Timeline for In which ways can the isogeny theorem fail for local fields?
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| Nov 23, 2011 at 23:22 | comment | added | fherzig | Dear Matt, Thank you very much for your comments! That's very nice. By the way, I think you are also assuming semisimplicity of Frobenius, which is of course expected. Regarding homological equivalence, aren't you implicitly assuming that the dimension of the span of the image of $CH^i$ in cohomology is the same as the rank of $CH^i$ modulo numerical equivalence? In other words, if cycle classes are l- or p-adically linearly dependent, are they rationally linearly dependent? Is this expected? Best, Florian | |
| Nov 18, 2011 at 3:49 | comment | added | Emerton | ... $1$ as a zero of the char. poly of Frob. is the same on the crystalline and $\ell$-adic cohomology. Thus the Tate conjecture holds for crystalline cohomology if and only if it holds for $\ell$-adic cohomology (assuming that both crystalline and $\ell$-adic homological equivalence coincide with numerical equivalence.) Regards, Matt | |
| Nov 18, 2011 at 3:47 | comment | added | Emerton | Dear Florian, Yes, the crystalline analogue of the Tate conjecture for smooth proper varieties over finite fields should hold. This might even follow from Katz--Messing (at least morally): there are cycle class maps into both crystalline cohomology and etale cohomology. If we assume that homological equivalence coincides for $\ell$-adic and crystalline cohomology (which would follow if we could prove that both are equal to numerical equivalence, say), then the rank of the image is the same, and is contained in the Frobenius fixed points. But Katz--Messing shows that the multiplicity of ... | |
| Feb 18, 2011 at 20:13 | comment | added | fherzig | ...it follows that there are non-zero Galois-invariant maps $T_p(E') \to T_p(E)$ (map to the unramified line in the split representation). But the elliptic curves cannot be isogenous, since $T_p(E)[1/p]$ is split and $T_p(E')[1/p]$ isn't. | |
| Feb 18, 2011 at 20:13 | comment | added | fherzig | I thought my argument explains it. The $p$-divisible group alone doesn't determine the elliptic curve, S-T is about deformations. Here's a simpler argument to see that it can't work in general: take an ordinary elliptic curve $E$ over $\mathbb Q_p$ with good ordinary reduction such that the Tate module is split (e.g., use S-T with $M^1$ that is $\phi$-stable, or take a CM elliptic curve defined by an imaginary quadratic field in which $p$ splits). Now take another elliptic curve $E'$ over $\mathbb Q_p$ with the same reduction and with $T_p(E')$ non-split. Since $T_p(E)$ is split,... | |
| Feb 18, 2011 at 19:23 | comment | added | Felipe Voloch | But in the ordinary case, the Tate modules (and the liftings) do vary in a continuous family parametrized by the Serre-Tate paramenter, which is a principal unit in the p-adics. So I am again confused. Maybe it has to do with the fact that the Serre-Tate parameter depends on fixing bases for the etale and connected parts of the Tate module. | |
| Feb 18, 2011 at 19:17 | comment | added | Felipe Voloch | Since the liftings of a fixed elliptic curve modulo p are uniquely characterized by their p-adic Tate module, I thought this would imply an isogeny theorem. Looks like I was wrong. | |
| Feb 18, 2011 at 15:53 | history | edited | fherzig | CC BY-SA 2.5 |
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| Feb 18, 2011 at 15:36 | history | edited | fherzig | CC BY-SA 2.5 |
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| Feb 18, 2011 at 3:27 | comment | added | fherzig | @Felipe: I'm glad it's helpful. Would you mind explaining the argument in the good ordinary case? I'm confused, because when I tried to think about it over $\mathbb Q_p$, it seemed similar to the good supersingular case (or rather, not different enough to see why the isogeny theorem holds). | |
| Feb 18, 2011 at 2:24 | history | edited | fherzig | CC BY-SA 2.5 |
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| Feb 18, 2011 at 1:35 | comment | added | Felipe Voloch | Thanks! That's very helpful. I think I was thinking of the ordinary case when I mentioned Serre-Tate in the question, I didn't really know what was happening in the supersingular case. Your final comment is right, in char $p$ one traditionally needs $l \ne p$. I don't know if there is a crystalline analogue. | |
| Feb 18, 2011 at 1:27 | history | edited | fherzig | CC BY-SA 2.5 |
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| Feb 18, 2011 at 0:55 | history | edited | fherzig | CC BY-SA 2.5 |
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| Feb 18, 2011 at 0:47 | history | answered | fherzig | CC BY-SA 2.5 |