[proofplan] We write the degree-$q$ part of the tangential Cauchy-Riemann complex as a closed Hilbert complex and use the assumed subelliptic estimate to convert graph-norm boundedness into compactness in $L^2$. This compactness gives finite-dimensionality of the harmonic space and, by contradiction on the orthogonal complement of the harmonic space, a coercive estimate for the quadratic form of $\Box_{b,q}$. The coercive estimate implies closed range for $\Box_{b,q}$ and hence the $L^2$ [Hodge decomposition](/theorems/2745) in degree $q$. Finally, the decomposition identifies smooth closed forms modulo smooth exact forms with smooth [harmonic representatives](/theorems/2747), using standard subelliptic regularity for the Kohn Laplacian to pass from weak $L^2$ solutions to smooth forms. [/proofplan]

[step:Set up the closed Hilbert complex in degree $q$] Define the Hilbert spaces \begin{align*} H_k:=L^2(M,\Lambda_b^{0,k}M) \end{align*} for $0\le k\le n$. Define the closed operators $A:=\bar\partial_{b,q}:\operatorname{Dom}(A)\subset H_q\to H_{q+1}$ and $B:=\bar\partial_{b,q-1}:\operatorname{Dom}(B)\subset H_{q-1}\to H_q$. At $q=n$ we set $A=0$, and at $q=0$ we set $B=0$. The Hilbert-space adjoint of $B$ is \begin{align*} B^*:\operatorname{Dom}(B^*)\subset H_q\to H_{q-1}. \end{align*} The degree-$q$ quadratic form domain is \begin{align*} V:=\operatorname{Dom}(A)\cap \operatorname{Dom}(B^*)\subset H_q, \end{align*} equipped with the graph norm \begin{align*} \|u\|_V:=\left(\|u\|_{L^2(M)}^2+\|Au\|_{L^2(M)}^2+\|B^*u\|_{L^2(M)}^2\right)^{1/2}. \end{align*} Since $A$ and $B^*$ are closed operators, $V$ is a [Hilbert space](/page/Hilbert%20Space) with this norm. The Kohn Laplacian in degree $q$ is the non-negative self-adjoint operator associated with the closed sesquilinear form $a:V\times V\to \mathbb C$ defined by \begin{align*} a(u,v):=(Au,Av)_{H_{q+1}}+(B^*u,B^*v)_{H_{q-1}}. \end{align*} Set $Q(u):=a(u,u)$ for $u\in V$. Thus \begin{align*} (\Box_{b,q}u,u)_{H_q}=Q(u) \end{align*} for every $u\in \operatorname{Dom}(\Box_{b,q})$. In particular, \begin{align*} \mathcal H_{b,L^2}^{0,q}(M)=\ker A\cap \ker B^*. \end{align*} Indeed, if $u\in \ker \Box_{b,q}$, then $0=(\Box_{b,q}u,u)_{H_q}=Q(u)$, so $Au=0$ and $B^*u=0$. Conversely, if $Au=0$ and $B^*u=0$, then $Q(u)=0$. Since $a$ is a non-negative closed sesquilinear form, the [Cauchy-Schwarz inequality](/theorems/432) for $a$ gives $|a(u,v)|^2\le Q(u)Q(v)=0$ for every $v\in V$. Hence $u$ lies in the operator domain of the self-adjoint operator associated with $Q$ and that operator annihilates $u$. [/step]

[step:Use the subelliptic estimate and Rellich compactness to get compactness in $L^2$]The assumed estimate says that there are constants $\varepsilon>0$ and $C>0$ such that every $u\in V$ satisfies \begin{align*} \|u\|_{H^\varepsilon(M)}\le C\left(\|Au\|_{L^2(M)}+\|B^*u\|_{L^2(M)}+\|u\|_{L^2(M)}\right). \end{align*} Let $(u_j)_{j=1}^{\infty}$ be a sequence in $V$ such that \begin{align*} \sup_{j\in\mathbb N}\|u_j\|_V<\infty. \end{align*} The subelliptic estimate gives \begin{align*} \sup_{j\in\mathbb N}\|u_j\|_{H^\varepsilon(M)}<\infty. \end{align*} Because $M$ is compact and $\varepsilon>0$, the Rellich compact embedding \begin{align*} H^\varepsilon(M,\Lambda_b^{0,q}M)\hookrightarrow L^2(M,\Lambda_b^{0,q}M) \end{align*} is compact. Therefore every $\|\cdot\|_V$-bounded sequence in $V$ has a subsequence converging strongly in $H_q$.[/step]

