Chapter 6 The Hodge Theorem

1.	If $\Delta$ operates on $E^p(M)$, then$$\Delta =\delta d+d\delta =(-1{)}^{n(p+2)+1}\ast d\ast d+(-1{)}^{n(p+1)+1}d\ast d\ast .$$(Be careful with the signs.) Therefore$$\ast \Delta =(-1{)}^{n(p+2)+1+(n-p)p}d\ast d+(-1{)}^{n(p+1)+1}\ast d\ast d\ast$$and$$\Delta \ast =(-1{)}^{n(n-p+2)+1}\ast d\ast d\ast +(-1{)}^{n(n-p+1)+1+p(n-p)}d\ast d.$$Since$$n(p+2)+1+(n-p)p\equiv n(n-p+1)+1+p(n-p)\mathrm{mod}2$$and$$n(p+1)+1\equiv n(n-p+2)+1\mathrm{mod}2,$$$\ast \Delta =\Delta \ast$ as desired.
2.	(a) Obviously $G$ is linear. To show that it is bounded, note that there exists a $c>0$ such that $\Vert \beta \Vert \le$ $c\Vert \Delta \beta \Vert$ for all $\beta \in (H^p{)}^{\perp }$ by 6.8(4). Hence if we let $G(\alpha )=\omega$, then$$\Vert G(\alpha )\Vert =\Vert \omega \Vert \le c\Vert \Delta \omega \Vert =c\Vert \alpha -H(\alpha )\Vert \le c\Vert \alpha \Vert .$$Thus $G$ is bounded. (b) For all $\alpha ,\beta \in (H^p{)}^{\perp }$, $\Delta G\alpha =\alpha ,\Delta G\beta =\beta$ by definition. Thus,$$⟨G\alpha ,\beta ⟩=⟨G\alpha ,\Delta G\beta ⟩=⟨\Delta G\alpha ,G\beta ⟩=⟨\alpha ,G\beta ⟩.$$ (c) Suppose $\{{\alpha }_n\}$ is a bounded sequence. Then $\{G{\alpha }_n\}$ is also bounded by (a), and $\{\Delta G{\alpha }_n\}$ is also bounded because$$\Vert \Delta G{\alpha }_n\Vert =\Vert {\alpha }_n-H({\alpha }_n)\Vert \le \Vert {\alpha }_n\Vert .$$Then use 6.6 to obtain the result.
3.	It suffices to show that$${\int }_{{\mathbb{R}}^n}{\left(1+x_1^2+\dots +x_n^2\right)}^{-k}d{x}_1\dots d{x}_n$$converges for $k\ge \left[\frac{n}{2}\right]+1$. By a change of variables $y_i=\frac{x_i}{\sqrt{1+x_n^2}}$ for $1\le i\le n-1$ and $y_n=x_n$, it is equal to$$\begin{array}{rl}& {\int }_{{\mathbb{R}}^n}{\left(1+y_n^2\right)}^{-k}{\left(1+y_1^2+\dots +y_{n-1}^2\right)}^{-k}{\left(1+y_n^2\right)}^{\frac{n-1}{2}}d{y}_1\dots d{y}_n\\ & ={\int }_{{\mathbb{R}}^n}{\left(1+y_1^2+\dots +y_{n-1}^2\right)}^{-k}{\left(1+y_n^2\right)}^{\frac{n-1}{2}-k}d{y}_1\dots d{y}_n\\ & ={\int }_{{\mathbb{R}}^{n-1}}{\left(1+y_1^2+\dots +y_{n-1}^2\right)}^{-k}d{y}_1\dots d{y}_{n-1}{\int }_{\mathbb{R}}{\left(1+y_n^2\right)}^{\frac{n-1}{2}-k}d{y}_n\end{array}$$(because $\frac{\partial (x_1,\dots ,x_n)}{\partial (y_1\dots ,y_n)}=(1+y_n^2{)}^{\frac{n-1}{2}}$). The former part converges by induction on $n$, and the latter one converges when $2k-(n-1)>1$, or $k>\frac{n}{2}$. It is equivalent to $k\ge \left[\frac{n}{2}\right]+1$ because $k$ is an integer.
