# An error in Gelbart's "Automorphic forms on adele groups"?

On p.46-47 of Gelbart's "Automorphic forms on adele groups", he tries to compare the Fourier coefficient of a modular form, with the Fourier coefficient of the "adelic version" of it.

Question: It seems like Lemma 3.6 should really be $y^{k/2} n a_n e^{-2\pi n y}$ instead of $a_n e^{-2\pi n y}$ when $n \in \mathbb{Z}$. Can anyone confirm?

Background for those who don't have the book at hand.

If $f(z)$ is a weight k modular form with respect to $SL_2(\mathbb{Z})$, we can reinterpret this as an adelic automorphic form $\phi_f$ on $G(\mathbb{Q})\backslash G(\mathbb{A})$ with trivial central character.

We can consider the Fourier coefficient of $\phi_f$ as follows. Fix $g \in GL_2(\mathbb{A})$, and consider the function $$x \to \phi_f(\begin{bmatrix}1 & x \\ 0 & 1 \end{bmatrix}g)$$ on $x \in \mathbb{Q} \backslash \mathbb{A}$. Pontryagin duality shows that the character group of $\mathbb{Q} \backslash \mathbb{A}$ is isomorphic to $\mathbb{Q}$. These characters $\chi_q(x)$ for $q \in \mathbb{Q}$, where $$\chi_q(x) = \prod_{p \leq \infty} \chi_p(qx_p)$$ Here $$\chi_{\infty}(u) = e^{2\pi i u}$$ for both $u \in \mathbb{R}$ and $\mathbb{Q}_p$, where in the latter case we take the implicitly used the maps $\mathbb{Q}_p \rightarrow \mathbb{Q}_p/\mathbb{Z}_p \cong \mathbb{Q}/\mathbb{Z}$.

We then have the $q$-th Fourier coefficient $$\int_{\mathbb{Q} \backslash \mathbb{A}} \phi_f \left(\begin{bmatrix} 1 & x \\ & 1 \end{bmatrix} g \right) \overline{\chi_q(x)}dx.$$

Fix $g = \begin{bmatrix}y & 0 \\ 0 & 1 \end{bmatrix} \in GL_2(\mathbb{R})$, embedded in $GL_2(\mathbb{A})$ in the $\infty$-factor. Lemma 3.6 in Gelbart's book is the computation of the Fourier coefficients of $\phi_f$ for this fixed $g$. He claims that the Fourier coefficient equals $$\begin{cases} a_n e^{-2 \pi n y} & \text{ if n \in \mathbb{Z}} \\ 0 & \text{ otherwise.} \end{cases}$$

What I think, if you would like to check.

On the fundamental domain $[0,1] \prod_{p < \infty} \mathbb{Z}_p$, it certainly seems true that $\phi_f\left(\begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}g\right)$ is factorizable in $x$ (i.e. it can be written as a product of functions in $x_p$ over $p \leq \infty$), which should be $$\phi_f\left(\begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}g\right) = (f(x_{\infty} + iy)y^{k/2}) \times 1 \times 1 \times \cdots$$ This is why I think there is a factor of $y^{k/2}$.

We can then compute the Fourier coefficient. The integral over $[0,1] \prod_{p < \infty} \mathbb{Z}_p$ again splits up into a product over the $p$'s. For $q = n \in \mathbb{N}$, Gelbart claims that the finite place all contributes 1, but it seems to be it should contribute $\frac{1}{|n|_p}$ at the $p$-th place instead, unless I am normalizing the measures wrongly. The product formula then gives an extra $n$ at the end.

So for the case $n \in \mathbb{Z}$, I think the answer should be $y^{k/2} n a_n e^{-2 \pi n y}$ instead.

I also tried to do the computation by factoring the integral. Here are my thoughts:

I agree that there should be a $y^{k/2}$ coming from the automorphy factor: $$j \left( \begin{pmatrix}y & 0\\ 0 & 1\end{pmatrix},i\right) = y^{-k/2}$$

The rest seemst to be fine but I think it is important to to choose the signs in the local components of the character $\tau : \mathbb{Q}\backslash \mathbb{A} \rightarrow \mathbb{C}^{\times}$ correctly. By this I mean that one has to take either

• $\tau_{\infty}(x_{\infty}) = e^{2 \pi i x_{\infty}}$ and $\tau_p(x_p) = e^{-2 \pi i \{x_p\}_p}$, OR
• $\tau_{\infty}(x_{\infty}) = e^{-2 \pi i x_{\infty}}$ and $\tau_p(x_p) = e^{2 \pi i \{x_p\}_p}$,

where $\{x_p\}_p$ denotes the $p$-adic fractional part of $x_p \in \mathbb{Q}_p$ defined by the formula $\{x_p\}_p = f$ if $x_p = z + f$ with $z \in \mathbb{Z}_p$ and $f \in \mathbb{Z}[\frac{1}{p}]$. Both of these choices have the desired effect that $\tau$ is indeed trivial on $\mathbb{Q}$.

Gelbart's seems to have chosen the first one. (The non-archimedean compnents of $\tau$ are not sufficiently precisely determined by the condition that they are trivial on $\mathbb{Z}_p$, in my opinion.)

In any case, using the fundamental domain argument I get, for $\xi \in \mathbb{Q}$, $y > 0$,

$$a_{\xi}(\begin{pmatrix}y & 1\\ 0 &1\end{pmatrix}) = \int_{x_{\infty} \in [0,1)}{y^{k/2}f(x_{\infty}+ iy)\overline{\tau_{\infty}(x_{\infty} \xi)}} \; \prod_{p < \infty}{\int_{x_p \in \mathbb{Z}_p}{\overline{\tau_p{(x_p \xi)}}}} \,.$$

Now if $\xi = n$ is an integer the product on the right is equals 1, because for each prime $p$, we integrate the function 1 over $\mathbb{Z}_p$ (which should be given volume 1). Indeed for every $x_p \in \mathbb{Z}_p$ the product $nx_{p}$ also belongs to $\mathbb{Z}_p$ hence has zero fractional part.

If $\xi$ is not an integer, there is some prime $p$ at which $\xi$ has strictly negative $p$-adic valuation, and then the character $x_p \mapsto \overline{\tau_p(x_p \xi)}$ is non-trivial on $\mathbb{Z}_p$ so it integrates to zero over $\mathbb{Z}_p$.

• Hi @m.s, thanks for your answer! It's been a long time since I last looked at this and I probably won't dive into this now, but it looks like when I was calculating the character $\tau_p$ in your notation, I mistakenly thought that $p$-valuation of $n$ would also affect the integral, that turns it into something like $\int_{n\mathbb{Z}_p} 1$. Your answer certainly looks right and thus I am accepting it. Thank you!
– Pig
Commented Nov 26, 2017 at 0:24