Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi! My background is in (complex) algebraic geometry. For the first time I'm considering varieties defined over $\mathbb{R}$ and I have some very basic questions about these. In particular I'm trying to understand MMP for real surfaces.

Let $X$ be a smooth projective surface over $\mathbb{R}$. We can consider its complexification $X_{\mathbb{C}}$, smooth complex projective surface.

Then Cartier divisors on $X$ correspond to Cartier divisors on $X_{\mathbb{C}}$ that are fixed by conjugation. Is intersection theory on $X$ just inherited from intersection theory on $X_{\mathbb{C}}$ via this identification? Is this the right way to look at intersection theory on $X$?

What is the relation between the canonical divisors $K_X$ and $K_{X_{\mathbb{C}}}$? Is it true that $K_{X_{\mathbb{C}}}=\overline{K_{X_{\mathbb{C}}}}$, the divisor defined by conjugated equations and it thus corresponds to the divisor $K_X$ on $X$?

Also, how to define blow-ups of points on $X$? The definition in Hartshorne cap II.7 works but I imagine one can find an easier definition.

Whatever reference about these and related basic facts would be appraciated.

share|improve this question
1  
You might find this question useful: mathoverflow.net/questions/89395/… –  Daniel Loughran May 17 '12 at 12:49
add comment

2 Answers

up vote 5 down vote accepted

I think a lot of your questions apply to arbitrary Galois field extensions, not just the field extension $\mathbb{R} \subset \mathbb{C}$. This might help clarify the problems you are having by putting them into a general context.

Let $E \subset F$ be a finite Galois extension of fields and let $X$ be a smooth variety over $E$. Then, the canonical divisor $K_X$ is always defined over the base field $E$. Moreover, it is well-behaved with respect to smooth base change, so that yes indeed the base change of $K_X$ to $X_F$ is the canonical divisor $K_{X_F}$ of $X_F$.

With regards to divisors, we have natural homomorphisms $\mathrm{Div}(X) \to \mathrm{Div}(X_F)$ and $\mathrm{Pic}(X) \to \mathrm{Pic}(X_F)$ given by base change. Moreover, in the case of surfaces this morphism respects the intersection pairing as your desire, as the intersection number of two divisors is defined geometrically. Also in the first case, it is true that the image consists of those divisors which are invariant under the action of $\mathrm{Gal}(F/E)$.

However, in general this does not hold for Picard groups. By which I mean, there may exist divisor classes which are Galois invariant, but nonetheless there does not exists a divisor in that class defined over $E$.

As an example, consider a conic $X$ defined over $E$ which has no rational points, such that $X_F$ has rational points. Then the natural map $\mathrm{Pic}(X) \to \mathrm{Pic}(X_F)$ corresponds to the inclusion $2\mathbb{Z} \to \mathbb{Z}$, as lowest degree of any divisor is $2$ (given by the anticanonical divisor). However, the action of $\mathrm{Gal}(F/E)$ on $\mathrm{Pic}(X_F)$ is trivial as it preserves the degree of a divisor.

As for blow-ups, one can define blow-ups of closed points in a similar manner to how one defines the blow-up of a rational point. Closed points correspond to Galois invariant collections of rational points on $X_F$, therefore the map given by blowing up each of these rational points is Galois invariant and so descends to a morphism defined over $E$.

A lot of these ideas can be found in Manin's book on cubic forms.

share|improve this answer
    
Dear Daniel, thanks a lot! Your answer is very helpful. So if I understand well to define the blow-up of X at p you blow-up X_F at all the points that form the Galois invariant collection corresponding to p. Now you have that the variety on F that you obtain, say Y, is actually a base change of a variety on E. In other words Y=Z_F for some Z. So you have that Bl_p X=Z. Am I wrong? –  Gianni Bello May 17 '12 at 13:16
    
Yes what you have written is correct. Note that in order to make this rigorous you need to use "Galois descent". Essentially, this says that varieties (and morphisms) which are invariant under the action of the Galois group in question are in fact defined over the ground field. –  Daniel Loughran May 17 '12 at 13:36
add comment

Hello. For a good reference, I suggest the following text of J. Kollar http://arxiv.org/abs/alg-geom/9712003

For me, the right way to look is to see real divisors as complex divisor invariant by the anti-holomorphic involution. The intersection form is just the same as the one which comes from the complex, as you mention.

For the blow-up of smooth points, you have either a blow-up of a real point or a blow-up of two imaginary conjugate points.

Note that taking the quadric of equation $w^2=x^2+y^2+z^2$ in the projective space, this admits a real Mori fibration onto a point. The real Picard group has dimension 1, whereas the complex one has dimension 2 (over the complex, it is equal to $\mathbb{P}^1\times\mathbb{P}^1$). In particular, there are more Mori fibrations over $\mathbb{R}$ than over $\mathbb{C}$. Another example is given by del Pezzo surfaces of degree $1$ or $2$ with real Picard group of dimension $1$.

share|improve this answer
    
Thank you. It seems like, from a Mori theory point of view, a real surface its "more simple" than its complexification, as the Picard number is in general smaller. Right? Do you also have an answer about the canonical divisor? –  Gianni Bello May 17 '12 at 11:26
    
Yes, it could be said to be "simpler" because the Picard number is smaller. There are "less" curves than on its complexification. But it does not mean at all that it is easier to understand the surface than its complexification... The canonical divisor is the same over R or C, because it is invariant. –  Jérémy Blanc May 30 '12 at 23:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.