A normal form of local anti-holomorphic involutions of $\mathbb C^2$?

Question

Suppose an anti-holomorphic involution $\sigma$ is defined in a neighbourhood of $0\in \mathbb C^2$. Suppose that $\sigma$ fixes a real two-dimensional surface $\Sigma$ containing $0$. Is it true that locally close to $0$ the involution $\sigma$ is holomorphically conjugate to the map $(z_1,z_2)\to (\bar z_1,\bar z_2)$? If not, is there some kind of classification of germs of such involutions?

Answer 1

I claim that the answer is positive in the sense that $\sigma$ is locally holomorphically conjugate to the standard antiholomorphic involution, assuming that when you say "a real surface" you mean "a totally real surface."

I will need:

Lemma. Let $S\subset {\mathbb C}^n$ be a totally-real real-analytic $k$-dimensional submanifold containing the origin $0\in {\mathbb C}^n$. Then there is a biholomorphic map $h: U\to V$ (where $U, V$ are neighborhoods of $0$ in ${\mathbb C}^n$) such that $h(S)\cap V= {\mathbb R}^k\cap V$, where ${\mathbb R}^k$ is the "standard" real subspace of ${\mathbb C}^n$, the real span of the standard basis vectors $e_1,...,e_k$.

Proof. Since $S$ is a real-analytic submanifold, there exists a local real-analytic parameterization $f: {\mathbb R}^k\to S$, $f(0)=0$ (in order to avoid complicated notation I will ignore the actual domain and codomain of $f$ which are neighborhoods of $0$ in ${\mathbb R}^k$ and ${\mathbb C}^n$ respectively). The map $f$ has real rank $k$ derivative at $0$. Let $F$ denote the complexification of $f$, i.e. holomorphic extension of $f$ to a small neighborhood of $0$ in ${\mathbb C}^k$, the complex span of the vectors $e_1,...,e_k$. (This extension exists since $f$ is real-analytic.)

The fact that $S$ is totally real implies that the complex derivative of $F$ has rank $k$ as well. Thus, $F$ extends (locally) to a biholomorphic map of open subsets $h^{-1}: V\to U$, sending ${\mathbb R}^k\cap V$ to $S\cap U$. qed

Now, suppose that $\sigma$ is an antiholomorphic involution of ${\mathbb C}^n$ fixing the origin, such that the derivative of $\sigma$ at $0$ is given by $$ \tau: (z_1,...,z_n)\mapsto (\bar z_1,...,\bar z_n). $$ (There is some linear algebra one has to do at this point, to ensure that all antiholomorphic linear involutions are conjugate via an element of $SL(n,C)$ to this standard involution.)

Since $\sigma$ is real-analytic, its fixed-point set $S$ is also a real-analytic submanifold of real dimension $n$. The assumption on the derivative of $\sigma$ implies that $S$ is totally real near the origin. Thus, by Lemma, after a local holomorphic change of coordinates, we can assume that $S= {\mathbb R}^n\cap W$, where $W$ is a neighborhood of the origin in ${\mathbb C}^n$.

The composition $g=\tau\circ \sigma$ still fixes $S\cap W$ pointwise but is now holomorphic. Since $S\cap W$ is totally-real and has real dimension $n$, it follows (by the uniqueness theorem for holomorphic functions) that $g$ is the identity map (near the origin). Hence, $\tau=\sigma$, thereby proving the claim.