[proofplan] The proof is local, because the vanishing of a smooth CR form can be checked after evaluating it on local sections of $T^{0,1}M$. We first recall that $\bar\partial_b$ is obtained by extending a CR antiholomorphic form to an ordinary complex differential form and then restricting its [exterior derivative](/theorems/1525) to $T^{0,1}M$ directions. Formal integrability is precisely what makes this restriction independent of the chosen extension. Once this is established, applying $\bar\partial_b$ twice is the same as restricting $d^2$ to $T^{0,1}M$, and the ordinary identity $d^2=0$ gives the result, with the top-degree cases handled by the zero target convention. [/proofplan] [step:Reduce the assertion to a local computation on antiholomorphic CR vector fields] Fix an integer $q$ with $0\le q\le m$ and let $\alpha\in C^\infty(M,\Lambda_b^{0,q}M)$. It is enough to show that, for every [open set](/page/Open%20Set) $U\subset M$ and every collection of local smooth sections \begin{align*} \bar L_0,\dots,\bar L_{q+1}\in C^\infty(U,T^{0,1}M), \end{align*} one has \begin{align*} (\bar\partial_b\bar\partial_b\alpha)(\bar L_0,\dots,\bar L_{q+1})=0 \end{align*} on $U$. When $q\ge m-1$, the bundle $\Lambda_b^{0,q+2}M$ is zero by convention, so the target of $\bar\partial_b\circ\bar\partial_b$ is the zero bundle and the asserted map is zero. Hence it remains to treat the case $0\le q\le m-2$. [/step] [step:Identify $\bar\partial_b$ with restricted exterior differentiation] Let $U\subset M$ be an open set over which $T^{0,1}M$ admits a smooth local frame. Let \begin{align*} \alpha_U:=\alpha|_U\in C^\infty(U,\Lambda_b^{0,q}M) \end{align*} denote the restriction of $\alpha$ to $U$. Choose an ordinary complex-valued smooth $q$-form \begin{align*} \widetilde\alpha\in C^\infty(U,\Lambda^q(\mathbb C T^*M)) \end{align*} whose restriction to $q$ arguments in $T^{0,1}M$ equals $\alpha_U$. Such an extension exists by choosing a smooth local complement to $T^{0,1}M$ in $\mathbb C TM$ and extending the alternating multilinear form arbitrarily on that complement. We claim that, for local sections $\bar L_0,\dots,\bar L_q\in C^\infty(U,T^{0,1}M)$, \begin{align*} (\bar\partial_b\alpha_U)(\bar L_0,\dots,\bar L_q)=(d\widetilde\alpha)(\bar L_0,\dots,\bar L_q). \end{align*} This is the defining local convention for the tangential Cauchy-Riemann operator on CR antiholomorphic forms. It remains only to justify that the right-hand side is independent of the chosen extension. Let \begin{align*} \widetilde\alpha_1,\widetilde\alpha_2\in C^\infty(U,\Lambda^q(\mathbb C T^*M)) \end{align*} be two extensions of $\alpha_U$, and set $\beta:=\widetilde\alpha_1-\widetilde\alpha_2$. Then $\beta(\bar K_1,\dots,\bar K_q)=0$ for all local sections $\bar K_1,\dots,\bar K_q$ of $T^{0,1}M$. The Koszul formula for $d\beta$ gives a sum of derivative terms and bracket terms. Each derivative term differentiates the zero function $\beta(\bar K_1,\dots,\bar K_q)$. Each bracket term contains an argument of the form $[\bar L_i,\bar L_j]$, which lies in $T^{0,1}M$ by formal integrability. Hence every term vanishes, and \begin{align*} (d\widetilde\alpha_1)(\bar L_0,\dots,\bar L_q)=(d\widetilde\alpha_2)(\bar L_0,\dots,\bar L_q). \end{align*} [guided] The point of this step is to make precise what “restricting $d$ to CR antiholomorphic directions” means. Start with an open set $U\subset M$ on which the bundle $T^{0,1}M$ has a smooth local frame, and write \begin{align*} \alpha_U:=\alpha|_U\in C^\infty(U,\Lambda_b^{0,q}M). \end{align*} A CR antiholomorphic $q$-form is only defined on $q$ arguments from $T^{0,1}M$. To apply the ordinary exterior derivative $d$, we first choose an ordinary complex-valued smooth $q$-form \begin{align*} \widetilde\alpha\in C^\infty(U,\Lambda^q(\mathbb C T^*M)) \end{align*} whose value on $T^{0,1}M$ agrees with $\alpha_U$. This extension exists because, locally, a vector bundle complement to $T^{0,1}M$ in $\mathbb