Algebraic surfaces and their (intrinsic) geometry

Question

Recently I began to consider algebraic surfaces, that is, the zero set of a polynomial in 3 (or more variables). My algebraic geometry background is poor, and I'm more used to differential and Riemannian geometry. Therefore, I'm looking for the relations between the two areas. I should also mention, that I'm interested in the realm of real surfaces, i.e. subsets of $\mathbb{R}^n$.

On my desk you could find the following books: Algebraic Geometry by Hartshorne, Ideals, Varieties, and Algorithms by Cox & Little & O'Shea, Algorithms in Real Algebraic Geometry by Basu & Pollack & Roy and A SINGULAR Introduction to Commutative Algebra by Greuel & Pfister. Unfortunately, neither of them introduced notions and ideas I'm looking for.

If I get it right, please correct me if I'm wrong, locally, around non-singular points, an algebraic surface behaves very nicely, for example, it is smooth. Here's the first question: is it locally (about non-singular point) a smooth manifold? Is it a Riemannian manifold, having, for instance, the metric induced from the Euclidean space?

Further questions I have are, for example:

1. Can I define geodesics (either in the sense of length minimizer or straight curves) in the non-singular areas of the surface? Can they pass singularities?
2. How about curvature? Is it defined for these objects?
3. Can we talk about convexity of subsets of the algebraic surface?
4. What other tools and term can be imported from differential/Riemannian geometry?

I will be grateful for any hint, tip and lead in the form of either answers to my questions, or references to books/papers which can be helpful, or any other sort of help.

Answer 1

It seems to me that your interest is not in algebraic geometry, but in the differential geometry of spaces defined by algebraic equations. An algebraic variety defined over $\mathbb R$ or $\mathbb C$ is a manifold away from the singularities. The singular set is is a proper closed subset, where closed means defined by some algebraic equations, so this set is actually lower dimensional than the original object, so you have a very nice big open subset where you have a manifold. The question of extending differential geometric constructions to singularities is in general a difficult one and is the focus of a lot of research. You might be able to get a better idea by looking at complex analytic geometry. In any case, you can obviously do any of those you ask on the smooth part, but it will not be algebraic geometry. But that's OK.

One possibility you could try is indeed looking at the resolution of singularities, do your Riemannian magic there and try to get bring the results back to the original space. I suspect that you don't know what a resolution of singularities is since it is actually an very specifically algebraic geometric notion. It is the following: Let $X$ be your starting object. A resolution of singularities is a morphism $\pi:\widetilde X\to X$ such that it is an isomorphism outside a smaller dimensional subspace of $X$. You can read more about these in Lectures on resolution of singularities by János Kollár. The difficulty will be in taking whatever you do on the resolution back to the original, but the good news is that it is differential geometry, so my suggestion would be the following: For now assume that such a $\pi$ exists and see if what you want to do you can on $\widetilde X$. If so, try to see if you can "push-forward" some of those results to $X$. Perhaps you will realize that "if only $\pi$ satisfied property $P$, then I could do this" and it is possible that $\pi$ does. SO, if you get to that point, then look at Kollár's book or come back to MO and ask more specific questions.

Answer 2

While a real/complex variety, being a subset of a real/complex vector space, will have a Riemannian/Kahler metric, this metric will not be defined up to isomorphism of the algebraic variety. For instance, if the equation defining it is linear, then the variety is isomorphic to a scaled-up version of itself.

So the induced metric exists, but depends on the embedding (which I guess is usually the case)

Since it is a Riemannian manifold, you can define curvature et. al. using the standard definitions. Some of these properties will be algebraic functioms, defined by polynomial functions or sections of algebraic vector bundles, and some will not. The main problem will be complex conjugation,

Geodesics, having real dimension 1, will not have a complex-algebraic description, and will probably fail to be real-algebraic curves. (Though they might be in some cases - e.g. the sphere.)

Answer 3

Any non-singular complex variety $V$ of dimension $n$ (in either affine space or projective space) can be endowed with the structure of a complex manifold of dimension $n$. Moreover as a submanifold of a Kahler manifold, it will also be Kahler.

The passage to a real manifold can be slightly subtle, as there are two approaches you can take. Firstly, you can simply consider the complex manifold as a real manifold of dimension $2n$, with a complex structure.

Alternatively, if the variety is defined over $\mathbb{R}$ (that is, it has equations with real coefficients), then you can look at the set of real points $V(\mathbb{R})$, which you can also endow with the struture of a real manifold. As a submanifold of a Riemannian manifold it is also a Riemannian manifold.

I should note that some funny things can happen though, for example even if $V$ is connected then $V(\mathbb{R})$ may not be.

If your variety has singularities, these always occur on a closed subset so you can just remove them to get a non-singular variety. Alternatively you can resolve them as Ariyan suggests.

Answer 4

Let $X$ be a connected normal projective C-scheme of dimension 2, i.e., an algebraic surface. The topology on an algebraic surface is the Zariski topology. But you can associate to $X$ its analytification. (See Hartshorne's appendix B or the wonderful SGA1 Exposé XII available on Arxiv.) Let $X^a$ be the analytification of $X$. Hartshorne explains that $X^a$ is a complex analytic variety (by definition) of dimension 2. So in general it won't be a complex analytic surface. More precisely, a complex manifold to me is a nonsingular complex analytic variety and if you look in SGA1 exposé XII you will see that $X$ having property P is equivalent to $X^a$ having property P. Here you can take P to be the property of nonsingular. (In fact, in SGA1 Exposé XII P is a property of a morphism such as flat smooth etale , etc.)

In any case, being nonsingular is a local property, i.e., it is a condition imposed on the local rings. (A variety is nonsingular if and only if its local rings are (local noetherian) regular rings.) Therefore, if you take a nonsingular point in $X$ you can find a neighborhood $U$ of $x$ which is nonsingular. (So this means that the local rings of $U$ are all regular.) The analytification of $U$ is a manifold in the above sense.

So the point I'm trying to make is that if you start with a nonsingular connected projective $\mathbf{C}$-scheme $X$ and take its analytification $X^a$ you can use all your knowledge of manifolds to work with $X^a$. Vaguely speaking, sometimes you can use results in the complex analytic world to deduce results for $X$ itself. (Think about comparison theorems for fundamental groups and cohomology or the theory of characteristic classes.)

I hope this helps a bit!

Ow btw, the notion of a complex analytic variety is explained more detailed in some article by Grauert I think. It is also explained in Hartshorne's appendix very briefly.

EDIT: I took a normal scheme above. There is a good reason for this. Namely, the singularities of $X$ will be closed points. That is, the singular locus of $X$ is closed and of codimension 2. Moreover, there is a nice theory of resolving these singularities. See for example the book Complex complex surfaces by Barth-Hulek-Peters-Ven. If you take a resolution of singularities $Y\longrightarrow X$ and endow $Y^a$ with the analytic topology you get a complex manifold. Sometimes it's nice to be able to work with $Y^a$.