[proofplan] We first translate genericity into a constant-rank statement for the real complex-tangent distribution $H(M):=TM\cap J(TM)$. The genericity condition makes a natural quotient map $TM\to T\mathbb C^N|_M/TM$ surjective, so its kernel $H(M)$ is a smooth real subbundle of rank $2(N-k)$. Applying the ambient projection onto $(1,0)$ vectors identifies $H(M)$ with the displayed bundle $T_{1,0}M$, giving smoothness and rank. Finally, the ambient type decomposition gives the zero-intersection condition, while formal integrability follows because brackets of ambient type $(1,0)$ fields remain type $(1,0)$ and brackets of vector fields tangent to $M$ remain tangent to $M$. [/proofplan]

[step:Extract the smooth real complex tangent distribution from genericity]Let $J:T\mathbb C^N\to T\mathbb C^N$ denote the standard complex structure, viewed as a smooth real bundle map. Since $M$ is an embedded real submanifold of real codimension $k$, the quotient bundle \begin{align*} Q:=(T\mathbb C^N|_M)/TM \end{align*} is a smooth real vector bundle over $M$ of rank $k$. Define the smooth real vector bundle morphism $A:TM\to Q$ by \begin{align*} A_p(v):=[Jv]\in T_p\mathbb C^N/T_pM \end{align*} for $p\in M$ and $v\in T_pM$. For each $p\in M$, the hypothesis \begin{align*} T_pM+J(T_pM)=T_p\mathbb C^N \end{align*} says exactly that every class in $T_p\mathbb C^N/T_pM$ has a representative of the form $Jv$ with $v\in T_pM$. Thus $A_p:T_pM\to Q_p$ is surjective for every $p\in M$. Therefore $\operatorname{rank}_{\mathbb R}A_p=k$ is constant, and the constant-rank theorem for smooth vector bundle morphisms gives that \begin{align*} H(M):=\ker A \end{align*} is a smooth real subbundle of $TM$. For each $p\in M$, \begin{align*} H_p(M)=\{v\in T_pM:Jv\in T_pM\}=T_pM\cap J(T_pM). \end{align*} Since $\operatorname{rank}_{\mathbb R}TM=2N-k$ and $\operatorname{rank}_{\mathbb R}A=k$, the [rank-nullity theorem](/theorems/916) gives \begin{align*} \operatorname{rank}_{\mathbb R}H(M)=2N-2k=2(N-k). \end{align*}[/step]

[guided]The main point is to prove smoothness, not only to compute dimensions at one point. We introduce a bundle map whose kernel is exactly the desired real complex-tangent distribution. Let $Q$ be the quotient bundle \begin{align*} Q:=(T\mathbb C^N|_M)/TM. \end{align*} Because $M\subset \mathbb C^N$ is a smooth embedded submanifold of real codimension $k$, each fiber $Q_p=T_p\mathbb C^N/T_pM$ is a real [vector space](/page/Vector%20Space) of dimension $k$, and these fibers form a smooth real vector bundle over $M$. Define $A:TM\to Q$ by \begin{align*} A_p(v):=[Jv]\in T_p\mathbb C^N/T_pM. \end{align*} This is smooth because $J$ is a smooth bundle endomorphism of $T\mathbb C^N$ and the quotient projection $T\mathbb C^N|_M\to Q$ is smooth. Now fix $p\in M$. The genericity hypothesis says \begin{align*} T_pM+J(T_pM)=T_p\mathbb C^N. \end{align*} Passing to the quotient by $T_pM$, this is precisely the assertion that every class in $T_p\mathbb C^N/T_pM$ is represented by some vector $Jv$ with $v\in T_pM$. Hence $A_p:T_pM\to Q_p$ is surjective. Since $\dim_{\mathbb R}Q_p=k$, the rank of $A_p$ is $k$ for every $p$. A smooth vector bundle morphism of constant rank has a smooth kernel subbundle. Therefore \begin{align*} H(M):=\ker A \end{align*} is a smooth real subbundle of $TM$. Its fiber is \begin{align*} H_p(M)=\{v\in T_pM:[Jv]=0\text{ in }T_p\mathbb C^N/T_pM\}. \end{align*} The equality $[Jv]=0$ means $Jv\in T_pM$, so \begin{align*} H_p(M)=T_pM\cap J(T_pM). \end{align*} Finally, rank-nullity applied to the surjective map $A_p:T_pM\to Q_p$ gives \begin{align*} \dim_{\mathbb R}H_p(M)=\dim_{\mathbb R}T_pM-\dim_{\mathbb R}Q_p=(2N-k)-k=2(N-k). \end{align*} Thus $H(M)$ is a smooth real subbundle of rank $2(N-k)$.[/guided]

