---
abstract: 'We prove a realization of the local Langlands correspondence for two-dimensional representations of a Weil group of conductor three in the cohomology of Lubin-Tate curves by a purely local geometric method.' 
author:
- Naoki Imai and Takahiro Tsushima
title: |
    Geometric realization of the local Langlands correspondence\
    for representations of conductor three
---

Introduction {#introduction .unnumbered}
============

Let $K$ be a non-archimedean local field with a finite residue field $k$ of characteristic $p$. Let $\mathfrak{p}$ be the maximal ideal of the ring of integers of $K$. We take an algebraic closure $K^{\mathrm{ac}}$ of $K$. Let $K^{\mathrm{ur}}$ be the maximal unramified extension of $K$ inside $K^{\mathrm{ac}}$. Let $\widehat{K}^{\mathrm{ac}}$ and $\widehat{K}^{\mathrm{ur}}$ denote the completion of $K^{\mathrm{ac}}$ and $K^{\mathrm{ur}}$ respectively.  For a natural number $n$, we write $\mathrm{LT}(\mathfrak{p}^n)$ the Lubin-Tate curve with full level $n$ structure over $\widehat{K}^{\mathrm{ur}}$. We write $W_K$ for the Weil group of $K$. Let $D$ be the central division algebra over $K$ of invariant $1/2$. Let $\ell$ be a prime number different from $p$. We take an algebraic closure $\overline{\mathbb{Q}}_{\ell}$ of $\mathbb{Q}_{\ell}$. Then the groups $W_K$, $\textit{GL}_2 (K)$ and $D^{\times}$ act on $\varinjlim_n H^1 _{\mathrm{c}} 
 (\mathrm{LT}(\mathfrak{p}^n)_{\widehat{K}^{\mathrm{ac}}}, 
 \overline{\mathbb{Q}}_{\ell} )$, and these actions partially realize the local Langlands correspondence and the local Jacquet-Langlands correspondence for $\textit{GL}_2$. This realization was proved by Carayol in [@CaNALT] using global automorphic arguments.  However there is no known proof using only a local geometric method. 

In this paper, we give a purely local proof of a realization of the local Langlands correspondence for two-dimensional $W_K$-representations of conductor three, using a calculation of a semi-stable reduction of a Lubin-Tate curve in [@ITstab3].  We note that three is the smallest integer which is a conductor of a primitive two-dimensional $W_K$-representation.  The calculation in [@ITstab3] is done by purely local methods.  One more ingredient of the proof is a result by Mieda in [@MieGeomJL], where a realization of the local Jacquet-Langlands correspondence is proved by purely local methods.  In [@ITstab3], the actions of $W_K$ and $D^{\times}$ on a Lubin-Tate curve are calculated in some sense.  Using the calculation, we can study representations of $W_K$ and $D^{\times}$ in the cohomology of Lubin-Tate curves.  On the other hand, we have already known relation between representations of $\textit{GL}_2 (K)$ and $D^{\times}$ in the cohomology by the result of [@MieGeomJL].  Therefore, we can study the relation between representations of $W_K$ and $\textit{GL}_2 (K)$ in the cohomology.  This enables us to show a realization of the local Langlands correspondence in the cohomology of Lubin-Tate curves. 

In the study of a realization of the local Langlands correspondence for $\textit{GL}_2$, the most difficult and interesting case is the dyadic case, which means the case where $p=2$. A proof in the case where $p$ is odd is similar and easier.  Therefore, we have decided to write a proof only in the dyadic case. 

In the dyadic case, the irreducible two-dimensional $W_K$-representations of conductor three are primitive.  Construction of the local Langlands correspondence for primitive representations in [@KuLC] uses Weil representations (cf.  [@BHLLC §12]).  On the other hand, our descriptions of representations of $\textit{GL}_2 (K)$ in the cohomology of Lubin-Tate curves are given by cuspidal types.  Therefore, it is a non-trivial problem to check that the described representations in the cohomology actually give the local Langlands correspondence. 

We explain the contents of this paper.  In Section \[LTsp\], we recall definitions of Lubin-Tate curves, and give an easy consequence of a result in [@MieGeomJL].  In Section \[sstred\], we recall a calculation of a semi-stable reduction of a Lubin-Tate curve in [@ITstab3]. 

In Section \[cohell\], we study cohomology of an elliptic curve, which appears in the semi-stable reduction of the Lubin-Tate curve.  The cohomology of this elliptic curve gives a primitive $W_K$-representation of conductor three.  We give also an explicit description of this primitive representation. 

In Section \[realcor\], we show that a correspondence of explicitly described representations appear in the cohomology of Lubin-Tate curves.  In Section \[LLCe\], we show that the correspondence obtained in Section \[realcor\] is actually the local Langlands correspondence by calculating epsilon factors.  In other word, we give a description of the local Langlands correspondence via cuspidal type for representation of conductor three.  To determine the sign of an epsilon factor, we calculate Artin map for a wildly ramified abelian extension with a non-trivial ramification filtration. 

Acknowledgment {#acknowledgment .unnumbered}
--------------

The authors thank Yoichi Mieda and Seidai Yasuda for helpful conversations. 

Notation {#notation .unnumbered}
--------

In this paper, we use the following notation.  For a field $F$, an algebraic closure of $F$ is denoted by $F^{\mathrm{ac}}$ and a separable closure of $F$ is denoted by $F^{\mathrm{sep}}$. Let $K$ be a non-archimedean local field.  Let $\mathcal{O}_K$ denote the ring of integers and $k$ the residue field of characteristic $p>0$. Let $\mathfrak{p}$ be the maximal ideal of $\mathcal{O}_K$. We fix a uniformizer $\varpi$ of $K$. Let $q=|k|$. For any finite extension $F$ of $K$, let $G_F$ denote the absolute Galois group of $F$, $W_F$ denote the Weil group of $F$ and $I_F$ denote the inertia subgroup of $W_F$. Let $K^{\mathrm{ur}}$ denote the maximal unramified extension of $K$ in $K^{\mathrm{ac}}$. The completion of $K^\mathrm{ac}$ and $K^\mathrm{ur}$ is denoted by $\widehat{K}^{\mathrm{ac}}$ and $\widehat{K}^{\mathrm{ur}}$ respectively.  We write $\mathcal{O}_{\widehat{K}^{\mathrm{ac}}}$ and $\mathcal{O}_{\widehat{K}^{\mathrm{ur}}}$ for the ring of integers of $\widehat{K}^{\mathrm{ac}}$ and $\widehat{K}^{\mathrm{ur}}$ respectively.  For an element $a \in \mathcal{O}_{\widehat{K}^{\mathrm{ac}}}$, we write $\bar{a}$ for the image of $a$ by the reduction map $\mathcal{O}_{\widehat{K}^{\mathrm{ac}}} \to k^\mathrm{ac}$. Let $v(\cdot)$ denote a valuation of $K^{\mathrm{ac}}$ such that $v(\varpi)=1$. Let $\lvert \cdot \rvert_K$ be the absolute value of $K$ such that $\lvert \varpi \rvert_K =q^{-1}$. For $a,b \in K^{\mathrm{ac}}$ and a rational number $\alpha \in 
\mathbb{Q}_{\geq 0}$, we write $a \equiv b \pmod{\alpha}$ if we have $v(a-b) \geq \alpha$. For a local ring $A$, the maximal ideal of $A$ is denoted by $m_A$. For an irreducible reduced curve $X$ over $k^{\mathrm{ac}}$, we denote by $X^c$ the smooth compactification of $X$, and the genus of $X$ means the genus of $X^{\mathrm{c}}$. For an affinoid $\mathbf{X}$, we write $\overline{\mathbf{X}}$ for its reduction.  The category of sets is denoted by $\mathbf{Set}$. For a representation $\pi$ of a group, the dual representation of $\pi$ is denoted by $\pi^*$. We fix $\varpi_0 \in K^{\mathrm{ac}}$ such that $\varpi_0 ^{6(q-1)} =\varpi$, and $\varpi^{m/(6(q-1))}$ denotes $\varpi_0 ^m$ for any integer $m$.

Lubin-Tate curve {#LTsp}
================

Let $n$ be a natural number.  We put $$K_1 (\mathfrak{p}^n) =\biggl\{ 
 \begin{pmatrix}
 a & b \\
 c & d
 \end{pmatrix} 
 \in \textit{GL}_2 (\mathcal{O}_K ) \ \bigg| \ 
 c \equiv 0 ,\ d \equiv 1 \bmod \mathfrak{p}^n \biggr\}.$$ In the following, we define the connected Lubin-Tate curve $\mathbf{X}_1 (\mathfrak{p}^n )$ with level $K_1 (\mathfrak{p}^n)$.

Let $\Sigma$ denote the unique (up to isomorphism) formal $\mathcal{O}_K$-module of dimension $1$ and height $2$ over $k^{\mathrm{ac}}$. Let $\mathcal{C}$ be the category of Noetherian complete local $\mathcal{O}_{\widehat{K}^{\mathrm{ur}}}$-algebras with residue field $k^{\mathrm{ac}}$. For a one-dimensional formal $\mathcal{O}_K$-module $\mathcal{F} =\operatorname{Spf}A[[X]]$ over $A \in \mathcal{C}$ and an element $a \in \mathcal{O}_K$, we write $[a]_{\mathcal{F}}(X) \in A[[X]]$ for the $a$-multiplication on $\mathcal{F}$. For a formal one-dimensional $\mathcal{O}_K$-module $\mathcal{F} =\operatorname{Spf}A[[X]]$ over $A \in \mathcal{C}$ and an $A$-valued point $P$ of $\mathcal{F}$, the corresponding element of $m_A$ is denoted by $x(P)$. We consider the functor $$\mathcal{A}_1 (\mathfrak{p}^n) \colon \mathcal{C} \to \mathbf{Set}; 
 A \mapsto [(\mathcal{F}, \iota, P)],$$ where $\mathcal{F}$ is a formal $\mathcal{O}_K$-module over $A$ with an isomorphism $\iota\colon \Sigma \simeq \mathcal {F} \otimes_A k^{\mathrm{ac}}$ and $P$ is a $\varpi^n$-torsion point of $\mathcal{F}$ such that $$\prod_{a \in \mathcal{O}_K /\mathfrak{p}^n} 
 \bigl( X- x([a]_{\mathcal{F}} (P)) \bigr) \biggm| 
 [\varpi^n ]_{\mathcal{F}}(X)$$ in $A[[X]]$. This functor is represented by a regular local ring $\mathcal{R}_1(\mathfrak{p}^n)$. We write $\mathfrak{X}_1(\mathfrak{p}^n)$ for $\operatorname{Spf}\mathcal{R}_1(\mathfrak{p}^n)$. Its generic fiber is denoted by $\mathbf{X}_1(\mathfrak{p}^n)$, which we call the connected Lubin-Tate curve with level $K_1 (\mathfrak{p}^n)$. The space $\mathbf{X}_1(\mathfrak{p}^n)$ is a rigid analytic curve over ${\widehat{K}^{\mathrm{ur}}}$. We can define the connected Lubin-Tate curve $\mathbf{X}(\mathfrak{p}^n)$ with full level $n$ structure by changing $P$ to be an $\mathcal{O}_K$-module homomorphism $\phi \colon (\mathcal{O}_K/\mathfrak{p}^n)^2 \to m_A$ such that $$\prod_{a \in (\mathcal{O}_K/\mathfrak{p}^n)^2}
 (X-\phi(a)) \biggm| 
 [\varpi^n ]_{\mathcal{F}}(X)$$ in $A[[X]]$. We can define also the Lubin-Tate curve $\mathrm{LT}(\mathfrak{p}^n)$ with full level $n$ structure by changing further $\mathcal{C}$ to be the category of $\mathcal{O}_{\widehat{K}^{\mathrm{ur}}}$-algebras where $\varpi$ is nilpotent, and $\iota$ to be a quasi-isogeny $\Sigma \otimes_{k^{\mathrm{ac}}} A/\mathfrak{p} A \to 
 \mathcal {F} \otimes_A A/\mathfrak{p} A$.  We consider $\mathbf{X}(\mathfrak{p}^n)$ and $\mathrm{LT}(\mathfrak{p}^n)$ as rigid analytic curves over ${\widehat{K}^{\mathrm{ur}}}$.

Let $D$ be the central division algebra over $K$ of invariant $1/2$. We write $\mathcal{O}_D$ for the ring of integers of $D$. For a positive integer $m$, let $K_m$ be the unramified extension of $K$ of degree $m$ and $k_m$ be the finite extension over $k$ of degree $m$. Let $\kappa \in \operatorname{Gal}(K_2/K)$ be the non-trivial element.  The ring $\mathcal{O}_D$ has the following description; $\mathcal{O}_D = \mathcal{O}_{K_2} \oplus \varphi \mathcal{O}_{K_2}$ with $\varphi^2=\varpi$ and $a \varphi=\varphi a^{\kappa}$ for $a \in \mathcal{O}_{K_2}$. We define an action of $\mathcal{O}_D$ on $\Sigma$ by $\zeta(X)=\bar{\zeta} X$ for $\zeta \in \mu_{q^2 -1} (\mathcal{O}_{K_2} )$ and $\varphi(X)=X^q$. Then this give an isomorphism $\mathcal{O}_D \simeq \operatorname{End}(\Sigma)$ by [@GHev Proposition 13.10].  By this isomorphism, we can define a left action of $\mathcal{O}_D^{\times}$ on $\mathbf{X}_1(\mathfrak{p}^n)$ and $\mathbf{X}(\mathfrak{p}^n)$.

Let $\ell$ be a prime number different from $p$. We take an algebraic closure $\overline{\mathbb{Q}}_{\ell}$ of $\mathbb{Q}_{\ell}$. We put $H^1 _{\mathrm{LT}} = 
 \varinjlim_n H^1 _{\mathrm{c}} 
 (\mathrm{LT}(\mathfrak{p}^n)_{\widehat{K}^{\mathrm{ac}}}, 
 \overline{\mathbb{Q}}_{\ell} )$.  Then we can define an action of $\mathit{GL}_2 (K) \times W_K \times D^{\times}$ on $H^1 _{\mathrm{LT}}$ (cf. [ @DaLTell 3.2, 3.3]). 

We write $\operatorname{Irr}(D^{\times},\overline{\mathbb{Q}}_{\ell})$ for the set of isomorphism classes of irreducible smooth representations of $D^{\times}$ over $\overline{\mathbb{Q}}_{\ell}$, and $\operatorname{Disc}(\textit{GL}_2 (K),\overline{\mathbb{Q}}_{\ell})$ for the set of isomorphism classes of irreducible discrete series representations of $\textit{GL}_2 (K)$ over $\overline{\mathbb{Q}}_{\ell}$. Let $\mathrm{JL} \colon 
\operatorname{Irr}(D^{\times},\overline{\mathbb{Q}}_{\ell}) 
 \to \operatorname{Disc}(\textit{GL}_2 (K),\overline{\mathbb{Q}}_{\ell})$ be the local Jacquet-Langlands correspondence.  We denote by $\mathrm{LJ}$ the inverse of $\mathrm{JL}$. Let $\operatorname{Nrd}_{D/K} \colon D^{\times} \to K^{\times}$ be the reduced norm map, and $\operatorname{Trd}_{D/K} \colon D\to K$ be the reduced trace map.  The following proposition follows from a result in [@MieGeomJL], which is proved by local methods.  We note that the characteristic of a local field is assumed to be zero in [@MieGeomJL], but the same proof works in the equal characteristic case. 

\[realJL\] For a cuspidal representation $\pi$ of $\textit{GL}_2 (K)$ over $\overline{\mathbb{Q}}_{\ell}$, we have $$\operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi) \simeq 
 \mathrm{LJ} (\pi)^{\oplus 2}$$ as representations of $D^{\times}$.

