My question is for which varieties over local fields is "independence of l" known for etale cohomology. Say $X/{\mathbb Q}_p$ is a complete non-singular variety and $W_l$ is the (complex) Weil-Deligne representation associated to its etale cohomology group $H^i(\bar X,{\mathbb Q}_l)$. Is it known that $W_l$ is independent of $l\>\>(\ne p)$

a) when $X$ is an abelian variety (with bad reduction)?

b) when $X$ is a variety with potentially good reduction?

c) when $X$ is a variety with potentially semistable reduction?

Any pointers to references or overviews would be very much appreciated!

(Edit: To answer jmc and David Loeffler: by "independence of l" I mean that $W_l$ and $W_{l'}$ have isomorphic Frobenius-semisimplifications. I'd be happy with that!)

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    $\begingroup$ Maybe mathoverflow.net/questions/189216/… is also relevant? $\endgroup$ – jmc Dec 29 '14 at 16:37
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    $\begingroup$ Dear Tim — What exactly do you mean with $W_{\ell}$ being independent of $\ell$? The same as in my question (linked in the above comment), or a stronger/weaker notion? $\endgroup$ – jmc Dec 30 '14 at 7:22
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    $\begingroup$ Is this even known for smooth proper varieties over $\mathbf{F}_p$? In this setting Gabber has shown that the characteristic polynomial of Frobenius is independent of $\ell$, but (if I remember correctly) semisimplicity of Frobenius is only known under the Tate conjecture, so the isomorphism class of $H^i(\overline{X}, \mathbf{Q}_\ell)$ might still depend on $\ell$. $\endgroup$ – David Loeffler Dec 30 '14 at 8:59
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    $\begingroup$ @jmc: Thank you for pointing this out, yes, it is certainly relevant! But your question is much harder. I get the impression that the two hardest obstacles for general varieties are Frobenius-semisimplicity (as David Loeffler notes), and the fact that general varieties are not known to have potentially semistable reduction. And replacing this by alterations is not strong enough to prove independence of $l$ for all $H^i$ separately. But I am restricting the question to those varieties for which these problems are not there. Does this make sense? $\endgroup$ – Tim Dokchitser Dec 30 '14 at 10:55
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    $\begingroup$ @David: Thanks! I forgot that Frobenius-semisimplicity is probably the hardest bit. But I am a bit confused - for (complete regular) varieties over ${\mathbb F}_p$, doesn't the independence of $l$ follow directly from the Weil conjectures? (So it is Deligne rather than Gabber?) $\endgroup$ – Tim Dokchitser Dec 30 '14 at 10:58

So maybe everything I'm about to say you already know, so apologies if I'm teaching my grandmother to suck eggs.

This is discussed a bit at the end of a paper of Fontaine "Representations $\ell$-adiques potentiellement semistables" (section 2.4) where he describes a notion of independence that (kind of) doesn't depend on base changing to $\mathbb{C}$. It's a bit hard to find, but it's in the "Periodes $p$-adiques" Asterisque volume (for some reason, I can't get onto MathSciNet today, so I can't give you better links).

UPDATE: The Imperial WiFi is now working properly, here's the MathSciNet link


From my understanding of it, he claims $\ell$-independence (in a reasonably strong form) for abelian varieties, curves (without conditions on the reduction), and anything with good reduction. As has been mentioned, the good reduction case is Deligne. (Also, as usual, the abelian variety case implies the case $i=1$ for any $X$, so you know $i=0,1,2d-1,2d$ when $X$ has dimension $d$.)

I'm not exactly sure how the proof goes for abelian varieties in general, but with semistable reduction this basically follows from the explicit description of the weight/monodromy filtration in SGA 7, Expose IX, together with the fact mentioned by Fontaine that 'compatibility' of Weil-Deligne reps follows from equality of characters on the graded pieces of monodromy.

I'm not sure about other cases you mention, like potentially good or potentially semistable reduction. If you knew the weight monodromy conjecture, then (I think) the weight spectral sequence would give you $\ell$-independence, again using the above check for 'compatibility' and the fact that you know $\ell$-independence for the $E_1$-page. But again, I'm not sure how to pass from semistable to potentially semistable.

Finally, when $X$ has dimension $2$, since you know $i=0,1,3,4$, if you know independence results for the whole cohomology $H^*_\mathrm{et}(X,\mathbb{Q}_\ell)$ then you know it for $i=2$. There are results to this effect here:


EDIT: On reflection, the claim that weight monodromy would imply $\ell$-independence for strictly semistable schemes is not true. What is true (and this is used in Saito's paper) is that for semistable schemes the existence of the weight spectral sequence implies that the alternating sum of traces of some element of the Weil group on $H^i(\overline{X},\mathbb{Q}_\ell)$ is independence of $\ell$. He then uses this and alterations to deduce independence of characteristic polynomials in dimension $\leq 2$.

  • $\begingroup$ Thank you Chris, that is very helpful! The abelian varieties case is clear now. Actually, could you explain a bit more how the weight monodromy conjecture would give l-independence? (Last but one paragraph) $\endgroup$ – Tim Dokchitser Jan 8 '15 at 10:50
  • $\begingroup$ Ah, actually reflecting on it a bit more, I think I got a bit carried away. What I was thinking was that if the weight-monodromy conjecture were true, then to know $\ell$-independence, you need to know $\ell$-independence of traces for everything appearing in the $E_2$ page of the weight spectral sequence (since these are then the graded pieces of the monodromy filtration). Now although you know independence of everything on the $E_1$ page, this doesn't imply $\ell$-independence for everything on the $E_2$ page (this was the mistake I made). $\endgroup$ – ChrisLazda Jan 8 '15 at 14:27
  • $\begingroup$ But what you can deduce is $\ell$-independence for the alternating sums of traces. $\endgroup$ – ChrisLazda Jan 8 '15 at 14:27
  • $\begingroup$ So, if I am reading this right, even for strictly semistable schemes l-independence is not known? $\endgroup$ – Tim Dokchitser Jan 9 '15 at 11:59
  • $\begingroup$ For the individual cohomology groups $H^i$, then no. $\endgroup$ – ChrisLazda Jan 12 '15 at 11:35

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