Obstruction for a real algebraic surface to be a complex algebraic curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:49:39Z http://mathoverflow.net/feeds/question/40615 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40615/obstruction-for-a-real-algebraic-surface-to-be-a-complex-algebraic-curve Obstruction for a real algebraic surface to be a complex algebraic curve Colin Tan 2010-09-30T12:56:06Z 2010-09-30T23:06:36Z <p>Not every real algebraic surface can be endowned a structure of a complex algebraic curve. The only obstruction I know is orientability. </p> <p>Are there any others?</p> http://mathoverflow.net/questions/40615/obstruction-for-a-real-algebraic-surface-to-be-a-complex-algebraic-curve/40619#40619 Answer by Jack Huizenga for Obstruction for a real algebraic surface to be a complex algebraic curve Jack Huizenga 2010-09-30T14:16:24Z 2010-09-30T23:06:36Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/40615/obstruction-for-a-real-algebraic-surface-to-be-a-complex-algebraic-curve/40622#40622 Answer by Bugs Bunny for Obstruction for a real algebraic surface to be a complex algebraic curve Bugs Bunny 2010-09-30T14:38:26Z 2010-09-30T14:38:26Z <p>If it is orientable, you have a complex structure and the field of meromorphic functions.</p> <p>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...</p> http://mathoverflow.net/questions/40615/obstruction-for-a-real-algebraic-surface-to-be-a-complex-algebraic-curve/40635#40635 Answer by Francesco Polizzi for Obstruction for a real algebraic surface to be a complex algebraic curve Francesco Polizzi 2010-09-30T15:50:43Z 2010-09-30T21:03:44Z <p>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.</p> <p>If one also considers the case of <em>singular</em> real algebraic surfaces, the situation is more involved and I don't know whether satisfactory results are known.</p> <p>Anyway, one obvious obstruction is the presence of <em>non-isolated</em> singularities, since every complex curve has only a finite number of singular points.</p> <p>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.</p>