Local property of real submanifold embedded in complex space Let $L$ be an embedded real $n$-dimensional manifold in $\mathbb C^n$, where $n\ge 2$. 
Given a point $p \in L$, does there exists a neighborhood $U$ of $p$ and holomorphic function $$\phi:U \rightarrow \mathbb C^n,$$ such $\phi$ is a diffeomoprhism onto its image and $$\phi(U \cap L) \subset \mathbb R^n \subset \mathbb C^n = \mathbb R^n  \oplus i \mathbb R^n?$$
My guess is this is not true in general. I would love to know why not and also what extra conditions, if any, would make it true.
 A: There's a fairly straightforward necessary and sufficient condition. First, a couple of definitions: 
A $k$-dimensional real submanifold $L\subset \mathbb C^n$ is said to be real analytic if for each $p\in L$ there is a neighborhood $U$ of $p$ and a real-analytic submersion $\phi\colon U\to \mathbb R^{2n-k}$ such that $L\cap U = \phi^{-1}(0)$. 
Such a submanifold is said to be totally real if for each $p\in L$, the real-linear map $J: \mathbb C^n\to \mathbb C^n$ obtained by multiplying by $i$ in complex coordinates satisfies $T_pL \cap J(T_pL) = \{0\}$. (In other words, there is no vector $v\in T_pL$ such that $iv$ is also tangent to $L$.)
Theorem. Let $L$ be an embedded real $n$-dimensional submanifold in $\mathbb C^n$, where $n\ge 2$. The following conditions are equivalent. 


*

*For each $p \in L$, there exists a neighborhood $U$ of $p$ and a holomorphic function $\phi:U \rightarrow \mathbb C^n$ such that $\phi$ is a diffeomoprhism onto its image and $\phi(L\cap U) = \phi(U)\cap \mathbb R^n$.

*$L$ is real-analytic and totally real.
Sketch of proof: Let's use coordinates $z^j = x^j + i y^j$ on $\mathbb C^n$. For necessity, note that both the real analytic and totally real conditions are satisfied by the submanifold $y^1=\dots = y^n = 0$, and both conditions are preserved by biholomorphic maps.
For sufficiency, suppose $L$ satisfies condition $2$, and let $p\in L$ be given. After applying a translation, we can assume $p$ is the origin. Let $(v_1,\dots,v_n)$ be a basis for the (real) tangent space $T_pL$. The fact that $T_pL$ is totally real ensures that the complex-linear map $A(z^1,\dots,z^n) = z^1v_1 + \dots z^n v_n$ has trivial kernel, and $A^{-1}$ takes $T_pL$ to the subspace defined by $y^1=\dots y^n=0$. Thus after applying $A^{-1}$, we may assume that $T_pL$ is that subspace. Now the real-analytic version of the implicit function theorem shows that there are neighborhoods $U_0,V_0$ of $0$ in $\mathbb R^n$ and a real-analytic map $f\colon U_0\to V_0$ such that $L\cap (U_0\oplus i V_0)$ is the set determined by the equations $y^j = f^j(x^1,\dots,x^n)$. The fact that $T_0L$ is the $y=0$ plane means that all first partial derivatives of each $f_j$ vanish at the origin. By real-analyticity, there is some neighborhood $U_1$ of $U_0$ in $\mathbb C^n$ to which the $f^j$'s extend as holomorphic functions, still denoted by $f^j$.
Define a map $F\colon U_1\to \mathbb C^n$ by $F^j(z) = z^j + if^j(z)$. The fact that the partial derivatives of $f^j$ vanish at the origin means that $dF_0$ is the identity map, so the holomorphic version of the inverse function theorem shows that $F$ is a biholomorphism from some smaller neighborhood of the origin to its image. You can check that $F$ takes a neighborhood of the origin in the set $\{y=0\}$ to a neighborhood of the origin in $L$. $\square$
A: As you correctly guessed, the answer to your question is "no" in general.
For example  in $\mathbb C^2$ with coordinates $z,w$ the real $2$- dimensional submanifold $L=\{w=0\}\subset \mathbb C^2$, which happens to be also a complex submanifold,  will be transported by any holomorphic isomorphism $\mathbb C^2 \stackrel \simeq \to  \mathbb C^2$ to a holomorphic submanifold  $\phi(L)\subset \mathbb C^2$.  Hence we will never have $\phi(L)\subset\mathbb R^2$.
(For ease of notations I have considered a global isomorphism $\phi$, but the argument works locally).
