Lemma 10.2.50149A5B39655

Proof. Throughout, we use the properties of the spaces $H^ r$ proved in lemma 10.2.1. We show that the inclusion $H^1\subset L^2$ is compact, prove Gårding's inequality, solve the equation $(\mathcal L+c)v=b$ weakly with the Riesz representation theorem, and obtain the regularity of the solution from lemma 10.2.4. This gives (3); we then prove (1), (4) and (2) in turn.

Step 1: the inclusion $H^1\subset L^2$ is compact. Let $(u_ j)$ be bounded in $H^1$, fix $k$, and put $v_ j=\chi _ ku_ j$, a coefficient vector in the $k$th chart that vanishes outside a fixed compact set $S$. For $R>0$, the part of $\widehat{v_ j}$ on $|\xi |>R$ has $L^2$ norm at most $R^{-1}\| v_ j\| _{H^1}\leq C/R$. The inverse Fourier transforms $w_{j,R}$ of the parts on $|\xi |\leq R$ are uniformly bounded with uniformly bounded derivatives, by the Cauchy–Schwarz inequality on the ball $|\xi |\leq R$, so by the Arzelà–Ascoli theorem (proposition 2.7.4) a subsequence converges uniformly on $S$. Since $v_ j$ vanishes outside $S$,

A diagonal subsequence over $R=1,2,\ldots $ is therefore Cauchy in $L^2$. Repeating this for the finitely many charts and using $u_ j=\sum _ k\chi _ ku_ j$ proves compactness.

Step 2: Gårding's inequality. We first compute $\mathcal L$ in a chart with real coordinates $x_1,\ldots ,x_{2n}$ and the coordinate frame $\mathrm dx_ I$ of the forms. Let $(g^{ij})$ be the inverse of the matrix $(g(\partial _ i,\partial _ j))$, so that $g^{ij}$ is the inner product of $\mathrm dx_ i$ and $\mathrm dx_ j$, write the pointwise inner product of forms as $\sum _{I,J}G_{IJ}u_ I\overline{v_ J}$ and the volume form as $\rho \, \mathrm dx$, and let $\epsilon _ i$ be the exterior multiplication by $\mathrm dx_ i$ and $\epsilon _ i^\dagger $ its pointwise adjoint.

If $e_1,\ldots ,e_{2n}$ is an orthonormal basis of the cotangent space at a point, then exterior multiplication $\epsilon (e_ a)$ by $e_ a$ and its adjoint $\epsilon (e_ a)^\dagger $, which maps $e_ a\wedge e_ K$ to $e_ K$ for $a\notin K$ and kills the basis forms $e_ J$ with $a\notin J$, satisfy $\epsilon (e_ a)\epsilon (e_ b)^\dagger +\epsilon (e_ b)^\dagger \epsilon (e_ a)=\delta _{ab}$. Indeed, for $a=b$, exactly one of the two terms maps a given basis form $e_ J$ to itself, namely the first if $a\in J$ and the second if $a\notin J$, and the other term kills it. For $a\neq b$, both terms kill $e_ J$ unless $b\in J$ and $a\notin J$; in that case, writing $e_ J=\pm e_ b\wedge e_ K$, they map $e_ b\wedge e_ K$ to $e_ a\wedge e_ K$ and to $\epsilon (e_ b)^\dagger (e_ a\wedge e_ b\wedge e_ K)=-e_ a\wedge e_ K$, respectively. Writing $\mathrm dx_ i=\sum _ aM_{ia}e_ a$, so that $\sum _ aM_{ia}M_{ja}=g^{ij}$, we obtain by bilinearity

In the coordinate frame $\mathrm du=\sum _ i\epsilon _ i\partial _ iu$, where $\partial _ i$ acts on the coefficients. Integrating by parts in $\int \sum _{I,J}G_{IJ}(\epsilon _ i\partial _ iu)_ I\overline{v_ J}\, \rho \, \mathrm dx$, we find $\mathrm d^*v=-\sum _ i\epsilon _ i^\dagger \partial _ iv+B_0v$, where $B_0$ is a matrix whose entries are linear combinations, with smooth coefficients, of first derivatives of the $G_{IJ}$ and of $\rho $. Since the $\epsilon _ i$ have constant coefficients in the coordinate frame, (10.2.6) gives

on each coefficient, where $B_1$ is a differential operator of order at most one with smooth coefficients. On functions $\mathrm d^*$ vanishes, so that

at a point where the first derivatives of the metric coefficients vanish, $Lf=-\sum _{i,j}g^{ij}\partial _ i\partial _ jf$ there.

