a$ and ${\mathrm{id}}_A$. b) For any $p\in M$ pick a convex neighborhood. A finite union of convex neighborhoods is again convex (check by definition), hence contractible.
7.	Let $U$ be a normal neighborhood of $p$ in $M$, that is, ${\exp }_p:V\to U$ is a diffeomorphism. Define $E_i({\exp }_p(v))=(d{\exp }_p{)}_v({\partial }_i)$, where $v\in V\subset T_{p}M$, ${\partial }_i=\frac{\partial }{\partial x_i}$, and $x_i,i=1,\cdots ,n$ are the standard Euclidean coordinates on $T_{p}M\simeq {\mathbb{R}}^n$. In other words, $(d{\exp }_p{)}_v$ maps ${\partial }_i$ at $v$ to $E_i$ at ${\exp }_p(v)$. For any $v=(v_i{)}_{i=1,\cdots ,n}\in V$, $\gamma (t)={\exp }_p(tv),t\in [0,1]$ is a geodesic. Then $({\nabla }_{{\gamma }^{\prime }}{\gamma }^{\prime })({\exp }_p(tv))=0$, where ${\gamma }^{\prime }=({\exp }_p(tv){)}^{\prime }=\sum _i{v}_i{E}_i({\exp }_p(tv))$. In particular let $t=0$, we get $({\nabla }_{{\gamma }^{\prime }}{\gamma }^{\prime })(p)=0$ where ${\gamma }^{\prime }=\sum _i{v}_i{E}_i(p)$, and consequently ${\nabla }_{E_i}{E}_j(p)=0$ by the arbitrariness of $v$.
8.	a) The first is because $$⟨\mathrm{grad}f,\sum _i{v}_i{E}_i⟩=df(\sum _i{v}_i{E}_i)=\sum _i{v}_i{E}_i(f)=⟨\sum _i{E}_i(f)E_i,\sum _i{v}_i{E}_i⟩.$$For the second, for $Y=\sum _j{g}_j{E}_j$, the trace of the mapping $$\sum _j{g}_j{E}_j\mapsto {\nabla }_{\sum _j{g}_j{E}_j}\sum _i{f}_i{E}_i=\sum _{i,j}{g}_j{E}_j(f_i)E_i$$is $\mathrm{div}X=\sum _i{E}_i{f}_i$. b) Apply a), as $\frac{\partial }{\partial x_1},\cdots ,\frac{\partial }{\partial x_n}$ is a geodesic frame.
10.	Differentiate $⟨\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}⟩$ with respect to $s$, obtaining $$\begin{array}{rl}\frac{d}{ds}⟨\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}⟩& =⟨\frac{D}{ds}\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}⟩+⟨\frac{\partial f}{\partial s},\frac{D}{ds}\frac{\partial f}{\partial t}⟩\\ & =0+⟨\frac{\partial f}{\partial s},\frac{D}{dt}\frac{\partial f}{\partial s}⟩\\ & =\frac{1}{2}\frac{d}{dt}⟨\frac{\partial f}{\partial s},\frac{\partial f}{\partial s}⟩=0,\end{array}$$we get $$⟨\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}⟩(s_0,t_0)=⟨\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}⟩(0,t_0)=0.$$

4Curvature

1.	a) Combine $${\nabla }_{X}Y+{\nabla }_{Y}X={\nabla }_{(X+Y)}(X+Y)-{\nabla }_{X}X-{\nabla }_{Y}Y=0$$and $${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y].$$ b) Using a) and the Jacobi identity, $$\begin{array}{rl}R(X,Y)Z& ={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z+{\nabla }_{[X,Y]}Z\\ & =\frac{1}{2}({\nabla }_Y[X,Z]-{\nabla }_X[Y,Z]+[[X,Y],Z])\\ & =\frac{1}{4}([Y,[X,Z]]-[X,[Y,Z]])+\frac{1}{2}[[X,Y],Z]\\ & =\frac{1}{4}[[X,Y],Z].\end{array}$$ c) By the equation (3) of Chapter 1 section 2 (p.40), we have $$\begin{array}{rl}K(\sigma )& =⟨R(X,Y)X,Y⟩=\frac{1}{4}⟨[[X,Y]X],Y⟩\\ & =-\frac{1}{4}⟨[X,Y],[Y,X]⟩=\frac{1}{4}\Vert [X,Y]{\Vert }^2.\end{array}$$
2.	a) $0=Z⟨X,X⟩=2⟨A_X(Z),X⟩$. b) Put $S$ to be the right side, that is to say, $$S=Z⟨{\nabla }_{Z}X,X⟩+⟨{\nabla }_Z{\nabla }_{X}X,Z⟩-⟨{\nabla }_X{\nabla }_{Z}X,Z⟩+⟨{\nabla }_{[X,Z]}X,Z⟩.$$Using the Killing equation $⟨{\nabla }_{Z}X,X⟩+⟨{\nabla }_{X}X,Z⟩=0$, the sum of first two terms becomes $$Z⟨{\nabla }_{Z}X,X⟩+⟨{\nabla }_Z{\nabla }_{X}X,Z⟩=-Z⟨{\nabla }_{X}X,Z⟩+⟨{\nabla }_Z{\nabla }_{X}X,Z⟩=-⟨{\nabla }_{X}X,{\nabla }_{Z}Z⟩;$$note that $⟨{\nabla }_{Z}X,Z⟩=0$, implying $X⟨{\nabla }_{Z}X,Z⟩=0$, and then the third term becomes $$-⟨{\nabla }_X{\nabla }_{Z}X,Z⟩=⟨{\nabla }_{Z}X,{\nabla }_{X}Z⟩;$$the fourth term is $$⟨{\nabla }_{[X,Z]}X,Z⟩=-⟨{\nabla }_{Z}X,[X,Z]⟩.$$Adding together, we get $$S=-⟨{\nabla }_{X}X,{\nabla }_{Z}Z⟩+⟨{\nabla }_{Z}X,{\nabla }_{Z}X⟩.$$Because of the Killing equation at $p$ and a), $$0=(⟨{\nabla }_{X}X,Y⟩+⟨{\nabla }_{Y}X,X⟩)(p)=⟨{\nabla }_{X}X,Y⟩(p)$$for any $Y$, it is inferred that ${\nabla }_{X}X(p)=0$, and therefore $$S(p)=⟨{\nabla }_{Z}X,{\nabla }_{Z}X⟩(p).$$
3.	That $A$ is anti–symmetric is simply $$⟨A(y),z⟩+⟨y,A(z)⟩=⟨{\nabla }_{Y}X,Z⟩+⟨Y,{\nabla }_{Z}X⟩=0,$$and that $A$ is an isomorphism is because by 2.b), $$⟨A(y),z⟩+⟨y,A(z)⟩(p)=\frac{1}{2}{Z}_p(Zf)+(X,Z,X,Z)(p)>0.$$As $-A=A^T$, taking determinant on both sides gives $$(-1{)}^{\dim E}\det A=\det A.$$This implies $\dim E=\dim M-1$ is even, which is a contradiction to $\dim M$ being even.
6.	a) $\nabla R=0$ implies that for any $W$, $$\begin{array}{rl}{\gamma }^{\prime }(R(X,Y,Z,W))& =R({\nabla }_{{\gamma }^{\prime }}X,Y,Z,W)+R(X,{\nabla }_{{\gamma }^{\prime }}Y,Z,W)\\ & +R(X,Y,{\nabla }_{{\gamma }^{\prime }}Z,W)+R(X,Y,Z,{\nabla }_{{\gamma }^{\prime }}W).\end{array}$$Substituting ${\nabla }_{{\gamma }^{\prime }}X={\nabla }_{{\gamma }^{\prime }}Y={\nabla }_{{\gamma }^{\prime }}Z=0$, we get $${\gamma }^{\prime }(R(X,Y,Z,W))=R(X,Y,Z,{\nabla }_{{\gamma }^{\prime }}W),$$which is $⟨{\nabla }_{{\gamma }^{\prime }}(R(X,Y)Z),W⟩=0$. b) For any curve $\gamma$, let $X,Y$ be two orthonormal parallel unit vector fields along $\gamma$. Then we have $$\frac{d}{dt}(R(X,Y,X,Y)(\gamma (t)))=⟨{\nabla }_{{\gamma }^{\prime }}(R(X,Y)X),Y⟩+⟨R(X,Y)X,{\nabla }_{{\gamma }^{\prime }}Y⟩=0.$$To put it another way, $R(X,Y,X,Y)$ is constant along $\gamma$. Therefore $R(X,Y,X,Y)$ is constant over $M$. c) Recall that $M$ has constant sectional curvature $K_0$ means $$R(X,Y,Z,W)=K_0(⟨X,W⟩⟨Y,Z⟩-⟨Y,W⟩⟨X,Z⟩).$$Then $\nabla R(X,Y,Z,W,V)=0$, i.e. $$\begin{array}{rl}V(R(X,Y,Z,W))& =R({\nabla }_{V}X,Y,Z,W)+R(X,{\nabla }_{V}Y,Z,W)\\ & +R(X,Y,{\nabla }_{V}Z,W)+R(X,Y,Z,{\nabla }_{V}W),\end{array}$$becomes the identity $$\begin{array}{rl}& V(⟨X,W⟩⟨Y,Z⟩-⟨Y,W⟩⟨X,Z⟩)\\ & =⟨{\nabla }_{V}X,W⟩⟨Y,Z⟩-⟨Y,W⟩⟨{\nabla }_{V}X,Z⟩\\ & +⟨X,W⟩⟨{\nabla }_{V}Y,Z⟩-⟨{\nabla }_{V}Y,W⟩⟨X,Z⟩\\ & +⟨X,W⟩⟨Y,{\nabla }_{V}Z⟩-⟨Y,W⟩⟨X,{\nabla }_{V}Z⟩\\ & +⟨X,{\nabla }_{V}W⟩⟨Y,Z⟩-⟨Y,{\nabla }_{V}W⟩⟨X,Z⟩.\end{array}$$

