Chapter 1 Manifolds

1.	Let $x=\left(x_1,\dots ,x_{n+1}\right)$. Then for $x\in S_n-n$, the map $p_n:S^d-n\to {\mathbb{R}}^n$ is defined as $p_n(x)=\frac{1}{1-x_{n+1}}\left(x_1,\dots ,x_n\right)$. Similarly, if $x\in S_n-s$, then $p_n(x)=\frac{1}{1+x_{n+1}}\left(x_1,\dots ,x_n\right)$. Therefore each function is continuous on its domain, respectively. Also on $p_s(S_n-n-p)$, $p_n\circ p_s^{-1}:p_s(S_n-n-p)\to p_n(S_n-n-p)\subset {\mathbb{R}}^n$ is defined as $p_n\circ p_s^{-1}(y_1,\dots ,y_n)=\frac{1}{k}(y_1,\dots ,y_n)$ $(k={\sum }_{i=1}^n{y}_i^2)$, which is smooth since $(0,\dots ,0)\notin p_s(S_n-n-p)$. Similarly, $p_s\circ p_n^{-1}$ is also smooth. Therefore, the maximal collection including these coordinate systems is indeed a differentiable structure on $S^d$.
2.	Let $f(t)=t^3$. Then ${\mathrm{id}}_{\mathbb{R}}\circ f^{-1}(t)=t^{\frac{1}{3}}$, which is not differentiable at $t=0$, and it means $\mathcal{F}\ne {\mathcal{F}}_1$. However, if we set $\phi :(\mathbb{R},\mathcal{F})\to \left(\mathbb{R},{\mathcal{F}}_1\right),t\mapsto t^{\frac{1}{3}}$, then ${\mathrm{id}}_{\mathbb{R}}\circ \phi \circ f^{-1}(t)=t$, which is a diffeomorphism. Therefore, $(\mathbb{R},\mathcal{F})$ and $\left(\mathbb{R},{\mathcal{F}}_1\right)$ are diffeomorphic.
3.	By Theorem 1.11, given an open cover $\left\{U_{\alpha }\right\}$, there exists a partition of unity $\left\{{\phi }_{\alpha }\right\}$ subordinate to the cover $\left\{U_{\alpha }\right\}$. If we set $V_{\alpha }={\phi }_{\alpha }^{-1}(\mathbb{R}\backslash \{0\})$, then it is also an open cover since for every $m\in M$, ${\sum }_{\alpha }{\phi }_{\alpha }(m)=1$, which means ${\phi }_{\alpha }(m)\ne 0$ for some $\alpha$, and $m\in V_{\alpha }$. Also, by definition of support, $\overline{V_{\alpha }}\subset U_{\alpha }$. Therefore $\left\{V_{\alpha }\right\}$ is the refinement satisfying the condition.
4.	Suppose two closed subsets $C,D\subset M$ are given. Then by regularity, $\forall c\in C$, $\exists$ open subset $U_c,V_c\subset M$ s.t. $c\in U_c,D\subset V_c$, and $U_c\cap V_c=\varnothing$. Then ${\left\{U_c\right\}}_{c\in C}\cup M\backslash C$ is an open cover of $M$. Hence, by paracompactness, there exists a refinement of ${\left\{U_c\right\}}_{c\in C}\cup M\backslash C$ which is locally finite. By discarding the open sets which do not intersect $C$, one can obtain an open cover of $C$ which is locally finite, say $\left\{{\tilde{U}}_{\beta }\right\}$. Then, $\forall d\in D,\exists \widetilde{V_d}$ s.t. ${\tilde{V}}_d\cap {\tilde{U}}_{\beta }\ne \varnothing$ for only finite ${\beta }^{\prime }$’s, which we can label as ${\beta }_1,\dots ,{\beta }_n$. Since $\left\{{\tilde{U}}_{\beta }\right\}$ is a refinement, one can find $c_i\in C$ for each ${\beta }_i$ s.t. ${\tilde{U}}_{{\beta }_i}\subset U_{c_i}$. If we set $W_d=\widetilde{V_d}\cap {\bigcap }_{i=1}^n{V}_{c_i}$, then one can easily see that $\forall \beta ,W_d\cap {\tilde{U}}_{\beta }=\varnothing$ and $W_d$ is an open neighborhood of $d$. If we set $U=\bigcup {\tilde{U}}_{\beta },W=\bigcup W_d$, then each open set is a neighborhood of $C$ and $D$ respectively which does not intersect each other. Since $C$ and $D$ are arbitrary, $M$ is normal.
