Chapter 4 Integration on Manifolds

1.	$(\Leftarrow )$ Let $\omega =d{r}_1\wedge \dots \wedge d{r}_{d+1}$ be the volume form on ${\mathbb{R}}^{d+1}$ and ${\omega }^{\prime }(x)=\delta f_x(i(N(x)){\omega }_{f(x)})$ for a smooth nowhere-vanishing normal vector field $N$ along $(X,f)$. Then it is non-vanishing $d$-form on $X$, since for $x\in X$ and $X_1,\dots ,X_d\in X_x$ which are linearly independent, ${\omega }_x^{\prime }(X_1,\dots ,X_d)\ne 0$. $(\Rightarrow )$ Let ${\omega }^{\prime }$ be a nowhere-vanishing $d$-form on $X$. For $x\in X$ choose $X_1,\dots ,X_d\in X_x$ which satisfies $\widetilde{X_1}\wedge \dots \wedge \widetilde{X_d}={\omega }_x$. Then we can pick $N(x)\in ({\mathbb{R}}^{d+1}{)}_{f(x)}$ s.t. $N(x)$ is orthogonal to $df(X_x)$ and $\omega (df(X_1),\dots ,df(X_d),N(x))=1$ where $\omega =d{r}_1\wedge \dots \wedge d{r}_{d+1}$. $N(x)$ is independent of the choice of $X_1,\dots ,X_d\in X_x$, since for $Y_1,\dots ,Y_d$ which also satisfies ${\tilde{Y}}_1\wedge \dots \wedge \widetilde{Y_d}={\omega }_x$, one can express $Y_i={\sum }_j{a}_{ij}{X}_j$, then using Exercise 2.10, we see that $\det a_{ij}=1$. Thus $\omega (df(Y_1),\dots ,df(Y_d),N(x))=1$ also. Then since $df(X_x{)}^{\perp }$ is $1$-dimensional, $N(x)$ which satisfies the condition above is uniquely determined. Also, it is by definition nowhere-vanishing. It remains to show that $N$ is smooth, but locally on the neighborhood of $x\in X$ we can choose smooth vector field $X_1,\dots ,X_d$ which satisfies $\widetilde{X_1}\wedge \dots \wedge \widetilde{X_d}={\omega }^{\prime }$ on the neighborhood, and $df(X_1),\dots ,df(X_d)$ is also smooth on the neighborhood of $f(x)$ over $f(X)$. Therefore locally $N$ is also smooth (locally you can use the Gram–Schmidt process, and scaling if necessary, then the result $N(x)$ would be a smooth function of the local coordinates of $x$).
2.	Taking a nowhere-vanishing $n$-form on ${\mathbb{P}}^n$ is equivalent to taking a nowhere-vanishing $n$-form on $S^n$ such that it is preserved by antipodal map. Now regard $S^n$ as imbedded to ${\mathbb{R}}^{n+1}$, and let $N$ be the outward unit normal vector field on $S^n$ and $\omega =d{r}_1\wedge \dots \wedge d{r}_{n+1}$. Then ${\omega }^{\prime }=i(N)\omega$ is nowhere-vanishing $n$-form on $S^n$. If $\phi :{\mathbb{R}}^{n+1}\to {\mathbb{R}}^{n+1},x\mapsto -x$, $\phi {|}_{S^n}$ is the antipodal map, and$$\delta {\phi }_x{\omega }_x^{\prime }=(-1{)}^{n}i(N(x)){\omega }_{-x}=(-1{)}^{n+1}i(N(-x){\omega }_{-x})=(-1{)}^{n+1}{\omega }_{-x}^{\prime }.$$Therefore, if $n$ is odd, $\phi$ is orientation-preserving, therefore ${\mathbb{P}}^n$ is orientable. But if $n$ is even, $\phi$ is orientation-reversing, so we cannot take a nowhere-vanishing $n$-form on $S^n$ which is preserved by antipodal map, which means ${\mathbb{P}}^n$ is not orientable.
