2
$\begingroup$

This is a short question with a lot of setup. I apologize in advance.

In Dan Bump's "Automorphic Forms and Representations," he constructs the L-function of a modular form via an "adelic Mellin transform."

$$Z(f,s) = \int_{A^\times/F^\times} f\begin{pmatrix} y & 0 \\ 0 & 1\end{pmatrix}|y|^{s-1/2} d^\times y.$$

He then pulls this apart into an integral on $A^\times$ using the Fourier transform $f(g) = \sum_{\alpha \in F^\times} W_f\left(\begin{pmatrix} \alpha & 0 \\ 0 & 1 \end{pmatrix}g\right)$, and computes the local integrals, showing that they agree with the Euler factors of the L-function of $f$.

I've used a slightly different method to compute this - Let $f$ be a modular form over $\mathbb{Q}$, for simplicity on $\Gamma_0(N)$ and with trivial nebentypus, and consider the adelic Mellin transform

$$Z(f,s) = \int_{A^\times/F^\times} f\begin{pmatrix} y & 0 \\ 0 & 1\end{pmatrix}|y|^{s-1/2} d^\times y.$$

Both $f\begin{pmatrix} y & 0 \\ 0 & 1\end{pmatrix}$ and $|y|^{s-1/2}$ are invariant under multiplication by $p$-adic units in the $p$ component. If we let $\mathcal{U}$ denote the subgroup of the ideles whose $p$-component is a $p$-adic unit for all $p$, then the integrand is a function of $A^\times/\mathbb{Q}^\times\mathcal{U} = \mathbb{R}_{>0}$, where each coset is represented by a unique idele $(t,1,1,\dots)$ for $t>0$. If we normalize the Haar measure $\mu^\times$ so that $\mu^\times(\mathcal{U}) = 1$, we can write

$$Z(f,s) = \int_{A^\times/F^\times} f\begin{pmatrix} y & 0 \\ 0 & 1\end{pmatrix}|y|^{s-1/2} d^\times y = \int_{A^\times/F^\times \mathcal{U}}\int_\mathcal{U} f\begin{pmatrix} au & 0 \\ 0 & 1\end{pmatrix}|au|^{s-1/2} d^\times u \, d^\times a = $$

$$ = \mu^\times(\mathcal{U})\int_{A^\times/F^\times\mathcal{U}} f\begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}|a|^{s-1/2} d^\times a = \int_{\mathbb{R}_{>0}}f\begin{pmatrix} t & 0 \\ 0 & 1\end{pmatrix}t^{s-1/2} d^\times t.$$

This is the standard real Mellin transform often used to construct the $L$-function of a modular form, up to a shift in $s$, which is great.

However, Bump then claims that we can twist by a Hecke character $\xi$, writing

$$Z(f,\xi,s) = \int_{A^\times/F^\times} f\begin{pmatrix} y & 0 \\ 0 & 1\end{pmatrix}\xi(y) |y|^{s-1/2} d^\times y.$$

He leaves it as an exercise to show that this gives the correct Euler factors after we use Fourier theory to write it as an integral over the whole idele group and compute the local integrals. I've done this, and agree that with this method, it works.

My method now becomes this: again specialize to the case where $f$ is a rational modular form on $\Gamma_0(N)$ with trivial nebentypus, and let $\xi \colon A^\times/\mathbb{Q}^\times \mathbb{R}_{>0} \mathcal{U}(m) \to \mathbb{C}^\times$ be a nontrivial Dirichlet character of conductor $m$. ($\mathcal{U}(m)$ is the neighborhood of $1$ consisting of all $p$-adic units $u \equiv 1 $ (mod $p^{ord_p(m)}\mathbb{Z}_p$).) Then the integrand is still invariant under the action of $\mathcal{U}(m)$, and $A^\times/\mathbb{Q}^\times\mathbb{R}_{>0}\mathcal{U}(m)$ is a finite ray class group. Thus if we choose representatives $r_1,\dots,r_h$ for it, we get representatives $\{r_it \mid i=1,\dots,h; t \in \mathbb{R}_{>0}\}$ for $A^\times/\mathbb{Q}\mathcal{U}(m)$.