C TM$ can be chosen, and an alternating form on $T^{0,1}M$ can be extended by specifying its values on the complementary directions. We then define the local value of $\bar\partial_b\alpha_U$ by \begin{align*} (\bar\partial_b\alpha_U)(\bar L_0,\dots,\bar L_q):=(d\widetilde\alpha)(\bar L_0,\dots,\bar L_q) \end{align*} for local sections $\bar L_0,\dots,\bar L_q\in C^\infty(U,T^{0,1}M)$. The only issue is whether this depends on the extension $\widetilde\alpha$. Suppose \begin{align*} \widetilde\alpha_1,\widetilde\alpha_2\in C^\infty(U,\Lambda^q(\mathbb C T^*M)) \end{align*} are two such extensions, and define their difference by $\beta:=\widetilde\alpha_1-\widetilde\alpha_2$. Then \begin{align*} \beta(\bar K_1,\dots,\bar K_q)=0 \end{align*} for every collection of local sections $\bar K_1,\dots,\bar K_q$ of $T^{0,1}M$, because the two extensions agree on exactly those arguments. Now evaluate $d\beta$ on $\bar L_0,\dots,\bar L_q$. The Koszul formula for the exterior derivative consists of terms in which one vector field differentiates the function obtained by inserting the remaining arguments into $\beta$, and terms in which $\beta$ is evaluated after replacing two arguments by their Lie bracket. The derivative terms vanish because they differentiate functions of the form \begin{align*} \beta(\bar L_0,\dots,\widehat{\bar L_i},\dots,\bar L_q), \end{align*} and these functions are identically zero. The bracket terms vanish because formal integrability gives \begin{align*} [\bar L_i,\bar L_j]\in C^\infty(U,T^{0,1}M) \end{align*} for every pair $i,j$, so $\beta$ is still being evaluated only on $T^{0,1}M$ arguments. Therefore \begin{align*} (d\beta)(\bar L_0,\dots,\bar L_q)=0. \end{align*} Since $d\beta=d\widetilde\alpha_1-d\widetilde\alpha_2$, this proves \begin{align*} (d\widetilde\alpha_1)(\bar L_0,\dots,\bar L_q)=(d\widetilde\alpha_2)(\bar L_0,\dots,\bar L_q). \end{align*} Thus the restricted exterior derivative is well-defined, and this is precisely the local operator $\bar\partial_b$. [/guided] [/step] [step:Apply ordinary nilpotence of the exterior derivative after the restriction is justified] Continue on the same open set $U\subset M$, and keep the extension \begin{align*} \widetilde\alpha\in C^\infty(U,\Lambda^q(\mathbb C T^*M)). \end{align*} From the previous step, \begin{align*} \bar\partial_b\alpha_U=(d\widetilde\alpha)|_{T^{0,1}M} \end{align*} as a CR antiholomorphic $(q+1)$-form. Therefore the ordinary $(q+1)$-form $d\widetilde\alpha$ is a valid local extension of $\bar\partial_b\alpha_U$. For local sections $\bar L_0,\dots,\bar L_{q+1}\in C^\infty(U,T^{0,1}M)$, the definition of $\bar\partial_b$ applied to the CR form $\bar\partial_b\alpha_U$ gives \begin{align*} (\bar\partial_b\bar\partial_b\alpha_U)(\bar L_0,\dots,\bar L_{q+1})=(d(d\widetilde\alpha))(\bar L_0,\dots,\bar L_{q+1}). \end{align*} The ordinary exterior derivative satisfies $d^2=0$, by the Koszul cancellation using the Lie bracket Jacobi identity. Hence \begin{align*} (d(d\widetilde\alpha))(\bar L_0,\dots,\bar L_{q+1})=0. \end{align*} Thus \begin{align*} (\bar\partial_b\bar\partial_b\alpha_U)(\bar L_0,\dots,\bar L_{q+1})=0 \end{align*} on $U$. [/step] [step:Conclude the global vanishing of the composed operator] The open set $U\subset M$ and the local sections $\bar L_0,\dots,\bar L_{q+1}$ were arbitrary. Since a section of $\Lambda_b^{0,q+2}M$ is determined locally by its values on local sections of $T^{0,1}M$, the equality \begin{align*} (\bar\partial_b\bar\partial_b\alpha)|_U=0 \end{align*} holds on every such open set $U$ when $0\le q\le m-2$. For $q=m-1$ and $q=m$, the target bundle $\Lambda_b^{0,q+2}M$ is zero by convention, so the composition also vanishes. Since $\alpha\in C^\infty(M,\Lambda_b^{0,q}M)$ was arbitrary, this proves \begin{align*} \bar\partial_b\circ\bar\partial_b=0:C^\infty(M,\Lambda_b^{0,q}M)\to C^\infty(M,\Lambda_b^{0,q+2}M) \end{align*} for every integer $q$ with $0\le q\le m$. [/step]