3.	At first find local vector fields $X_1,\dots ,X_n$ near $p\in X$ s.t. $X_{1,p},\dots ,X_{n,p}$ is orthonormal, and let $L(m)=\{g(X_{i,m},X_{j,m}){\}}_{i,j}$ where $g$ is the given Riemannian metric. Then near $p\in X$, it is a smooth function to the set of $n\times n$ matrix. Also $L(p)=I$. Therefore locally $L$ is positive definite and also symmetric by definition. Thus there is $R$ s.t. $R^2=L$, which is also smooth and symmetric. Define $(Y_1,\dots ,Y_n{)}^t=R^{-1}(X_1,\dots ,X_n{)}^t$. Then we see that $g(Y_i,Y_j)={\delta }_{ij}$, which means that they are local orthonormal frame fields.
4.	Use the notation on the problem. Let $V=f_1{e}_1+\dots +f_n{e}_n$. Then $\tilde{V}=f_1{\omega }_1+\dots +f_n{\omega }_n$, so$$\ast \tilde{V}=f_1{\omega }_2\wedge \dots \wedge {\omega }_n-\dots +(-1{)}^{n-1}{f}_n{\omega }_1\wedge \dots \wedge {\omega }_{n-1}.$$Therefore on $\partial D$,$$\ast \tilde{V}=f_1{\omega }_2\wedge \dots \wedge {\omega }_n$$since ${\omega }_1{|}_{\partial D}=0$. On the other hand, $⟨V,\vec{n}⟩=f_1$ and ${\omega }_2\wedge \dots \wedge {\omega }_n$ is the volume form on $\partial D$. Thus$${\int }_{\partial D}\ast \tilde{V}={\int }_{\partial D}{f}_1.$$Since$${\int }_D\mathrm{div}V={\int }_D\ast d\ast \tilde{V}={\int }_{D}d\ast \tilde{V}={\int }_{\partial D}\ast \tilde{V}$$by Stokes’ theorem, the result follows.
5.	$$\begin{array}{rl}-{\int }_{D}f\Delta g& ={\int }_{D}f(\ast d\ast dg)={\int }_{D}f(d\ast dg)\\ & ={\int }_{D}d(f\ast dg)-{\int }_{D}df\wedge \ast dg\\ & ={\int }_{D}d(f\ast dg)-{\int }_D⟨df,dg⟩.\end{array}$$Therefore$$\begin{array}{rl}{\int }_D⟨\mathrm{grad}f,\mathrm{grad}g⟩-{\int }_{D}f\Delta g& ={\int }_D⟨\widetilde{df},\widetilde{dg}⟩-{\int }_{D}f\Delta g={\int }_{D}d(f\ast dg)\\ & ={\int }_{\partial D}f\ast dg={\int }_{\partial D}\ast (fdg).\end{array}$$Using the proof of Exercise 4,$$\ast (fdg)=⟨\widetilde{fdg},\vec{n}⟩=f\frac{\partial g}{\partial n}$$on $\partial D$. Therefore Green’s 1st equation follows. For the 2nd equation, interchange $f$ and $g$ and subtract from the former one.
6.	If we establish the latter equation, the former one directly follows. For the latter one, since the vector space of $1$-forms is $1$-dimensional, we only need to show that they have the same nonzero value for some arbitrary $n$-tuples of vectors. If we put $X_1,\dots ,X_n$, the left side becomes $\omega (X_1,\dots ,X_n{)}^2$, which is the same as ${\tilde{X}}_1\wedge \dots \wedge {\tilde{X}}_n(X_1,\dots ,X_n)=\det \{⟨X_i,X_j⟩\}$ since $\omega$ is the volume form.
