Tag Info

Hot answers tagged

34

I would say the answer to both questions is no. In fact, abelian varieties should be an "easy" case. For example, it is known that for abelian varieties (but not other varieties), the variational Hodge conjecture implies the Hodge conjecture. It is disconcerting that we can't prove the Hodge conjecture even for abelian varieties, even for abelian varieties ...


30

There are several different ways to see this. Here is one: Let $G$ be our irreducible projective algebraic group variety over the field $k$, with identity element $e$. The group $G$ acts on itself by conjugation, and this action fixes $e$. Thus this induces a $k$-linear action of $G$ on the local ring $\mathcal O_e,$ and hence on the ...


27

It's a result related in spirit to Minkowski's theorem that $\mathbb Q$ admits no non-trivial unramified extensions. If $A$ is an abelian variety over $\mathbb Q$ with everywhere good reduction, then for any integer $n$ the $n$-torsion scheme $A[n]$ is a finite flat group scheme over $\mathbb Z$. Although this group scheme will be ramified at primes $p$ ...


21

There are no rational curves in an abelian variety, this is much stronger than not being uniruled. If there is a map $P^1 \to A$, $A$ abelian, the map would factor through the Albanese variety of $P^1$, by definition. However, for curves, the Albanese is the Jacobian (from general theory of the Jacobian) and the Jacobian of $P^1$ is a point.


20

This is false even for elliptic curves over $\mathbb{C}$. This was proved by T. Shioda in "Some remarks on abelian varieties" J. Fac. Sci. Univ. Tokyo Sect. IA Math. 24 (1977), no. 1, 11-21, http://repository.dl.itc.u-tokyo.ac.jp/dspace/bitstream/2261/6164/1/jfs240102.pdf.


18

Here is another construction, followed by some comments on how to solve the existence problem in general. If $A$ is a $g$-dimensional principally polarized abelian variety over $\mathbf{C}$ with $\operatorname{End} A = \mathbf{Z}$, and $G$ is a finite subgroup whose order $n$ is not a $g$-th power, then $B:=A/G$ is an abelian variety that admits no ...


18

There's a more down to earth way to deal with this, which is already explained in Mumford's GIT: make an fppf (or even etale) surjective base change to acquire a section, use that to define the principal polarization, and then show it is independent of the choice. (Short reason: varying the choice amounts to a morphism from the smooth proper curve to a Hom ...


18

A result of Kawamata (Kawamata, Yujiro, Characterization of abelian varieties. Compositio Mathematica, 43 no. 2 (1981), p. 253-276) implies that, under your assumptions, $X$ is birational to an abelian variety (in fact you just need the Kodaira dimension of $X$ to be zero and the irregularity to be equal to the dimension of $X$). Once you know that $X$ is ...


17

Over $\mathbb C$ you can argue as follows. Suppose you have a morphism $\mathbb P^1(\mathbb C) \to A $ ($A$= abelian variety ). Since $\mathbb P^1(\mathbb C) $ is simply connected , the morphism lifts to the universal cover of $A$, affine space $\mathbb C^n$. But since $\mathbb P^1(\mathbb C)$ is complete and connected, the lift to affine space must be ...


15

Here is Ribet's proof (expanding on Ulrich's comment): Let $G_K:=Gal(\bar{K} / K)$ and $V_l:=T_l(E)\otimes \mathbb{Q}_l$. The image $\rho_{l,E}(G)$ is a closed subgroup of the $l$-adic Lie group $\text{Aut}(V_l(E)) \cong \text{GL}_{2}(\mathbb{Q}_l)$ and is therefore a Lie subgroup of $\text{Aut}(V_l(E))$. Its Lie algebra $\mathfrak{g}_l$ is a subalgebra of ...


15

Yes. Take $f\in O_K$ a uniformizing element of some prime $\mathfrak p$. Consider the hyperelliptic curve defined by the equation $$y^2=x^{2g+1}+f.$$ Then this curve doesn't have semi-stable reduction at $\mathfrak p$. In fact, this equation defines a proper regular model of the curve over the localization $O_{K, \mathfrak p}$ and this model is minimal ...


15

I've always meant to sit down and figure out some examples. OK, got it. I think the following works over any field (including finite fields and numbers fields) and so must be standard (unless I've overlooked something). Let $E$ and $E'$ be non-isogenous elliptic curves over a field $k$ (i.e., no nonzero maps between them over $k$), and $G$ a group scheme of ...


14

In fact, it is no for completely elementary reasons. If $A$ is simple and $B\subset A^n$ is an abelian variety with $\dim B < g$, then $Hom(B,A^n)=Hom(B,A)^n$ is necessarily zero. So $B=0$.


13

In general, if an abelian variety $A$ contains an abelian subvariety $B\subseteq A$, then $A$ contains another abelian subvariety $B'\subseteq A$ such that $A$ is isogenous to $B\times B'$. This is Poincaré's reducibility theorem. (See also Poincaré's complete reducibility theorem, same book, next page). An abelian variety is called simple if it does not ...


