Timeline for Motives and homotopy theories of algebraic varieties
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2019 at 21:06 | history | bounty ended | CommunityBot | ||
| May 17, 2019 at 21:06 | vote | accept | CommunityBot | ||
| May 17, 2019 at 13:53 | comment | added | Will Sawin | @schematic_boi Unconditionally this is not known and there could be more information lurking in the category of motives. But this is not a good reason to believe that there is lots of useful information hidden in the category of motives, and to go looking for it! Disproving the conjectures might be even harder than proving them. | |
| May 17, 2019 at 13:52 | comment | added | Will Sawin | @schematic_boi You have to distinguish in this setting between what is true and what can be proved. Conditionally on conjectures the category of motives is very well-behaved and thus provides a very bad model for homotopy of algebraic varieties (which makes sense as it is intended to provide a model for abelianized homotopy theory). in particular, conditional on conjectures, the category of pure motives can be almost completely understood by studying one of the the realization functors. | |
| May 17, 2019 at 13:50 | comment | added | Will Sawin | @schematic_boi Yes, the first claim is conditional on the Kunneth type standard conjecture (to define the projectors) and applies e.g. to Chow motives modulo numerical equivalence. The second claim is conditional on the Hodge or Tate conjecture. | |
| May 17, 2019 at 13:27 | comment | added | user138661 | @WillSawin could you clarify? Do I understand correctly that given a smooth projective scheme over an algebraically closed field, we have a $\mathbb{Q}$-motive $M$ including cohomology in all degrees and motives $H^i M$ which only include cohomology in $i$-th degree, and you claim that $M\approx \oplus_{i\geq 0} H^i M$? Next question: if we have two such motives $H^i M_1$ and $H^i M_2$, between which we do not have a map, but we know that under all realization functors the respective vector spaces are isomorphic, does that imply that there exists an isomorphism $H^i M_1\rightarrow H^i M_2$? | |
| May 17, 2019 at 11:45 | comment | added | Will Sawin | @user25309 Yes, good examples. If I remember correctly an Enriques surface can be obtained as an order $2$ logarithmic transformation of a rational elliptic surface (i.e. the unramified double cover of the Enriques is a ramified double cover of the rational elliptic surface, ramified at two elliptic fibers). The graph of this relation gives the relevant correspondence to give an explicit isomorphism of their motives. | |
| May 17, 2019 at 7:30 | comment | added | user25309 | An example in the spirit of this answer: a rational elliptic surface and an Enriques surface have the same motive with rational coefficients (1+10L+L^2), but have étale fundamental groups of order 1 and 2 respectively, similarly P^2 blow up in 8 points and a classical Godeaux surface have the same motives (1+9L+L^2) but étale fundamental groups of order 1 and 5 respectively, see pdfs.semanticscholar.org/7bbd/… and references there. | |
| May 16, 2019 at 18:52 | comment | added | Will Sawin | @schematic_boi Under the Kunneth type standard conjecture and Conjecture d, a motive is just isomorphic to the sum of its cohomology groups as motives, and so two motives with isomorphic cohomology groups are isomorphic. Under the Hodge or Tate conjecture, the same is true when viewing the cohomology groups as Hodge structures / Galois representations. | |
| May 16, 2019 at 18:49 | comment | added | Will Sawin | @schematic_boi Any algebraic surface has an open set with a finite-to-one map to projective space. This gives a correspondence, giving a map of motives. For a fake projective plane, it should be easy to see that this correspondence gives an isomorphism on any reasonable cohomology theory - we just have to check that some degrees are nonzero. I don't know if it can be shown to give an isomorphism on an arbitrary Weil cohomology theory, except over finite fields. | |
| May 16, 2019 at 18:14 | comment | added | user138661 | Dr. Sawin, could you clarify "isomorphic cohomology"? Do you mean "isomorphic cohomology with respect to an arbitrary Weil cohomology theory"? If you mean that, how do I see that that is true? Then do we know that there is, in fact, a map of motives inducing these isomorphisms on cohomologies (like you know, two topological manifolds can have isomorphic homotopy groups in all degrees and still fail to be homotopy equivalent)? | |
| May 16, 2019 at 12:51 | history | answered | Will Sawin | CC BY-SA 4.0 |