4.	$$\sum _{[\alpha ]=0}^t\Vert D^{\alpha }\phi {\Vert }^2=\sum _{[\alpha ]=0}^t\sum _{\xi }{\xi }^{2\alpha }|{\phi }_{\xi }{|}^2.$$Thus we only need to show that there exists $c>0$ which depends on $t$ and $n$ such that$$c(1+|\xi {|}^2{)}^t\le \sum _{[\alpha ]=0}^t{\xi }^{2\alpha }\le (1+|\xi {|}^2{)}^t$$for all $\xi$. The second inequality is clear because each term in $\sum _{[\alpha ]=0}^t{\xi }^{2\alpha }$ is in the expansion of $(1+|\xi {|}^2{)}^t$ with coefficients $\ge 1$. Also, if we set $c=n^{-t}$, then it is easy to see that the first inequality holds for similar reasons.
5.	Since both are defined by pointwise inner products, there exists a positive-definite Hermitian matrix $M_x$ for each $x\in {\mathbb{R}}^n$ such that $\phi {\circ }^{\prime }\bar{\psi }=\phi \circ \overline{A\psi }$. Also it is smooth because the volume form on $M$ is smooth.
6.	I will deal only with the case of $\alpha =fd{x}_1\wedge \dots \wedge d{x}_p$. The general case is similar. By definition,$$d\alpha =(-1{)}^p\sum _{i=p+1}^n\frac{\partial f}{\partial x_i}d{x}_1\wedge \dots \wedge d{x}_p\wedge d{x}_i,$$and$$\ast \alpha =fd{x}_{p+1}\wedge \dots \wedge d{x}_n$$(assuming $\ast 1=d{x}_1\wedge \dots \wedge d{x}_n$). Also$$\delta \alpha =(-1{)}^{n(p+1)+1}\ast d\ast \alpha =(-1{)}^{n(p+1)+1}\sum _{i=1}^p(-1{)}^{p(i-1)+(p-1)(p-i)}\frac{\partial f}{\partial x_i}d{x}_1\wedge \dots \wedge {\widehat{dx}}_i\wedge \dots \wedge d{x}_p.$$Furthermore, by tedious calculation, we get$$\Delta \alpha =-\sum _{i=1}^n\frac{{\partial }^{2}f}{\partial x_i^2}d{x}_1\wedge \dots \wedge d{x}_p.$$
7.	By 6.16(14),$$-\frac{{\partial }^2\phi }{\partial x\partial y}=\sum _{(a,b)\in {\mathbb{Z}}^2}ab{\phi }_{(a,b)}{e}^{i(ax+by)}.$$Thus$${\Vert \frac{{\partial }^2\phi }{\partial x\partial y}\Vert }^2=\sum _{(a,b)\in {\mathbb{Z}}^2}{a}^2{b}^2|{\phi }_{(a,b)}{|}^2$$by the Parseval identity. Similarly, we get$$\Vert \Delta \phi {\Vert }^2=\sum _{(a,b)\in {\mathbb{Z}}^2}(a^2+b^2{)}^2|{\phi }_{(a,b)}{|}^2.$$Thus the given inequality follows from $a^2{b}^2\le \frac{(a^2+b^2{)}^2}{4}$.
8.	If $\{f_n\}$ is bounded in $C^1$, then it is uniformly bounded and equicontinuous by the assumption, thus by the Arzelà–Ascoli theorem it has a convergent subsequence.
