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I am currently reading Garrett's book "Holomorphic Hilbert Modular Forms". But I meet trouble at the starting line. Let $F = \mathbb{Q}(\sqrt D)$ be a real quadratic field, $u= a + b\sqrt{D}\in F$, and $z = (z_1,z_2)\in \mathbb{H} \times \mathbb{H}$ where $\mathbb{H}$ denotes the upper half-plane. Then what is the definition of $\operatorname{Tr}(uz)$?

The book says "$\operatorname{Tr}$ is the $\mathbb{C}$-linear extension to $\mathbb{C}^2 \rightarrow \mathbb{C}$ of the Galois trace $F \rightarrow \mathbb{Q}$." But I don't understand what this means.

I know the Galois trace is $\operatorname{Tr}(u) = 2a$. Moreover, I found an explicit expression in the second page of Luo - Moments of the central L-values of the Asai lifts: $$ \operatorname{Tr}(uz) = \frac{a + b\sqrt{D}}{\sqrt{D}}z_1 - \frac{a - b\sqrt{D}}{\sqrt{D}}z_2. $$

Could someone please explain how this expression fits into the definition of the trace used in the context of Hilbert Modular Forms?

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  • $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. Regarding high-level tags, see meta.mathoverflow.net/q/1075 $\endgroup$ Commented Nov 5, 2024 at 3:49

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If $F$ is any totally real field of degree $r$ over $\mathbb Q$ and if $\tau=(\tau_1,\dotsc,\tau_r)\in \mathbb H^r$, by definition it is usual to define $\operatorname{Tr}(u\tau)=u^{(1)}\tau_1+\cdots u^{(r)}\tau_r$, where the $u^{(i)}$ denote the $r$ embeddings of $F$ into $\mathbb R$. However, the Fourier expansion at infinity of a holomorphic Hilbert modular form is of the form $$f(\tau)=a(0)+\sum_{0\ll u\in{\frak d}^{-1}}a(u)e^{2\pi i\operatorname{Tr}(u\tau)}$$ where the sum is over totally positive elements of the codifferent.

In the special case when the codifferent is a principal ideal generated by some element $\delta\in O_F$, as in the case of quadratic fields, one can write $u=v/\delta$ and now sum on $v\in O_F$ instead. In that case $\operatorname{Tr}(u\tau)=\operatorname{Tr}(v\tau/\delta)$. But it is a bad idea to define the trace specifically for the case of quadratic fields (where $\delta=\sqrt{D}$) as the reference that you mention seems to do, since it does not seem to be generalizable to arbitrary totally real fields.

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  • $\begingroup$ Thank you for your clear response! It has helped me understand the concept much better. Additionally, I have had the pleasure of studying your textbook 'Number Theory' in a discussion group, and it has been a valuable resource for our learning. (^▽^) $\endgroup$ Commented Nov 5, 2024 at 4:52
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$\DeclareMathOperator\Tr{Tr}$I think that it means that we regard $F \otimes_{\mathbb Q} \mathbb C$ as a 2-dimensional vector space over $\mathbb C$ by action on the second factor. Then $\Tr((a + b\sqrt D) \otimes z)$ equals $z\Tr(a + b\sqrt D) = 2a z$, by definition.

If we define $F \otimes_{\mathbb Q} \mathbb C \to \mathbb C^2$ by $v \otimes z \mapsto (v z, \overline v z)$, where $\overline\cdot$ is the Galois conjugation of $F/\mathbb Q$, then we may transport the natural $\mathbb C$-linear action of $F$ on the source to such an action on the target. Specifically, $(z_1, z_2) \in \mathbb C^2$ is the image of $$\frac1 2(1 \otimes z_1 + \sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(1 \otimes z_2 - \sqrt D \otimes z_2\sqrt D^{-1}),$$ so that $u(z_1, z_2)$ is the image of $$\frac1 2(u \otimes z_1 + u\sqrt D \otimes z_1\sqrt D^{-1}) + \frac1 2(u \otimes z_2 - u\sqrt D \otimes z_2\sqrt D^{-1}),$$ whose trace in the sense above is $$\frac1 2(z_1 + z_2)\Tr(u) + \frac1{2\sqrt D}(z_1 - z_2)\Tr(u\sqrt D).$$ If $u$ equals $a + b\sqrt D$, then this last equals $$ \frac1 2(z_1 + z_2)(2a) + \frac1{2\sqrt D}(z_1 - z_2)(2b D) = %(z_1 + z_2)a + (z_1 - z_2)b\sqrt D = (a + b\sqrt D)z_1 + (a - b\sqrt D)z_2. %\tag{$*$}\label{481799_star} $$ This agrees with the formula in @HenriCohen's answer, since, in the notation there, we can number the real embeddings so that $u^{(1)} = a + b\sqrt D$ and $u^{(2)} = a - b\sqrt D$. It is off by a factor of $\sqrt D$ from the explicit expression in the reference. I did not find the formula there on a quick scan, so cannot explain the discrepancy, but it could come simply from a different identification of $F \otimes_{\mathbb Q} \mathbb C$ with $\mathbb C^2$.

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  • $\begingroup$ Thank you very much for your detailed answer! It's quite creative, and I haven't seen anything similar before. $\endgroup$ Commented Nov 5, 2024 at 4:59

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