10.2 The trace equation and global potentials308B81D53CDF

Proof. In this proof, the constants $C$ depend on $r$, $l$ and the data indicated, but not on the forms or functions involved. We first compare the coefficient vectors of a form in two charts, and then prove (1)–(5) in turn.

(a) A change of charts. Let $\kappa \colon U\to \mathbb R^{2n}$ and $\kappa '\colon U'\to \mathbb R^{2n}$ be charts of $X$ with frames of the bundle of forms over $U$ and $U'$, let $K\subset U$ be compact, and let $W\subset U'$ be an open set whose closure is a compact subset of $U'$. For a smooth form $w$ that vanishes outside $K$, let $w_\kappa $ and $w_{\kappa '}$ be its coefficient vectors in the two charts and frames, the first one extended by zero to $\mathbb R^{2n}$. We claim that

with $C$ depending on the charts, frames, $K$ and $W$. Indeed, on $\kappa '(U\cap U')$ we have $w_{\kappa '}(y)=A(y)\, w_\kappa (\tau (y))$, where $\tau =\kappa \circ \kappa '^{-1}$ and $A$ is the smooth matrix of the change of frames. By the chain rule, the Leibniz rule and induction on $|\alpha |$,

where the matrices $c_{\alpha \beta }$ are polynomials in the derivatives of order at most $|\alpha |$ of $A$ and of $\tau $; they are smooth on $\kappa '(U\cap U')$ and therefore bounded on its compact subset $\kappa '(\overline W\cap K)$. Since $w$ vanishes on the open set $X\setminus K$, all derivatives of $w_{\kappa '}$ vanish on $\kappa '(W\setminus K)$, and every point of $\kappa '(W)$ lies in this set or in $\kappa '(W\cap K)$. Hence, for $|\alpha |\leq r$,

by the change of variables $x=\tau (y)$, whose Jacobian determinant $|\det D\tau ^{-1}|$ is bounded on the compact set $\kappa (\overline W\cap K)$. This proves (10.2.2). We also use the Leibniz rule in the following form. If $M$ is a smooth matrix-valued function on an open set $V\subset \mathbb R^{2n}$ whose derivatives of order at most $r$ are bounded on $V$, then $\| Mv\| _{(r),V}\leq C\| v\| _{(r),V}$ for smooth $v$ on $V$, with $C$ depending on these bounds. If $v$ vanishes outside a compact set $L\subset V$, it suffices that $M$ be smooth on $V$, the bounds being taken on $L$, and then $\| Mv\| _{(r)}\leq C\| v\| _{(r)}$ for the extensions by zero.

(b) Proof of (1). For a test function $v$ on $\mathbb R^{2n}$ we have $\widehat{\partial ^\beta v}(\xi )=(2\pi \mathrm i\xi )^\beta \widehat v(\xi )$, and $(1+|\xi |^2)^ r=\sum _{|\beta |\leq r}c_\beta |\xi ^\beta |^2$ with positive multinomial coefficients $c_\beta $. Hence Plancherel's theorem (theorem 2.9.4) gives

and the norm of $H^ r$ is equivalent on smooth forms to $\bigl (\sum _ k\| \chi _ ku\| _{(r)}^2\bigr )^{1/2}$. Let charts with domains $U'_ m$, frames over them, and cutoffs $\chi '_ m$ with $\sum _ m\chi '_ m=1$ be other choices, and choose open sets $W_ m$ with $\operatorname{Supp}\chi '_ m\subset W_ m$ and $\overline{W_ m}$ compact in $U'_ m$. For a smooth form $u$ we have $\chi '_ mu=\sum _ k\chi '_ m\chi _ ku$, and the form $\chi '_ m\chi _ ku$ vanishes outside the compact set $\operatorname{Supp}\chi _ k$ and outside $\operatorname{Supp}\chi '_ m\subset W_ m$. By (10.2.2), with $K=\operatorname{Supp}\chi _ k$ and $W=W_ m$, its norm $\| \cdot \| _{(r)}$ in the $m$th new chart is at most $C$ times its norm in the $k$th old chart, and by the Leibniz rule, applied to the factor $\chi '_ m$ on the compact set $\kappa _ k(\operatorname{Supp}\chi _ k)$, where $\kappa _ k$ is the $k$th chart, the latter is at most $C\| \chi _ ku\| _{(r)}$. Summing over $k$ and $m$, we see that the new norm is bounded by a constant times the old one, and by symmetry the two norms are equivalent.

If $M$ is a smooth function or endomorphism, then $\chi _ kMu=M\chi _ ku$ vanishes outside $\operatorname{Supp}\chi _ k$, so the Leibniz rule gives $\| \chi _ kMu\| _{(r)}\leq C\| \chi _ ku\| _{(r)}$. Finally, $\| \chi _ k\phi \| _{(r)}$ is bounded, by the Leibniz rule and since $\kappa _ k(\operatorname{Supp}\chi _ k)$ has finite measure, by a constant times the suprema on $\operatorname{Supp}\chi _ k$ of the derivatives of order at most $r$ of the coefficients of $\phi $; hence $\| \phi \| _{H^ r}\leq C\| \phi \| _{C^ r}$.

