In which ways can the isogeny theorem fail for local fields? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:54:22Z http://mathoverflow.net/feeds/question/53014 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53014/in-which-ways-can-the-isogeny-theorem-fail-for-local-fields In which ways can the isogeny theorem fail for local fields? Felipe Voloch 2011-01-24T00:26:02Z 2011-02-18T15:53:19Z <p>Fix a field $K$ with absolute Galois group $G$. By an isogeny theorem over $K$, I mean the statement that the map $\operatorname{Hom}(A,B)\otimes\mathbb{Z}_l \to \operatorname{Hom}_G(T_l A, T_l B)$ is an isomorphism, where $A,B$ are abelian varieties over $K$, and $T_l A$ is the Tate module of $A$. Such a statement was proved for finite fields by Tate, for global function fields by Zarhin and for number fields by Faltings. </p> <p>I'm interested in the case where $K$ is a $p$-adic field. The statement is then generally false, but is sometimes true. It holds e.g. when $A,B$ are elliptic curves with bad reduction and $l = p$ (Serre) or when $A,B$ have the same (good) reduction and again $l=p$ (Serre-Tate). It certainly fails if $A,B$ have good reduction and $l \ne p$. </p> <p>What I don't have is a clear picture of the various possibilities for the reductions and for which cases the statement holds or fails. So that's the question.</p> http://mathoverflow.net/questions/53014/in-which-ways-can-the-isogeny-theorem-fail-for-local-fields/53074#53074 Answer by Mikhail Bondarko for In which ways can the isogeny theorem fail for local fields? Mikhail Bondarko 2011-01-24T15:53:49Z 2011-01-24T15:53:49Z <p>The main types of the reduction are: good, semi-stable, and bad one.:) If you consider the product of abelian varieties, then its reductions is the 'worst' of those of its 'components'; so in the higher-dimensional case there are more chances that a result will hold for abelian varieties of good reduction.</p> <p>Yet you could also consider the following types of reduction: additive, toroidal (of multiplicative type?), or multiplicative one. It seems that for the latter type of varieties a p-adic uniformization (generalizing the term 'Tate curve') exists; see section 4 of <a href="http://www.google.com/url?sa=t&amp;source=web&amp;cd=9&amp;ved=0CEwQFjAI&amp;url=http%3A%2F%2Fwww.math.psu.edu%2Fpapikian%2FResearch%2FRAG.pdf&amp;ei=-p09TduGPM7rsgapvNTzBg&amp;usg=AFQjCNEASA40t7BL6JapVzmjdVja8ss02w&amp;sig2=xmCTgc5rcZmLfGdwPImtQw" rel="nofollow">http://www.google.com/url?sa=t&amp;source=web&amp;cd=9&amp;ved=0CEwQFjAI&amp;url=http%3A%2F%2Fwww.math.psu.edu%2Fpapikian%2FResearch%2FRAG.pdf&amp;ei=-p09TduGPM7rsgapvNTzBg&amp;usg=AFQjCNEASA40t7BL6JapVzmjdVja8ss02w&amp;sig2=xmCTgc5rcZmLfGdwPImtQw</a> Then it is fairly easy to calculate the $p$-adic Tate modules and verify the corresponding case of the conjecture (I suspect it would be true). Furthemore, it seems not impossible to carry over this calculation to the case of potentially multiplicative reduction.</p> http://mathoverflow.net/questions/53014/in-which-ways-can-the-isogeny-theorem-fail-for-local-fields/53114#53114 Answer by Tim Dokchitser for In which ways can the isogeny theorem fail for local fields? Tim Dokchitser 2011-01-24T19:54:40Z 2011-01-25T12:46:10Z <p>I believe for $l\ne p$ the theorem is 'always' false, in the sense that for every positive-dimensional abelian variety $A$ one can find many $B$s for which your map is not onto, independently of the reduction type:</p> <p>Fix $A$ and take any non-constant family of abelian varieties in which $A$ is a fibre, e.g. some neighbourhood of $A$ in the moduli space (choosing a polarization or whatever). Then all nearby fibers $A'$ in the family have the same $l$-adic representation, $V_l A\cong V_l A'$. This is a special case of a very general theorem of Mark Kisin that $l$-adic representations are locally constant in families in the $p$-adic topology (Math. Z. 230, 569-593 (1999), <a href="http://www.springerlink.com/content/44x7kn31jga7rlf7/" rel="nofollow">http://www.springerlink.com/content/44x7kn31jga7rlf7/</a>, esp. lines 4-5 from the bottom). So there are uncountably many abelian varieties with the same $l$-adic representation as $A$, but only countably many isogenous ones, contradicting surjectivity.