Everywhere locally isomorphic abelian varieties - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:53:31Z http://mathoverflow.net/feeds/question/10033 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10033/everywhere-locally-isomorphic-abelian-varieties Everywhere locally isomorphic abelian varieties Chandan Singh Dalawat 2009-12-29T13:03:50Z 2009-12-30T14:16:20Z <p>Is there a standard example of two abelian varieties $A$, $B$ over some number field $k$ which are $k_v$-isomorphic for every place $v$ of $k$ but not $k$-isomorphic ?</p> http://mathoverflow.net/questions/10033/everywhere-locally-isomorphic-abelian-varieties/10044#10044 Answer by Idoneal for Everywhere locally isomorphic abelian varieties Idoneal 2009-12-29T16:20:35Z 2009-12-29T16:40:47Z <p>Selmer's curve $3X^3+4y^3+5z^3=0$ is a non-example (see the comment below) but somwhat relevant. See Theorem 1 in Mazur's article titled <a href="http://www.ams.org/bull/1993-29-01/S0273-0979-1993-00414-2/S0273-0979-1993-00414-2.pdf" rel="nofollow">ON THE PASSAGE FROM LOCAL TO GLOBAL IN NUMBER THEORY</a>.</p> http://mathoverflow.net/questions/10033/everywhere-locally-isomorphic-abelian-varieties/10052#10052 Answer by Felipe Voloch for Everywhere locally isomorphic abelian varieties Felipe Voloch 2009-12-29T17:32:12Z 2009-12-29T17:32:12Z <p>If $A,B$ are as stated, then $B$ must be a twist of $A$ which is everywhere locally trivial, so $B$ gives a class in $H^1(k,G)$ (where $G$ is the automorphism group of $A$), which is everywhere locally trivial. So, pick a group $G$ that you know has everywhere locally trivial but globally non-trivial class in $H^1(k,G)$ and make it act on an abelian variety. For instance you can make the group act on a curve and therefore on its jacobian. </p> <p>As for your actual question, if there is a "standard" such example, I guess the answer is no.</p> http://mathoverflow.net/questions/10033/everywhere-locally-isomorphic-abelian-varieties/10055#10055 Answer by David Speyer for Everywhere locally isomorphic abelian varieties David Speyer 2009-12-29T18:52:27Z 2009-12-30T14:16:20Z <p>Here's a slight variant of <a href="http://mathoverflow.net/questions/10033/everywhere-locally-isomorphic-abelian-varieties/10052#10052" rel="nofollow">Felipe Voloch's answer</a>, for those who don't have a favorite group cohomology class. Let $C$ be an abelian variety over $\mathbb{Q}$. Suppose that all the $\overline{\mathbb{Q}}$ automorphisms of $C$ are defined over $\mathbb{Q}$ and let $P$ be this automorphism group.</p> <p>Take two classes in $H^1(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), P)$ which are distinct, but become equal in $H^1(\mathrm{Gal}(\overline{\mathbb{Q}_v}/\mathbb{Q}_v), P)$ for every $v$. The corresponding twists of $C$ should give you the examples you want. </p> <p>How have I made things easier? Because I made the action of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on $P$ trivial, I can describe the group cohmology explicitly as $$H^1(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), P) \cong \mathrm{Hom}(\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), P)/P.$$ Here $P$ acts by conjugation on the target. </p> <p>Since $P$ is finite, any of these Hom's factor through $\mathrm{Gal}(K/\mathbb{Q})$ for some finite extension $K/\mathbb{Q}$.</p> <p>So we are now reduced to the following: We must find finite groups $G$ and $P$, an extension $K/\mathbb{Q}$ with Galois group $G$, an abelian variety with automorphism group $P$ and two maps $\alpha$, $\beta: G \to P$ such that</p> <ul> <li>$\alpha$ and $\beta$ are not conjugated to each other by any element of $P$ but</li> <li>when we restrict to any decomposition subgroup group, $\alpha$ and $\beta$ become conjugate.