(c) Proof of (2). For a smooth form $u$ we have $|u|\leq \sum _ k|\chi _ ku|$ pointwise, since $\chi _ k\geq 0$ and $\sum _ k\chi _ k=1$. On the compact sets $\operatorname{Supp}\chi _ k$ the pointwise norm of forms is comparable with the Euclidean norm of the coefficient vectors, and $\mathrm dV$ with Lebesgue measure in the charts; hence $\| u\| _{L^2}\leq C\sum _ k\| \chi _ ku\| _{(0)}\leq C\| u\| _{H^ r}$, and conversely $\| \chi _ ku\| _{(0)}\leq C\| u\| _{L^2}$ because $0\leq \chi _ k\leq 1$. Therefore the identity map of smooth forms extends to a bounded map $j$ from $H^ r$ to the space of $L^2$ forms. Let $u\in H^ r$ be the limit in $H^ r$ of smooth forms $u_ i$. By (1), for every $k$ and $|\alpha |\leq r$ the functions $\partial ^\alpha (\chi _ ku_ i)$ form a Cauchy sequence in $L^2(\mathbb R^{2n})$, with a limit $g_{k,\alpha }$, while $\chi _ ku_ i\to \chi _ kj(u)$ in $L^2$. For every test function $\phi $ on $\mathbb R^{2n}$,

so $g_{k,\alpha }$ is the distributional derivative $\partial ^\alpha (\chi _ kj(u))$. Hence these derivatives are square integrable, and $\sum _ k\| \chi _ kj(u)\| _{(r)}^2=\lim _ i\sum _ k\| \chi _ ku_ i\| _{(r)}^2$, which by (1) lies between two positive multiples of $\lim _ i\| u_ i\| _{H^ r}^2=\| u\| _{H^ r}^2$. In particular, $j(u)=0$ implies $u=0$, so $j$ is injective, and we identify $u$ with $j(u)$. This proves the first assertion, the necessity of the condition and the equivalence of the norms.

Conversely, let $u$ be an $L^2$ form such that every $\chi _ ku$ has square integrable distributional derivatives of order at most $r$, and let $\rho _\varepsilon $ be the mollifier of section 2.5. Since $\chi _ ku$ vanishes outside the compact set $\kappa _ k(\operatorname{Supp}\chi _ k)$, the functions $v_{k,\varepsilon }=\rho _\varepsilon *(\chi _ ku)$ are smooth and, for small $\varepsilon $, vanish outside a fixed compact subset $L_ k$ of $\kappa _ k(U_ k)$, where $U_ k$ is the domain of the $k$th chart; moreover $\partial ^\alpha v_{k,\varepsilon }=\rho _\varepsilon *\partial ^\alpha (\chi _ ku)$ by proposition 2.5.3, and this converges to $\partial ^\alpha (\chi _ ku)$ in $L^2$ as $\varepsilon \to 0$, by lemma 2.5.1, for $|\alpha |\leq r$. Let $w_{k,\varepsilon }$ be the smooth form on $X$ whose coefficient vector in the $k$th chart and frame is $v_{k,\varepsilon }$, extended by zero outside $U_ k$, and put $u_\varepsilon =\sum _ kw_{k,\varepsilon }$. Choose open sets $W_ m$ with $\operatorname{Supp}\chi _ m\subset W_ m$ and $\overline{W_ m}$ compact in $U_ m$. By (10.2.2), with $K=\kappa _ k^{-1}(L_ k)$ and $W=W_ m$, and by the Leibniz rule for the factor $\chi _ m$, we have $\| \chi _ m(w_{k,\varepsilon }-w_{k,\varepsilon '})\| _{(r)}\leq C\| v_{k,\varepsilon }-v_{k,\varepsilon '}\| _{(r)}$. By (1), $(u_\varepsilon )$ is therefore a Cauchy sequence in $H^ r$ as $\varepsilon \to 0$ along a sequence, and it converges in $L^2$ to $\sum _ k\chi _ ku=u$, by the comparison of pointwise norms on the compact sets $\kappa _ k^{-1}(L_ k)\cup \operatorname{Supp}\chi _ k\subset U_ k$. Hence $u\in H^ r$.

For $r=0$ the condition holds for every $L^2$ form, so $H^0$ is the space of all $L^2$ forms, and its norm is equivalent to the $L^2$ norm by the first paragraph. Other choices of the charts, frames and cutoffs give, by (1), equivalent norms on smooth forms, hence the same Cauchy sequences and the same limits in $L^2$; so they give the same subspace $H^ r$ with an equivalent norm. If $r\leq r'$, then $\| \phi \| _{H^ r}\leq \| \phi \| _{H^{r'}}$ for smooth $\phi $, so every Cauchy sequence in $H^{r'}$ is a Cauchy sequence in $H^ r$ with the same limit in $L^2$; hence $H^{r'}\subset H^ r$.