7.	Since $\bar{\lambda }(\sigma \tau )=\bar{\lambda }(\sigma )\bar{\lambda }(\tau )$, we only need to show that it is ${\mathcal{C}}^{\infty }$ at $e$. On the local chart of $e$ with coordinate functions $x_1,\dots ,x_n$ centered at $e$, $\omega$ can be expressed as $f(x_1,\dots ,x_n)d{x}_1\wedge \dots \wedge d{x}_n$ where $f$ is ${\mathcal{C}}^{\infty }$. Since $\bar{\lambda }(\sigma ){\omega }_e=\delta r_{\sigma }({\omega }_{\sigma })$,$$\begin{array}{rl}\bar{\lambda }(\sigma )f(0,\dots ,0)d{x}_1\wedge \dots \wedge d{x}_n& =f(x_1(\sigma ),\dots ,x_n(\sigma ))d(x_1\circ r_{\sigma })\wedge \dots \wedge d(x_n\circ r_{\sigma })\\ & =f(x_1(\sigma ),\dots ,x_n(\sigma ))\frac{\det \{⟨d{x}_i\circ r_{\sigma },d{x}_j\circ r_{\sigma }⟩\}}{\det \{⟨d{x}_i,d{x}_j⟩\}}d{x}_1\wedge \dots \wedge d{x}_n,\end{array}$$so$$\bar{\lambda }(\sigma )=\frac{f(x_1(\sigma ),\dots ,x_n(\sigma ))}{f(0,\dots ,0)}\frac{\det \{⟨d{x}_i\circ r_{\sigma },d{x}_j\circ r_{\sigma }⟩\}}{\det \{⟨d{x}_i,d{x}_j⟩\}}$$which is ${\mathcal{C}}^{\infty }$.
8.	(This is one of the characteristics of the Haar measure on a unimodular Lie group. Here is the sketch of the proof.) Since$$\delta r_{\sigma }\delta \alpha (\omega )=\delta \alpha \delta l_{{\sigma }^{-1}}(\omega )=\delta \alpha (\omega ),$$it is right invariant. Let ${\omega }^{\prime }=\delta \alpha (\omega )$ and think of $\mu ,{\mu }^{\prime }$ as measures on $G$ corresponding to $\omega$ and ${\omega }^{\prime }$ respectively. Then we can define $f\mapsto {\int }_{G}f{\mu }^{\prime }$ which is a well–defined right–invariant integral on $G$. Now consider$${\int }_{G\times G}f(\tau \sigma )\mu (\sigma ){\mu }^{\prime }(\tau );$$if we first integrate with respect to $\mu$ it gives $\left({\int }_{G}f\omega \right)\left({\int }_G{\omega }^{\prime }\right)$. And if we integrate with respect to ${\mu }^{\prime }$ first, we get ${\int }_{G}f{\omega }^{\prime }$. By Fubini’s theorem, they must be the same. Also,$${\int }_G{\omega }^{\prime }={\int }_G\delta \alpha (\omega )={\int }_{\alpha (G)}\omega =1,$$so ${\int }_{G}f\omega ={\int }_{G}f{\omega }^{\prime }$. Since ${\int }_{G}f{\omega }^{\prime }={\int }_{\alpha (G)}(f\circ \alpha )\omega$, the result follows.
9.	$(\Rightarrow )$ Since$$⟨d{l}_{\sigma }X,d{l}_{\sigma }Y⟩=⟨d{r}_{\tau }X,d{r}_{\tau }Y⟩=⟨X,Y⟩,$$$\mathrm{Ad}(G)$ in $\mathrm{Aut}(g)$ is actually in $O(g)$. Since $O(g)$ is compact, the closure of $\mathrm{Ad}(G)$ is also compact. $(\Leftarrow )$ Give to $g$ a non-degenerate inner product. Let $\ell \in \mathrm{Ad}(G)$. Since $\mathrm{Ad}(G)$ is a subgroup of $\mathrm{Aut}(g)$ and has compact closure, ${\ell }^n$ should be bounded for all $n\in \mathbb{Z}$. It is satisfied only when $\Vert \ell \Vert =1$, so all elements of $\mathrm{Ad}(G)$ preserve the inner product. Now for $\sigma \in G$ and $X,Y\in G_{\sigma }$, define$$⟨X,Y{⟩}_{\sigma }={⟨d{l}_{{\sigma }^{-1}}X,d{l}_{{\sigma }^{-1}}Y⟩}_e.$$Then by definition it is left–invariant, and since$$\begin{array}{rl}⟨d{r}_{\tau }X,d{r}_{\tau }Y{⟩}_{\sigma \tau }& =⟨d{l}_{{\tau }^{-1}{\sigma }^{-1}}d{r}_{\tau }X,d{l}_{{\tau }^{-1}{\sigma }^{-1}}d{r}_{\tau }Y{⟩}_e\\ & =⟨dA{d}_{{\tau }^{-1}}d{l}_{{\sigma }^{-1}}X,dA{d}_{{\tau }^{-1}}d{l}_{{\sigma }^{-1}}Y{⟩}_e\\ & =⟨d{l}_{{\sigma }^{-1}}X,d{l}_{{\sigma }^{-1}}Y{⟩}_e=⟨X,Y{⟩}_{\sigma },\end{array}$$it is also right–invariant.
