[proofplan]
The proof is the Frobenius argument behind Levi-flatness. We first record the bracket interpretation of the Levi form: vanishing of the Levi form is exactly the assertion that the mixed brackets of local CR vector fields have no normal component. This makes the real CR tangent distribution $H(M)$ involutive, so the smooth Frobenius theorem produces real leaves of dimension $2n-2$. Since $H(M)$ is invariant under the ambient complex structure $J$, these leaves are immersed complex hypersurfaces. Conversely, if such complex leaves exist, then CR brackets remain tangent to the leaves, so the normal component defining the Levi form vanishes.
[/proofplan]
[step:Dispose of the zero-dimensional CR case]
Assume first that $n=1$. Then $M\subset\mathbb C$ is a smooth real hypersurface, so $T_pM$ is a real line for every $p\in M$. Since $J(T_pM)$ is the real line obtained by rotating $T_pM$ by $\pi/2$, one has
\begin{align*}
H_p(M)=T_pM\cap J(T_pM)=\{0\}.
\end{align*}
For every local defining function $\rho$, the two-form $d\theta_\rho$ restricts to the zero [bilinear form](/page/Bilinear%20Form) on $H(M)$ because $H_p(M)=\{0\}$ at every point $p\in M$. The distribution $H(M)$ is the zero distribution, and the local foliation by points has tangent space $\{0\}$ at every point. Thus the statement holds for $n=1$. For the rest of the proof assume $n\ge 2$.
[/step]
[step:Relate Levi-flatness to involutivity through the contact form]
Let $p\in M$. Choose an open neighbourhood $V\subset\mathbb C^n$ of $p$ and a real-valued function $\rho\in C^\infty(V;\mathbb R)$ such that
\begin{align*}
M\cap V=\{z\in V:\rho(z)=0\}
\end{align*}
and $d\rho_q\ne 0$ for every $q\in M\cap V$. Define the real one-form
\begin{align*}
\theta_\rho:T(M\cap V)&\to\mathbb R
\end{align*}
\begin{align*}
v&\mapsto d\rho(Jv).
\end{align*}
For every $q\in M\cap V$, its kernel is
\begin{align*}
\ker (\theta_\rho)_q=T_qM\cap J(T_qM)=H_q(M).
\end{align*}
Indeed, if $v\in T_qM$, then $\theta_\rho(v)=0$ exactly when $Jv\in\ker d\rho_q=T_qM$. Here $\Gamma(H(M)|_{M\cap V})$ denotes the space of smooth real local sections of the vector bundle $H(M)|_{M\cap V}$.
The Levi form of $M$ determined by $\rho$ is the Hermitian form on $H_q(M)$ represented, in real notation, by the restriction of $d\theta_\rho$ to $H_q(M)$. Thus Levi-flatness near $p$ is equivalent to
\begin{align*}
d\theta_\rho(X,Y)=0
\end{align*}
for all smooth local sections $X,Y\in\Gamma(H(M)|_{M\cap V})$. Since $\theta_\rho(X)=\theta_\rho(Y)=0$, Cartan's [exterior derivative](/theorems/1525) formula gives
\begin{align*}
d\theta_\rho(X,Y)=-\theta_\rho([X,Y]).
\end{align*}
Consequently Levi-flatness is equivalent to $\theta_\rho([X,Y])=0$ for all local sections $X,Y$ of $H(M)$, that is, to the real distribution $H(M)$ being involutive.
[guided]
We need a version of Levi-flatness that talks directly about real brackets, because the Frobenius theorem applies to a real distribution. Fix $p\in M$ and choose a local defining function $\rho:V\to\mathbb R$ with $M\cap V=\rho^{-1}(\{0\})$ and $d\rho\ne0$ on $M\cap V$. The natural one-form that detects the missing direction inside $T(M\cap V)$ is
\begin{align*}
\theta_\rho:T(M\cap V)&\to\mathbb R
\end{align*}
\begin{align*}
v&\mapsto d\rho(Jv).