</li> </ul> <p><hr></p> <p>Take $G=(\mathbb{Z}/2)^2$ and $P=S_6 \times (\mathbb{Z}/2)$. We will not use the $(\mathbb{Z}/2)$ factor at all in the following; the reason it is there is that the automorphism group of an abeliabn variety always contains a central involution, namely $-1$. Feel free to think of $P$ as $S_6$.</p> <p>Take $K/\mathbb{Q}$ to be any biquadratic extension in which no prime is completely ramified. This condition assures that no decomposition group is the whole of $G$. Let $\alpha$ send the generators of $G$ to the elements $(12)(56)$ and $(34)(56)$ of $S_6$. Let $\beta$ send the generators of $G$ to $(12)(34)$ and $(13)(24)$. Then $\alpha$ and $\beta$ are not conjugate in $S_6$, but they become conjugate when restricted to any of the three cyclic subgroups.</p> <p>The one missing step is to construct an abelian variety with automorphism group $S_6 \times (\mathbb{Z}/2)$, and all automorphisms defined over $\mathbb{Q}$. <strike>Dror Spieser, in the comments, points out that we can just take the restriction of scalars of an elliptic curve (without CM) defined over an $S_6$ extension of $\mathbb{Q}$. </strike> I still don't have a good construction of this but, thanks to Bjorn's answer, I don't need one.</p> http://mathoverflow.net/questions/10033/everywhere-locally-isomorphic-abelian-varieties/10088#10088 Answer by Bjorn Poonen for Everywhere locally isomorphic abelian varieties Bjorn Poonen 2009-12-30T02:23:56Z 2009-12-30T05:19:10Z <p>(If you upvote this answer, please consider upvoting the answers by Felipe Voloch and David Speyer too, since this answer builds on their ideas.)</p> <p>The smallest examples are in dimension $2$. Let $E$ be any elliptic curve over $\mathbf{Q}$ without complex multiplication, e.g., $X_0(11)$. We will construct two twists of $E^2$ that are isomorphic over $\mathbf{Q}_p$ for all $p \le \infty$ but not isomorphic over $\mathbf{Q}$.</p> <p>Let $K:=\mathbf{Q}(\sqrt{-1},\sqrt{17})$. Let $G:=\operatorname{Gal}(K/\mathbf{Q}) = (\mathbf{Z}/2\mathbf{Z})^2$. Let $\alpha \colon G \to \operatorname{GL}_2(\mathbf{Z}) = \operatorname{Aut}(E^2)$ be a homomorphism sending the two generators to the reflections in the coordinate axes of $\mathbf{Z}^2$, and let $A$ be the $K/\mathbf{Q}$-twist of $E^2$ given by $\alpha$. Define $\beta$ and $B$ similarly, but with the lines $y=x$ and $y=-x$ in place of the coordinate axes. The representations $\alpha$ and $\beta$ of $G$ on $\mathbf{Z}^2$ are not conjugate: only the former is such that the lattice vectors fixed by nontrivial elements of $G$ generate all of $\mathbf{Z}^2$. Thus $A$ and $B$ are not isomorphic over $\mathbf{Q}$.</p> <p>On the other hand, every decomposition group $D_p$ in $G$ is smaller than $G$ since $-1$ is a square in $Q_{17}$ and $17$ is a square in $\mathbf{Q}_2$. Also, the restrictions of $\alpha$ and $\beta$ to any proper subgroup of $G$ are conjugate: any <em>single</em> line spanned by a primitive vector in $\mathbf{Z}^2$ can be mapped to any other by an element of $\operatorname{GL}_2(\mathbf{Z})$. Thus $A$ and $B$ become isomorphic after base extension to $\mathbf{Q}_p$ for any $p \le \infty$. $\square$</p> <p><strong>Remark:</strong> The abelian surfaces $A$ and $B$ constructed above are <em>isogenous</em> even over $\mathbf{Q}$, because the $\mathbf{Z}^2$ with one Galois action can be embedded into the $\mathbf{Z}^2$ with the other Galois action: rotate $45^\circ$ and dilate.</p> <p><strong>Remark:</strong> The nonexistence of examples in dimension $1$ follows from these two well-known facts:</p> <p>1) Twists of an elliptic curve over a field $k$ of characteristic $0$ are classified by $H^1(k,\mu_n)=k^\times/k^{\times n}$ where $n$ is 2, 4, or 6. </p> <p>2) If $n&lt;8$, the map $k^\times/k^{\times n} \to \prod_v k_v^\times/k_v^{\times n}$ is injective. </p> <p>[<strong>Edit:</strong> This answer was edited to simplify the construction and to add those remarks at the end.]</p>