Take the 2-minute tour ×

I've read in (abstracts of) papers that there are abelian varieties over fields of positive characteristic that admit no prinicipal polarization. Apparently its not the easiest thing to find an example of, but I was thinking it should be much easier over the complex numbers.

All abelian varieties of dimension 1 are elliptic curves which always have a principal polarization. So any example would have to be at least two dimensional. So my question is given an abelian variety with a polarization, is there a good way of telling if there is or isn't a principal polarization?

Or if that's in general a difficult question, are there some relatively simple examples where you can really see that there are or aren't any principal polarizations?

share|improve this question

2 Answers 2

up vote 20 down vote accepted

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 principal polarization.

Proof: If $B$ had a principal polarization, its pullback to $A$, given by the composition $A \to B \to B' \to A' \simeq A$ (where $A'$ is the dual of $A$ and so on) would be an endomorphism of degree $n^2$. But this endomorphism is multiplication-by-$m$ for some integer $m$, which has degree $m^{2g}$. So $n$ would have to be a $g$-th power. $\square$

To complete this answer, observe that most abelian varieties $A$ over $\mathbf{C}$ satisfy $\operatorname{End} A=\mathbf{Z}$. An explicit example is the Jacobian of the hyperelliptic curve that is the smooth projective model of the affine curve $$y^2= a_{2g+1} x^{2g+1} + \cdots + a_1 x + a_0$$ where $a_{2g+1},\ldots,a_0 \in \mathbf{C}$ are algebraically independent over $\mathbf{Q}$.


1) Of course there is no reason to restrict to $\mathbf{C}$. For instance, one can find examples over $\mathbf{Q}$ by using the fact that the endomorphism ring injects into the endomorphism ring of the reduction modulo any prime of good reduction, and combining this information for several primes.

2) For an arbitrary abelian variety $A$, if you are given a polarization $A \to A'$, then if there is a principal polarization, following the first by the inverse of the second would give you an endomorphism of $A$. So one way of answering the existence question is to determine the endomorphism ring of $A$ and to study those endomorphisms that factor through your given polarization. (That's not quite sufficient, but it gives you an idea of the complexity of the problem since determining the endomorphism ring can be rather difficult.)

share|improve this answer

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 prime order $p$ over $k$ which occurs inside both $E$ and $E'$. Fix such embeddings. (e.g., $E$ and $E'$ over a number field with split $p$-torsion and $G = \mathbf{Z}/p\mathbf{Z}$ embedded in each.) Embed $G$ diagonally into $E \times E'$, and let $A = (E \times E')/G$.

I claim that any polarization of $A$ over $k$ has degree divisible by $p$. (Thus, this gives examples of abelian surfaces without principal polarization, and also in char. $p > 0$ examples with no separable polarization by using non-isogenous ordinary elliptic curves.) Suppose to the contrary, and let $\phi:A \rightarrow A^{\rm{t}}$ denote the symmetric isogeny that "is" such a polarization (I'm never sure if we should call $\phi$ the polarization, or be more symmetric and call $(1,\phi)^{\ast}(\mathcal{P}_A)$ on $A \times A^{\rm{t}}$ the polarization), so ${\rm{deg}}(\phi) = d^2$ where $p$ doesn't divide $d$.

Due to the definition of $A$, the map $j:E \rightarrow A$ induced via inclusion into the first factor of $E \times E'$ followed by projection has trivial kernel and so is a closed subvariety. There is also the dual map $j^{\rm{t}}:A^{\rm{t}} \rightarrow E^{\rm{t}}$, and by general theorem of polarizations the composite map $$f:E \stackrel{j}{\rightarrow} A \stackrel{\phi}{\rightarrow} A^{\rm{t}} \stackrel{j^{\rm{t}}}{\rightarrow} E^{\rm{t}}$$ is a symmetric isogeny that "is" a polarization of $E$. In particular, it must have square degree. We likewise get a symmetric isogeny $f':E' \rightarrow {E'}^{\rm{t}}$ that "is" a polarization of $E'$.

Consider the quotient map $q:E \times E' \rightarrow A$ which is an isogeny of degree $p$. The dual map $$q^{\rm{t}}:A^{\rm{t}} \rightarrow (E \times E')^{\rm{t}} \simeq E^{\rm{t}} \times {E'}^{\rm{t}}$$ is also an isogeny of the same degree, and the composite map $$E \times E' \stackrel{q}{\rightarrow} A \stackrel{\phi}{\rightarrow} A^{\rm{t}} \stackrel{q^{\rm{t}}}{\rightarrow} E^{\rm{t}} \times {E'}^{\rm{t}}$$ must be a direct product of a pair of maps $E \rightarrow E^{\rm{t}}$ and $E' \rightarrow {E'}^{\rm{t}}$ since $E$ and $E'$ are assumed to not be $k$-isogenous. These maps must respectively be $f$ and $f'$ as defined above. In particular, since the degree of $\phi$ is not divisible by $p$ but the degrees of $q$ and $q^{\rm{t}}$ are each equal to $p$, and moreover each of $f$ and $f'$ has square degree, we conclude that one of $f$ or $f'$ has degree not divisible by $p$ and the other has degree divisible by $p^2$. But an elliptic curve has a unique polarization of each square degree $m^2$, namely the composite of $[m]$ with the canonical autoduality (with the right sign to get the ampleness condition). In particular, the kernel is the $m$-torsion. It follows that one of $f$ or $f'$ has kernel with trivial $p$-part, and the other has kernel containing the entire $p$-torsion.

Now $E$ and $E'$ are each naturally abelian subvarieties of $A$, and $\phi$ is an isomorphism on $p$-divisible groups (since degree is prime to $p$), so the $p$-parts of the respective kernels of $f$ and $f'$ must be where each meets the $p$-part of the kernel of $q^{\rm{t}}$. For one of $f$ or $f'$, this is the entire $p$-torsion of the corresponding elliptic curve, and in particular contains $G$. But $E$ and $E'$ in $A$ meet in exactly $G$ (scheme-theoretically) due to the construction of $A$, whence $G$ sits in both kernels. But one of the kernels has trivial $p$-part. Contradiction.

share|improve this answer

Your Answer


