7
$\begingroup$

Further to my question,

A Naive Question on Mixed Motives and Mixed Hodge Structures

that has received very good replies and suggestions, and I really appreciate it. I am going to ask a question on generalised Hodge conjecture which is closely related to last one. From conjecture 3.22 of Marc Levine's Mixed Motives in K-theory hand book

https://www.uni-due.de/~bm0032/publ/MixMotKHB.pdf

it conjectures a functor from Nori's mixed motives to rational MHS, \begin{equation} H:\mathcal{MM}_{\text{Nori}}(k,\mathbb{Q})\rightarrow \text{MHS}_{\mathbb{Q}} \end{equation} is fully faithful, which could be seen as the generalisation of Hodge conjecture. My first question is why it is Nori's mixed motives that fits into this conjecture?

If $H$ is conjectured to be fully-faithful, then (I guess) the derived functor of $H$ (by abuse of notation) \begin{equation} DH:D^b(\mathcal{MM}_{\text{Nori}}(k,\mathbb{Q}))\rightarrow D^b(\text{MHS}_{\mathbb{Q}}) \end{equation} is also fully faithful. In D. Harrer's PhD thesis, Comparison of the Categories of Motives defined by Voevodsky and Nori

https://arxiv.org/ftp/arxiv/papers/1609/1609.05516.pdf

From main theorem, 7.4.17, there exists a realization functor \begin{equation} R_{\text{Nori}}: DM_{gm}(k,\mathbb{Q})\rightarrow D^b(\mathcal{MM}_{\text{Nori}}(k,\mathbb{Q})) \end{equation} I guess the composition of $R_{\text{Nori}}$ and $DH$ \begin{equation} DH \circ R_{\text{Nori}}:DM_{gm}(k,\mathbb{Q}) \rightarrow D^b(\text{MHS}_{\mathbb{Q}}) \end{equation} is the usual Hodge realisation functor of Voevodsky's motive. My second question is, is there a generalised Hodge conjecture stated using Voevodsky's category instead of Nori's mixed motive? e.g. like $DH \circ R_{\text{Nori}}$ is fully faithful?

Any references, comments and answers will be fully faithfully appreciated!

$\endgroup$
3
  • 1
    $\begingroup$ I think $R_{\textrm{Nori}}$ is itself conjectured to be fully faithful, which would imply that the composition is fully faithful as well. $\endgroup$
    – Will Sawin
    Jun 1, 2017 at 15:58
  • $\begingroup$ Do you know any references for this? $\endgroup$
    – Wenzhe
    Jun 1, 2017 at 15:59
  • 1
    $\begingroup$ As I have already told you, $MHS$ and its derived category is definitely not the "optimal" category here. The reason is that there exist non-trivial extensions of pure Hodge structures of the same weight, whereas pure motives should form a semi-simple category. You should look at graded polarizable Hodge categories after all. $\endgroup$ Jun 1, 2017 at 17:05

1 Answer 1

6
$\begingroup$

Here are a series of comments which might help.

  1. "Why it is Nori's mixed motives that fits into this conjecture?". The usual Hodge conjecture is known to be equivalent to the full-faithfulness of the realization from pure homological motives to Hodge structures. Suppose that one sought an extension to the mixed setting, where the target is $MHS_\mathbb{Q}$. Then one would like a source which is (expected to be) abelian together with an (exact) realization functor to $MHS$. Nori's category is abelian etc. and it may be the "right" category of mixed motives, so it seems natural to use (at least to me). A side remark: Nori's Hodge conjecture is really more of an analogue rather than a strict generalization of Hodge (it doesn't imply it with unless one also assume's Grothendieck's standard conjectures or something like it).
  2. "..then (I guess) the derived functor of H ... is also fully faithful. " I don't think this follows or is even reasonable to expect. There are no higher $Ext$'s beyond $Ext^1$ in $MHS$, whereas there should be on the other side. Similar objections would apply to conjecturing your $DH\circ R_{Nori}$ is fully-faithful.
$\endgroup$
1
  • 1
    $\begingroup$ Another reason to use Nori's motives here is that constructing realizations for them is easy (as far as I remember).:) $\endgroup$ Jun 1, 2017 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.