Let $\omega_{\pi}$ be the central character of $\pi$. We take $c_{\pi} \in \overline{\mathbb{Q}}_{\ell}$ such that $c_{\pi}^2 =\omega_{\pi} (\varpi)$. We define a character $\zeta_{\pi}$ of $\textit{GL}_2 (K)$ by $\zeta_{\pi} (g) =c_{\pi}^{v(\det (g))}$, and a character $\xi_{\pi}$ of $D^{\times}$ by $\xi_{\pi} (d) =c_{\pi}^{v(\operatorname{Nrd}_{D/K} (d))}$ for $d \in D^{\times}$. We put $$H^i _{\mathrm{LT},\varpi} = 
 \varinjlim_n H^i _{\mathrm{c}} 
 \bigl( (\mathrm{LT}(\mathfrak{p}^n) /
 \varpi^{\mathbb{Z}})_{\widehat{K}^{\mathrm{ac}}}, 
 \overline{\mathbb{Q}}_{\ell} \bigr)$$ for non-negative integer $i$. Then we have $$\operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi) \simeq 
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT},\varpi} , 
 \pi \otimes \zeta_{\pi} ^{-1} ) \otimes \xi_{\pi}$$ as representations of $D^{\times}$ by arguments in [@StrJLLT 3.3].  In the Grothendieck group of finite-dimensional smooth representations of $D^{\times}$, we have $$\sum_{i,j\geq 0} (-1)^{i+j-1} 
 \operatorname{Ext}^j _{GL_2 (K)/\varpi^{\mathbb{Z}}} 
 (H^i _{\mathrm{LT},\varpi} , 
 \pi \otimes \zeta_{\pi} ^{-1} ) \otimes \xi_{\pi} 
 =
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT},\varpi} , 
 \pi \otimes \zeta_{\pi} ^{-1} ) \otimes \xi_{\pi}$$ by [@MieNoncuspLT Theorem 3.7] and the injectivity of cuspidal representations.  Therefore, the claim follows from [@MieGeomJL Definition 6.2 and Theorem 6.6], because $\mathrm{LJ}(\pi \otimes \zeta_{\pi} ^{-1}) \otimes \xi_{\pi} 
 = \mathrm{LJ}(\pi)$.

Semi-stable reduction of $\mathbf{X}_1(\mathfrak{p}^3)$ {#sstred}
=======================================================

From now on, we assume that $p=2$. The dual graph of a semi-stable reduction of $\mathbf{X}_1(\mathfrak{p}^3)$ in the case where $q$ is even is the following:\

where $k_2 ^{\times} = 
 \{ \zeta_1 ,\ldots ,\zeta_{q^2 -1} \}$, $k^{\times} = 
 \{ \zeta'_1 ,\ldots ,\zeta'_{q-1} \}$, $\overline{\mathbf{Y}}_{1,2}$ and $\overline{\mathbf{Y}}_{2,1}$ are defined by $x^q y-xy^q=1$, $\overline{\mathbf{Z}}^{\mathrm{c}}_{1,1}$ and $\overline{\mathbf{P}}^{\mathrm{c}}_{\zeta}$ are isomorphic to $\mathbb{P}^1_{k^{\mathrm{ac}}}$, and $\overline{\mathbf{X}}_{\zeta,\zeta'}$ are defined by $z^2+z=w^3$.

For a finite extension $K'$ of $K$, let $\mathrm{Art}_{K'} \colon  {K'}^{\times} 
\stackrel{\sim}{\to} W^{\mathrm{ab}}_{K'}$ be the Artin reciprocity map normalized so that the image by $\mathrm{Art}_{K'}$ of a uniformizer is a lift of the geometric Frobenius.  We define a homomorphism $\lvert \cdot \rvert \colon W_K \to \mathbb{Q}_{>0}$ by the composition $$W_K \twoheadrightarrow W^{\mathrm{ab}}_K 
 \xrightarrow{\mathrm{Art}_K^{-1}} K^{\times} 
 \xrightarrow{\lvert \cdot \rvert_K} \mathbb{Q}_{>0}.$$ We put $\mathcal{S}=k_2 ^{\times} \times k^{\times}$ and $$(W_K \times D^{\times})^0 = \bigl\{ (\sigma,d) \in W_K \times D^{\times} 
 \bigm| \lvert \operatorname{Nrd}_{D/K} (d) \rvert_K \cdot 
 \lvert \sigma \rvert =1 \bigr\}.$$ Then $(W_K \times D^{\times})^0$ acts on $\mathbf{X}_1(\mathfrak{p}^3)$. It induces an action of $(W_K \times D^{\times})^0$ on $\coprod_{(\zeta,\zeta') \in \mathcal{S}} 
 \overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'}$. For $\sigma \in W_K$, let $r_{\sigma}$ be an integer such that $\lvert \sigma \rvert =q^{-r_{\sigma}}$. Let $\mathcal{O}_D ^{\times} \rtimes W_K$ be a semidirect product where $\sigma \in W_K$ acts on $\mathcal{O}_D ^{\times}$ by $d \mapsto \varphi^{r_{\sigma}} d \varphi^{-r_{\sigma}}$. Then we have an isomorphism $$\label{0isom}
 \mathcal{O}_D ^{\times} \rtimes W_K \simeq 
 (W_K \times D^{\times})^0 ; 
 (d,\sigma) \mapsto (\sigma, d \varphi^{-r_{\sigma}} ).$$ By this isomorphism $\mathcal{O}_D ^{\times} \rtimes W_K$ acts on $\coprod_{(\zeta,\zeta') \in \mathcal{S}} 
 \overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'}$. We will describe this action. 

For $d \in \mathcal{O}_D^{\times}$, we put $\kappa_1 (d)=\bar{d}_1$ and $\kappa_2 (d)=-\bar{d}_2 /\bar{d}_1^q$, where $d=d_1 +\varphi d_2$ with $d_1 \in \mathcal{O}_{K_2}^{\times}$ and $d_2 \in \mathcal{O}_{K_2}$. We take $(\zeta,\zeta') \in \mathcal{S}$. We put $f_d =\operatorname{Tr}_{k_2/\mathbb{F}_2}
 ( \zeta^{1-q}  \zeta'^{-2} \kappa_2 (d))$ for $d \in \mathcal{O}_D ^{\times}$. We briefly recall the definition of $\zeta_{3,\sigma}$, $\nu_{\sigma}$ and $\mu_{\sigma}$ for $\sigma \in W_K$ from [@ITstab3 Section 6.2.2].  Consult there for detailed discussions.  Let $\hat{\zeta}' \in \mu_{q-1}(K)$ be the lift of $\zeta'$. We choose $\zeta'' \in \mu_{3(q-1)} (K^{\mathrm{ur}})$ such that $\zeta''^3=\hat{\zeta}'^4$. We take $\delta \in K^{\mathrm{ac}}$ such that $\delta^4 -\delta=1/(\zeta''\varpi^{1/3})$ and $\delta \varpi^{1/12} \equiv \hat{\zeta}' \zeta''^{-1} 
 \pmod{0}$. We take $\theta \in K^{\mathrm{ac}}$ such that $\theta^2 -\theta =\delta^3$. Note that $v(\delta) =-1/12$, $v(\theta) =-1/8$ and $\delta \in K(\zeta'' \varpi^{1/3} ,\theta)$.

Let $\sigma \in W_K$ in this paragraph.  We put $\zeta_{3,\sigma} =\sigma(\zeta''\varpi^{1/3})/(\zeta''\varpi^{1/3}) 
 \in \mu_3 (K^{\mathrm{ur}})$. We take $\nu_{\sigma} \in \mu_3(K^{\mathrm{ur}})\cup\{0\}$ such that $\sigma(\delta) \equiv 
 \zeta_{3,\sigma} ^{-1} (\delta +\nu_{\sigma}) \pmod{5/6}$. We choose $\zeta_3 \in \mu_3(K^{\mathrm{ur}})$ such that $\zeta_3 \neq 1$. Then, we can take $\mu_{\sigma} \in \mu_3(K^{\mathrm{ur}})\cup\{0\}$ such that $\mu_{\sigma} \equiv 
 \sigma (\theta)-\theta +\nu_{\sigma}^2 \delta +\nu_{\sigma}^3 
 +\sigma(\zeta_3) -\zeta_3 \pmod{0+}$.

We put $\lambda_{\sigma} =\sigma(\varpi^{1/(2(q-1))} )/\varpi^{1/(2(q-1))} 
 \in \mu_{2(q-1)}(K^{\mathrm{ac}})$ for $\sigma \in W_K$. We define a character $\lambda \colon W_K \to k^{\times}$ by $\lambda (\sigma)=\bar{\lambda}_{\sigma}$. We put $$Q=
 \Biggl\{
 g(\alpha, \beta, \gamma)=
 \begin{pmatrix}
 \alpha & \beta &  \gamma \\
 & \alpha^2 &  \beta^2\\
 & & \alpha
 \end{pmatrix}
 \in \textit{GL}_3 (\mathbb{F}_4) \ \Bigg| \ 
 \alpha\gamma^2
 +\alpha^2 \gamma=\beta^3 
 \Biggr\}.$$ We note that $|Q|=24$. Let $Q \rtimes \mathbb{Z}$ be a semidirect product, where $r \in \mathbb{Z}$ acts on $Q$ by $g(\alpha, \beta, \gamma) \mapsto 
 g(\alpha^{q^r} , \beta^{q^r} , \gamma^{q^r} )$.  Let $k_2 ^{\times} \rtimes \operatorname{Gal}(k_2 /k)$ be a semidirect product with a natural action of $\operatorname{Gal}(k_2 /k)$ on $k_2 ^{\times}$. We consider $\mathbb{F}_4$ as a subfield of $k_2 \subset k^{\mathrm{ac}}$. Let $\operatorname{Fr}_q$ be the $q$-th power Frobenius map on $k^{\mathrm{ac}}$. Then $(Q \rtimes \mathbb{Z}) \times (k_2 ^{\times} \rtimes \operatorname{Gal}(k_2 /k))$ acts on $\coprod_{(\zeta,\zeta') \in \mathcal{S}} 
 \overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'}$ as a scheme over $k$, where the action of $\bigl( (g(\alpha, \beta, \gamma),r), 
 (a,\operatorname{Fr}_q ^b) \bigr) \in 
 (Q \rtimes \mathbb{Z}) \times (k_2 ^{\times} \rtimes \operatorname{Gal}(k_2 /k))$ induces the isomorphism $$\overline{\mathbf{X}}_{\zeta, \zeta'} \to 
 \overline{\mathbf{X}}_{a \zeta^{q^b}, \zeta'}; 
 (z,w) \mapsto 
 \bigl(z^{q^{-r}} +\alpha^{-1} \beta w^{q^{-r}} +\alpha^{-1} \gamma, 
 \alpha(w^{q^{-r}} +(\alpha^{-1} \beta )^2) \bigr),$$ where we describe a bijection on $k^{\mathrm{ac}}$-valued points. 

\[descact\] The action of $\mathcal{O}_D ^{\times} \rtimes W_K$ on $\coprod_{(\zeta,\zeta') \in \mathcal{S}} 
 \overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'}$ gives the homomorphism $$\begin{aligned}
 \Xi_{\zeta'} \colon \mathcal{O}_D ^{\times} \rtimes W_K &\to 
 (Q \rtimes \mathbb{Z}) \times (k_2 ^{\times} \rtimes \operatorname{Gal}(k_2 /k)) ;\\ 
 (d,\sigma) &\mapsto 
 \Bigl( \bigl( 
 g(1,0,f_d) g(\bar{\zeta}_{3,\sigma}, \bar{\zeta}_{3,\sigma}^2 
 \bar{\nu}_{\sigma}^2, 
 \bar{\zeta}_{3,\sigma} \bar{\mu}_{\sigma}), r_{\sigma} \bigr), 
 (\kappa_1 (d) \bar{\lambda}_{\sigma} ,\operatorname{Fr}_q ^{-r_{\sigma}} ) \Bigr).  \end{aligned}$$

This follows from [@ITstab3 Proposition 5.4 and Proposition 6.12]. 

Let $\Theta_{\zeta'} \colon W_K \to 
 Q \rtimes \mathbb{Z}$ be the composite of $\Xi_{\zeta'} |_{W_K}$ with the projection to $Q \rtimes \mathbb{Z}$. By [@ITstab3 Proposition 6.13], the map $\Theta_{\zeta'}$ gives an isomorphism $W(K^{\mathrm{ur}} (\varpi^{1/3}, \theta)/K) \simeq Q \rtimes \mathbb{Z}$ and a finite extension of $K$ inside $K^{\mathrm{ur}} (\varpi^{1/3}, \theta)$ corresponds to a finite index subgroup of $Q \rtimes \mathbb{Z}$.

Cohomology of elliptic curve {#cohell}
============================

Let $\ell$ be an odd prime number.  We fix an algebraic closure $\overline{\mathbb{Q}}_{\ell}$ of $\mathbb{Q}_{\ell}$. In the sequel, we consider representations of groups over $\overline{\mathbb{Q}}_{\ell}$. We put $Q_8 =\{ g(1,\beta,\gamma) \in Q \}$, which is a normal subgroup of $Q$ of order $8$. Let $C_4 \subset Q_8$ be the cyclic subgroup of order $4$ generated by $g(1,1,\gamma)$ for $\gamma \in \mathbb{F}^{\times}_{4} \backslash \{1\}$. Let $Z \subset C_4$ be the subgroup consisting of $g(1,0,\gamma)$ with $\gamma^2+\gamma=0$, which is the center of $Q$. We take a faithful character $\phi$ of $C_4$. By [@BHLLC 22.2 Lemma], there exists a unique irreducible two-dimensional representation $\tau$ of $Q$ such that $$\label{qq}
 \tau |_{Z} \simeq (\phi |_{Z} )^{\oplus 2},\ 
 \operatorname{Tr}\tau(g(\alpha,0,0)) =-1$$ for $\alpha \in \mathbb{F}^{\times}_{4} \backslash \{1\}$. Let $C_3 \subset Q$ be the cyclic subgroup of order $3$ consisting of $g(\alpha,0,0)$ with $\alpha \in \mathbb{F}^{\times}_{4}$. Then we have $$\label{tau}
 \tau =\operatorname{Ind}_{C_4}^Q \phi - 
 \operatorname{Ind}_{Z \times C_3}^Q (\phi |_Z \otimes 1_{C_3} )$$ by [@BHLLC 16.4 Lemma 2.(4)] and a proof of [@BHLLC 22.2 Lemma]. 