</p> <p>This is just an extension of what you mentioned in the question: in the case of good reduction, $V_l A$ is a trivial inertia module so it is determined as a Galois representation by the characteristic polynomial of Frobenius. So, e.g., in a family of elliptic curves if $E$ and $E'$ are close enough so that they have the same reduction mod $p$, then $V_l E\cong V_l E'$; but, of course, there are uncountably many such 'nearby' $j$-invariants in any non-isotrivial family, so for any $E$ you can find lots of $E'$s for which the theorem fails.</p> <p>(I don't know what happens for $l=p$, so this is only a partial answer, but it won't fit in a comment box.)</p> http://mathoverflow.net/questions/53014/in-which-ways-can-the-isogeny-theorem-fail-for-local-fields/55808#55808 Answer by fherzig for In which ways can the isogeny theorem fail for local fields? fherzig 2011-02-18T00:47:17Z 2011-02-18T15:53:19Z <p>I think the statement can fail in the case of elliptic curves of good reduction even when $l = p$. But then your comment on Serre-Tate theory confused me for a little while! (A discussion with Jared Weinstein helped me clear it up.) The post ended up getting long as a result; it's partly for my own reference.</p> <p>Recently FC pointed out the following to me: if $E$ over $\mathbb Q_p$ is an elliptic curve of good supersingular reduction and $p > 3$, then $T_p(E)$ (up to isomorphism) does not depend on $E$.</p> <p>First, $a_p = 0$ by the Weil bounds. The $p$-adic Galois representation $\rho := T_p(E) \otimes \mathbb Q_p$ is crystalline (due to good reduction) of Hodge-Tate weights 0, 1 with (crystalline) $a_p = 0$. In particular, it is non-ordinary so $\rho$ is (even residually) absolutely irreducible. It's a basic fact in $p$-adic Hodge theory that any 2-dim. absolutely irreducible $G_{\mathbb Q_p}$-representation with distinct Hodge-Tate weights is uniquely determined by $a_p$. (This is nowadays an exercise about weakly admissible filtered $\phi$-modules.) Finally, as $\rho$ is residually irreducible, it follows that the lattice $T_p(E)$ is unique up to homothety (in particular, up to isomorphism).</p> <p>Just as Tim notes, isogeny classes are countable. But there are uncountably many elliptic curves with good supersingular reduction. (Fix a supersingular one over $\mathbb F_p$ and consider lifts of the $j$-invariant.)</p> <p>I wonder how general a phenomenon this is that the Galois representation doesn't vary. Kisin's result no longer seems to apply in this $l = p$ case (non-torsion coefficients). </p> <hr> <p>Now, here is why there is no contradiction with Serre-Tate theory. First, it is true that any two of the elliptic curves above have isomorphic $p$-divisible groups over $\mathbb Z_p$: the reason is that $T_p(E)$ (with the Galois action) precisely tells us what $E[p^\infty]$ is over $\mathbb Q_p$, and that Tate's full-faithfulness theorem says that this determines $E[p^\infty]$ over $\mathbb Z_p$. Besides, the special fibre $E/\mathbb F_p$ is uniquely determined up to isogeny, because we know $a_p$. [There's even an isogeny over $\mathbb F_p$, e.g., because the Dieudonne module is the same (see below), but there must be an elementary way to see that.]</p> <p>Serre-Tate theory says the following (specialised to our situation). The category of elliptic curves over $\mathbb Z_p$ is equivalent to the following category $C$: the objects consist of triples $(\mathcal G, \overline E, f : \mathcal G \times \mathbb F_p \to \overline E[p^\infty])$, where $\mathcal G$ is a $p$-divisible group over $\mathbb Z_p$, $\overline E$ is an elliptic curve over $\mathbb F_p$ and $f$ is an isomorphism of $p$-divisible groups over $\mathbb F_p$. Morphisms consist of a map of $p$-divisible groups over $\mathbb Z_p$ and a map of elliptic curves over $\mathbb F_p$ that are compatible with the isomorphisms in the special fibre. In this equivalence, an elliptic curve $E/\mathbb Z_p$ maps to the triple $(E[p^\infty], E \times \mathbb F_p, can)$, where $can$ is the canonical isomorphism.