5.	(a) If $\vec{a}=(a_1,\dots ,a_d)$ and $\vec{v}=(v_1,\dots ,v_d)$, $\tilde{\psi }\circ {\tilde{\phi }}^{-1}(\vec{a},\vec{v})=\left(\psi \circ {\phi }^{-1}(\vec{a}),\mathcal{J}(\psi \circ {\phi }^{-1})(\vec{v})\right)$, which is ${\mathcal{C}}^{\infty }$. (b) $\tilde{\phi }:{\pi }^{-1}(U)\to {\mathbb{R}}^{2d}$ is a bijection between ${\pi }^{-1}(U)$ and $\phi (U)\times {\mathbb{R}}^d$, which is open in ${\mathbb{R}}^{2d}$. Also, for any $(V,\psi ),{\tilde{\psi }}^{-1}(W^{\prime })\cap {\pi }^{-1}(U)$ is of the form ${\tilde{\phi }}^{-1}(W)$, which is ${\tilde{\phi }}^{-1}(\tilde{\phi }\circ {\tilde{\psi }}^{-1}(\psi (U\cap V)\times {\mathbb{R}}^d\cap W^{\prime }\left)\right)$. Therefore, on the defined topology, $\tilde{\phi }$ is a homeomorphism and that $T(M)$ is a $2d$-dimensional locally Euclidean space directly follows. For second countability, since $M$ is second countable, we can choose countable such $(U,\phi )$ that cover $T(M)$. Since ${\mathbb{R}}^{2d}$ is also second countable, it follows that $T(M)$ is second countable. (c) By (a) and (b), it directly follows.
6.	By Theorem 1.30, Corollary (a), $\psi$ is a diffeomorphism if and only if $d\psi$ is surjective everywhere. Suppose that $d\psi$ is not surjective at $m\in M$. Since it is nonsingular, $\dim M=p<\dim N=d$. Pick a coordinate system on $N$, $(U,\phi )$, such that $\phi (U)={\mathbb{R}}^d$. Then $\phi \circ \psi :{\psi }^{-1}(U)\to {\mathbb{R}}^d$ is surjective. Also, by the second countability of $M$, one can find countable coordinate systems of ${\psi }^{-1}(U)$, say ${\left\{\left(V_{\alpha },f_{\alpha }\right)\right\}}_{\alpha \in A}$. Then ${\mathbb{R}}^d={\bigcup }_{\alpha \in A}\phi \circ \psi \circ f_{\alpha }^{-1}({\mathbb{R}}^p)$. Using Proposition 1.35, since $\phi \circ \psi \circ f_{\alpha }^{-1}$ is an immersion, one can obtain a countable cover $\left\{U_{\beta }\right\}$ of ${\mathbb{R}}^p$ such that $\phi \circ \psi \circ f_{\alpha }^{-1}(U_{\beta })$ is a subset of a slice of a open set in ${\mathbb{R}}^d$, which is nowhere dense. Therefore, by the Baire category theorem, ${\bigcup }_{\alpha \in A}\phi \circ \psi \circ f_{\alpha }^{-1}({\mathbb{R}}^p)$ cannot be all of ${\mathbb{R}}^d$, contradicting the assumption. Therefore, $\psi$ is a diffeomorphism.