Let $f$ be the degree of the extension $k$ over $\mathbb{F}_2$. We choose $(-2)^{1/2} \in \overline{\mathbb{Q}}_{\ell}$. We define a two-dimensional representation $\tau'_q$ of $Q \times 2\mathbb{Z} \subset Q \rtimes \mathbb{Z}$ so that $\tau'_q |_Q \simeq \tau$ and $(g(1,0,0),2) \in Q \times 2\mathbb{Z}$ acts by a scalar multiplication $(-2)^{-f}$. In the following, we will define a two-dimensional representation $\tau_q $ of $Q \rtimes \mathbb{Z}$ such that $\tau_q |_{Q \times 2\mathbb{Z}} \simeq \tau'_q$. We define a character $\phi_0$ of $Q \rtimes \mathbb{Z}$ by sending $(g,n)$ to $(-2)^{fn/2} q^{-n}$.

First, we consider the case where $f$ is even.  Then $Q \rtimes \mathbb{Z}$ is the direct product $Q \times \mathbb{Z}$. We define $\tau_q$ so that $(g(1,0,0),1) \in Q \times \mathbb{Z}$ acts by a scalar multiplication $(-2)^{-f/2}$. We define a character $\phi_1$ of $C_4 \times \mathbb{Z}$ by $\phi_1 (g,n) =\phi (g) \phi_0 (g,n)$. Then we have $\tau_q |_{Q_8 \times \mathbb{Z}} \simeq 
 \operatorname{Ind}_{C_4 \times \mathbb{Z}}^{Q_8 \times \mathbb{Z}} \phi_1$.

Next, we consider the case where $f$ is odd.  We assume that $f$ is odd, until the end of Lemma \[chartau\].  Let $C \subset Q_8 \rtimes \mathbb{Z}$ be the subgroup which consists of $(g(1,\beta,\gamma),n)$ satisfying $g(1,\beta,\gamma) \in C_4$ if $n$ is even, and $g(1,\beta,\gamma) \notin C_4$ if $n$ is odd.  We note that the index of $C$ in $Q_8 \rtimes \mathbb{Z}$ is two.  We take $\eta \in \overline{\mathbb{Q}}_{\ell}$ satisfying $\eta^2 +(-2)^{(f+1)/2} \eta +q =0$. We note that $\eta^4 =-q^2$. We define a character $\phi_1$ of $C$ by sending $(g(1,\bar{\zeta}_3,\bar{\zeta}_3),1)$ to $\eta q^{-1}$ and $(g(1,0,0),2)$ to $-q^{-1}$. We note that $(g(1,\bar{\zeta}_3,\bar{\zeta}_3),1)$ and $(g(1,0,0),2)$ generates $C$ as a group. 

We define a character $\phi_2$ of $(Z \times C_3) \rtimes \mathbb{Z} \subset Q \rtimes \mathbb{Z}$ so that $\phi_2 |_{Z \times C_3} =\phi|_Z \otimes 1_{C_3}$ and $\phi_2$ sends $(g(1,0,0),1)$ to $(-2)^{f/2} q^{-1}$. We put $C_6 =Z \times C_3$.

\[Psisurj\] The representation $\operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2$ is irreducible, and there is a surjective homomorphism $$\Psi \colon 
 \operatorname{Ind}_{C}^{Q \rtimes \mathbb{Z}} \phi_1 
 \longrightarrow 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 .$$

First we note that $\operatorname{Ind}_{C}^{Q \rtimes \mathbb{Z}} \phi_1$ and $\operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2$ are semisimple representations, because their twists by $\phi_0 ^{-1}$ factor through representations of the finite group $Q \rtimes \mathbb{Z}/2\mathbb{Z}$.

For the first statement, it suffices to show $\dim \operatorname{Hom}_{Q \rtimes \mathbb{Z}} ( 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 ,
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2) =1$. We have $$\label{FrQC_6}
 \operatorname{Hom}_{Q \rtimes \mathbb{Z}} \bigl( 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 ,
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 \bigr) 
 \simeq 
 \operatorname{Hom}_{C_6 \rtimes \mathbb{Z}} \bigl( \phi_2 , 
 \operatorname{Res}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 \bigr)$$ by the Frobenius reciprocity.  We have $$\begin{aligned}
 &(C_6 \rtimes \mathbb{Z}) \backslash (Q \rtimes \mathbb{Z}) / 
 (C_6 \rtimes \mathbb{Z}) = 
 \{ [(g(1,0,0),0)], [(g(1,1,\bar{\zeta}_3),0)]\} \\ 
 &(C_6 \rtimes \mathbb{Z}) \cap \bigl( (g(1,1,\bar{\zeta}_3),0) 
 (C_6 \rtimes \mathbb{Z}) (g(1,1,\bar{\zeta}_3),0)^{-1} \bigr) = 
 Z \rtimes \mathbb{Z} \subset Q \rtimes \mathbb{Z} \end{aligned}$$ where $[\cdot]$ denotes an equivalent class.  Hence, we obtain $$\label{decQC_6}
 \operatorname{Res}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 
 \simeq 
 \phi_2 \oplus 
 \operatorname{Ind}_{Z \rtimes \mathbb{Z}}^{C_6 \rtimes \mathbb{Z}} \phi'_2$$ where $\phi'_2 ((g,n)) =\phi_2 \bigl((g(1,1,\bar{\zeta}_3),0)^{-1} 
 (g,n) (g(1,1,\bar{\zeta}_3),0) \bigr)$ for $(g,n) \in Z \rtimes \mathbb{Z}$. We have $$\label{FrC_6Z}
 \operatorname{Hom}_{C_6 \rtimes \mathbb{Z}} \bigl( 
 \phi_2 , 
 \operatorname{Ind}_{Z \rtimes \mathbb{Z}}^{C_6 \rtimes \mathbb{Z}} \phi'_2 \bigr) 
 \simeq 
 \operatorname{Hom}_{Z \rtimes \mathbb{Z}} ( \phi_2 |_{Z \rtimes \mathbb{Z}}, 
 \phi'_2 ) =0$$ by the Frobenius reciprocity.  By , and , we have $$\dim \operatorname{Hom}_{Q \rtimes \mathbb{Z}} ( 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 ,
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2) =1.$$ For the second statement, it suffices to show $\dim \operatorname{Hom}_{Q \rtimes \mathbb{Z}} ( 
 \operatorname{Ind}_C^{Q \rtimes \mathbb{Z}} \phi_1 ,
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2) =1$. This follows from $$C\backslash (Q \rtimes \mathbb{Z}) / 
 (C_6 \rtimes \mathbb{Z}) = 
 \{ [(g(1,0,0),0)] \}$$ by the similar arguments as above. 

We define $\tau_q$ by the kernel of $\Psi$. Then we have $\tau_q |_{Q \times 2\mathbb{Z}} \simeq \tau'_q$ by the definition of $\tau'_q$ and .  In particular, $\tau_q$ is irreducible. 

\[quadind\] We have an isomorphism $\tau_q |_{Q_8 \rtimes \mathbb{Z}} \simeq 
 \operatorname{Ind}_C ^{Q_8 \rtimes \mathbb{Z}} \phi_1$.

By Lemma \[Psisurj\], we see that $\operatorname{Ind}_C ^{Q \rtimes \mathbb{Z}} \phi_1$ is a direct sum of a two-dimensional irreducible representation and a four-dimensional irreducible representation.  Hence, we see that $\operatorname{Ind}_C ^{Q_8 \rtimes \mathbb{Z}} \phi_1$ is irreducible.  We have $$\begin{aligned}
 &\dim \operatorname{Hom}_{Q_8 \rtimes \mathbb{Z}} 
 (\operatorname{Ind}_C ^{Q_8 \rtimes \mathbb{Z}} \phi_1 , 
 \operatorname{Res}_{Q_8 \rtimes \mathbb{Z}} ^{Q \rtimes \mathbb{Z}} 
 \operatorname{Ind}_C ^{Q \rtimes \mathbb{Z}} \phi_1) 
 =\dim 
 \operatorname{Hom}_{Q \rtimes \mathbb{Z}} 
 (\operatorname{Ind}_C ^{Q \rtimes \mathbb{Z}} \phi_1 , 
 \operatorname{Ind}_C ^{Q \rtimes \mathbb{Z}} \phi_1) =2 \\
 &
 \dim \operatorname{Hom}_{Q_8 \rtimes \mathbb{Z}} 
 (\operatorname{Ind}_C ^{Q_8 \rtimes \mathbb{Z}} \phi_1 , 
 \operatorname{Res}_{Q_8 \rtimes \mathbb{Z}} ^{Q \rtimes \mathbb{Z}} 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}} ^{Q \rtimes \mathbb{Z}} \phi_2) 
 =\dim 
 \operatorname{Hom}_{Q \rtimes \mathbb{Z}} 
 (\operatorname{Ind}_C ^{Q \rtimes \mathbb{Z}} \phi_1 , 
 \operatorname{Ind}_C ^{Q \rtimes \mathbb{Z}} \phi_2) =1 \end{aligned}$$ by the proof of Lemma \[Psisurj\].  Therefore, we obtain the claim by the definition of $\tau_q$.

For a representation $\pi$ of a group $G$ on $V$ and $g \in G$, the trace of $g$ on $V$ is denoted by $\operatorname{tr}(g;\pi)$ or $\operatorname{tr}(g;V)$, and the determinant of $g$ on $V$ is denoted by $\det (g;\pi)$ or $\det (g;V)$

\[chartau\] The representation $\tau_q$ is the unique representation satisfying $\tau_q |_{Q \times 2\mathbb{Z}} \simeq \tau'_q$ and $$\label{trtau}
 \operatorname{tr}\bigl( (g(1,\bar{\zeta}_3,\bar{\zeta}_3), 1) ; 
 \tau_q \bigr) =-(-2)^{\frac{f+1}{2}} q^{-1} .$$

There are two representations of $Q \rtimes \mathbb{Z}$ extending $\tau'_q$, and one is a twist of another by the character $Q \rtimes \mathbb{Z} \to 
 \overline{\mathbb{Q}}_{\ell}^{\times} ;\ 
 (g,n) \mapsto (-1)^n$. Therefore, the uniqueness in the statement follows. 

It remains to show that $\tau_q$ satisfies .  Let $C'$ be the subgroup of $C$ generated by $(g(1,\bar{\zeta}_3,\bar{\zeta}_3),1)$, and $C''$ be the subgroup of $C$ generated by $(g(1,0,1),4)$. We have $C'' \subset C'$, because $(g(1,0,1),4)=(g(1,\bar{\zeta}_3,\bar{\zeta}_3),1)^4$.

Then we have $$\begin{aligned}
 &C' \backslash (Q \rtimes \mathbb{Z}) /C =
 \{ [(g(1,0,0),0)], [(g(1,\bar{\zeta}_3,\bar{\zeta}_3),0)], 
 [(g(\bar{\zeta}_3,0,0),0)] \}, \\ 
 &C' \cap \bigl( (g(\bar{\zeta}_3,0,0),0) C 
 (g(\bar{\zeta}_3,0,0),0)^{-1} \bigr)= 
 C'' \subset Q \rtimes \mathbb{Z}.  \end{aligned}$$ Hence, we obtain $$\label{RIphi1}
 \operatorname{Res}_{C'}^{Q \rtimes \mathbb{Z}}
 \operatorname{Ind}_{C}^{Q \rtimes \mathbb{Z}} \phi_1 
 \simeq 
 \phi_1 |_{C'} \oplus \phi'_1 \oplus 
 \operatorname{Ind}_{C''}^{C'} (\phi_1 |_{C''} )$$ where $\phi'_1 ((g,n)) =\phi_1 \bigl(
 (g(1,\bar{\zeta}_3,\bar{\zeta}_3),0)^{-1} 
 (g,n) (g(1,\bar{\zeta}_3,\bar{\zeta}_3),0) \bigr)$ for $(g,n) \in C'$. Therefore we have $$\label{trphi1}
 \operatorname{tr}\bigl(
 (g(1,\bar{\zeta}_3 ,\bar{\zeta}_3), 1) ; 
 \operatorname{Ind}_{C}^{Q \rtimes \mathbb{Z}} \phi_1 \bigr) = 
 \frac{\eta}{q} -\frac{\eta^3}{q^2} =\frac{-(-2)^{\frac{f+1}{2}}}{q}$$ using $$(g(1,\bar{\zeta}_3,\bar{\zeta}_3),0)^{-1} 
 (g(1,\bar{\zeta}_3 ,\bar{\zeta}_3), 1) 
 (g(1,\bar{\zeta}_3,\bar{\zeta}_3),0) 
 =(g(1,\bar{\zeta}_3,\bar{\zeta}_3),1)^3 
 (g(1,0,0),-2).$$ By $C' \backslash (Q \rtimes \mathbb{Z})/(C_6 \rtimes \mathbb{Z}) 
 =\{ [(g(1,0,0),0)] \}$ and $C' \cap (C_6 \rtimes \mathbb{Z} )=C''$, we obtain $$\label{RIphi2}
 \operatorname{Res}_{C'}^{Q \rtimes \mathbb{Z}} 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} \phi_2 
 \simeq \operatorname{Ind}_{C''}^{C'} (\phi_2 |_{C''} ).$$ Therefore we have $$\label{trphi2}
 \operatorname{tr}\bigl(
 (g(1,\bar{\zeta}_3 ,\bar{\zeta}_3), 1) ; 
 \operatorname{Ind}_{C_6 \rtimes \mathbb{Z}}^{Q \rtimes \mathbb{Z}} 
 \phi_2 \bigr) =0.$$ The equation follows from and . 

\[dettau\] We have $\det \bigl( (g,n);\tau_q \bigr) =q^{-n}$ for $(g,n) \in Q \rtimes \mathbb{Z}$.

First we are going to show that $\det \tau =1$. We see that $\det \tau$ factors through $Q/Q_8$, because $Q/Q_8$ is the maximal abelian quotient of $Q$. By , we know that $\tau$ is self-dual.  Hence, the character of $Q/Q_8$ induced from $\det \tau$ is trivial.  Therefore, we have $\det \tau =1$. Then we see that $\det \tau_q$ factors through $\mathbb{Z}$, because $\det \tau =1$.

First, we assume that $f$ is even.  Then we have $\det \bigl( (g(1,0,0),1); \tau_q \bigr) =q^{-1}$, and this shows the claim. 

Next, we assume that $f$ is odd.  Then, it suffices to show that $\det \bigl( (g(1,\bar{\zeta}_3,\bar{\zeta}_3),1) ; 
 \tau_q \bigr) =q^{-1}$. This follows from and , because $\Psi$ is surjective. 