5Jacobi Fields

3.	Assume that $a\in C(p)$, then the existence of a non-trivial Jacobi field $J$ along the geodesic $\gamma :[0,a]\to M$, with $\gamma (0)=p,J(0)=J(a)=0$. Using the Jacobi equation, we have $$\frac{d}{dt}⟨\frac{DJ}{dt},J⟩=-({\gamma }^{\prime },J,{\gamma }^{\prime },J)+\Vert \frac{DJ}{dt}{\Vert }^2\ge 0.$$But $⟨\frac{DJ}{dt},J⟩=0$ at $t=0,a$, then $⟨\frac{DJ}{dt},J⟩\equiv 0$. Since $\frac{d}{dt}⟨J,J⟩=2⟨\frac{DJ}{dt},J⟩\equiv 0$, we get $\Vert J{\Vert }^2\equiv \text{constant}$, and hence $\equiv 0$, a contradiction.
4.	From Example 2.3, the Jacobi field $J_1$ along $\gamma$ satisfying $J_1(0)=0,J_1^{\prime }(0)=\frac{u_0}{|u_0|}$, is given by $$J_1(t)=\frac{\sin (t\sqrt{-b})}{\sqrt{-b}}w(t).$$In addition, from Corollary 2.5, $$J_1(\ell )=(d{\exp }_p{)}_{\ell {\gamma }^{\prime }(0)}(\ell w(0)).$$It follows that $$J(\ell )=v=(d{\exp }_p{)}_{\ell {\gamma }^{\prime }(0)}(u_0)=J_1(\ell )\frac{|u_0|}{|\ell w(0)|}=J_1(\ell )\frac{|u_0|}{\ell }.$$In addition, since $|w(t)|=|w(0)|=1$, $$1=|v|=|J(\ell )|=|J_1(\ell )|\frac{|u_0|}{\ell }=\frac{\sin (\ell \sqrt{-b})}{\sqrt{-b}}\frac{|u_0|}{\ell },$$we have $$\frac{|u_0|}{\ell }=(\frac{\sin (\ell \sqrt{-b})}{\sqrt{-b}}{)}^{-1},$$which implies what was asserted.
6.	a) By Proposition 2.9, ${\exp }_p:B_{\delta }(0)\subset T_{p}M\to B_{\delta }(p)\subset M$ is a diffeomorphism onto an open subset $\bar{U}$ of $M$. To make the polar coordinates $(\rho ,\theta )$ into a coordinate system indeed, an entire radius $-\rho v(0),0\le \rho <\delta$ in $B_{\delta }(0)$ must be removed, and correspondingly the ray ${\exp }_p(-\rho v(0)),0\le \rho <\delta$ in $\bar{U}$, forming the required $U$. b) $f(0,\theta ),-\pi <\theta <\pi$ is the point $p$ instead of a curve, so $g_{12}(0,\theta )=0$. Considering the unit–speed geodesic $f(\rho ,{\theta }_0),0<\rho <\delta$ where $\rho$ varies and ${\theta }_0$ is fixed, we get $g_{11}=1$, and ${\nabla }_{v}v=0$ for $v=\frac{\partial f}{\partial \rho }(\rho ,{\theta }_0)$, the latter implying $$2g_{12,1}-g_{11,2}=0.$$So $g_{12,1}=0$, and consequently $g_{11}=0$. $g_{22}=|\frac{\partial f}{\partial \theta }{|}^2$ is too simple to be explained. c) On one hand, according to b) and Corollary 2.10, $$\sqrt{g_{22}}=\left|\frac{\partial f}{\partial \theta }\right|=|J(\rho )|=\rho -\frac{1}{6}K(p){\rho }^3+o({\rho }^3);$$on the other hand, the Taylor expansion yields $$\sqrt{g_{22}}=[\sqrt{g_{22}}+(\sqrt{g_{22}}{)}_{\rho }\rho +\frac{1}{2}(\sqrt{g_{22}}{)}_{\rho \rho }{\rho }^2+\frac{1}{6}(\sqrt{g_{22}}{)}_{\rho \rho \rho }{\rho }^3]{|}_{(0,0)}+o({\rho }^3).$$Comparing the coefficients, we get $$\sqrt{g_{22}}=0,(\sqrt{g_{22}}{)}_{\rho }=1,(\sqrt{g_{22}}{)}_{\rho \rho }=0,(\sqrt{g_{22}}{)}_{\rho \rho \rho }=-K(p)$$at $(0,0)$, and then $$(\sqrt{g_{22}}{)}_{\rho \rho }=[(\sqrt{g_{22}}{)}_{\rho \rho }+(\sqrt{g_{22}}{)}_{\rho \rho \rho }\rho ]{|}_{(0,0)}+o(\rho )=-K(p)\rho +o(\rho ).$$ d)$$\underset{\rho \to 0}{\lim }\frac{(\sqrt{g_{22}}{)}_{\rho \rho }}{\sqrt{g_{22}}}=\underset{\rho \to 0}{\lim }\frac{-K(p)\rho +o(\rho )}{\rho +o(\rho )}=-K(p).$$
7.	Using Exercise 6 c), we have $$L_r={\int }_0^{2\pi }\left|\frac{\partial f}{\partial \theta }\right|d\theta ={\int }_0^{2\pi }(-K(p)\rho +o({\rho }^3))d\theta =2\pi r-\frac{\pi }{3}K(p)r^3+o(r^3).$$