10.	(a) Since ${\mathbb{R}}^n\cong D(0,1)\subset {\mathbb{R}}^n$, the singular homology groups of the two must be the same. Then by the de Rham theorem, it is also the same as the de Rham cohomology groups. By Poincaré Lemma, it must be zero for given conditions. (b) By the de Rham theorem, it suffices to show that the $0$-th singular homology group is $\mathbb{R}$. But since it is isomorphic to the free $\mathbb{R}$-module generated by the path-components, the result follows.
11.	By the de Rham theorem, it suffices to calculate the singular homology groups. But it can be easily seen that ${}_{\infty }{H}_0(M;\mathbb{R})=\mathbb{R},{}_{\infty }{H}_1(M;\mathbb{R})=\mathbb{R},{}_{\infty }{H}_n(M;\mathbb{R})=0$ for $n\ge 2$.
12.	$d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^k\alpha \wedge d\beta =0$. If $\beta =d\gamma$, then $d(\alpha \wedge \gamma )=d\alpha \wedge \gamma +(-1{)}^k\alpha \wedge \beta$, so $\alpha \wedge \beta$ is exact.
13.	By Stokes’ theorem,$${\int }_z\alpha ={\int }_{\mathrm{int}z}d\alpha =-8{\int }_{\mathrm{int}z}dx\wedge dy=-8.$$
14.	It is closed by direct calculation. Also $d(\sin xy+x^2)=\alpha$, it is exact. Therefore,$${\int }_z\alpha ={\int }_{\mathrm{int}z}ddf=0.$$
15.	It is closed by direct calculation. Also,$${\int }_{S^1}\alpha ={\int }_{S^1}\frac{xdy-ydx}{2\pi }={\int }_D\frac{dx\wedge dy}{\pi }=1.$$If $\delta i(\alpha )$ is exact, then its integral on $S^1$ must be zero since $S^1$ is a compact manifold without boundary. Therefore $\delta i(\alpha )$ is not exact. Also, if $\alpha$ is exact, then $\delta i(\alpha )$ is also exact since $\delta$ and $d$ commute. Therefore, $\alpha$ is also not exact.
16.	(a) It is because ${}_{\infty }{H}_1(S^2;\mathbb{R})=0$, and by the de Rham theorem. (b) By direct calculation. (c)$${\int }_{S^2}\sigma =3\mathrm{Vol}(S^2),$$so it is not zero. If it is exact it must be zero since $S^2$ is a compact manifold without boundary. (d)$$\ast \alpha =\frac{{\sum }_{i=1}^n(-1{)}^{i-1}{r}_{i}d{r}_1\wedge \dots \wedge {\widehat{dr}}_i\wedge \dots \wedge d{r}_n}{({\sum }_{i=1}^n{r}_i^2{)}^{\frac{n}{2}}}.$$It is closed by calculation similar to (b). (e)$${\int }_{S^{n-1}}\ast \alpha =n\mathrm{Vol}(S^{n-1}),$$so it is not zero, and $\ast \alpha$ is not exact.
17.	$H_{\text{de Rham}}^1(S^2;\mathbb{R})=0\ne {\mathbb{R}}^2=H_{\text{de Rham}}^1(T^2;\mathbb{R})$, using the singular homology theory and de Rham theorem.
