# Characters and weights of Hilbert modular forms

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?

• 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. Sep 17, 2019 at 18:16
• 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 $\mathbb{Q}$). Sep 18, 2019 at 1:44
• 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 :) Sep 18, 2019 at 8:15

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 $$k$$.