I am asking a question which looks very elementary to experts.

Let $F$ be a number field and $\mathbb{A}_F$ its adele ring. Let $\omega$ be a unitary central character of $GL_2(\mathbb{A}_F)$,

$X_{GL_2}=GL_2(F) \backslash GL_2(\mathbb{A}_F)$, and

$C^{\infty}_{\omega}(X_{GL_2})=\{f:X_{GL_2} \to \mathbb{C} \ | \ f(zg)=\omega(z)f(g) \text{ for $z\in$ center of $GL_2$ } \text{ and } \ g\mapsto f(xg) \text { is } C^{\infty}(F_{\infty}) \text{ for each } x\in X_{GL_2}\}$.

Define a projection map $P:C^{\infty}_{\omega}(X_{GL_2})\to C^{\infty}_{\omega}(X_{GL_2})$ by $$Pf(x):=\int_{h \in SL_2(F)\backslash SL_2(\mathbb{A}_F)}f(hx)\,dh.$$ Then how can we prove $$Pf(x)=\sum_{\chi^2=\omega} \chi(x) \cdot \left(\int_{PGL_2(F)\backslash PGL_2(\mathbb{A}_F)} f(y)\overline{\chi(y)}\,dy\right)?$$

Here, $\chi$ is a unitary character of $\mathbb{A}_F^{\times}/F^{\times}$, and $\chi^2=\omega$ if we think of $\chi$ as a character of $X_{GL_2}$ via $\chi(g)=\chi(\det g)$ for $g\in GL_2(\mathbb{A}_F)$.

Any comments will be highly appreciated!


It is clear from the definition and right-invariance of the measure $dh$ that $Pf$ is left-invariant under $SL_2(\mathbb{A}_F)$. As Paul Garrett kindly explained, $Pf$ is also left-invariant under $GL_2(F)$, because $f$ is left-invariant under $GL_2(F)$ and $GL_2(F)$ normalizes the coset space $SL_2(F)\backslash SL_2(\mathbb{A}_F)$. As a result, for any $g\in GL_2(\mathbb{A}_F)$, we have $$ Pf(g)=Pf\left(g\begin{pmatrix}\det g^{-1}&\\&1\end{pmatrix}\begin{pmatrix}\det g&\\&1\end{pmatrix}\right)=Pf\left(\begin{pmatrix}\det g&\\&1\end{pmatrix}\right).$$ This motivates the definition $$\widetilde{Pf}(a):=Pf\left(\begin{pmatrix}a&\\&1\end{pmatrix}\right),\qquad a\in\mathbb{A}_F^\times.$$ The function $\widetilde{Pf}$ is clearly invariant under $F^\times$. Moreover, it satisfies the identity $$\tag{$\ast$}\widetilde{Pf}(ab^2)=\omega(b)\widetilde{Pf}(a),\qquad a,b\in\mathbb{A}_F^\times,$$ since $$Pf\left(\begin{pmatrix}ab^2&\\&1\end{pmatrix}\right)=Pf\left(\begin{pmatrix}b&\\&b\end{pmatrix}\begin{pmatrix}a&\\&1\end{pmatrix}\right)=\omega(b)\,Pf\left(\begin{pmatrix}a&\\&1\end{pmatrix}\right).$$ Now let us fix a decomposition $\mathbb{A}_F^\times=\mathbb{R}_{>0}\times \mathbb{A}_F^1$ and focus on the compact quotient $\mathbb{A}_F^1/F^\times$. We have a spectral decomposition $$ \widetilde{Pf}(a)=\sum_\chi c_\chi \chi(a),\qquad a\in\mathbb{A}_F^1,$$ where $\chi$ runs through the (discrete set of) characters of $\mathbb{A}_F^1/F^\times$. Then ($\ast$) yields, by the uniqueness of the spectral coefficients, that $c_\chi\chi^2(b)=c_\chi\omega(b)$ holds for any $b\in\mathbb{A}_F^1$. That is, $c_\chi=0$ unless $\chi^2=\omega$ on $\mathbb{A}_F^1$. Any such $\chi$ extends uniquely to a character of $\mathbb{A}_F^\times/F^\times$ maintaining the property $\chi^2=\omega$, and one can verify, using ($\ast$) and existence of square-roots in $\mathbb{R}_{>0}$, that these extended idele class characters satisfy $$ \widetilde{Pf}(a)=\sum_{\chi^2=\omega} c_\chi \chi(a),\qquad a\in\mathbb{A}_F^\times.$$ The coefficients $c_\chi$ can be evaluated by the orthogonality of characters, \begin{align*} c_\chi&=\int_{\mathbb{A}_F^1/F^\times}\widetilde{Pf}(a)\,\overline{\chi(a)}\,da\\ &=\int_{\mathbb{A}_F^1/F^\times}\int_{h \in SL_2(F)\backslash SL_2(\mathbb{A}_F)}f\left(h\begin{pmatrix}a&\\&1\end{pmatrix}\right)\overline{\chi(a)}\,dh\,da\\ &=\int_{GL_2(F)\backslash GL_2^{(1)}(\mathbb{A}_F)}f(g)\,\overline{\chi(\det g)}\,dg, \end{align*} where $GL_2^{(1)}(\mathbb{A}_F)$ denotes the subgroup of adelic matrices whose determinant lies in $\mathbb{A}_F^1$. The result is immediate from here, since $f(g)\,\overline{\chi(\det g)}$ as a function on $GL_2(\mathbb{A}_F)$ is invariant under the center (recalling that $\chi^2=\omega$).

