# A small question on Carayol's paper

This question might be silly but really confuses me...

On page 31 of his paper: Sur la mauvaise réduction des courbes de Shimura , Carayol gave a modular interpretation of the actions of $G'(\mathbb A_f)$ on the projective system {$M'_{K'}$}. (In fact he defined the actions of a bigger group.)

We use the same notations and assumptions as in Carayol's paper. Take $\gamma\in G'(\mathbb A_f)$, and suppose $\gamma^{-1}$ maps $V_{\hat{\mathbb Z}}$ to $V_{\hat{\mathbb Z}}$, then $\exists m\in \mathbb Z_{\geq 1}$ such that $\gamma V_{\hat{\mathbb Z}}\subset \frac{1}{m} V_{\hat{\mathbb Z}}$. And $\exists ! f \in F^{*+}$ s.t $\nu'_1(\gamma)\in f\cdot \hat{\mathbb Z}^{*}$.

Let $(A, \tau, \theta, k)$ be a point in $M'_{K'}$, i.e $k: \hat{T} A \rightarrow V_{\hat{\mathbb Z}}$ a symplectic isomorphism. Thus we can identify $\gamma (V_{\hat{\mathbb Z}})/ V_{\hat{\mathbb Z}}$ as a subgroup (denoted by $\Lambda$ ) of $A_m$. Then we define $A'=A/{\Lambda}$, which is naturally an abelian variety with $\mathcal O_D$ structure.

Then now we need to define a "right" polarization on $A'$. In his paper, Carayol wrote "...Le schéma abélien dual $\hat{A'}$ de $A'$ s'identifie à un quotient $\hat{A}/\hat{\Lambda}$ où le dual $\hat{\Lambda}$ de $\Lambda$ contient $f^{-1} \theta({\Lambda})$..." Shouldn't $\hat{A}=\hat{A'}/{\hat{\Lambda}}$? I'm very confused here.

Besides, since we have $$0\rightarrow A_m/\Lambda \rightarrow A' \rightarrow A \rightarrow 0,$$ we can identify $\hat{A'}$ with $\hat{A}/(A_m/\Lambda)^{\wedge}$. Then consider the composition $$A\xrightarrow{f^{-1}\theta} \hat{A} \rightarrow \hat{A}/(A_m/\Lambda)^{\wedge}=\hat{A'},$$ we can get a "polarization" $\theta': A'\rightarrow \hat{A'}$. But since $m$ could be arbitrarily large, we could not expect $deg(\theta')=deg(\theta)$. However, in this moduli problem, I think the degree of the polarization is fixed, because of the symplectic isomorphism $\hat{T} A \rightarrow V_{\hat{\mathbb Z}}$? So how could we define a polarization $\theta'$ of $A'$, such that $deg(\theta')=deg(\theta)$?

Thanks.

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