@oxeimon This is a bit late to the party, but you can, and do, define level structures over $\mathrm{Spec}(\mathbb{Z})$. These are the so-called Drinfeld level structures (see Katz/Mazur). The idea is that one must taken into account the scheme-theoretic mass of the torsion, opposed to the physical mass which suffices when one can guarantee that $E[N]$ is étale (i.e. over $\mathbb{Z}[\frac{1}{N}]$).

Isn't $\pi_1^\mathrm{ab}(\mathrm{Spec}(\mathcal{O}_K))$ the narrow class group of $K$, not the class group? In the link you provided it's fine since that person states it for a quadratic imaginary field where the two notions coincide.

Thanks for the nice explicit example Donu! By the way, you mentioned above that you thought that 2,3 were open. Do you still believe that? I haven't been able to find any explicit answers online, but that doesn't mean much. There is the book 'Fundamental Groups of Compact Kahler Manifolds', but I don't feel like reading hundreds of pages, and can't find a super relevant proposition. Thanks again!

I should've just looked in your thesis. :)

@potentiallydense I may be the potentially dense one. :) Thanks for pointing that out.

$g^\ast\mathbb{Q}_\ell=\mathbb{Q}_\ell$ identification if we just think of the $G_K$-representation $H^i(X_{\overline{K}},\mathbb{Q}_\ell)$ as the $\overline{\eta}$-stalk (corresponding to the choice of $\overline{K}$) of the smooth $\mathbb{Q}_\ell$-sheaf $R^if_\ast\mathbb{Q}_\ell$ (where $f$ is the structure morphism $X\to\mathrm{Spec}(K)$). Is there something similar here? Sorry for beating a dead horse, and thanks!

I apologize for being annoying. As you can see, this is a purely silly notational question. So, we want a canonical isomorphism $\overline{g}(\overline{x}}=\overline{x}$. I guess this is because $\overline{g}\circ\overline{x}=\overline{x}\circ\overline{g}$ since $\overline{x}$ lies over a $k$-point (where on the right $\overline{g}$ acts on $\mathrm{Spec}(\overline{k})$). And then we just identify this RHS with $\overline{x}$ via $\overline{g}^{-1}$? Is that right? Also, is there a way of phrasing this without having to make such an identification? For example, we can ignore the

Just to be explicit. Literally taking stalks gives me an isomorphism $\sigma(g)_{\overline{x}}:(\overline{g}^\ast (R\Psi\overline{\mathbb{Q}_\ell}))_{\overline{x}}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$. But, this is just an isomorphism $(R\Psi\overline{\mathbb{Q}_\ell})_{\overline{g}(\overline{x})}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$. NOT an isomorphism $(R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$ which is what I would want to have an action. Like I said, maybe this is just some 'canonical identification