Proof. When \(F\) is nonarchimedean, this is proved in [12] and [1]. For archimedean \(F,\) this is computed in [1]. ◻

Proof. Since \(\mathrm{O}_{V_\ell}(F)\) acts transitively on \(X_\ell^\circ(F)\) and \(C^\infty(X_\ell(F)^1)\subset \mathcal{S}(X_\ell^\circ(F))\) is stable under \(N_{\ell+1}(F)\), it suffices to show \(C^\infty(X_\ell(F)^1)\) is an irreducible \(N_{\ell+1}(F)\rtimes \mathrm{O}_{V_\ell}(\mathcal{O})\)-module. Let \(0 \neq f \in C^\infty(X_{\ell}(F)^1)\) and let \(W\) be the vector space spanned by the \(N_{\ell+1}(F)\rtimes \mathrm{O}_{V_\ell}(\mathcal{O})\)-translates of \(f.\) Let \(x \in X_{\ell}(F)^1.\) The group \(\mathrm{O}_{V_\ell}(\mathcal{O})\) acts transitively on \(X_{\ell}(F)^1.\) Thus using the action of \(\mathrm{O}_{V_\ell}(\mathcal{O})\) there is an element \(f' \in W\) such that \(f'(x) \neq 0.\) By Fourier inversion, the functions \[\begin{align} \xi\mapsto \psi(v\xi^t),\quad \xi\in X_\ell(F)^1 \end{align}\] ranging over all \(v\in V_{\ell}(F)\) generate \(C^\infty(V_\ell(\mathcal{O}))|_{X_\ell(F)^1}\). Thus \(\mathbb{1}_{x+\varpi^n V_\ell(\mathcal{O})}|_{X_\ell^1(F)}\cdot f' \in W\) for all \(n.\) We deduce that \(W\) contains \(\mathbb{1}_{x+\varpi^nV_{\ell}(\mathcal{O})}\) for all \(n\) sufficiently large and all \(x.\) Thus \(W=C^\infty(X_\ell(F)^1).\) ◻

Proof. Let \(f\in \mathcal{S}(V_\ell(F)\oplus F^2)\) and \(h\in \mathrm{SL}_2(F).\) We have \[\begin{align} Z_\ell(r_\ell(h)f,s)&=\int_{N(F)\backslash \mathrm{SL}_2(F)} e^{H(gh^{-1})(2-\ell-s)}r_\ell(g)f(0_{\underline{2\ell}},0,1)d\dot{g}\\ &= \int_K \int_{F^\times}|t|^se^{H(kh^{-1})(2-\ell-s)}r_\ell(k)f(0_{\underline{2\ell}},0,t)dtdk\\ &=Z_\ell(f,s)+\int_K\int_{F^\times} |t|^s\left(e^{H(kh^{-1})(2-\ell-s)}-1\right)r_\ell(k)f(0_{\underline{2\ell}},0,t) dtdk. \end{align}\] Note that the integrand is zero when \(s=2-\ell.\) Therefore, the latter term is holomorphic at \(s=2-\ell,\) and it vanishes at \(s=2-\ell\) if \(\zeta(s)\) is holomorphic at \(s=2-\ell\). This implies \(\tilde{c}_\ell(f)=\tilde{c}_\ell(r_\ell(h)f).\) ◻

Proof. Let \(f\in \mathcal{S}(V_{\ell+1}(F)).\) We may assume \(f\) is \(r_\ell(K)\)-invariant. Then for \(x\in X_{\ell-1}^\circ(F),\) \[\begin{align} I(\tilde{d}_\ell(f))(x)=\int_{F^\times}\left(\int_{F} |t|^{\ell-2} f(tx,0,ty,0,0)\psi(y)dy\right) d^\times t. \end{align}\] For \(a\in F\) consider the function \[\begin{align} f_{x,a}(t):=\int_{F} |t|^{\ell-2} f(tx,0,ty,0,t^{-1}a)\psi(y)dy. \end{align}\] It is continuous in \(a,\) so \[I(\tilde{d}_\ell(f))(x)=\int_{F^\times}\lim_{|a|\to 0} f_{x,a}(t)d^\times t.\] We claim the limit and the integral can be interchanged. Assuming the claim, we have \[\begin{align} I(\tilde{d}_\ell(f))(x)&=\lim_{|a|\to 0}\int_{F^\times}f_{x,a}(t)d^\times t =\lim_{|a|\to 0}\int_{F^\times}\int_{F} |t|^{\ell-2} f(tx,0,ty,0,t^{-1}a)\psi(y)dyd^\times t. \end{align}\] For \(a\neq 0,\) since \(f\) is Schwartz, the integral over \(y\) and \(t\) are absolutely convergent. Thus by the Fubini-Tonelli theorem and a change of variables \(t\mapsto ta,\) it equals \[\begin{align} &\lim_{|a|\to 0}\int_{F}\int_{F^\times} |t|^{\ell-2} f(tx,0,ty,0,t^{-1}a)\psi(y)d^\times tdy=\lim_{|a|\to 0}\int_{F} |a|^{\ell-2}I(f)(ax,0,ay) \psi(y)dy. \end{align}\]

To justify the claim, observe that \[\begin{align} |f_{x,a}(t)|\le |t|^{\ell-3}f'(tx,t^{-1}) \end{align}\] for some \(f'\in \mathcal{S}(F^{2\ell-1})\) depending only on \(f\). Since \(t\mapsto |t|^{\ell-3}f'(tx,t^{-1})\) is a function in \(L^1(F^\times, d^\times t),\) the claim follows from the Lebesgue’s dominated convergence theorem. ◻

Proof. Consider \[\begin{align} W:=\mathcal{S}(V_\ell(F)-\{0\})\otimes \mathcal{S}(F^2-\{0\}). \end{align}\] Note that \(I(f)(0)=0\) for \(f\in W.\) The group \(\mathrm{O}_{V_\ell}(F)\) stabilizes \(W\) and acts transitively on \(X_\ell^\circ(F).\) Therefore, if \(I(\widetilde{\mathcal{F}}_{X_\ell}(f))(x_0)=0\) for all \(f\in W,\) then \(I(\widetilde{\mathcal{F}}_{X_\ell}(f))=0\) for all \(f\in W\) by Proposition 5. However, \(\mathcal{F}_{\wedge}(\mathcal{S}(F^2-\{0\}))\) has codimension \(1\) in \(\mathcal{S}(F^2),\) so one can easily find \(f\in W\) such that \(I(\widetilde{\mathcal{F}}_{X_\ell}(f))\neq 0.\) ◻

Proof. Suppose \(\widetilde{\mathcal{F}}_{X_\ell}(\ker I) \leq \ker I.\) Since \(\widetilde{\mathcal{F}}_{X_\ell}\) has order \(2,\) one has \(\widetilde{\mathcal{F}}_{X_\ell}(\ker I)=\ker I\) and the lemma follows. Thus it suffices to show if \(f\in\mathcal{S}(V_{\ell+1}(F))\) satisfies \(I(f)=0\) then \(I(\widetilde{\mathcal{F}}_{X_\ell}(f))=0.\)

Let \(E\) be a number field such that \(E_v=F\) for some place \(v\). Let \(f_v\in \mathcal{S}(V_{\ell+1}(F))\) such that \(I(f_v)=0\). Choose two finite places \(v_1, v_2\) distinct from \(v\) and functions \(f_{v_i} \in \mathcal{S}(V_{\ell+1}(E_{v_i}))\) such that \(f_{v_1}\in W_\ell(E_{v_1}),\) \(\widetilde{\mathcal{F}}_{X_\ell}(f_{v_2}) \in W_{\ell}(E_{v_2})\) and \[I(f_{v_1})(0)=I(\widetilde{\mathcal{F}}_{X_\ell}(f_{v_2}))(0)=0.\] Choose \(f^{vv_1v_2} \in \mathcal{S}(V_{\ell+1}(\mathbb{A}_E^{vv_1v_2}))\). By the fact that \(\tilde{d}_\ell(W_\ell(E_{v_i}))=0\) and Lemma 12 below, terms in Theorem 9 involving \(\tilde{c}_i, \tilde{d}_{\ell,i}\) vanish, and we have \[\begin{align} \label{eq:sumvanish} \sum_{\xi\in X_\ell^\circ(E)} I(\widetilde{\mathcal{F}}_{X_\ell}(f_{v}f_{v_1}f_{v_2}f^{vv_1v_2}))(\xi)=\sum_{\xi\in X_\ell^\circ(E)} I(f_{v}f_{v_1}f_{v_2}f^{vv_1v_2})(\xi)=0. \end{align}\tag{8}\] Let \(\xi_0 \in X_\ell^\circ(E).\) By Lemma 10 we can choose \(f_{v_1}\) as above so that \(I(\widetilde{\mathcal{F}}_{X_\ell}(f_{v_1}))(\xi_0)\neq 0.\) Since \(X_\ell(E)\) is discrete in \(X_\ell(\mathbb{A}_E)\), we may choose \(f_{v_2}\) and \(f^{vv_1v_2}\) such that 8 equals to \(I(\widetilde{\mathcal{F}}_{X_\ell}(f_v))(\xi_0)\). Thus \(I(\widetilde{\mathcal{F}}_{X_\ell}(f_{v}))\) vanishes on \(X_\ell^\circ(E).\) As \(X_\ell^\circ(E)\) is dense in \(X_\ell^\circ(F),\) the assertion follows by continuity. ◻

Proof. Let \(f\in \mathcal{S}(V_{\ell+1}(F))\). Note that \(I\) and \(Z_{\ell}\) are by definition \(r_\ell(\mathrm{SL}_2(\mathcal{O}))\)-invariant, so we may assume \(f\) is \(r_\ell(\mathrm{SL}_2(\mathcal{O}))\)-invariant. By the Iwasawa decomposition, we have for \(x\in X_\ell^\circ(F)\) \[\begin{align} I(f)(x)&=\int_{F^\times} |t|^{2-\ell} f(t^{-1}x,0,t)d^\times t, \end{align}\] and \[\begin{align} Z_\ell(f,s)=\int_{F^\times} f(0_{\underline{2\ell}},0,t)|t|^sd^\times t. \end{align}\] In the computation of the integrals above, \(\psi\) only affects the value \(\mathrm{vol}(\mathcal{O}^\times)\) in the intermediate computation, but not the conclusion. Therefore, for simplicity we assume \(\psi\) is unramified and thus \(\mathrm{vol}(\mathcal{O}^\times)=1\).