7.	(a) For two differentiable structures on $A$, $\mathcal{F}$ and ${\mathcal{F}}^{\prime }$, ${\mathrm{id}}_A:(A,\mathcal{F})\to \left(A,{\mathcal{F}}^{\prime }\right)$ is continuous, hence ${\mathcal{C}}^{\infty }$ by Theorem 1.32. Since its inverse is also ${\mathcal{C}}^{\infty }$, it is a diffeomorphism: that is, two structures are the same. (b) Let $(\tilde{A},{\iota }^{\prime })$ be $A$ equipped with a manifold structure possibly different from the induced structure form $M$. Then ${\mathrm{id}}_A:\tilde{A}\to A$ is continuous and hence ${\mathcal{C}}^{\infty }$ by Theorem 1.32. Also it is nonsingular since $d{\iota }^{\prime }=d(\iota \circ {\mathrm{id}}_A)$ is. Therefore, by Exercise 6, it is a diffeomorphism, which means that a manifold structure on $\tilde{A}$ is equal to that on $A$.
8.	Use notations of Theorem 1.37. Since $\left\{\frac{\partial f_i}{\partial s_j}{|}_{\left(r_0,s_0\right)}\right\}$ is nonsingular (i.e., invertible), by continuity, there exists an open set $(r_0,s_0)\in V_0\times W_0\subset U$ such that $df$ is surjective. Also, $(r_0,s_0)\in f^{-1}(0)$, hence nonempty. Therefore, by Theorem 1.38, $P=f{|}_{V_0\times W_0}^{-1}(0)$ is an imbedded submanifold of $V_0\times W_0$ of dimension $c-d$. Now, let ${\pi }_1:V_0\times W_0\to V_0$ be the first projection. Then, ${\pi }_1\circ \iota :P\to V_0$ is ${\mathcal{C}}^{\infty }$, and $d({\pi }_1\circ \iota ):P_{\left(r_0,s_0\right)}\to V_{0r_0}$ is an isomorphism. $(\because P_{(r_0,s_0)}$ can be identified with $\ker df\subset V_0\times W_{0(r_0,s_0)}$, and ${df|}_{W_{0s_0}}$ is an isomorphism, $V_0\times W_{0(r_0,s_0)}=\ker df\oplus W_{s_0}=V_{0r_0}\oplus W_{0s_0}$. Hence $\ker df\to V_{0r_0}$ is an isomorphism also.) Therefore, by Theorem 1.30, one can find an open set $(r_0,s_0)\in V\times W\subset V_0\times W_0$ such that ${\pi }_1\circ {\iota }_{P\cap V\times W}^{-1}:V\to P\cap V\times W$ is an one-to-one onto ${\mathcal{C}}^{\infty }$ map. Therefore, if we set $g={\pi }_2\circ \iota \circ ({\pi }_1\circ \iota {|}_{P\cap V\times W}^{-1}):V\to W$ (where ${\pi }_2$ is the second projection), then it is the map satisfying the condition.
9.	$\frac{\partial f}{\partial x}=3x^2+y,\frac{\partial f}{\partial y}=x+3y^2$. Therefore, $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$ if and only if $(x,y)=(0,0),\left(-\frac{1}{3},-\frac{1}{3}\right)$. Therefore, by Theorem 1.38, $f^{-1}\left(f\left(\frac{1}{3},\frac{1}{3}\right)\right)$ is an imbedded submanifold, since $(0,0),(-\frac{1}{3},-\frac{1}{3})\notin f^{-1}\left(f\right(\frac{1}{3},\frac{1}{3}\left)\right)$. $f^{-1}(f(0,0))$ is not a manifold since $(0,0)$ is a node. If $p=\left(-\frac{1}{3},-\frac{1}{3}\right)$, then $f(x,y)=f(\frac{1}{3},\frac{1}{3})$ iff$$x^3+xy+y^3-\frac{1}{27}=0$$iff $$\left(x+y-\frac{1}{3}\right)\left(x^2+y^2+\frac{1}{9}+\frac{x}{3}+\frac{y}{3}-xy\right)=0$$iff $x+y=\frac{1}{3}$, since$$x^2+y^2+\frac{1}{9}+\frac{x}{3}+\frac{y}{3}-xy=\frac{1}{2}\left((x+\frac{1}{3}{)}^2+(y+\frac{1}{3}{)}^2+(x-y{)}^2\right).$$Therefore $f^{-1}\left(f\right(-\frac{1}{3},-\frac{1}{3}\left)\right)$ is a line, which is an imbedded submanifold.