Finally, let $W$ be open with closure compact in a chart with a frame. For smooth $u$, we have $u=\sum _ m\chi _ mu$, and (10.2.2), applied to $w=\chi _ mu$ with $K=\operatorname{Supp}\chi _ m$, gives $\| u\| _{(r),W}\leq C\sum _ m\| \chi _ mu\| _{(r)}\leq C\| u\| _{H^ r}$. If $u\in H^ r$ is the limit in $H^ r$ of smooth forms $u_ i$, the coefficient vectors of the $u_ i$ on $W$ form a Cauchy sequence for $\| \cdot \| _{(r),W}$ and converge in $L^2(W)$ to that of $u$. Testing against test functions on $W$ as above, we find that their derivatives of order at most $r$ converge in $L^2(W)$ to those of the coefficient vector of $u$, and the bound passes to the limit.

(d) Proof of (3). Let first $v$ be a smooth $\mathbb C^ N$-valued function with compact support, and let $|\alpha |\leq l$. The inversion formula of theorem 2.9.4, applied to $\partial ^\alpha v$, whose Fourier transform is $(2\pi \mathrm i\xi )^\alpha \widehat v$, and the Cauchy–Schwarz inequality give

where the first integral on the right is finite, by polar coordinates, because $2(r-l)>2n$. By (10.2.3), $\sup |\partial ^\alpha v|\leq C\| v\| _{(r)}$. If $v$ is only compactly supported with $\| v\| _{(r)}<\infty $, then, as in (c), the mollifications $v_\varepsilon =\rho _\varepsilon *v$ are smooth with compact support and $\| v_\varepsilon -v\| _{(r)}\to 0$ as $\varepsilon \to 0$. Along a sequence $\varepsilon _ i\to 0$, the estimate shows that the derivatives of order at most $l$ of the $v_{\varepsilon _ i}$ converge uniformly; hence the $v_{\varepsilon _ i}$ converge uniformly to a function of class $C^ l$ whose derivatives of order at most $l$ are bounded by $C\| v\| _{(r)}$. This function agrees with $v$ almost everywhere, since $v_{\varepsilon _ i}\to v$ in $L^2$ and a subsequence converges almost everywhere.

Now let $u$ be a smooth form. On $\operatorname{Supp}\chi _ k$, the coefficient vector of $u$ in the $k$th chart and frame is the sum over $m$ of the coefficient vectors of the forms $\chi _ mu$. We apply the formula for $\partial ^\alpha w_{\kappa '}$ in (a) with $\kappa $ the $m$th chart, $\kappa '$ the $k$th chart and $K=\operatorname{Supp}\chi _ m$. The derivatives of the coefficient vector of $\chi _ mu$ in the $k$th chart vanish outside $\kappa _ k(\operatorname{Supp}\chi _ m)$, and the coefficients $c_{\alpha \beta }$ are bounded on the compact set $\kappa _ k(\operatorname{Supp}\chi _ k\cap \operatorname{Supp}\chi _ m)$. Hence the derivatives of order at most $l$ of this coefficient vector are bounded on $\kappa _ k(\operatorname{Supp}\chi _ k)$ by $C$ times the suprema of the derivatives of order at most $l$ of the coefficient vector of $\chi _ mu$ in the $m$th chart, and the Euclidean estimate and (1) give $\| u\| _{C^ l}\leq C\sum _ m\| \chi _ mu\| _{(r)}\leq C\| u\| _{H^ r}$.

If $u\in H^ r$ is the limit in $H^ r$ of smooth forms $u_ i$, then $\| u_ i-u_{i'}\| _{C^ l}\leq C\| u_ i-u_{i'}\| _{H^ r}$, so in every chart the coefficients of the $u_ i$ and their derivatives of order at most $l$ converge uniformly on $\operatorname{Supp}\chi _ k$, and hence on the open sets $\{ \chi _ k>0\} $, which cover $X$. The limit is a form of class $C^ l$ with $\| \cdot \| _{C^ l}\leq C\| u\| _{H^ r}$, and it agrees with $u$ almost everywhere, because $u_ i\to u$ in $L^2$. Finally, if $u$ lies in every $H^ r$, then for each $l$ it agrees almost everywhere with a form of class $C^ l$; these continuous forms agree almost everywhere, hence everywhere, so $u$ agrees almost everywhere with a smooth form.