9.	The differential operators given in the following 4 problems are all elliptic. Therefore we can deduce that a solution exists if and only if the given function is in the orthogonal complement of the kernel of the formal adjoint. (a) Adjoint: $D^{\ast }(u)=-u^{\prime }$ (b) Adjoint: $D^{\ast }(u)=-u^{\prime }-u$ (c) Adjoint: $D^{\ast }(u)=u^{\prime \prime }$ (d) Adjoint: $D^{\ast }(u)=u^{\prime \prime }+\pi u$
10.	By the Hodge decomposition theorem, we may represent any form $\omega$ as$$\omega =\delta \alpha +d\beta +\gamma$$with $\gamma$ harmonic. If it is closed, then $d\omega =d\delta \alpha =0$. Since $⟨\delta \alpha ,\delta \alpha ⟩=⟨d\delta \alpha ,\alpha ⟩=0$, we see that $\delta \alpha =0$. Hence $\omega =d\beta +\gamma$ and the result follows.
11.	$(\Rightarrow )$ Since $\mathcal{P}$ is a (infinite dimensional) vector space, the element $u=\sum _{\xi }{u}_{\xi }{e}^{ix\cdot \xi }$ as a formal sum is uniquely determined. It remains to show that in fact $u\in H_{-\infty }$. By 6.22(2), we know that$$|l(\phi )|\le \mathrm{const}\sum _{[\alpha ]\le k}\Vert D^{\alpha }\phi {\Vert }_{\infty }\le \mathrm{const}\Vert \phi {\Vert }_N$$for some $N\gg 0$. In other words,$$\left|\sum _{\xi }{u}_{\xi }{\phi }_{\xi }\right|\le \mathrm{const}\Vert \phi {\Vert }_N.$$By squaring both sides, it becomes$${\left|\sum _{\xi }{u}_{\xi }{\phi }_{\xi }\right|}^2\le \mathrm{const}\Vert \phi {\Vert }_N^2.$$Since multiplying $e^{i\theta }$ $(\theta \in \mathbb{R})$ does not change the absolute values of ${\phi }_{\xi }$, we may assume that $u_{\xi }\Vert {\phi }_{\xi }$. Then$$\sum _{\xi }|u_{\xi }{|}^2|{\phi }_{\xi }{|}^2\le {\left(\sum _{\xi }|u_{\xi }||{\phi }_{\xi }|\right)}^2={\left|\sum _{\zeta }{u}_{\xi }{\phi }_{\xi }\right|}^2\le \mathrm{const}\Vert \phi {\Vert }_N^2.$$Since $\mathcal{P}$ is dense in $H_{-N}$, the inequality above is also true when $\phi \in H_{-N}$. Now set$$|{\phi }_{\xi }|=(1+|\xi {|}^2{)}^{-\frac{N+M}{2}}$$for some $M>\left[\frac{n}{2}\right]+1$. Then $\phi \in H_{-N}$ by 6.16(7), and the inequality says that$$\sum _{\xi }|u_{\xi }{|}^2|{\phi }_{\xi }{|}^2=\sum _{\xi }|u_{\xi }{|}^2(1+|\xi {|}^2{)}^{-N-M}<\infty ,$$that is to say $u\in H_{-N-M}$. $(\Leftarrow )$ If $u\in H_{-\infty }$, there exists $s<0$ such that $u\in H_s$. Then$$\begin{array}{rl}|l(\phi )|& =|⟨u,\phi {⟩}_0|\le \Vert u{\Vert }_s\Vert \phi {\Vert }_{-s}\\ & \le \text{const}\sqrt{\sum _{[\alpha ]\le -s}\Vert D^{\alpha }\phi {\Vert }^2}\\ & \le \text{const}\sqrt{\sum _{[\alpha ]\le -s}\Vert D^{\alpha }\phi {\Vert }_{\infty }^2}\\ & \le \text{const}\sum _{[\alpha ]\le -s}\Vert D^{\alpha }\phi {\Vert }_{\infty }\end{array}$$as desired.
12.	It suffices to show that if ${\int }_M\alpha =0$, then it is an exact form. $E^n=\mathrm{im}d\oplus H_n$ because $\mathrm{im}\delta =0$. Let $\alpha =d\gamma +\epsilon$ with $\epsilon$ harmonic. Since $[\alpha ]\mapsto {\int }_M\alpha$ is the natural isomorphism, $[\alpha ]=[\epsilon ]=0$. Hence by 6.11, $\epsilon =0$. In other words, $\alpha$ is exact.
