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?

  • 3
    $\begingroup$ 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. $\endgroup$ 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
  • 1
    $\begingroup$ 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 :) $\endgroup$ Sep 18, 2019 at 8:15

1 Answer 1


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 $k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.