</p> <p>Suppose we have two elliptic curves $E_1$, $E_2$ over $\mathbb Z_p$ as above (i.e., supersingular reduction and $p > 3$). Then we know that $E_1[p^\infty] \cong E_2[p^\infty]$ and that $E_1 \times \mathbb F_p$ is isogenous to $E_2 \times \mathbb F_p$. But in order to apply the theorem of Serre-Tate and deduce the existence of an isogeny $E_1 \to E_2$ we need to know that we can choose a map $\alpha : E_1[p^\infty] \to E_2[p^\infty]$ of $p$-divisible groups over $\mathbb Z_p$ and a map $\beta : E_1 \times \mathbb F_p \to E_2 \times \mathbb F_p$ of elliptic curves such that $\alpha \times \mathbb F_p = \beta[p^\infty]$ as map of $p$-divisible groups $E_1[p^\infty] \to E_2[p^\infty]$ over $\mathbb F_p$.</p> <p>The problem is that we have very little flexibility: there are only countably many homomorphisms $E_1 \times \mathbb F_p \to E_2 \times \mathbb F_p$. But it's hard to lift a map of $p$-divisible groups over $\mathbb F_p$ to a map of $p$-divisible groups over $\mathbb Z_p$:</p> <p>The category of $p$-divisible groups over $\mathbb Z_p$ is equivalent, by results of Fontaine-Laffaille, to the category of quadruples $(M,M^1,\phi,\phi^1)$, where $M$ is a finite free $\mathbb Z_p$-module, $M^1$ a $\mathbb Z_p$-direct summand ("Hodge filtration"), and $\phi : M \to M$, $\phi^1 : M^1 \to M$ are $\mathbb Z_p$-linear maps such that $p\phi^1 = \phi|_{M^1}$ and $\phi(M) + \phi^1(M^1) = M$. [In our case $M$ is of rank 2 and $M^1$ of rank 1.] The Dieudonne module of the special fibre of the given $p$-divisible group is recovered as follows (reference?): we need to give a $\mathbb Z_p[F,V]/(FV-p)$-module $D$ that is finite free as $\mathbb Z_p$-module. We take $D := M$, $F := \phi$, and $V := pF^{-1}$ (this is defined on $M[1/p]$ initially, but the final condition on $M$ forces it to preserve $M$). In other words, one forgets the Hodge filtration. So lifting $\beta[p^\infty]$ to $\mathbb Z_p$ is difficult because a given map of Dieudonne modules doesn't need to preserve the Hodge filtration. (The countable flexibility we have with $\beta$ is not enough to move a given line to any other. Note that $M^1 \otimes \mathbb F_p$ is uniquely determined as the kernel of $\phi \otimes \mathbb F_p$; but apart from that $M^1$ is free to vary.)</p> <p>[Finally, a small comment/question: I think Tate's result over finite fields assumes that $l$ is prime to the characteristic. <a href="http://www.jmilne.org/math/articles/1971a.pdf" rel="nofollow">Milne-Waterhouse (1971)</a> have a version for $l = p$ where $T_l$ is replaced by the Dieudonné module. In this case $T_p E$ is just the first crystalline cohomology group. Is there a general crystalline analogue of the Tate conjecture?]</p> <p><strong>Added:</strong> I think the isogeny theorem can also fail in the case of good ordinary reduction, using the same argument. Consider again elliptic curves over $\mathbb Q_p$, now of good ordinary reduction. Let's restrict to those elliptic curves of a fixed reduction $\overline E$ over $\mathbb F_p$. We have $a_p \ne 0$, and let's assume for simplicity that the two roots of $x^2-a_p x + p$ are in $\mathbb Z_p$, so they are $u$, $pv$ with $u$, $v \in \mathbb Z_p^\times$. Then there's a basis of $M$ (the Fontaine-Laffaille object associated to $E[p^\infty]$) such that $\phi = \begin{bmatrix}u &amp; \\ &amp; pv\end{bmatrix}$. The possible choices of $M^1$ are precisely the lines generated by the vector $\begin{bmatrix} px\\ 1\end{bmatrix}$, where $x \in \mathbb Z_p$. The isomorphism class of $M$ (and thus of the Galois representation $T_p E$) only depends on the valuation of $x$, because a diagonal change of basis leaves $\phi$ invariant. But there are uncountably many $x$ of any fixed positive valuation (and thus deformations of $\overline E$ to $E$). By the above argument, these cannot all be isogenous.</p>