10.	Since $f(M)$ is compact, $\exists m\in M$ s.t. $|f(m)|$ is the biggest. If $df:M_m\to {\mathbb{R}}_{f(m)}^n$ is nonsingular, then by Theorem 1.30, Corollary (a), $\exists U\subset f(M)$ neighborhood of $f(m)$, which contradicts the maximality of $|f(m)|$.
11.	Not closed: $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)\subset \mathbb{R},g(x)=\tan x$. Not imbedded: Second Example of 1.28, $g(x)=x$. To prove the converse: if there is a disk $D(x,r)\subset {\mathbb{R}}^n$, one can construct a ${\mathcal{C}}^{\infty }$ function which is nonzero exactly on $D(x,r)$ and zero elsewhere. Since $M$ is locally Euclidean, there exists a basis $\left\{B_{\alpha }\right\}$ of $M$ such that $\forall B_{\alpha },\exists g_{\alpha }$ s.t. $g_{\alpha }^{-1}(\mathbb{R}\backslash \{0\})=B_{\alpha }$. Then by the assumption there is a ${\mathcal{C}}^{\infty }$ function $f_{\alpha }$ on $N$ such that $f_{\alpha }\circ \psi =g_{\alpha }$. Since $f_{\alpha }^{-1}(\mathbb{R}\backslash \{0\})\cap \psi (M)=\psi (B_{\alpha })$, $\psi (B_{\alpha })$ is open in $\psi (M)$ with relative topology, which means $\psi$ is an imbedding. Therefore we can say $M\subset N$. If $M$ is not closed in $N$, then there is $n\in \bar{M}\backslash M$ and $(U,\phi )$ a coordinate system of $n$, and there is a ${\mathcal{C}}^{\infty }$ function $g$ on $M$ such that$$g(m)=\frac{1}{\phi (m)-\phi (n)}$$on the neighborhood of $n$ on $M$. However, since $n$ is a limit point of $M$, it cannot be extended to all of $N$. Therefore, $M$ should be closed in $N$.
12.	(a) Since $df=\sum 2r_{i}d{r}_i$, it is surjective except at the origin. Therefore one can use Theorem 1.38. Also, since it is a submanifold with the relative topology, it should be the same as 1.5(d) by 1.33(b). (b) By following the argument, we need only to show that $d{\psi }_I$ is surjective. Since $\psi (A)=({\sum }_k{a}_{ik}{a}_{jk}{)}_{ij}$, $d{\psi }_A=({\sum }_k(a_{jk}d{a}_{ik}+a_{ik}d{a}_{jk}){)}_{ij}$. $\therefore d{\psi }_I={\left(d{a}_{ij}+d{a}_{ji}\right)}_{ij}$. But, since $Gl(n,\mathbb{R}{)}_I\cong M_{n\times n}(\mathbb{R})$, it is surjective onto ${\mathrm{Sym}}_n(\mathbb{R}{)}_I\cong {\mathrm{Sym}}_n(\mathbb{R})$.