13.	Let ${\phi }_n=\sum _{a\in \mathbb{Z}}{u}_a^n{e}^{iax}$, where$$u_a^n=\{\begin{array}{ll}\frac{1}{a^2}& \text{ if }|a|<n\\ 0& \text{ otherwise }\end{array}.$$Then$$\Vert {\phi }_n{\Vert }^2=\sum _{a\in \mathbb{Z}}|u_a^n{|}^2<\sum _{a\in \mathbb{Z}}\frac{1}{a^4}$$and$$\Vert \Delta {\phi }_n{\Vert }^2=\sum _{a\in \mathbb{Z}}{a}^2|u_a^n{|}^2<\sum _{a\in \mathbb{Z}}\frac{1}{a^2}$$are both bounded. Also, ${\phi }_n$ is a finite sum, so ${\phi }_n\in \mathcal{P}$. However, its limit$$\phi =\sum _{a\in \mathbb{Z}}\frac{1}{a^2}{e}^{iax}\notin H_3.$$Thus it is not in $\mathcal{P}$.
14.	The statement can be proved by following the proof of 6.32. The only property of $\Delta$ used here is that $\Delta$ is an elliptic operator. Also, by a “Reduction to the Periodic Case” argument, we only need to show this locally.
15.	It is elliptic because $x+iy\ne 0$ for $(x,y)\ne (0,0)$. Hence every holomorphic function is ${\mathcal{C}}^{\infty }$. Also every holomorphic function is harmonic because$$\Delta =\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}=\left(\frac{\partial }{\partial x}-i\frac{\partial }{\partial y}\right)\left(\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}\right).$$If a holomorphic function $f$ has compact support, then by Green’s 1st identity,$${\int }_{\partial D}f\frac{\partial f}{\partial n}={\int }_D⟨\mathrm{grad}f,\mathrm{grad}f⟩-{\int }_{D}f\Delta f$$for every disc $D$. Since $\Delta f=0$, if we take a sufficiently large disc $D$, then it becomes$${\int }_D⟨\mathrm{grad}f,\mathrm{grad}f⟩=0.$$By letting the radius of $D$ go to infinity, we see that $${\int }_{\mathbb{C}}⟨\mathrm{grad}f,\mathrm{grad}f⟩=0,$$and it means $\mathrm{grad}f=0$ everywhere, or $f$ is constant. But $f$ has compact support, so $f$ has to be zero everywhere.
16.	(a) If $\Delta u=\lambda u$ and $u\ne 0$,$$\lambda ⟨u,u⟩=⟨\lambda u,u⟩=⟨\Delta u,u⟩=⟨(d\delta +\delta d)u,u⟩=⟨du,du⟩+⟨\delta u,\delta u⟩\ge 0.$$Thus $\lambda \ge 0$. (b) If not, it contradicts 6.6 if we have an orthonormal sequence. (c) If not, it also contradicts 6.6 if we have an orthonormal sequence. (d) If $\Delta u=\lambda u$ and $\Delta v={\lambda }^{\prime }v$, then$$\lambda ⟨u,v⟩=⟨\Delta u,v⟩=⟨u,\Delta v⟩={\lambda }^{\prime }⟨u,v⟩.$$Thus $⟨u,v⟩=0$. (e)$$⟨{\psi }_j,G{\psi }_j⟩=⟨\Delta G{\psi }_j,G{\psi }_j⟩=⟨dG{\psi }_j,dG{\psi }_j⟩+⟨\delta G{\psi }_j,\delta G{\psi }_j⟩\ge 0.