Proof. First assume \(v\) is a real place. Let us use the corresponding real embedding to identify \(F_v\) with \(\mathbb{R}\). Then there is \(s\in\mathbb{C}\) such that \(\omega(y)=y^s\) for all \(y\in F_v^{\times,\circ}=\mathbb{R}_{>0}.\) Let \(X=\begin{pmatrix} 1 & \\ & \ddots &\\ & & 1 \end{pmatrix}\) as an element in the centre of the complexified universal enveloping algebra of \(G(F_v)\). By definition of the Lie algebra action, for all \(u\in V_{sm}\), \[X\cdot u = \frac{d}{dt}\Big|_{t=0}\rho(e^{Xt})u=\frac{d}{dt}\Big|_{t=0} e^{st}u=su.\] Thus, the infinitesimal character determines \(s\) and hence \(\omega|_{F_v^{\times,\circ}}\).

Now, assume \(v\) is a complex place. Let \(\sigma_1,\sigma_2:F\to \mathbb{C}\) be the two complex embeddings corresponding to \(v\). Use the same notation for the induced isomorphisms \(F_v\xrightarrow{\sim}\mathbb{C}.\) Then there are \(s_1,s_2\in\mathbb{C}\) with \(s_1-s_2\in\mathbb{Z}\) such that \(\omega(y)=\sigma_1(y)^{s_1}\sigma_2(y)^{s_2}\) for all \(y\in F_v^{\times}\;(\cong\;\mathbb{C}^\times).\) For \(x\in F_v\), let \(X=\begin{pmatrix} x & \\ & \ddots &\\ & & x \end{pmatrix}\) as an element in the centre of the complexified universal enveloping algebra of \(G(F_v)\). By definition of the Lie algebra action, for all \(u\in V_{sm}\), \[X\cdot u = \frac{d}{dt}\Big|_{t=0}\rho(e^{Xt})u=\frac{d}{dt}\Big|_{t=0} e^{s_1\sigma_1(x)t+s_2\sigma_2(x)t}u=(s_1\sigma_1(x)+s_2\sigma_2(x))u.\] Taking \(x=1\) and \(x=\sigma_1^{-1}(i)\) gives us two equations that allows us to solve for \(s_1,s_2\). Thus, the infinitesimal character determines \(s_1,s_2\) and hence \(\omega|_{F_v^{\times}}\). ◻

Proof. Let \(M_{\lambda_v}:=\begin{cases} M_{\lambda_{\tau},\mathbb{C}} &\text{if v is real}\\ M_{\lambda_{\tau},\mathbb{C}}\otimes_\mathbb{C}M_{\lambda_{\bar\tau},\mathbb{C}} &\text{if v is complex} \end{cases}\), where \(\tau\) is an embedding \(F\to \mathbb{C}\) whose associated place is \(v\). Then the infinitesimal character of \(\pi_v\) is the same as that of \(M_{\lambda_v}^\vee\). We want to show that the restriction of the central character of \(\pi_v\) to \(F_v^{\times,\circ}\) is that of \(M_{\lambda_v}^\vee=M_{\lambda_v^\vee}\). This follows from 1. ◻

Proof. Let \[\begin{align} G^1(F_\infty)&:=\left\{(g_v)\in G(F_\infty): \prod_{v\mid\infty}|\det g_v|_{F_v}=1\right\},\\ G^1(F_v)&:=\{g_v\in G(F_v): |\det g_v|_v=1\},\\ H&:=\left\{(a_v)\in \prod_{v\mid \infty}\mathbb{R}_{>0}:\prod_v |a_v|_{F_v}=1\right\}\subset G(F_\infty), \end{align}\] where we view \(\prod_{v\mid \infty}\mathbb{R}_{>0}\) as a subgroup of the centre of \(G(F_\infty)\) via the diagonal embedding. Let \(\mathfrak g^1_\infty, \mathfrak g^1_v, \mathfrak h\) be the corresponding Lie algebras. Let \[K^1_v:= \begin{cases} O_n(\mathbb{R}) \text{ if v is real}\\ U_n(\mathbb{R}) \text{ if v is complex} \end{cases},\] which is a subgroup of \(G^1(F_v)\), so \(\mathfrak k^1_v :=Lie(K^1_v) \subset \mathfrak g^1_v.\)

From the Lie group decompositions \(G(F_\infty)= G^1(F_\infty) \times A_G\) and \(K_\infty= \prod_{v\mid \infty}K^1_v \times A_G\), we get corresponding decompositions of Lie algebras, which in turns give \[\label{eq:32lie32alg32decomposes} \mathfrak g/\mathfrak k_\infty = \mathfrak g^1_\infty /\prod_{v\mid \infty}\mathfrak k^1_v\tag{1}\] as \(\mathbb{C}[K_\infty]\)-modules, where \(K_\infty\) acts by conjugation.

Similarly, from the Lie group decomposition \(G^1(F_\infty) =\prod_{v\mid \infty}G^1(F_v) \times H\), we get a corresponding Lie algebra decomposition for \(\mathfrak g^1_\infty\), which we can substitute to equation 1 to get \[\label{eq:32lie32alg32decomposes32v2} \mathfrak g/\mathfrak k_\infty = \left(\prod_{v\mid \infty}\frac{\mathfrak g^1_v}{\mathfrak k^1_v}\right) \times \mathfrak h\tag{2}\] as \(\mathbb{C}[K_\infty]\)-modules. Note that the action of \(K_\infty\) on \(\mathfrak h\) is trivial.

The relative Lie algebra complex computing \(H^{i}(\mathfrak g, K_\infty^0,\pi_\infty\otimes_\mathbb{C}M_{\lambda,\mathbb{C}})\) is by definition the \(i\)-th cohomology of \[\begin{align} &\;\;(\wedge^* (\mathfrak g/\mathfrak k_\infty)^\vee \otimes_\mathbb{R}(\pi_\infty)_{K_\infty\text{-fin}} \otimes_\mathbb{C}M_{\lambda,\mathbb{C}})^{K_\infty^\circ}\\ = &\left(\wedge^* \left(\prod_{v\mid \infty}\frac{\mathfrak g^1_v}{\mathfrak k^1_v}\right)^\vee\otimes_\mathbb{R}\wedge^*\mathfrak h^\vee \otimes_\mathbb{R}(\pi_\infty)_{K_\infty\text{-fin}} \otimes_\mathbb{C}M_{\lambda,\mathbb{C}}\right)^{K_\infty^\circ} \end{align}\] where \((\pi_\infty)_{K\text{-fin}}\) means the \(K_\infty\)-finite vectors of \(\pi_\infty\). As the action of \(K_\infty\) on \(\mathfrak h\) is trivial, we can pull out that factor from the cohomology and get \[H^{i}(\mathfrak g, K_\infty^0,\pi_\infty\otimes_\mathbb{C}M_{\lambda,\mathbb{C}}) = \bigoplus_{a+b=i} H^a\left(\left(\wedge^* \left(\prod_{v\mid \infty}\frac{\mathfrak g^1_v}{\mathfrak k^1_v}\right)^\vee\otimes_\mathbb{R}(\pi_\infty)_{K_\infty\text{-fin}} \otimes_\mathbb{C}M_{\lambda,\mathbb{C}}\right)^{K_\infty^\circ}\right) \otimes_\mathbb{R}\wedge^b\mathfrak h^\vee .\] Since \(A_G\) acts trivially on \(\wedge^* \left(\prod_{v\mid \infty}\frac{\mathfrak g^1_v}{\mathfrak k^1_v}\right)^\vee\) and \(\pi_\infty\otimes_\mathbb{C}M_{\lambda,\mathbb{C}}\) by 2, we can replace \(K_\infty^\circ\) on the right hand side by \(\prod_{v\mid \infty}K_v^1\). Then by definition \[H^{i}(\mathfrak g, K_\infty^0,\pi_\infty\otimes_\mathbb{C}M_{\lambda,\mathbb{C}}) = \bigoplus_{a+b=i}H^{a}\left(\prod_{v\mid\infty}\mathfrak g^1_v, \prod_{v\mid\infty}\mathfrak k^1_v,\pi_\infty\otimes_\mathbb{C}M_{\lambda,\mathbb{C}}\right) \otimes_\mathbb{R}\wedge^b\mathfrak h^\vee,\] which by Künneth formula [16] equals \[\bigoplus_{a_1+\dots+a_{m}+b=i}H^{a_1}\left(\mathfrak g^1_v,\mathfrak k^1_v,\pi_v\otimes_\mathbb{C}M_{\lambda_v,\mathbb{C}}\right)\otimes_\mathbb{C}\cdots\otimes_\mathbb{C}H^{a_m}\left(\mathfrak g^1_v,\mathfrak k^1_v,\pi_v\otimes_\mathbb{C}M_{\lambda_v,\mathbb{C}}\right) \otimes_\mathbb{R}\wedge^b\mathfrak h^\vee,\] where \(m:=r_1+r_2\), \(M_{\lambda_v,\mathbb{C}}:=\begin{cases} M_{\lambda_{\tau},\mathbb{C}} &\text{if v is real}\\ M_{\lambda_{\tau},\mathbb{C}}\otimes_\mathbb{C}M_{\lambda_{\bar\tau},\mathbb{C}} &\text{if v is complex} \end{cases}\), and \(\tau\) is an embedding \(F\to \mathbb{C}\) whose associated place is \(v\). The result now follows from [9], 3, and the fact that \(\mathfrak h\cong \mathbb{R}^{r_1+r_2-1}\). ◻

Proof. Let \(H_2^*=H^*(\mathfrak g,K_\infty^\circ,L^2_d(\mathrm{GL}_n(F)\backslash \mathrm{GL}_n(\mathbb{A}_F),\chi)_{sm}\otimes_\mathbb{C}M_{\mu,\mathbb{C}}),\) where \(()_{sm}\) stands for the smooth vectors. We have a Hilbert space decomposition \[L^2_d(\mathrm{GL}_n(F)\backslash \mathrm{GL}_n(\mathbb{A}_F),\chi)\cong\widehat\bigoplus_{\Pi\in \Pi_d}m_d(\Pi)\Pi.\] By the multiplicity one theorem for the discrete spectrum proved by Moeglin and Waldspurger [17], we have \(m_d(\Pi)=1\) for all such \(\Pi\). We have a corresponding decomposition for the cohomology [18]: \[\label{eq:H95232decomp} H_2^*\cong \bigoplus_{\Pi\in \Pi_d} H^*(\mathfrak g,K_\infty^\circ,\Pi_\infty\otimes_\mathbb{C}M_{\mu,\mathbb{C}})\otimes_\mathbb{C}(\Pi^\infty)^U.\tag{5}\] Each \(\Pi^\infty\) is an irreducible admissible representation of \(\mathrm{GL}_n(\mathbb{A}_F^\infty)\), so it is factorisable, so \((\Pi^\infty)^U=(\Pi_S)^{U_S}\otimes_\mathbb{C}(\Pi^{\infty,S})^{U^S}\) is isomorphic to a direct sum of simple modules of \(\mathcal{H}_\mathbb{C}\). It follows that \(H_2^*\) is a semisimple \(\mathcal{H}_\mathbb{C}\)-module.