13.	(a) If we set $X=\sum a_i\frac{\partial }{\partial x_i},Y=\sum b_i\frac{\partial }{\partial x_i}$ locally as Proposition 1.43(b),$$[X,Y]=\sum _i\sum _j\left(a_i\frac{\partial b_j}{\partial x_i}-b_i\frac{\partial a_j}{\partial x_i}\right)\frac{\partial }{\partial x_j}.$$Therefore by 1.43(b), it is a smooth vector field. (b) Using the same notation as above,$$\begin{array}{rl}[fX,gY]& =\sum _i\sum _j\left(f{a}_i\frac{\partial g{b}_j}{\partial x_i}-g{b}_i\frac{\partial f{a}_j}{\partial x_i}\right)\frac{\partial }{\partial x_j}\\ & =\sum _i\sum _j\left(fg{a}_i\frac{\partial b_j}{\partial x_i}-fg{b}_i\frac{\partial a_j}{\partial x_i}+f{a}_i{b}_j\frac{\partial g}{\partial x_i}-g{a}_j{b}_i\frac{\partial f}{\partial x_i}\right)\frac{\partial }{\partial x_j}\\ & =fg[X,Y]+f(Xg)Y-g(Yf)X.\end{array}$$(c)$$[X,Y]=\sum _i\sum _j\left(a_i\frac{\partial b_j}{\partial x_i}-b_i\frac{\partial a_j}{\partial x_i}\right)\frac{\partial }{\partial x_j}=-\sum _i\sum _j\left(b_i\frac{\partial a_j}{\partial x_i}-a_i\frac{\partial b_j}{\partial x_i}\right)\frac{\partial }{\partial x_j}=-[Y,X].$$(d) Let $Z=\sum c_i\frac{\partial }{\partial x_i}$. Then,$$\begin{array}{rl}[[X,Y],Z]=& \sum _i\sum _j\sum _k(\left(a_i\frac{\partial b_j}{\partial x_i}-b_i\frac{\partial a_j}{\partial x_i}\right)\frac{\partial c_k}{\partial x_j}\\ & -c_j\left(\frac{\partial a_i}{\partial x_j}\frac{\partial b_k}{\partial x_i}+a_i\frac{{\partial }^2{b}_k}{\partial x_j\partial x_i}-\frac{\partial b_i}{\partial x_j}\frac{\partial a_k}{\partial x_i}-b_i\frac{{\partial }^2{a}_k}{\partial x_j\partial x_i}\right))\frac{\partial }{\partial x_k}\end{array}$$and by direct calculation, the Jacobi identity gets proved.
14.	No. If $X_x=\exp (x)\frac{\partial }{\partial r}$ and ${\gamma }_0(t)$ is the integral curve at $0\in \mathbb{R}$, then by definition, $\frac{d{\gamma }_0}{dt}(t)=\exp ({\gamma }_0(t))$ and ${\gamma }_0(0)=0$. By solving the differential equation, ${\gamma }_0(t)=-\log (-t+1)$, and it is not defined on $(-\infty ,\infty )$.
15.	Use the equation of Exercises 13(a) above.
16.	Consider the first example of 1.31 and a curve along the horizontal part of the submanifold. It is clear that on $0\in {\mathbb{R}}^2$, the tangent vector of the integral curve should be horizontal, whereas $N_0$ consists of vertical vectors.
17.	By 1.48(f), ${\bigcup }_{t>0}{\mathcal{D}}_t={\bigcup }_{t<0}{\mathcal{D}}_t=M$, thus by the compactness of $M$, $\exists t_0>0$ s.t. ${\mathcal{D}}_{t_0}={\mathcal{D}}_{-t_0}=M$. Thus by 1.48(g), $X_{t_0}$ is an endomorphism, and $\forall n\in \mathbb{N}$, $X_{t_0}^n$ is defined on all of $M$. Therefore by 1.48(h), $\forall N>0,{\mathcal{D}}_N=M$. Likewise, $\forall N<0,{\mathcal{D}}_N=M$. Since it is clear that $D_0=M$, $\mathrm{M}$ is complete.
18.	If it is one-to-one, the image should be infinite. Therefore, there exist $a,b,c\in \mathrm{im}f$ s.t. $a<b<c$. By erasing $b$, $a$ and $c$ can be separated by open sets $(-\infty ,b)\cap \mathrm{im}f$ and $(b,\infty )\cap \mathrm{im}f$, hence $\mathrm{im}f$ is disconnected, whereas ${\mathbb{R}}^2\backslash f^{-1}(b)$ is connected since $f^{-1}(b)$ consists of one point. This contradicts the continuity of $f$.