Let $\mathcal{E}$ be the elliptic curve over $\mathbb{F}_2$ defined by $z^2+z=w^3$. Then $(g(\alpha, \beta, \gamma),r) \in Q \rtimes \mathbb{Z}$ acts on $\mathcal{E}_{k^{\mathrm{ac}}}$ by $(z,w) \mapsto 
 \bigl(z^{q^{-r}} +\alpha^{-1} \beta w^{q^{-r}} +\alpha^{-1} \gamma, 
 \alpha(w^{q^{-r}} +(\alpha^{-1} \beta )^2)\bigr)$. The action of $Q \rtimes \mathbb{Z}$ gives a representation $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})$ of $Q \rtimes \mathbb{Z}$ by the pullback by the inverse.  For a representation $V$ of $Q \rtimes \mathbb{Z}$ and an integer $m$, we write $V(m)$ for the twist of $V$ by the character $Q \rtimes \mathbb{Z} \ni (g,n) \mapsto q^{-mn}$.

\[cohtau\] We have an isomorphism $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1) \simeq \tau_q$ as representations of $Q \rtimes \mathbb{Z}$.

Let $\mathrm{fr}_{2^m}$ be the $2^m$-th power geometric Frobenius map of $\mathcal{E}_{k^{\mathrm{ac}}}$ for any positive integer $m$. Then we have $\operatorname{tr}(\mathrm{fr}_2 ^* ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) )=0$ and $\operatorname{tr}(\mathrm{fr}_4 ^* ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) )=-4$ by $\lvert \mathcal{E}(\mathbb{F}_2) \rvert =3$ and $\lvert \mathcal{E}(\mathbb{F}_4) \rvert =9$. Hence we obtain $\operatorname{tr}(\mathrm{fr}_q ^* ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) )= 
 ((-2)^{1/2})^f +(-(-2)^{1/2})^f$.

As $Q$-representations, $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) \simeq \tau$ by [@ITstab3 Lemma 7.7].  Then we have $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1) \simeq \tau_q '$ as representations of $Q \times 2\mathbb{Z}$, because $$\begin{aligned}
 \operatorname{tr}((g(1,0,0), 2) ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1) ) 
 &= 
 \operatorname{tr}((g(1,0,0), -2)^* ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) ) 
 q^{-2} \\ 
 &=
 \operatorname{tr}(\mathrm{fr}_{q^2} ^* ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) ) q^{-2} 
 = 2(-2)^{-f} \end{aligned}$$ for $(g(1,0,0), 2) \in Q \times \mathbb{Z}$.

First, we consider the case where $f$ is even.  Then $Z \times \mathbb{Z}$ acts on $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1)$ by scalar multiplications.  Hence we have $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1) \simeq \tau_q$, because $$\operatorname{tr}((g(1,0,0), 1) ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1) ) =
 \operatorname{tr}(\mathrm{fr}_q ^* ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) ) q^{-1} 
 = 2(-2)^{-f/2}$$ for $(g(1,0,0), 1) \in Q \times \mathbb{Z}$.

Next, we consider the case where $f$ is odd.  It suffices to consider the case where $f=1$, because the claimed isomorphisms for general cases are induced from the isomorphism for this case by the group homomorphism $Q \rtimes \mathbb{Z} \to Q \rtimes \mathbb{Z}; 
 (g,n) \mapsto (g,fn)$.

We assume that $f=1$. It suffices to show that $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1)$ satisfies the conditions in Lemma \[chartau\].  We have already checked the first condition.  Let $\operatorname{Fr}_{2,\mathcal{E}}$ be the absolute $2$-th power Frobenius map on $\mathcal{E}_{k^{\mathrm{ac}}}$. By the Lefschetz trace formula, we have $$\begin{aligned}
\label{LTFe}
 2+1 -&\operatorname{tr}\bigl( (g(1,\bar{\zeta}_3 ,\bar{\zeta}_3), 1); 
 H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell}) \bigr) = 
 \bigl\lvert \{ P \in \mathcal{E}(k^{\mathrm{ac}}) \mid 
 \bigl((g(1,\bar{\zeta}_3,\bar{\zeta}_3),1)^{-1} \circ \operatorname{Fr}_{2,\mathcal{E}} \bigr) 
 P=P \} \bigr\rvert \\ 
 &= \bigl\lvert \{ (z,w) \in k^{\mathrm{ac}} \times k^{\mathrm{ac}} \mid 
 z^2+z=w^3,\ z=z^2+\bar{\zeta}_3^2 w^2 +\bar{\zeta}_3,\ 
 w=w^2+\bar{\zeta}_3 \} \bigr\rvert +1 \\ 
 &= \bigl\lvert \{ (z,w) \in k^{\mathrm{ac}} \times k^{\mathrm{ac}} \mid 
 z^2+z=w^3,\ w^3 +\bar{\zeta}_3^2 w^2 +\bar{\zeta}_3 =0,\ 
 w^2+w+\bar{\zeta}_3 =0 \} \bigr\rvert +1 
 =1.  \end{aligned}$$ Therefore, the condition is satisfied for $H^1(\mathcal{E}_{k^{\mathrm{ac}}},\overline{\mathbb{Q}}_{\ell})(1)$.

Let $\tau_{\zeta'}$ be the representation of $W_K$ induced from the $(Q \rtimes \mathbb{Z})$-representation $\tau_q$ by $\Theta_{\zeta'}$. We say that a continuous two-dimensional representation $V$ of $W_K$ over $\overline{\mathbb{Q}}_{\ell}$ is primitive, if there is no pair of a quadratic extension $K'$ and a continuous character $\chi$ of $W_{K'}$ such that $V \simeq \operatorname{Ind}_{W_{K'}}^{W_K} \chi$. The representation $\tau_{\zeta'}$ is primitive of Artin conductor $3$ by [@ITstab3 Lemma 7.8]. 

Realization of correspondence {#realcor}
=============================

Let $\chi_2 \colon \mathbb{F}_2 \to \overline{\mathbb{Q}}_{\ell}^{\times}$ be the non-trivial character.  We put $$\label{JPdef}
 \mathfrak{J} =\biggl\{ 
 \begin{pmatrix}
 a & b \\
 c & d
 \end{pmatrix} 
 \in M_2 (\mathcal{O}_K ) \ \bigg| \ 
 c \equiv 0 \bmod \mathfrak{p} \biggr\},\ 
 \mathfrak{P} =\biggl\{ 
 \begin{pmatrix}
 a & b \\
 c & d
 \end{pmatrix} 
 \in \mathfrak{J} \ \bigg| \ 
 a \equiv d \equiv 0 \bmod \mathfrak{p} \biggr\},$$ and $U_{\mathfrak{J}}^1 =1+ \mathfrak{P} \subset M_2 (\mathcal{O}_K )$.  We put $L=K(\varphi) \subset D$, and consider $L$ as a $K$-subalgebra of $M_2 (K)$ by the embedding $$\label{embL}
 L \hookrightarrow M_2 (K) ; \varphi \mapsto 
 \begin{pmatrix}
 0 & 1 \\
 \varpi & 0
 \end{pmatrix}.$$ We put $U_D ^1 =1+\varphi \mathcal{O}_D$. We take an additive character $\psi_K \colon K \to \overline{\mathbb{Q}}_{\ell}^{\times}$ such that $\psi_K (a) =(\chi_2 \circ 
 \operatorname{Tr}_{k/\mathbb{F}_2} )(\bar{a})$ for $a \in \mathcal{O}_K$.

For a finite abelian group $A$, the character group $\operatorname{Hom}_{\mathbb{Z}} (A,\overline{\mathbb{Q}}_{\ell} ^{\times})$ is denoted by $A^{\vee}$. Let $\zeta' \in k^{\times}$, $\chi \in (k^{\times})^{\vee}$ and $c \in \overline{\mathbb{Q}}_{\ell}^{\times}$. We define a character $\Lambda_{\zeta',\chi,c} \colon L^{\times} U_{\mathfrak{J}}^1 
 \to \overline{\mathbb{Q}}_{\ell}^{\times}$ by $\Lambda_{\zeta',\chi,c} (\varphi)=-c$, $\Lambda_{\zeta',\chi,c} (a) =\chi (\bar{a})$ for $a \in \mathcal{O}_L ^{\times}$ and $$\Lambda_{\zeta',\chi,c} (x)=(\psi_K \circ 
 \operatorname{tr})(\hat{\zeta}'^{-2} \varphi^{-1}(x-1))$$ for $x \in U_{\mathfrak{J}}^1$. We put $\pi_{\zeta',\chi,c} = 
 \mathrm{c\mathchar`-Ind}_{L^{\times} U_{\mathfrak{J}}^1}^{\textit{GL}_2(K)} 
 \Lambda_{\zeta',\chi,c}$. Next, we define a character $\theta_{\zeta',\chi,c} \colon L^{\times} U_D ^1 
 \to \overline{\mathbb{Q}}_{\ell}^{\times}$ by $\theta_{\zeta',\chi,c} (\varphi)=c$, $\theta_{\zeta',\chi,c} (a) =\chi (\bar{a})$ for $a \in \mathcal{O}_L ^{\times}$ and $\theta_{\zeta',\chi,c} (d) =(\chi_2 \circ \operatorname{Tr}_{k_2 /\mathbb{F}_2}) 
 ({\zeta'}^{-2} \kappa_2 (d))$ for $d \in U_D ^1$. We put $\rho_{\zeta', \chi,c} = 
 \operatorname{Ind}_{L^{\times} U_D ^1}^{D^{\times}} \theta_{\zeta', \chi,c}$.

\[expJL\] For $\zeta' \in k^{\times}$, $\chi \in (k^{\times})^{\vee}$ and $c \in \overline{\mathbb{Q}}_{\ell}^{\times}$, we have $\mathrm{JL} (\rho_{\zeta', \chi,c}) =\pi_{\zeta',\chi,c}$.

This follows from [@BHLLC 56.5], because $$(\psi_K \circ \operatorname{Trd}_{D/K} )(\hat{\zeta}'^{-2} \varphi^{-1}(d-1)) 
 = 
 (\chi_2 \circ \operatorname{Tr}_{k_2 /\mathbb{F}_2}) 
 ({\zeta'}^{-2} \kappa_2 (d))$$ for $d \in U_D ^1$.

For $c \in \overline{\mathbb{Q}}_{\ell}^{\times}$, let $\phi_c \colon W_K \to \overline{\mathbb{Q}}_{\ell}^{\times}$ be the character defined by $\phi_c (\sigma)=c^{r_{\sigma}}$. For $\zeta' \in k^{\times}$, $\chi \in (k^{\times})^{\vee}$ and $c \in \overline{\mathbb{Q}}_{\ell}^{\times}$, we put $\tau_{\zeta', \chi, c} =
 \tau_{\zeta'} \otimes (\chi \circ \lambda) \otimes \phi_c$. For a representation $V$ of a Weil group and an integer $m$, we write $V(m)$ for the $m$-times Tate twist of $V$.

\[realLLC\] For $\zeta' \in k^{\times}$, $\chi \in (k^{\times})^{\vee}$ and $c \in \overline{\mathbb{Q}}_{\ell}^{\times}$, we have $$\operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) \simeq 
 \tau_{\zeta', \chi, c} \otimes \rho_{\zeta', \chi,c}$$ as representations of $W_K \times D^{\times}$.

Let $\zeta' \in k^{\times}$, $\chi \in (k^{\times})^{\vee}$ and $c \in \overline{\mathbb{Q}}_{\ell}^{\times}$. By Proposition \[realJL\] and Proposition \[expJL\], we know that $$\operatorname{Hom}_{D^{\times}} \bigl( \rho_{\zeta', \chi,c} , 
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) \bigr) 
 \simeq \tau'$$ for some two-dimensional $W_K$-representation $\tau'$. First, we will show that $\tau' =\tau_{\zeta', \chi, c}$.