6Isometric Immersions

1.	a) Trivially verified by definition or local expressions. b) Suppose that ${\gamma }_p=\{p\}\times \gamma$ is a geodesic of $(M_2{)}_p$, naturally identified with $\gamma$ which is a geodesic of $M_2$. That is to say, ${\nabla }_{{\gamma }^{\prime }}^2{\gamma }^{\prime }=0$. Then using a), we too have ${\nabla }_{{\gamma }_p^{\prime }}{\gamma }_p^{\prime }={\nabla }_0^{1}0+{\nabla }_{{\gamma }^{\prime }}^2{\gamma }^{\prime }=0$. c) Easy to see that $x,y$ commute, and so do ${\nabla }_x^1,{\nabla }_y^2$. Hence $R(x,y)=[{\nabla }_x,{\nabla }_y]-{\nabla }_{[x,y]}=0$, and then $K(\sigma )=0$.
2.	$d{\mathbf{x}}_p:T_p{\mathbb{R}}^2\to T_{\mathbf{x}(p)}{\mathbb{R}}^4$, given by $$\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-\sin \theta & 0\\ \cos \theta & 0\\ 0& -\sin \phi \\ 0& \cos \phi \end{array}\right),$$is injective, and the composite $\pi \circ d{\mathbf{x}}_p:T_p{\mathbb{R}}^2\to T_{\mathbf{x}(p)}{\mathbb{R}}^4\to T_{\mathbf{x}(p)}{S}^3$ ($\pi$ is the projection) is also injective, since the vector $d{\mathbf{x}}_p(v)$ is perpendicular to $\overrightarrow{Op}=\frac{1}{\sqrt{2}}(\cos \theta ,\sin \theta ,\cos \phi ,\sin \phi )$, and as a result, $d{\mathbf{x}}_p(v)$ lies in $T_{\mathbf{x}(p)}{S}^3$ for $v\in T_p{\mathbb{R}}^2$ and then $\pi =\mathrm{id}$. The zero sectional curvature is simply from $\mathbf{x}({\mathbb{R}}^2)=S^1\left(\frac{1}{\sqrt{2}}\right)\times S^1\left(\frac{1}{\sqrt{2}}\right)$ and Exercise 1.c).
3.	Use the fact that $N$ is totally geodesic in $M$ iff all geodesics in $N$ are also geodesics in $M$.
4.	For $x=x_1+x_2,y=y_1+y_2\in T_{p_1}{N}_1\oplus T_{p_2}{N}_2$, from Exercise 1.a) we have $$B(x,y)={\overline{\nabla }}_{\overline{X}}\overline{Y}-{\nabla }_{X}Y={\overline{\nabla }}_{{\overline{X}}_1}^1{\overline{Y}}_1+{\overline{\nabla }}_{{\overline{X}}_2}^2{\overline{Y}}_2-{\nabla }_{X_1}^1{Y}_1-{\nabla }_{X_2}^2{Y}_2=B_1(x_1,y_1)+B_2(x_2,y_2).$$Then for $\eta ={\eta }_1+{\eta }_2\in (T_{p_1}{N}_1{)}^{\top }\oplus (T_{p_2}{N}_2{)}^{\top }$, we have$$H_{\eta }(x,y)=H_{{\eta }_1}(x_1,y_1)+H_{{\eta }_2}(x_2,y_2),$$so $H_{{\eta }_1},H_{{\eta }_2}$ vanish implies that $H_{\eta }$ vanishes.
5.	Let $X_1+Y_1,X_2+Y_2\in \mathcal{X}(S^2)\oplus \mathcal{X}(S^2)=\mathcal{X}(S^2\times S^2)$, then $$R(X_i,X_j,X_k,Y_{\ell })=R(X_i,X_j,Y_k,Y_{\ell })=R(X_i,Y_j,Y_k,Y_{\ell })=0$$for all $i,j,k,\ell \in \{1,2\}$. The reason is that $g_{X_i{Y}_j}=0$ for all $i,j\in \{1,2\}$, and thus ${\nabla }_{X_i}{Y}_j=0$. Expand the curvature below, we have $$R(X_1+Y_1,X_2+Y_2,X_1+Y_1,X_2+Y_2)=R(X_1,X_2,X_1,X_2)+R(Y_1,Y_2,Y_1,Y_2)\ge 0$$as $K(S^2)=1>0$. Let $i$ be the inclusion from $S^1$ into an equator of $S^2$. By Exercise 4, $(i,i):T^2=S^1\times S^1\hookrightarrow S^2\times S^2$ is totally geodesic.
7.	For any $p\in M$ and $\eta \in (T_{p}M{)}^{\perp }$, $H_{\eta }(X,Y)=⟨B(X,Y),\eta ⟩\equiv 0$, which is to say, $B(X,Y)={\overline{\nabla }}_{\overline{X}}\overline{Y}-{\nabla }_{X}Y\equiv 0$.
8.	a) Note that $e_1=\sqrt{2}\frac{\partial \mathbf{x}}{\partial \theta },e_2=\sqrt{2}\frac{\partial \mathbf{x}}{\partial \phi }$, and therefore $e_1,e_2\in T_p{T}^2$. And check without effort that $e_1,e_2,n_1,n_2$ form an orthonormal basis of $T_p{\mathbb{R}}^4$. b) Extend $e_1\in T_p{T}^2$ to $\sqrt{2}(-x_2{\partial }_1+x_1{\partial }_2)\in T_p{\mathbb{R}}^2$, and $e_2\in T_p{T}^2$ to $\sqrt{2}(-x_4{\partial }_3+x_3{\partial }_4)\in T_p{\mathbb{R}}^2$. As ${\overline{\nabla }}_{X}Y=\sum _{k}X(y_k){\partial }_k$ for $Y=\sum _k{y}_k{\partial }_k$ in ${\mathbb{R}}^4$, we have $$\begin{array}{rl}{\overline{\nabla }}_{e_1}{e}_1& =2([(-x_2{\partial }_1+x_1{\partial }_2)(-x_2)]{\partial }_1+[(-x_2{\partial }_1+x_1{\partial }_2)(x_1)]{\partial }_2)\\ & =2(-x_1{\partial }_1-x_2{\partial }_2)=-n_1+n_2,\end{array}$$and ${\overline{\nabla }}_{e_1}{e}_2=0$, ${\overline{\nabla }}_{e_2}{e}_1=0$, ${\overline{\nabla }}_{e_2}{e}_2=2(-x_3{\partial }_3-x_4{\partial }_4)=-n_1-n_2$. Then $$\begin{array}{l}⟨S_{n_1}{e}_1,e_1⟩=⟨{\overline{\nabla }}_{e_1}{e}_1,n_1⟩=-1,\\ ⟨S_{n_1}{e}_1,e_2⟩=⟨{\overline{\nabla }}_{e_1}{e}_2,n_1⟩=0\end{array}\}\Rightarrow S_{n_1}{e}_1=-e_1;$$$$\begin{array}{l}⟨S_{n_1}{e}_2,e_1⟩=⟨{\overline{\nabla }}_{e_2}{e}_1,n_1⟩=0,\\ ⟨S_{n_1}{e}_2,e_2⟩=⟨{\overline{\nabla }}_{e_2}{e}_2,n_1⟩=-1\end{array}\}\Rightarrow S_{n_1}{e}_2=-e_2;$$$$\begin{array}{l}⟨S_{n_2}{e}_1,e_1⟩=⟨{\overline{\nabla }}_{e_1}{e}_1,n_2⟩=1,\\ ⟨S_{n_2}{e}_1,e_2⟩=⟨{\overline{\nabla }}_{e_1}{e}_2,n_2⟩=0\end{array}\}\Rightarrow S_{n_2}{e}_1=e_1;$$$$\begin{array}{l}⟨S_{n_2}{e}_2,e_1⟩=⟨{\overline{\nabla }}_{e_2}{e}_1,n_2⟩=0,\\ ⟨S_{n_2}{e}_2,e_2⟩=⟨{\overline{\nabla }}_{e_2}{e}_2,n_2⟩=-1\end{array}\}\Rightarrow S_{n_2}{e}_2=-e_2.$$c) Since $T_p{S}^3(1)\oplus \mathbb{R}{n}_1=T_p{\mathbb{R}}^4$ and $T_p{T}^2\oplus \mathbb{R}{n}_2=T_p{S}^3(1)$, the mean curvature of $\mathbf{x}$ is $\mathrm{trace}{S}_{n_2}=1-1=0$.
9.	Follow the hint, and a visualization is offered here.
10.	Differentiating $⟨S_{\eta }X,Y⟩=⟨X,S_{\eta }Y⟩$ with respect to $V$, we obtain $$⟨{\nabla }_V(S_{\eta }X),Y⟩+⟨S_{\eta }X,{\nabla }_{V}Y⟩=⟨{\nabla }_{V}X,S_{\eta }Y⟩+⟨X,{\nabla }_V(S_{\eta }Y)⟩.$$Using the fact that $$\begin{array}{rl}({\nabla }_V{S}_{\eta })(X,Y)& =(\nabla S_{\eta })(X,Y,V)\\ & =V(S_{\eta }(X,Y))-S_{\eta }(X,{\nabla }_{V}Y)-S_{\eta }({\nabla }_{V}X,Y)\\ & =V⟨S_{\eta }X,Y⟩-⟨S_{\eta }X,{\nabla }_{V}Y⟩-⟨{\nabla }_{V}X,S_{\eta }Y⟩\\ & =⟨{\nabla }_V(S_{\eta }X),Y⟩-⟨{\nabla }_{V}X,S_{\eta }Y⟩\end{array}$$and the previous expression, we obtain that $$({\nabla }_V{S}_{\eta })(X,Y)=({\nabla }_V{S}_{\eta })(Y,X).$$
11.	a) By definition, $$\begin{array}{rl}\overline{△}f& =\mathrm{div}\mathrm{grad}f\\ & =\text{trace of the linear mapping }Y\mapsto {\nabla }_Y\mathrm{grad}f\\ & =\text{trace }\mathrm{Hess}f.\end{array}$$b) Use ${\overline{\nabla }}_{X}Y-{\overline{\nabla }}_{Y}X=XY-YX$ and $$\begin{array}{rl}⟨(\mathrm{Hess}f)Y,X⟩& =⟨{\overline{\nabla }}_Y\mathrm{grad}f,X⟩\\ & =Y⟨\mathrm{grad}f,X⟩-⟨\mathrm{grad}f,{\overline{\nabla }}_{Y}X⟩\\ & =Y(df(X))-df({\overline{\nabla }}_{Y}X)\\ & =(YX-{\overline{\nabla }}_{Y}X)f.\end{array}$$c) Follow the hint, where $⟨{\overline{\nabla }}_{\eta }\eta ,\eta ⟩=\eta ⟨\eta ,\eta ⟩/2=0$. d) We need to verify that $\mathrm{grad}f(p)\perp T_{p}M$. In fact, for any $v\in T_{p}M$, let $\gamma$ be a curve in $M$ with $\gamma (0)=p,{\gamma }^{\prime }(0)=v$, and then $f\circ \gamma =0$. It follows that $⟨\mathrm{grad}f(p),v⟩=d{f}_p(v)=\frac{d}{dt}{|}_{t=0}(f\circ \gamma )=0$.