Choose \(A\in \mathbb{Z}_{\ge 0}\) such that \[\begin{align} f(v+w)=f(v),\quad &\forall\, v\in V_{\ell+1}(F), w\in \varpi^A V_{\ell+1}(\mathcal{O}),\\ f(v)=0, \quad &\forall v\notin \varpi^{-A}V_{\ell+1}(\mathcal{O}). \end{align}\] Let \(x\in X_\ell(F)^1\). For any \(n\ge 2A-1\) \[\begin{align} \label{eq:exp} \begin{aligned} I(f)(\varpi^n x)&=\sum_{j=-A}^{\infty} q^{j(\ell-2)}f(\varpi^{n-j}x,0,\varpi^j)\\ &=\sum_{j=-A}^{A-1} q^{j(\ell-2)}f(0_{\underline{2\ell}},0,\varpi^j)+\sum_{j=A}^{n-A} q^{j(\ell-2)}f(0)+\sum_{j=n-A+1}^{n+A} q^{j(\ell-2)}f(\varpi^{n-j}x,0,0). \end{aligned} \end{align}\tag{9}\] For \(\ell\neq 2,\) this is \[\begin{align} &\sum_{j=-A}^{A-1} q^{j(\ell-2)}f(0_{\underline{2\ell}},0,\varpi^j)+\frac{q^{A(\ell-2)}}{1-q^{\ell-2}}f(0)-\frac{q^{(n-A+1)(\ell-2)}}{1-q^{\ell-2}}f(0)\\ &+q^{(n-A+1)(\ell-2)}\sum_{j=0}^{2A-1} q^{j(\ell-2)}f(\varpi^{A-j-1}x,0,0). \end{align}\] Suppose \(\ell\ge 3\) and \(f\in W_\ell'\). By taking \(n\) large we have \[\begin{align} I(f)(0)=I(f)(\varpi^n x)=\frac{q^{A(\ell-2)}}{1-q^{\ell-2}}f(0)+\sum_{j=-A}^{A-1} q^{j(\ell-2)}f(0_{\underline{2\ell}},0,\varpi^j). \end{align}\] On the other hand, since \(\zeta(s)\) is holomorphic at \(s=2-\ell,\) \[\begin{align} \tilde{c}_\ell(f)=Z_{\ell}(f,2-\ell)&=\int_{F^\times}f(0_{\underline{2\ell}},0,a)|a|^s d^\times a \bigg |_{s=2-\ell}\\ &=\int_{|a|\le q^{-A}}f(0)|a|^s d^\times a \bigg |_{s=2-\ell}+\sum_{j=-A}^{A-1} q^{j(\ell-2)}f(0_{\underline{2\ell}},0,\varpi^j)\\ &=\frac{q^{A(\ell-2)}}{1-q^{\ell-2}}f(0)+\sum_{j=-A}^{A-1} q^{j(\ell-2)}f(0_{\underline{2\ell}},0,\varpi^j)\\ &=I(f)(0). \end{align}\] The same argument applies when \(\ell=1\). This proves (i) and (iii).

Now assume \(\ell=2\). Then \[\begin{align} Z_2(f,s)=q^{-As}\zeta(s)f(0)+\sum_{j=-A}^{A-1} q^{-sj}f(0_{\underline{4}},0,\varpi^j), \end{align}\] so \(\tilde{c}_2(f)=f(0).\) We can write 9 as \[\begin{align} \label{eq:forell612} \begin{aligned} I(f)(\varpi^nx)&=n\tilde{c}_2(f)+(-2A+1)f(0)+\sum_{j=-A}^{A-1} f(0_{\underline{4}},0,\varpi^j)+\sum_{j=-A+1}^A f(\varpi^{-j}x,0,0)\\ &=:n\tilde{c}_2(f)+\tilde{a}_2(f)(x). \end{aligned} \end{align}\tag{10}\] Since both \(I(f)\) and \(\tilde{c}_2(f)\) do not depend on the choice of \(A\), \(\tilde{a}_2(f)\) is well-defined. If \(I(f)=0,\) then clearly both \(\tilde{c}_2(f)\) and \(\tilde{a}_2(f)\) are zero. This completes the proof. ◻

Proof. For the first statement, we can assume \(\psi\) is unramified. Let \[\begin{align} \mathcal{S}:=\mathcal{S}(X_\ell^\circ(F))+\mathcal{F}\left(\mathcal{S}(X_\ell^\circ(F))\right) <L^2(X_{\ell}(F)). \end{align}\] By the proof of [12] \(\mathcal{S}\) is a smooth \(\mathrm{O}_{V_{\ell+1}}(F)\)-subrepresentation of \(\sigma'_\ell,\) and it is irreducible by Lemma 4. We claim \(\mathcal{S}\) contains \(b_\ell.\) Assume the claim for the moment. Since \(b_{\ell}=I(\mathbb{1}_{V_{\ell+1}(\mathcal{O})}),\) \(b_\ell\) is spherical with respect to both \(\sigma_\ell\) and \(\sigma'_\ell.\) Using the Iwasawa decomposition and the fact that unramified irreducible smooth representations are determined by spherical vectors, we deduce that \(\mathcal{S}=\mathcal{S}(X_\ell(F))\) as \(\mathrm{O}_{V_{\ell+1}}(F)\)-representations. In particular, \(\mathcal{S}_{\mathrm{ES}}(X_\ell(F))\le \mathcal{S}\) for \(\ell\ge 3\) by Theorem 3. This is also true for \(\ell=2\) by looking at the asymptotics of \(b_2\) (cf. [1]). It follows that \[\begin{align} \mathcal{S}=I(W_\ell)+\mathcal{F}(I(W_\ell))=I(W_\ell)+I(\widetilde{\mathcal{F}}_{X_{\ell}}(W_\ell))=I(\widetilde{\mathcal{S}}(X_{\ell}(F))). \end{align}\] Therefore, the map \(I\) is an \(\mathrm{O}_{V_{\ell+1}}(F)\)-equivariant surjection onto \(\mathcal{S}(X_\ell(F))\). In particular, \(\mathcal{F}=\mathcal{F}_{X_\ell}\). To see \(b_\ell\in \mathcal{S}\), it suffices to show the existence of functions \(f_1,f_2\in \mathcal{S}(F^2)\) satisfying the following: \[\begin{align} f_1+\mathcal{F}_\wedge(f_2)=\mathbb{1}_{\mathcal{O}^2}, \textrm{ and } f_i(0)=0, \,\,I(\mathbb{1}_{V_\ell(\mathcal{O})}\otimes f_i)(0)=0, \,\,\textrm{ for } i=1,2. \end{align}\] Here \(I(\mathbb{1}_{V_{\ell}(\mathcal{O})} \otimes f_i)(0)\) is defined because \(f_i(0)=0.\) One can find such a pair in the subspace \(\langle \mathbb{1}_{\varpi^j\mathcal{O}^2}: j \in \mathbb{Z}\rangle\subset\mathcal{S}(F^2),\) for example.

For the second statement, suppose \(\ell\ge 3\) and \(\psi\) is arbitrary. By Lemma 12(i) \(\tilde{c}_\ell\) induces an \(\mathrm{O}_{V_\ell}(F)\)-equivariant linear functional \[\begin{align} \tilde{c}_\ell':\widetilde{\mathcal{S}}(X_\ell(F))/I^{-1}(\mathcal{S}(X_\ell^\circ(F)))\tilde{\,\longrightarrow\,}\mathcal{S}(X_\ell(F))/\mathcal{S}(X_\ell^\circ(F))\,\longrightarrow\,\mathbb{C}. \end{align}\] Since \(\mathcal{S}(X_\ell(F))/\mathcal{S}(X_\ell^\circ(F))\cong \mathbb{C}\oplus \mathcal{S}(X_{\ell-1}(F))\) by Theorem 3, and \(\tilde{c}_\ell'\) and \(c_\ell\) agree on \(\mathbb{C}\) by Lemma 12(i), we conclude \(c_\ell\circ I=I\circ \tilde{c}_\ell\).

