Real submanifold of a complex manifold is a complex submanifold?

Question

The statement in the title is obviously wrong, with an easy counter example being any real submanifold of odd dimension.

My question is more specific: say I have a complex manifold $X$ with a real submanifold $M\cong S^2$, where $S^2$ is the $2$-dimensional sphere with its usual differentiable structure. Does it follow that $M$ is a complex submanifold of $X$?

I feel that this is true, since the real structure of $S^2$ coincides with its usual complex structure. But I'm not exactly sure since I wasn't able to produce a clean argument for this.

Answer 1

Definitely false. Think of $\Bbb C^2 = \{(z_1,z_2)\}$, and write $z_j=x_j+\sqrt{-1}y_j$. Consider the $(x_1,x_2)$-plane. This is not a complex subspace of $\Bbb C^2$, so certainly not a complex submanifold.

You can compactify this to $S^2\subset\Bbb CP^2$, if you so desire.

Answer 2

To add to Ted's excellent answer: No connected, compact holomorphic manifold of positive dimension embeds holomorphically in $\mathbf{C}^{n}$: A holomorphic function on a compact manifold is constant by the maximum principle, and each component function of a holomorphic embedding is holomorphic by definition.

So...if you think of all the ways a $2$-sphere can be smoothly (or even real-analytically) embedded in a complex Euclidean space (e.g., in a local coordinate chart for $X$), you get a better sense of how few two-spheres are liable to be holomorphically embedded in $X$.

In $\mathbf{CP}^{2}$, for example, a holomorphically-embedded $2$-sphere turns out to be either a projective line or a conic; that's a finite-dimensional family in an infinite-dimensional sea of real-analytic $2$-spheres.