Suppose that we have a mixed motif $M$ with only two (subsequent) weights, and have a subobject for each weight factor. How we can control whether there exists a submotif of $M$ with these two factors? It seems that here the existence of a mixed Hodge substructure with the corresponding weight factors in the singular cohomology of $M$ is not sufficient, since the corresponding mixed motivic $Ext^1$ does not inject into its Hodge 'realization'. Is this correct? What happens here if the base field is a number field? Are there any examples here that are well understood (certainly, the matter is very conjectural)?
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asked Nov 17, 2010 at 21:57
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$\begingroup$ Dear Mikhail, Could you explain in more detail what you mean by having "a subobject for each weight factor"? Based on the question in your second sentence, I guess you don't mean submotives. By the way, are you sure the $Ext^1$s don't inject? (The Hodge conjecture says something like the realization map induces an isomorphism on Hom's, and then perhaps general nonsense implies that you have an injection on $Ext^1$s? At least in the case of $Ext^1$ of the trivial motive by the $H^1$ of an abelian variety, the realization map sends the Mordell--Weil group into the $\mathbb C$ points of ... $\endgroup$– EmertonCommented Nov 17, 2010 at 22:55
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$\begingroup$ ... the abelian variety, and this is an injection. Might the more general case be similar?) $\endgroup$– EmertonCommented Nov 17, 2010 at 22:55
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$\begingroup$ Sorry, it may be that what I suggested above about being an injection in general is nonsense; if so, ignore it. Hopefully I can get my thoughts straight and try to say something with more confidence (and more chance of being correct). $\endgroup$– EmertonCommented Nov 17, 2010 at 23:02
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$\begingroup$ If you assume that the Hodge realization is fully faithful and its image is stable by subojects, then the injectivity of $Ext^1_{MM(F)}(\mathbb{Q},M) \to Ext^1_{MHS}(\mathbb{Q}^H,M^H)$ for $M$ simple follows by the tannakian formalism. Indeed the assumption is equivalent to saying that you have a surjective map between the tannakian fundamental groups $MHS$ and $MM(F)$. $\endgroup$– AFKCommented Nov 17, 2010 at 23:38
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$\begingroup$ So you also have a surjection $Gr^W H_1(\mathfrak{u}_{MHS}) \to Gr^W H_1(\mathfrak{u}_{MM(F)})$ between abelianizations of their pro-unipotent radicals. And this induces the dual to the map between $Ext^1$'s as $\mathfrak{u}_{MM(F)} = \bigoplus_M Ext^1(\mathbb{Q},M)^\vee \otimes M$ where $M$ ranges over a class of representative of simple objects. $\endgroup$– AFKCommented Nov 17, 2010 at 23:38
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