[guided]The role of the subelliptic estimate is to turn control of the graph norm into control of a positive Sobolev norm. We start with a sequence \begin{align*} (u_j)_{j=1}^{\infty}\subset V \end{align*} satisfying \begin{align*} \sup_{j\in\mathbb N}\|u_j\|_V<\infty. \end{align*} By the definition of $\|\cdot\|_V$, this means that the three quantities \begin{align*} \|u_j\|_{L^2(M)},\qquad \|Au_j\|_{L^2(M)},\qquad \|B^*u_j\|_{L^2(M)} \end{align*} are uniformly bounded in $j$. The assumed subelliptic estimate applies to every $u_j\in V$ and gives \begin{align*} \|u_j\|_{H^\varepsilon(M)}\le C\left(\|Au_j\|_{L^2(M)}+\|B^*u_j\|_{L^2(M)}+\|u_j\|_{L^2(M)}\right). \end{align*} The right-hand side is uniformly bounded, so \begin{align*} \sup_{j\in\mathbb N}\|u_j\|_{H^\varepsilon(M)}<\infty. \end{align*} Now we use compactness. Since $M$ is compact and $\varepsilon>0$, the Rellich [compactness theorem](/theorems/2748) for Sobolev spaces on compact manifolds says that the inclusion \begin{align*} H^\varepsilon(M,\Lambda_b^{0,q}M)\hookrightarrow L^2(M,\Lambda_b^{0,q}M) \end{align*} is compact. Its hypotheses are satisfied because $M$ is a compact smooth manifold, the bundle $\Lambda_b^{0,q}M$ is smooth and finite rank, and the Sobolev norms are defined using any auxiliary smooth atlas and bundle metric. Hence a bounded sequence in $H^\varepsilon(M,\Lambda_b^{0,q}M)$ has a subsequence converging strongly in $L^2(M,\Lambda_b^{0,q}M)$. Applying this to $(u_j)_{j=1}^{\infty}$ proves that every graph-norm bounded sequence in $V$ has a strongly convergent subsequence in $H_q$.[/guided]

[step:Prove that the harmonic space is finite-dimensional] Let \begin{align*} \mathcal H:=\mathcal H_{b,L^2}^{0,q}(M)=\ker A\cap \ker B^*. \end{align*} If $h\in \mathcal H$, then $Ah=0$ and $B^*h=0$, so the subelliptic estimate gives \begin{align*} \|h\|_{H^\varepsilon(M)}\le C\|h\|_{L^2(M)}. \end{align*} Thus the closed unit ball of $\mathcal H$ in $H_q$ is bounded in $H^\varepsilon(M,\Lambda_b^{0,q}M)$ and is therefore compact in $H_q$ by Rellich compactness. A [normed vector space](/page/Normed%20Vector%20Space) whose closed unit ball is compact is finite-dimensional. Hence $\mathcal H_{b,L^2}^{0,q}(M)$ is finite-dimensional. [/step]

[step:Derive the coercive estimate on the harmonic orthogonal complement] We claim that there exists a constant $c>0$ such that every \begin{align*} u\in V\cap \mathcal H^\perp \end{align*} satisfies \begin{align*} \|u\|_{L^2(M)}^2\le c\left(\|Au\|_{L^2(M)}^2+\|B^*u\|_{L^2(M)}^2\right). \end{align*} Suppose not. Then for each $j\in\mathbb N$ there exists $u_j\in V\cap \mathcal H^\perp$ such that \begin{align*} \|u_j\|_{L^2(M)}=1 \end{align*} and \begin{align*} \|Au_j\|_{L^2(M)}^2+\|B^*u_j\|_{L^2(M)}^2\le \frac{1}{j}. \end{align*} The sequence $(u_j)_{j=1}^{\infty}$ is bounded in $V$, so after passing to a subsequence there exists $u\in H_q$ such that \begin{align*} u_j\to u \end{align*} strongly in $H_q$. Since $A$ and $B^*$ are closed and $Au_j\to 0$, $B^*u_j\to 0$ in $L^2$, we have $u\in \ker A\cap \ker B^*=\mathcal H$. Since $\mathcal H^\perp$ is closed in $H_q$ and each $u_j\in \mathcal H^\perp$, we also have $u\in \mathcal H^\perp$. Hence $u\in \mathcal H\cap \mathcal H^\perp=\{0\}$. But strong convergence in $H_q$ and $\|u_j\|_{L^2(M)}=1$ imply \begin{align*} \|u\|_{L^2(M)}=1, \end{align*} contradicting $u=0$. Therefore the coercive estimate holds. [/step]

