The proof of that the Riemann sphere is not a complex submanifold of $\mathbb{C}^n$

140 Views Asked by At

As is well known, according to the maximal principle we can easily conclude that every compact connected complex manifold of $\mathbb{C}^n$ degenerate to a point. From the above point of view, we can easily conclude that the Riemann sphere $\mathbb{C}P^1$ is not a complex submanifold of $\mathbb{C}^n$. Actually the submanifold of a stein manifold is again a stein manifold and thereby not compact. However intuitively the Riemann sphere lies in $\mathbb{C}^2$, since $\mathbb{C}P^1=S^2$ lies in $\mathbb{R}^3\subset \mathbb{C}^2$. So the sole cause of the phenomenon must be that the inclusion $ \mathbb{C}P^1\hookrightarrow \mathbb{C}^2$ is not an holomorphically embedding map.

My question: How prove directly that the inclusion $ \mathbb{C}P^1\hookrightarrow \mathbb{C}^2$ is not a holomorphically embedding map.?

Any idea is appreciated. Thanks a lot.

1

There are 1 best solutions below

0
On BEST ANSWER

The map $\Bbb{P}^1\to S^2$ is not holomorphic.

If $f: \Bbb{P}^1\to \Bbb{C}^n$ is holomorphic then $f(\Bbb{P}^1)$ is compact so $f([z:1])$ is a bounded entire function thus it is constant.