By a result of Borel [9], we have injections \[\label{eq:332injections} H^*_{cusp}(X_U,M_{\mu,\mathbb{C}}) \hookrightarrow H^*_{!}(X_U,M_{\mu,\mathbb{C}}) \hookrightarrow \tilde{H}^*_2\tag{6}\] where \(\tilde{H}^*_2\) is the image of \(H^*_2\to H^*(X_U,M_{\mu,\mathbb{C}}).\) Submodules of semisimple modules are semisimple, so \(H^*_{!}(X_U,M_{\mu,\mathbb{C}})\) is a semisimple \(\mathcal{H}_\mathbb{C}\)-module and part [semisimple] follows.

For part [isotypic], recall that discrete automorphic representations are isobaric (as they are Speh representations), so they satisfy strong multiplicity one, so \(H_2^*[(\pi^{\infty,S})^{U^S}]=H^*(\mathfrak g,K_\infty^\circ,\pi_\infty\otimes_\mathbb{C}M_{\mu,\mathbb{C}})\otimes_\mathbb{C}(\pi^\infty)^U\) by 5 . But this is also the \((\pi^{\infty,S})^{U^S}\)-isotypic component of the cuspidal cohomology by equation 4 . Part [isotypic] now follows from 6 .

For part [reduced], assume \(U^S:=\mathrm{GL}_n(\prod_{v\notin S\cup\{\infty\}}\mathcal{O}_{F_v})\). By Satake isomorphism, \(\mathcal{H}_\mathbb{C}\) is commutative. Thus, \((\Pi^{\infty,S})^{U^S}\) is \(0\)-dimensional or \(1\)-dimensional. Hence each element of \(\mathcal{H}_\mathbb{C}\) acts on \((\Pi^{\infty,S})^{U^S}\) by a scalar. The result now follows from part [semisimple]. ◻

Proof. For [M95lambda], we first consider a slightly more general setup. Suppose we have a product of commutative rings \(A\times B\) acting on \(N\). We can decompose \(N\) into \(N_1\times N_2\) accordingly. It is then clear that \(N_1=N[\ker \pi_A]\), where \(\pi_A:A\times B\to A\) is the first projection.

In our case, taking \(A\times B=E\times T_E^c\) and \(N=M_E\) shows \(e M_E=M_E[\ker\lambda_E]\). Hence \(M_\lambda=e M_E \cap M=\{m\in M : m\otimes 1\in M_E \text{ is annihilated by }(\ker\lambda)\otimes_\mathcal{O}E\}=M[\ker\lambda]\) since \(M\) is a free \(\mathcal{O}\)-module.

For [wiles32and32tu], note that under the \(\mathcal{O}\)-module isomorphism \(T_E\cong E\times T^c_E\), the \(\mathcal{O}\)-module \(T^\lambda=eT\) corresponds to \(\lambda(T)\times 0=\mathcal{O}\times 0\) while \(T_\lambda=T[\ker\lambda]\) corresponds to \(\lambda(T[\ker\lambda])\times 0\). Hence, \(C_0^\lambda(T)=\mathcal{O}/\lambda(T[\ker\lambda])\) and \(\eta_\lambda(M)=\lambda(T[\ker\lambda])=\eta_\lambda\). ◻

Proof. We know that \(M^\lambda/M_\lambda\) is a finite torsion \(\mathcal{O}\)-module, so \(rk_\mathcal{O}(M^\lambda)=rk_\mathcal{O}(M_\lambda)=1\). Thus, there exists \(m\in M\) such that \(eM=\mathcal{O}em\) as \(\mathcal{O}\)-module. Then we have a surjection of \(\mathcal{O}\)-modules \[\begin{align} C_0^\lambda(T)=\frac{eT}{T[\ker\lambda]} &\twoheadrightarrow C_0^\lambda (M)=\frac{eM}{M[\ker\lambda]}\\ x&\mapsto xm. \end{align}\] The lemma now follows from the observation that \(C_0^\lambda(T)=\mathcal{O}/ \eta_\lambda\). ◻

Proof. It is easy to see that \([\,,\,]\) extends to an \(E\)-bilinear perfect pairing \([\,,\,]:(M_1)_E\times (M_2)_E \to E\). Let \(em_1\in eM_1\) (with \(m_1\in M_1\)) and \(\tilde{e}m_2\in M_2\cap \tilde{e}M_2\) (with \(m_2\in M_2\)). Then \([em_1,\tilde{e}m_2]=[em_1+(1-e)m_1,\tilde{e}m_2]=[m_1,\tilde{e}m_2]\in \mathcal{O}\) because \((m_1,\tilde{e}m_2)\in M_1\times M_2\). A symmetric consideration shows \([\,,\,]\) induces a map \[\langle\,,\,\rangle:C_0^\lambda(M_1)\times C_0^{\tilde{\lambda}}(M_2) \to E/\mathcal{O}.\]

To show \(\langle\,,\,\rangle\) is non-degenerate, we let \(m_1\in M_1\). Suppose that \(\langle em_1,-\rangle\in \mathop{\mathrm{Hom}}_\mathcal{O}(C_0^{\tilde{\lambda}}(M_2),E/\mathcal{O})\) is zero. This means that for all \(m_2\in M_2\), \[[em_1,\tilde{e}m_2]=[em_1,\tilde{e}m_2+(1-\tilde{e})m_2]=[em_1,m_2]\in\mathcal{O}\] so \([em_1,-]\in \mathop{\mathrm{Hom}}_\mathcal{O}(M_2,\mathcal{O})\). By perfectness, there exists \(n\in M_1\) such that \([em_1,-]=[n,-]\), so \(n=em_1\in eM_1\cap M_1=(M_1)_\lambda\) by perfectness again, so \(em_1=0\in C_0^\lambda(M_1)\). By symmetry, \(\langle\, , \, \rangle\) is non-degenerate and hence perfect. Also, \(C_0^\lambda(M_1)\cong \mathop{\mathrm{Hom}}_\mathcal{O}(C_0^{\tilde{\lambda}}(M_2),E/\mathcal{O})\cong C_0^{\tilde{\lambda}}(M_2).\)

For part [pair32basis], let \((\varpi^a):=\eta_\lambda(M_1)=\eta_{\tilde{\lambda}}(M_2)\). By lemma 6, \(C_0^\lambda(M_1)\) and \(C_0^{\tilde{\lambda}}(M_2)\) are free \(\mathcal{O}/(\varpi^a)\)-modules of rank 1, with respective bases \(b_1,b_2\) say. By part [E47O], we have isomorphism \[\begin{align} \frac{\mathcal{O}}{(\varpi^a)} \cong \frac{\mathcal{O}}{(\varpi^a)}b_1\xrightarrow{\sim} \mathop{\mathrm{Hom}}_\mathcal{O}\left(\frac{\mathcal{O}}{(\varpi^a)}b_2,E/\mathcal{O}\right)&\xrightarrow{\sim}\frac{\varpi^{-a}\mathcal{O}}{\mathcal{O}}\xrightarrow{\cdot\varpi^a}\frac{\mathcal{O}}{(\varpi^a)}\\ f&\mapsto f(b_2). \end{align}\] This isomorphism maps \(1\) to \([b_1,b_2]\varpi^a \pmod {\varpi^a}\), so \([b_1,b_2]\varpi^a\in \mathcal{O}^\times\). As \(\mathcal{O}b_i/\mathcal{O}\delta_i \cong \mathcal{O}/(\varpi^a)\), we know \(\delta_i\in \varpi^a b_i\mathcal{O}^\times\). We deduce that \([\delta_1,\delta_2]\in \varpi^a\mathcal{O}^\times\), as desired. ◻

Proof. Given \(\lambda\), we can decompose \(T\otimes_\mathcal{O}E\cong E\times T_E^c\) and get an idempotent \(e\in T\otimes_\mathcal{O}E\) as before. Define \(T^c:=im(T\to T\otimes_\mathcal{O}E\to T_E^c).\)

We assume that \(\lambda'\) is as in the statement of the lemma. We claim that \(\lambda'\) factors through \(T\to T^c\). As \(\mathcal{O}_L\subset \mathcal{O}_L\otimes_\mathcal{O}E\), it suffices to show \[\lambda'_E:=\lambda'\otimes E: T\otimes_\mathcal{O}E\to \mathcal{O}_L\otimes_\mathcal{O}E\] factors through \(T^c_E\). Note that \(\mathcal{O}_L\otimes_\mathcal{O}E=L\) is an integral domain while \(T\otimes_\mathcal{O}E\cong E\times T_E^c\), so \(\lambda'_E\) must factor through \(E\) or \(T_E^c\). Since \(\lambda'_E\) is an \(E\)-algebra homomorphism and \(\lambda'_E \neq \lambda_E\), \(\lambda'_E\) cannot factor through \(E\). Hence \(\lambda'_E\) must factor through \(T_E^c\), as claimed. Suppose \(\eta_\lambda=\mathcal{O}\). This means \(eT\cap T = eT\), i.e. \(T\supset eT\). In particular, \((1,0)=e\cdot (1,1)\in T\). Then \[\begin{align} 1&=\lambda(1,0)\\ &\equiv\lambda'(1,0)\pmod{\varpi_L}\\ &\equiv 0\pmod\varpi \end{align}\] where the last equality holds since \(\lambda'\) factors through \(T^c\). We get a contradiction, so \(\eta_\lambda\neq \mathcal{O}\).