We are left to show \(\varepsilon(\ell-2,\psi)\, d_\ell\circ I=I\circ \tilde{d}_\ell\). By Theorem 3 \(d_{\ell}\circ I\) induces an \(\mathrm{O}_{V_\ell}(F)\)-equivariant isomorphism \[\begin{align} \iota:\widetilde{\mathcal{S}}(X_\ell(F))/W'_\ell \tilde{\,\longrightarrow\,} \mathcal{S}(X_\ell(F))/\mathcal{S}_{\mathrm{ES}}(X_\ell(F))\tilde{\,\longrightarrow\,}\mathcal{S}(X_{\ell-1}(F)). \end{align}\] Thus by Corollary 8 we have an \(\mathrm{O}_{V_\ell}(F)\)-equivariant surjection \[\begin{align} J:\mathcal{S}(X_{\ell-1}(F))\xrightarrow{\iota^{-1}} \widetilde{\mathcal{S}}(X_\ell(F))/W'_\ell \xrightarrow{\,\,\tilde{d}_\ell\,\,} \widetilde{\mathcal{S}}(X_{\ell-1}(F))/\tilde{d}_\ell(W'_\ell)\xrightarrow{\,\,I\,\,} \mathcal{S}(X_{\ell-1}(F)). \end{align}\] As \(\mathcal{S}(X_{\ell-1}(F))\) is an irreducible \(\mathrm{O}_{V_\ell}(F)\)-representation, we conclude that there is a nonzero constant \(C_\psi\) such that \(C_\psi d_\ell\circ I=I\circ \tilde{d}_\ell.\) When \(\psi\) is unramified, \(C_\psi=1\) because \(J(b_{\ell-1})=b_{\ell-1}\) by our choice of \(d_\ell\).

For general \(\psi,\) let \(f\in \mathcal{S}(V_{\ell+1}(F))\) be invariant under \(r_\ell(\mathrm{SL}_2(\mathcal{O})).\) Let us compute \[\begin{align} \tilde{c}_{\ell-1}\circ \tilde{d_\ell}(f)=c_{\ell-1}\circ I\circ \tilde{d}_\ell(f)=C_\psi c_{\ell-1}\circ d_\ell\circ I(f). \end{align}\] By 1 \[\begin{align} Z_{\ell-1}(\tilde{d_\ell}(f),s)&=\int_{F^\times} |t|^s\mathcal{F}_2(f)(\underline{0}_{2\ell-2},0,t,0,0) d^\times t\\ &=\frac{\epsilon(1-s,\psi)\zeta(s)}{\zeta(1-s)}\int_{F^\times} |t|^{1-s} f(\underline{0}_{2\ell-2},0,t,0,0)d^\times t. \end{align}\] For \(\mathrm{Re}(s)<1,\) by Lebesgue’s dominated convergence theorem \[\begin{align} \int_{F^\times} |t|^{1-s} f(\underline{0}_{2\ell-2},0,t,0,0)d^\times t&=\lim_{|a|\to 0}\int_{F^\times} |t|^{1-s} f(\underline{0}_{2\ell-2},0,t,0,t^{-1}a)d^\times t\\ &=\lim_{|a|\to 0} |a|^{1-s}\int_{F^\times} |t|^{1-s} f(\underline{0}_{2\ell-2},0,ta,0,t^{-1})d^\times t. \end{align}\] Therefore, \[\begin{align} \tilde{c}_{\ell-1}\circ \tilde{d_\ell}(f)=\varepsilon(\ell-2,\psi)\lim_{|a|\to 0} |a|^{\ell-2}I(f)(\underline{0}_{2\ell-2},0,a)\begin{cases} \frac{\zeta(3-\ell)}{\zeta(\ell-2)} & \textrm{if } \ell>3,\\ \zeta(1)^{-1} & \textrm{if }\ell=3. \end{cases} \end{align}\] By comparing terms, we have \(C_\psi=\varepsilon(\ell-2,\psi).\) ◻

Proof. For the first assertion, we may assume \(\psi\) is unramified. Assume \(f\in \mathcal{S}(V_2(F))\) is \(r_1(\mathrm{SL}_2(\mathcal{O}))\)-invariant. Let \(x\in X_1(F)^1.\) Using the notation in the proof of Lemma 12, for \(n\) large by 9 \[\begin{align} \label{eq:2comp} \begin{aligned} I(f)(\varpi^n x)=&I(f)(0)-q^{-n}\Big(\frac{q^{A-1}}{1-q^{-1}}f(0)-q^{A-1}\sum_{j=0}^{2A-1} q^{-j}f(\varpi^{A-j-1}x,0,0)\Big). \end{aligned} \end{align}\tag{11}\] Since \(f\) is \(r_1(\mathrm{SL}_2(\mathcal{O}))\)-invariant, by 11 we have \[\begin{align} I(f)(\varpi^nx)=\begin{cases} I(f)(\varpi^n(1,0)) &\textrm{ if } x=(a,0), a\in \mathcal{O}^\times,\\ I(f)(\varpi^n(0,1))& \textrm{ if } x=(0,a), a\in \mathcal{O}^\times. \end{cases} \end{align}\] Define \[\begin{align} (d_1',d_1)(I(f))&:=\frac{q^n}{\zeta(-1)}\big(I(f)(\varpi^n(1,0))-I(f)(0), I(f)(\varpi^n(0,1))-I(f)(0)\big). \end{align}\] This is well-defined by 11 . We define \(c_1(I(f)):=\frac{I(f)(0)}{\zeta(1)}\) so \(c_1(I(f))=\tilde{c}_1(f)\) by Lemma 12(iii). The first assertion follows by definition.

For the second assertion, by Corollary 8 we have an \(\mathrm{SO}_{V_1}(F)\)-equivariant surjection \[\begin{align} J:\mathbb{C}^2 \tilde{\,\longrightarrow\,} \mathcal{S}(X_{1}(F))/\mathcal{S}_{\mathrm{ES}}(X_{1}(F))\tilde{\,\longrightarrow\,} \widetilde{\mathcal{S}}(X_1(F))/W_1'\xrightarrow{\tilde{d}_1} \widetilde{\mathcal{S}}(X_0(F))/\tilde{d}_1(W_1')\xrightarrow{\,\,I\,\,} \mathbb{C}. \end{align}\] Since \(\tilde{d}_1\circ \tilde{\sigma}_1{ \left(\begin{smallmatrix} a & \\ & a^{-1} \end{smallmatrix}\right)} =|a|^{-1}\tilde{d}_1,\) we have \(I\circ \tilde{d}_1=C_\psi d_1\circ I\) for some nonzero constant \(C_\psi\). When \(\psi\) is unramified, we have \(\zeta(2)d_1(b_1)=\zeta(2)=I(\tilde{d}_1(\mathbb{1}_{V_2(\mathcal{O})})),\) so \(C_\psi=\zeta(2)\). By Lemma 7, for \(f\in \widetilde{\mathcal{S}}(X_1(F))\) \[\begin{align} I(\tilde{d}_1(f))=I(\tilde{d}_1(f))(0)&=\lim_{|a|\to 0}\int_{F} |a|^{-1}I(f)(0,ay)\psi(y)dy. \end{align}\] Let \(z\in F^\times\) such that \(\psi_u(y):=\psi(zy)\) is unramified. Let \(d_u y\) be the self-dual Haar measure on \(F\) with respect to \(\psi_u\). By changing variables \(y\mapsto zy\) and \(a\mapsto az^{-1}\), we have \[\begin{align} I(\tilde{d}_1(f))=|z|^{3/2}\lim_{|a|\to 0}\int_{F} |a|^{-1}I(f)(0,ay)\psi_u(y)d_u y. \end{align}\] Note that \(|z|^{3/2}=\varepsilon(-1,\psi)\). By comparing terms \(C_\psi=\varepsilon(-1,\psi)\zeta(2).\)

Finally, for \(f\in \widetilde{\mathcal{S}}(X_1(F))\) by Fourier inversion \[\begin{align} \tilde{d}_1(f)(0)&=\int_{F} f(0,y,0,0)dy\\ &=\int_{F^2}\left(\int_{F} f(t,y,0,0)\psi(xt)\right)dxdy\\ &=\int_{F^2}\left(\int_{F} \left(\tilde{\sigma}_1{ \left(\begin{smallmatrix} 0 & 1\\ 1 & 0 \end{smallmatrix}\right)} f\right)(y,t,0,0)\psi(xt)\right)dxdy\\ &=\int_{F^2}\tilde{d}_1\circ \tilde{\sigma}_1{ \left(\begin{smallmatrix} 0 & 1\\ 1 & 0 \end{smallmatrix}\right)} (f)(y,x)dxdy \end{align}\] By our choice of Haar measures, this is \[\begin{align} &\frac{1}{\zeta(2)\varepsilon(0,\psi)}\int_{N(F)\backslash \mathrm{SL}_2(F)}\tilde{d}_1\circ \tilde{\sigma}_1{ \left(\begin{smallmatrix} 0 & 1\\ 1 & 0 \end{smallmatrix}\right)} (f)((0,1)g)d\dot{g}\\ &=\frac{1}{\zeta(2)\varepsilon(0,\psi)}I\circ\tilde{d}_1\circ \tilde{\sigma}_1{ \left(\begin{smallmatrix} 0 & 1\\ 1 & 0 \end{smallmatrix}\right)} (f)\\ &=\frac{\varepsilon(-1,\psi)}{\varepsilon(0,\psi)}d_{1}\circ I\circ \tilde{\sigma}_1{ \left(\begin{smallmatrix} 0 & 1\\ 1 & 0 \end{smallmatrix}\right)} (f)\\ &=\varepsilon(-1/2,\psi)d_{1}'\circ I(f). \end{align}\] ◻

Proof. By Proposition 13, \(\mathcal{S}(X_2(F))=I(\widetilde{\mathcal{S}}(X_2(F)).\) Taking germs at the origin of functions in \(\mathcal{S}(X_{2}(F))\) gives rise to an exact sequence of smooth \(\mathrm{SL}_2^2(F)\)-representations \[\begin{align} \label{eq:2exact} 0\,\longrightarrow\,\mathcal{S}(X_{2}^\circ(F))\,\longrightarrow\,\mathcal{S}(X_{2}(F))&\longrightarrow \pi''\,\longrightarrow\,0. \end{align}\tag{12}\] By Lemma 12(ii), the map to \(\pi''\) is given explicitly by \[I(f) \longmapsto \tilde{a}_2(f)(\cdot)-\tilde{c}_2(f) \log_{q}(|\cdot|).\] Note that both \(\tilde{a}_2(f)\) and \(\tilde{c}_2(f)\) only depend on \(I(f),\) so this map is well-defined.