We put $H^1 _{\mathbf{X}} = 
 \varinjlim_n H^1 _{\mathrm{c}} 
 (\mathbf{X}(\mathfrak{p}^n)_{\widehat{K}^{\mathrm{ac}}}, 
 \overline{\mathbb{Q}}_{\ell} )$ and $$(\textit{GL}_2 (K) \times W_K \times D^{\times})^0 
 = \bigl\{ (g,\sigma,d) \in \textit{GL}_2 (K) \times W_K \times D^{\times} 
 \bigm| \lvert \det (g)^{-1} \operatorname{Nrd}_{D/K} (d) \rvert_K \cdot 
 \lvert \sigma \rvert =1 \bigr\}.$$ Then we have $$\begin{aligned}
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) &\simeq 
 \operatorname{Hom}_{GL_2 (K)} \bigl( 
 \mathrm{c\mathchar`-Ind}
 _{(\textit{GL}_2 (K) \times W_K \times D^{\times})^0 }
 ^{\textit{GL}_2 (K) \times W_K \times D^{\times}} 
 H^1 _{\mathbf{X}} ,\pi_{\zeta', \chi,c} \bigr) \\ 
 &\subset 
 \operatorname{Hom}_{K_1(\mathfrak{p}^3)} \bigl( 
 \mathrm{c\mathchar`-Ind}
 _{K_1(\mathfrak{p}^3) \times (W_K \times D^{\times})^0 }
 ^{K_1(\mathfrak{p}^3) \times W_K \times D^{\times}} 
 H^1 _{\mathbf{X}} ,\pi_{\zeta', \chi,c} \bigr) \\ 
 &\subset 
 \operatorname{Hom}_{\overline{\mathbb{Q}}_{\ell}} \bigl( 
 \mathrm{c\mathchar`-Ind}
 _{(W_K \times D^{\times})^0 }
 ^{W_K \times D^{\times}} 
 H^1 _{\mathrm{c}} 
 (\mathbf{X}_1 (\mathfrak{p}^3)_{\widehat{K}^{\mathrm{ac}}}, 
 \overline{\mathbb{Q}}_{\ell} ) , \overline{\mathbb{Q}}_{\ell} \bigr), \end{aligned}$$ where the last inclusion follows by taking the $K_1(\mathfrak{p}^3)$-invariant part, because the conductor of $\pi_{\zeta', \chi,c}$ is three.  Hence, we obtain $$\begin{aligned}
 \tau' &\simeq \operatorname{Hom}\bigl( \rho_{\zeta', \chi,c} , 
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) 
 \bigr) \notag \\ 
 &\subset 
 \operatorname{Hom}_{D^{\times}} \bigl( \rho_{\zeta', \chi,c} ,
 \bigl( \mathrm{c\mathchar`-Ind}
 _{(W_K \times D^{\times})^0 }
 ^{W_K \times D^{\times}} 
 H^1 _{\mathrm{c}} 
 (\mathbf{X}_1 (\mathfrak{p}^3)_{\widehat{K}^{\mathrm{ac}}}, 
 \overline{\mathbb{Q}}_{\ell} ) \bigr)^* \bigr) \label{tau'incl} \\ 
 &\simeq 
 \operatorname{Hom}_{D^{\times}} \bigl( 
 \mathrm{c\mathchar`-Ind}
 _{(W_K \times D^{\times})^0 }
 ^{W_K \times D^{\times}} 
 H^1 _{\mathrm{c}} 
 (\mathbf{X}_1 (\mathfrak{p}^3)_{\widehat{K}^{\mathrm{ac}}}, 
 \overline{\mathbb{Q}}_{\ell} ) , 
 \rho_{\zeta', \chi,c} ^* \bigr) \notag \\ 
 &\simeq 
 \operatorname{Hom}_{D^{\times}} \Bigl( 
 \mathrm{c\mathchar`-Ind}
 _{(W_K \times D^{\times})^0 }
 ^{W_K \times D^{\times}} 
 \bigl( \bigoplus_{\zeta \in k_2 ^{\times}} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )^* (-1) \bigr) , 
 \rho_{\zeta', \chi,c} ^* \Bigr), \notag \end{aligned}$$ where the last isomorphism follows from [@ITstab3 Proposition 7.3, Proposition 7.9 and Theorem 7.16] by studying only $\mathcal{O}_D ^{\times}$-actions.  As vector spaces, the last space is isomorphic to $$\operatorname{Hom}_{D^{\times}} \Bigl( 
 \mathrm{c\mathchar`-Ind}
 _{\mathcal{O}_D ^{\times}} ^{D^{\times}} 
 \bigl( \bigoplus_{\zeta \in k_2 ^{\times}} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )^* \bigr) , 
 \rho_{\zeta', \chi,c} ^* \Bigr) 
 \simeq 
 \operatorname{Hom}_{\mathcal{O}_D ^{\times}} \bigl( 
 \bigoplus_{\zeta \in k_2 ^{\times}} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )^* , 
 \rho_{\zeta', \chi,c} ^* |_{\mathcal{O}_D ^{\times}} \bigr)$$ which is two-dimensional by [@ITstab3 Proposition 7.9].  Hence, the inclusion in is an equality.  Therefore it suffices to show that there is a non-trivial homomorphism $$\mathrm{c\mathchar`-Ind}
 _{(W_K \times D^{\times})^0 }
 ^{W_K \times D^{\times}} \bigl( 
 \bigoplus_{\zeta \in k_2 ^{\times}} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )^* (-1) \bigr) \longrightarrow 
 \tau_{\zeta', \chi,c}^* \otimes \rho_{\zeta', \chi,c}^*$$ as representations of $W_K \times D^{\times}$. By the Frobenius reciprocity, this is equivalent to give a non-trivial homomorphism $$(\tau_{\zeta', \chi,c} \otimes \rho_{\zeta', \chi,c} )
 |_{(W_K \times D^{\times})^0} 
 \longrightarrow 
 \bigoplus_{\zeta \in k_2 ^{\times}} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )(1)$$ as representations of $(W_K \times D^{\times})^0$. We put $$(\mathcal{O}_D ^{\times} \rtimes W_K)^0 = 
 \{ (d,\sigma) \in \mathcal{O}_D ^{\times} \rtimes W_K \mid 
 \kappa_1 (d) \bar{\lambda}_{\sigma} =1\}$$ and consider this group as a subgroup of $(W_K \times D^{\times})^0$ by the isomorphism .  Then we have $$\bigoplus_{\zeta \in k_2 ^{\times}} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )(1) 
 \simeq 
 \operatorname{Ind}_{(\mathcal{O}_D ^{\times} \rtimes W_K)^0 }
 ^{(W_K \times D^{\times})^0} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{1,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )(1),$$ because, by Proposition \[descact\], the action of $\mathcal{O}_D ^{\times} \rtimes W_K$ on $\coprod_{(\zeta,\zeta') \in \mathcal{S}} 
 \overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'}$ permutes the connected components transitively, and $(\mathcal{O}_D ^{\times} \rtimes W_K)^0$ is the stabilizer of the connected component $\overline{\mathbf{X}}^{\mathrm{c}}_{1,\zeta'}$. Hence, we have $$\begin{aligned}
 \operatorname{Hom}_{(W_K \times D^{\times})^0} & \bigl( 
 (\tau_{\zeta',\chi,c} \otimes \rho_{\zeta',\chi,c} ) 
 |_{(W_K \times D^{\times})^0} , 
 \bigoplus_{\zeta \in k_2 ^{\times}} H^1 
 (\overline{\mathbf{X}}^{\mathrm{c}}_{\zeta,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )(1) \bigr) \\ 
 &\simeq 
 \operatorname{Hom}_{(\mathcal{O}_D ^{\times} \rtimes W_K)^0} \bigl( 
 (\tau_{\zeta', \chi,c} \otimes \rho_{\zeta', \chi,c})
 |_{(\mathcal{O}_D ^{\times} \rtimes W_K)^0} , 
 H^1 (\overline{\mathbf{X}}^{\mathrm{c}}_{1,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )(1) \bigr).  \end{aligned}$$ Since $\tau_{\zeta', \chi,c} \otimes \rho_{\zeta', \chi,c} \simeq 
 \operatorname{Ind}_{W_K \times L^{\times} U_D ^1} ^{W_K \times D^{\times}} 
 (\tau_{\zeta',\chi,c} \otimes \theta_{\zeta',\chi,c})$ and $(\mathcal{O}_D ^{\times} \rtimes W_K)^0 \subset 
 W_K \times L^{\times} U_D ^1$, we have a non-trivial homomorphism $$(\tau_{\zeta', \chi,c} \otimes \rho_{\zeta', \chi,c} ) 
 |_{(\mathcal{O}_D ^{\times} \rtimes W_K)^0} 
 \longrightarrow 
 (\tau_{\zeta',\chi,c} \otimes \theta_{\zeta',\chi,c}) 
 |_{(\mathcal{O}_D ^{\times} \rtimes W_K)^0}$$ by the Frobenius reciprocity.  Hence, it suffices to show there is a non-trivial homomorphism $$(\tau_{\zeta',\chi,c} \otimes \theta_{\zeta',\chi,c}) 
 |_{(\mathcal{O}_D ^{\times} \rtimes W_K)^0} 
 \longrightarrow 
 H^1 (\overline{\mathbf{X}}^{\mathrm{c}}_{1,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )(1)$$ as representations of $(\mathcal{O}_D ^{\times} \rtimes W_K)^0$. We put $W'_K =\{ (\lambda_{\sigma}^{-1} ,\sigma ) \in 
 (\mathcal{O}_D ^{\times} \rtimes W_K)^0 \mid 
 \sigma \in W_K \}$. We consider $U_D ^1$ as a subgroup of $(\mathcal{O}_D ^{\times} \rtimes W_K)^0$ by identifying $d \in U_D ^1$ with $(d,1) \in (\mathcal{O}_D ^{\times} \rtimes W_K)^0$. Then we have an isomorphism $$(\tau_{\zeta',\chi,c} \otimes \theta_{\zeta',\chi,c}) 
 |_{W'_K} 
 \longrightarrow 
 H^1 (\overline{\mathbf{X}}^{\mathrm{c}}_{1,\zeta'} , 
 \overline{\mathbb{Q}}_{\ell} )(1)|_{W'_K}$$ as representation of $W'_K$ by Proposition \[cohtau\] and the definition of $\tau_{\zeta',\chi,c}$ and $\theta_{\zeta',\chi,c}$. This isomorphism is compatible with the action of $U_D ^1$ by Proposition \[descact\] and by Proposition \[cohtau\].  Then this is an isomorphism as representations of $(\mathcal{O}_D ^{\times} \rtimes W_K)^0$, because $(\mathcal{O}_D ^{\times} \rtimes W_K)^0$ is generated by $W'_K$ and $U_D ^1$. Thus we have proved that $$\label{homisom}
 \operatorname{Hom}_{D^{\times}} \bigl( \rho_{\zeta', \chi,c} , 
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) \bigr) 
 \simeq \tau_{\zeta', \chi, c}.$$

By , we see that $\operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c})$ is an irreducible representation of $W_K \times D^{\times}$. Let $\xi_c \colon D^{\times} \to \overline{\mathbb{Q}}_{\ell}^{\times}$ be the character defined by $\xi_c (d)=c^{v(\operatorname{Nrd}_{D/K}(d))}$. By , we have $$\begin{aligned}
 \operatorname{Hom}_{D^{\times}} \bigl( \rho_{\zeta', \chi,c} \otimes 
 \xi^{-1}_c , 
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) 
 \otimes \phi^{-1}_{c(-2)^{-f/2}} 
 &\otimes (\chi \circ \lambda)^{-1} \otimes \xi^{-1}_c \bigr) \\ 
 &\simeq \tau_{\zeta', \chi, c} 
 \otimes \phi^{-1}_{c(-2)^{-f/2}} 
 \otimes (\chi \circ \lambda)^{-1}.  \end{aligned}$$ Then we see that $\tau_{\zeta', \chi, c} \otimes \phi^{-1}_{c(-2)^{-f/2}}
 \otimes (\chi \circ \lambda)^{-1}$ and $\rho_{\zeta', \chi,c} \otimes \xi^{-1}_c$ factor through representations of $Q \rtimes \mathbb{Z}/2\mathbb{Z}$ and $D^{\times}/(\varpi^{\mathbb{Z}} (1+\varpi\mathcal{O}_D))$ respectively.  Hence, the $(W_K \times D^{\times})$-representation $\operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) 
 \otimes \phi^{-1}_{c(-2)^{-f/2}} 
 \otimes (\chi \circ \lambda)^{-1} \otimes \xi^{-1}_c$ factors through a representation of the finite group $(Q \rtimes \mathbb{Z}/2\mathbb{Z}) \times 
 \bigl( D^{\times}/(\varpi^{\mathbb{Z}} (1+\varpi\mathcal{O}_D))
 \bigr)$. Then we have $$\begin{aligned}
 \operatorname{Hom}_{GL_2 (K)} (H^1 _{\mathrm{LT}} ,\pi_{\zeta', \chi,c}) 
 \otimes \phi^{-1}_{c(-2)^{-f/2}} 
 &\otimes (\chi \circ \lambda)^{-1} \otimes \xi^{-1}_c \\ 
 &\simeq \bigl( \tau_{\zeta', \chi, c} 
 \otimes \phi^{-1}_{c(-2)^{-f/2}} 
 \otimes (\chi \circ \lambda)^{-1} \bigr) 
 \otimes (\rho_{\zeta', \chi,c} \otimes 
 \xi^{-1}_c )\end{aligned}$$ because an irreducible representation of a product of two finite groups is isomorphic to a tensor product of irreducible representations of the two groups.  Therefore, we have the claim. 

Local Langlands correspondence {#LLCe}
==============================

In this section, we prove that the correspondence in Theorem \[realLLC\] actually gives the local Langlands correspondence. 

We write $\mathcal{G}_2(K,\overline{\mathbb{Q}}_{\ell})$ for the set of equivalent classes of two-dimensional semisimple Weil-Deligne representations of $W_K$ over $\overline{\mathbb{Q}}_{\ell}$, and $\mathrm{Irr} (\textit{GL}_2 (K), \overline{\mathbb{Q}}_{\ell})$ for the set of equivalent classes of irreducible smooth representations of $\textit{GL}_2 (K)$. For $\pi \in \mathrm{Irr} (\textit{GL}_2 (K), \overline{\mathbb{Q}}_{\ell})$, let $\omega_{\pi}$ denote the central character of $\pi$. Let $\mathrm{LL}_{\ell} \colon 
 \mathcal{G}_2(K,\overline{\mathbb{Q}}_{\ell}) \to 
 \mathrm{Irr} (\textit{GL}_2 (K), \overline{\mathbb{Q}}_{\ell})$ be the $\ell$-adic Langlands correspondence (c.f. [ @BHLLC 35.1]).  If we take an isomorphism $\iota \colon \overline{\mathbb{Q}}_{\ell} \simeq \mathbb{C}$, then ${}^{\iota} \tau$ and ${}^{\iota} \pi$ denote the representations over $\mathbb{C}$ associated to $\tau$ and $\pi$ by $\iota$ respectively for $\tau \in \mathcal{G}_2(K,\overline{\mathbb{Q}}_{\ell})$ and $\pi \in \mathrm{Irr} (\textit{GL}_2 (K), \overline{\mathbb{Q}}_{\ell})$. We use similar notations also over a finite extension of $K$.

\[ellLLC\] The $\ell$-adic Langlands correspondence $\mathrm{LL}_{\ell}$ satisfy that $\omega_{\mathrm{LL}_{\ell}(\tau)} \circ \mathrm{Art}_K ^{-1} = 
 (\det \tau ) \otimes \lvert \cdot \rvert^{-1}$ for $\tau \in \mathcal{G}_2(K,\overline{\mathbb{Q}}_{\ell})$. If we take an isomorphism $\iota \colon \overline{\mathbb{Q}}_{\ell} \simeq 
 \mathbb{C}$, then we have $$\varepsilon ({}^{\iota} \tau ,s, \psi) = 
 \varepsilon ({}^{\iota} \mathrm{LL}_{\ell}(\tau) ,s+1/2, \psi)$$ for any non-trivial additive character $\psi \colon K \to \mathbb{C}^{\times}$.

For a finite extension $K'$ of $K$, we define an additive character $\psi_{K'} \colon K' \to \overline{\mathbb{Q}}_{\ell}^{\times}$ by $\psi_{K'} =\psi_K \circ \operatorname{Tr}_{K'/K}$, and let $v_{K'}$ be the normalized discrete valuation that sends a uniformizer to $1$.

We take $\zeta' \in k^{\times}$. We simply write $\pi_{\zeta'}$, $\Lambda_{\zeta'}$ and $\tau_{\zeta'}$ for $\pi_{\zeta',1,1}$, $\Lambda_{\zeta',1,1}$ and $\tau_{\zeta',1,1}$ respectively.  We put $F=K(\zeta'' \varpi^{1/3})$ and $L'=F(\varphi)$. We define $\mathfrak{J}_F ,\mathfrak{P}_F \subset M_2 (\mathcal{O}_F)$ similarly to $\mathfrak{J}$ and $\mathfrak{P}$ as in .  We put $U_{\mathfrak{J}_F}^i =1+\mathfrak{P}_F ^i$ for any positive integer $i$. We consider $L'$ as a $F$-subalgebra of $M_2 (F)$ similarly as .  We put $\epsilon_{F/K} =(-1)^f$.

Let $\pi_{F,\zeta'}$ be the tame lifting of $\pi_{\zeta'}$ to $F$. See [@BHLLC 46.5 Definition] for the tame lifting.  We define a character $\Lambda_{F,\zeta'} \colon 
 {L'}^{\times} U_{\mathfrak{J}_F} ^2 \to 
 \overline{\mathbb{Q}}_{\ell}^{\times}$ by $\Lambda_{F,\zeta'} (x) =\epsilon_{F/K} ^{v_{L'} (x)} 
 \Lambda_{\zeta'} (\operatorname{Nr}_{L'/L} (x))$ for $x \in L'$ and $\Lambda_{F,\zeta'} (x) = 
 (\psi_F \circ \operatorname{tr}) 
 (\hat{\zeta}'^{-2} \varphi^{-1} (x-1) )$ for $x \in U_{\mathfrak{J}_F} ^2$. Then we have $\pi_{F,\zeta'} =
 \mathrm{c\mathchar`-Ind}_{{L'}^{\times} U_{\mathfrak{J}_F}^2} 
 ^{\textit{GL}_2(F)} \Lambda_{F,\zeta'}$ by [@BHLLC 46.3 Proposition] and the construction of the tame lifting. 