Now let $u$ be smooth with compact support in the chart. Since $\langle \mathcal Lu,u\rangle =\| \mathrm du\| ^2+\| \mathrm d^*u\| ^2$, integrating the second-order term by parts once gives

where $|E(u)|\leq C\| u\| _{L^2}\| u\| _{H^1}$, with $C$ depending on a compact set containing the support of $u$. For $\zeta =(\zeta _{jI})\in \mathbb C^{2n\times N}$, where $N$ is the number of multi-indices $I$, write $\zeta _ j=(\zeta _{jI})_ I$ and $G(\zeta _ j,\zeta _ i)=\sum _{I,J}G_{IJ}\zeta _{jI}\overline{\zeta _{iJ}}$. Diagonalizing the positive definite real symmetric matrix $(g^{ij})$ by an orthogonal matrix $O$, we get $\sum _{i,j}g^{ij}G(\zeta _ j,\zeta _ i)=\sum _ ad_ aG(\eta _ a,\eta _ a)$ with eigenvalues $d_ a>0$ and $\eta _ a=\sum _ jO_{aj}\zeta _ j$; hence this Hermitian form is positive definite, uniformly on compact subsets of the chart, and the integral is at least $\lambda \| \nabla u\| _{L^2}^2$ for some $\lambda >0$. Applying this to the $\chi _ ku$, whose images under $\mathrm d$ and $\mathrm d^*$ differ from $\chi _ k\, \mathrm du$ and $\chi _ k\, \mathrm d^*u$ by terms bounded by $C|u|$, summing over $k$, using $\sum _ k\chi _ k^2\leq 1$ and absorbing the error terms by $ab\leq \tau a^2+b^2/(4\tau )$ with small $\tau >0$, we obtain

for smooth forms $u$ on $X$, and hence for all $u\in H^1$, since both sides are continuous for the norm of $H^1$.

Step 3: weak solutions. Fix $c>0$ and put $P=\mathcal L+c$, with the quadratic form $Q(v,w)=\langle \mathrm dv,\mathrm dw\rangle +\langle \mathrm d^*v,\mathrm d^*w\rangle +c\langle v,w\rangle $. Since $\mathrm d$ and $\mathrm d^*$ are first-order operators with smooth coefficients, $|Q(v,w)|\leq C\| v\| _{H^1}\| w\| _{H^1}$ for smooth forms, so $Q$ extends to a bounded sesquilinear form on $H^1$, and by (10.2.8) the norm $Q(u,u)^{1/2}$ is equivalent to the norm of $H^1$. For $b\in H^{-1}$, the map $w\mapsto \langle b,w\rangle $ is a bounded conjugate-linear functional on $H^1$, so the Riesz representation theorem (proposition 6.2.1) gives a unique $v\in H^1$ with $Q(v,w)=\langle b,w\rangle $ for all $w\in H^1$, and $\| v\| _{H^1}\leq C\| b\| _{H^{-1}}$. For smooth $\phi $ we have $Q(v,\phi )=\langle v,P\phi \rangle $, first for smooth $v$ and then for all $v\in H^1$, since both sides are continuous in $v$ for the norm of $H^1$; hence $Pv=b$ as distributional forms. We write $v=P^{-1}b$.

