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Not every real algebraic surface can be endowned a structure of a complex algebraic curve. The only obstruction I know is orientability.

Are there any others?

flag	asked Sep 30 2010 at 12:56 Colin Tan 4,191●13●43

Jack Huizenga · Accepted Answer · 2010-09-30 23:06:36Z UTC

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If you're not requiring any compatibility criterion between the real and complex structure, then the only obstruction is in fact orientability. Every smooth projective real algebraic surface is a smooth compact real 2-manifold (without boundary). If it's orientable, it must then be a surface of genus $g$ for some $g$. But every surface of genus $g$ admits a complex structure, and every Riemann surface is algebraic.

I don't study real algebraic geometry much, but I'm not aware of a good compatibility condition to impose on your complex structure. If you've got something in mind, let me know.

edited Sep 30 2010 at 23:06

answered Sep 30 2010 at 14:16

Jack Huizenga
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One could ask the surface to be the Weil restriction of the curve -- in this case another obstruction would be that the surface must split geometrically as a product of two same-genus curves. – Dustin Clausen Sep 30 2010 at 14:41

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@Dustin Clausen: Not just same genus, they actually have to be conjugate to each other, right? For instance, a surface that splits as a product of two non-isomorphic real curves of the same genus won't do. – t3suji Sep 30 2010 at 16:05

t3suji - I agree, thanks! And I suppose that now we have not just an obstruction, but actually an equivalent condition. – Dustin Clausen Sep 30 2010 at 16:13

@ Jack Complex does not automatically mean algebraic. It is clearly true for compact surfaces but I doubt this for non-compact surfaces. Can you provide a reference? – Bugs Bunny Sep 30 2010 at 19:08

I meant to include the word "projective," which would have addressed this. Answer edited. – Jack Huizenga Sep 30 2010 at 23:08

Francesco Polizzi · Answer 2 · 2010-09-30 21:03:44Z UTC

As observed in some of the previous comments, every closed (=compact, without boundary) orientable real 2-manifold admits a complex structure. So in the smooth case orientability is essentially the only obstruction.

If one also considers the case of singular real algebraic surfaces, the situation is more involved and I don't know whether satisfactory results are known.

Anyway, one obvious obstruction is the presence of non-isolated singularities, since every complex curve has only a finite number of singular points.

For instance, take $X:=S^1 \times C$, where $C \subset \mathbb{RP}^2$ is the nodal real cubic of equation $y^2z=x^3+x^2z$. The singular locus of $X$ is isomorphic to $S^1$, so $X$ surely cannot be endowed with the structure of a complex algebraic curve.

Bugs Bunny · Answer 3 · 2010-09-30 14:38:26Z UTC

If it is orientable, you have a complex structure and the field of meromorphic functions.

Putting my ears into the firing line, I suggest that something should go wrong with the transcendence degree of the field of meromorphic functions. If it is 1, you can consider DVR-s that will give you a compact algebraic curve, and I see no reason for the original curve not to be a subset. If it is more than 1 the surface cannot be algebraic...

Obstruction for a real algebraic surface to be a complex algebraic curve

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