Since \(\tilde{a}_2(f) \in C^\infty(\mathbb{P}X_2(F))=\mathrm{Ind}_{B}^{\mathrm{SL}_2}(1_{-1})\otimes \mathrm{Ind}_{B}^{\mathrm{SL}_2}(1_{-1}),\) one has an exact sequence of smooth \(\mathrm{SL}_2^2(F)\)-representations \[\begin{align} \label{no:split} \begin{aligned} 0\,\longrightarrow\,\pi'\,\longrightarrow\,\pi''&\,\longrightarrow\,\mathbb{C}\,\longrightarrow\,0 \end{aligned} \end{align}\tag{13}\] where \(\pi'\) is a subrepresentation of \(\mathrm{Ind}_{B}^{\mathrm{SL}_2}(1_{-1})\otimes \mathrm{Ind}_{B}^{\mathrm{SL}_2}(1_{-1}).\) Here the second map is determined by the requirement that \(-\tilde{c}_2(f)\log_{q}(|\cdot|)\) is sent to \(\tilde{c}_2(f).\)

To see \(\mathbb{C}\otimes \mathbb{C}, \mathrm{St}\otimes \mathbb{C}, \mathbb{C}\otimes \mathrm{St}\) are composition factors of \(\pi'\), by symmetry and Lemma 15 it suffices to show there is \(f\in \mathcal{S}(V_{3}(F))\) such that \(\tilde{a}_2(f)\) is nonconstant and \(\tilde{c}_2(f)=0.\) We may assume \(\psi\) is unramified. Consider \(\tilde{b}_2':= \mathbb{1}_{\varpi \mathcal{O}\times \varpi^{-1}\mathcal{O}}\otimes \mathbb{1}_{\mathcal{O}^4}\). It is \(r_{2}(\mathrm{SL}_2(\mathcal{O}))\)-invariant. Then by 10 \(\tilde{a}_2(\mathbb{1}_{V_3(\mathcal{O})}-\tilde{b}_2')\) is nonconstant and \(\tilde{c}_2(\mathbb{1}_{V_3(\mathcal{O})}-\tilde{b}_2')=1-1=0\).

Let \(\pi:=\pi''/(\mathbb{C}\otimes \mathbb{C})\). To complete the proof of (i) we are left with showing that \(\pi'\) is a proper subrepresentation of \(\mathrm{Ind}_{B}^{\mathrm{SL}_2}(1_{-1})\otimes \mathrm{Ind}_{B}^{\mathrm{SL}_2}(1_{-1})\). For \(f\in \mathcal{S}(X_{2}(F))\), let \[\begin{align} f_s(x):&=\int_{F^\times}|a|^{s+1}f(ax) d^\times a,\quad x\in X_{2}^\circ(F) \end{align}\] be the Mellin transform of \(f\) along the quasicharacter \(1_{s}\otimes 1_{s}\) (extended by meromorphic continuation). We claim that \(f\in \mathcal{S}(X_{2}(F))\) only if \(M_{w_0}(s)\otimes M_{w_0}(s)(f_s)\) is holomorphic at \(s=-1\). Let \(\tilde{f}\in \widetilde{\mathcal{S}}(X_2(F))\) such that \(I(\tilde{f})=f.\) For \(x\in X_{2}(F)^1,\) the Laurent series expansion of \(f_s(x)\) at \(s=-1\) is \[\begin{align} \tilde{c}_2(\tilde{f})\zeta(s+1)^2+(\tilde{a}_2(\tilde{f})(x)-\tilde{c}_2(\tilde{f}))\zeta(s+1)+O(1). \end{align}\] Note that \(\tilde{a}_2(\mathbb{1}_{V_3(\mathcal{O})})=1\) is a constant function, and thus \(M_{w_0}(s)\otimes M_{w_0}(s)((b_2)_s)\) is holomorphic at \(s=-1\) by Lemma 15. By Lemma 4 and the Iwasawa decomposition \(\pi''\cong \mathcal{S}(X_{2}(F))/\mathcal{S}(X_{2}^\circ(F))\) is generated by (the image of) \(b_2\) as an \(\mathrm{\mathrm{SL}}^2_{2}(F)\)-representation, so the claim follows. Now if \(\pi'\) is not proper, then by Lemma 15 there is \(\tilde{f}\in \widetilde{\mathcal{S}}(X_{2}(F))\) such that \(M_{w_0}(-1)\otimes M_{w_0}(-1)(\tilde{a}_2(\tilde{f}))\) is nonzero, and thus \(M_{w_0}(s)\otimes M_{w_0}(s)(I(f)_s)\) has a simple pole at \(s=-1\), which is a contradiction.

Finally, to prove (ii) by Corollary 8 we have a \(P_{2}(F)\)-equivariant surjection \[\begin{align} \tilde{J}:\pi\cong \mathcal{S}(X_{2}(F))/\mathcal{S}_{\mathrm{ES}}(X_{2}(F))\cong \widetilde{\mathcal{S}}(X_{2}(F))/W_2'\xrightarrow{\,\,\tilde{d}_2\,\,} \widetilde{\mathcal{S}}(X_{1}(F))/\tilde{d}_2(W_2')\xrightarrow{\,\, I\,\,} \mathcal{S}(X_{1}(F)). \end{align}\] Since \(\pi\) as a \(P_2(F)\)-representation is of length \(3\) whose maximal semisimple quotients are finite-dimensional, \(\tilde{J}\) must be injective and hence an isomorphism. Let \(J:\pi\tilde{\longrightarrow} \mathcal{S}(X_1)\) be the unique isomorphism such that \(J(d_2(b_2))=b_1\). When \(\psi\) is unramified, \(J=\tilde{J}.\) For general \(\psi,\) by Lemma 7 for \(f\in \widetilde{\mathcal{S}}(X_{2}(F))\) and \(x\in X_{1}^\circ(F),\) \[\begin{align} I(\tilde{d}_2(f))(x)=\lim_{|a|\to 0} \int_{F} I(f)(ax,0,ay) \psi(y)dy. \end{align}\] Say \(x=(0,x_1)\). Then we have \[\begin{align} I(\tilde{d}_2(f))(0,x_1)&=\lim_{|a|\to 0} \int_{F} I(f)(0,a,0,ax_1^{-1}y) \psi(y)dy\\ &=|x_1|\lim_{|a|\to 0} \int_{F} I(f)(0,a,0,ay) \psi(x_1y)dy. \end{align}\] Since \(I(\tilde{d}_2(f))\in \mathcal{S}(X_1(F)),\) by Proposition 14 the above function in \(x_1\) is a function in \(\mathcal{S}(F)+|\cdot| \mathcal{S}(F)\). Let \(z\in F^\times\) such that \(\psi_u(y)=\psi(zy)\) is unramified. Let \(d_u y\) be the self-dual Haar measure on \(F\) with respect to \(\psi_u\). Then \[\begin{align} &\lim_{|x_1|\to 0}I(\tilde{d}_2(f))(0,x_1)\\ &=|z|^{-1/2}\lim_{|x_1|\to 0}|x_1|\lim_{|a|\to 0} \int_{F} I(f)(0,a,0,ay) \psi_u(x_1z^{-1}y)d_uy\\ &=|z|^{1/2}\lim_{|x_1|\to 0}|x_1|\lim_{|a|\to 0} \int_{F} I(f)(0,a,0,ay) \psi_u(x_1y)d_uy. \end{align}\] By comparing terms, we have \(\tilde{J}=|z|^{1/2}J=\varepsilon(0,\psi)J.\) ◻

Proof. By the computation in the proposition above, \(c_2\) is equal to \(c_1\circ d_2\) up to a nonzero scalar. Thus, it suffices to check the identity for \(f=b_2\). The statement is independent of \(\psi,\) so we can assume \(\psi\) is unramified. By Proposition 14 and Proposition 16 we have \[\begin{align} c_2(b_2)=1=c_1(b_1)=c_1(d_2(b_2)). \end{align}\] ◻