We will describe the restriction of $\tau_{\zeta'}$ to $W_F$. The field $F$ corresponds to the subgroup $Q_8 \rtimes \mathbb{Z}$ of $Q \rtimes \mathbb{Z}$.

First, we consider the case where $f$ is even.  We put $h_0 (x)=x^2-x$. Then we have $h_0 (\delta^2 -\delta) \equiv 1/(\zeta'' \varpi^{1/3}) \pmod{3/4}$. Hence we can take $\delta_2 \in F(\delta)$ such that $h_0 (\delta_2)=1/(\zeta'' \varpi^{1/3})$ and $\delta_2 \equiv \delta^2 -\delta \pmod{3/4}$ by Newton’s method.  Similarly, we can take $\delta_4 \in F(\delta)$ such that $\delta_4 ^2 - \delta_4=\delta_2$ and $\delta_4 \equiv \delta \pmod{3/4}$. Then we have $F(\delta_4) =F(\delta)$. Further, we can take $\theta_2 \in F(\theta)$ such that $\theta_2 ^2 -\theta_2 =\delta_4 ^3$ and $\theta_2 \equiv \theta \pmod{7/12}$. We have $F(\theta_2) =F(\theta)$. Then $F(\delta_2)$ corresponds to the subgroup $C_4 \times \mathbb{Z}$ of $Q \times \mathbb{Z}$.

Next, we consider the case where $f$ is odd.  We put $h_1 (x)=x^2-x +1$. Then we have $h_1 (\delta^2 -\delta +\zeta_3) \equiv 
 1/(\zeta'' \varpi^{1/3}) \pmod{3/4}$. Hence we can take $\delta_2 \in F(\zeta_3, \delta)$ such that $h_1 (\delta_2)=1/(\zeta'' \varpi^{1/3})$ and $\delta_2 \equiv \delta^2 -\delta +\zeta_3 \pmod{3/4}$ by Newton’s method.  Similarly, we can take $\delta_4 \in F(\zeta_3 ,\delta)$ such that $\delta_4 ^2 - \delta_4 +\zeta_3 =\delta_2$ and $\delta_4 \equiv \delta \pmod{3/4}$. Then we have $F(\zeta_3 ,\delta_4) =F(\zeta_3 ,\delta)$. Further, we can take $\theta_2 \in F(\zeta_3, \theta)$ such that $\theta_2 ^2 -\theta_2=\delta_4 ^3$ and $\theta_2 \equiv \theta \pmod{7/12}$. We have $F(\zeta_3,\theta_2) =F(\zeta_3,\theta)$. Then $F(\delta_2)$ corresponds to the subgroup $C$ of $Q \rtimes \mathbb{Z}$.

We put $E=F(\delta_2)$. Let $\phi_{\zeta'}$ be the character of $W_E$ induced from $\phi_1$ by $\Theta_{\zeta'}$. Then we have $\tau_{\zeta'} |_{W_F} \simeq 
 \operatorname{Ind}_{W_E}^{W_F} \phi_{\zeta'}$. We consider $\phi_{\zeta'}$ as a character of $E^{\times}$ by the Artin reciprocity map $\mathrm{Art}_E$. For a finite extension $K'$ of $K$ and integer $i$, we write $\mathfrak{p}_{K'}$ for the maximal ideal of $\mathcal{O}_{K'}$, and put $U_{K'}^i =1+\mathfrak{p}_{K'}^i$. Let $E_m$ be the unramified extension over $E$ of degree $m$ for a positive integer $m$. Let $\varkappa_{E/F}$ be the character of $F^{\times}$ with kernel $\operatorname{Nr}_{E/F} (E^{\times})$.

For a Galois group $G$ of a finite Galois extension of a non-Archimedean field, let $G_{(s)}$ and $G^{(t)}$ be the ramification subgroups of $G$ with lower numbering and upper numbering. 

\[phivarkappa\] We have $\phi_{\zeta'} (1+x) =\psi_E (\delta_2 ^3 x)$ for $x \in \mathfrak{p}_E ^2$ and $\varkappa_{E/F} (1+y) =
 \psi_F ((\zeta''\varpi^{1/3})^{-1} y)$ for $y \in \mathfrak{p}_F$.

We prove the first statement only in the case where $f$ is odd.  It is easier to prove the first statement in the case where $f$ is even. 

We put $G=\operatorname{Gal}(E_2 (\theta)/E)$. For $\sigma \in I_E$, we can show that $$v \biggl( \sigma\biggl(\frac{\delta}{\theta}\biggl) 
 -\frac{\delta}{\theta} \biggr) =
 \begin{cases}
 \frac{1}{12} &\quad 
 \textrm{if $\zeta_{3,\sigma}=1$, $\nu_{\sigma} =1$,}\\ 
 \frac{1}{6} &\quad 
 \textrm{if $\zeta_{3,\sigma}=1$, $\nu_{\sigma}=0$, $\mu_{\sigma} =1$} 
 \end{cases}$$ by the definition of $\zeta_{\sigma}$, $\nu_{\sigma}$ and $\mu_{\sigma}$. Then we have $\operatorname{Gal}(E_2 (\theta)/E_2) =G_{(0)} =G_{(1)} \supset 
 \operatorname{Gal}(E_2 (\theta)/E_2 (\delta)) =G_{(2)} =G_{(3)} \supset 
 \{1 \}=G_{(4)}$ and $$G^{(t)} = 
 \begin{cases}
 \operatorname{Gal}(E_2 (\theta)/E_2 ) &\quad 
 \textrm{if $0 \leq t \leq 1$,}\\ 
 \operatorname{Gal}(E_2 (\theta)/E_2 (\delta)) &\quad 
 \textrm{if $1 < t \leq 2$,}\\ 
 \{ 1 \} &\quad 
 \textrm{if $2 < t$.} 
 \end{cases}$$ Then the restriction of $\phi_{\zeta'}$ to $U_E ^2$ is the homomorphism $$U_E ^2 \twoheadrightarrow 
 U_E ^2 /(U_E ^3 \operatorname{Nr}_{E_2(\theta)/E}(U_{E(\theta)}^3)) 
 \xrightarrow{\sim} 
 \operatorname{Gal}(E_2 (\theta)/E_2 (\delta)) 
 \simeq Z \xrightarrow{\phi |_Z} 
 \overline{\mathbb{Q}}_{\ell}^{\times}$$ by [@SerCL XV §2 Corollaire 3 au Théorèm 1].  We define $N_2 \colon k_2 \to k$ by $N_2 (x) =\operatorname{Tr}_{k_2/k}(x)^2 +\operatorname{Tr}_{k_2/k}(x)$. Then we can check that $\operatorname{Nr}_{E_2(\theta)/E} \colon 
 U_{E_2(\theta)}^3 /U_{E_2(\theta)}^4 \to 
 U_E ^2 /U_E ^3$ becomes $N_2 \colon k_2 \to k$ under the identifications $$U_{E_2(\theta)}^3 /U_{E_2(\theta)}^4 \simeq k_2 ; 
 1+\theta^{-1} x \mapsto \bar{x} 
 \textrm{ and } 
 U_E^2 /U_E^3 \simeq k ; 
 1+\delta_2^{-2} x \mapsto \bar{x}.$$ Therefore we have $\phi_{\zeta'} (1+x) =(\chi_2 \circ \operatorname{Tr}_{k/\mathbb{F}_2} ) 
 (\overline{\delta_2 ^2x} )$, because $\operatorname{Im}N_2 =\operatorname{Ker}\operatorname{Tr}_{k/\mathbb{F}_2}$. Since we have $(\chi_2 \circ \operatorname{Tr}_{k/\mathbb{F}_2} ) 
 (\bar{x} ) =\psi_E (\delta_2 x)$, the first statement follows. 

We can prove the second statement similarly. 

\[detNr\] We consider $\hat{\zeta}'^{-2} \varphi^{-1}$ as an element of $\textit{GL}_2 (F)$ by the embedding .  Then we have $$\begin{aligned}
{2}
 \det (\hat{\zeta}'^{-2} \varphi^{-1} ) &\equiv \operatorname{Nr}_{E/F} (\delta_2 ^3) 
 & \quad &\mod U_F ^1 , \\ 
 \operatorname{tr}(\hat{\zeta}'^{-2} \varphi^{-1} ) &\equiv 
 (\zeta'' \varpi^{\frac{1}{3}})^{-1} + \operatorname{Tr}_{E/F} (\delta_2 ^3 ) 
 & \quad &\mod \mathcal{O}_F .  \end{aligned}$$

We can check the claim by easy calculations. 

Let $\mathcal{E}'$ be the elliptic curve over $\mathbb{F}_2$ defined by $z^2 +z =w^3 +w$. Let $\eta_2, \eta'_2 \in \overline{\mathbb{Q}}_{\ell}$ be the roots of $x^2 +2x +2 =0$.

\[ratell\] We have $\lvert \mathcal{E} (\mathbb{F}_q) \rvert 
 =q+1 -((-2)^{1/2})^f -(-(-2)^{1/2})^f$ and $\lvert \mathcal{E}' (\mathbb{F}_q) \rvert 
 =q+1 -\eta_2^f -{\eta'_2}^f$.

We have proved $\operatorname{tr}(\mathrm{fr}_q ^* ;H^1(\mathcal{E}_{k^{\mathrm{ac}}},
 \overline{\mathbb{Q}}_{\ell}) )= 
 ((-2)^{1/2})^f +(-(-2)^{1/2})^f$ in the first paragraph of the proof of Proposition \[cohtau\].  The first claim follows from this.  The second claim is proved similarly. 

If $f$ is odd, then the map $\Theta_{\zeta'}$ induces an isomorphism $W(E^{\mathrm{ur}}(\theta)/E) \simeq C$, and we write $\mathfrak{a}_E$ for the composite $$E^{\times} \xrightarrow{\mathrm{Art}_E} W_E^{\mathrm{ab}} 
 \twoheadrightarrow W(E^{\mathrm{ur}}(\theta)/E) \simeq C.$$

\[reclaw\] We assume that $f$ is odd.  Let $n_f$ and $m_f$ be the integers such that $1 \leq n_f, m_f \leq 2$, $n_f \equiv (f+1)/2 \bmod 2$ and $m_f \equiv (f^2+7)/8 \bmod 2$. Then we have $\mathfrak{a}_E (\delta_2)=
 (g(1,\bar{\zeta}_3 ^{2n_f} ,\bar{\zeta}_3 ^{m_f}),-1)$.

Let $\overline{C}$ be the image of $C$ in $Q_8 \rtimes (\mathbb{Z}/2\mathbb{Z})$. Then $\overline{C}$ is a cyclic group of order $8$. Let $\overline{\mathfrak{a}}_E$ be the composite of $\mathfrak{a}_E$ with the natural projection $C \to \overline{C}$. It suffices to show that $\overline{\mathfrak{a}}_E (\delta_2 )
 =(g(1,\bar{\zeta}_3 ^{2n_f} ,\bar{\zeta}_3 ^{m_f}),1)$, because we know that the second component of $\mathfrak{a}_E (\delta_2 )$ is $-1$. We note that the isomorphism $W(E^{\mathrm{ur}}(\theta)/E) \simeq C$ induces $\operatorname{Gal}(E_2 (\theta_2) /E) \simeq \overline{C}$. By this isomorphism, we consider $(g(1,\bar{\zeta}_3 ^{2n_f} ,\bar{\zeta}_3 ^{m_f}),1)$ as an element of $\operatorname{Gal}(E_2 (\theta_2) /E)$.

We write $E_{(0)}$ for $E$ in the mixed characteristic case, and $E_{(p)}$ for $E$ in the equal characteristic case.  We use similar notations for other fields and elements of the fields.  Then we have the isomorphism $$E_{(0)}^{\times} /U^3_{E_{(0)}} \simeq 
 E_{(p)}^{\times} /U^3_{E_{(p)}} 
 ;\ 
 \hat{\xi}_0 +\hat{\xi}_1 \delta_{2,(0)}^{-1} 
 +\hat{\xi}_2 \delta_{2,(0)}^{-2} \mapsto 
 \xi_0 +\xi_1 \delta_{2,(p)}^{-1} 
 +\xi_2 \delta_{2,(p)}^{-2}$$ where $\xi_0, \xi_1, \xi_2 \in k \subset E_{(p)}$ and $\hat{\xi}_0, \hat{\xi}_1, \hat{\xi}_2$ are their lifts in $\mu_{q-1} (E_{(0)} ) \cup \{0\}$. This isomorphism induces an isomorphism $$\bigl( \operatorname{Gal}(E_{(0)}^{\mathrm{sep}}/E_{(0)}) / 
 \operatorname{Gal}(E_{(0)}^{\mathrm{sep}}/E_{(0)})^{(3)} \bigr)^{\mathrm{ab}} 
 \simeq 
 \bigl( \operatorname{Gal}(E_{(p)}^{\mathrm{sep}}/E_{(p)}) / 
 \operatorname{Gal}(E_{(p)}^{\mathrm{sep}}/E_{(p)})^{(3)} \bigr)^{\mathrm{ab}}$$ by [@Delp0 (3.5.2)].  It further induces an isomorphism $\operatorname{Gal}\bigl( E_{2,(0)}(\theta_{2,(0)})/E_{(0)} \bigr) \simeq 
 \operatorname{Gal}\bigl( E_{2,(p)}(\theta_{2,(p)})/E_{(p)} \bigr)$. Then we have a commutative diagram $$\begin{aligned}
 &\xymatrix{
 E_{(0)}^{\times} /U^3_{E_{(0)}}
 \ar@{->}[r]^{\mathrm{Art}_{E_{(0)}} \hspace*{3em}} 
 \ar@{-}[d]^{\!  \rotatebox{90}{$\sim$}}  & 
 \operatorname{Gal}\bigl( E_{2,(0)}(\theta_{2,(0)})/E_{(0)} \bigr) 
 \ar@{->}[r]^{\hspace*{4.5em} \sim} 
 \ar@{-}[d]^{\!  \rotatebox{90}{$\sim$}}  & 
 \overline{C} 
 \ar@{=}[d] 
 \\
 E_{(p)}^{\times} /U^3_{E_{(p)}}
 \ar@{->}[r]^{\mathrm{Art}_{E_{(p)}} \hspace*{3em}} & 
 \operatorname{Gal}\bigl( E_{2,(p)}(\theta_{2,(p)})/E_{(p)} \bigr) 
 \ar@{->}[r]^{\hspace*{4.5em} \sim} & 
 \overline{C} 
 }\end{aligned}$$ by [@Delp0 (3.6.1)] and the construction of the isomorphisms.  Therefore, it suffices to show that $\overline{\mathfrak{a}}_E (\delta_2 )
 =(g(1,\bar{\zeta}_3 ^{2n_f} ,\bar{\zeta}_3 ^{m_f}),1)$ in the equal characteristic case. 