\end{align*}
For $v\in T_qM$, the condition $\theta_\rho(v)=0$ says $d\rho_q(Jv)=0$, which is equivalent to $Jv\in T_qM$ because $\ker d\rho_q=T_qM$. Hence
\begin{align*}
\ker (\theta_\rho)_q=H_q(M).
\end{align*}
The Levi form can be written in this real notation as the restriction of $d\theta_\rho$ to the $J$-invariant distribution $H(M)$. Therefore $M$ is Levi-flat exactly when
\begin{align*}
d\theta_\rho(X,Y)=0
\end{align*}
for all smooth local sections $X,Y\in\Gamma(H(M)|_{M\cap V})$. To convert this into a bracket statement, use Cartan's formula for the exterior derivative of a one-form:
\begin{align*}
d\theta_\rho(X,Y)=X(\theta_\rho(Y))-Y(\theta_\rho(X))-\theta_\rho([X,Y]).
\end{align*}
Since $X$ and $Y$ are sections of $H(M)=\ker\theta_\rho$, the first two terms vanish. Thus
\begin{align*}
d\theta_\rho(X,Y)=-\theta_\rho([X,Y]).
\end{align*}
So Levi-flatness says precisely that $\theta_\rho([X,Y])=0$. Together with the fact that brackets of vector fields on $M$ remain tangent to $M$, this means $[X,Y]\in H(M)$.
[/guided]
[/step]
[step:Use Levi-flatness to make $H(M)$ involutive]
Assume that $M$ is Levi-flat. Let $U\subset M\cap V$ be a sufficiently small coordinate neighbourhood on which $H(M)$ is spanned by smooth real vector fields. Let $X,Y\in\Gamma(H(M)|_U)$ be arbitrary smooth local sections. Since $X$ and $Y$ are vector fields on the smooth manifold $U$, their Lie bracket $[X,Y]$ is a smooth vector field on $U$, hence $[X,Y]_q\in T_qM$ for every $q\in U$.
By the contact-form characterization from the previous step, Levi-flatness gives
\begin{align*}
\theta_\rho([X,Y])=0
\end{align*}
on $U$. Since $H(M)|_U=\ker\theta_\rho$, it follows that
\begin{align*}
[X,Y]\in\Gamma(H(M)|_U).
\end{align*}
Thus $H(M)$ is an involutive smooth real distribution on $U$.
[guided]
Let $X,Y\in\Gamma(H(M)|_U)$ be smooth local sections. Their bracket is taken as the ordinary Lie bracket of vector fields on the real manifold $U\subset M$, so no ambient extension is needed at this stage. Because $[X,Y]$ is again a vector field on $U$, it is already tangent to $M$.
The remaining point is to prove that $[X,Y]$ has no component in the characteristic real direction complementary to $H(M)$ inside $TM$. The one-form $\theta_\rho=d\rho\circ J$ detects exactly that component, because $H(M)|_U=\ker\theta_\rho$. From the previous step, Levi-flatness gives
\begin{align*}
d\theta_\rho(X,Y)=0.
\end{align*}
Cartan's formula and $\theta_\rho(X)=\theta_\rho(Y)=0$ give
\begin{align*}
0=d\theta_\rho(X,Y)=-\theta_\rho([X,Y]).
\end{align*}
Thus $\theta_\rho([X,Y])=0$. Since $[X,Y]$ is tangent to $M$ and lies in the kernel of $\theta_\rho$, we obtain
\begin{align*}
[X,Y]\in\Gamma(H(M)|_U).
\end{align*}
This proves involutivity of the real CR tangent distribution.