Proof. We may assume \(f\) is \(r_2(\mathrm{SL}_2(\mathcal{O}))\)-invariant. Using the computation in Lemma 12 and the notation therein, we have \[\begin{align} \label{eq:2asy} \begin{aligned} \zeta(s)^{-1}Z_{2}(f,s)&=\mathrm{vol}(\mathcal{O}^\times)\left(q^{-As}f(0)+(1-q^{-s})\sum_{j=-A}^{A-1}q^{-js}f(0_{\underline{4}},0,\varpi^j)\right)\\ &=\mathrm{vol}(\mathcal{O}^\times)f(0)-s\mathrm{vol}(\mathcal{O}^\times)(\log q)\big(Af(0)-\sum_{j=-A}^{A-1}f(0_{\underline{4}},0,\varpi^j)\big)+O_{f}(s^2) \end{aligned} \end{align}\tag{15}\] as \(s \to 0.\) On the other hand, by 1 \[\begin{align} Z_1(\tilde{d}_2(f),s+1)=\varepsilon(-s,\psi)\frac{\zeta(1+s)}{\zeta(-s)}\int_{F^\times} f(0_{\underline{2}},0,t,0,0)|t|^{-s}d^\times t, \end{align}\] so \[\begin{align} \frac{Z_{1}(\tilde{d}_2(f), s+1)}{\varepsilon(-s,\psi)\zeta(s+1)}&=(1-q^s)\int_{|t|> q^{-A}} f(\underline{0}_2,0,t,0,0)|t|^{-s} d^\times t+\mathrm{vol}(\mathcal{O}^\times)f(0)q^{As}. \end{align}\] The coefficient of \(s\) in the Taylor series expansion of this function at \(s=0\) is \[\begin{align} \mathrm{vol}(\mathcal{O}^\times)(\log q)Af(0)-\log q \int_{|t|>q^{-A}} f(0_{\underline{2}},0,t,0,0)d^\times t. \end{align}\] Consequently, by 10 and Proposition 16 \[\begin{align} &\frac{d}{ds} \left(-\frac{Z_{2}(f,s)}{\zeta(s)}+\frac{Z_{1}(\tilde{d}_2(f), s+1)}{\varepsilon(-s,\psi)\zeta(s+1)}\right)\Bigg|_{s=0}\\ &=\mathrm{vol}(\mathcal{O}^\times)(\log q)\big(Af(0)-\sum_{j=-A}^{A-1}f(0_{\underline{4}},0,\varpi^j)\big)\\ &+\mathrm{vol}(\mathcal{O}^\times)(\log q)(Af(0)-\sum_{j=-A+1}^{A} f(0_{\underline{2}},0,\varpi^{-j},0,0))\\ &=(\log q)(\tilde{c}_2(f)-\tilde{a}_2(f)(0:0:0:1))\\ &=(\log q)(c_2(I(f))-a_2(I(f))). \end{align}\] ◻

Proof. Let \(f\in \mathcal{S}(V_{\ell+1}(F)).\) We may assume \(f\) is \(r_\ell(K)\)-invariant. Therefore, for \(x\in X_\ell^\circ(F)\) \[\begin{align} I(f)(x)&=\int_{F^\times} |t|^{2-\ell} f(t^{-1}x,0,t)d^\times t \end{align}\] and \[\begin{align} Z_\ell(f,s)=\int_{F^\times} f(0_{\underline{2\ell}},0,t)|t|^sd^\times t. \end{align}\] Both integral converges absolutely when \(\ell=s=1,\) in which case we can take \(x=0\). This proves (iii).

Suppose \(\ell\ge 3\). Since these two operators are continuous in \(f\), to compute both terms we now assume \(f(v,0,t)=f_1(v)f_2(t)\) for some \(f_1\in \mathcal{S}(F^{2\ell})\) and \(f_2\in \mathcal{S}(F)\). We will prove the lemma for \(F=\mathbb{R}.\) The case \(F=\mathbb{C}\) is analogous by identifying \(\mathbb{C}=\mathbb{R}^2,\) so we leave it to the reader.

Choose \(r>0\) and complex numbers \(b_0,\ldots, b_{\ell-2}\) such that \[\begin{align} \bigg|f_2(t)-\sum_{n=0}^{\ell-2}b_nt^n\bigg|\le |t|^{\ell-2+1/4} \textrm{ for all } |t|\le r. \end{align}\] Then for \(a\in F^\times\) \[\begin{align} I(f)(ax)&=\int_{|t|>r} |t|^{2-\ell}f_1(t^{-1}ax)f_2(t) d^\times t+\int_{|t|\le r} |t|^{2-\ell}f_1(t^{-1}ax)\left(f_2(t)-\sum_{n=0}^{\ell-2} b_nt^n\right) d^\times t\\ &+\sum_{n=0}^{\ell-2}\int_{|t|\le r} b_nt^n|t|^{2-\ell} f_1(t^{-1}ax)d^\times t. \end{align}\] The first two terms are absolutely convergent and bounded above by a constant independent of \(a\) and \(x\). For the last term, by changing variables we obtain \[\begin{align} \sum_{n=0}^{\ell-2} b_n a^n|a|^{2-\ell}\int_{|t|\le |a|^{-1}r}t^n|t|^{2-\ell} f_1(t^{-1}x)d^\times t. \end{align}\] Therefore, if \(f\in W_\ell'\) so that \(I(f)(0)=\lim_{|a|\to 0}I(f)(ax)\) exists, each summand above is zero except possibly for \(n=\ell-2\). In this case, we arrive at \[\begin{align} b_{\ell-2}\mathrm{sgn}^{\ell}(a)\int_{|t|\le |a|^{-1}r} \mathrm{sgn}^\ell(t)f_1(t^{-1} x)d^\times t. \end{align}\] This term must be zero if \(\ell\) is odd. For \(\ell\) even, it must be \(f_1(0)=0.\)

When \(\ell\) is odd, we have \[\begin{align} I(f)(0)&=f_1(0)\left(\int_{|t|>r} |t|^{2-\ell}f_2(t) d^\times t+\int_{|t|\le r} |t|^{2-\ell}\left(f_2(t)-b_{\ell-2}t^{\ell-2}\right) d^\times t\right). \end{align}\] On the other hand, for any \(\ell\ge 3\) we have \[\begin{align} Z_\ell(f,s)=f_1(0)\left(\int_{|t|>r} f_2(t)|t|^{s}d^\times+\int_{|t|>r} (f_2(t)-b_{\ell-2}t^{\ell-2})|t|^sd^\times t+\int_{|t|<r} b_{\ell-2}t^{\ell-2}|t|^s d^\times t\right). \end{align}\] This is zero when \(\ell\) is even. When \(\ell\) is odd, the last integral vanishes, so \(I(f)(0)=Z_\ell(f,2-\ell).\) This proves (i).

Suppose \(\ell=2\). Let \(x\in X_2(F)^1\) and \(|a|<1\). Write \[\begin{align} I(f)(ax)&=\int_{|t|>1} f(t^{-1}ax,0,t)d^\times t+ \int_{|t|\le 1} f(t^{-1}ax,0,t)-f(t^{-1}ax,0,0)d^\times t\\ &+\int_{|t|\le |a|^{-1}} f(t^{-1}x,0,0)d^\times t. \end{align}\] The last term can be written as \[\begin{align} &\int_{|t|< 1} f(t^{-1}x,0,0)d^\times t+\int_{1\le |t|\le |a|^{-1}} f(t^{-1}x,0,0)d^\times t\\ &=\int_{|t|> 1} f(tx,0,0)d^\times t+\int_{1\ge |t|\ge|a|} f(tx,0,0)-f(0)d^\times t -\frac{\mathrm{vol}(K_{\mathbb{G}_m})}{[F:\mathbb{R}]}f(0)\log |a|. \end{align}\] Observe that \[\begin{align} \lim_{|a|\to 0} I(f)(ax)+\frac{\mathrm{vol}(K_{\mathbb{G}_m})}{[F:\mathbb{R}]}f(0)\log |a| \end{align}\] is well-defined, and we denote this as \(\tilde{a}_2(f)(x)\). Explicitly, \[\begin{align} \label{eq:a2:arch} \begin{aligned} \tilde{a}_2(f)(x)&=\int_{|t|>1} f(0_{\underline{4}},0,t)d^\times t+ \int_{|t|\le 1} f(0_{\underline{4}},0,t)-f(0)d^\times t\\ &+\int_{|t|> 1} f(tx,0,0)d^\times t+\int_{|t|\le 1} f(tx,0,0)-f(0)d^\times t. \end{aligned} \end{align}\tag{16}\]

On the other hand, \[\begin{align} \label{eq:Z2compute} Z_2(f,s)=\int_{|t|\ge 1} f(0_{\underline{4}},0,t)|t|^s d^\times t+\int_{|t|<1} (f(0_{\underline{4}},0,t)-f(0))|t|^s d^\times t+\int_{|t|<1} f(0)|t|^s d^\times t. \end{align}\tag{17}\] The first two integrals are absolutely convergent for \(\mathrm{Re}(s)>-1/2\). The last term above is \(\frac{1}{[F:\mathbb{R}]s}\mathrm{vol}(K_{\mathbb{G}_m})f(0),\) so \(\tilde{c}_2(I(f))=\frac{\mathrm{vol}(K_{\mathbb{G}_m})}{[F:\mathbb{R}]}f(0).\) This justifies (ii). ◻

Proof. We have \[\begin{align} Z_{\ell}(\tilde{b}_\ell,s)=\int_{F^\times} e^{-\pi [F:\mathbb{R}]|t|_\mathbb{R}^2}|t|^s d^\times t=\zeta(s). \end{align}\] Assume \(\ell\) is even and \(F=\mathbb{R}\). Then \[\begin{align} \tilde{c}_\ell(\tilde{b}_\ell)=\mathrm{Res}_{s=2-\ell} \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)=2\pi^{\frac{\ell-2}{2}}\frac{(-1)^{\frac{\ell-2}{2}}}{\left(\frac{\ell-2}{2}\right)!}. \end{align}\] On the other hand by [16], the coefficient of \(-\log |x|_\mathbb{R}=-\log |x|\) in the asymptotic expansion of \(b_\ell(x)=2\pi^{\frac{\ell-2}{2}}\widetilde{K}_{\frac{\ell-2}{2}}(2\pi|x|_\mathbb{R})\) is \[\begin{align} c_\ell(b_\ell)=-2\pi^{\frac{\ell-2}{2}}\frac{(-1)^{\ell/2}}{\Gamma(\ell/2)}=\tilde{c}_\ell(\tilde{b}_\ell). \end{align}\]