Conversely, suppose \(\eta_\lambda\neq \mathcal{O}\). The key observation is that \(\mathcal{O}/\eta_\lambda=\mathcal{O}\otimes_T T^c\), because \(\ker(T\twoheadrightarrow T^c)=T\cap eT\). Thus we have an \(\mathcal{O}\)-algebra homomorphism \[f:T^c\twoheadrightarrow \mathcal{O}\otimes_T T^c=\mathcal{O}/\eta_\lambda \twoheadrightarrow \mathcal{O}/\varpi.\] We get a commutative diagram \[\begin{tikzcd} & {\mathcal{O}} \\ T && {\mathcal{O}/\varpi} \\ & {T^c} \arrow[two heads, from=1-2, to=2-3] \arrow[two heads, from=2-1, to=1-2] \arrow[two heads, from=2-1, to=3-2] \arrow["f"', two heads, from=3-2, to=2-3] \end{tikzcd}\] We want to lift \(f\). There is a classical argument due to Deligne and Serre. Let \(\mathfrak{m}^c:=\ker f\). Consider the structure map \(\mathcal{O}\to T^c\). As \(\varpi \in \mathfrak{m}^c\), the prime ideal \(\mathfrak{m}^c\) lies above \(\varpi\mathcal{O}\). Note that multiplication by \(\varpi\) is invertible in \(T_E\), so \(T_E\) and hence \(T^c\) is \(\mathcal{O}\)-torsion free. Thus, \(T^c\) is a finite free \(\mathcal{O}\)-module. By flatness, the going down property holds, so there is a prime ideal \(\mathfrak p^c\subset \mathfrak{m}^c\) lying above \((0)\), so \(\mathcal{O}\hookrightarrow T^c/\mathfrak p^c\). This is a finite extension, so \[L:= Frac(T^c/\mathfrak p^c)\] is a finite extension of \(E\). We know \(T^c/\mathfrak p^c\) is finite over \(\mathcal{O}\) and hence integral over \(\mathcal{O}\), so \[T^c/\mathfrak p^c\subset \mathcal{O}_L.\] We want to show that this inclusion is a local homomorphism of local rings. As \(\mathcal{O}\) is a Henselian ring and \(T^c/\mathfrak p^c\) is an integral domain, \(T^c/\mathfrak p^c\) is a local ring. Note \(\mathfrak{m}':=\varpi_L\mathcal{O}_L \cap T^c/\mathfrak p^c\) is a prime ideal of \(T^c/\mathfrak p^c\) lying above \(\varpi\mathcal{O}\) (since its pullback to \(\mathcal{O}\) is the preimage of \(\varpi_L \mathcal{O}_L\) under the \(\mathcal{O}\)-algebra map \(\mathcal{O}\to T^c/\mathfrak p^c \to \mathcal{O}_L\)). By the incomparability theorem for injective integral ring extensions, \(\mathfrak{m}'\) is a maximal ideal of \(T^c/\mathfrak p^c\). Hence, the fact that \(T^c/\mathfrak p^c\) is a local ring implies \(\mathfrak{m}'=\mathfrak{m}^c/\mathfrak p^c\). This gives us another commutative diagram \[\begin{tikzcd} & {T^c/\mathfrak m^c} & {\mathcal{O}_L/\varpi_L} \\ {T^c} & {T^c/\mathfrak p^c} & {\mathcal{O}_L} \arrow[from=1-2, to=1-3] \arrow["f", two heads, from=2-1, to=1-2] \arrow[two heads, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=1-2] \arrow[hook, from=2-2, to=2-3] \arrow[two heads, from=2-3, to=1-3] \end{tikzcd}\] Combining the two diagrams gives us \[\begin{tikzcd} & {\mathcal{O}} \\ T &&& {\mathcal{O}_L/\varpi_L} \\ & {T^c} & {\mathcal{O}_L} \arrow[from=1-2, to=2-4] \arrow["\lambda", from=2-1, to=1-2] \arrow[from=2-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=2-4] \end{tikzcd}\] Denote the bottom map \(T\to\mathcal{O}_L\) by \(\lambda'\). It remains to show \(\lambda \neq \lambda'\) as maps to \(\mathcal{O}_L\). If this is false, then for all \((x,y)\in T\subset \mathcal{O}\times T^c\), we have \(x-y\in \mathfrak p^c\) by construction of \(\lambda'\). In particular, if \((x,y)\in T\cap eT\), then \(y=0\) and \(x\in \mathcal{O}\cap \mathfrak p^c=(0)\) as \(\mathfrak p^c\) lies above \((0)\). This means \(T\cap eT=0\), which is false as \(\varpi^a e\in T\cap eT\) for \(a\in \mathbb{Z}\) sufficiently large. ◻

Proof. In this proof, we may sometime abuse notation and regard a \(\mathbb{C}\)-vector space as an \(\mathcal{O}\)-module via the map \(\iota:\overline{\mathbb{Q}}_p\xrightarrow{\sim} \mathbb{C}\). Let \(\mathcal{H}:=\mathcal{H}(G^S,U^S)\).

As \(\mathrm{GL}_n(\mathbb{A}_F^S)\) acts on the \(\overline{\mathbb{Q}}_p\)-vector space \(\iota^{-1}\pi^\infty\), we know \(\mathbb{T}^S\) acts on the \(U^S\)-invariant \((\iota^{-1}\pi^\infty)^{U^S}\) and hence also on \((\iota^{-1}\pi^\infty)^{U}\). Moreover, for all finite \(v\notin S\), \(\pi_v\) is unramified so \(\pi_v^{U_v}\) is a one dimensional \(\overline{\mathbb{Q}}_p\)-vector space. It follows that each element of \(\mathbb{T}^S\) acts by a scalar on \((\iota^{-1}\pi^\infty)^{U}\), so we get an \(\mathcal{O}\)-algebra homomorphism \[\label{eq:eigenvalue} \mathbb{T}^S \to \overline{\mathbb{Q}}_p\tag{10}\] sending an element to its eigenvalue.

Note that \(\mathbb{T}^S(H^*_!(X_U,M_{\mu,\mathcal{O}}))\) acts on \(H^*_!(X_U,M_{\mu,\mathcal{O}})\otimes_{\mathcal{O}}\mathbb{C}\cong H^*_!(X_U,M_{\mu,\mathbb{C}})\). Since \(\pi\) is cohomological of weight \(\mu\), we have \(\mathcal{H}\)-equivariant injections \[(\pi^\infty)^U \hookrightarrow H^*_{cusp}(X_U,M_{\mu,\mathbb{C}})\hookrightarrow H^*_!(X_U,M_{\mu,\mathbb{C}}).\] Pick any non-zero \(x\in (\pi^\infty)^U\) and let \(y\) be its image under this injection. For all \(t\in \mathbb{T}^U\), its eigenvalue on \((\iota^{-1}\pi^\infty)^U\) only depends on how it acts on \(y\) and this is determined by the image of \(t\) in \(\mathbb{T}^S(H^*_!(X_U,M_{\mu,\mathcal{O}}))\). It follows that 10 factors through \(\mathbb{T}^S(H^*_!(X_U,M_{\mu,\mathcal{O}})).\) It remains to show that the image of 10 lies in \(\mathcal{O}\).

We first show that its image lies in \(E\). By [9], there is a \(\mathrm{GL}_n(\mathbb{A}_F^\infty)\)-stable \(\mathbb{Q}(\pi)\)-vector subspace \(W\) of \(\pi^\infty\) such that \(\pi^\infty=W\otimes_{\mathbb{Q}(\pi)}\mathbb{C}.\) Then \((\pi^\infty)^U=W^U\otimes_{\mathbb{Q}(\pi)}\mathbb{C}.\) Let \(h\in \mathcal{H}\). We have already seen that it acts by a scalar on \((\pi^\infty)^U\), so the same is true for \(W^U\). As \(W^U\) is a \(\mathbb{Q}(\pi)\)-vector space, the scalar must lies in \(\mathbb{Q}(\pi)\). Hence the image of 10 lies in \(\iota^{-1}(\mathbb{Q}(\pi))\subset E\).

Finally, by the existence of Borel-Serre compactification of \(X_U\), we know that \(H^*_!(X_U,M_{\mu,\mathcal{O}})\) is a finite \(\mathcal{O}\)-module, so \[H^*_!(X_U,M_{\mu,\mathcal{O}})/\mathcal{O}\text{-torsion}\] is a finite free \(\mathcal{O}\)-module stable under \(\mathcal{H}\). Pick an \(\mathcal{O}\)-basis \(\mathcal{B}\) for this module. We can then express the action of \(\mathcal{H}\) on \(H^*_!(X_U,M_{\mu,\mathcal{O}})/\mathcal{O}\text{-torsion}\) by matrices with entries in \(\mathcal{O}\). We can view \(\mathcal{B}\) as a \(\mathbb{C}\)-basis for \(H^*_!(X_U,M_{\mu,\mathbb{C}})\). Then the action of \(\mathcal{H}\) on \(H^*_!(X_U,M_{\mu,\mathbb{C}})\) is given by the same matrices. Hence each eigenvalue of \(\mathcal{H}\) on this space is a root of a monic polynomial over \(\mathcal{O}\). We know they lie in \(E\) by the previous paragraph, so they lie in \(\mathcal{O}\) as \(\mathcal{O}\) is integrally closed. ◻

Proof. Let \(H:=H^*_!(X_U,M_{\mu,\mathcal{O}})\). By the existence of Borel-Serre compactification of \(X_U\), \(H\) is a finite \(\mathcal{O}\)-module, so \(\mathbb{T}^S(H)\) is a finite \(\mathcal{O}\)-algebra. We let \[\Lambda':\mathbb{T}^S(H) \to \mathcal{O}\] be the map induced by \(\Lambda\), which exists by lemma 9, and let \(q:\mathbb{T}^S\to \mathbb{T}^S(H)\) be the quotient map, so \(\Lambda=\Lambda'\circ q\). Then \(\mathfrak{m}\supset \ker q\), so \(q(\mathfrak{m})\) is a maximal ideal of \(\mathbb{T}^S(H)\) and hence \[\mathbb{T}^S(H)_\mathfrak{m}= \mathbb{T}^S(H)_{q(\mathfrak{m})}\] is a finite local \(\mathcal{O}\)-algebra. The same is thus true for \(T\). As \(T\) is a finitely generated torsion-free \(\mathcal{O}\)-module and \(\mathcal{O}\) is a PID, \(T\) is flat over \(\mathcal{O}\). Hence \(T\hookrightarrow T\otimes_\mathcal{O}\mathbb{C}\). As \(\mathbb{C}\) is \(\mathcal{O}\)-flat, \(T\otimes_\mathcal{O}\mathbb{C}=\mathbb{T}^S_\mathbb{C}(H^*_!(X_U,M_{\mu,\mathbb{C}}))_\mathfrak{m},\) which is reduced by 4. Thus, \(T\), which injects into \(T\otimes_\mathcal{O}\mathbb{C}\), is also reduced.