[/guided]
[/step]
[step:Integrate $H(M)$ and identify the leaves as complex hypersurfaces]
For $n\ge2$, the bundle $H(M)$ is smooth of constant real rank $2n-2$. Indeed, locally $M$ is the regular level set of $\rho$, so $TM=\ker d\rho|_{TM}$ has real rank $2n-1$, and $H(M)=\ker\theta_\rho$ inside $TM$. The one-form $\theta_\rho$ is not identically zero on $TM$: if it vanished at $q$, then $J(T_qM)\subset T_qM$, which would make the odd-dimensional real [vector space](/page/Vector%20Space) $T_qM$ invariant under the complex structure $J$, impossible. Hence $\ker\theta_\rho$ has rank $2n-2$.
Since $H(M)$ is involutive, the smooth Frobenius theorem applies to the smooth constant-rank distribution $H(M)|_U$. After possibly shrinking $U$, it gives a smooth foliation of $U$ whose leaves are immersed real submanifolds $S\subset U$ satisfying
\begin{align*}
T_qS=H_q(M)
\end{align*}
for every $q\in S$.
Because $H_q(M)$ is invariant under $J$ for every $q\in U$, the restriction
\begin{align*}
J_S:TS&\to TS
\end{align*}
of the ambient tensor $J$ is a smooth almost complex structure on each leaf $S$. The inclusion map
\begin{align*}
\iota:S&\to\mathbb C^n
\end{align*}
is an immersion and satisfies $d\iota_q\circ J_S=J\circ d\iota_q$ for every $q\in S$. Since $J$ on $\mathbb C^n$ is the standard integrable complex structure and the tangent bundle $TS$ is $J$-invariant, the local coordinate restrictions supplied by the holomorphic Frobenius or complex-submanifold criterion make $S$ an immersed complex submanifold. Its real dimension is $2n-2$, so its complex dimension is $n-1$. Thus every leaf is an immersed complex hypersurface of $\mathbb C^n$.
[guided]
We first check the hypotheses needed for Frobenius. The distribution must be smooth and have constant rank. Locally $M$ is the regular level set of the smooth defining function $\rho$, so $TM$ has real rank $2n-1$. Inside $TM$, the CR distribution is
\begin{align*}
H(M)=\ker\theta_\rho.
\end{align*}
The one-form $\theta_\rho$ has rank one on $TM$: if $\theta_\rho$ vanished on all of $T_qM$, then $d\rho_q(Jv)=0$ for every $v\in T_qM$, so $J(T_qM)\subset T_qM$. That would make the odd-dimensional real vector space $T_qM$ invariant under $J$, impossible because a real vector space carrying a complex structure has even real dimension. Therefore $H_q(M)=\ker(\theta_\rho)_q$ has real dimension $2n-2$ at every $q$.
The previous step proved involutivity: for smooth local sections $X,Y\in\Gamma(H(M)|_U)$, one has $[X,Y]\in\Gamma(H(M)|_U)$. Thus the smooth Frobenius theorem applies to the smooth constant-rank distribution $H(M)|_U$. After shrinking $U$ if necessary, Frobenius gives a smooth foliation of $U$ by immersed real submanifolds $S$ satisfying
\begin{align*}
T_qS=H_q(M)
\end{align*}
for every point $q\in S$.
It remains to explain why these real Frobenius leaves are complex hypersurfaces, not merely real submanifolds. By definition,
\begin{align*}
H_q(M)=T_qM\cap J(T_qM),
\end{align*}
so $J(H_q(M))=H_q(M)$. Hence the ambient complex structure restricts to a smooth bundle map
\begin{align*}
J_S:TS&\to TS
\end{align*}
on each leaf. The inclusion
\begin{align*}
\iota:S&\to\mathbb C^n
\end{align*}
is an immersion, and the equality $T_qS=H_q(M)$ gives
\begin{align*}
d\iota_q\circ J_S=J\circ d\iota_q
\end{align*}
for every $q\in S$. Since the ambient $J$ is the standard integrable complex structure and $TS$ is $J$-invariant, the holomorphic Frobenius or complex-submanifold criterion applies: a smooth immersed real submanifold of $\mathbb C^n$ with $J$-invariant tangent spaces is an immersed complex submanifold. The real dimension of $S$ is $2n-2$, so its complex dimension is $n-1$. Thus each leaf is an immersed complex hypersurface of $\mathbb C^n$.