  • $\begingroup$ GH from MO, Thank you for great answer. It helped me a lot. But there are two points I did not understand well. 1. You wrote "$$ \widetilde{Pf}(a)=\sum_\chi c_\chi \chi(a),\qquad a\in\mathbb{A}_F^1,$$ where $\chi$ runs through the (discrete set of) characters of $\mathbb{A}_F^1/F^\times$." It looks that it comes from something like Fourier transfor on compact group. But since I don't know much about this, would recommend a book or paper which deals this material? $\endgroup$ – Monty Jan 17 '18 at 9:30
  • $\begingroup$ 2. I am wondering if $GL_2^{(1)}(\mathbb{A}_F)/Z(\mathbb{A}_F)$ is isomorphic to $PGL_2(\mathbb{A}_F)$? Because it looks that you are claiming the last integral over $GL_2$ you wrote is equal to the integral over $PGL_2$. Would you please shed me a light on these two points? I am always thank you very much! $\endgroup$ – Monty Jan 17 '18 at 9:30
  • $\begingroup$ @Monty: For the harmonic analysis used here, I recommend Weil: Basic number theory and Ramakrishnan-Valenza: Fourier analysis on number fields. Also, the group $GL_2^{(1)}(\mathbb{A}_F)/Z^{(1)}(\mathbb{A}_F)$ is isomorphic to $PGL_2(\mathbb{A}_F)$. To see this, consider the natural projection $GL_2(\mathbb{A}_F)\to PGL_2(\mathbb{A}_F)$, and restrict it to $GL_2^{(1)}(\mathbb{A}_F)$. This restriction is surjective, and its kernel is $Z^{(1)}(\mathbb{A}_F)$, done. $\endgroup$ – GH from MO Jan 17 '18 at 23:04
  • $\begingroup$ Thank you again for your reply! By your precious comment, I took a look on Ramakrishnan's book and found that the Fourier inversion formula is what you applied. But I have still three questions and would like to ask them to you if you don't mind. To apply fourier inversion formula, the function $\widetilde{Pf}$ should be of positive type according to Thm 3.9. But I can't check it. How did you check it? Or is there alternative condition to apply Fourier inversion formula, like smoothness? $\endgroup$ – Monty Jan 19 '18 at 7:11
  • $\begingroup$ Secondly, you wrote $\chi:\mathbb{A^{1}_F}/F^{\times} \to \mathbb{C}^{\times}$ can be uniquely lifted to the character of $\mathbb{A_F}^{\times}/F^{\times}$ satisfying $$ \widetilde{Pf}(a)=\sum_\chi c_\chi \chi(a),\qquad a\in\mathbb{A}_F^{\times}$$. I am wondering that how one can uniquely extend it to character of $\mathbb{A_F}^{\times}/F^{\times}$? $\endgroup$ – Monty Jan 19 '18 at 7:13

Your Answer

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

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