# Non-compactness of Riemann Surfaces in $\mathbb{C}^{2}$.

I stumbled upon the following remark somewhere and unfortunately lacks a proof. "Every Riemann Surface in $\mathbb{C}^{2}$ in non-compact". A complex algebraic curve given as a zero locus of polynomial $P \in \mathbb{C}[x,y]$, apparently is not compact (since it's not bounded). But what about other Riemann Surfaces in $\mathbb{C}^{2}$, can you give me a reference or a proof of the aforementioned statement? Moreover, does this sentence mean that every Riemannian Surface cannot admit a holomorphic embedding into $\mathbb{C}^{2}$ even if it is compact on its own as a topological space?

P.S. If I have said something completely wrong or mistaken please do let me know. I'm not very familiar with all these things. Thank you!

• A smooth embedding would not be a problem, as $\mathbb{C}^2$ is diffeomorphic to $\mathbb{R}^4$. I think the problem here is that you want a holomorphic embedding. – Daniel Robert-Nicoud Nov 28 '16 at 12:37
• you're right! I just edited it! – user321268 Nov 28 '16 at 12:39
• There are no non-constant maps $X\to\mathbb{C}^2$ if $X$ is compact. Indeed, one would then obtain a non-constant coordinate function $X\to\mathbb{C}$ but every holomorphic function on $X$ is constant (recall that the image of a non-constant map is both open (open mapping theorem), connected, and closed (since it's compact!) which implies it's $\mathbb{C}$ which is not compact). – Alex Youcis Nov 28 '16 at 12:46
• Thank you Alexis! You're right, it was almost trivial! Perhaps you can write it as answer I think! – user321268 Nov 28 '16 at 13:00
• Possible duplicate of Embedding of Kähler manifolds into $\Bbb C^n$ – Andrew D. Hwang Dec 19 '16 at 13:05