Thursday, December 06th, 2012 | Author: Konrad Voelkel
I want to explain a particularly easy example of a motivic cellular decomposition: That of -dimensional projective space. The discussion started with cohomology (part 1) and in this part 2, we discuss intersection-theoretic and bundle-theoretic invariants. In part 3 we will see the motivic stuff.
The invariants discussed in this article are closely related and I wrote an article about the divisorial jungle before.
Divisor class group, Picard group
Since projective space is an integral scheme of finite type, smooth over the base, we have four different isomorphic characterizations of the Picard group. It is the group of invertible sheaves (lines bundles) with tensor product, up to isomorphism . It is isomorphic to the Cartier class group of Cartier divisors modulo principal divisors, which is isomorphic to the sheaf cohomology group . It is the Class group of Weil divisors modulo principal divisors.
Any hypersurface of degree in projective -space is linearly equivalent to the Weil divisor , where is a (generic) hyperplane. In particular, all hyperplanes are linearly equivalent to each other. The degree of is , so the degree homomorphism is an isomorphism. In particular, is non-trivial. One can show by first proving for the divisor of a non-zero rational function and then writing for some rational functions, so .
The line bundles on , for a field, are generated (up to isomorphism) by the tautological bundle , its dual and their tensor powers , which are all called twisting sheaves. The usual isomorphism identifies with .
The Chow ring of is the graded ring whose -th graded component is the Chow group of codimension cycles up to rational equivalence, with ring structure coming from the intersection product. Here, we consider Chow groups with integer coefficients.
Let us first look at . Can we come up with any cycles, hopefully non-trivial ones? In dimension , that is codimension , we have the points which are all rationally equivalent, since any two points in can be joined by a via the map (check ). In dimension , that is codimension , there is and that's the only closed subvariety, since is irreducible. In higher or lower codimensions, there is nothing.
The trick that shows that two points on are rationally equivalent actually works for hyperplanes in , and we have seen (in the section above) that one can show (using linear equivalence of divisors) that these are all classes of cycles. Since rational equivalence of codimension 1 cycles coincides with linear equivalence of divisors, this shows that , for the class of a hyperplane (in fact, a point).
The structure of looks less clear at first, but one can show that it is generated by any codimension linear subspace via the following "excision lemma":
Lemma: Let be a closed immersion, the open complement, then the sequence is exact for all (where the arrows are pushforward by and , so we need to take the grading by dimension).
We apply this with , , a hyperplane, and then . Then we use for to get that is a surjection. In fact, also is a surjection for , and because of we can use induction. A small remark: the excision lemma can be improved to a long exact localization/Gysin sequence for higher Chow groups, and the next term on the left would be , which gives a much smoother proof. The localization sequence, on the other hand, is very very hard to prove (at least that's what Voevodsky writes, so I believe it). To see that there are no relations in (for ), i.e. with the class of a codimension linear subspace, we look at the cases (since the case is clear and we have seen the case above, that was the divisor-setting). Any relation would be of the form for some rational functions. Let and the projection to a linear -dimensional subspace disjoint from . Then is proper, so there is proper pushforward which can not annull because of compatibility with the degree morphism (and has degree over ), but has to annul .
We see that two hyperplanes in in general position intersect to a linear subspace of codimension , and in general , so we have
From the general theory, we can use smoothness of over a perfect field to deduce that the Chern character with rational coefficients is an isomorphism. I think this also works for over , since only regularity is used. However, the rational coefficients are quite unsatisfying. At least from we can guess how might look like. Or can we? Actually it is way more complicated than the rationalized picture, and I think the general answer is still unknown!
Serre proved a theorem, that every coherent sheaf on admits a surjective map of some for positive . This shows that is generated by the set of all . Furthermore, one can show that there is a relation in between any set of coherent sheaves on , by analyzing the Koszul complex on .
For a rank vector bundle on a quasiprojective variety , one can look at the projective bundle which is fiber-wise a projective space of the fiber. On the total space one has canonically the tautological bundle and its tensor powers , which generate (for ) . One can take a vector bundle , pull it back to and twist it (i.e. tensor it with some ). This gives functors .
The projective bundle theorem for algebraic K-Theory says that these functors induce an equivalence and is a ring isomorphism.
If we apply this to a trivial rank vector bundle , where , we see that .
In particular, . But the higher algebraic K-groups of the integers are not known! One might naively guess that it becomes better by taking a field as base, but fields (even algebraically closed ones) have a very rich higher K-theory, too.
Still, the projective bundle theorem shows us that the K-Theory of projective space isn't really complicated, since all the complexity lies in the K-Theory of the base.
By the way, I realise during writing this that I would love to hear something about other isomorphism invariants I'm missing, like the Kodaira dimension, existence of a spin structure on the associated complex manifold, volume of the Fubini-Study metric, geodesic completeness, Lusternik-Schnirelman category, whatever... I fell in love with the idea of computing everything of . Please tell me your favourite isomorphism invariant of or or in the comments!