Now we can pull apart the integral as before, using the fact that the integrand is invariant under the action of $\mathcal{U}(m)$, and $\xi$ doesn't depend on the real component.

$$Z(f,\xi,s) = \int_{A^\times/F^\times} f\begin{pmatrix} y & 0 \\ 0 & 1\end{pmatrix}\xi(y)|y|^{s-1/2} d^\times y = $$

$$ = \int_{A^\times/F^\times \mathcal{U}(m)}\int_{\mathcal{U}(m)} f\begin{pmatrix} au & 0 \\ 0 & 1\end{pmatrix}\xi(au)|au|^{s-1/2} d^\times u \, d^\times a = $$

$$ = \mu^\times(\mathcal{U}(m))\int_{A^\times/F^\times\mathcal{U}(m)} f\begin{pmatrix} a & 0 \\ 0 & 1\end{pmatrix}\xi(a)|a|^{s-1/2} d^\times a = $$

$$ = \mu^\times(\mathcal{U}(m)) \sum_{i=1}^h\int_{\mathbb{R}_{>0}}f\begin{pmatrix} r_it & 0 \\ 0 & 1\end{pmatrix}\xi(r_it)|r_it|^{s-1/2} d^\times t.$$

We find that $\xi$ doesn't depend on $t$, and then that the rest of the integral doesn't depend on $r_i$ by using the fact that $d^\times t$ is a Haar measure. Thus after we pull $\xi$ out, each integral with $r_i$ included is just $Z(f,s)$. Then we can pull these apart:

$$ Z(f,\xi,s) = \mu^\times(\mathcal{U}(m))Z(f,s) \sum_{i=1}^h \xi(r_i). $$

But this is a constant multiple of the sum of the values of a nontrivial character $\xi$, so we get $0$.

This is obviously wrong, but I can't figure out where. I've checked to make sure that absolute convergence isn't an issue (Bump claims that the "adelic Mellin transform" integral is convergent for all $s$, and multiplying by a unitary character shouldn't change that; on top of this, we should always be able to increase the real part of $s$ to be large enough so that we have absolute convergence, and then analytically continue the $0$ we get to the rest of the complex plane), and I've specialized to $\mathbb{Q}$ so that the consideration of the units is easier (though it's still what I expect for no character, and $0$ when a nontrivial character is included, even if we move up to bigger fields). I've even done this for the simpler case of $\zeta(s)$, to the same result. There's some step that is wrong when a nontrivial character is inserted that either isn't wrong when there is no character, or the actions that I've taken look valid, but only produce the correct results by accident when the character is trivial. Can anyone see what I'm missing?

$\endgroup$
  • $\begingroup$ How did you replace r_i t with t in the final integral? I agree that you can scale down by |r_i| but, correct me if I’m wrong, r_i is an idele and you’re integrating over \R^+. $\endgroup$ – alpoge Jun 6 at 21:21
  • $\begingroup$ I chose an ideal (N) corresponding to the class of r_i in the ray class group, divided through by N (diagonally, as an element of the rationals), and was left with the idele (t/N,1,1,...) instead of r_i t. Then I just translated the real integral by N, and since it's against the Haar measure, nothing changed. I'll make sure to check again that this is valid. $\endgroup$ – Jon Aycock Jun 6 at 21:30
  • $\begingroup$ In fact, the existence of such an N is more than I want. I should probably be able to divide through by r_i itself if I'm careful enough with what I mean by the Haar measure on the quotient, or even to choose different representatives for the set of positive reals inside A^x/(Q^x U). That way the argument generalizes to fields that aren't PIDs. $\endgroup$ – Jon Aycock Jun 6 at 22:15
  • $\begingroup$ Have you looked at this paper of Booker and Krishnamurthy? I think they do some explicit calculations along the lines that you want. doi.org/10.1112/S0010437X10005087 $\endgroup$ – Peter Humphries Jun 6 at 22:40
  • $\begingroup$ Skimming that paper, it looks like they do a lot of things close to what I'm asking. But I think their explicit calculations are all done locally after expanding the integral to A^x, and what they do on the idele class group never involves a character. I'll keep looking to see if it's any help though, thanks for the source! $\endgroup$ – Jon Aycock Jun 6 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.