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Let $M$ be a complex manifold, $N$ is a smooth immersed submanifold of $M$. If $T_p M$ is invariant under the multiplication by $i$ for any $p\in M$, then can we conclude that $N$ is a complex immersed submanifold of $M$?

Since $C^1$ property somehow means analytic property in complex setting, can we drop the assumption to that $N$ is merely a $C^1$ submanifold?

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Let $f : N \to M$ denote the immersion. Since $f_*TN$ is invariant under $I$ where $I$ is the underlying almost complex structure of $M$, it induces an almost complex structure $I'$ on $N$. Applying Newlander-Nirenberg theorem, we see that since $I$ is integrable, $I'$ is also integrable. Thus $I'$ is a complex structure on $N$, and $f: (N,I') \to (M,I)$ is holomorphic.

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  • $\begingroup$ Thank you for the answer. Can we get a similar result when $f$ is merely a $C^r (r\ge1)$ immersion ? $\endgroup$ Jun 8, 2021 at 8:49
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    $\begingroup$ @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. $\endgroup$
    – Ben McKay
    Jun 8, 2021 at 9:49
  • $\begingroup$ @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. $\endgroup$ Jun 8, 2021 at 14:27
  • $\begingroup$ @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. $\endgroup$
    – Ben McKay
    Jun 8, 2021 at 14:51
  • $\begingroup$ (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. $\endgroup$
    – Ben McKay
    Jun 8, 2021 at 14:52

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