$$For the latter part, we need to show that $l((\Delta -1/\eta {)}^{\ast }\alpha )=0$. It is equivalent to$$\underset{j\to \infty }{\lim }\eta ⟨G{\phi }_j,(\Delta -1/\eta {)}^{\ast }\alpha ⟩=0\iff \underset{j\to \infty }{\lim }\eta ⟨(\Delta -1/\eta )G{\phi }_j,\alpha ⟩=0\iff \underset{j\to \infty }{\lim }⟨\eta {\phi }_j-G{\phi }_j,\alpha ⟩=0,$$which is true. Also it is not trivial; take $N$ such that $\Vert G{\phi }_N-\eta {\phi }_N\Vert <\epsilon$ and $\Vert G{\phi }_i-G{\phi }_N\Vert <\epsilon$ for $i>N$. Since $\Vert {\phi }_N\Vert =1$ and by the inequality$$|⟨v,w⟩|\ge \Vert v{\Vert }^2-|⟨v,v-w⟩|\ge \Vert v\Vert (\Vert v\Vert -\Vert v-w\Vert ),$$we see that $|⟨G{\phi }_i,\eta {\phi }_N⟩|\ge \eta (\eta -2\epsilon )>0$ for sufficiently small $\epsilon >0$. Thus $l({\phi }_N)\ne 0$. (f) It is the same as (e). (g) $\alpha -\sum _{i=1}^k⟨\alpha ,u_i⟩u_i\in (H^p{)}^{\perp }$, so there exists $\beta \in (H^p{)}^{\perp }$. If $k+1\le j$,$$⟨\alpha ,u_j⟩=⟨\alpha -\sum _{i=1}^k⟨\alpha ,u_i⟩u_i,u_j⟩=⟨G\beta ,u_j⟩=⟨\beta ,G{u}_j⟩=\frac{1}{{\lambda }_j}⟨\beta ,u_j⟩.$$Thus$$⟨\alpha ,u_j⟩u_j=\frac{1}{{\lambda }_j}⟨\beta ,u_j⟩u_j=G(⟨\beta ,u_j⟩u_j).$$Then it follows from it and the definition of $\lambda$. ${\lambda }_n\to \infty$ because of (c). (h) Locally it is true, and by the compactness of $M$, we can find $c$ and $k$ such that it is still true globally. Then it follows from (g).
17.	By the Hodge decomposition, $E^p(M)=\mathrm{im}\Delta \oplus H^p=\mathrm{im}{\Delta }^2\oplus V\oplus H^p$ with $V=\ker {\Delta }^2\cap \mathrm{im}\Delta$. If $\Delta \alpha \in V$, $⟨\Delta \alpha ,\Delta \alpha ⟩=⟨{\Delta }^2\alpha ,\alpha ⟩=0$, so $\Delta \alpha =0$. Thus $V=0$ and $E^p(M)=\mathrm{im}{\Delta }^2\oplus H^p$. It means ${\Delta }^2\alpha =\beta$ is solvable if and only if $\beta \perp H^p$.
18.	We can assume the matrix consisting of the coefficients of the highest order terms is symmetric because $\frac{\partial }{\partial x_i}$ and $\frac{\partial }{\partial x_j}$ commutes. Also, $L$ is elliptic at $x$ if for all $v\in {\mathbb{R}}^n\backslash \{0\}$, $v^t(a_{ij}(x){)}_{ij}v>0$, which means that $(a_{ij}(x){)}_{ij}$ is positive definite. But for $\square$, the corresponding matrix $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ is not positive definite, so it is not elliptic. For example, if we let$$u(x,y)=\{\begin{array}{ll}{x}^3+3x{y}^2& \text{ if }x\ge |y|\\ & \\ 3x^{2}y+y^3& \text{ if }-t\le x\le t\\ & \\ -3x^{2}y-y^3& \text{ if }t\le x\le -t\\ & \\ -x^3-3x{y}^2& \text{ if }x\le |t|\end{array}$$it is easy to see that $\square u=0$, but $u\notin {\mathcal{C}}^3$.
19.	Since $\Delta +c$ is elliptic, if $\beta \perp \ker (\Delta +c)$ then the equation can be solved. But $(\Delta +c)\alpha =0\iff \Delta \alpha =-c\alpha \iff \alpha =0$ by assumption, so it can be solved for all $\beta \in E^p(M)$. (In fact, the minimal eigenvalue of $\Delta$ is $0$.)
