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Are there any good way to understand $k$-polarized Abelian surfaces? I am aware that if $A \cong \mathbb{C}^2/\Gamma$ is $k$-polarized, the lattice $\Gamma$ can be taken of the form $$ \begin{bmatrix} 1 & 0 & \tau_1 & \tau_2\\ 0 & k & \tau_3 & \tau_4 \end{bmatrix} $$ over $\mathbb{Z}$ (think of $\mathbb{C}^2\cong \mathbb{R}^4$) such that the imaginary part of $ \begin{bmatrix} \tau_1 & \tau_2\\ \tau_3 & \tau_4 \end{bmatrix} $ is positive definite. Are there any other good way to see $k$-polarized Abelian surfaces?

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up vote 5 down vote accepted

It is well known that any polarized abelian variety is isogenous to a principally polarized one. More precisely, given any polaized abelian variety $(A, \, L)$ there exists a principally polarized abelian variety $(B,\, \Theta)$ and an isogeny $u \colon A \longrightarrow B$ such that $L=u^* \Theta$.

Then any polarized abelian surface $(A, \, L)$, where $L$ is of type $(1, \,k)$, admits an isogeny of degree $k$ over a principally polarized abelian surface $(B, \, \Theta)$ which is compatible with the polarizations. In other words, $(A, \, L)$ is an étale cover (necessarily Galois, with abelian Galois group) of $(B, \, \Theta)$.

For instance, $A$ admits a polarization of type $(1, \,2)$ if and only if it is an étale double cover of a principally polarized abelian surface.

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Thanks for the answer. This gives an interesting way to see $k$-polarized Abelian varieties. Can I then ask why we are happy if they reduce to principal polarized ones? In other words, why principally polarized nes are interesting? – user2013 Jun 30 '13 at 1:24
Principally polarized abelian varieties are easier to understand and in some cases can be completely classified. For instance, a pp abelian surface is either the Jacobian of a genus $2$ curve or a product of elliptic curves; a pp abelian threefold is either the Jacobian of a genus $3$ curve or a product of an elliptic curve and a pp abelian surface or the product of three elliptic curves (see Birkhenake-Lange, Corollary 11.8.2 page 341). – Francesco Polizzi Jun 30 '13 at 10:05
I see. Your answer open a new way to see abelian varieties. Thank you very much, Francesco. – user2013 Jun 30 '13 at 12:20

I don't know what your background is in abelian varieties, but there are several equivalent ways of talking about $k$-polarized abelian varieties (I assume that you're speaking of a polarized abelian variety of type $(1\hspace{0.1cm}1\cdots1\hspace{0.1cm}k)$?). The following are equivalent:

  1. $(A,L)$ is $k$-polarized (where $L$ is an ample line bundle)

  2. The map $A\to\mbox{Pic}^0(A)$ where $x\mapsto t_x^*L\otimes L^{-1}$ is an isogeny with kernel isomorphic to $(\mathbb{Z}/k\mathbb{Z})^2$. Over $\mathbb{C}$, if $A\simeq\mathbb{C}^n/\Lambda$ and if $H$ is a positive definite hermitian form such that $\Im H(\Lambda\times\Lambda)\subseteq\mathbb{Z}$ that induces the polarization, then the previous condition turns into $\{z\in\mathbb{C}^n:\Im H(z,\Lambda)\subseteq\mathbb{Z}\}/\Lambda\simeq(\mathbb{Z}/k\mathbb{Z})^2$.

One way that these appear in nature is the following:

Let $(X,\Theta)$ be a principally polarized abelian variety of dimension $n$, and assume that there exists an elliptic curve $E\subseteq X$. By translating, we can see $E$ as an abelian subvariety of $X$. Let $(E\cdot\Theta)=k$ (that is, $\Theta|_E$ is a divisor on $E$ of degree $k$). There exists a complementary abelian subvariety $Z\subseteq X$ such that $Z\cap E$ is finite, $\dim Z=n-1$, and $h^0(\Theta|_E)=h^0(\Theta|_Z)$. Now, complementary abelian subvarieties have the same exponent, which means in this case that $(Z,\Theta|_Z)$ is of type $(1\hspace{0.1cm}1\cdots1\hspace{0.1cm}k)$. So basically, this type of polarized abelian variety arises naturally in ppavs that contain an elliptic curve.

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Thank you for the new insight. – user2013 Jul 29 '13 at 23:59

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