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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!

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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 $\chi^2=\omega$).

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  • $\begingroup$ 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

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