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In very short:

When proving the equivalence of intersection theory constructed through (Milnor) K-sheaves and their product vs. defining the product via Serre's local multiplicity formula + moving, I don't see any way of doing this proof not needing an ENORMOUS amount of general intersection theory stuff.

If I only want to intersect codim 1 cycles, and intersection theory is really much easier in this case, does anybody have an idea of how to bypass as much pages of Fulton, Gersten, Grayson as possible?

(as a specific example, does anybody know how to bypass using somewhere in the middle the deformation-to-normal-cone construction of the Chow product? Or have at least a suggestion for me how I could try to go directly to Serre's?)

In a little bit longer (if you are interested):

One can express the intersection multiplicity of two (properly intersecting) algebraic cycles through Serre's formula with alternating Tor's (see for example http://en.wikipedia.org/wiki/Serre%27s_multiplicity_conjectures). That's great and nice for computation.

On the other hand, Bloch-Quillen tells me (in any nice situation) CH^l = H^l (X, K_l), where K_l is the l-th K-theory sheaf (Zariski site). For l=1 this comes down to CH^1 = H^1(X, O_X*), the classical relation between Cartier and Weil divisor class groups.

Using Bloch-Quillen (in a version for Milnor K, that's known) and an old result of Grayson (I believe) which says that the sheaf cohomology product construction and the Chow ring product up to sign give the same, I get the same by considering either

H^1 (O*) $\otimes$ ... $\otimes$ H^1(O*) ---> H^n($K^M_n$)

(using sheaf cohom. product + naive 'symbol concatenation' product on Milnor K ring)

as well as

CH^1 (X) $\otimes$ ... $\otimes$ CH^1 (X) ---> Z.

Fine.

However, to construct the former map, no big theory or anything is needed: Product in sheaf. cohom. is easy; all else needed would be to make explicit the map

H^n($K^M_n$) ---> Z.

This is a bit cumbersome, but writing out some explicit sheaf cohomology resolution (say Godement or Cech or whatever you like) and using the flasque (Milnor K version) of the Gersten resolution for $K^M_n$, one ends up having to follow a zig-zag in a bicomplex.

Along this zig-zag one accumulates plenty of Milnor K residue maps $\partial_v$ (which constitute the Gersten differential and are induced into the bicomplex), and maybe it is not even so important to unwind all these things in full detail, but at the end one gets a terrible expression (allow me to keep this name for below) involving piles over piles of Milnor K residue maps and a vast sum over all scheme points (but only finitely many non-zero summands, so it's okay).

Anyway, still I may just keep this more-or-less explicit formula. This way, I could theoretically construct the map

H^1 (O*) $\otimes$ ... $\otimes$ H^1(O*) ---> Z

without using any theory basically, just use the explicit formula of the Godement resolution to get the product of 1-cycles and apply the [terrible formula] coming from the unwinding of the Bloch-Quillen iso [to be totally precise, to get to Z, we first unwind Bloch-Quillen H^n(K^M_n) to CH^n(X) and then make explicit the pushforward to the base field scheme, this gets us to CH_0 of Spec k and that's canonically iso to Z, this is the map we unwind here].

(say we ignore here the question to show that the resulting thing is indeed a well-defined map on cohomology classes)

Now my problem is the following: In principle, expressing the intersection product

CH^1 (X) $\otimes$ ... $\otimes$ CH^1 (X) ---> Z

(and the above discussion implies that this [terrible-formula]-given version is the same) is really not so hard. One the one hand, intersection just with divisors is much easier than the general case, and also one can construct it quite quickly using Euler characteristics for example (see for example the first pages of Debarre's book on higher-dim. alg. geometry).

Hence, my stupid self would have expected (and this triggered this whole activity somehow) that when working out the explicit [terrible formula] as mentioned above from the Bloch-Quillen perspective, it would maybe be actually not so hard to identify it as such an Euler characteristic / Serre's multiplicity formula. Why not?

However, all the contrary is the case. The formula is super-terrible and involves plenty of Milnor K-residue maps (so in the surface case that would be tame symbols essentially), and it is not at all clear how that could ever be directly linked with Euler chars., Hilbert polynomials, Serre's formula or whatever other approach to intersections one may have in mind (or at least I have in mind).

So I dare to ask, is truth really as depressing as having to admit that even in this simple intersect-only-divisors situation, there is no simpler way of linking the [terrible-formula] with (say) Serre's formula than going through the whole story of Bloch-Quillen and a reasonable amount of Fulton's book?

Please excuse my stupidity and give me some idea/approach how to shortcut Bloch-Quillen/Fulton. Is there maybe some approach using directly derived tensor products of O* sheaves, something like that seems to happen in Deligne's "symbole modéré" paper (and the tame symbol appears in the 2-dim case of [terrible formula]), it comes from Milnor K residue, but it seemed (but maybe I was blind) not to indicate how to connect this to geometry).

Ahhhh, it's a hard life.

share|improve this question
    
I'll just note: writing in full sentences in the title is strongly encouraged, even this makes it seem horrifyingly long. –  Ben Webster Feb 8 '10 at 4:33
    
I should say, though, otherwise its a very nice question. –  Ben Webster Feb 8 '10 at 4:34
    
I like this (somewhat old) question and will try to think about it. I just wanted to mention that there is a good treatment of intersection theory "the Milnor K-theory way" in the book "The Algebraic and geometric theory of quadratic forms" by Karpenko and Merkurjev. Unfortunately, they do not compare with Serre's formula. –  Simon Pepin Lehalleur Aug 9 '10 at 15:26
    
The book is by Elman-Karpenko-Merkurjev, sorry. –  Simon Pepin Lehalleur Aug 9 '10 at 15:26

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