[step:Identify the complex subbundle $T_{1,0}M$ with the $(1,0)$ part of $H(M)$] Let $T\mathbb C^N\otimes_{\mathbb R}\mathbb C$ denote the complexified ambient tangent bundle. The complex-linear extension of $J$ has eigenbundles $T_{1,0}\mathbb C^N=\{Z:JZ=iZ\}$ and $T_{0,1}\mathbb C^N=\{Z:JZ=-iZ\}$. Because $H_p(M)=T_pM\cap J(T_pM)$, the restriction $J_H:=J|_{H(M)}$ is a smooth bundle endomorphism of $H(M)$ with $J_H^2=-\operatorname{id}_{H(M)}$. Thus $H(M)$ is a smooth complex vector bundle when multiplication by $i$ is defined to be $J_H$. Define the smooth real bundle map $\Phi:H(M)\to T\mathbb C^N|_M\otimes_{\mathbb R}\mathbb C$ by \begin{align*} \Phi_p(v):=v-iJv. \end{align*} It is complex-linear when $H(M)$ is regarded as a complex vector bundle through multiplication by $i$ equal to $J_H$. Because $v\in H_p(M)$ implies $v\in T_pM$ and $Jv\in T_pM$, we have \begin{align*} \Phi_p(v)\in T_pM\otimes_{\mathbb R}\mathbb C. \end{align*} Also \begin{align*} J\Phi_p(v)=Jv+i v=i(v-iJv), \end{align*} so $\Phi_p(v)\in T_{1,0,p}\mathbb C^N$. Hence \begin{align*} \Phi_p(H_p(M))\subset T_{1,0,p}\mathbb C^N\cap (T_pM\otimes_{\mathbb R}\mathbb C). \end{align*} Conversely, let \begin{align*} Z\in T_{1,0,p}\mathbb C^N\cap (T_pM\otimes_{\mathbb R}\mathbb C). \end{align*} Write $Z=a+ib$ with $a,b\in T_pM$. Since $JZ=iZ$, comparison of real and imaginary parts gives \begin{align*} Ja=-b,\qquad Jb=a. \end{align*} Thus $a\in T_pM$ and $Ja=-b\in T_pM$, so $a\in H_p(M)$, and \begin{align*} Z=a+ib=a-iJa=\Phi_p(a). \end{align*} Therefore \begin{align*} T_{1,0,p}M=\Phi_p(H_p(M)). \end{align*} Since $\Phi_p(v)=0$ implies $v=iJv$, with the left side real and the right side purely imaginary in the complexification, we have $v=0$. Thus $\Phi_p$ identifies the real vector space $H_p(M)$ with the underlying real vector space of $T_{1,0,p}M$. Consequently $T_{1,0}M$ is a smooth complex subbundle of $TM\otimes_{\mathbb R}\mathbb C$ and \begin{align*} \operatorname{rank}_{\mathbb C}T_{1,0}M=N-k. \end{align*} [/step]

[step:Separate $T_{1,0}M$ from its conjugate] Define \begin{align*} T_{0,1}M:=\overline{T_{1,0}M}\subset TM\otimes_{\mathbb R}\mathbb C. \end{align*} Since complex conjugation on $TM\otimes_{\mathbb R}\mathbb C$ interchanges the ambient eigenbundles $T_{1,0}\mathbb C^N$ and $T_{0,1}\mathbb C^N$, we have \begin{align*} T_{0,1,p}M=T_{0,1,p}\mathbb C^N\cap (T_pM\otimes_{\mathbb R}\mathbb C) \end{align*} for every $p\in M$. The ambient complexified tangent space has the direct-sum decomposition \begin{align*} T_p\mathbb C^N\otimes_{\mathbb R}\mathbb C=T_{1,0,p}\mathbb C^N\oplus T_{0,1,p}\mathbb C^N. \end{align*} Hence \begin{align*} T_{1,0,p}\mathbb C^N\cap T_{0,1,p}\mathbb C^N=\{0\}. \end{align*} Since $T_{1,0,p}M\subset T_{1,0,p}\mathbb C^N$ and $T_{0,1,p}M\subset T_{0,1,p}\mathbb C^N$, it follows that \begin{align*} T_{1,0,p}M\cap T_{0,1,p}M=\{0\}. \end{align*} Therefore \begin{align*} T_{1,0}M\cap \overline{T_{1,0}M}=\{0\}. \end{align*} [/step]

