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Jun 9, 2021 at 2:32 vote accept Chicken feed
Jun 8, 2021 at 14:52 comment added Ben McKay (2) Now on $N$, $w$ is a $C^1$ function of $z$. But each tangent space is complex linear, so $dw$ is a complex linear function of $dz$, i.e. the Cauchy--Riemann equations are satisfied.
Jun 8, 2021 at 14:51 comment added Ben McKay @Chickenfeed: Since the tangent space is $I$-invariant, it is a complex linear subspace. Pick one point $n_0\in N$. Take holomorphic local coordinates $z^{\mu},w^{\nu}$ so that $f_*T_{n_0} N$ is the complex linear subspace $w=0$ at the origin of coordinates. So at $n_0$, the real and imaginary parts of the various $dz^{\mu}$ are linearly independent, and so also nearby. So we can replace $n_0$ with a neighborhood on which the real and imaginary parts of $z$ are local coordinates on $N$. Replace $N$ by a small neighborhood of $n_0$ on which they are global coordinates.
Jun 8, 2021 at 14:27 comment added Chicken feed @Ben McKay:Thank you for the answer. But how to gurantee that there is such a holomorphic coordinate in which the immersed image is local the graph of some coordinates as $C1$ functions of others. Did you use the invariant property of the tangent space? Since $N$ can be odd dimentional without this condition.
Jun 8, 2021 at 9:49 comment added Ben McKay @Chickenfeed: yes. In any holomorphic coordinates in which the immersed image is local the graph of some coordinates as $C^1$ functions of others, the image (of any open subset on which $f$ is an embedding) is a $C^1$ solution of the Cauchy-Riemann equations, so has image a complex submanifold to which $f$ immerses, so the complex structure pulls back and so on.
Jun 8, 2021 at 8:49 comment added Chicken feed Thank you for the answer. Can we get a similar result when $f$ is merely a $C^r (r\ge1)$ immersion ?
Jun 8, 2021 at 4:11 history answered HYL CC BY-SA 4.0