Welcome to MathOverflow

Question

3

1

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 4,211●13●43

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

1

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
3,082●1●5●18

3

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

2

@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

Bugs Bunny · Answer 2 · 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...

Francesco Polizzi · Answer 3 · 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.

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

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.