Suppose \(F=\mathbb{C}\) and \(\ell\ge 2\). Then \[\begin{align} \tilde{c}_\ell(\tilde{b}_\ell)=\mathrm{Res}_{s=2-\ell} 2(2\pi)^{-s}\Gamma(s)=2(2\pi)^{\ell-2}\frac{(-1)^{\ell-2}}{(\ell-2)!}. \end{align}\] In the asymptotic expansion of \(b_\ell(x)=4(2\pi)^{\ell-2}\widetilde{K}_{\ell-2}(4\pi |x|_\mathbb{R})\), the coefficient of \(-\log |x|=-2\log |x|_\mathbb{R}\) is \[\begin{align} \tilde{c}_\ell(b_\ell)=-2(2\pi)^{\ell-2}\frac{(-1)^{\ell-1}}{\Gamma(\ell-1)}=c_\ell(b_\ell). \end{align}\] ◻

Proof. Suppose \(\ell\ge 3\). Let \(\mathcal{F}\) be the operator in Lemma 11. We claim \(\mathcal{F}(f)=\mathcal{F}_{X_\ell}(f)\) for \(f\in \mathcal{S}_{\mathrm{ES}}(X_\ell(F))\). Assuming the claim, we have as vector spaces \[\begin{align} I(\widetilde{\mathcal{S}}(X_\ell(F)))&=I(W_\ell)+I(\widetilde{\mathcal{F}}_{X_\ell}(W_\ell))=\mathcal{S}_{\mathrm{ES}}(X_\ell(F))+\mathcal{F}(\mathcal{S}_{\mathrm{ES}}(X_\ell(F)))\\ &=\mathcal{S}_{\mathrm{ES}}(X_\ell(F))+\mathcal{F}_{X_\ell}(\mathcal{S}_{\mathrm{ES}}(X_\ell(F)))=\mathcal{S}(X_\ell(F)). \end{align}\] The map \(I: \widetilde{\mathcal{S}}(X_\ell(F))\longrightarrow \mathcal{S}(X_\ell(F))\) is continuous by the closed graph theorem, so \(I(\widetilde{\mathcal{S}}(X_\ell(F)))=\mathcal{S}(X_\ell(F))\) as Fréchet spaces by the open mapping theorem. Now as each \(f\in \mathcal{S}(X_\ell(F))\) is of the form \(f=f_1+\mathcal{F}(f_2)\) for some \(f_1,f_2\in \mathcal{S}_{\mathrm{ES}}(X_\ell(F)),\) we have \[\begin{align} \mathcal{F}(f)&=\mathcal{F}(f_1+\mathcal{F}(f_2))=\mathcal{F}(f_1)+f_2=\mathcal{F}_{X_\ell}(f_1)+f_2\\ &=\mathcal{F}_{X_\ell}(f_1+\mathcal{F}_{X_\ell}(f_2))=\mathcal{F}_{X_\ell}(f_1+\mathcal{F}(f_2))=\mathcal{F}_{X_\ell}(f). \end{align}\] Therefore, \(\mathcal{F}=\mathcal{F}_{X_\ell}\) and thus \(I\) is \(\mathrm{O}_{V_{\ell+1}}(F)\)-equivariant.

To prove the claim, we use a global argument. If \(F=\mathbb{R}\) (resp. \(\mathbb{C}\)), choose \(E=\mathbb{Q}\) (resp. \(\mathbb{Q}[i]\)). Let \(f_\infty\in \mathcal{S}_{\mathrm{ES}}(X_\ell(F))\) and \(\xi\in X_\ell^\circ(E).\) Since the local theory over nonarchimedean local fields agrees by results in §5, as argued in Lemma 11 we can choose finite places \(v_1,v_2,\) and \(f_{v_1}\in \mathcal{S}(X_\ell^\circ(E_{v_1})),\) \(f_{v_2}\in \mathcal{F}_{X_\ell}(\mathcal{S}(X_\ell^\circ(E_{v_2})))\) and \(f^{\infty v_1 v_2}\in \mathcal{S}(X_\ell(\mathbb{A}_E^{\infty v_1v_2}))\) so that additional terms in Theorem 1 and Theorem 9 vanish, \(\mathcal{F}_{X_\ell}(f_{v_1}f_{v_2}f^{\infty v_1v_2})(\xi)\neq 0\), and \[\begin{align} \mathcal{F}_{X_\ell}(f_\infty)\mathcal{F}_{X_\ell}(f_{v_1}f_{v_2}f^{\infty v_1v_2})(\xi)=\sum_{x\in X_\ell^\circ(E)}f_\infty f_{v_1}f_{v_2}f^{\infty v_1v_2}(x)=\mathcal{F}(f_\infty)\mathcal{F}_{X_\ell}(f_{v_1}f_{v_2}f^{\infty v_1v_2})(\xi). \end{align}\] Therefore, \(\mathcal{F}_{X_\ell}(f_\infty)(\xi)=\mathcal{F}(f_\infty)(\xi)\) for \(\xi\in X_\ell^\circ(E).\) Since \(X_\ell^\circ(E)\) is dense in \(X_\ell^\circ(F),\) the claim follows by continuity.

For \(\ell=2,\) by Theorem 3 and Corollary 8 we have a continuous map \[\begin{align} J:\mathcal{S}(X_{2}(F))\longrightarrow \mathcal{S}(X_{3}(F))/\mathcal{S}_{\mathrm{ES}}(X_{3}(F)) &\cong \widetilde{\mathcal{S}}(X_{3}(F))/W_3'\\ &\xrightarrow{\,\,\tilde{d}_3\,\,} \widetilde{\mathcal{S}}(X_{2}(F))/\tilde{d}_3(W_3')\xrightarrow{\,\,I\,\,} I(\widetilde{\mathcal{S}}(X_{2}(F))). \end{align}\] We may assume \(\psi(t)=e^{2\pi i\mathrm{tr}_{F/\mathbb{R}}(t)}.\) Then \(J(b_2)=b_2\) and \(J\circ \sigma_2=\sigma_2'\circ J\). Since \(\mathcal{S}(X_{2}(F))\) is irreducible, by the Iwasawa decomposition \(J\) is the inclusion map. Therefore, \(\mathcal{F}\) and \(\mathcal{F}_{X_{2}}\) agree on \(\mathcal{S}(X_{2}(F))\). In particular, \(\mathcal{F}\) is unitary and thus \(\mathcal{F}=\mathcal{F}_{X_{2}}\) on \(L^2(X_{2}(F)).\) Since \(I(\widetilde{\mathcal{S}}(X_{2}(F)))\) is contained in the set of smooth vectors \(\mathcal{S}(X_{2}(F))\) in \(L^2(X_{2}(F)),\) we have \(J\) is the identity map. This justifies the first statement.

The second assertion can be proved similarly as in Proposition 13 using Theorem 3, Lemma 19 and Lemma 20. We leave it to the reader. ◻

Proof. Let \(f\in \mathcal{S}(V_2(F))\) that is \(r_1(K)\)-invariant. By 1 , we have \[\begin{align} I(\tilde{d}_1(f))&=\int_{F^\times}\int_F|t|^{2}f(0,y,0,0)\psi(ty)dy d^\times t\\ &=\zeta(2)\varepsilon(-1,\psi)\lim_{s\to -1}\frac{1}{\zeta(s)}\int_{F^\times} |t|^{s}f(0,t,0,0) d^\times t. \end{align}\] Since \(f\) is \(r_1(K)\)-invariant, the expression above is invariant under changing variables \(t\mapsto tc\) for \(c\in K_{\mathbb{G}_m}.\) Thus we may assume \(f(0,\cdot,0,\cdot)\) is invariant under \(K_{\mathbb{G}_m}\) in both entries.

Assume \(F=\mathbb{R}\). Then \[\begin{align} &\int_{F^\times} |t|^{s}f(0,t,0,0) d^\times t\Bigg|_{s=-1}\\ &=\int_{|t|\ge 1} |t|^{-1}f(0,t,0,0)d^\times t-\mathrm{vol}(K_{\mathbb{G}_m})f(0)+\int_{|t|<1} |t|^{-1}(f(0,t,0,0)-f(0))d^\times t\\ &= \int_{F^\times} |t|^{-1}f(0,t,0,0)-|t|^{-1}f(0) d^\times t. \end{align}\] Here we have used the fact that \(f(0,t,0,0)-f(0)=O(|t|^2)\) for \(|t|<r\) small. Write this as \[\begin{align} \int_{F^\times} \lim_{|a|\to 0} |t|^{-1}\left(f(0,t,0,at^{-1})-f(0,0,0,at^{-1})\right) d^\times t. \end{align}\] Then we can apply Lebesgue’s dominated convergence theorem to see this is \[\begin{align} &\lim_{|a|\to 0}\int_{F^\times} |t|^{-1}\left(f(0,t,0,at^{-1})-f(0,0,0,at^{-1})\right) d^\times t=\lim_{|a|\to 0}\frac{1}{|a|} \left(I(f)(0,a)-I(f)(0)\right), \end{align}\] where the last equality follows from changing variables \(t\mapsto ta.\) We define \[\begin{align} d_1(I(f)):=\frac{1}{\zeta(-1)}\lim_{|a|\to 0}\frac{1}{|a|} \left(\frac{I(f)(0,a)+I(f)(0,-a)}{2}-I(f)(0)\right). \end{align}\]