(e) Proof of (4). Let $u\in H^{-r}$. Since smooth forms are dense in $H^ r$, the bounded functional $\phi \mapsto \langle u,\phi \rangle $ on smooth forms extends uniquely to a bounded conjugate-linear functional on $H^ r$, with the same norm. Conversely, let $\ell $ be a bounded conjugate-linear functional on $H^ r$. Since $*$ is a real bundle isomorphism with smooth coefficients, the formula $T(\alpha )=\ell (\overline{*^{-1}\alpha })$ defines a complex-linear functional $T$ on smooth forms. If $\alpha $ is supported in a compact subset $K$ of a chart, then, by (1) and the chain rule on the compact sets $K\cap \operatorname{Supp}\chi _ k$, we have $|T(\alpha )|\leq \| \ell \| \, \| *^{-1}\overline\alpha \| _{H^ r}\leq C\| *^{-1}\overline\alpha \| _{C^ r}\leq C_ K\sum _ Jp_{K,r}(\alpha _ J)$ in the notation of section 5.3. Hence the components of $T$ of the various degrees are currents, and $T$ is a distributional form $u$ with $\langle u,\phi \rangle =u(*\overline\phi )=\ell (\phi )$ for smooth $\phi $. Thus $u\in H^{-r}$ and $\| u\| _{H^{-r}}=\| \ell \| $. It is unique, because $*$ maps the smooth forms onto the smooth forms, so that a distributional form is determined by its pairings with smooth forms.

For a smooth function or endomorphism $M$, with pointwise adjoint $M^\dagger $, we have $\langle Mu,\phi \rangle =\langle u,M^\dagger \phi \rangle $ for distributional forms $u$ and smooth $\phi $, so $|\langle Mu,\phi \rangle |\leq \| u\| _{H^{-r}}\| M^\dagger \phi \| _{H^ r}\leq C\| u\| _{H^{-r}}\| \phi \| _{H^ r}$ by (1). The inclusion $H^{-r}\subset H^{-r'}$ for $r\leq r'$ follows from $\| \phi \| _{H^ r}\leq \| \phi \| _{H^{r'}}$, and by (1) other choices of the charts, frames and cutoffs change the norm of $H^ r$ on smooth forms, and hence $H^{-r}$, only up to equivalence.

The definition of $H^ r$, $r\geq 0$, does not involve $\omega $. Let $\omega '$ be another Kähler form, with pointwise inner product $\langle \cdot ,\cdot \rangle '$ and volume form $\mathrm dV'$. At each point, the Hermitian form $\langle a,b\rangle '\, \mathrm dV'/\mathrm dV$ on forms is $\langle a,Sb\rangle $ for a unique endomorphism $S$, which is positive definite, hence invertible, and depends smoothly on the point. For smooth $u$ and $\phi $ we then have $u\wedge *'\overline\phi =\langle u,\phi \rangle '\, \mathrm dV'=\langle u,S\phi \rangle \, \mathrm dV=u\wedge *\overline{S\phi }$, where $*'$ is the Hodge star of $\omega '$; since $u$ is arbitrary, $*'\overline\phi =*\overline{S\phi }$. Hence the pairing defined by $\omega '$ is $u(*'\overline\phi )=\langle u,S\phi \rangle $ for every distributional form $u$. Since multiplication by $S$ and by $S^{-1}$ is bounded on $H^ r$ by (1), the condition defining $H^{-r}$ is the same for both pairings, and the two norms are equivalent.

(f) Proof of (5). Let $u$ be a distributional form. Since $\sum _ k\chi _ k=1$, we have $\langle u,\phi \rangle =\sum _ ku(\chi _ k\, {*\overline\phi })$ for smooth $\phi $. The forms $\chi _ k\, {*\overline\phi }$ are supported in the compact sets $\operatorname{Supp}\chi _ k$ of the charts, so the definition of currents (section 5.3) gives integers $m_ k$ and constants $C_ k$ with $|u(\chi _ k\, {*\overline\phi })|\leq C_ k\sum _ Jp_{\operatorname{Supp}\chi _ k,m_ k}\bigl ((\chi _ k\, {*\overline\phi })_ J\bigr )$. By the Leibniz rule, and since $*$ and the change from the coordinate frames to the given frames have smooth coefficients, the right side is at most $C\| \phi \| _{C^ l}$ with $l=\max _ km_ k$. By (3), $\| \phi \| _{C^ l}\leq C\| \phi \| _{H^ r}$ for $r>l+n$, and therefore $u\in H^{-r}$. $\square$

Proof. In this proof $H^ j$ stands for $H^ j(\mathbb R^{2n})$. We prove the case $k=0$ with difference quotients and then argue by induction on $k$. Let $0\leq \chi \leq 1$, let $h_0$ be half the distance from $\operatorname{Supp}\chi $ to $\partial U$, and for a coordinate direction $e_ l$ and $0<|h|<h_0$ put $\delta _ hv(x)=(v(x+he_ l)-v(x))/h$. For smooth $f$ we have $\delta _ hf(x)=\int _0^1\partial _ lf(x+the_ l)\, \mathrm dt$, so the Cauchy–Schwarz inequality in $t$ and Fubini's theorem bound the $L^2$ norm of $\delta _ hf$ on a set by that of $\partial _ lf$ on a slightly larger set; by mollification the same holds for $f\in H^1$. The function $w=-\delta _{-h}(\chi ^2\delta _ hv)$ lies in $H^1$ and has compact support in $U$. Its mollifications are test functions converging to $w$ in $H^1$ (lemma 2.5.1), and both sides of the equation are continuous in $w$ for the norm of $H^1$, so the equation also holds for $w$.