[step:Prove formal integrability by combining ambient type closure with tangency]Let $U\subset M$ be an [open set](/page/Open%20Set). For a smooth vector bundle $E\to U$, write $\Gamma(U,E)$ for the space of smooth sections of $E$ over $U$. Let \begin{align*} X,Y\in \Gamma(U,T_{1,0}M) \end{align*} be smooth local sections. We must show that \begin{align*} [X,Y]\in \Gamma(U,T_{1,0}M). \end{align*} Fix $p\in U$. The assertion is local on $U$, so replace $U$ by a smaller neighbourhood of $p$ in $M$. Choose embedded-submanifold coordinates on an ambient open set $\widetilde U\subset\mathbb C^N$ in which $M\cap\widetilde U$ is a coordinate slice and the smaller $U$ equals $M\cap\widetilde U$. In these coordinates, a smooth section along $M\cap\widetilde U$ of a restricted ambient vector bundle is written in a smooth ambient frame with coefficient functions on the coordinate slice, and each coefficient function extends smoothly to $\widetilde U$ by making it independent of the [normal coordinates](/theorems/2713). Since $X$ and $Y$ are smooth sections of the restricted ambient bundle $T_{1,0}\mathbb C^N|_U$, extend them locally to smooth ambient vector fields \begin{align*} \widetilde X,\widetilde Y\in \Gamma(\widetilde U,T_{1,0}\mathbb C^N) \end{align*} whose restrictions to $U$ are $X$ and $Y$. In standard holomorphic coordinates $(z_1,\dots,z_N):\mathbb C^N\to\mathbb C^N$, write \begin{align*} \widetilde X=\sum_{j=1}^N a_j\frac{\partial}{\partial z_j},\qquad \widetilde Y=\sum_{j=1}^N b_j\frac{\partial}{\partial z_j}, \end{align*} where $a_j,b_j:\widetilde U\to\mathbb C$ are smooth functions. A direct computation of the Lie bracket gives \begin{align*} [\widetilde X,\widetilde Y]=\sum_{j=1}^N\left(\widetilde X(b_j)-\widetilde Y(a_j)\right)\frac{\partial}{\partial z_j}. \end{align*} Thus $[\widetilde X,\widetilde Y]$ is again a section of $T_{1,0}\mathbb C^N$. It remains to prove tangency to $M$. Let \begin{align*} I_M(\widetilde U):=\{f\in C^\infty(\widetilde U;\mathbb C):f|_{M\cap \widetilde U}=0\} \end{align*} be the ideal of smooth complex-valued functions vanishing on $M\cap\widetilde U$. In embedded-submanifold coordinates, a vector field along $M$ is tangent to $M$ exactly when it annihilates $I_M(\widetilde U)$ after restriction to $M$. Since $X$ and $Y$ are tangent to $M$, for every $f\in I_M(\widetilde U)$ the restrictions $(\widetilde Xf)|_U$ and $(\widetilde Yf)|_U$ vanish. Because $\widetilde X|_U=X$ is tangent to $M$ and $\widetilde Yf$ vanishes on $M\cap\widetilde U$, the restriction $(\widetilde X(\widetilde Yf))|_U$ vanishes. Likewise, because $\widetilde Y|_U=Y$ is tangent to $M$ and $\widetilde Xf$ vanishes on $M\cap\widetilde U$, the restriction $(\widetilde Y(\widetilde Xf))|_U$ vanishes. Therefore \begin{align*} ([\widetilde X,\widetilde Y]f)|_U=(\widetilde X(\widetilde Yf)-\widetilde Y(\widetilde Xf))|_U=0. \end{align*} This ideal criterion for tangency shows that $[\widetilde X,\widetilde Y]|_U$ is tangent to $M$. Consequently \begin{align*} [\widetilde X,\widetilde Y]|_U\in \Gamma(U,T_{1,0}\mathbb C^N|_U)\cap \Gamma(U,TM\otimes_{\mathbb R}\mathbb C)=\Gamma(U,T_{1,0}M). \end{align*} Because $\widetilde X|_U=X$ and $\widetilde Y|_U=Y$, the commutator derivation $[\widetilde X,\widetilde Y]|_U$ acts on every smooth function on $U$ as the intrinsic commutator $[X,Y]$. The preceding ideal calculation shows that this commutator is tangent to $M$, so the restricted ambient bracket is exactly the intrinsic bracket $[X,Y]$. Hence $T_{1,0}M$ is formally integrable.[/step]