We assume that the characteristic of $E$ is $p$. We define the central division algebra $D_g$ over $E$ of degree $64$ by $D_g \simeq \bigoplus_{i=0} ^7 E_2 (\theta_2) s^i$ where $s^8 =\delta_2$ and $s a s^{-1} =
 (g(1,\bar{\zeta}_3 ^{2n_f} ,\bar{\zeta}_3 ^{m_f}),1) (a)$ for $a \in E_2 (\theta_2)$. Let $\sigma_q \in \operatorname{Gal}(E_8 /E)$ be the lift of $\operatorname{Fr}_q$. We define the central division algebra $D_{\sigma}$ over $E$ of degree $64$ by $D_{\sigma} \simeq \bigoplus_{i=0} ^7 E_8 t^i$ where $t^8 =\delta_2$ and $t a t^{-1} =\sigma_q (a)$ for $a \in E_8$. By the construction of the Artin reciprocity map, it suffices to show $D_g \simeq D_{\sigma}$ to prove the claim.  To show this isomorphism, it suffices to find $s' ,\delta'_4 ,\theta'_2 \in D_{\sigma}$ such that $$\begin{aligned}
 &{s'}^8=\delta_2,\quad 
 {\delta'_4}^2 -\delta'_4 +\zeta_3 =\delta_2,\quad 
 {\theta'_2}^2 -\theta'_2 ={\delta'_4}^3,\quad 
 \delta'_4 \theta'_2 =\theta'_2 \delta'_4 ,\\ 
 &s' \zeta_3 {s'}^{-1} =\zeta_3 ^2,\quad 
 s' \delta'_4 {s'}^{-1} = \delta'_4 +\zeta_3 ^{n_f},\quad 
 s' \theta'_2 {s'}^{-1} = \theta'_2 + 
 \zeta_3 ^{2n_f} \delta'_4 +\zeta_3 ^{m_f}.  \end{aligned}$$

We put $s' =t$. Then ${s'}^8=\delta_2$ and $s' \zeta_3 {s'}^{-1} =\zeta_3 ^2$. We take $a_0 \in \mu_{q^4 -1} (E_4)$ such that $a_0 ^2 -a_0 =\zeta_3$. We put $\delta'_4 =a_0 +t^2 +t^4$. Then we can check that ${\delta'_4}^2 -\delta'_4 +\zeta_3 = \delta_2$ using $t^2 a_0 t^{-2} = a_0 + 1$. We can check also that $t \delta'_4 t^{-1} = \delta'_4 + \zeta_3 ^{n_f}$ using $t a_0 t^{-1} = a_0 + \zeta_3 ^{n_f}$.

We take $b_0 \in \mu_{q^8 -1} (E_8)$ and $b_4 \in \mu_{q^4 -1} (E_4)$ such that $b_0 ^2 -b_0 =a_0 \zeta_3^2 +\zeta_3$ and $b_4 ^2=a_0$. We put $\theta'_2 =b_0 +(a_0 +\zeta_3)t^2 +b_4 t^4 +t^6$. Then we can check that ${\theta'_2}^2 -\theta'_2 ={\delta'_4}^3$, $\delta'_4 \theta'_2 =\theta'_2 \delta'_4$ and $t \theta'_2 t^{-1} = \theta'_2 + 
 \zeta_3 ^{2n_f} \delta'_4 +\zeta_3 ^{m_f}$ using $t b_0 t^{-1} =b_0 +a_0 \zeta_3 ^{2n_f} + \zeta_3 ^{m_f}$ and $t b_4 t^{-1} =b_4 +\zeta_3^{2n_f}$. Therefore, we have proved the claim. 

In the next proposition, we show that $\tau_{\zeta'} |_{W_F}$ corresponds to $\pi_{F,\zeta'}$ by the local Langlands correspondence by calculating epsilon factors.  We will show a correspondence over $K$ in Theorem \[expLLC\] using this correspondence over $F$.

\[indLLC\] We have $\mathrm{LL}_{\ell} (\tau_{\zeta'} |_{W_F}) =\pi_{F,\zeta'}$.

We put $\mathrm{LL}_{\ell} (\tau_{\zeta'}) =\pi'_{\zeta'}$ and $\mathrm{LL}_{\ell} (\tau_{\zeta'} |_{W_F}) =\pi'_{F,\zeta'}$. We want to show that $\pi'_{F,\zeta'} = \pi_{F,\zeta'}$.

By Lemma \[phivarkappa\], Lemma \[detNr\] and [@BHLLC 44.7 Proposition], the representation $\pi'_{F,\zeta'}$ contains the ramified simple stratum $(\mathfrak{J}_F, 3, \hat{\zeta}'^{-2} \varphi^{-1} )$.  Then the representation $\pi'_{\zeta'}$ contains the ramified simple stratum $(\mathfrak{J}, 1, \hat{\zeta}'^{-2} \varphi^{-1} )$ by the construction of $\pi'_{\zeta'}$ in [@BHLLC 50.3].  Therefore we have $\pi'_{\zeta'} =
 \mathrm{c\mathchar`-Ind}_{L^{\times} U_{\mathfrak{J}}^1} 
 ^{\textit{GL}_2(K)} \Lambda'_{\zeta'}$ for a character $\Lambda'_{\zeta'} \colon 
 L^{\times} U_{\mathfrak{J}} ^1 \to 
 \overline{\mathbb{Q}}_{\ell}^{\times}$ such that $\Lambda'_{\zeta'} =\Lambda_{\zeta'}$ on $U_{\mathfrak{J}} ^1$.

Let $1_F$ denote the trivial character of $W_F$. We put $\varkappa_{F/K} =\det \operatorname{Ind}_{W_F}^{W_K} 1_F$. Then $\varkappa_{F/K} |_{W_F} =1_F$ if $f$ is even, and $\varkappa_{F/K} |_{W_F}$ is the unramified character of order two if $f$ is odd.  Hence, the definition of $\epsilon_{F/K}$ in [@BHLLC 46.3] coincides with that in this paper.  By [@BHLLC 46.3 Proposition], we have $\pi'_{F,\zeta'} =
 \mathrm{c\mathchar`-Ind}_{{L'}^{\times} U_{\mathfrak{J}_F}^2} 
 ^{\textit{GL}_2(F)} \Lambda'_{F,\zeta'}$ for a character $\Lambda'_{F,\zeta'} \colon 
 {L'}^{\times} U_{\mathfrak{J}_F} ^2 \to 
 \overline{\mathbb{Q}}_{\ell}^{\times}$ such that $\Lambda'_{F,\zeta'} =\Lambda_{F,\zeta'}$ on $U_{\mathfrak{J}_F} ^2$ and $\Lambda'_{F,\zeta'} (x) =\epsilon_{F/K} ^{v_{L'} (x)} 
 \Lambda'_{\zeta'} (\operatorname{Nr}_{L'/L} (x))$ for $x \in L'$. Hence we have $\Lambda'_{F,\zeta'} (x)=1$ for $x \in U_{L'} ^1$, because $\Lambda'_{\zeta'} (x)=1$ for $x \in U_L ^1$. Then we see that $\Lambda'_{F,\zeta'} =\Lambda_{F,\zeta'}$ on $x \in U_{L'} ^1 F^{\times} U_{\mathfrak{J}_F} ^2$, because $\Lambda'_{F,\zeta'} =\Lambda_{F,\zeta'}$ on $F^{\times}$ by Remark \[ellLLC\] and Lemma \[dettau\]. 

We define $\kappa_F \colon F^{\times} \to \overline{\mathbb{Q}}_{\ell}^{\times}$ by $\kappa_F (x)=(-1)^{v_F (x)}$. Since $\Lambda'_{F,\zeta'} =\Lambda_{F,\zeta'}$ on $x \in U_{L'} ^1 F^{\times} U_{\mathfrak{J}_F} ^2$, we know that $\Lambda'_{F,\zeta'} =\Lambda_{F,\zeta'}$ or $\Lambda'_{F,\zeta'} =\Lambda_{F,\zeta'} \otimes (\kappa_F \circ \det)$. We take an isomorphism $\iota \colon \overline{\mathbb{Q}}_{\ell} \simeq \mathbb{C}$. Then, to show $\Lambda'_{F,\zeta'} =\Lambda_{F,\zeta'}$, it suffices to show that $\varepsilon ({}^{\iota} \pi'_{F,\zeta'}, 1/2, \iota \circ \psi_F) = 
 \varepsilon ({}^{\iota} \pi_{F,\zeta'}, 1/2, \iota \circ \psi_F)$. We note that we have already known this equality up to sign. 

In the sequel of this proof, we identify $\overline{\mathbb{Q}}_{\ell}$ and $\mathbb{C}$ by $\iota$, and omit to write $\iota$. By [@BHLLC 25.5 Corollary], we obtain $\varepsilon (\pi_{F,\zeta'}, 1/2, \psi_F) 
 =-\epsilon_{F/K}$ using that $5$ is the least integer $m \geq 0$ such that $U_{\mathfrak{J}_F} ^{m+1} \subset \operatorname{Ker}\Lambda_{F,\zeta'}$. On the other hand, we have $\varepsilon (\pi'_{F,\zeta'}, 1/2, \psi_F) 
 =\varepsilon (\tau_{\zeta'}|_{W_F}, 0, \psi_F) 
 =q^{3/2} \varepsilon (\tau_{\zeta'}|_{W_F}, 1/2, \psi_F)$, because the Artin conductor of $\tau_{\zeta'}|_{W_F}$ is three.  Hence, it suffices to show that $\varepsilon (\tau_{\zeta'}|_{W_F},1/2,\psi_F) =-\epsilon_{F/K} q^{-3/2}$.

Let $\lambda_{E/F} (\psi_F)$ be the local constant of $E$ over $F$ with respect to $\psi_F$. Let $1_E$ denote the trivial character of $W_E$. Then we have $$\lambda_{E/F} (\psi_F) = 
 \varepsilon (\operatorname{Ind}_{W_E}^{W_F} 1_E, 1/2,\psi_F) 
 \varepsilon (1_E, 1/2,\psi_E)^{-1} = 
 \varepsilon (\varkappa_{E/F}, 1/2,\psi_F) = 
 \varkappa_{E/F}(\zeta''\varpi^{1/3}) 
 =\epsilon_{F/K}$$ by Lemma \[phivarkappa\] and [@BHLLC 23.6 Proposition], because we have $$\operatorname{Nr}_{E/F} (\delta_2)=
 \begin{cases}
 -1/(\zeta''\varpi^{1/3}) &\quad \textrm{if $f$ is even}, \\ 
 -1/(\zeta''\varpi^{1/3}) +1&\quad \textrm{if $f$ is odd}.  
 \end{cases}$$ Therefore, it suffices to show that $\varepsilon (\phi_{\zeta'}, 1/2, \psi_E) =-q^{-3/2}$, because we have $\varepsilon (\tau_{\zeta'}|_{W_F}, 1/2, \psi_F) = 
 \varepsilon (\phi_{\zeta'}, 1/2, \psi_E) 
 \lambda_{E/F} (\psi_F)$.

We define $\psi'_E$ by $\psi'_E (x)=\psi_E (\delta_2 x)$ for $x \in E^{\times}$. Then $\psi_E '$ has level one, and we have $$\begin{aligned}
 \varepsilon (\phi_{\zeta'}, 1/2, \psi_E) = 
 \phi_{\zeta'} (\delta_2)^{-1} \varepsilon (\phi_{\zeta'}, 1/2, \psi'_E ) &= 
 q^{-1/2} \phi_{\zeta'} (\delta_2)^{-1} 
 \sum_{y \in U^1_E / U^2_E} \phi_{\zeta'} (\delta_2^2 y)^{-1} 
 \psi'_E (\delta_2^2 y) \\ 
 &=
 q^{-1/2} \phi_{\zeta'} (\delta_2^3)^{-1} 
 \sum_{x \in \mathfrak{p}_E / \mathfrak{p}^2_E} 
 \phi_{\zeta'} (1+x)^{-1} 
 \psi_E (\delta_2^3 x) \end{aligned}$$ by [@BHLLC 23.5 Lemma 1, (23.6.2) and (23.6.4)] and $\psi_E (\delta_2^3)=1$. Therefore it suffices to show that $$\label{aim}
 \phi_{\zeta'} (\delta_2^3)^{-1} 
 \sum_{x \in \mathcal{O}_E / \mathfrak{p}_E} 
 \phi_{\zeta'} (1+\delta_2^{-1}x)^{-1} 
 \psi_E (\delta_2^2 x) = -q^{-1}.$$ Note that we have already known this equality up to sign. 