19.	(1.60$\Leftarrow$1.61) Follow the remark of 1.61. Then,$$\left[Y_a,Y_b\right]=\sum _{j=1}^{d-c}\left(\frac{\partial f_{jb}\circ \gamma }{\partial y_a}-\frac{\partial f_{ja}\circ \gamma }{\partial y_b}\right)\frac{\partial }{\partial y_{c+j}}+\sum _{j=1}^{d-c}\sum _{k=1}^{d-c}\left(\frac{\partial f_{jb}\circ \gamma }{\partial y_{c+k}}{f}_{ka}-\frac{\partial f_{ja}\circ \gamma }{\partial y_{c+k}}{f}_{kb}\right)\frac{\partial }{\partial y_{c+j}}.$$Since $\mathcal{D}$ is involutive, $\left[Y_a,Y_b\right]\in \mathcal{D}$. But the expression above does not include $\frac{\partial }{\partial y_i}$ terms for $1\le i\le c$, it must be zero. Therefore, the assumption 1.61(2) is satisfied. Then we obtain a map $\alpha$ as 1.61(3). Since $a(r_0,s)=s$, $(\mathrm{id},\alpha {)}^{-1}$ is ${\mathcal{C}}^{\infty }$ local homeomorphism, so shrinking the neighborhood if necessary, we obtain a local chart $\phi =(\mathrm{id},\alpha {)}^{-1}\circ \gamma :W\to U\times V$ and $\left(z_1,\dots ,z_d\right)=(\mathrm{id},\alpha {)}^{-1}\circ \left(y_1,\dots ,y_d\right)$ is coordinate functions. Also, if $1\le i\le c$, $$\frac{\partial }{\partial z_i}=\sum _{j=1}^d\frac{\partial y_j}{\partial z_i}\frac{\partial }{\partial y_j}=\frac{\partial }{\partial y_i}+\sum _{j=1}^{d-c}{f}_{ji}\circ \gamma \frac{\partial }{\partial y_{c+j}}=Y_i$$by 1.61(4) and (6). Therefore, if we let $I$ be a submanifold defined by $z_j=$ constant for $c+1\le j\le d$, it is an integral manifold of $\mathcal{D}$. (1.60$\Rightarrow$1.61) Let $X_i=\frac{\partial }{\partial r_i}+{\sum }_{j=1}^n{b}_{ji}\frac{\partial }{\partial s_j}$ for $1\le i\le n$ and $\mathcal{D}$ be a distribution generated by $X_i$’s. Then by similar calculation above, $\left[X_a,X_b\right]=0\in \mathcal{D}$ by assumption 1.61 (2). Thus by 1.60, there exists a cubic coordinate system $(W,\phi )$ with coordinate functions $\left(x_1,\dots ,x_m,y_1,\dots ,y_n\right)$ defined as the theorem. Let us define $\alpha$ on the neighborhood of $(0,0)$ s.t. $y_j(r,{\alpha }_s(r))=y_j(r_0,s),\forall 1\le j\le n$ by implicit function theorem. We can define this because $X_1,\dots ,X_m,\frac{\partial }{\partial s_1},\dots ,\frac{\partial }{\partial s_1}$ is a basis of tangent space and $X_i(y_j)=0$ since $y_j$’s are constant on integral manifolds of $\mathcal{D}$, which means $\left\{\frac{\partial y_j}{\partial s_k}\right\}$ should be nonsingular. Then clearly ${\alpha }_s(r_0)=s$ and$$\frac{\partial y_j}{\partial r_i}+\sum _{k=1}^n\frac{\partial y_j}{\partial s_k}\frac{\partial {\alpha }_{s,k}}{\partial r_i}=0.$$Also, $X_i(y_j)=0$ means$$\frac{\partial y_j}{\partial r_i}+\sum _{k=1}^n{b}_{ki}\frac{\partial y_j}{\partial s_k}=0.$$Since $\left\{\frac{\partial y_j}{\partial s_k}\right\}$ is nonsingular, condition (4) is satisfied.
20.	Let $M=\mathbb{R},N=\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ and regard $N$ as a submanifold imbedded in $M$ and $\phi$ as an inclusion. If $X(t)=\tan (t)\frac{\partial }{\partial r}$, then it satisfies the assumption since $\phi$ is one-to-one, but it cannot be extended to any vector field on $M$.
21.	It is clearly one-to-one, ${\mathcal{C}}^{\infty }$, and nonsingular. To simplify the calculation, let us think the torus as $T={\mathbb{R}}^2/\mathbb{Z}\times \mathbb{Z}$ and $\phi :\mathbb{R}\to T,x\mapsto (x,\alpha x)$. Then $\mathrm{im}\phi \cap \{(0,y)\in T\}$ is dense in $\{(0,y)\in T\}$, since $\alpha$ is irrational. Also, given $x_0$, $\mathrm{im}\phi \cap \left\{\left(x_0,y\right)\in T\right\}$ is dense in $\left\{\left(x_0,y\right)\in T\right\}$ since it is the transition of $\mathrm{im}\phi \cap \{(0,y)\in T\}$. Since $x_0$ is arbitrary, the image of $\phi$ is dense in $T$.