Now suppose \(F=\mathbb{C}.\) Let \[\begin{align} b_{1}:=\frac{\partial^2}{\partial t\partial \overline{t}}f(0,t,0,0)\bigg|_{t=0}. \end{align}\] Then using \(f(0,t,0,0)=f(0)+b_1|t|+O(|t|^2)\) for \(|t|\) small, we have \[\begin{align} &\lim_{s\to -1}\frac{1}{\zeta(s)}\int_{F^\times} |t|^{s}f(0,t,0,0) d^\times t\\ &=\lim_{s\to -1} \frac{1}{\zeta(s)}\int_{|t|<1} |t|^{s}f(0,t,0,0) d^\times t\\ &=\lim_{s\to -1} \frac{1}{\zeta(s)}\int_{|t|<1} |t|^{s}\left(f(0,t,0,0)-f(0)-b_1|t|\right)+f(0)|t|^s+b_1|t|^{s+1} d^\times t\\ &=\frac{1}{2\mathrm{Res}_{s=-1}\zeta(s)} b_1. \end{align}\] On the other hand, using again the fact that \(f(0,a,0,t^{-1})-f(0,0,0,t^{-1})=O(|a|)g(t^{-1})\) for some \(g\in \mathcal{S}(F)\) for \(0<|a|<1\) sufficiently small, we have \[\begin{align} &\frac{1}{-|a|\log |a|}\left(I(f)(0,a)-I(f)(0\right)) \\&=\frac{1}{-|a|\log |a|}\int_{F^\times}|t|^{-1}\left(f(0,at,0,t^{-1})-f(0,0,0,t^{-1})\right) d^\times t\\ &=\frac{1}{-\log |a|}\int_{F^\times}|t|^{-1}\left(f(0,t,0,at^{-1})-f(0,0,0,at^{-1})\right) d^\times t\\ &=\frac{1}{-\log |a|}\int_{|a|<|t|<1}|t|^{-1}\left(f(0,t,0,at^{-1})-f(0,0,0,at^{-1})\right) d^\times t+O\left(\frac{1}{-\log |a|}\right). \end{align}\] The limit as \(|a|\to 0\) exists, which equals \[\begin{align} &\lim_{|a|\to 0}\frac{1}{-\log |a|}\int_{|a|<|t|<1}\frac{\partial^2}{\partial v_2\partial \overline{v}_2}f(0,0,0,at^{-1})d^\times t\\ &=\lim_{|a|\to 0}\frac{1}{-\log |a|}\int_{|a|<|t|<1}\frac{\partial^2}{\partial v_2\partial \overline{v}_2}f(0,0,0,t)d^\times t=b_1/2. \end{align}\] We define \[\begin{align} d_1(I(f)):=\frac{1}{\mathrm{Res}_{s=-1} \zeta(s)}\lim_{|a|\to 0}\frac{1}{-|a|\log |a|} \left(\frac{\int_{K_{\mathbb{G}_m}}I(f)(0,ac)dc}{\mathrm{vol}(K_{\mathbb{G}_m})}-I(f)(0)\right). \end{align}\] Then in both cases, by the definition of \(d_1\) we have \(I\circ \tilde{d}_1=\varepsilon(-1,\psi)\zeta(2)d_1\circ I.\)

The map \(d_1'\) can be defined analogously by considering the asymptotic expansion of \(I(f)(a,0)\). The same proof as in Proposition 14 implies \(\mathrm{ev}\circ\tilde{d}_1=\varepsilon(-1/2,\psi)d_1'\circ I.\) Define \[\begin{align} c_1(I(f))&:=\frac{1}{\zeta(1)}I(f)(0). \end{align}\] Then \(c_1(I(f))=\tilde{c}_1(f)\) by Lemma 19(iii). This completes the proof. ◻

Proof. Arguing as in Proposition 16, by Lemma 19(ii) and Proposition 21, by taking germs at the origin of functions in \(\mathcal{S}(X_{2}(F))\) we have a natural map \[\begin{align} \mathcal{S}(X_{2}(F))/\mathcal{S}(X_{2}^\circ(F))&\longrightarrow \pi''\\ I(f)&\mapsto \tilde{a}_2(f)(\cdot)-\tilde{c}_2(f)\log|\cdot|. \end{align}\] The representation \(\pi''\) fits into the following nonsplit exact sequence of \(\mathrm{O}_{V_2}(F)\)-representations \[\begin{align} &0 \longrightarrow \pi' \longrightarrow \pi''\longrightarrow \mathbb{C}\longrightarrow 0,\\ & 0 \longrightarrow \mathbb{C}\longrightarrow \pi'\longrightarrow \mathrm{Ind}_{\mathrm{SO}_{V_2}}^{\mathrm{O}_{V_2}}(\mathrm{St} \otimes \mathbb{C})\longrightarrow 0. \end{align}\] Take \(\pi:=\pi''/\mathbb{C}\otimes\mathbb{C}.\) By the differential action of \(\mathrm{Lie}\,\mathrm{O}_{V_3}(F)\) on the minimal representation [13], [14], we have \(\mathcal{S}(X_2(F))\) is stable under multiplication by the coordinate functions \(\mathbb{C}[\mathrm{Res}_{F/\mathbb{R}}X_{2}],\) so we can extend the results by tensoring \(\mathbb{C}[[\mathrm{Res}_{F/\mathbb{R}}X_{2}]]\) at the germs.

By the asymptotics in Lemma 19(ii) and well-known results on principal series of \(\mathrm{SL}_2(F)\), the space of continuous \(\mathrm{O}_{V_2}(F)\)-invariant linear functionals \(\mathcal{S}(X_2(F))/\mathcal{S}(X_2^\circ(F))\rightarrow \mathbb{C}\) is one-dimensional (see also the remark below). Therefore, \(\tilde{c}_2(f)=c_2(I(f))\) by Lemma 20. This proves (i).

For (ii) by Lemma 7 for \(f\in \widetilde{\mathcal{S}}(X_{2}(F))\) and \(x\in X_{1}^\circ(F),\) \[\begin{align} \label{eq:compute} I(\tilde{d}_2(f))(x)=\lim_{|a|\to 0} \int_{F} I(f)(ax,0,ay) \psi(y)dy. \end{align}\tag{18}\] This vanishes if \(f\in W_{2}'\) by Corollary 8. Furthermore, if \(I(f)(v)=O(|v|^{\epsilon})\) as \(|v|\to 0\) for some \(\epsilon>0,\) then since \(I(f)(x,0,y)\) as a function in \(y\) behaves like a Schwartz function in \(\mathcal{S}(F)\) as \(|y|\to \infty\), we have for \(\epsilon/(\epsilon+1)>\delta>0\) \[\begin{align} &\int_{F} I(f)(ax,0,ay) \psi(y)dy\\ &=|a|^{-1}\left(\int_{|y|\le |a|^{1-\delta}} I(f)(ax,0,y)\psi(a^{-1}y)dy+\int_{|a|^{1-\delta}<|y|} I(f)(ax,0,y)\psi(a^{-1}y)dy\right)\\ &=O(|a|^{-\delta+\epsilon(1-\delta)}) \end{align}\] converges to \(0\) as \(|a|\to 0.\) Thus \(I(\tilde{d}_2(f))=0,\) and the map \(I\circ \tilde{d}_2\) factors through \[\begin{align} \widetilde{\mathcal{S}}(X_2(F))\xrightarrow{\quad I\quad} \mathcal{S}(X_2(F))\xrightarrow{\quad d_2\quad } \pi. \end{align}\] Hence we have an induced \(P_2(F)\)-equivariant surjection \(\tilde{J}:\pi\to \mathcal{S}(X_1(F)).\)

Elements in \(\sigma\) can be viewed as functions on \((\mathbb{P}^1(F)\times \{\mathrm{pt}\})\cup (\{\mathrm{pt}\}\times\mathbb{P}(F))\). In particular, \(\sigma\) contains a \(P_2(F)\)-stable subspace \(\sigma'\) homeomorphic to \(\mathcal{S}((N(F)-\{0\})^2)=\mathcal{S}((F^\times)^2)\). Arguing similarly as in Lemma 4, by Fourier inversion and the fact that \(N_2(F)\) acts trivially on \(\sigma/\sigma',\) we have \(\sigma'\) is the unique (topologically) irreducible \(P_2(F)\)-subrepresentation of \(\sigma\). Furthermore, by 18 \(\tilde{J}(\sigma')=\mathcal{S}((F^\times)^2).\) Thus \(\tilde{J}\) is injective and hence an isomorphism. When \(\psi(t)=e^{2\pi i\mathrm{tr}_{F/\mathbb{R}}(t)},\) \(\tilde{J}(d_2(b_2))=b_1.\) For general \(\psi,\) the argument in Proposition 16 applies. ◻

Proof. We may assume \(\psi(t)=e^{2\pi i\mathrm{tr}_{F/\mathbb{R}}(t)}.\) By the computation in Lemma 20, \[\begin{align} c_2(b_2)= 2=2c_{1}(b_1)=2c_1(d_2(b_2)). \end{align}\] Since both \(c_2\) and \(c_1\circ d_2\) are continuous \(\mathrm{O}_{V_2}(F)\)-equivariant linear functionals on \(\mathcal{S}(X_2(F))/\mathcal{S}(X_2^\circ(F))\), we have \(c_2=2c_1\circ d_2.\) ◻