Discrete integration by parts, $\int f\, \overline{\delta _{-h}g}=-\int (\delta _ hf)\, \overline g$, turns the principal term into $\int \sum \delta _ h(a^{ij}_{IJ}\partial _ jv_ I)\, \overline{\partial _ i(\chi ^2\delta _ hv_ J)}$. Here $\delta _ h(a\, \partial _ jv)=a(\cdot +he_ l)\, \partial _ j\delta _ hv+(\delta _ ha)\, \partial _ jv$ and $\partial _ i(\chi ^2\delta _ hv)=\chi ^2\partial _ i\delta _ hv+2\chi \, \partial _ i\chi \, \delta _ hv$. The ellipticity condition at $x+he_ l$ bounds the main term $\int \sum a^{ij}_{IJ}(x+he_ l)\chi ^2\partial _ j\delta _ hv_ I\overline{\partial _ i\delta _ hv_ J}$ from below by $\lambda \| \chi \nabla \delta _ hv\| _{L^2}^2$. The remaining terms, the lower-order terms and the right-hand side are estimated with $|\delta _ ha|\leq C$, with $\| \delta _ hv\| _{L^2(\operatorname{Supp}\chi )}\leq \| \partial _ lv\| _{L^2(U)}$, and with $\| w\| _{L^2}\leq \| \partial _ l(\chi ^2\delta _ hv)\| _{L^2}\leq \| \chi \nabla \delta _ hv\| _{L^2}+C\| v\| _{H^1(U)}$, where the difference quotients are bounded as above. This gives

and hence $\| \chi \nabla \delta _ hv\| _{L^2}\leq C(\| F\| _{L^2(U)}+\| v\| _{H^1(U)})$, independently of $h$.

If $\phi \in C^\infty _ c$ is supported in the interior of $\{ \chi =1\} $, then $-\int \partial _ jv\, \partial _ l\phi =\lim _{h\to 0}\int (\delta _ h\partial _ jv)\, \phi $ is bounded by this constant times $\| \phi \| _{L^2}$. By the Riesz representation theorem in $L^2$ (proposition 6.2.1), applied to the extension of $\phi \mapsto -\int \partial _ jv\, \partial _ l\phi $ to $L^2$, the distributional derivatives $\partial _ l\partial _ jv$ are square integrable there, with the same bound. Since $\chi $ may be chosen equal to one near the support of any given cutoff, this proves the lemma for $k=0$.

Suppose the lemma holds for some $k\geq 0$, let $F\in H^{k+1}(U)$, and fix $\chi $ and an open set $U'\Subset U$ containing $\operatorname{Supp}\chi $. By the lemma for $k$, $v\in H^{k+2}(U')$, with norm at most $C(\| F\| _{H^ k(U)}+\| v\| _{H^1(U)})$. Fix $m$ and test the equation with $-\partial _ mw$, $w\in C^\infty _ c(U')$. Integrating by parts in $x_ m$, we find that $u=\partial _ mv\in H^1(U')$ satisfies $\mathcal A(u,w)=\int \sum _ JF'_ J\overline{w_ J}\, \mathrm dx$ for all $w\in C^\infty _ c(U')$, where, with the indices $I,J$ suppressed,

Since $v\in H^{k+2}(U')$, we have $F'\in H^ k(U')$ with $\| F'\| _{H^ k(U')}\leq C(\| F\| _{H^{k+1}(U)}+\| v\| _{H^{k+2}(U')})$. The lemma for $k$ on $U'$ gives $\chi \, \partial _ mv\in H^{k+2}$ for every $m$, hence $\chi v\in H^{k+3}$ with the asserted bound. $\square$

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. $\square$

Proof. For smooth real functions $f,g$ put $B(f,g)=\int _ X\mathrm{d}f\wedge \mathrm{d}^{\mathrm c}g\wedge A$. Only the $(1,1)$-part of $\mathrm df\wedge \mathrm d^{\mathrm c}g$ contributes, and this part equals $\frac{\mathrm i}{4\pi }(\partial f\wedge \bar\partial g+\partial g\wedge \bar\partial f)$, which is symmetric in $f$ and $g$; in particular $\mathrm{d}f\wedge \mathrm{d}^{\mathrm c}f$ has $(1,1)$-part $\frac{\mathrm i}{2\pi }\partial f\wedge \bar\partial f$. Thus $B$ is symmetric. Since $A$ is closed, Stokes' formula (proposition 5.2.6) applied to $\mathrm d(g\, \mathrm d^{\mathrm c}f\wedge A)$ gives

