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

Here's a problem that may ultimately require just simple algebraic-geometry skills to be solved, or perhaps it's very deep and will never be solved at all. From the comments, some literature and my memory it appears this was posed by Grothendieck as part of the big program of motives.

Consider classes of complex algebraic varieties X modulo relations

    [X] - [Y] = [X\Y], 
    [X x Y] = [X] x [Y], 

Also, if you're familiar with taking inverse of an affine line, let's do that too: $$ \exists \mathbb A^{-1}\quad \text{such that}\quad [\mathbb A] \cdot [\mathbb A^{-1}] = [\mathbb A^0].$$

(+ if you want, you can also take idempotent completion and formal completion by A^-1).

It's not hard to see that you can add (formally) and multiply (geometric product as above) those things, so they form a ring. Let's denote this ring  Mot (It's actually very close to what Grothendieck called baby motives.)

And for things that form a ring you can study their Spec. For example, you can talk about points of the ring — each point is by definition a homomorphism to complex numbers.

Question: what are the properties of Spec Mot? How to describe its points?

For example, one point is Euler characteristics $\chi \in \text{Spec}\,\mathbf{Mot}$, since it's additive and multiplicative (it's even integral!) Any other homomorphism to complex numbers is thus sometimes called generalized Euler characteristics.

There's also a plane there given by mixed Hodge polynomials (that is, polynomials whose coefficients are weighted Hodge numbers $h^{p,q}_k$), since Hodge polynomial at a given point satisfies those relations too (see the references below).

As Ben says below, things would become even more interesting if we considered this ring for schemes over $\mathbb Z$, because then each $q$ would give a generalized Euler characteristic $\chi_q$ that counts points of $X(\mathbb F_q).$

Are there any other points? Any more information?

share|improve this question
    
ilya- you're not so far from 250. I agree with you that requirement is a little low at the moment, but once we get more users, it will be easy to get 100 reputation from a good answer, and 250 won't seem so bad. –  Ben Webster Oct 11 '09 at 23:23
    
The ring, before inverting the Lefschetz motive L=[A], is usually called the Grothendieck ring of varieties. It is known that it has zero-divisors (at least in char. 0). A presentation of this in terms of generators and relations has been given by F. Bittner (arxiv:math/0111062v1). I think that it is common to pass to certain completions of the ring with respect to the Lefschetz motive. –  David Rydh Oct 12 '09 at 18:34
    
By the way, I think you've made a bit too strong a claim about Hodge numbers. Think about A^1 versus A^1 minus a point unioned with a point. You're right that weighted Euler characteristic is invariant under scissors congruence, but I'm not sure if you can get much more out of Hodge theory –  Ben Webster Oct 12 '09 at 19:13
    
See the references below. –  Ilya Nikokoshev Dec 31 '09 at 18:36
add comment

4 Answers

I think I'll be collecting references I found in this answer, rather then in the original (already large) post:

It's a community wiki — feel free to add!

share|improve this answer
add comment

This ring is very important for motivic integration; so it might be useful for you to read surveys on this subject.

Yet I would say that this ring is too large and complicated. A reasonable factor-ring of it is K_0 of Chow motives. If you take Chow motives with rational coefficients then as a group it (conjecturally!) would be a free abelian group with generators being isomorphism classes of indecomposable numerical motives.

You could also be interested in weight complexes: see H. Gillet, C. Soule, Descent, motives and K-theory, J. Reine Angew. Math. 478 (1996) or my own paper http://arxiv.org/abs/math/0601713

share|improve this answer
add comment

One interesting fact about Spec M is that it isn't integral; i.e., the ring M has zero divisors. This was proved by Poonen in 2002:

"The Grothendieck ring of varieties is not a domain"

Re points of Spec M: I suppose if you considered varieties over R instead of C, you would in addition have the map sending X to the Euler characteristic of X(R), though I've never seen this used.

Update: Oh, I've never seen this used because it's totally wrong. For instance, A^0(R) and A^1(R) have Euler characteristic 1 but P^1(R) doesn't have Euler characteristic 2. I think the mod-2 Euler characteristic would probably be OK here.

share|improve this answer
    
Yes, there are many interesting directions one might explore over different fields! I think it's quite straightforward that Mot is not a domain over C, yes. –  Ilya Nikokoshev Oct 17 '09 at 9:32
3  
That's just because you're measuring real Euler characteristic wrong. A better, for certain purposes, Euler characteristic is given by Schanuel in MR0842922 and other articles. For finite cell complexes (like CW complexes, but don't require that each cell have compact closure), the formula is just the alternating sum of the number of cells at each dimension. So e.g. R has one 1-cell and nothing else: Schanuel's euler characteristic is -1 in this case. –  Theo Johnson-Freyd Oct 17 '09 at 19:46
    
Cool! So does this actually give a homomorphism from K_0(Var/R) to Z? –  JSE Oct 18 '09 at 0:01
2  
Incidentally, another article by Schanuel, namely MR1173024, is more mathematical, and certainly more interesting for this discussion. For example, he computes the Burnside rings for various geometric categories, mostly those whose objects comprise the Boolean ring generated by positive solutions to polynomials/R (resp linear functions) and whose maps are piecewise polynomial (resp affine). He also discusses the case of varieties/C, but remarks that the Burnside ring is too complicated to actually compute (he computes a useful quotient). –  Theo Johnson-Freyd Oct 18 '09 at 0:50
add comment

If you considered varieties over Z instead of over C, you would have homomorphisms given by counting points over all the different finite fields.

share|improve this answer
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.