# Smooth submanifold of a complex manifold with invariant tangent space under multiplication by $i$

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?

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.
• 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 8:49
• @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 9:49
• @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 14:27
• @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 14:51
• (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 14:52