Proof. Note that for \(F\) real or complex, \[\begin{align} \frac{d}{ds}\zeta(s)^{-1}\bigg|_{s=0}=\lim_{s\to 0} \frac{1}{s\zeta(s)}=\frac{1}{2}. \end{align}\] We can assume \(f\) is \(r_2(K)\)-invariant. By 17 we have \[\begin{align} \frac{d}{ds}\frac{Z_2(f,s)}{\zeta(s)}\bigg|_{s=0}&=\frac{1}{2}\frac{d^2}{d^2s}\zeta(s)^{-1}\bigg|_{s=0}f(0)\frac{\mathrm{vol}(K_{\mathbb{G}_m})}{[F:\mathbb{R}]}\\ &+\frac{1}{2}\left(\int_{|t|\ge1} f(0_{\underline{4}},0,t) d^\times t+\int_{|t|<1} (f(0_{\underline{4}},0,t)-f(0)) d^\times t\right). \end{align}\] On the other hand, by 1 \[\begin{align} Z_1(\tilde{d}_2(f),s+1)&=\varepsilon(-s,\psi)\frac{\zeta(1+s)}{\zeta(-s)}\int_{F^\times} f(0_{\underline{2}},0,t,0,0)|t|^{-s}d^\times t. \end{align}\] Thus similarly \[\begin{align} &\frac{d}{ds} \frac{Z_1(\tilde{d}_2(f),s+1)}{\varepsilon(-s,\psi)\zeta(s+1)}\Bigg|_{s=0}\\ &=-\frac{1}{2}\frac{d^2}{ds^2}{\zeta(s)}^{-1}\bigg|_{s=0}f(0)\frac{\mathrm{vol}(K_{\mathbb{G}_m})}{[F:\mathbb{R}]}\\ &-\frac{1}{2}\left(\int_{|t|>1} f(0_{\underline{2}},0,t,0,0)d^\times t+\int_{|t|\le 1} (f(0_{\underline{2}},0,t,0,0)-f(0))d^\times t\right). \end{align}\] By the proof of Lemma 19 and 16 , \(\tilde{c}_2(f)=f(0)\frac{\mathrm{vol}(K_{\mathbb{G}_m})}{[F:\mathbb{R}]}\) and \[\begin{align} \tilde{a}_2(f)(0:0:0:1)=&\int_{|t|>1} f(0_{\underline{4}},0,t)d^\times t+ \int_{|t|\le 1} f(0_{\underline{4}},0,t)-f(0)d^\times t\\ &+ \int_{|t|< 1} f(0_{\underline{2}},0,t^{-1},0,0)d^\times t+\int_{1\ge |t|} (f(0_{\underline{2}},0,t,0,0)-f(0))d^\times t. \end{align}\] Therefore, by the definition of \(c_2\) and \(a_2\) we have \[\begin{align} &\frac{d}{ds} \left(-\frac{Z_{2}(f,s)}{\zeta(s)}+\frac{Z_{1}(\tilde{d}_2(f), s+1)}{\varepsilon(-s,\psi)\zeta(s+1)}\right)\Bigg|_{s=0}=-\frac{d^2}{d^2s}\zeta(s)^{-1}\bigg|_{s=0}c_2(I(f))-\frac{1}{2}a_2(I(f)). \end{align}\] ◻

Proof. Only the identity \(\tilde{c}_{\ell}=\zeta(2-\ell)c_\ell\circ I\) for \(\ell\ge 3\) is unclear. Suppose \(f=\otimes f_v\in \mathcal{S}(V_\ell(\mathbb{A}_E)\oplus \mathbb{A}_E^2).\) Since \[\begin{align} \frac{Z_\ell(f,s)}{\zeta(s)}=\prod_{v} \frac{Z_\ell(f_v,s)}{\zeta_v(s)} \end{align}\] whose factors are \(1\) at almost all \(v\), we have \[\begin{align} \frac{\tilde{c}_\ell(f)}{\zeta(2-\ell)}= \left(\prod_{v\in S} \tilde{c}_{\ell,v}(f_v)\frac{\zeta_v(s)^{-1}}{s+\ell-2}\right)\bigg|_{s=2-\ell}\left(\prod_{v\not\in S} \frac{\tilde{c}_{\ell,v}(f_v)}{\zeta_v(2-\ell)}\right) \end{align}\] where \(S\) is the set of infinite places such that \(\zeta(s)\) has a (simple) pole at \(s=2-\ell.\) By the local computations in §5 and §6, this is \[\begin{align} \left(\prod_{v\in S} c_{\ell,v}(I(f_v))\frac{\zeta_v(s)^{-1}}{s+\ell-2}\right)\bigg|_{s=2-\ell}\left(\prod_{v\not\in S} \frac{c_{\ell,v}(I(f_v))}{\zeta_v(2-\ell)}\right)=c_\ell(f). \end{align}\] The identity for general \(f\) follows by continuity of \(c_\ell\) and linearity. ◻

Proof of Theorem 2. By Theorem 9 and Theorem 26 it remains to show for \(f=\otimes f_v\in \mathcal{S}(V_3(\mathbb{A}_E))\) \[\begin{align} &\tilde{c}_2(f)+\tilde{c}_1(\tilde{d}_{2}(f))=2^{1-|\infty|}\kappa' c_2(I(f))+2^{1-|\infty|}|D|^{\frac{1}{2}}\kappa a_2(I(f)) \end{align}\] where \[\begin{align} \kappa'=\frac{d}{ds}s\zeta(s)\bigg|_{s=0}. \end{align}\] We begin by unwinding the definition of \(\tilde{c}_1\) and \(\tilde{c}_2\). Recall \(\kappa=\mathrm{Res}_{s=1} \zeta(s).\) Using the functional equation \(\zeta(s)=\epsilon(s)\zeta(1-s),\) we have by Proposition 16 and Proposition 23 \[\begin{align} \tilde{c}_2(f)&= \left(\frac{d}{ds} s\zeta(s)\frac{Z_{2}(f,s)}{\zeta(s)}\right)\Bigg|_{s=0}\\ &=\big(\mathrm{Res}_{s=0}\zeta(s)\big)\left(\frac{d}{ds}\frac{Z_2(f,s)}{\zeta(s)}\right)\Bigg|_{s=0}+\kappa'\frac{Z_2(f,s)}{\zeta(s)}\Bigg|_{s=0}\\ &=-|D|^{1/2}\kappa\left(\frac{d}{ds}\frac{Z_2(f,s)}{\zeta(s)}\right)\Bigg|_{s=0}+2^{-|\infty|}\kappa' c_2(I(f)). \end{align}\] Similarly, by Proposition 14, Proposition 16, Proposition 22 and Proposition 23 \[\begin{align} \tilde{c}_1(\tilde{d}_{2}(f))&= \left(\frac{d}{ds} s\varepsilon(-s)\zeta(1+s)\frac{Z_1(\tilde{d}_2(f),s+1)}{\varepsilon(-s)\zeta(1+s)}\right)\Bigg|_{s=0}\\ &=|D|^{1/2}\kappa\left(\frac{d}{ds}\frac{Z_1(\tilde{d}_2(f),s+1)}{\varepsilon(-s)\zeta(1+s)}\right)\Bigg|_{s=0}+\left(\frac{d}{ds}s\zeta(-s)\right)\bigg|_{s=0} \left(\frac{Z_1(\tilde{d}_2(f),s+1)}{\varepsilon(-s)\zeta(1+s)}\right)\Bigg|_{s=0}\\ &=|D|^{1/2}\kappa\left(\frac{d}{ds}\frac{Z_1(\tilde{d}_2(f),s+1)}{\varepsilon(-s)\zeta(1+s)}\right)\Bigg|_{s=0}+\kappa' c_1(d_2(I(f))). \end{align}\] We claim \[\begin{align} a_2(I(f))&=\frac{d}{ds}\left(-\frac{Z_2(f,s)}{\zeta(s)}+\frac{Z_1(\tilde{d}_2(f),s+1)}{\varepsilon(-s)\zeta(1+s)}\right)\Bigg|_{s=0}. \end{align}\] Since \(c_2=2^{|\infty|}c_1\circ d_2\), the assertion follows from the claim.

Let \(S\) be a finite set of places including \(\infty\) such that for \(v\not\in S,\) \(f_v=\mathbb{1}_{V_3(\mathcal{O}_v)}\) and \(\psi_v\) is unramified. Then \[\begin{align} -\frac{Z_2(f,s)}{\zeta(s)}+\frac{Z_1(\tilde{d}_2(f),s+1)}{\varepsilon(-s)\zeta(1+s)}=-\prod_{v\in S}\frac{Z_2(f_v,s)}{\zeta_v(s)}+\prod_{v\in S}\frac{Z_1(\tilde{d}_2(f_v),s+1)}{\varepsilon(-s,\psi_v)\zeta_v(1+s)}. \end{align}\] By the product rule and the identity \[\begin{align} \frac{Z_2(f_v,s)}{\zeta_v(s)}\bigg|_{s=0}= 2^{-\delta_{v|\infty}}c_{2,v}(I(f_v))= c_{1,v}\circ d_{2,v}(I(f_v))=\frac{Z_1(\tilde{d}_2(f_v),s+1)}{\varepsilon(-s,\psi_v)\zeta_v(1+s)}\bigg|_{s=0} \end{align}\] used above, the claim follows from Lemma 18 and Lemma 25. This completes the proof. ◻