22.	Let $t_0$ be a point satisfying $\dot{\gamma }(t_0)=0$. Then, the constant map whose image is $\gamma (t_0)$ is also an integral curve. Therefore, by Theorem 1.48(c), $\gamma$ should be a constant map.
23.	Let $\left\{U_{\alpha }\right\}$ be an open cover of $M$ such that $\forall \alpha ,\left(U_{\alpha },{\psi }_{\alpha }\right)$ is a chart. By Theorem 1.11, there exists a partition of unity $\left\{{\phi }_{\alpha }\right\}$ subordinate to the cover. Also, on each chart, there exists a positive definite inner product induced by the standard inner product on the Euclidean space, say $⟨,{⟩}_{\alpha }$, and it is smooth on each chart. If we set $⟨,{⟩}_m={\sum }_{\alpha }⟨,{⟩}_{\alpha ,m}{\phi }_{\alpha }(m)$, then it is an inner product which makes $M$ a Riemannian manifold.
24.	(a) $(\Rightarrow )$ Obvious. $(\Leftarrow )\forall (U,\phi ),\forall (V,\psi )$ charts of $M$ and $N$ respectively, $(U\times V,\phi \times \psi )$ is a chart of $M\times N$. Then $(\phi \times \psi )\circ \alpha =\left(\phi \circ {\pi }_1\circ \alpha ,\psi \circ {\pi }_2\circ \alpha \right)$ is ${\mathcal{C}}^{\infty }$ if ${\pi }_1\circ \alpha ,{\pi }_2\circ \alpha$ are ${\mathcal{C}}^{\infty }$. (b) Let $m\in U,n\in V$, and $(x_1,\dots ,x_s),(y_1,\dots ,y_t)$ be the local coordinates of $U,V$ respectively. Then$$(M\times N{)}_{(m,n)}=\underset{i}{⨁}\mathbb{R}\frac{\partial }{\partial x_i}{|}_{(m,n)}\underset{j}{⨁}\mathbb{R}\frac{\partial }{\partial y_j}{|}_{(m,n)}\cong M_m\oplus N_n.$$ (c) On $U\times V,\tilde{X}={\sum }_i{a}_i(x_1,\dots ,x_s)\frac{\partial }{\partial x_i}$ and $\tilde{Y}={\sum }_j{b}_j(y_1,\dots ,y_t)\frac{\partial }{\partial y_j}$. Therefore, by direct calculation, $[\tilde{X},\tilde{Y}]=0$. (d) Let$$v=\sum _i{c}_i\frac{\partial }{\partial x_i}{|}_{(m_0,n_0)}+\sum _j{d}_j\frac{\partial }{\partial y_j}{|}_{(m_0,n_0)}\in (M\times N{)}_{\left(m_0,n_0\right)}.$$Then$$d{\pi }_1(v)=\sum _i{c}_i\frac{\partial }{\partial x_i}{|}_{m_0},d{\pi }_2(v)=\sum _j{d}_j\frac{\partial }{\partial y_j}{|}_{n_0}.$$Also,$$\frac{\partial f\circ i_{n_0}}{\partial x_i}{|}_{m_0}=\frac{\partial f}{\partial x_i}{|}_{(m_0,n_0)},\frac{\partial f\circ i_{m_0}}{\partial y_j}{|}_{n_0}=\frac{\partial f}{\partial y_j}{|}_{(m_0,n_0)}.$$Summing up,$$v(f)=\sum _i{c}_i\frac{\partial f}{\partial x_i}{|}_{(m_0,n_0)}+\sum _j{d}_j\frac{\partial f}{\partial y_j}{|}_{(m_0,n_0)}=v_1(f\circ i_{n_0})+v_2(f\circ i_{m_0}).$$