On the compact manifold there are constants $0<c\leq C$ with $c\, \omega \leq \omega _ j\leq C\omega $ for all $j$, by lemma 6.14.1. For $(1,1)$-forms $\gamma _ j,\gamma '_ j$ we have the telescoping identity

Applying it with $\gamma _ j=\omega _ j$, $\gamma '_ j=c\, \omega $ and with $\gamma _ j=C\omega $, $\gamma '_ j=\omega _ j$, we see that the forms $A-c^{n-1}\omega ^{n-1}$ and $C^{n-1}\omega ^{n-1}-A$ are sums of products of semipositive $(1,1)$-forms. Such products, and their products with the semipositive form $\mathrm i\partial f\wedge \bar\partial f$, are strongly positive (proposition 5.3.2). Hence $B(f,f)$ lies between $c^{n-1}$ and $C^{n-1}$ times $\int _ X\frac{\mathrm i}{2\pi }\partial f\wedge \bar\partial f\wedge \omega ^{n-1}$. At a point, if $\theta _1,\ldots ,\theta _ n$ is a unitary coframe, that is, $\omega =\mathrm i\sum _ j\theta _ j\wedge \bar\theta _ j$, and $\partial f=\sum _ ja_ j\theta _ j$, then $\mathrm i\partial f\wedge \bar\partial f\wedge \omega ^{n-1}/(n-1)!=\sum _ j|a_ j|^2\omega ^ n/n!$ and $|\mathrm df|^2=2\sum _ j|a_ j|^2$; so the last integral is $\frac{(n-1)!}{4\pi }\| \mathrm df\| _{L^2}^2$. By lemma 10.2.5(4), $B$ is therefore an equivalent inner product on the closed subspace of real functions of mean zero in $H^1$.

The sesquilinear extension $B_{\mathbb C}(f_1+\mathrm if_2,g_1+\mathrm ig_2)=B(f_1,g_1)+B(f_2,g_2)+\mathrm i\bigl (B(f_2,g_1)-B(f_1,g_2)\bigr )$ to complex-valued functions is, by the symmetry of $B$, a Hermitian form with $B_{\mathbb C}(f,f)=B(f_1,f_1)+B(f_2,f_2)$, hence an equivalent inner product on the complex Hilbert space of functions of mean zero in $H^1$. The Riesz representation theorem (proposition 6.2.1), applied to the bounded conjugate-linear functional $g\mapsto -\int _ X\overline g\, \eta $, gives $F$ in this space with $B_{\mathbb C}(F,g)=-\int _ X\overline g\, \eta $ for all $g$. For real $g$ the right side is real and $B_{\mathbb C}(F,g)=B(\operatorname{Re}F,g)+\mathrm iB(\operatorname{Im}F,g)$, so $f=\operatorname{Re}F$ is a real function of mean zero in $H^1$ with $B(f,g)=-\int _ Xg\eta $ for every real mean-zero $g\in H^1$. Since $\int _ X\eta =0$ and $B(f,1)=0$, the identity holds for all $g\in H^1$; this is the equation $\mathrm{dd}^{\mathrm{c}}f\wedge A=\eta $ in the weak sense.

It remains to prove that $f$ is smooth. In a chart, with $\eta =\tilde\eta \, \mathrm dx$, the weak equation reads $\mathcal A(f,w)=-\int \tilde\eta \, \overline w\, \mathrm dx$ for $w\in C^\infty _ c$, where $\mathcal A(f,w)=B(f,\overline w)=\int \sum _{i,j}a^{ij}\partial _ jf\, \overline{\partial _ iw}\, \mathrm dx$ and the real symmetric matrix $(a^{ij})$ is the density of the quadratic form $\mathrm df\wedge \mathrm d^{\mathrm c}f\wedge A$. By the bounds just proved, $(a^{ij})$ is uniformly positive definite on compact subsets of the chart. This is a form of the kind considered in lemma 10.2.4, with $N=1$, and since $\tilde\eta $ is smooth, that lemma, applied with the cutoffs $\chi _ k$ on relatively compact open subsets of the charts, shows that every $\chi _ kf$ has square integrable distributional derivatives of all orders. Hence $f$ lies in every $H^ r$ by lemma 10.2.1(2), and it is smooth by lemma 10.2.1(3). If $f$ and $f'$ are two solutions, then $B(f-f',f-f')=0$, so $f-f'$ is constant on the connected manifold $X$, and it vanishes since it has mean zero. $\square$