7Complete Manifolds; Hopf–Rinow and Hadamard Theorems

1.	Take $M$ to be a non–convex subset of $N={\mathbb{R}}^n$.
7.	For any curve $\gamma$ joining $p,q\in M$, $f\circ \gamma$ is a curve joining $f(p),f(q)\in \bar{M}$, then $$\int |{\gamma }^{\prime }|\ge c\int |df({\gamma }^{\prime })|=c\int |(f\circ \gamma {)}^{\prime }|\ge cd(f(p),f(p^{\prime })).$$Taking infimum on $\gamma$, we obtain that $$d(p,q)\ge cd(f(p),f(q)).$$We are now equipped to verify that $M$ is complete as a metric space. Let $\{p_n\}$ be a Cauchy sequence in $M$, that is, $d(p_n,p_m)\to 0$. Then $d(f(p_n),f(p_m))\to 0$ too. As $\bar{M}$ is complete, $\{f(p_n)\}$ converges to some point in $\bar{M}$, written $f(p)$, for $f$ is surjective. Then $\{p_n\}$ converges to $p$, as $f$ is a diffeomorphism.
8.	Were $f$ not injective, i.e. $f(p)=f(q)$ for two distinct points $p,q\in M$, then for a geodesic $\gamma$ joining $p$ to $q$, $f\circ \gamma$ is a loop. Because $f$ is a local isometry, $f\circ \gamma$ is a geodesic too. Take any point $r$ different from $f(p)$ in the image of $f\circ \gamma$, then there are two geodesics joining $f(p)$ to $r$, a contradiction. Next we show that $f$ is surjective. For any $s\in \bar{M}$, let $\bar{\gamma }$ be the geodesic joining $f(p)$ to $s$, where $f(p)$ is any point in the image of $f$. Let $\gamma$ be the geodesic starting from $p$ with initial speed $v$ such that $d{f}_p(v)={\bar{\gamma }}^{\prime }(0)$. Then $\bar{\gamma }$ is precisely $f\circ \gamma$ as both start from $f(p)$ with the same initial speed, and it follows that $(f\circ \gamma )(l(\bar{\gamma }))=s$, where owing to the completeness of $M$, $\gamma$ is defined on the entire $[0,\infty )$, and in particular at $l(\bar{\gamma })$.
9.	The length is $${\int }_{\epsilon }^1\sqrt{\frac{1}{y}}dy=2\sqrt{y}{|}_{\epsilon }^1=2-2\sqrt{\epsilon }\to 2.$$Note that $x=0,\epsilon \le y\le 1$ is a geodesic that does not extend beyond $2$, and therefore $M$ is not complete.
10.	Method I. As illustrated by Chapter 3 Example 3.10 (p.73), all the geodesics in the Lobatchevski plane are of the form $x=x_0,y>0$ or $x=x_0+r\cos \theta ,y=r\sin \theta ,\theta \in (0,\pi )$. Their lengths are respectively $${\int }_0^{\infty }\frac{dy}{y}=\infty \text{or}{\int }_{\curvearrowleft }\frac{\sqrt{d{x}^2+d{y}^2}}{y}={\int }_0^{\pi }\frac{rd\theta }{r\sin \theta }\ge {\int }_0^{\pi }\frac{d\theta }{\min (\theta ,\pi -\theta )}=\infty .$$ Method II. Exhaust ${\mathbb{R}}_{+}^2$ by a sequence of compact sets $K_n=[-n\ln n,n\ln n]\times [1/n,n]$. Let $p=(0,1)$, then for $q_n\notin K_n$, it is not hard to show that $$d(p,q_n)\ge \min (\frac{n\ln n}{n},{\int }_{1/n}^1\frac{dy}{y},{\int }_1^n\frac{dy}{y})=\ln n\to \infty .$$