Note that \(q(\mathfrak{m})=\ker(\Lambda'\mod \varpi)\). It follows that \(\Lambda'\) induces an \(\mathcal{O}\)-algebra map \(\mathbb{T}^S(H^*_!(X_U,M_{\mu,\mathcal{O}}))_{q(\mathfrak{m})}\to \mathcal{O}\) and hence an \(\mathcal{O}\)-algebra map \(\lambda:T\to \mathcal{O}\), i.e. a commutative diagram \[\begin{tikzcd} T & \mathcal{O}\\ \mathcal{O} \arrow["\lambda", from=1-1, to=1-2] \arrow[hook, from=2-1, to=1-1] \arrow[from=2-1, to=1-2] \end{tikzcd}\] It follows that \(\lambda^{-1}(\varpi \mathcal{O})\) is a prime ideal of \(T\) lying above \(\varpi\mathcal{O}\) (i.e. the preimage of \(\lambda^{-1}(\varpi \mathcal{O})\) along the structure map \(\mathcal{O}\to T\) is \(\varpi\mathcal{O}\)). Since \(T\) is a local \(\mathcal{O}\)-algebra, the maximal ideal of \(T\) also lies above \(\varpi\mathcal{O}\). By the incomparability theorem for injective integral ring extensions, we deduce that \(\lambda^{-1}(\varpi \mathcal{O})\) is the maximal ideal of \(T\). Thus, \(\lambda\) is a local homomorphism. ◻

Proof. Note that \(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\) is finite dimensional as a \(\overline{\mathbb{Q}}_p\)-vector space, so it is an Artinian ring. In particular, it is the finite product of its localisations at maximal ideas. Hence we get a corresponding descomposition \[V=\oplus_{\mathfrak p\in \mathrm{mSpec}\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)} V_\mathfrak p.\] By [19], \(V_\mathfrak p=V[\mathfrak p^\infty]\). By Zariski’s lemma, \(\mathrm{mSpec}\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V) = \{\ker \alpha |\; \alpha:\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\to\overline{\mathbb{Q}}_p \text{ is a \overline{\mathbb{Q}}_p-algebra homomorphism}\}.\) We thus get the desired decomposition of \(V\).

To see that \((\alpha|_{\mathbb{T}^S})^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p})= \mathfrak{n}\), note that \(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V) = \mathbb{T}^S(H^i(X_U,M_{\mu})_\mathfrak{n}) \otimes_{\mathcal{O}} \overline{\mathbb{Q}}_p\) by \(\mathcal{O}\)-flatness of \(\overline{\mathbb{Q}}_p\). Thus, \(\alpha\) gives rise to an \(\mathcal{O}\)-algebra homomorphism \(\mathbb{T}^S(H^i(X_U,M_{\mu})_\mathfrak{n}) \to \overline{\mathbb{Q}}_p\). The image of the this homomorphism necessarily lies in \(\mathcal{O}_{\overline{\mathbb{Q}}_p}\) because \(\mathbb{T}^S(H^i(X_U,M_{\mu})_\mathfrak{n})\) is finitely generated as an \(\mathcal{O}\)-module. By the same reasoning as 5, \(\mathbb{T}^S(H^i(X_U,M_{\mu})_\mathfrak{n})=\mathbb{T}^S(H^i(X_U,M_{\mu}))_\mathfrak{n}.\) Thus, we get an \(\mathcal{O}\)-algebra homomorphism \(a:\mathbb{T}^S(H^i(X_U,M_{\mu}))_\mathfrak{n}\to \mathcal{O}_{\overline{\mathbb{Q}}_p}\). We have \(a^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p})= \mathfrak{n}\mathbb{T}^S(H^i(X_U,M_{\mu}))_\mathfrak{n}\) by the going up theorem [20]. This proves \((\alpha|_{\mathbb{T}^S})^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p})= \mathfrak{n}\).

Part [decomp32of32W] can be proved in the same way using 4[semisimple]. One can also prove it in a more straightforward manner.

For part [item:32isotypic], we know \(T[\ker\lambda]\) is a finite free \(\mathcal{O}\)-module since it is torsion-free and finitely generated. To find its rank, note that \(\ker\lambda\) is the image of \(\ker\Lambda=(t-\Lambda(t):t\in \mathcal{\mathbb{T}}^S)\) under the projection \(q:\mathbb{T}^S\twoheadrightarrow T\). As \(T\) is Noetherian, we can pick \(t_1,\dots,t_n\in \mathbb{T}^S\) such that \(q((t_i-\Lambda(t_i):1\le i\le n))=\ker\lambda\). Then \[H[\ker\lambda]=H[\{t_i-\Lambda(t_i):1\le i\le n\}]=\ker(H\xrightarrow{m\mapsto (t_i-\Lambda(t_i)m)} H^n).\] Taking kernel commutes with flat base change, so \[\begin{align} H[\ker\lambda]\otimes_\mathcal{O}\mathbb{C}&= (H^i_!(X_U,M_{\mu,\mathcal{O}})_\mathfrak{m}[\epsilon]\otimes_\mathcal{O}\mathbb{C})[\ker\lambda]\nonumber\\ &= H^i_!(X_U,M_{\mu,\mathbb{C}})_\mathfrak{m}[\epsilon][\ker\lambda] \nonumber\\ &=H^i_!(X_U,M_{\mu,\mathbb{C}})[\epsilon][\ker\lambda]\tag{12}\\ &=H^i_!(X_U,M_{\mu,\mathbb{C}})[(\pi^{\infty,S})^{U^S}][\epsilon]\nonumber\\ &=H^i(\mathfrak g,K_\infty^\circ,\pi_\infty\otimes_\mathbb{C}M_{\mu,\mathbb{C}})[\epsilon]\otimes_\mathbb{C}(\pi^\infty)^U.\tag{13} \end{align}\] To see that equation 12 holds, note that \(\mathcal{O}\) is a Henselian local ring, so \(\mathbb{T}^S(H^i_!(X_U,M_{\mu,\mathcal{O}})) = \oplus_\mathfrak{n}\mathbb{T}^S(H^i_!(X_U,M_{\mu,\mathcal{O}}))_\mathfrak{n}\), where \(\mathfrak{n}\) runs over the maximal ideal of \(\mathbb{T}^S\) that contains \(\ker(\mathbb{T}^S \twoheadrightarrow \mathbb{T}^S(H^i_!(X_U,M_{\mu,\mathcal{O}})))\). Thus, \(H^i_!(X_U,M_{\mu,\mathbb{C}}) = \oplus_{\mathfrak{n}} H^i_!(X_U,M_{\mu,\mathbb{C}})_\mathfrak{n}\). By part [decomp32of32W], if \(\mathfrak{n}\neq \mathfrak{m}\), then \(H^i_!(X_U,M_{\mu,\mathbb{C}})_\mathfrak{n}[\ker\lambda]=0\). Equation 13 holds by lemma 4. By 3 and the choice of \(U\), 13 is a one dimensional \(\mathbb{C}\)-vector space. This proves part [item:32isotypic].

The first part of [eta32contains32eta] now follows from lemma 6. The remaining part is easy. ◻

Proof. It is enough to show the analogous fact after base changing from \(\mathcal{O}\) to \(\mathbb{C}\) and working at a single place. Let \(v\notin S\) be a finite place, \(G=\mathrm{GL}_n(F_v)\), \(K=\mathrm{GL}_n(\mathcal{O}_{F_v})\), \(V=\pi_v\).

It is well known that the restriction map induces an isomorphism \[\label{eq:res} (\widetilde{V})^K \xrightarrow{\sim} \widetilde{V^K}\tag{15}\] so in particular \((\widetilde{V})^K\) is 1-dimensional.