13

Let $\chi$ be a one-dimensional geometric (in the sense of FM) $p$-adic Galois representation of $G_K$ and let $\psi$ be the Hecke character of $K$ associated to $\chi$ by class field theory. The fact that $\chi$ is de Rham (=pst) at all primes above $p$ imples that $\psi$ is an algebraic Hecke character. Generally, the only algebraic Hecke characters of $K$ ...


13

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: 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 ...


13

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. Recently FC pointed out the following to me: if ...


13

Incidentally, as I posted this question someone who knew the answer wandered into my office. The map $M_g \to A_g$ factors through the moduli space $\tau_g$ of pairs (A,P,L) where A is an abelian variety, P is an A torsor, and L is an ample line bundle on P which is geometrically a principal polarization. The map $M_g \to \tau_g$ is given by $C \mapsto ...


12

Here's a related question on which there has been much work. If you actually need an answer to your question for some other reason and you follow this up, I'd be interested to see where it goes. Let $X$ be some moduli space of abelian varieties (so, for $g>1$, there will be extra structure beyond what you have mentioned, for example we'll also be ...


12

This is an immediate consequence of Theorem 2.11 and Proposition 2.9 in Deligne's 'Hodge cycles on abelian varieties' (notes by Milne available here: http://www.jmilne.org/math/Books/DMOS.pdf 2.11 shows that every Hodge cycle on an abelian variety $A$ over a field $k$ embeddable in $\mathbb{C}$ is absolutely Hodge, and 2.9 shows that all absolutely Hodge ...


12

Hi, this is actually a part of my Ph.D. thesis. I am going to discuss it in 6 months. Here you can find a preprint of the work with my advisor http://arxiv.org/abs/1010.4483. It is not the final version, so there could be some minor mistakes. We have proved that, with two ramification point, the Prym map is generically injective when g is greater or equal ...


12

An argument due to T. Saito goes as follows: Let p be a prime of good reduction for the abelian variety $A$ over a number field $K$. Consider the p-adic logarithm on $A(\bar{K_p})$; this vanishes precisely on the torsion points. Since $A(\bar{K})$ is dense in $A(\bar{K}_p)$ and the p-adic logarithm is not identically zero, $A(\bar{K})$ contains non-torsion ...


12

Yes. This follows from the main result of the following paper of Zarhin. MR0885780 (88h:14046) Zarhin, Yu. G. Endomorphisms and torsion of abelian varieties. Duke Math. J. 54 (1987), no. 1, 131–145. His result, specialized to the $K$-simple case, is the following (fantastic) theorem. Let $A$ be a $K$-simple abelian variety defined over a number ...


12

The lattice $L_{K3}=H^2(K3,\mathbb Z)$ is $2E_8+3U$, with $E_8$ negative definite and $U$ the hyperbolic lattice for the bilinear form $xy$. It is unimodular and has signature $(3,19)$. The 16 (-2)-curves $E_i$ form a sublattice $16A_1$ of determinant $2^{16}$. It is not primitive in $L_{K3}$. The primitive lattice $K$ containing it is computed as follows. ...


12

Yes. In zero characteristic the image of an isogeny of elliptic curves is determined up to isomorphism by its kernel. Your isogeny has the same kernel as the doubling map from $E$ to itself.


12

Let me put it this way, Tate's conjecture for abelian varieties is known to imply the Hodge conjecture for abelian varieties, and the last is very much open for this class. For the implication, see the article by Deligne and Milne on "Hodge cycles on abelian varieties" (you can get a copy off of Milne's website). There are lot's of interesting cases known ...


11

Abelian varieties over the rationals are modular if and only if they are of "$GL_2$"-type, which is a notion introduced by Ribet who proved that this statement is a consequence of Serre's conjecture which, as you know, has since been proved. Here is a link to Ribet's paper: http://math.berkeley.edu/~ribet/Articles/korea.pdf Generalizing the statement of ...


11

Not in general. For instance, if $A$ is simple then this is not possible. In fact, any isogeny $$f \colon E_1 \times \cdots \times E_n \longrightarrow A$$ would give a dual isogeny $$f^{\ast} \colon A^{\ast} \longrightarrow E_1^{\ast} \times \cdots \times E_n^{\ast}$$ and the pullback via $f^*$ of any elliptic curve in $E_1^{\ast} \times \cdots \times ...


11

My next door neighbor (in the math department) is a Hodge theorist, and I have never heard her say that abelian varieties are the hardest case of the Hodge conjecture. However, they are certainly a demonstrably "rich" case of the Hodge conjecture. My neighbor did tell me once that the Hodge conjecture is presumably true "most of the time" even for compact ...


11

The class of Abelian varieties is the simplest class of varieties where the Hodge conjecture is not known to be true. So naturally a certain amount of effort is directed toward them. However, it's not clear me that one can make an obvious reduction from more general smooth projective varieties to Abelian varieties. The reason I'm skeptical is because the ...



Only top voted, non community-wiki answers of a minimum length are eligible