The following is a question I posted about a week ago on Maths stackexchange there, but it didn't bring any discussion nor comment. For this reason I am posting it here also.

Let $X$ be an abelian variety of dimension $g$ over an algebraically closed field $k$ of characteristic different from $2$, and consider $\lambda:X\rightarrow \hat X$ a polarization of degree $d$. Assume that $d$ is prime to the characteristic of $k$. Then it is known that the kernel $\mathrm{Ker}(\lambda)$ is an étale, constant group scheme over $k$. Moreover because $\lambda$ is symmetric, its kernel also has the structure of a symplectic module. We deduce the existence of a unique sequence of integers $d_1|\ldots |d_n$ such that $d_1\geq 2$ and $$\mathrm{Ker}(\lambda)\simeq \left( \mathbb Z/d_1\mathbb Z \times \ldots \times \mathbb Z/d_n\mathbb Z\right)^2$$ as group schemes over $k$. (In particular, $d$ is the square of the product of all the $d_i$'s).

On many occasions in the litterature, I see that the integer $n$ is taken to be equal to the dimension $g$ of $X$, up to adding some $1$ at the beginning of the sequence $(d_1,\ldots ,d_n)$. We then call $D = (d_1,\ldots ,d_g)$ the type of the polarization. I am all fine with that when $n\leq g$, but wouldn't it be possible for $n$ to actually be bigger than $g$ in the first place ? Am I missing something obvious ?

The only way I can think of relating the kernel of $\lambda$ with the dimension of $X$ would be by mean of the Tate module, attached to any prime $l$ different from the characteristic of $k$. Indeed, this module $\mathrm T_l(X)$ has rank $2g$ over $\mathbb Z_l$, and it is equipped with the Weil symplectic pairing which involves the polarization $\lambda$. Considering the restricted product of these modules, we obtain a symplectic space over the ring $\mathbb A_f^p$ of finite adèles away from $p$. In PEL moduli problems, we impose the condition that this symplectic pairing should also have type $D$, ie. it should be represented by the matrix $\left( \begin{matrix} 0 & \mathrm{Diag}(D) \\ -\mathrm{Diag}(D) & 0 \end{matrix} \right)$ in some appropriate basis. This also suggests that $n$ shouldn't be bigger than $g$, but I have been failing to write down convincing arguments to show it.

Some references where $n$ is taken to be the dimension $g$ of $X$ without any specific explanation:

  • Genestier and Ngo's lecture on Shimura varieties, available here. See the definition of the moduli problem in 2.3 page 13. The condition (3) implicitly implies that $n=g$. This moduli problem corresponds to that studied in Mumford's GIT, where no such condition was imposed to my understanding.
  • Olsson's workshop notes on abelian varieties, available here. See remark 6.13.
  • Hulek and Sankaran's paper on the geometry of Siegel modular varieties, available here. See p.93 (ie. p.5 of the pdf). In the case of abelian varieties over $\mathbb C$ described as projective tori, the definition of a polarization seems to be slightly adapted, and there it is clear that the number of integers in the sequence $(d_1,\ldots,d_n)$ is precisely the dimension of $X$.

1 Answer 1


Let $\lambda: A\rightarrow A^{\vee}$ be any polarization of degree prime to the characteristic, not necessarily self-dual. There exists an $\lambda^{\vee} : A^{\vee}\rightarrow A$ such that $\lambda^{\vee}\circ \lambda = [n]$ for some integer $n$ where $n$ is invertible in $k$. So $\ker(\lambda) \subset A[n] \simeq \left(\mathbb{Z}/n\mathbb{Z} \right)^{2g}$. By linear algebra over $\mathbb{Z}$, any subgroup of a finite group generated by $2g$ elements is generated by at most $2g$ elements, and we can choose the generators to be compatible with a given symplectic form. This proves the claim.

  • 1
    $\begingroup$ And so it was as simple as that !! Thank you very muchc Jef L, I am relieved by this very clean answer. It helped a lot $\endgroup$
    – Suzet
    Jun 7, 2020 at 21:20

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