## Hot answers tagged abelian-varieties

38

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 ...

32

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 $p$ ...

31

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 (finite-...

23

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.

21

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 ...

20

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 ...

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.

19

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 ...

19

Really this is mostly just consolidating what has been (implicitly) said in the comments and cleaning it up a bit (e.g. using the Chow ring instead of singular cohomology), but might as well make it an answer...
Let $A$ be an abelian variety of dimension $d$. We prove that $A$ cannot embed in $\mathbb{P}^{2d-1}$ and can only embed in $\mathbb{P}^{2d}$ if $d=...

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

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 ...

16

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 ...

15

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 (...

15

Let us work over $\mathbb{C}$.
The inclusion $u \colon B \to A$ induces a surjection $\hat{u} \colon A^{\vee} \to B^{\vee}$.
By general facts on Abelian varieties, the kernels of $u$ and $\hat{u}$ have the same number of connected components. Since $u$ is injective, its kernel is trivial, so it follows $\ker \hat{u}=(\ker \hat{u})_0$; in other words $\ker \...

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 ...

14

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 $K$ ...

14

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 $...

14

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 PoincarĂ©'s reducibility theorem. (See also PoincarĂ©'s complete reducibility theorem, same book, next page).
An abelian variety is called simple if it does not ...

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

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 ...

13

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. ...

13

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 ...

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

Weil introduced the term "polarization" in connection with his study of abelian varieties with complex multiplication. His definition is slightly different from what one sees today; one might call it a polarization up to isogeny instead of a polarization. One can find a discussion in Weil's article "On the theory of complex multiplication" ([1955d] in volume ...

13

Another proof that $L = \,\overline{\bf \!Q\!}\,$:
Clearly $L$ is contained in $\,\overline{\bf \!Q\!}\,$,
so we need only show $L$ contains every algebraic number $x \notin \bf Q$.
Let $P(X)$ be the minimal polynomial of $x$. If $\deg P$ is odd,
then the class of $((x,0)) - (\infty)$ is a $2$-torsion point on
the Jacobian of the elliptic or hyperelliptic ...

12

The construction of an Albanese scheme and an Albanese map for proper and geometrically irreducible schemes over a perfect field goes back to the work of Chevalley, to this talk of Serre, and to Grothendieck, e.g. Theorem 3.3. in FGA Exp.VI . The idea is always to look at the dual of an appropriate component of the Picard scheme. One just has to pile enough ...

12

Let $C$ be the connected component of the identity in $B$. Then $C$ is a projective group variety, hence an abelian variety; let it have dimension $d$. Let $B/C=G$, a finite group. Then the number of $\ell$-torsion points of $B$ is at most $\left|G\right|\cdot\ell^{2d}$. For large $\ell$, this is less than $\ell^{2n}$ if $d$ is not $n$.

12

Yes, an abelian variety $A$ contains no rational curves.
Suppose not and let $f: \mathbb P^1 \to A$ be a non-constant morphism.
If $f$ is inseparable then it must be the composition of some power of Frobenius of $\mathbb P^1$ with a non-constant separable map $g: \mathbb P^1 \to A$.
Thus we may assume that $f$ is separable, i.e., $df : T \mathbb P^1 \...

12

In general it could happen that the Albanese variety does not admit a principal polarization at all. For instance the Albanese variety of an abelian variety is the Abelian variety itself. So choose $X$ to be some abelian variety that has no principal polarization and you will get an example.
On the other hand it can happen that the Albanese variety is ...

12

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 ...

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