[step:Use the coercive estimate to prove closed range of $\Box_{b,q}$] Let \begin{align*} T:=\Box_{b,q}: \operatorname{Dom}(\Box_{b,q})\subset H_q\to H_q. \end{align*} Because $T$ is self-adjoint and non-negative, \begin{align*} \ker T=\mathcal H \end{align*} and $T$ maps $\operatorname{Dom}(T)\cap \mathcal H^\perp$ into $\mathcal H^\perp$. For $u\in \operatorname{Dom}(T)\cap \mathcal H^\perp$, the coercive estimate and the identity $Q(u)=(Tu,u)_{H_q}$ give \begin{align*} \|u\|_{L^2(M)}^2\le c(Tu,u)_{H_q}. \end{align*} By the Cauchy-Schwarz inequality in the Hilbert space $H_q$, \begin{align*} |(Tu,u)_{H_q}|\le \|Tu\|_{L^2(M)}\|u\|_{L^2(M)}. \end{align*} If $u\ne 0$, division by $\|u\|_{L^2(M)}$ gives \begin{align*} \|u\|_{L^2(M)}\le c\|Tu\|_{L^2(M)}. \end{align*} The same inequality is immediate when $u=0$. Now let $(Tu_j)_{j=1}^{\infty}$ be a convergent sequence in $\operatorname{Range}(T)$. For each initial preimage $w_j\in\operatorname{Dom}(T)$, replace $w_j$ by $u_j:=w_j-P_{\mathcal H}w_j$, where $P_{\mathcal H}:H_q\to\mathcal H$ is the [orthogonal projection](/theorems/437). Since $T$ annihilates $\mathcal H$, we have $Tu_j=Tw_j$, and \begin{align*} u_j\in \operatorname{Dom}(T)\cap \mathcal H^\perp. \end{align*} The preceding inequality applied to $u_j-u_k$ shows that $(u_j)_{j=1}^{\infty}$ is Cauchy in $H_q$. Let $u\in \mathcal H^\perp$ be its $H_q$ limit. Since $T$ is closed and $Tu_j$ converges in $H_q$, we have $u\in \operatorname{Dom}(T)$ and \begin{align*} Tu=\lim_{j\to\infty}Tu_j. \end{align*} Thus every [limit point](/page/Limit%20Point) of $\operatorname{Range}(T)$ lies in $\operatorname{Range}(T)$, so $\operatorname{Range}(\Box_{b,q})$ is closed. [/step]

[step:Obtain the $L^2$ Hodge decomposition in degree $q$] Since $\Box_{b,q}$ is self-adjoint and has closed range, the Hilbert-space closed range identity $\overline{\operatorname{Range}(T)}=(\ker T)^\perp$ for closed densely defined operators gives \begin{align*} H_q=\ker \Box_{b,q}\oplus \operatorname{Range}(\Box_{b,q}). \end{align*} Thus every $f\in H_q$ has a unique decomposition \begin{align*} f=h+\Box_{b,q}u \end{align*} with \begin{align*} h\in \mathcal H \end{align*} and \begin{align*} u\in \operatorname{Dom}(\Box_{b,q})\cap \mathcal H^\perp. \end{align*} This is the degree-$q$ $L^2$ [Hodge decomposition](/theorems/3941). [/step]

[step:Represent each smooth closed class by a harmonic form] Let \begin{align*} f\in C^\infty(M,\Lambda_b^{0,q}M) \end{align*} satisfy \begin{align*} \bar\partial_b f=0. \end{align*} Using the $L^2$ Hodge decomposition, write \begin{align*} f=h+r \end{align*} with $h\in \mathcal H$ and $r\in \operatorname{Range}(\Box_{b,q})\subset \mathcal H^\perp$. Since $h\in\mathcal H$ and the smooth subelliptic regularity hypothesis gives $h\in C^\infty(M,\Lambda_b^{0,q}M)$, the form \begin{align*} r=f-h \end{align*} belongs to $C^\infty(M,\Lambda_b^{0,q}M)\cap\ker\bar\partial_b$ and is orthogonal to $\mathcal H$. By the smooth subelliptic regularity and solvability hypothesis in degree $q$, applied to this smooth closed form $r$, there exists \begin{align*} v\in C^\infty(M,\Lambda_b^{0,q-1}M) \end{align*} such that \begin{align*} \bar\partial_{b,q-1}v=r=f-h. \end{align*} Therefore $f$ and $h$ define the same smooth Kohn-Rossi cohomology class. This proves surjectivity of the harmonic-to-cohomology map. [/step]

[step:Show that the harmonic representative is unique in each smooth cohomology class] Suppose \begin{align*} h\in \mathcal H \end{align*} maps to the zero class in $H_{KR}^{0,q}(M)$. Then there exists \begin{align*} v\in C^\infty(M,\Lambda_b^{0,q-1}M) \end{align*} such that \begin{align*} h=\bar\partial_{b,q-1}v=Bv. \end{align*} The nilpotence assumption $\bar\partial_{b,q}\bar\partial_{b,q-1}=0$ ensures that $\operatorname{im}\bar\partial_{b,q-1}\subseteq\ker\bar\partial_{b,q}$, so the quotient defining $H_{KR}^{0,q}(M)$ is well-defined. Since $h\in \mathcal H=\ker B^*\cap \ker A$, we have $B^*h=0$. Taking the $H_q$ [inner product](/page/Inner%20Product) of $h=Bv$ with $h$ and using the definition of the Hilbert-space adjoint gives \begin{align*} \|h\|_{L^2(M)}^2=(Bv,h)_{H_q}=(v,B^*h)_{H_{q-1}}=0. \end{align*} Thus $h=0$. Hence the harmonic-to-cohomology map is injective. Combining injectivity with the surjectivity proved above, the natural map \begin{align*} \mathcal H_{b,L^2}^{0,q}(M)\to H_{KR}^{0,q}(M) \end{align*} is an isomorphism. The smooth subelliptic regularity hypothesis in the statement also shows that every element of $\mathcal H_{b,L^2}^{0,q}(M)$ is smooth. This completes the proof. [/step]