Let \(g\in G\). By the same proof as [21], there exist \(g_1,\dots,g_m\in G\) such that \(K g K= \sqcup_i g_i K = \sqcup K g_i\). For all \(f\in (\widetilde{V})^K\) and \(v\in V^K\), \[\begin{align} ([KgK]f)(v)=(\sum g_i\cdot f)(v)=f(\sum g_i^{-1}v) = f([Kg^{-1}K]v)=\lambda([Kg^{-1}K])f(v). \end{align}\] By 15 , \([KgK]f=\lambda([Kg^{-1}K])f\), as desired. ◻

Proof. As stated in [23], we have by Verdier duality an isomorphism \[RHom_\mathcal{O}(R\Gamma_c(X_U,M_\mu),\mathcal{O}) \cong R\Gamma(X_U,M_\mu^\vee)[d]\] in the derived category of \(\mathcal{O}\)-modules \(D(\mathcal{O})\). It follows from one of the spectral sequences for \(Ext\) that we have a spectral sequence \[E_2^{i,j}=Ext_\mathcal{O}^i(H^{-j}_c(X_U,M_\mu),\mathcal{O})\Rightarrow E^{i+j}=H^{i+j+d}(X_U,M_\mu^\vee).\] Since \(\mathcal{O}\) is a PID, the only non-zeros terms lie in \(\{(i,j):0\le i \le 1, -d\le j\le 0\}\). For all \(j\in\mathbb{Z}\), we have an exact sequence \[0\to E_2^{1,-j-1} \to E^{-j} \to E_2^{0,-j} \to 0\] i.e. \[0 \to Ext_\mathcal{O}^1(H^{j+1}_c(X_U,M_\mu),\mathcal{O}) \to H^{d-j}(X_U,M_\mu^\vee) \xrightarrow{f} \mathop{\mathrm{Hom}}_\mathcal{O}(H^{j}_c(X_U,M_\mu),\mathcal{O}) \to 0.\] Since \(H^{j+1}_c(X_U,M_\mu)\) is finitely generated over \(\mathcal{O}\) by the existence of Borel-Serre compactification, the second term is \(\mathcal{O}\)-torsion. On the other hand, the 4th term is \(\mathcal{O}\)-torsion free. It follows that \(\ker f\) is precisely the \(\mathcal{O}\)-torsion of \(H^{d-j}(X_U,M_\mu^\vee)\), so \(f\) induces an isomorphism \[\tilde{f}:H^{d-j}(X_U,M_\mu^\vee)/(\mathcal{O}-\text{tors}) \xrightarrow{\sim} \mathop{\mathrm{Hom}}_\mathcal{O}(H^{j}_c(X_U,M_\mu)/(\mathcal{O}-\text{tors}),\mathcal{O}).\] We have a pairing \[\begin{align} H^{d-j}(X_U,M_\mu^\vee)/(\mathcal{O}-\text{tors}) &\times H^{j}_c(X_U,M_\mu)/(\mathcal{O}-\text{tors}) \to \mathcal{O}\\ (a&,b)\mapsto \tilde{f}(a)(b). \end{align}\] This is a perfect pairing because both \(H^{d-j}(X_U,M_\mu^\vee)/(\mathcal{O}-\text{tors})\) and \(H^{j}_c(X_U,M_\mu)/(\mathcal{O}-\text{tors})\) are finite free \(\mathcal{O}\)-modules and \(\tilde{f}\) is an isomorphism. It is well-known that this is given by the cup product. The last assertion about the action of the Hecke algebra follows from [23]. ◻

Proof.

We shall show the first one. The second one is similar and easier. To show this, we let \(A:= H^b(X_U,M_{\mu,\overline{\mathbb{Q}}_p})_{\mathfrak{m}_1}, B:= H^{t}_c(X_U,M_{\mu,\overline{\mathbb{Q}}_p}^\vee)_{\mathfrak{m}_2}\). Suppose \([A,B]\neq 0\). By 11[decomp32of32V] and its analogue for the compactly supported cohomology, there exist \(a\in A[(\ker \alpha)^\infty], b\in B[(\ker \beta)^\infty]\) such that \([a,b]\neq 0\), where \(\alpha,\beta:\mathbb{T}^S_{\overline{\mathbb{Q}}_p} \to \overline{\mathbb{Q}}_p\) are \(\overline{\mathbb{Q}}_p\)-algebra homomorphisms with \((\alpha|_{\mathbb{T}^S})^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p})= \mathfrak{m}_1\) and \((\beta|_{\mathbb{T}^S})^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p})= \mathfrak{m}_2\). Let \(t\in \mathbb{T}^S\). Then \(t-\alpha(t)\in\ker\alpha\), so \((t-\alpha(t))^r a=0\) for some \(r\ge 1\). For all \(y\in B[(\ker \beta)^\infty]\), we have \[0 = [(t-\alpha(t))^r a,y] = [x, (\theta(t)-\alpha(t))^r y]\] Thus, \((\theta(t)-\alpha(t))^r\) is not surjective as an \(\overline{\mathbb{Q}}_p\)-linear endomorphism of \(B[(\ker \beta)^\infty]\), so it is not injective, so \((\theta(t)-\alpha(t))^r c=0\) for some \(c\in B[(\ker \beta)^\infty]\setminus\{0\}\). By the definition of \(B\), there exists \(s\ge 1\) such that \((\theta(t)-\beta(\theta(t)))^s c=0\). Since \(c\neq 0\), we must have \(\alpha(t) = \beta(\theta(t))\). Thus, \(\alpha|_{\mathbb{T}^S}=\beta|_{\mathbb{T}^S}\circ\theta\). It follows that \(\mathfrak{m}_1 = (\alpha|_{\mathbb{T}^S})^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p}) = (\beta|_{\mathbb{T}^S}\circ\theta)^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p}) = \theta(\mathfrak{m}_2)\).

Using this claim, we shall now deduce the first part of [item:32pp] by a similar argument to [1] under our assumption that \(H^b(\partial X_U,M_\mu)_\mathfrak{m}\) is \(\mathcal{O}\)-torsion free.

We first note that the base change of 17 to \(E\) \[\label{E32perf32pair} [\,,\,]:H^b_!(X_U,M_{\mu,E})_\mathfrak{m}\times H^{t}_!(X_U,M_{\mu,E}^\vee)_{\tilde{\mathfrak{m}}} \to E\tag{18}\] is a perfect pairing, because this is non-degenerate by Poincaré duality and the claim proved above. Similarly, \[\label{E32perf32pair2} [\,,\,]:H^b(X_U,M_{\mu,E})_\mathfrak{m}\times H^{t}_c(X_U,M_{\mu,E}^\vee)_{\tilde{\mathfrak{m}}} \to E\tag{19}\] is a perfect pairing.

Next, note that since \(\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}\) and \(\overline{H}^{t}_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak{m}}}\) are both finite free \(\mathcal{O}\)-modules, to show that 17 is a perfect pairing, it is enough to show that the induced map \[\label{eq:32perf32pair} \overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}\to \mathop{\mathrm{Hom}}_{\mathcal{O}}(\overline{H}^{t}_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak{m}}} , \mathcal{O})\tag{20}\] is an isomorphism. To see injectivity, note that \(\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}\hookrightarrow \overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}\otimes_{\mathcal{O}}E = H^b_!(X_U,M_{\mu,E})_\mathfrak{m}\) and similarly for the right hand side of 20 . Thus, we deduce the injectivity of 20 from the non-degeneracy of 18 .

To check surjectivity of 20 , let \(f\in \mathop{\mathrm{Hom}}_{\mathcal{O}}(\overline{H}^{t}_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak{m}}} , \mathcal{O})\). Analogous to 16 , \(\overline{H}^{t}_!(X_U,M_\mu^\vee)\) is the direct sum of its localisations at maximal ideals of \(\mathbb{T}^S\). We extend \(f\) to \(f': \overline{H}^{t}_!(X_U,M_\mu^\vee) \to \mathcal{O}\) by setting it to be \(0\) on other summands. Let \(q: \overline{H}^{t}_c(X_U,M_\mu^\vee) \twoheadrightarrow \overline{H}^{t}_!(X_U,M_\mu^\vee)\) be the natural map. By 6, there exists \(x\in \overline{H}^b(X_U,M_\mu)\) such that \[f'\circ q=[x,-].\] By the claim, we can assume that \(x\) lies in the direct summand \(\overline{H}^b(X_U,M_\mu)_\mathfrak{m}\) of \(\overline{H}^b(X_U,M_\mu)\). On the other hand, by the perfectness of 20 , there exists \(z\in H^b_!(X_U,M_{\mu,E})_\mathfrak{m}\) such that \(f_E = [z,-]\), where \(f_E\) is the base change of \(f\) along \(\mathcal{O}\to E\). Then \[f'_E\circ q_E = [z,-]\] as maps \(H^{t}_c(X_U,M_{\mu,E}^\vee) \to E\). By the perfectness of 19 , \(x_E = z\), where \(x_E\) is the image of \(x\) in \(H^b(X_U,M_{\mu,E})_\mathfrak{m}\). In particular, \[x_E \in H^b_!(X_U,M_{\mu,E})_\mathfrak{m}.\] Recall that there is a long exact sequence \[\dots\to H^i_c(X_U,M_\mu)\to H^i(X_U,M_\mu) \to H^i(\partial X_U,M_\mu)\to \dots\] We lift \(x\) to \(H^b(X_U,M_\mu)_{\mathfrak{m}}\) and consider its image \(y\) in \(H^b(X_U,M_\mu)_{\mathfrak{m}}/H^b_!(X_U,M_\mu)_{\mathfrak{m}} \subset H^b(\partial X_U,M_\mu)_{\mathfrak{m}}\). By what we have just shown, \(y_E \in H^b(\partial X_U,M_{\mu,E})_{\mathfrak{m}}\) is \(0\). Since \(H^b(\partial X_U,M_\mu)_{\mathfrak{m}}\) is \(\mathcal{O}\)-torsion-free, \(y=0\). Thus, \(x \in \overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}\).

For the second part of [item:32pp], note that we have a decomposition \[\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}= \bigoplus_{\epsilon_1\in \widehat{K_\infty/K_\infty^\circ}}\overline{H}^b_!(X_U,M_\mu)_{\mathfrak{m}}[\epsilon_1]\] since \(p>2\). The perfectness then follows from the first part and the proof of [1]. (The proof there works here in view of the decomposition of \(H^{i}(\mathfrak g, K_\infty^0,\pi_\infty\otimes_\mathbb{C}M_{\mu,\mathbb{C}})\) in the last part of the proof of 3.)

