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A special case of locally symmetric spaces is the moduli space of abelian varieties of a given dimension $g$ (over a given base field $k$), lets call it $\mathcal{A}_g$ and is a $k$-scheme. For any field extension $l/k$, we have (I think) $$\mathcal{A}_{g,l}=\mathcal{A}_g\times_{k} l.$$ In particular if $l/k$ is say étale, then the moduli space of abelian varieties (as a scheme over $l$) is an étale cover of the moduli space of abelian varieties over $k$ (I'm not sure about this and apologies if this is wrong).

However some locally symmetric spaces we care about in number theory are of course not schemes but "just" manifolds, however still dependent on a base field, i.e. $$X_k:=Gl_n(k)\backslash [Gl_n(k\otimes \mathbb{R})/\mathbb{R}_{>0}K_{\infty}\times Gl_n(\mathbb{A}_{k,f})/K].$$ My question is if there exists a similar phenomenon that there exists a naturual map $X_l\to X_k$ which is a covering?

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  • $\begingroup$ It's a completely different situation. For example, if you take $n=2$ and $k$ a totally real number field, you get Hilbert modular varieties. $\endgroup$
    – user483792
    Commented Aug 9, 2022 at 12:03
  • $\begingroup$ @anon I don't quite see the relation between your comment and the question if changing the base field for locally symmetric spaces gives covers, could you elaborate? (I'm well aware that locally symmetric spaces are not all $\mathcal{A}_g$ if that is what you are hinting at) $\endgroup$
    – curious math guy
    Commented Aug 9, 2022 at 12:45
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    $\begingroup$ Continuing with $n=2$, $X_k$ is a moduli variety for abelian varieties with (real) multiplication by $k$. If $K$ is a totally real field containing $k$, then there is a map $X_k\to X_K$, namely, tensor the abelian variety with $K$ (over $k$). I don't see any obvious map going the other way. $\endgroup$
    – user483792
    Commented Aug 9, 2022 at 12:52
  • $\begingroup$ @anon So in particular your argument would say that I was wrong to say there is a map $\mathcal{A}_{g,l}\to \mathcal{A}_{g,k}$, right? (My apologies for the mistake in that case) $\endgroup$
    – curious math guy
    Commented Aug 9, 2022 at 16:36
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    $\begingroup$ There's a map $\mathcal{A}_{g,l}\to \mathcal{A}_{g,k}$ but that's not what you are asking for. In terms of Shimura varieties you are changing the group defining the variety. That has nothing to do with changing the base field of the variety. $\endgroup$
    – user483792
    Commented Aug 9, 2022 at 18:43

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