8Spaces of Constant Curvature

1.

a) Recall that by Chapter 4 Corollary 3.5 (p.96), the metric having constant curvature $K$ is equivalent to $R_{ijij}=-R_{ijji}=K$ for all $i\ne j$ and $R_{ijk\ell }=0$ in the other cases. From the calculation at the end of p.161 and $$f_i=\frac{F_i}{F},f_{ii}=\frac{F_{ii}F-F_i^2}{F^2},$$we get $$R_{ijij}=-\sum _{\ell }(F_{\ell }{)}^2+(F_{ii}+F_{jj})F;$$from the formula (2) on the same page, $$R_{ijk\ell }=R_{ijk}^s\frac{{\delta }_{s\ell }}{F^2}=\frac{R_{ijk}^{\ell }}{F^2}=\frac{(-f_k{f}_j-f_{kj}){\delta }_{i\ell }+(f_i{f}_k+f_{ik}){\delta }_{j\ell }}{F^2}=\frac{-F_{kj}{\delta }_{i\ell }+F_{ik}{\delta }_{j\ell }}{F^4},$$taking $i\ne j=k=\ell$, we get $F_{ij}=0$ provided that $R_{ijk\ell }=0$. b) i. $F_{ij}=0$ for $i\ne j$ implies that $F_i$ depends only on $x_i$ for all $1\le i\le n$. Then for $x=(x_1,\cdots ,x_n)$, $$F(x)=F(0)+{\int }_0^{(x_1,0,\cdots ,0)}{F}_1+\cdots +{\int }_{(x_1,\cdots ,x_{n-1},0)}^{(x_1,\cdots ,x_{n-1},x_n)}{F}_n:=G_1(x_1)+\cdots +G_n(x_n).$$ii. Note that $(F_{ii}{)}_j=(F_{ij}{)}_i=0$, denote $F_{ii}\equiv 2a$, where each $a$ for $i=1,\cdots ,n$ coincides due to the second equation in a). Then we write $$F=a{x}_i^2+b_i{x}_i+c_i(x_1,\cdots ,{\hat{x}}_i,\cdots ,x_n),$$where the $\hat{}$ means $h_i$ is a function independent of $x_i$. Further, express $F$ as $$F=\sum _i(a{x}_i^2+b_i{x}_i)+c(x_1,\cdots ,x_n)$$and make a comparison ($n$ comparisons indeed), we find that $c$ does not depend on $x_i$ for each $i=1,\cdots ,n$ indeed. Therefore $$F=\sum _i(a{x}_i^2+b_i{x}_i)+c,$$and substituting into the second equation in a) yields $$(\sum _i(a{x}_i^2+b_i{x}_i)+c)4a=K+\sum _{i=1}^n(2a{x}_i+b_i{)}^2.$$ c) This is a statement, instead of a problem. d) For any $x=(x_1,\cdots ,x_n)\in {\mathbb{R}}^n$, the distance between $0$ and $x$ is bounded by the length of the curve $\gamma =\{tx:0\le t\le 1\}$, which is $$\begin{array}{rl}d(0,x)& ={\int }_{\gamma }\Vert {\gamma }^{\prime }{\Vert }_g={\int }_{\gamma }\frac{1}{1+\frac{K}{4}\Vert tx{\Vert }^2}\Vert x\Vert \\ & ={\int }_0^1\frac{d(t\Vert x\Vert )}{1+{\left(\frac{\sqrt{K}}{2}t\Vert x\Vert \right)}^2}\\ & =\frac{2}{\sqrt{K}}\arctan (\frac{\sqrt{K}}{2}\Vert x\Vert )<\frac{\pi }{\sqrt{K}}.\end{array}$$Hence $({\mathbb{R}}^n,g)$ is bounded, and also closed as a space. But ${\mathbb{R}}^n$ is non–compact (compactness is a topological property), violating b) of Hopf–Rinow Theorem. Then $({\mathbb{R}}^n,g)$ is not complete.

13.

It is clear that $\sigma (\gamma (t))=\gamma (-t)$, as both are geodesics with the same starting point $p$ and intitial speed $-{\gamma }^{\prime }(t)$. To prove that $d{\sigma }_{\gamma (t)}X(\gamma (t))$ is parallel along $\gamma (t)$, compare $$\frac{d}{dt}⟨X,Y{⟩}_{\gamma (t)}=⟨X,{\nabla }_{\dot{\gamma }}Y{⟩}_{\gamma (t)}+⟨{\nabla }_{\dot{\gamma }}X,Y{⟩}_{\gamma (t)}=⟨X,{\nabla }_{\dot{\gamma }}Y{⟩}_{\gamma (t)}$$and $$\begin{array}{rl}\frac{d}{dt}⟨d\sigma (X),d\sigma (Y){⟩}_{\sigma (\gamma (t))}& =⟨d\sigma (X),{\nabla }_{d\sigma (\dot{\gamma })}d\sigma (Y){⟩}_{\sigma (\gamma (t))}+⟨{\nabla }_{d\sigma (\dot{\gamma })}X,d\sigma (Y){⟩}_{\sigma (\gamma (t))}\\ & =⟨X,{\nabla }_{\dot{\gamma }}Y{⟩}_{\gamma (t)}+⟨{\nabla }_{d\sigma (\dot{\gamma })}X,d\sigma (Y){⟩}_{\sigma (\gamma (t))},\end{array}$$we get the last term is $0$, and consequently ${\nabla }_{d\sigma (\dot{\gamma })}X(\sigma (\gamma (t)))=0$ by the arbitrariness of $Y$ (and that $d\sigma$ is surjective). Note that at the colored “$=$”, $d\sigma ({\nabla }_{\dot{\gamma }}Y)={\nabla }_{d\sigma (\dot{\gamma })}d\sigma (Y)$ is used (check it using (9) of Chapter 2 Theorem 3.6, p.55).

