Timeline for Should the etale cohomology of a smooth projective variety (over rationals) be semi-simple; why?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Feb 27, 2013 at 19:25 | vote | accept | Mikhail Bondarko | ||
| Feb 27, 2013 at 19:25 | vote | accept | Mikhail Bondarko | ||
| Feb 27, 2013 at 19:25 | |||||
| Feb 27, 2013 at 15:16 | history | edited | Daniel Miller | CC BY-SA 3.0 |
improved / fixed some of the LaTeX
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| Feb 2, 2013 at 19:40 | answer | added | anon | timeline score: 8 | |
| Feb 2, 2013 at 13:07 | comment | added | naf | Minor correction: In my previous comment $p$ should be $l$ (an arbitrary prime) and the extension is non-split -- the image of the Galois group is non-abelian -- even with $\mathbb{Z}$ replaced by $\mathbb{Q}$. | |
| Feb 2, 2013 at 12:35 | comment | added | naf | Mikhail: The Galois representation for elliptic curves with multiplicative reduction is easy to understand using the theory of Tate curves. One gets that the Tate module (which is equal to $H^1$ upto twist) in the the case of split multiplicative reduction is a (non-trivial) extension of $\mathbb{Z}_p$ by $\mathbb{Z}_p(1)$. (See, for example, Silverman's book "Advanced Topics in the Arithemetic of Elliptic Curves" p. 452) | |
| Feb 2, 2013 at 10:15 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
Upd. added.; added 26 characters in body
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| Feb 2, 2013 at 9:43 | comment | added | user30035 | oops I meant to continue Funny question, because input is local and output is global. If memory serves, Emerton proved that if you assume something slightly stronger, namely that $a_p$ is a $P$-adic unit for all $P$ above $p$, and that all the local reps are decomposable, then $f$ has CM, assuming the standard conjectures. One last thing: if $E$ is an elliptic curve over $Q_p$ with good ordinary reduction then the $p$-adic Tate module is in practice always indecomposable when $E$ has no CM, so there's another counterexample to the orig q, but I'm a bit unsure as to whether this is proved. | |
| Feb 2, 2013 at 9:40 | comment | added | user30035 | PS Mikhail: here's a problem that is open (although some variant of it does follow from the standard conjectures). Take a modular form of weight $k\geq2$ and coefficient field $K$. Take a prime $p$ not dividing the level of $f$, and a prime $P$ of $K$ above $p$. Assume $a_p$, the $p$th Fourier coefficient of $f$, is a $P$-adic unit. Consider the Galois representation to $GL(2,K_P)$ attached to $f$ and restrict it to the abs Gal gp of $Q_p$. Is it true (there's computational evidence) that this rep is indecomposable iff $f$ has no CM? Funny | |
| Feb 2, 2013 at 9:37 | comment | added | user30035 | PS ulrich's comment should be an answer, and if this doesn't solve the real question for you then perhaps you might want to edit to explain what the question now becomes. | |
| Feb 2, 2013 at 9:32 | comment | added | user30035 | Mikhail: it's almost always indecomposable in practice (you can usually check this by looking at the mod $p$ or mod $p^2$ torsion; already this usually is indecomposable). But I think that the general question about whether the extension is split is subtle. The extension class might be related to some sort of subtle invariant of the curve (e.g. maybe the log of some $L$-invariant?) so this might actually be some subtle question in transcendence theory! | |
| Feb 2, 2013 at 8:00 | comment | added | Mikhail Bondarko | Dear ulrich: is the corresponding Galois representation necessarily indecomposable? | |
| Feb 2, 2013 at 6:41 | comment | added | naf | Semi-simplicity certainly does not hold for arbitrary $K$: for example, it does not hold for elliptic curves with multiplicative reduction over a $p$-adic field $K$ | |
| Feb 2, 2013 at 5:47 | comment | added | anon | According to the Tate conjecture, l-adic realization gives an equivalence of categories from motives tensor $Q_l$ to the category of l-adic Galois representations generated by the cohomology of smooth projective algebraic varieties over $Q$. The standard conjectures imply the first is semisimple, hence also the second. | |
| Feb 2, 2013 at 3:50 | comment | added | Donu Arapura | Dear Mikhail and Joel, I realize I was a bit hasty in my comment, but there is no way to edit it. | |
| Feb 2, 2013 at 3:38 | comment | added | Mikhail Bondarko | Dear Donu: the Tate conjecture certainly yields that a direct summand of a motivic representation is motivic. Yet why can you prove anything about arbitrary subrepresentations? | |
| Feb 2, 2013 at 0:39 | comment | added | Joël | @Donu: This is not exactly true. If you consider the motive $M$ of rank 2 attached to an elliptic curve $E$ over a quadratic imaginary field $K$ with complex multiplication by $K$ defined over $K$, then $M_{\mathbb Q_l}$ will have sub-representations of dimension that are not motivic for half of the $l$'s, namely those which are split in $K$. Now your argument will work I think after extending the field of coefficients. Since semi-simplicity is invariant by extension of fields of the coefficients (which is not the case of course of irreducibility), your and Will's argument is saved. | |
| Feb 1, 2013 at 23:01 | comment | added | Donu Arapura | Mikhail, if you assume Tate, as Will suggested, then the subrepresentations would be motivic. Also, regarding 2, there is a conjecture that the image of $Gal(\matbb{Q})$ in $GL(H^i(X_{et}, \mathbb{Q}_\ell)$ is the $\mathbb{Q}_\ell$ points of the Mumford-Tate group of the Hodge structure on $H^i(X)$. If this were true then semisimplicity would follow from Hodge theory, but establishing this would perhaps be harder than establishing semisimplicity directly. | |
| Feb 1, 2013 at 22:35 | comment | added | Mikhail Bondarko | Why does the semi-simplicity of motives imply the semi-simplicity of the corresponding representations? Representations could have subobjects that are not 'motivic'. | |
| Feb 1, 2013 at 22:12 | comment | added | Will Sawin | I think the standard conjectures + the Tate conjecture imply this, because they imply that motives are semisimple, and galois representations from abelian varieties are motives. | |
| Feb 1, 2013 at 21:53 | history | asked | Mikhail Bondarko | CC BY-SA 3.0 |