Step 4: interior regularity. We apply lemma 10.2.4 to $P$. In a chart, with the notation of Step 2, one integration by parts in the expression of $\mathcal L$ shows that $\langle Pv,w\rangle =\mathcal A(v,w)$ for smooth $v$ and $w$, the latter with compact support in the chart, where $\mathcal A$ is a form as in lemma 10.2.4 with principal coefficients $a^{ij}_{IJ}=g^{ij}G_{IJ}\rho $ and smooth lower-order coefficients. These principal coefficients satisfy the ellipticity condition on compact subsets of the chart, as shown in Step 2. Both sides are continuous in $v\in H^1$, so the equation $Q(v,w)=\langle b,w\rangle $ of Step 3 reads $\mathcal A(v,w)=\int \sum _ JF_ J\overline{w_ J}\, \mathrm dx$ with $F_ J=\rho \sum _ IG_{IJ}b_ I$. Let $b\in H^ r$, $r\geq 0$, and $v=P^{-1}b$. Applying lemma 10.2.4 with $\chi =\chi _ k$ on relatively compact open subsets of the finitely many charts, and using lemma 10.2.1(2), we obtain $\| v\| _{H^{r+2}}\leq C(\| b\| _{H^ r}+\| v\| _{H^1})$, and since $\| v\| _{H^1}\leq C\| b\| _{H^{-1}}\leq C\| b\| _{L^2}$,

In particular, if $b$ is smooth, then $P^{-1}b$ lies in every $H^ r$ and is therefore smooth by lemma 10.2.1(3), and $PP^{-1}b=b$; and if $u$ is smooth, then $P^{-1}Pu=u$, because $Q(u,w)=\langle Pu,w\rangle $ for $w\in H^1$ and the solution in Step 3 is unique. For smooth $b$ and $\phi $, with $\psi =P^{-1}\phi $, we get $\langle P^{-1}b,\phi \rangle =\langle P^{-1}b,P\psi \rangle =Q(P^{-1}b,\psi )=\langle b,P^{-1}\phi \rangle $.

We therefore extend $P^{-1}$ to all distributional forms by $\langle P^{-1}u,\phi \rangle =\langle u,P^{-1}\phi \rangle $ for smooth $\phi $. If $u\in H^{-s}$ with $s\geq 2$, then $|\langle u,P^{-1}\phi \rangle |\leq \| u\| _{H^{-s}}\| P^{-1}\phi \| _{H^ s}\leq C\| u\| _{H^{-s}}\| \phi \| _{H^{s-2}}$, so $P^{-1}u\in H^{-s+2}$ and $P^{-1}$ is bounded from $H^ r$ to $H^{r+2}$ for $r\leq -2$; for $u\in H^{-1}$ the new definition agrees with that of Step 3, by the same computation with $b=u$. Since $P$ is formally self-adjoint, $\langle PP^{-1}u,\phi \rangle =\langle u,P^{-1}P\phi \rangle =\langle u,\phi \rangle $ and $\langle P^{-1}Pu,\phi \rangle =\langle u,PP^{-1}\phi \rangle =\langle u,\phi \rangle $ for every distributional form $u$, so $P$ is a bijection of the space of distributional forms onto itself with inverse $P^{-1}$. Conversely, $P$ is bounded from $H^{r+2}$ to $H^ r$ for every integer $r$: for $r\geq 0$ because it is a differential operator of order two with smooth coefficients, for $r=-1$ because $|\langle Pv,\phi \rangle |=|Q(v,\phi )|\leq C\| v\| _{H^1}\| \phi \| _{H^1}$, and for $r\leq -2$ by transposition, since $|\langle Pu,\phi \rangle |=|\langle u,P\phi \rangle |\leq C\| u\| _{H^{r+2}}\| \phi \| _{H^{-r}}$. Hence $P$ maps $H^{r+2}$ isomorphically onto $H^ r$. As $c>0$ was arbitrary, this proves (3).

Step 5: harmonic forms. If $\mathcal Lu=0$ distributionally, then $Pu=cu$, so $u=cP^{-1}u$. By lemma 10.2.1(5), $u\in H^{-r}$ for some $r$; iterating this identity raises the Sobolev order by two at each step, so $u$ lies in every $H^ r$ and is smooth by lemma 10.2.1(3). For smooth harmonic $u$, (10.2.8) gives $\| u\| _{H^1}\leq C\| u\| _{L^2}$. An $L^2$ limit of harmonic forms is distributionally harmonic, hence smooth and harmonic, so the closed unit ball of the space of harmonic forms in $L^2$ is bounded in $H^1$ and therefore compact by Step 1. A normed space with compact closed unit ball is finite dimensional: an infinite orthonormal sequence would have no convergent subsequence. This proves (1).