20.	The first assertion is true because $l_{\sigma }$ are isometries, thus they commute with $d$ and $\delta$. Then the second assertion directly follows from it. The third assertion is that ${\phi }_i\circ l_{\sigma }$ is continuous as a function of $\sigma$, which is also true by the properties of Lie group.
21.	The formal adjoint exists by argument similar to 6.24 (integration by parts) and the fact that the integral of exact forms over $M$ is zero. Also, the proof of 6.5 and 6.6 is still valid because it is valid locally, and we can glue the solution with respect to the transition maps of vector bundles. Then (a) follows from 6.6, and (b) follows from arguments similar to 6.8.
22.	For $n$ sufficiently large, or $n>10$, then$$\Vert {\phi }_n{\Vert }_{\infty }=\phi (0,0)=\log \log n.$$On the other hand, by 6.16(15),$$\mathrm{const}\Vert {\phi }_n{\Vert }_1^2\le \Vert {\phi }_n{\Vert }^2+{\Vert \frac{\partial {\phi }_n}{\partial x}\Vert }^2+{\Vert \frac{\partial {\phi }_n}{\partial y}\Vert }^2.$$Since $\{{\phi }_n\}$ is monotonically increasing, $$\Vert {\phi }_n{\Vert }^2\to {\int }_{r\le \frac{1}{2}}{\left(\log \log \frac{1}{r}\right)}^{2}dxdy<\infty$$by the monotone convergence theorem. Also,$$\frac{\partial {\phi }_n}{\partial x}=\frac{x}{r\left(\frac{1}{n}+r\right)\log \left(\frac{1}{n}+r\right)}$$and we can use a similar argument to see that$${\Vert \frac{\partial {\phi }_n}{\partial x}\Vert }^2\to {\int }_{r\le \frac{1}{2}}{\left(\frac{x}{r^2\log r}\right)}^{2}dxdy=\frac{\pi }{\log 2}.$$Similarly,$${\Vert \frac{\partial {\phi }_n}{\partial y}\Vert }^2\to \frac{\pi }{\log 2}.$$Thus $\Vert {\phi }_n{\Vert }_1$ is bounded and there is no constant $c$ satisfying the given condition.
23.	By the assumption, there exists a solution $u$ of $Lu=f$. Also, since ${\mathcal{C}}^{\infty }(M)=\mathrm{im}{L}^{\ast }\oplus \ker L$, we may take $u=L^{\ast }v\in \mathrm{im}{L}^{\ast }$. Since $f=f\circ \gamma =Lu\circ \gamma =L(u\circ \gamma )$, $u\circ \gamma$ is also a solution. Meanwhile, $L(\beta \circ \gamma )=0\iff L\beta \circ \gamma =0\iff L\beta =0$, ${\phi }_{\gamma }:{\mathcal{C}}^{\infty }(M)\to {\mathcal{C}}^{\infty }(M),\beta \mapsto \beta \circ \gamma$ preserves $\ker L$. Also, ${\phi }_{\gamma }$ is an isomorphism (with the inverse ${\phi }_{{\gamma }^{-1}}$) and preserves the inner product (because it preserves the volume form), we conclude that ${\phi }_{\gamma }$ also preserves $\mathrm{im}{L}^{\ast }$. Thus $u\circ \gamma \in \mathrm{im}{L}^{\ast }$. Write $u\circ \gamma =L^{\ast }{v}^{\prime }$. Then$$L{L}^{\ast }v=L{L}^{\ast }{v}^{\prime }\Rightarrow L{L}^{\ast }(v-v^{\prime })=0\Rightarrow 0=⟨L{L}^{\ast }(v-v^{\prime }),v-v^{\prime }⟩=⟨L^{\ast }(v-v^{\prime }),L^{\ast }(v-v^{\prime })⟩\Rightarrow L^{\ast }v=L^{\ast }{v}^{\prime },$$we conclude that $u=u\circ \gamma$, or $u$ is invariant under $\gamma$.