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I'm trying to learn about weights of Hilbert modular forms, viewing the weights as characters $Res_{\mathcal{O}_F/\mathbb{Q}}\mathbb{G}_m \to \mathbb{G}_m$. For simplicity, assume that our totally real field $F$ is Galois over $\mathbb{Q}$. I'm pretty sure that for parallel weight $k$, the character is just $\left(Norm_{\mathcal{O}_F/\mathbb{Z}}\right)^k$, and that after base changing to $\mathbb{R}$ we get weights $(k_\sigma)_{\sigma \colon F \to \mathbb{R}}$ which correspond to $n \otimes r \mapsto \prod_\sigma (\sigma (n) \otimes r)^{k_\sigma}$.

Are the characters corresponding to non-parallel weights defined over $\mathcal{O}_F$, or do we have to base change all the way up to $\mathbb{R}$? If they're defined over $\mathcal{O}_F$, how are they defined? If they aren't, does this break integrality/rationality arguments?

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    $\begingroup$ I think what one does is, for $(k_\sigma)_\sigma \in \mathbb{Z}^\Sigma$, one sends $n$ to $\prod_\sigma \sigma(n)^{k_\sigma}$ where $\Sigma$ is the set of embedding of $F$ into $\mathbb{R}$. This clearly preserves the integrality. If you then take $k$ to be parallel you then see that this is just the $k$-th power of the norm. $\endgroup$
    – Leray Jenkins
    Sep 17, 2019 at 18:16
  • $\begingroup$ Right, we can clearly do this after base changing to $\mathbb{R}$. I guess we do get that $\chi(n)$ is in $\overline{\mathbb{Z}}$, the integral closure of $\mathbb{Z}$ in $\mathbb{R}$, so there is some integrality preserved. Is the character defined over $\overline{\mathbb{Z}} \subset \mathbb{R}$? If so, it shouldn't be too hard to have it be defined over $\mathcal{O}_F$ (since $F$ is assumed Galois over $\mathbb{Q}$). $\endgroup$
    – Jon Aycock
    Sep 18, 2019 at 1:44
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    $\begingroup$ One thing I forgot to mention, but this is slightly tangential, is that usually weights for Hilbert modular forms also have a parity condition, i.e. you require all the $k_\sigma$ to have the same parity. This is to make sure your spaces of Hilbert modular forms aren't empty for silly reasons. In any case, I think you got the idea :) $\endgroup$
    – Leray Jenkins
    Sep 18, 2019 at 8:15

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Taking Leray Jenkins' hint that integrality is preserved, I think I've figured it out. Let $F$ be a totally real field of degree $[F:\mathbb{Q}] = d$, Galois over $\mathbb{Q}$. Non-parallel weights correspond not to characters $\operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m \to \mathbb{G}_m$, but to characters $\operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m \to \operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m$. Parallel weights simply have their image in the subgroup $\mathbb{G}_m \subset \operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m$.

For non-parallel weights, we fix a base embedding $\sigma_0 \colon F \to \mathbb{R}$. This identifies $\Sigma$, the set of real embeddings of $F$, with $\operatorname{Gal}(F/\mathbb{Q})$ by $g \leftrightarrow \sigma_0 \circ g^{-1}$. Over $\mathbb{R}$, we consider $(k_\sigma)_{\sigma \in \Sigma}$, corresponding to the character $\chi \colon \lambda \mapsto \prod_{\sigma} \sigma(\lambda)^{k_\sigma}$. We build $\chi_{\sigma_0} \colon \operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m \to \operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m$ so that $\sigma_0(\chi_{\sigma_0}) = \chi$. Define $g_0^\sigma$ so that $\sigma(\lambda) = \sigma_0(g_0^\sigma (\lambda))$. Then we fulfill that requirement with

$$ \chi_{\sigma_0}(\lambda) = \prod_{\sigma \in \Sigma} (g_0^\sigma \cdot \lambda)^{k_\sigma}. $$

We have the character $\chi_{\sigma_0} \colon \operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m \to \operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m$, defined over $\mathbb{Z}$. It depends on the choice of $\sigma_0$, and the weight $(k_{g^{-1} \cdot \sigma})_\sigma$, along with the choice $g \cdot \sigma_0$ as our base (possibly $g^{-1}\cdot \sigma_0$) will give the same integral character, and so the same modular forms; $g$ or $g^{-1}$ should give an algebraic isomorphism between the relevant spaces of modular forms over $\mathbb{R}$, meaning this shouldn't matter.

This gives us an algebraic character for our weight ($\chi_{\sigma_0}$ is defined over $\mathbb{Z}$), and landing the characters in $\operatorname{Res}_{\mathcal{O}_F/\mathbb{Z}}\mathbb{G}_m$ rather than just $\mathbb{G}_m$ is okay since the values of Hilbert modular forms defined over $R$ are allowed to be in $\mathcal{O}_F \otimes_\mathbb{Z} R$ anyway. One can check that this reduces to the character $(Norm_{\mathcal{O}_F/\mathbb{Z}})^k$ for the case of parallel weight $k$.

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