Proof. Choose holomorphic coordinates $z$ centered at $x$. Since the matrix of $\sigma $ at $x$ is positive definite Hermitian, after a linear change of coordinates $\sigma =\frac{\mathrm i}{2}\sum _{k,l}g_{k\bar l}\, \mathrm dz_ k\wedge \mathrm d\bar z_ l$ with $g_{k\bar l}(0)=\delta _{kl}$; moreover $\overline{g_{l\bar k}}=g_{k\bar l}$, since $\sigma $ is real. The $(2,1)$-part of $\mathrm d\sigma =0$ is $\partial \sigma =\frac{\mathrm i}{2}\sum _{k,l,m}\partial _ mg_{k\bar l}\, \mathrm dz_ m\wedge \mathrm dz_ k\wedge \mathrm d\bar z_ l=0$, which gives $\partial _ mg_{k\bar l}=\partial _ kg_{m\bar l}$.

By the holomorphic inverse function theorem (proposition 5.2.4(1)), the equations $z_ l=w_ l-\frac12\sum _{k,m}\partial _ mg_{k\bar l}(0)w_ kw_ m$ define holomorphic coordinates $w$ near $0$, since the Jacobian matrix of this map at $0$ is the identity. Substituting $\mathrm dz_ l=\mathrm dw_ l-\sum _{k,m}\partial _ mg_{k\bar l}(0)w_ m\, \mathrm dw_ k$, which uses the symmetry in $k$ and $m$, and $g_{k\bar l}(z)=\delta _{kl}+\sum _ m\partial _ mg_{k\bar l}(0)w_ m+\sum _ m\partial _{\bar m}g_{k\bar l}(0)\bar w_ m+O(|w|^2)$, we find that the coefficient of $\frac{\mathrm i}{2}\mathrm dw_ k\wedge \mathrm d\bar w_ l$ in $\sigma $ is $\delta _{kl}$ plus terms of order two in $w$: the terms $\sum _ m\partial _ mg_{k\bar l}(0)w_ m$ of $g_{k\bar l}$ cancel against the correction in $\mathrm dz_ k$, and their conjugates $\sum _ m\partial _{\bar m}g_{k\bar l}(0)\bar w_ m=\sum _ m\overline{\partial _ mg_{l\bar k}(0)}\, \bar w_ m$ cancel against the correction in $\mathrm d\bar z_ l$. Hence the first derivatives of the new coefficients vanish at $0$. $\square$

Proof. Both sides of each identity are differential operators, or bundle maps, with smooth coefficients, acting on distributions and currents as explained at the beginning of this section. Hence it suffices to verify the identities for smooth $f$ and $T$, at an arbitrary point. By lemma 10.2.11, there are holomorphic coordinates centered at the point in which $\omega =\frac{\mathrm i}{2}\sum _ j\mathrm dz_ j\wedge \mathrm d\bar z_ j$ at the point and the first derivatives of the coefficients of $\omega $, hence of the metric coefficients, vanish there. Thus (10.2.7) gives $Lf=-\sum _ j(\partial _{x_ j}^2+\partial _{y_ j}^2)f=-4\sum _ j\partial _ j\partial _{\bar j}f$ there, while $\theta _ j=\mathrm dz_ j/\sqrt2$ is a unitary coframe and

Hence $\Lambda \mathrm{dd}^{\mathrm{c}}f=\pi ^{-1}\sum _ jf_{j\bar j}=-Lf/(4\pi )$. For the second identity, diagonalize the Hermitian matrix of $T$ by a unitary change of coframe; by linearity it suffices to treat $T=\mathrm i\theta _1\wedge \bar\theta _1$. Write $\sqrt2\theta _ j=e_{2j-1}+\mathrm ie_{2j}$ with real covectors $e_ a$; they form an orthonormal basis, positively oriented since $\omega ^ n/n!=e_1\wedge \cdots \wedge e_{2n}$, and $\omega _ j=\mathrm i\theta _ j\wedge \bar\theta _ j=e_{2j-1}\wedge e_{2j}$. By the explicit formula for $*$, the star of $\omega _1$ is $\omega _2\wedge \cdots \wedge \omega _ n$. On the other hand $\omega ^{n-1}/(n-1)!$ is the sum of all products of $n-1$ distinct $\omega _ j$, while $\omega _1\wedge \omega ^{n-2}/(n-2)!$ is the sum of those products that contain $\omega _1$; their difference is exactly $\omega _2\wedge \cdots \wedge \omega _ n$, and $\Lambda \omega _1=1$.