9Variations of Energy

1.	Suppose that $\{q_n\}$ is a sequence of points in $N$ such that $d(p_0,q_n)\to d(p_0,N)$, then it follows from the boundness and thus compactness of $\{q_n\}$ that there is a cluster point $q_0$, and $q_0\in N$ as $N$ is closed. Let $\gamma :[0,\ell ]\to M$ be a minimizing geodesic joining $p_0$ and $q_0$. Consider the variational fields $V(t)$ of variations keeping the starting point $p_0$ fixed yet with varying end points in $N$. As $\ddot{\gamma }=0$ and $V(0)=0$, we have $$0=\frac{1}{2}{E}^{\prime }(0)=-{\int }_0^{\ell }⟨V,\ddot{\gamma }⟩-⟨V(0),\dot{\gamma }(0)⟩+⟨V(\ell ),\dot{\gamma }(\ell )⟩=⟨V(\ell ),\dot{\gamma }(\ell )⟩,$$then $\dot{\gamma }(\ell )$ is perpendicular to all $V(\ell )\in T_{q_0}N$.
2.	Suppose, to the contrary, that $K(x,y)\ge \frac{1}{{\kappa }^2}>0$ on $\{(x,y):x^2+y^2\ge R^2\}$ for a certain $R>0$. As will be soon shown, $\mathrm{diam}{\mathbb{R}}^2\le 2\pi \kappa +\mathrm{diam}{B}_R$, where $B_R=\{(x,y):x^2+y^2\le R^2\}$ is compact hence bounded. Then the whole ${\mathbb{R}}^2$ is bounded and hence compact, a contradiction. For any $p,q\in {\mathbb{R}}^2$, join them by a minimizing geodesic $\gamma$ and establish a uniform upper bound of its length. If $\gamma \subset B_R$, then $\ell (\gamma )\le \mathrm{diam}{B}_R$; if $\gamma \subset B_R^c$, then $\ell (\gamma )\le \pi \kappa$ as the proof of Bonnet–Myers theorem (theorem 3.1) indicates. Things are worse if $p\notin B_R,q\in B_R$, in which case $\ell (\gamma )\le \pi \kappa +\mathrm{diam}{B}_R$; the worst case is when $\gamma$ intersects $\{(x,y):x^2+y^2=R^2\}$ twice (had it crossed the boundary more than twice, it would not be the shortest), and $\ell (\gamma )\le 2\pi \kappa +\mathrm{diam}{B}_R$.
3.	Suppose that the geodesic is of length $\ell$, then $$\begin{array}{rl}0\le \sum _{j=1}^{n-1}{E}_j^{\prime \prime }(0)& \le {\int }_0^{\ell }{\sin }^2\frac{\pi s}{\ell }[(n-1)(\frac{\pi }{\ell }{)}^2-(n-1)(a+\frac{df}{ds})]ds\\ & =(n-1){\int }_0^{\ell }{\sin }^2\frac{\pi s}{\ell }\left[\right(\frac{\pi }{\ell }{)}^2-a-\frac{df}{ds}]ds\\ & =(n-1)\left[\frac{\ell }{2}\right[(\frac{\pi }{\ell }{)}^2-a]+{\int }_0^{\ell }\frac{\pi }{\ell }\sin \frac{2\pi s}{\ell }fds]\\ & \le (n-1)(\frac{{\pi }^2}{2\ell }-\frac{a\ell }{2}-2c).\end{array}$$So $$\ell \le \frac{2c+\sqrt{4c^2+a{\pi }^2}}{a}.$$
4.	Suppose that $P$ is the parallel transport along $\gamma$ from $\gamma (0)$ to $\gamma (2\pi )$ ($=\gamma (0)$), where $\gamma$ is assumed to be parametrized by $[0,2\pi )$. $P$ preserves the inner product and is thus orthogonal; $M$ is orientable and $n$ is even, so $\det P=1=(-1{)}^n$. Applying Lemma 3.8 (p.203), $P$ leaves invariant some nonzero $v\perp \dot{\gamma }(0)$, that is, $Pv=v\in T_{\gamma (2\pi )}M=T_{\gamma (0)}M$. Then $V(t):=P_{\gamma (t)}$ is a vector field along the closed curve $\gamma$, and $Pv=v$ ensures that it is well–defined. Now $$\frac{1}{2}{E}_V^{\prime \prime }(0)=-{\int }_0^{2\pi }⟨V,R(\dot{\gamma },V)\dot{\gamma }⟩<0.$$Note that though a variation that generates $V$ is improper, the last four terms of (5) of this chapter (p.199) all offset as $\gamma$ is a closed curve.
5.	a) Existence is from compactness, and using Exercise 1 twice we get $\gamma \perp N_1,\gamma \perp N_2$. b) Denote $\nabla ,{\nabla }^{(1)},{\nabla }^{(2)}$ the Riemann connection on $M,N_1,N_2$, and write $h^{\prime }$ for $\frac{\partial h}{\partial s}$. We have $$\begin{array}{rl}\frac{1}{2}{E}^{\prime \prime }(0)& =I_{\ell }(V,V)+⟨\frac{D}{ds}\frac{\partial h}{\partial s},\dot{\gamma }⟩(0,\ell )-⟨\frac{D}{ds}\frac{\partial h}{\partial s},\dot{\gamma }⟩(0,0)\\ & =I_{\ell }(V,V)+⟨{\nabla }_{h^{\prime }}{h}^{\prime },\dot{\gamma }⟩(0,\ell )-⟨{\nabla }_{h^{\prime }}{h}^{\prime },\dot{\gamma }⟩(0,0)\\ & =I_{\ell }(V,V)+⟨B(h^{\prime },h^{\prime }),\dot{\gamma }⟩(0,\ell )-⟨B(h^{\prime },h^{\prime }),\dot{\gamma }⟩(0,0)\\ & =I_{\ell }(V,V)+⟨V(\ell ),S_{\dot{\gamma }(\ell )}^{(2)}V(\ell )⟩-⟨V(0),S_{\dot{\gamma }(0)}^{(1)}V(0)⟩.\end{array}$$Here we used $h^{\prime }(\ell )\in T_{p_2}{N}_2,h^{\prime }(0)\in T_{p_1}{N}_1$ where $p_1,p_2$ are the two endpoints of $\gamma$, and hence ${\nabla }_{h^{\prime }(\ell )}^{(2)}{h}^{\prime }(\ell )\in T_{p_2}{N}_2\perp \dot{\gamma }(\ell ),{\nabla }_{h^{\prime }(0)}^{(1)}{h}^{\prime }(0)\in T_{p_1}{N}_1\perp \dot{\gamma }(0)$.
6.	a) (i) Since $\ddot{\gamma }\equiv 0,V(0)=0$, $$\frac{1}{2}{E}^{\prime }(s)=-{\int }_0^{\ell }⟨V,\ddot{\gamma }⟩-⟨V(0),\dot{\gamma }(0)⟩+⟨V(s),\dot{\gamma }(s)⟩=⟨V(s),\dot{\gamma }(s)⟩=⟨{\sigma }_s^{\prime }(d(s)),\dot{\gamma }(s)⟩.$$(ii) Put $V(t)=\frac{\partial h}{\partial s}(s_0,t)$. Then $$\frac{1}{2}{E}^{\prime \prime }(s_0)=I_{d(s_0)}(V,V)-⟨\frac{D}{ds}\frac{\partial h}{\partial s},\dot{\gamma }⟩(s_0,0)+⟨\frac{D}{ds}\frac{\partial h}{\partial s},\dot{\gamma }⟩(s_0,d(s_0))=I_{d(s_0)}(V,V)\ge 0,$$where we used $\frac{\partial h}{\partial s}(s_0,0)=0$ as $h(s_0,0)={\sigma }_{s_0}(0)\equiv p$, and $\frac{D}{ds}\frac{\partial h}{\partial s}(s_0,d(s_0))=\frac{D}{ds}\frac{\partial \gamma }{\partial s}(s_0)=0$ as $\gamma$ is a geodesic. Suppose that $E^{\prime \prime }(s_0)=0$, then combined with $K\le 0$ we get $\dot{V}(t)\equiv 0$, so $V(t)\equiv 0$ as a result of $V(0)=0$. b) By Hadamard’s theorem (3.1 of Chapter 7), ${\exp }_p:T_p\tilde{M}\to \tilde{M}$ is a diffeomorphism. Hence every geodesic joining $p$ to any point $\gamma (s)$ of $\gamma$ is a curve in the variation $h$ (where $s$ is assumed to range over the entire $(-\infty ,\infty )$). Use the property of convex functions. c) When $K\le 0$ is dropped, let $\tilde{M}$ be a sphere, $p$ be the north pole and $\gamma$ be its equator, then there are infinitely many perpendicular lines of minimal length. When simple connectedness is dropped, let $\tilde{M}$ be an infinite cylinder and $p$ be any point, and $\gamma$ be a straight line passing through the antipodal point of $p$, then there are two lines reaching minimal length.