First, we consider the case where $f$ is even.  Then we have $\phi_{\zeta'} (\delta_2^3) =(-2)^{3f/2}$ by $\operatorname{Nr}_{E(\theta_2)/E} (\theta_2)=-\delta_2^3$ and Lemma \[phivarkappa\].  Hence, it suffices to show $\sum_{x \in \mathcal{O}_E / \mathfrak{p}_E} 
 \phi_{\zeta'} (1+\delta_2^{-1}x)^{-1} 
 \psi_E (\delta_2^2 x) = -(-2)^{f/2}$. For $\xi \in k$, the Teichmüller lift of $\xi$ is denoted by $\hat{\xi}$. We have $$\label{evenphi}
 \phi_{\zeta'} \bigl( 1+(\hat{\xi}^2 +\hat{\xi}^4)\delta_2 ^{-1}+ 
 (\hat{\xi} +\hat{\xi}^2 +\hat{\xi}^3 )\delta_2 ^{-2} \bigr) =1,\ 
 \phi_{\zeta'} \bigl( 1+(\hat{\xi}^2 +\hat{\xi}^4) \delta_2 ^{-2}
 \bigr) =1$$ for $\xi \in k$, since $$\begin{aligned}
 \operatorname{Nr}_{E(\theta_2)/E} (1+\hat{\xi}\delta_4 \theta_2 ^{-1})
 &\equiv 1+(\hat{\xi}^2 +\hat{\xi}^4)\delta_2 ^{-1}+ 
 (\hat{\xi} +\hat{\xi}^2 +\hat{\xi}^3 )\delta_2 ^{-2} 
 \pmod{1/2}, \\ 
 \operatorname{Nr}_{E(\theta_2)/E} (1+\hat{\xi} \delta_4^2 \theta_2 ^{-2}) 
 &\equiv 1+(\hat{\xi}^2 +\hat{\xi}^4) \delta_2 ^{-2} 
 \pmod{1/2}.  \end{aligned}$$ Therefore we see that $\phi_{\zeta'} (1+\delta_2^{-1}x) =\pm \sqrt{-1}$ for $x \in \mathcal{O}_E$ if $\bar{x} \neq \xi^2 +\xi^4$ for any $\xi \in k$. Then we have $$\sum_{x \in \mathcal{O}_E / \mathfrak{p}_E} 
 \phi_{\zeta'} (1+\delta_2^{-1}x)^{-1} 
 \psi_E (\delta_2^2 x) = 
 \frac{1}{2} \sum_{\xi \in k} 
 \phi_{\zeta'} \bigl( 1+(\hat{\xi}^2 +\hat{\xi}^4)\delta_2^{-1} 
 \bigr)^{-1} 
 \psi_E \bigl( \delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4) \bigr) ,$$ since we have already known that $\sum_{x \in \mathcal{O}_E / \mathfrak{p}_E} 
 \phi_{\zeta'} (1+\delta_2^{-1}x)^{-1} 
 \psi_E (\delta_2^2 x) = \pm (-2)^{f/2} \in \mathbb{Q}$. We have $$\sum_{\xi \in k} 
 \phi_{\zeta'} \bigl( 1+(\hat{\xi}^2 +\hat{\xi}^4) \delta_2^{-1} 
 \bigr)^{-1} 
 \psi_E \bigl( \delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4) \bigr) 
 = \sum_{\xi \in k} 
 \phi_{\zeta'} \bigl( 1+
 \hat{\xi}^3 \delta_2^{-2} \bigr) 
 \psi_E \bigl( \delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4) \bigr),$$ since $$\phi_{\zeta'} \bigl( 1+(\hat{\xi}^2 +\hat{\xi}^4) \delta_2^{-1} 
 \bigr)^{-1} = 
 \phi_{\zeta'} \bigl( 1+
 (\hat{\xi}+\hat{\xi}^2 +\hat{\xi}^3 ) 
 \delta_2^{-2} \bigr) = 
 \phi_{\zeta'} \bigl( 1+
 \hat{\xi}^3 \delta_2^{-2} \bigr)$$ by .  Further we have $$\begin{aligned}
 \sum_{\xi \in k} 
 \phi_{\zeta'} \bigl( 1+
 \hat{\xi}^3 
 \delta_2^{-2} 
 \bigr)
 \psi_E \bigl( \delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4) \bigr) 
 &= \sum_{\xi \in k} 
 \chi_2 (\operatorname{Tr}_{k/\mathbb{F}_2} (\xi^3 +\xi^2 +\xi^4)) \\ 
 &= \sum_{\xi \in k} 
 \chi_2 (\operatorname{Tr}_{k/\mathbb{F}_2} (\xi^3)) =-2(-2)^{f/2} \end{aligned}$$ by Lemma \[phivarkappa\], because $\operatorname{Ker}\operatorname{Tr}_{k/\mathbb{F}_2} = 
 \{ \xi +\xi^2 \mid \xi \in k \}$ and $$\lvert \{ (x,y) \in k^2 \mid x^2 +x =y^3 \} \rvert = 
 \lvert \mathcal{E} (\mathbb{F}_q ) \rvert -1 
 = q -2(-2)^{f/2}$$ by Lemma \[ratell\].  Thus we have the claim in the case where $f$ is even. 

Next we consider the case where $f$ is odd.  We define $n_f$ and $m_f$ as in Lemma \[reclaw\].  We treat only the case where $n_f =m_f =1$. The other cases are proved similarly. 

We assume that $n_f =m_f =1$, which is equivalent to that $f \equiv 1 \bmod 8$. By Lemma \[reclaw\] and the definition of $\phi_{\zeta'}$, we have $\phi_{\zeta'} (\delta_2) =q/\eta$. Hence, to prove , it suffices to show $\sum_{x \in \mathcal{O}_E / \mathfrak{p}_E} 
 \phi_{\zeta'} (1+\delta_2^{-1}x)^{-1} 
 \psi_E (\delta_2^2 x) = -q^2 \eta^{-3}$. We have $$\begin{aligned}
 &\phi_{\zeta'} \Bigl( 
 1+\operatorname{Tr}_{E_2/E} (\hat{\xi}^2 +\hat{\xi}^4)\delta_2 ^{-1}+ 
 \bigl( \operatorname{Tr}_{E_2/E}(\hat{\xi} +\hat{\xi}^2 \zeta_3^2 
 +\hat{\xi}^3 +\hat{\xi}^4 \zeta_3 ) + 
 \operatorname{Nr}_{E_2/E}(\hat{\xi}^2 +\hat{\xi}^4) \bigr) \delta_2 ^{-2} \Bigr) =1, 
 \label{oddphi1} \\ 
 &\phi_{\zeta'} \bigl( 
 1+\operatorname{Tr}_{E_2/E} (\hat{\xi}^2 +\hat{\xi}^4) \delta_2 ^{-2} 
 \bigr) =1 \label{oddphi2}\end{aligned}$$ for $\xi \in k_2$, since $$\begin{aligned}
 \operatorname{Nr}_{E_2(\theta_2)/E} (1+\hat{\xi}\delta_4 \theta_2 ^{-1})
 \equiv &1+\operatorname{Tr}_{E_2/E} (\hat{\xi}^2 +\hat{\xi}^4)\delta_2 ^{-1} \\ 
 &+ \bigl( \operatorname{Tr}_{E_2/E}(\hat{\xi} +\hat{\xi}^2 \zeta_3^2 
 +\hat{\xi}^3 +\hat{\xi}^4 \zeta_3 ) + 
 \operatorname{Nr}_{E_2/E}(\hat{\xi}^2 +\hat{\xi}^4) \bigr) \delta_2 ^{-2} 
 \pmod{1/2}, \\ 
 \operatorname{Nr}_{E_2(\theta_2)/E} (1+\hat{\xi} \delta_4^2 \theta_2 ^{-2}) 
 \equiv &1+\operatorname{Tr}_{E_2/E} (\hat{\xi}^2 +\hat{\xi}^4) \delta_2 ^{-2} 
 \pmod{1/2}.  \end{aligned}$$ By , we have $$\label{oddphi2'}
 \phi_{\zeta'} ( 
 1+(\hat{\xi} +\hat{\xi}^2 ) \delta_2 ^{-2} ) =1$$ for $\xi \in k$, because $\operatorname{Tr}_{k_2/k}$ is surjective and $\operatorname{Tr}_{k_2/k} (\xi^2 +\xi^4) = 
 \operatorname{Tr}_{k_2/k} (\xi)^2 +\operatorname{Tr}_{k_2/k} (\xi)^4$ for $\xi \in k_2$. Then, by and , we have $$\label{oddphi1'}
 \phi_{\zeta'} ( 
 1+(\hat{\xi}^2 +\hat{\xi}^4 ) \delta_2 ^{-1} +
 (\hat{\xi} +\hat{\xi}^3 ) \delta_2 ^{-2} ) =1$$ for $\xi \in k$, because $$\begin{aligned}
 \operatorname{Tr}_{k_2/k} (\xi^3) +\operatorname{Nr}_{k_2/k} (\xi^2 +\xi^4) =
 \operatorname{Tr}_{k_2/k} (\xi)^3 + 
 \bigl( \operatorname{Nr}_{k_2/k} &(\xi^2) + \xi^{q+1}\operatorname{Tr}(\xi) \bigr) \\ 
 &+ 
 \bigl( \operatorname{Nr}_{k_2/k} (\xi^2) + \xi^{q+1}\operatorname{Tr}(\xi) \bigr)^2 \end{aligned}$$ for $\xi \in k_2$. We have $\phi_{\zeta'} (\delta_2 ^2 +\delta_2 +1) =-q$, since $\operatorname{Nr}_{E_2(\theta_2)/E} (\theta_2/\delta_4) =
 \delta_2 ^2 +\delta_2 +1$. Hence, we obtain $\phi_{\zeta'} (1+\delta_2 ^{-1} +\delta_2 ^{-2})=-\eta^2 q^{-1}$. Therefore we have $$\phi_{\zeta'} 
 (1+ (\hat{\xi}^2 +\hat{\xi}^4 +1)\delta_2^{-1} )^{-1} 
 \psi_E (\delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4 +1)) 
 = -\frac{q}{\eta^2} 
 \phi_{\zeta'} 
 (1+ (\hat{\xi}^2 +\hat{\xi}^4 )\delta_2^{-1} )^{-1} 
 \psi_E (\delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4 )),$$ since we have $\phi_{\zeta'} (1+\delta_2^{-2}) = 
 \psi_E (\delta_2)=\psi_E (\delta_2^2)$ by Lemma \[phivarkappa\].  Then we have $$\begin{aligned}
 \sum_{x \in \mathcal{O}_E / \mathfrak{p}_E} 
 \phi_{\zeta'} (1+\delta_2^{-1}x)^{-1} 
 \psi_E (\delta_2^2 x) =
 \frac{1}{2} \Bigl( 1-\frac{q}{\eta^2} \Bigr) 
 \sum_{\xi \in k} 
 \phi_{\zeta'} 
 (1+(\hat{\xi}^2 +\hat{\xi}^4)\delta_2^{-1} )^{-1} 
 \psi_E (\delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4)), \end{aligned}$$ because $\xi \in \operatorname{Ker}\operatorname{Tr}_{k/\mathbb{F}_2} 
 =\{ {\xi'}^2 +{\xi'}^4 \mid \xi' \in k \}$ if and only if $\xi +1 \notin \operatorname{Ker}\operatorname{Tr}_{k/\mathbb{F}_2}$. Therefore, it suffices to show that $$\sum_{\xi \in k} 
 \phi_{\zeta'} 
 (1+(\hat{\xi}^2 +\hat{\xi}^4)\delta_2^{-1} )^{-1} 
 \psi_E (\delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4)) = 
 -(-2)^{(f+1)/2}.$$ On the other hand, by and Lemma \[phivarkappa\], we have $$\begin{aligned}
 \sum_{\xi \in k} 
 \phi_{\zeta'} 
 (1+(\hat{\xi}^2 +\hat{\xi}^4)\delta_2^{-1} )^{-1} 
 \psi_E (\delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4)) &= 
 \sum_{\xi \in k} 
 \phi_{\zeta'} 
 (1+(\hat{\xi} +\hat{\xi}^3 )\delta_2^{-2} ) 
 \psi_E (\delta_2^2 (\hat{\xi}^2 +\hat{\xi}^4)) \\ 
 &= \sum_{\xi \in k} 
 \chi_2 (\operatorname{Tr}_{k/\mathbb{F}_2} (\xi +\xi^3)) = 
 -(-2)^{(f+1)/2} \end{aligned}$$ because $\operatorname{Ker}\operatorname{Tr}_{k/\mathbb{F}_2} = 
 \{ \xi +\xi^2 \mid \xi \in k \}$ and $$\lvert \{ (x,y) \in k^2 \mid x^2 +x =y^3 +y \} \rvert = 
 \lvert \mathcal{E}'(\mathbb{F}_q ) \rvert -1 
 = q -(-2)^{(f+1)/2}$$ by Lemma \[ratell\] under the assumption $f \equiv 1 \bmod 8$. Thus we have proved the claim. 

\[expLLC\] For $\zeta' \in k^{\times}$, $\chi \in (k^{\times})^{\vee}$ and $c \in \overline{\mathbb{Q}}_{\ell}^{\times}$, we have $\mathrm{LL}_{\ell} (\tau_{\zeta', \chi, c}) = 
 \pi_{\zeta', \chi, c}$.

We may assume that $\chi=1$ and $c=1$ by character twists.  By [@BHLLC 50.3 Lemma 1 and (50.3.2)] and Proposition \[indLLC\], it suffices to show that $\Lambda_{\zeta'}|_{K^{\times}} \circ \mathrm{Art}_K ^{-1} 
 =(\det \tau_{\zeta'}) \otimes \lvert \cdot \rvert^{-1}$, $\Lambda_{\zeta'} ^3 |_{U_{\mathfrak{J}}^1} = 
 \Lambda_{F, \zeta'} |_{U_{\mathfrak{J}}^1}$ and $\Lambda_{\zeta'} ^3 (x)= \epsilon_{F/K} ^{v_{L'} (x)} 
 \Lambda_{F, \zeta'} (x)$ for $x \in L^{\times}$. The first equality follows from Lemma \[dettau\] and Remark \[ellLLC\], and the third equality follows from the definition of $\Lambda_{F, \zeta'}$.

We are going to show the second equality.  We put $$U'_{\mathfrak{J}} =\biggl\{ 
 \begin{pmatrix}
 a & b \\
 c & d
 \end{pmatrix} 
 \in \mathfrak{J} \ \bigg| \ 
 a \equiv d \equiv 1 ,\ 
 b \equiv 0 \bmod \mathfrak{p} \biggr\}, \ 
 U''_{\mathfrak{J}} =\biggl\{ 
 \begin{pmatrix}
 1 & b \\
 0 & 1 
 \end{pmatrix} 
 \in \mathfrak{J} \biggr\}.$$ Then we have $\Lambda_{\zeta'} ^3 |_{U'_{\mathfrak{J}}} = 
 \Lambda_{F, \zeta'} |_{U'_{\mathfrak{J}}}$ by the definition of $\Lambda_{\zeta'}$ and $\Lambda_{F, \zeta'}$, because $U'_{\mathfrak{J}} \subset U_{\mathfrak{J}_F}^2$. On the other hand, we have $$\begin{aligned}
 \Lambda_{F, \zeta'} 
 \biggl( 
 \begin{pmatrix}
 1 & b \\
 0 & 1 
 \end{pmatrix} \biggr) 
 &= 
 \Lambda_{F, \zeta'} 
 \biggl( \varphi 
 \begin{pmatrix}
 1 & b \\
 0 & 1 
 \end{pmatrix} \varphi^{-1} \biggr) 
 = 
 \Lambda_{F, \zeta'} 
 \biggl( 
 \begin{pmatrix}
 1 & 0 \\
 b\varpi & 1 
 \end{pmatrix} \biggr) = 
 \psi_F (\hat{\zeta}'^{-2} b) \\
 &= \psi_K (\hat{\zeta}'^{-2} b)^3 
 = 
 \Lambda_{\zeta'} 
 \biggl( 
 \begin{pmatrix}
 1 & 0 \\
 b\varpi & 1 
 \end{pmatrix} \biggr)^3 
 = 
 \Lambda_{\zeta'} 
 \biggl( \varphi 
 \begin{pmatrix}
 1 & b \\
 0 & 1 
 \end{pmatrix} \varphi^{-1} \biggr)^3 
 =
 \Lambda_{\zeta'} 
 \biggl( 
 \begin{pmatrix}
 1 & b \\
 0 & 1 
 \end{pmatrix} \biggr)^3 \end{aligned}$$ for $b \in \mathcal{O}_K$. Therefore we have the claim, because $U_{\mathfrak{J}}^1$ is generated by $U'_{\mathfrak{J}}$ and $U''_{\mathfrak{J}}$.

[SGA]{}

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Naoki Imai\
Research Institute for Mathematical Sciences, Kyoto University, Kyoto 606-8502 Japan\
naoki@kurims.kyoto-u.ac.jp\
Takahiro Tsushima\
Faculty of Mathematics, Kyushu University, 744 Motooka, Nishi-ku, Fukuoka city, Fukuoka 819-0395, Japan\
tsushima@math.kyushu-u.ac.jp\