18.	(a) Since $H_1(M;\mathbb{R})=0$. (b) $\sigma$ in Exercise 16(b) satisfies the condition. (c) Since $H_2(D;\mathbb{R})=0$ and $0\ne [\sigma ]\in H_2(M;\mathbb{R})$.
19.	Let $t$ be a variable corresponding to the interval $(-\epsilon ,1+\epsilon )$ on $M\times (-\epsilon ,1+\epsilon )$. For a $k$-form $\omega ={\omega }_1+dt\wedge {\omega }_2$ on $M\times (-\epsilon ,1+\epsilon )$, $i\left(\frac{\partial }{\partial t}\right)\omega ={\omega }_2$ is a $(k-1)$-form, and its expression does not contain $dt$ term. Therefore, we can define$${\int }_0^{1}i\left(\frac{\partial }{\partial t}\right)\omega dt={\int }_0^1{\omega }_{2}dt$$which is a $(k-1)$-form on $M$. Denote it $D_k(\alpha )$. For convenience of the notation, denote $d_M$ as a differential operator on $M$, and $d=d_M+d_t$ as a differential operator on $M\times (-\epsilon ,1+\epsilon )$. Then$$\begin{array}{rl}{D}_{k}d(\omega )& ={\int }_0^{1}i\left(\frac{\partial }{\partial t}\right)(d{\omega }_1-dt\wedge d{\omega }_2)dt\\ & ={\int }_0^1{L}_{\frac{\partial }{\partial t}}({\omega }_1)-d_M{\omega }_{2}dt,\end{array}$$and$$d{D}_k(\omega )=d_M{\int }_0^1{\omega }_{2}dt={\int }_0^1{d}_M{\omega }_{2}dt.$$Therefore$$d{D}_k(\omega )+D_{k}d(\omega )=i_1^{\ast }{\omega }_1-i_0^{\ast }{\omega }_1=i_1^{\ast }\omega -i_0^{\ast }\omega$$by definition of Lie derivative and $i_0,i_1$ ($i_0^{\ast }(dt\wedge {\omega }_2)=i_1^{\ast }(dt\wedge {\omega }_2)=0$). It means $i_0^{\ast }$ and $i_1^{\ast }$ are the same map on the cohomology level. Since $f=F\circ i_0$ and $g=F\circ i_1$, $f^{\ast }=i_0^{\ast }\circ F^{\ast }$ and $g^{\ast }=i_1^{\ast }\circ F^{\ast }$ also induce the same cohomology maps.
20.	(a) Since the vector space of $n$-form on $M$ is $1$-dimensional on each fiber, we only need to show that $\omega =\delta f(i(\vec{n})(d{r}_1\wedge \dots \wedge d{r}_{n+1}))$ satisfies $\omega (v_1,\dots ,v_n)=1$ for each point $m\in M$ and $v_1,\dots ,v_n\in M_m$ which is an oriented orthonormal basis of $M_m$. Since$$\omega (v_1,\dots ,v_n)=d{r}_1\wedge \dots \wedge d{r}_{n+1}(\vec{n},df(v_1),\dots ,df(v_n))=1$$by assumption, the result follows. (b) From vector calculus we obtain$$\vec{n}=\frac{1}{\sqrt{f_x^2+f_y^2+1}}\left(-f_x\frac{\partial }{\partial x}-f_y\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right).$$Therefore,$$\begin{array}{rl}\omega & =\delta \phi (i(\vec{n})(dx\wedge dy\wedge dz))\\ & =\delta \phi \left(\frac{1}{\sqrt{f_x^2+f_y^2+1}}\left(-f_{x}dy\wedge dz+f_{y}dx\wedge dz+dx\wedge dy\right)\right)\\ & =\frac{1}{\sqrt{f_x^2+f_y^2+1}}\left(-f_{x}dy\wedge df+f_{y}dx\wedge df+dx\wedge dy\right)\\ & =\frac{1}{\sqrt{f_x^2+f_y^2+1}}\left(f_x^2+f_y^2+1\right)dx\wedge dy=\left(\sqrt{f_x^2+f_y^2+1}\right)dx\wedge dy.\end{array}$$