10The Rauch Comparison Theorem

2.	Take $K_0>0$ arbitrarily small in Klingenberg’s lemma and show that if $M$ is simply connected, there exists a unique geodesic joining the two points $p,q\in M$. This is because were there to exist two such geodesics ${\alpha }_0$ and ${\alpha }_1$, they would be homotopic by the simple connectedness, and by Klingenberg’s lemma, there would exist a sequence of curves ${\alpha }_t$ in this homotopy of lengths $\ge \pi /\sqrt{K_0}$, which would be arbitrarily big; however, $\ell ({\alpha }_t)$ as a continuous function of $t\in [0,1]$ is bounded, a contradiction. This asserts that ${\exp }_p:T_{p}M\to M$ is a one-one map. That is, it has an inverse ${\exp }_p^{-1}$. It suffices to show that ${\exp }_p$ is a local diffeomorphism then, as this will imply that ${\exp }_p^{-1}$ is continuous locally, and hence globally. This is a consequence of Rauch’s Theorem, in combination with Chapter 5 Proposition 3.5 (p.117) (and Chapter 0 Theorem 2.10, p.10).
3.	Note that $\tilde{\gamma }(t)=tv$ is a geodesic on $T_{p}M$, and $\tilde{J}(t)=tw$ is a Jacobi field along $\tilde{\gamma }$; further, $J(t)=(d{\exp }_p{)}_{tv}(tw)$ is a Jacobi field along $\gamma$. As $$J(0)=\tilde{J}(0)=0,J^{\prime }(0)={\tilde{J}}^{\prime }(0)=w,{\gamma }^{\prime }(0)={\tilde{\gamma }}^{\prime }(0)=v,$$and $K\le 0=\tilde{K}$, we have $|J(1)|\ge |\tilde{J}(1)|$, which is $|(d{\exp }_p{)}_v(w)|\ge |w|$.
5.	a) For all $t\in [0,\ell ]$, integration by parts gives $$\begin{array}{rl}0& ={\int }_0^t\tilde{f}(f^{\prime \prime }+Kf)-f({\tilde{f}}^{\prime \prime }+\tilde{K}\tilde{f})\\ & =[\tilde{f}{f}^{\prime }-f{\tilde{f}}^{\prime }{]}_0^t+{\int }_0^t(K-\tilde{K})f\tilde{f}.\end{array}$$From the initial conditions $f(0)=0$ and $f^{\prime }(0)=1$, $f$ is positive in a neighborhood of $0$. Assume that $f(t_1)=0$ for certain $t_1<t_0$. Then take $t=t_1$ and get $$0=(\underset{>0}{\underbrace{\tilde{f}(t_1)}}\underset{\le 0}{\underbrace{f^{\prime }(t_1)}}-\underset{=0}{\underbrace{f(t_1)}}{\tilde{f}}^{\prime }(t_1))-(\underset{=0}{\underbrace{\tilde{f}(0)}}{f}^{\prime }(0)-\underset{=0}{\underbrace{f(0)}}{\tilde{f}}^{\prime }(0))+{\int }_0^{t_1}(\underset{\le 0}{\underbrace{K-\tilde{K}}})\underset{>0}{\underbrace{f\tilde{f}}},$$forcing $K\equiv \tilde{K}$. Then $f$ and $\tilde{f}$ satisfy exactly the same ode, with the same initial conditions $f(0)=\tilde{f}(0)=0$ and $f^{\prime }(0)={\tilde{f}}^{\prime }(0)=1$, and for this reason $f\equiv \tilde{f}$. b) Again, examine $$0=(\tilde{f}(t)f^{\prime }(t)-f(t){\tilde{f}}^{\prime }(t))-(\underset{=0}{\underbrace{\tilde{f}(0)}}{f}^{\prime }(0)-\underset{=0}{\underbrace{f(0)}}{\tilde{f}}^{\prime }(0))+{\int }_0^t(\underset{\le 0}{\underbrace{K-\tilde{K}}})\underset{>0}{\underbrace{f\tilde{f}}}.$$It follows that $\tilde{f}{f}^{\prime }-f{\tilde{f}}^{\prime }\ge 0$, that is, $(f/\tilde{f}{)}^{\prime }=(\tilde{f}{f}^{\prime }-f{\tilde{f}}^{\prime })/({\tilde{f}}^2)\ge 0$. Conclude by $$f/\tilde{f}\ge \underset{t\to 0}{\lim }(f/\tilde{f})=\underset{t\to 0}{\lim }(f^{\prime }/{\tilde{f}}^{\prime })=1.$$For equality, $\tilde{f}{f}^{\prime }-f{\tilde{f}}^{\prime }\equiv 0$ forces $K\equiv \tilde{K}$ as a). To verify the Theorem of Rauch in dimension two as an application, first reduce the Jacobi fields in question to normal Jacobi fields as the proof in the book does, and hence suppose that $J(t)=f(t)e_2(t)$ where $e_2(t)$ is the parallel transport of a unit vector $e_2\in T_{\gamma (0)}M$ with $e_2\perp {\gamma }^{\prime }(0)$. Then $\frac{D^{2}J}{d{t}^2}=f^{\prime \prime }{e}_2$, so the Jacobi equation is equivalent to $$⟨f^{\prime \prime }{e}_2+R({\gamma }^{\prime },f{e}_2){\gamma }^{\prime },{\gamma }^{\prime }⟩=0$$and $$⟨f^{\prime \prime }{e}_2+R({\gamma }^{\prime },f{e}_2){\gamma }^{\prime },e_2⟩=0,$$since $\{{\gamma }^{\prime },e_2\}$ spans $T_{\gamma (t)}M$. The first equation trivially holds, and the second is $f^{\prime \prime }+Kf=0$. Similarly for $\tilde{J}(t)$.
6.	Suppose that $\tilde{f}(t)\ne 0$ for all $t\in [0,t_0]$. Using the expression from the last exercise, $${\int }_0^{t_0}\underset{\le 0}{\underbrace{(K-L)}}\underset{<0}{\underbrace{f\tilde{f}}}+\underset{>0}{\underbrace{\tilde{f}(t_0)}}\underset{\ge 0}{\underbrace{f^{\prime }(t_0)}}-\underset{>0}{\underbrace{\tilde{f}(0)}}\underset{\le 0}{\underbrace{f^{\prime }(0)}}=0.$$This forces $K\equiv L$, as well as $f^{\prime }(0)=f^{\prime }(t_0)=0$. Then the initial conditions $f(0)=f^{\prime }(0)=0$ imply that $f\equiv 0$, contradicting that $J\ne 0$ on $(0,t_0)$.
7.	A correction: it should be $$|{\int }_t^{\infty }K(s)ds|\le \frac{1}{4(t+1)}.$$ a)$${\omega }^2\le (\frac{1}{4(t+1)}+\frac{1}{4(t+1)}{)}^2=(-\frac{1}{4(t+1)}{)}^{\prime }=-{\omega }^{\prime }-K.$$b) ${\tilde{f}}^{\prime }=\omega \tilde{f}$, and ${\tilde{f}}^{\prime \prime }={\omega }^{\prime }\tilde{f}+\omega {\tilde{f}}^{\prime }=({\omega }^{\prime }+{\omega }^2)\tilde{f}=-L\tilde{f}$. $\tilde{f}(0)=\exp 0=1$. c) $-L={\omega }^{\prime }+{\omega }^2\le -K$, which is $K\le L$. Then the oscillation theorem of Sturm yields that $\tilde{f}$ has a zero on $[0,s_0]$. But $\tilde{f}>0$.