[guided]We now verify the condition that makes the CR structure integrable: sections of $T_{1,0}M$ must be closed under the Lie bracket. Let $U\subset M$ be open, and let \begin{align*} X,Y\in \Gamma(U,T_{1,0}M) \end{align*} be smooth local sections. Fix $p\in U$. The argument is local on $U$, so we replace $U$ by a smaller neighbourhood of $p$ in $M$. Choose an ambient open neighbourhood $\widetilde U\subset \mathbb C^N$ with embedded-submanifold coordinates in which $M\cap\widetilde U$ is a coordinate slice and the smaller $U$ equals $M\cap\widetilde U$. In these coordinates, coefficient functions of sections along the slice extend smoothly to $\widetilde U$ by keeping them constant in the normal coordinate directions. Since $X$ and $Y$ are restrictions of sections of the smooth ambient bundle $T_{1,0}\mathbb C^N$, we may extend them locally to ambient smooth type $(1,0)$ vector fields \begin{align*} \widetilde X,\widetilde Y\in \Gamma(\widetilde U,T_{1,0}\mathbb C^N). \end{align*} First we check the type. In standard holomorphic coordinates $(z_1,\dots,z_N):\mathbb C^N\to\mathbb C^N$, every smooth type $(1,0)$ vector field is a smooth linear combination of the coordinate fields $\partial/\partial z_j$. Thus there are smooth functions $a_j,b_j:\widetilde U\to\mathbb C$ such that \begin{align*} \widetilde X=\sum_{j=1}^N a_j\frac{\partial}{\partial z_j},\qquad \widetilde Y=\sum_{j=1}^N b_j\frac{\partial}{\partial z_j}. \end{align*} Using the definition of the Lie bracket as the commutator of derivations, and using that the coordinate fields $\partial/\partial z_j$ commute with one another, we get \begin{align*} [\widetilde X,\widetilde Y]=\sum_{j=1}^N\left(\widetilde X(b_j)-\widetilde Y(a_j)\right)\frac{\partial}{\partial z_j}. \end{align*} This is again a smooth linear combination of the fields $\partial/\partial z_j$, so it is again of ambient type $(1,0)$. Second we check tangency to $M$. Let \begin{align*} I_M(\widetilde U):=\{f\in C^\infty(\widetilde U;\mathbb C):f|_{M\cap \widetilde U}=0\} \end{align*} be the ideal of smooth complex-valued functions vanishing on $M$. A vector field along $M$ is tangent to $M$ exactly when it sends every function in this vanishing ideal to a function that also vanishes on $M$. Since $X$ and $Y$ are tangent to $M$, for every $f\in I_M(\widetilde U)$ we have \begin{align*} (\widetilde Xf)|_U=0,\qquad (\widetilde Yf)|_U=0. \end{align*} Now use tangency one more time. The function $\widetilde Yf$ vanishes on $M\cap\widetilde U$, and $\widetilde X|_U=X$ is tangent to $M$, so $(\widetilde X(\widetilde Yf))|_U=0$. Similarly, $\widetilde Xf$ vanishes on $M\cap\widetilde U$, and $\widetilde Y|_U=Y$ is tangent to $M$, so $(\widetilde Y(\widetilde Xf))|_U=0$. Applying the commutator to such an $f$, we obtain \begin{align*} ([\widetilde X,\widetilde Y]f)|_U=(\widetilde X(\widetilde Yf)-\widetilde Y(\widetilde Xf))|_U=0. \end{align*} Thus $[\widetilde X,\widetilde Y]|_U$ is tangent to $M$. We have shown both required properties: the bracket is of ambient type $(1,0)$ and tangent to $M$. Therefore \begin{align*} [\widetilde X,\widetilde Y]|_U\in \Gamma(U,T_{1,0}\mathbb C^N|_U)\cap \Gamma(U,TM\otimes_{\mathbb R}\mathbb C)=\Gamma(U,T_{1,0}M). \end{align*} The restricted bracket is the intrinsic bracket $[X,Y]$ on $U$: for every smooth function $g:U\to\mathbb C$, extend $g$ locally to a smooth function $\widetilde g:\widetilde U\to\mathbb C$; then $X(g)=(\widetilde X\widetilde g)|_U$ and $Y(g)=(\widetilde Y\widetilde g)|_U$, so the commutator derivation induced on $U$ is $[X,Y]$. The ideal calculation above proves that this commutator is tangent to $M$, hence $[X,Y]$ is a section of $T_{1,0}M$, which is precisely formal integrability.[/guided]

[step:Conclude that the induced bundle is an abstract CR structure] We have proved that $T_{1,0}M$ is a smooth complex subbundle of $TM\otimes_{\mathbb R}\mathbb C$, that \begin{align*} T_{1,0}M\cap \overline{T_{1,0}M}=\{0\}, \end{align*} and that its smooth local sections are closed under the Lie bracket. These are exactly the defining properties of an abstract CR structure. Therefore the displayed bundle $T_{1,0}M$ defines an abstract CR structure on $M$. [/step]