[/guided]
[/step]
[step:Use a complex foliation tangent to $H(M)$ to force the Levi form to vanish]
Conversely, assume that for every $p\in M$ there is a neighbourhood $U\subset M$ of $p$ foliated by immersed complex hypersurfaces whose tangent spaces are $H(M)|_U$. Let $\rho:V\to\mathbb R$ be a local defining function for $M$ on an open neighbourhood $V\subset\mathbb C^n$ of a point of $U$, and define $\theta_\rho=d\rho\circ J$ on $T(M\cap V)$.
Let $X,Y\in\Gamma(H(M)|_U)$ be smooth real local sections. Since the leaves have tangent spaces $H(M)|_U$, the restrictions of $X$ and $Y$ to each leaf are tangent to that leaf. The Lie bracket of vector fields tangent to a smooth leaf is again tangent to that leaf, so
\begin{align*}
[X,Y]_q\in T_qS=H_q(M)
\end{align*}
for every $q\in U$ lying on the leaf $S$. Therefore
\begin{align*}
\theta_\rho([X,Y])=0.
\end{align*}
Cartan's formula, together with $\theta_\rho(X)=\theta_\rho(Y)=0$, gives
\begin{align*}
d\theta_\rho(X,Y)=-\theta_\rho([X,Y])=0.
\end{align*}
Thus the Levi form vanishes identically near every point of $M$. Hence $M$ is Levi-flat.
[guided]
Now assume the foliation exists and fix a point near which it is defined. Choose a local defining function $\rho:V\to\mathbb R$ and set $\theta_\rho=d\rho\circ J$ on $T(M\cap V)$. To prove Levi-flatness in the sense of the statement, we must show that
\begin{align*}
d\theta_\rho(X,Y)=0
\end{align*}
for arbitrary smooth real local sections $X,Y\in\Gamma(H(M)|_U)$.
Let $S$ be the leaf through a point $q\in U$. The hypothesis says that the tangent spaces of the leaves are exactly the CR tangent spaces:
\begin{align*}
T_qS=H_q(M).
\end{align*}
Therefore the restrictions of $X$ and $Y$ to $S$ are vector fields tangent to the smooth manifold $S$. The Lie bracket of two vector fields tangent to a fixed smooth submanifold is again tangent to that submanifold, so
\begin{align*}
[X,Y]_q\in T_qS=H_q(M).
\end{align*}
Since $H(M)=\ker\theta_\rho$, this gives
\begin{align*}
\theta_\rho([X,Y])=0.
\end{align*}
Finally use Cartan's formula for the exterior derivative of a one-form:
\begin{align*}
d\theta_\rho(X,Y)=X(\theta_\rho(Y))-Y(\theta_\rho(X))-\theta_\rho([X,Y]).
\end{align*}
Because $X$ and $Y$ are sections of $H(M)=\ker\theta_\rho$, the first two terms vanish. The bracket term also vanishes by the previous paragraph. Hence
\begin{align*}
d\theta_\rho(X,Y)=0.
\end{align*}
This is exactly the local Levi-flatness condition from the theorem statement.
[/guided]
[/step]
[step:Conclude the equivalence]
The forward direction constructs, near every point of a Levi-flat hypersurface, a Frobenius foliation tangent to $H(M)$ and then identifies its leaves as immersed complex hypersurfaces of $\mathbb C^n$. The converse shows that any such local complex foliation forces all mixed CR bracket normal components, and therefore the Levi form, to vanish. Together with the zero-dimensional case $n=1$, this proves the stated equivalence.
[/step]
