# The terribleness of higher dimensional algebraic geometry

So, I was talking to Mikhail Khovanov at tea today (there are advantages to being at IAS), when the following question “How is it that one could take a ‘nice’ variety (where for now, we’ll let ‘nice’ mean ‘rational’), take a quotient by a free finite group action, and end up with a variety which is ‘bad’ (i.e. not rational)? Or is such even possible?”

I insist it is, having come across examples in my research, but Mikhail seemed to be unconvinced, and I couldn’t summon the energy to come up with a good example (also, let’s honest, sometimes when one thinks something is true, but has a seed of doubt that one might somehow be wrong, it’s hard to argue forcefully. It’s so much less embarrassing when you turn out to wrong if you were tentative).

Thinking about this while making dinner, it occurred to me that this is just another instance of mathematicians refusing to believe that higher dimensional algebraic geometry is really is bad as it is. It’s like the femme fatale (or handsome bad boy) who breaks our heart over and over, but who we can’t bring ourselves to believe the fundamental badness of.

So let’s be honest with ourselves, higher dimensional algebraic geometry is a terrifying place, nothing like the world of curves (which, let’s be honest, is what most of us having mind when thinking about algebraic geometry). It even has a Murphy’s law!

Anyone out there have good tidbits illuminating the horribleness of algebraic varieties that are explainable (if not in full detail) over tea?

## 6 thoughts on “The terribleness of higher dimensional algebraic geometry”

1. A similar idea to the Murphy’s Law thing is the non-smoothness of the moduli stack of of rank n vector bundles on a higher dimensional variety. Toen has some talk notes that explain this:
http://www.picard.ups-tlse.fr/~toen/essen.pdf
However, he uses it as justification for a richer notion of moduli space which I think might also resolve Murphy’s law. Maybe higher dimensional varieties are just misunderstood…?

2. In characterist 0 one has the Chevalley Shepard-Baron Todd theorem – let $G$ be a finite group acting on the polynomial ring $k[x_1,\ldots,x_n]$. Then the ring of invariants is a polynomial algebra iff for each stablizer $G_Q$ is generated by pseudo-reflections relative to $Q$ (relative means that the fixed point locus contains $Q$). $\tau$ is a pseudo reflection if its fixed point locus is codimension one.

So for example the quotient of $\A^n$ by a free action is rational. The full theorem doesn’t hold in char p, but for a free action it does.

3. On correction: by generated’ I mean largest *normal* subgroup containing all pseudo-reflections relative to $Q$’ (so that trivial stablizers are generated by pseudo-reflections).

A neat example of this is $D_4$ acting on $A^2$.

4. David Speyer says:

Ben — regarding your original question, the standard example of a variety which is a finite quotient of a rational variety, but not rational itself, is a degree three hypersurface in P^4. I don’t think this is a quotient by a free action, but you could always remove points to make it free. I suspect you could also make it free by blowing up, but I haven’t checked this.

Unfortunately, the construction of a finite covering of the threefold by a rational surface lacks in explicitness. Here is how it goes: Let X be a degree 3 hypersurface in P^4 over the complex numbers. First, we give a rational map from P^2 \times P^2 to X. Then we argue that the restriction of this map to a generic degree (1,1) hypersurface is finite.

I’ll just write out the first step, I am copying Kollar math.AG/0005146 , who says that he is copying Segre. Let p be a point of X, let H_p be the hyperplane tangent to X at p and let C_p=X intersect H_p. So C_p is a cubic in P^3 with a singularity at p. A generic line through p and in H will intersect C_p with multiplicity 2 at p, so it must meet C_p in one other point. This gives us a rational map P^2–> C_p where P^2 is the lines through p. Let q be another point of X, and similarly define C_q. Again, we have a rational map P^2 –> C_q. Now, if u and v are generic points of C_p and C_q respectively, then the line through u and v must meet X at one more point. This gives a rational map C_p x C_q –> X and hence P^2 x P^2 –> X.

For a proof that X is not rational, see
MR0302652 (46 #1796)
Clemens, C. Herbert; Griffiths, Phillip A.
The intermediate Jacobian of the cubic threefold.
Ann. of Math. (2) 95 (1972), 281–356.

5. Lots of examples come from the “Noether problem.” Given a finite group G with a representation, we have an action of G on projective space, which can be made free by stripping out some discriminant locus. If the quotient X were a rational variety over Q, then there would be tons of points in X(Q), whose preimages in P^N would give lots of extensions of Q with Galois group G. So if X were always rational, the inverse Galois problem wouldn’t be as hard as it is! I’m writing this w/o MathSciNet access, but there are papers of Lenstra and Saltman that speak to specific examples of non-rationality. I think Swan and Voskrenskii gave the first such examples. (I’m not sure which of these papers are concerned with rationality over Q and which are over an algebraically closed field.)

6. Jason Starr says:

I don’t see how irrationality of a variety is at all related to “terribleness”. If anything, I would think your interaction with Prof. Khovanov demonstrates that unirationality is more relevant for many purposes than is rationality (except, as Jordan points out, for arithmetic questions like existence/abundance of rational points). If I recall correctly, the book of Corti-Reid-Pukhlikov at least implicitly suggests the emphasis on rationality is historical rather than because of strong intrinsic importance.