Step 6: the scalar equation. We first prove (4). If it fails, there are scalar functions $v_ j\in H^1$ of mean zero with $\| v_ j\| _{L^2}=1$ and $\mathrm dv_ j\to 0$ in $L^2$. Since $\mathrm d^*$ vanishes on functions, (10.2.8) shows that $(v_ j)$ is bounded in $H^1$, so by Step 1 a subsequence converges in $L^2$ to some $v$ of norm one and mean zero, and $\mathrm dv=\lim _ j\mathrm dv_ j=0$ in the sense of distributions. In every chart $v$ is locally constant by corollary 2.5.5, so $v$ is constant on the connected manifold $X$, which contradicts $\| v\| _{L^2}=1$ and mean zero. This proves (4).

Consequently $\| \mathrm dv\| _{L^2}$ is an equivalent Hilbert norm on the closed subspace of $H^1$ of functions of mean zero. Let $a$ be smooth of mean zero. The Riesz representation theorem (proposition 6.2.1) gives $f$ in this subspace with $\langle \mathrm df,\mathrm dw\rangle =\langle a,w\rangle $ for all $w$ of mean zero, hence for all $w\in H^1$, since both sides vanish for constant $w$. Testing with $w=f$ gives $\| \mathrm df\| ^2=\langle a,f\rangle \leq \| a\| \, \| f\| \leq C\| a\| \, \| \mathrm df\| $, hence $\| f\| _{H^1}\leq C\| a\| _{L^2}$. With $c=1$ the function $f$ satisfies $Q(f,w)=\langle a+f,w\rangle $ for all $w\in H^1$, so $f=P^{-1}(a+f)$ by the uniqueness in Step 3; if $f\in H^ r$, $r\geq 0$, then Step 4 gives $f\in H^{r+2}$ with $\| f\| _{H^{r+2}}\leq C(\| a\| _{H^ r}+\| f\| _{H^ r})$. By induction $f$ lies in every $H^ r$, so it is smooth, and $\| f\| _{H^{r+2}}\leq C_ r\| a\| _{H^ r}$ for every $r\geq 0$.

Let $\Pi a=\langle a,1\rangle /\int _ X\mathrm dV$ be the average of $a$, and let $\mathcal G$ be the operator $a\mapsto f$ applied to $a-\Pi a$, for smooth $a$. Then $\| \mathcal Ga\| _{H^{r+2}}\leq C_ r\| a\| _{H^ r}$ for $r\geq 0$, and $L\mathcal G=1-\Pi $ by construction. If $a$ is smooth, then $La$ has mean zero, because $\langle La,1\rangle =\langle a,L1\rangle =0$, and both $\mathcal GLa$ and $a-\Pi a$ are smooth solutions $f$ of mean zero of $Lf=La$. The difference $f_0$ of two such solutions satisfies $\int _ X|\mathrm df_0|^2\, \mathrm dV=\langle Lf_0,f_0\rangle =0$, so it is constant, and it vanishes since it has mean zero; hence $\mathcal GL=1-\Pi $. For smooth $a$ and $b$, since $\mathcal Ga$ has mean zero and $L$ is formally self-adjoint,

so $\mathcal G$ is formally self-adjoint, and so is $\Pi $. Extend $\mathcal G$ to distributions by transposition, $\langle \mathcal Ga,\chi \rangle =\langle a,\mathcal G\chi \rangle $. This defines a distribution: if $a\in H^{-s}$ with $s\geq 2$, then $|\langle a,\mathcal G\chi \rangle |\leq C\| a\| _{H^{-s}}\| \chi \| _{H^{s-2}}$. The two identities $L\mathcal G=\mathcal GL=1-\Pi $ remain valid, by transposition of the same identities for smooth functions. Hence $f=\mathcal Ga$ solves $Lf=a$ for every distribution $a$ of mean zero, and $\langle \mathcal Ga,1\rangle =\langle a,\mathcal G1\rangle =0$. If $Lf=0$ for a distribution $f$, then $\mathcal Lf=0$, so $f$ is smooth by Step 5, and $\int _ X|\mathrm df|^2\, \mathrm dV=\langle Lf,f\rangle =0$ shows that $f$ is constant; this proves uniqueness. If $a$ is real, then $\overline f$ is another solution of mean zero, since $L$ has real coefficients, so $f$ is real. $\square$