# About the type of a polarization of an abelian variety

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

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.

• 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 Jun 7 '20 at 21:20