Let $A/k$ be an abelian variety with real multiplication by some ring of integers $\mathcal O \subset F$. Let $n$ be an integer prime to the characteristic of $k$.

We have the standart $e_n$ pairing $A[n] \times A^\vee[n] \to \mu_n$. It satisfies $e_n(a \cdot x, y) = e_n(x, a \cdot y)$ for any $a \in \mathcal O$.

It is desirable to "extend" this pairing to an $\mathcal O$-linear pairing with respect to the $\mathcal O$-action on $A[n] \times A^\vee[n]$.

In Rapoprt's "Compactifications de l'espace de modules de Hilbert-Blumenthal" 1.21 we find such an extension:

$e_{n_{\mathcal O}} \colon A[n] \times A^\vee[n] \to (\mathcal D^{-1}/n\mathcal D^{-1})(1)$

where $\mathcal D$ is the different of $F$ and with $(1)$ we denote the Tate-twist. Concerning the definition of the pairing it is only said that $Tr(e_{n_{\mathcal O}}) = e_n$.

Can someone make this definition more explicit?

  • 3
    $\begingroup$ Isn't it just nothing to do with abelian varieties and just a case of unravelling some algebra and using that $Hom_{\mathbf{Z}/n}(\mathcal{O}/n,\mathbf{Z}/n)=\mathcal{D}^{-1}/n$? No time to do the exercise but hopefully shouldn't be hard. Think of $e$ as a map from $A^\vee[n]$ to $Hom_{\mathbf{Z}/n}(A[n],\mathbf{Z}/n)(1)$. $\endgroup$ Jun 3, 2011 at 18:05

1 Answer 1


Yes, it's just a linear algebra stuff. Probably, it's more convenient to look at the Weil pairing between the $\ell$-adic Tate modules of $A$ and its dual, taking into account that these modules are free $O\otimes Z_{\ell}$-modules of the same rank.

From the linear algebra point of view the situation is as follows.

Let $e: M \times N \to P$ be a perfect pairing of free $Z_{\ell}$-modules $M$ and $N$ of the same rank that takes values in a free $Z_{\ell}$-module $P$ of rank $1$. Assume also that $M$ and $N$ are free $O\otimes Z_{\ell}$-modules of the same rank. Then we get a natural $Z_{\ell}$-linear map

$M \times N \to Hom_{Z_{\ell}}(O\otimes Z_{\ell}, P),$ $(m,n) \mapsto [a \mapsto e(ax,y)]$ for all $a \in O$. Taking into account that $$Hom_{Z_{\ell}}(O\otimes Z_{\ell}, Z_{\ell})=D^{-1}\otimes Z_{\ell}$$ (via the trace map), we get the natural pairing

$$M \times N \to P \otimes_{Z_{\ell}}[D^{-1}\otimes Z_{\ell}].$$ Here $M$ and $N$ are the Tate modules of $A$ and its dual while $P$ is $Z_{\ell}(1)$ and $e$ is the Weil pairing.


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