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Does anyone know of a reference for the following fact?

Let $M_g$ denote the moduli stack of genus g curves, let $A_g$ denote the moduli stack of abelian varieties, and let $U_g \rightarrow A_g$ denote the universal abelian variety. For any basepoint $[C] \in M_g$, there is a representation $\pi_1(M_g, [C]) \rightarrow Sp_{2g}(\mathbb Z)$ by sending a loop to a homeomorphism of the curve C. On the other hand, if one applies the torelli map, sending a curve to its Jacobian, $\tau: M_g \rightarrow A_g$, one can consider the Galois representation $\rho_{U_g}: \pi_1^{et}(A_g, \overline{[\tau(C)]}) \rightarrow Sp_{2g}(\widehat{\mathbb Z})$ (the monodromy of the moduli stack of abelian varieties, given by action on the $\ell$ torsion).

Why do these two representations agree? That is, why does the square, with vertical maps given by profinite completion, commute?

$\require{AMScd}$ \begin{CD} \pi_1(M_g, [C]) @>>> Sp_{2g}(\mathbb Z) \\ @VVV @VVV \\ \pi_1^{et}(A_g, \overline{[\tau(C)]}) @>>> Sp_{2g}(\widehat{\mathbb Z}) \end{CD}

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    $\begingroup$ I think this is just an elementary statement that if a group $\Gamma$ acts on $\mathbb {Z}^{2g}$, then the profinite completion $\widehat{\Gamma}$ acts on the profinite completion $\widehat{\mathbb{Z}}^{2g}$. $\endgroup$
    – Venkataramana
    Commented Jul 12, 2016 at 3:06
  • $\begingroup$ @Venkataramana: that doesn't seem like the hard part of the question. Of course if we profinitely complete the first representation we get something that looks like the second representation. But why is it in fact the second representation? $\endgroup$
    – Qiaochu Yuan
    Commented Jul 12, 2016 at 3:37
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    $\begingroup$ @Qiaochu Yuan: the action on $l$ division points is exactly the action on the pro-l completion of the fundamental group of the abelian variety in question. $\endgroup$
    – Venkataramana
    Commented Jul 12, 2016 at 3:39
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    $\begingroup$ For a smooth curve $C$, there are canonical isomorphisms $H^1(C,\mathbb{Z})\otimes \mathbb{Z}_{\ell}\rightarrow H^1_{et}(C,\mathbb{Z}_{\ell})$ and $H^1_{et}(JC,\mathbb{Z}_{\ell})\rightarrow H^1_{et}(C,\mathbb{Z}_{\ell})$. $\endgroup$
    – abx
    Commented Jul 12, 2016 at 6:54
  • $\begingroup$ @abx That seems to answer it since all the representations are given by the action of $\pi_1$ on $H^1$, and then one can pass between the topological $H^1$ and etale $H^1$. $\endgroup$
    – Aaron Landesman
    Commented Jul 12, 2016 at 21:13

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