Part [item:32theta32isom] follows from part [item:32pp] and the same argument as [23], namely the commutativity of the diagram \[\begin{tikzcd} {\mathbb{T}^S} & {End_{\mathcal{O}}(\overline{H}^b_!(X_U,M_\mu)_{\mathfrak m})} \\ {\mathbb{T}^S} & {End_{\mathcal{O}}(\overline{H}^t_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak m}})} \arrow[from=1-1, to=1-2] \arrow["\theta"', from=1-1, to=2-1] \arrow["{\text{transpose}}", from=1-2, to=2-2] \arrow[from=2-1, to=2-2] \end{tikzcd}\] ◻

Proof. Note that \(\int_{A_G G(F)\backslash G(\mathbb{A}_F)} = \int_{Z(\mathbb{A}_F) G(F)\backslash G(\mathbb{A}_F)} \int_{A_G G(F)\backslash Z(\mathbb{A}_F)G(F)}\), where \(Z\) is the centre of \(G\). Also, \(A_G G(F)\backslash Z(\mathbb{A}_F)G(F) = F^\times \backslash \mathbb{A}^1_F\), where \(\mathbb{A}^1_F:=\{x\in \mathbb{A}_F: |x|_{\mathbb{A}_F}=1\}.\) It follows that \[\langle \phi,\tilde{\phi} \rangle = vol(F^\times\backslash \mathbb{A}_F^1)\int_{Z(\mathbb{A}_F)G(F)\backslash G(\mathbb{A}_F)} \phi(g)\tilde{\phi}(g) dg.\] The result now follows from [1]. They obtained their result by relating the Petersson inner product with \(L(1,\pi,\mathop{\mathrm{Ad}}^0)\) by the Rankin-Selberg method and using the fact that \(L(s,\pi\times \tilde{\pi})=\tilde{\zeta}_F(s)L(s,\pi,\mathop{\mathrm{Ad}}^0).\) ◻

Proof. This is more or less what [1] obtained, except that our space \(X_U\) is different from the locally symmetric spaces they used. As in [1], we have \[\begin{align} [\mathfrak p_{\pi,b,\epsilon}\vartheta^\circ_{b,\epsilon},\mathfrak p_{\tilde{\pi},t,\tilde{\epsilon}}\tilde{\vartheta}^\circ_{t,\tilde{\epsilon}}] &= \frac{1}{vol(U)} \int_{G(F) \backslash G(\mathbb{A}_F)/K_\infty^\circ U} \varsigma\\ &= \frac{1}{vol(U)}\int_{G(F) \backslash G(\mathbb{A}_F)/A_G U} \varsigma\\ &= \int_{G(F) \backslash G(\mathbb{A}_F)/A_G} \varsigma\\ &= \frac{L(1,\pi,\mathop{\mathrm{Ad}}^0)}{\alpha_F \,\mathfrak p_{ram}(\pi)\mathfrak p_\infty(\pi)} \end{align}\] where \(\varsigma\) has the same meaning as that in [1] and in the last equality we used 13 instead of [1]. Dividing both sides by \(\mathfrak p_{\pi,b,\epsilon}\mathfrak p_{\tilde{\pi},t,\tilde{\epsilon}}\) gives the result. ◻

Proof. Since \(H^b(\partial X_U,M_\mu)_\mathfrak{m}\) is \(\mathcal{O}\)-torsion free, the cup product gives a perfect pairing \[[\,,\,]:\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}[\epsilon] \times \overline{H}^{t}_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak{m}}}[\tilde{\epsilon}] \to \mathcal{O}\] by 1.²⁵ We would like to apply 7. The conditions in part (b) of that lemma is satisfied by 11 and the fact that \((\pi^\infty)^U\) and \((\tilde{\pi}^\infty)^U\) are one dimensional.

Recall that \(\lambda:T\to \mathcal{O}\) is the Hecke eigensystem for \(\pi\). This factors through \(\mathbb{T}^S(\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m})\) because the action of the Hecke algebra on \(\overline{H}^*_!(X_U,M_\mu)_\mathfrak{m}\) preserves degree and \(\pi\) is isomorphic to a submodule of \(\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}\otimes_\mathcal{O}\mathbb{C}\).

Similar to how we defined the idempotent \(e\in T_E\) using \(\lambda:T\to \mathcal{O}\), we can define an idempotent \(e'\in \mathbb{T}^S(\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m})_E\) using the induced map on the quotient \(\mathbb{T}^S(\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m})\to \mathcal{O}\). It is clear from the definitions of \(e\) and \(e'\) that \(e'\) is the image of \(e\) under \(T_E \twoheadrightarrow \mathbb{T}^S(\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m})_E\).

The same argument (with \(H^b_!(X_U,M_\mu)_\mathfrak{m}\) replaced by \(H^t_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak{m}}}\)) works for the contragredient \(\tilde{\pi}\), \(\tilde{\lambda}\), \(\tilde{e}, \tilde{e}'\) and we know \(\tilde{e}'\) is the image of \(\tilde{e}\) under \(\tilde{T}_E \twoheadrightarrow \mathbb{T}^S(\overline{H}^t_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak{m}}})_E\).

By 12, we know \(\tilde{\lambda}=\lambda\circ\theta\). It follows from definition and part (b) of 1 that \(\theta(e')=\tilde{e}'\). Thus, for all \(x\in\overline{H}^b_!(X_U,M_\mu)_\mathfrak{m}[\epsilon]\), \(y \in \overline{H}^{t}_!(X_U,M_\mu^\vee)_{\tilde{\mathfrak{m}}}[\tilde{\epsilon}]\), we have \[[ex,y]=[e'x,y]=[x,\theta(e')y]=[x,\tilde{e}' y]=[x,\tilde{e} y].\] From this²⁶ and the fact that \(e,e'\) are idempotents, we deduce that all the conditions of 7 are satisfied.

By 7 and 14, \[\eta_{\pi,b,\epsilon}=\eta_{\tilde{\pi},t,\tilde{\epsilon}}= [\vartheta^\circ_{b,\epsilon},\tilde{\vartheta}^\circ_{t,\tilde{\epsilon}}] =(L^{alg}(1,\pi,\mathop{\mathrm{Ad}}^0,\epsilon)).\] ◻

Proof. Abusing notation, we shall identify \(\overline{\mathbb{Q}}_p\) with \(\mathbb{C}\) using our fixed isomorphism \(\iota\). Note that \(\pi'\cong \pi\) if and only if \(\Lambda'=\Lambda\) by the strong multiplicity one theorem (where we regard \(\Lambda\) as having codomain in \(\overline{\mathbb{Q}}_p\)).

Suppose \(\varpi\mid L^{alg}(1,\pi,\mathop{\mathrm{Ad}}^0,\epsilon)\). By 8 and 6, \(\eta_\pi\neq \mathcal{O}\). By 4 there exists a \(\overline{\mathbb{Q}}_p\)-algebra homomorphism \(\lambda':T\otimes_\mathcal{O}\overline{\mathbb{Q}}_p \to \overline{\mathbb{Q}}_p\) such that \(\lambda'\neq \lambda\otimes_{\mathcal{O}}\overline{\mathbb{Q}}_p\) and \(|\lambda(t)-\lambda'(t)|<1\) for all \(t\in T\). We have a natural map \[\mathbb{T}^S(H^*_!(X_U,M_\mu))\otimes_\mathcal{O}\overline{\mathbb{Q}}_p = \mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\to T\otimes_\mathcal{O}\overline{\mathbb{Q}}_p,\] where \(V:=H^*_!(X_U,M_{\mu,\overline{\mathbb{Q}}_p})\) and \(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V):= im(\mathbb{T}^S\otimes_\mathcal{O}\bar\mathbb{Q}_p \to \mathop{\mathrm{End}}_{\bar\mathbb{Q}_p}(V))\). Composing this with \(\lambda'\), we get a \(\overline{\mathbb{Q}}_p\)-algebra homomorphism \[f:\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\to \overline{\mathbb{Q}}_p.\] Let \(\mathfrak{n}:=\ker f\). Then \[V_\mathfrak{n}\neq 0\] because \(Supp_{\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)}(V)= \{\mathfrak p\in Spec(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)): \mathfrak p\supset Ann_{\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)}(V)\}=Spec(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)).\) By [19], \(V[\mathfrak{n}]\neq 0\). By 4 part [semisimple], there is a discrete automorphic representation \(\pi'\) such that \((\pi'^{\infty,S})^{U^S}\) is isomorphic to a sub-\(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\)-module of \(V\) and \((\pi'^{\infty,S})^{U^S}[\mathfrak{n}]\neq 0.\) Note that \((\pi'^{\infty,S})^{U^S} = \otimes'_v (\pi'_v)^{U_v}\) is one dimensional over \(\overline{\mathbb{Q}}_p\), so \((\pi'^{\infty,S})^{U^S}[\mathfrak{n}]= (\pi'^{\infty,S})^{U^S}.\) The Hecke eigensystem attached to \(\pi'\) has the desired property.

Suppose \(H^b_!(X_U,M_{\mu,\mathcal{O}})_\mathfrak{m}[\epsilon]/\mathcal{O}\text{-torsion}\) is a free \(T\)-module, say, of rank \(d\), and there is a \(\pi'\) satisfying the statement of the corollary. Note that \(d\ge 1\) by 11[item:32isotypic]. By freeness, \(\eta_{\pi,b,\epsilon}=\eta_\pi^d\), so by 8, it suffices to show that \(\eta_\pi\neq \mathcal{O}\). By 8, it suffices to show that there is an \(\mathcal{O}\)-algebra homomorphism \(\lambda':T\to \overline{\mathbb{Q}}_p\) with \(\lambda\neq \lambda'\) and \(|\lambda(t)-\lambda'(t)|<1\) for all \(t\in T\). Equivalently, we need to show that there is an \(\mathcal{O}\)-algebra homomorphism \(\Lambda'':\mathbb{T}^S\to \overline{\mathbb{Q}}_p\) that factors through \(\mathbb{T}^S(H^*_!(X_U,M_{\mu}))\) with \(\Lambda\neq \Lambda''\) and \(|\Lambda(t)-\Lambda''(t)|<1\) for all \(t\in \mathbb{T}^S\), because any such \(\Lambda''\) necessarily factors through \(T\). For this, we can take \(\Lambda''\) to be the Hecke eigensystem attached to \(\pi'\). ◻

Proof. The key input for the first part is [14], which states that for every smooth \(\mathcal{O}[U_S]\)-module \(A\) that is finite as an \(\mathcal{O}\)-module, \(H^*(\partial X_U,A)_\mathfrak{m}=0.\) They proved this using the fact that \(\partial X_U\) admits a stratification with strata indexed by conjugacy classes of proper parabolic subgroups of \(\mathrm{GL}_n\) and by an in depth analysis of the cohomology of each stratum.