11The Morse Index Theorem

2.	Expand $f$ and $f^{\prime }$ into the Fourier series $$f(t)=\sum _{n=1}^{\infty }{a}_n\sin nt,f^{\prime }(t)=\sum _{n=1}^{\infty }n{a}_n\cos nt.$$Then the inequality follows from $${\int }_0^{\pi }{f}^2=\frac{\pi }{2}\sum _{n=1}^{\infty }{a}_n^2,{\int }_0^{\pi }{f}^{\prime 2}=\frac{\pi }{2}\sum _{n=1}^{\infty }{n}^2{a}_n^2.$$Yet we need to justify that differentiating $f$ by terms does produce $f^{\prime }$, which relies on the decay of the coefficients $a_n$. Indeed, integrating by parts, we find $$\frac{\pi }{2}{a}_n={\int }_0^{\pi }f\sin nt={\int }_0^{\pi }{f}^{\prime }\frac{\cos nt}{n}=-{\int }_0^{\pi }{f}^{\prime \prime }\frac{\sin nt}{n^2}.$$Then $$\begin{array}{rl}& (\sum _{n=1}^{\infty }n{a}_n{)}^2\le \sum _{n=1}^{\infty }(n^2{a}_n{)}^2\sum _{n=1}^{\infty }\frac{1}{n^2}\\ & =\left(\frac{\pi }{2}{)}^{-2}\sum _{n=1}^{\infty }\right({\int }_0^{\pi }{f}^{\prime \prime }\sin nt{)}^2\frac{{\pi }^2}{6}=\frac{2}{3}{\int }_0^{\pi }{f}^{\prime \prime 2}<\infty ,\end{array}$$yielding the uniform convergence of the series of $f^{\prime }$, and consequently the justification.
3.	It follows that $K(t)\equiv 0$ for $t\in (0,\pi )$. Let $t\to 0$, we get $K(p)=1$. Then use Theorem 4.1 of Chapter 8 (p.163).
4.	${\phi }^{\prime }(0)=0,{\phi }^{\prime \prime }(0)=-a(0)<0$ implies that ${\phi }^{\prime }(t_0)<0$ for some $t_0>0$. Suppose that $\phi$ has no positive zero, then ${\phi }^{\prime \prime }=-a\phi \le 0$ on $(0,+\infty )$. Hence ${\phi }^{\prime }(t)\le {\phi }^{\prime }(t_0)$ for $t>t_0$. But then $\phi (t)\le \phi (t_0)+(t-t_0){\phi }^{\prime }(t_0)\to -\infty$ as $t\to +\infty$, a contradiction.
5.	Since $\phi (-t_1)=\phi (t_2)=0$, and ${\phi }_Y\ge K\ge a$ with strict inequality at $0$, $$\begin{array}{rl}{I}_{[-t_1,t_2]}(X,X)& ={\int }_{-t_1}^{t_2}{\phi }^{\prime 2}-{\phi }^2{\phi }_Y\\ =(\phi {\phi }^{\prime }){|}_{-t_1}^{t_2}& -{\int }_{-t_1}^{t_2}\phi {\phi }^{\prime \prime }+{\phi }^2{\phi }_Y<-{\int }_{-t_1}^{t_2}\phi {\phi }^{\prime \prime }+{\phi }^{2}a=0.\end{array}$$
6.	From Exercise 5, there exists $t_0>0$ and a vector field $Y$ along the segment $\gamma ([-t_0,t_0])$ such that $I(Y,Y)<0$. Let $f$ be a proper variation that generates $Y$, then $\frac{1}{2}{E}^{\prime \prime }(0)=I(Y,Y)<0$, which is, $\gamma ([-t_0,t_0])$ does not minimize the length between its endpoints. If $K\ge 0$ instead, $M=\mathbb{R}\times S^{n-1}$ violates the theorem. Remark: Cf. [YS] 1.2 (p.13-7) for the proof of the splitting theorem (Cheeger–Gromoll, 1971) below. Let $M$ be an $n$-dimensional complete Riemannian manifold with $\mathrm{Ric}(M)\ge 0$. Suppose that $M$ has a line. Then $M$ is isometric to $\mathbb{R}\times N$, where $N$ is an $(n-1)$-dimensional Riemannian manifold with the product metric.

References

[Chen]	陈维桓, 李兴校. 黎曼几何引论 (上册) . 北京大学数学教学系列丛书. 北京大学出版社, 2002.
[dC]	M.P. do Carmo. Differential Geometry of Curves and Surfaces. Dover Publications Inc., 2016.
[YS]	丘成桐, 孙理察. 微分几何讲义. 高等教育出版社, 2018.