If $\mathrm dT=0$ and $\Lambda T$ is constant, the second identity and $\mathrm d\omega =0$ give $\mathrm d*T=0$, hence $\mathrm d^*T=-*\mathrm d*T=0$, and $\mathcal LT=\mathrm d\mathrm d^*T+\mathrm d^*\mathrm dT=0$. In dimension one a $(1,1)$-current is $T=(\Lambda T)\omega $, so $*T=\Lambda T$ is constant and the same conclusion holds. $\square$

Proof. Put $a=\int _ XT\wedge \omega ^{n-1}/\int _ X\omega ^ n$. Since $T\wedge \omega ^{n-1}=(\Lambda T)\omega ^ n/n$, the distribution $\Lambda T-na$ has mean zero, and lemma 10.2.5 gives a real distribution $f$ of mean zero with $Lf=-4\pi (\Lambda T-na)$. By lemma 10.2.12, $\Lambda \mathrm{dd}^{\mathrm{c}}f=\Lambda T-na$, so the closed current $U=T-a\omega -\mathrm{dd}^{\mathrm{c}}f$ has $\Lambda U=0$. By lemma 10.2.12 again, $U$ is harmonic, and by lemma 10.2.5 it is smooth. The Kähler form is harmonic as well, since $\mathrm d\omega =0$ and $*\omega =\omega ^{n-1}/(n-1)!$ is closed. Hence $H=a\omega +U$ and $f$ give the decomposition (10.2.16). If $T$ is smooth, then $\Lambda T$ is smooth, and so is $f$ by lemma 10.2.5.

An exact harmonic form $B$ vanishes: if $B=\mathrm dS$ for a current $S$, then

since $\mathrm d*B=0$. If $T=H'+\mathrm{dd}^{\mathrm{c}}f'$ is a second decomposition, then $H-H'=\mathrm d\, \mathrm d^{\mathrm c}(f'-f)$ is exact and harmonic, hence zero. Then $\mathrm{dd}^{\mathrm{c}}(f-f')=0$, so $L(f-f')=0$ by lemma 10.2.12; thus $f-f'$ is constant by lemma 10.2.5, and it vanishes because both have mean zero. The same argument proves the exactness criterion: if $T$ is exact, so is $H=T-\mathrm d\, \mathrm d^{\mathrm c}f$, whence $H=0$; conversely, if $H=0$, then $T=\mathrm d\, \mathrm d^{\mathrm c}f$.

Suppose now that $T\geq 0$. By theorem 5.4.3, theorem 5.9.1, lemma 6.14.2, every point of $X$ has a coordinate ball on which $T=\mathrm{dd}^{\mathrm{c}}u$ with $u$ plurisubharmonic and $H=\mathrm{dd}^{\mathrm{c}}h$ with $h$ smooth. Then $\mathrm{dd}^{\mathrm{c}}(f+h-u)=0$, so the scalar Laplacian of the distribution $f+h-u$ vanishes, and Weyl's lemma (theorem 3.1.11) shows that it is a smooth pluriharmonic function $p$. Thus $f=u-h+p$ is, on the ball, the class of a plurisubharmonic function minus a smooth function. On the overlap of two balls, the two local representatives $u_1-h_1+p_1$ and $u_2-h_2+p_2$ agree almost everywhere; after adding $h_1-p_1$ to both, both become plurisubharmonic, because $h_1-p_1-h_2+p_2$ is smooth with $\mathrm{dd}^{\mathrm{c}}(h_1-p_1-h_2+p_2)=H-H=0$, and plurisubharmonic functions that agree almost everywhere agree everywhere (theorem 5.9.1). Hence the local representatives define a global quasi-plurisubharmonic representative of $f$. It is the only one, since two quasi-plurisubharmonic functions that agree almost everywhere agree everywhere (lemma 6.14.4(1)). $\square$

Proof. Let $T\geq 0$ be closed with $\{ T\} =\{ \theta \} $. Write $T=H+\mathrm{dd}^{\mathrm{c}}f$ as in (10.2.16); since $\{ T\} =\{ \theta \} $, also $\theta =H+\mathrm{dd}^{\mathrm{c}}g$ with $g$ smooth. By theorem 10.2.15, $f$ has a canonical quasi-plurisubharmonic representative, and the difference $\varphi $ of this representative and $g$ satisfies $T=\theta +\mathrm{dd}^{\mathrm{c}}\varphi $; thus $\varphi \in \mathrm{PSH}(X,\theta )$. Conversely, if $\varphi \in \mathrm{PSH}(X,\theta )$, then $\theta +\mathrm{dd}^{\mathrm{c}}\varphi $ is a closed positive current, and it lies in $\{ \theta \} $ because adding $\mathrm{dd}^{\mathrm{c}}$ of a distribution does not change the class. If $\theta +\mathrm{dd}^{\mathrm{c}}\varphi =\theta +\mathrm{dd}^{\mathrm{c}}\varphi '$ with $\varphi ,\varphi '\in \mathrm{PSH}(X,\theta )$, then $\varphi -\varphi '$ is constant almost everywhere by the uniqueness part of theorem 10.2.15, say equal to $c$, and the quasi-plurisubharmonic functions $\varphi $ and $\varphi '+c$, which agree almost everywhere, agree everywhere by lemma 6.14.4(1). $\square$