Recall that by [14], \(M_\mu \otimes_\mathcal{O}\mathcal{O}/\varpi=M_{\mu,\mathcal{O}/\varpi},\) which receives an action of \(\mathrm{GL}_n(\mathcal{O}/\varpi)\), compatible with that of \(\mathrm{GL}_n(\mathcal{O})\). Thus \(M_{\mu,\mathcal{O}/\varpi}\) is a smooth \(\mathcal{O}[U_S]\)-module that is finite as \(\mathcal{O}\)-module, so \[H^*(\partial X_U,M_{\mu,\mathcal{O}/\varpi})_\mathfrak{m}=0.\] We get the desired result by considering the long exact sequence associated to \[0\to M_\mu \xrightarrow{\varpi} M_\mu \to M_{\mu,\mathcal{O}/\varpi} \to 0\] and applying the Nakayama lemma to the finite \(\mathcal{O}\)-module \(H^*(\partial X_U,M_{\mu})_\mathfrak{m}=0.\)

Part [two32equal] now follows from the long exact sequence \[\dots\to H^i_c(X_U,M_\mu) \to H^i(X_U,M_\mu) \to H^i(\partial X_U,M_\mu)\to \dots\]

For part [alt32T], note that \(T^S(H)_\mathfrak{m}/\mathcal{O}\text{-tors}\cong \mathbb{T}^S(\overline{H_\mathfrak{m}})\) for \(H\in\{H^*(X_U,M_{\mu}),H^*_!(X_U,M_{\mu})\}\) by the same proof as 5

For part [all32equal], it is proved in [23] using Franke’s decomposition of \(H^*(X_U,M_{\mu,\mathbb{C}})\) via automorphic forms that \(H^*(X_U,M_{\mu,\mathbb{C}})_\mathfrak{m}=H^*_{cusp}(X_U,M_{\mu,\mathbb{C}})_\mathfrak{m}.\) As \(H^*_{!}(X_U,M_{\mu,\mathbb{C}})_\mathfrak{m}\) is always sandwiched between these two groups, these groups are all equal. (The first equality also follows from part [two32equal].) ◻

Proof. The proof is just a slight variation of that of 2. Abusing notation, we shall identify \(\overline{\mathbb{Q}}_p\) with \(\mathbb{C}\) using our fixed isomorphism \(\iota\). Suppose \(\varpi\mid L^{alg}(1,\pi,\mathop{\mathrm{Ad}}^0,\epsilon)\). By 8 and 6, \(\eta_\pi\neq \mathcal{O}\). By 4 there exists a \(\overline{\mathbb{Q}}_p\)-algebra homomorphism \(\lambda':T\otimes_\mathcal{O}\overline{\mathbb{Q}}_p \to \overline{\mathbb{Q}}_p\) such that \(\lambda'\neq \lambda\otimes_{\mathcal{O}}\overline{\mathbb{Q}}_p\) and \(|\lambda(t)-\lambda'(t)|<1\) for all \(t\in T\). We have a natural map \[\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\to T\otimes_\mathcal{O}\overline{\mathbb{Q}}_p=\mathbb{T}^S_{\bar\mathbb{Q}_p}(H^*_{cusp}(X_U,M_{\mu,\bar\mathbb{Q}_p})_\mathfrak{m}),\] where the last equality is by 15, \(V:=H^*_{cusp}(X_U,M_{\mu,\overline{\mathbb{Q}}_p})\), and \(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V):= im(\mathbb{T}^S\otimes_\mathcal{O}\bar\mathbb{Q}_p \to \mathop{\mathrm{End}}_{\bar\mathbb{Q}_p}(V))\). Composing this with \(\lambda'\), we get a \(\overline{\mathbb{Q}}_p\)-algebra homomorphism \[f:\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\to \overline{\mathbb{Q}}_p.\] Let \(\mathfrak{n}:=\ker f\). Then \[V_\mathfrak{n}\neq 0\] because \(Supp_{\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)}(V)= \{\mathfrak p\in Spec(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)): \mathfrak p\supset Ann_{\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)}(V)\}=Spec(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)).\) By [19], \(V[\mathfrak{n}]\neq 0\). Thus, there is a cohomological cuspidal automorphic representation \(\pi'\) of weight \(\iota\mu\) such that \((\pi'^{\infty,S})^{U^S}\) is isomorphic to a sub-\(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(V)\)-module of \(V\) and \((\pi'^{\infty,S})^{U^S}[\mathfrak{n}]\neq 0.\) Note that \((\pi'^{\infty,S})^{U^S}= \otimes'_v (\pi'_v)^{U_v}\) is one dimensional over \(\overline{\mathbb{Q}}_p\), so \((\pi'^{\infty,S})^{U^S}[\mathfrak{n}]= (\pi'^{\infty,S})^{U^S}.\) The Hecke eigensystem attached to \(\pi'\) has the desired property.

When \(H^b_!(X_U,M_{\mu,\mathcal{O}})_\mathfrak{m}[\epsilon]/\mathcal{O}\text{-torsion}\) is a free \(T\)-module, the converse follows from 2 because if \(\pi'\) is a cohomological cuspidal automorphic representation \(\pi'\) of weight \(\iota\mu\) of \(\mathrm{GL}_n(\mathbb{A}_F)\) with \((\pi')^U\neq 0\), then \(\pi'\) is a discrete automorphic representation and \(0\neq H^b_{cusp}(X_U,M_{\mu,\mathbb{C}})[(\pi'^{\infty,S})^{U^S}] \subset H^b_!(X_U,M_{\mu,\mathbb{C}})[(\pi'^{\infty,S})^{U^S}]\). ◻

Proof. This is well-known so we will just sketch a proof. Note that \(\mathfrak p/\mathfrak p^2\) is a finitely generated \(R_{\mathcal{S}}/\mathfrak p=\mathcal{O}\)-module, so any \(\mathcal{O}\)-algebra homomorphism \(\mathfrak p/\mathfrak p^2 \to E/\mathcal{O}\) has image contained in \(\varpi^{-m}\mathcal{O}/\mathcal{O}\) for some \(m\ge 1\). Thus, if we know \[\mathop{\mathrm{Hom}}_\mathcal{O}(\mathfrak p/\mathfrak p^2, \varpi^{-m}\mathcal{O}/\mathcal{O}) \cong H^1_{\mathcal{S}}(\mathop{\mathrm{Ad}}\rho \otimes_\mathcal{O}\varpi^{-m}\mathcal{O}/\mathcal{O})\] for all \(m\ge 1\), then taking colimit will give the desired result. This follows from the chain of isomorphisms \[\begin{align} &\;\;\;H^1_{\mathcal{S}}(\mathop{\mathrm{Ad}}\rho \otimes_\mathcal{O}\varpi^{-m}\mathcal{O}/\mathcal{O})\nonumber\\ \cong &\;\{\text{liftings } G_{F,S} \to \mathrm{GL}_n(\mathcal{O}_m) \text{ of }\rho\text{ of type }\mathcal{S}\}/\ker(\mathrm{GL}_n(\mathcal{O}_m)\to \mathrm{GL}_n(\mathcal{O})) \tag{22}\\ \cong &\;\{\text{deformations } G_{F,S} \to \mathrm{GL}_n(\mathcal{O}_m) \text{ of }[\rho]\text{ of type }\mathcal{S}\}\tag{23}\\ \cong &\;\{f\in\mathop{\mathrm{Hom}}_\mathcal{O}(R_{\mathcal{S}},\mathcal{O}_m): f\pmod\epsilon=\theta\}\nonumber\\ \cong &\;\mathop{\mathrm{Hom}}_\mathcal{O}(\mathfrak p/\mathfrak p^2, \varpi^{-m}\mathcal{O}/\mathcal{O}).\nonumber \end{align}\] In 22 , \(\ker(\mathrm{GL}_n(\mathcal{O}_m)\to \mathrm{GL}_n(\mathcal{O}))\) acts on \(\{\text{liftings } G_{F,S} \to \mathrm{GL}_n(\mathcal{O}_m) \text{ of }\rho\text{ of type }\mathcal{S}\}\) by conjugation. In 23 , by deformations of \([\rho]\), we mean deformations of \(\bar\rho\) whose pushforward to \(\mathrm{GL}_n(\mathcal{O})\) is \([\rho]\). To show the bijectivity of 22 and 23 , note that the centralizer of the image of \(\rho\) in \(\mathrm{GL}_n(\mathcal{O})\) is \(\mathcal{O}^\times\) by [30] and \(\bar\rho\) is absolutely irreducible. The other steps are easy. ◻

Proof. The key to proving the existence³² of \(\rho_\mathfrak{m}\) is to use [29] and Carayol’s lemma.

Since \(\mathfrak{m}\) is non-Eisenstein, \(H^*_!(X_U,M_{\mu,\overline{\mathbb{Q}}_p})_\mathfrak{m}=H^*_{cusp}(X_U,M_{\mu,\overline{\mathbb{Q}}_p})_\mathfrak{m}\) by 15. Let \(W=H^*_{cusp}(X_U,M_{\mu,\overline{\mathbb{Q}}_p})_\mathfrak{m}\). By (the proof of) 11[decomp32of32W], \[\label{decomp32of32W322} W = \oplus_{\alpha: \mathbb{T}^S_{\overline{\mathbb{Q}}_p}(W) \to \overline{\mathbb{Q}}_p} W[\ker\alpha]\tag{25}\] where \(\alpha\) runs over \(\overline{\mathbb{Q}}_p\)-algebra homomorphisms \(\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(W)\to\overline{\mathbb{Q}}_p\). Any such \(\alpha\) satisfies \((\alpha|_{\mathbb{T}^S})^{-1}(\mathfrak{m}_{\overline{\mathbb{Q}}_p})= \mathfrak{m}\). By equation 3 and [16], \[H^*_{cusp}(X_U,M_{\mu,\mathbb{C}})= \oplus_{\pi'} H^*(\mathfrak g,K_\infty^\circ , M_{\mu,\mathbb{C}}\otimes_\mathbb{C}\pi'_\infty)\otimes_\mathbb{C}(\pi'^\infty)^U,\] where \(\pi'\) runs over all cohomological cuspidal automorphic representations of \(\mathrm{GL}_n(\mathbb{A}_F)\) of weight \(\mu\) with \(\pi'^U\neq 0\). It follows that any \(\alpha\) as above gives rise to a cohomological cuspidal automorphic representation \(\pi_{\alpha}\) of \(\mathrm{GL}_n(\mathbb{A}_F)\) of weight \(\mu\) such that \(\pi_{\alpha}^U\neq 0\) and \(t\in\mathbb{T}^S\) acts on \(\pi^U\) by \(\alpha(t)\).

We have a \(\overline{\mathbb{Q}}_p\)-algebra homomorphism \[\begin{align} h:T\otimes_\mathcal{O}\overline{\mathbb{Q}}_p= \mathbb{T}^S_{\overline{\mathbb{Q}}_p}(W) &\to \prod_{\alpha:\mathbb{T}^S_{\overline{\mathbb{Q}}_p}(W) \to \overline{\mathbb{Q}}_p} \overline{\mathbb{Q}}_p\\ t &\mapsto (\alpha(t))_{\alpha}. \end{align}\] By 25 , \(h\) is injective. If \(h\) is not surjective, then its image is a proper \(\overline{\mathbb{Q}}_p\)-subalgebra of \(\prod_{\alpha}\overline{\mathbb{Q}}_p\), so, by a simple arguement, there exist \(\beta,\gamma: \mathbb{T}^S_{\overline{\mathbb{Q}}_p}(W) \to \overline{\mathbb{Q}}_p\) such that \(\beta\neq \gamma\) and the image is contained in \(\{z\in\prod_\alpha \overline{\mathbb{Q}}_p: z_\beta = z_\gamma\}.\) This is impossible by the strong multiplicity one theorem. Hence, \(h\) is an isomorphism.

By [29], for every \(\alpha\), there is a unique continuous semisimple representation \[\rho_{\pi_\alpha}: \mathop{\mathrm{Gal}}(\overline{F}/F) \to \mathrm{GL}_n(\overline{\mathbb{Q}}_p)\] such that for all \(v\not\in S\), \(\rho_{\pi_\alpha}\) is unramified at \(v\) and \(\rho_{\pi_\alpha}(\mathop{\mathrm{Frob}}_v)\) has characteristic polynomial \(\alpha(P_v(X))\), where \(\alpha(P_v(X))\) is the polynomial obtained by applying \(\alpha\) to every coefficient of \(P_v(X):= X^n-T_{v,1}X^{n-1}+\dots+(-1)^i q_v^{i(i-1)/2}T_{v,i}X^{n-i}+\dots+ (-1)^n q_v^{n(n-1)/2}T_{v,n}\). By the Baire category theorem and the compactness of \(\mathop{\mathrm{Gal}}(\overline{F}/F)\), after conjugating \(\rho_{\pi_\alpha}\) by elements of \(\mathrm{GL}_n(\overline{\mathbb{Q}}_p)\), we can assume that \(\rho_{\pi_\alpha}(\mathop{\mathrm{Gal}}(\overline{F}/F)) \subset \mathrm{GL}_n(\mathcal{O}_L)\) for every \(\alpha\), where \(L\) is a finite extension of \(E\). See [30] for the proof of an analogous result. By the Brauer-Nesbitt theorem, every residual Galois representation \(\overline{\rho}_{\pi_\alpha}: \mathop{\mathrm{Gal}}(\overline{F}/F) \to \mathrm{GL}_n(k_L)\), where \(k_L\) is the residue field of \(L\), is isomorphic to \(\overline{\rho}_\pi\). We can conjugate \(\rho_{\pi_\alpha}\) by elements of \(\mathrm{GL}_n(\mathcal{O}_L)\) to ensure that all these residual Galois representations are equal to \(\overline{\rho}_\pi\), not just conjugate to it. By enlarging \(L\) if necessary, we can also assume \(h(T)\subset \prod_\alpha \mathcal{O}_L\).

Let \[\rho=(\rho_{\pi_\alpha}) : \mathop{\mathrm{Gal}}(\overline{F}/F) \to \mathrm{GL}_n(T\otimes_\mathcal{O}\overline{\mathbb{Q}}_p) = \prod_\alpha \mathrm{GL}_n(\overline{\mathbb{Q}}_p).\] For every \(v\not\in S\), \(\rho\) is unramified at \(v\) and \(\rho(\mathop{\mathrm{Frob}}_v)\) has characteristic polynomial \(P_v(X).\)

Let \(A:=\{(x_\alpha)\in\prod_\alpha \mathcal{O}_L: \bar x_\alpha \text{ are all equal and lies in }\mathcal{O}/\varpi\}\), where \(\bar x_\alpha\) is the reduction mod \(\mathfrak{m}_L\) of \(x\). This is a local ring with maxial ideal \(\mathfrak{m}_A:=\{(x_\alpha)\in\prod_\alpha \mathcal{O}_L: \bar x_\alpha =0 \text{ for all }\alpha\}\) and residue field \(\mathcal{O}/\varpi\). This is also finitely generated as an \(\mathcal{O}\)-module, so it is a complete Noetherian ring.³³ Thus, \(A\in CNL_\mathcal{O}\) (the category of complete Noetherian local \(\mathcal{O}\)-algebra with residue field \(\mathcal{O}/\varpi\)).

In the remaining part, let us denote the image of \(\mathfrak{m}\) in \(T\) also as \(\mathfrak{m}\). Note that \(T\subset A\) and \(T\in CNL_\mathcal{O}\). Also, \(T\cap \mathfrak{m}_A = \mathfrak{m}\) by, so \(T\hookrightarrow A\) is continuous. This is a closed map of topological spaces, because \(T\) is compact³⁴ and \(A\) is Hausdorff. In particular, \(T\) is a closed subset of \(A\).

Note that \(\rho(\mathop{\mathrm{Gal}}(\overline{F}/F)) \subset \mathrm{GL}_n(A)\), so we can view \(\rho\) as a representation \[\rho : \mathop{\mathrm{Gal}}(\overline{F}/F) \to \mathrm{GL}_n(A).\] The residual Galois representation \(\bar\rho:\mathop{\mathrm{Gal}}(\overline{F}/F) \to \mathrm{GL}_n(A/\mathfrak{m}_A)\) is absolutely irreducible because \(\mathfrak{m}\) is non-Eisenstein. We know that \(\mathrm{tr}(\rho(\mathop{\mathrm{Frob}}_v))= \pm T_{v,1} \in T\) for all \(v\not\in S\). By Chebotarev density theorem and the fact that \(T\subset A\) is a closed subset, \(\mathrm{tr}\rho(\mathop{\mathrm{Gal}}(\overline{F}/F)) \subset T\). By Carayol’s lemma ([30]), there exists \(g\in \mathrm{GL}_n(A) \to \mathrm{GL}_n(\mathcal{O}/\varpi)\) such that \(g \rho(\mathop{\mathrm{Gal}}(\overline{F}/F)) g^{-1} \subset \mathrm{GL}_n(T)\). Then \(g\rho g^{-1}\) is unramified outside \(S\), so it gives rise to \[\rho_\mathfrak{m}:G_{F,S} \to \mathrm{GL}_n(T)\] such that for all \(v\notin S\) of \(F\), the characteristic polynomial of \(\rho_\mathfrak{m}(\mathop{\mathrm{Frob}}_v)\) is \(P_v(X)\). This proves the first part.

For the second part, assume that \(\rho_\mathfrak{m}\) is of type \(\mathcal{S}\). Its strict equivalence class induces an \(\mathcal{O}\)-algebra homomorphism \(f:R_{\mathcal{S}}\to T\). We know that \(\mathbb{T}^S\) is generated by \[\{T^i_v,(T^n_v)^{-1}: v\notin S, 1\le i\le n\}\] as an \(\mathcal{O}\)-algebra³⁵. For all \(g\in G_{F,S}\), every coefficient of the characteristic polynomial of \(\rho_\mathfrak{m}(g)\) is in the image of \(R_{\mathcal{S}}\to T\). Taking \(\mathop{\mathrm{Frob}}_v^{\pm 1}\), we know \(f\) is surjective.

If \(H^1_{\mathcal{S}}(\mathop{\mathrm{Ad}}\rho \otimes_\mathcal{O}E/\mathcal{O})\) is infinite, then equation 24 is trivial. Suppose \(H^1_{\mathcal{S}}(\mathop{\mathrm{Ad}}\rho \otimes_\mathcal{O}E/\mathcal{O})\) is finite. Let \(\theta=\lambda\circ f\) and \(\mathfrak p:=\ker\theta\). It follows from 16 that \(\mathfrak p/\mathfrak p^2\) is also finite and \[\#H^1_{\mathcal{S}}(\mathop{\mathrm{Ad}}\rho \otimes_\mathcal{O}E/\mathcal{O}) = \#(\mathfrak p/\mathfrak p^2).\] By [31], \[\#(\mathfrak p/\mathfrak p^2) \ge \#(\mathcal{O}/\eta_{R_{\mathcal{S}}})\] where \(\eta_{R_{\mathcal{S}}}:=\theta(Ann_{R_{\mathcal{S}}}(\mathfrak p))\). By [31], \[\#(\mathcal{O}/\eta_{R_{\mathcal{S}}}) \ge \#(\mathcal{O}/\eta_T),\] where \(\eta_T=\lambda(Ann_T(\ker\lambda))\), which is the same as \(\eta_\pi\) in the previous subsections. By 6, 15, and 8, \[\#(\mathcal{O}/\eta_\pi) \ge \#(\mathcal{O}/\eta_{\pi,b,\epsilon}) \ge \# (\mathcal{O}/L^{alg}(1,\pi,\mathop{\mathrm{Ad}}^\circ,\epsilon)).\] We get the desired inequality. ◻

Proof. This follows from the proof of 13 and by replacing the use of [29] by [7]. ◻