**Proof.** By Corollary 4.13, $\mathbb{E} = \mathbb{F}^{\Sigma\bar{\rho}}$. First we show that $\bar{\rho}$ can be conjugated to land in $ZGL_2(\mathbb{E})$. Let $\sigma \in \text{Gal}(\mathbb{F}/\mathbb{E})$ be a generator and $\eta$ a character such that $(\sigma, \eta) \in \Sigma_{\bar{\rho}}$. Then there is some $x \in \text{GL}_2(\mathbb{F})$ such that for all $g \in \Pi$, we have $\sigma\bar{\rho}(g) = x^{-1}\eta(g)\bar{\rho}(g)x$. By a theorem of Serge Lang [Lan56, Corollary to Theorem 1], it follows that there is some $y \in \text{GL}_2(\mathbb{F})$ such that $x = \sigma^{yy^{-1}}$. Thus $\sigma(y^{-1}\bar{\rho}(g)y) = \eta(g)(y^{-1}\bar{\rho}(g)y)$. Replacing $\bar{\rho}$ by its conjugate by $y$, we have that the projective image of $\bar{\rho}$ is fixed by $\text{Gal}(\mathbb{F}/\mathbb{E})$, and hence the image of $\bar{\rho}$ lands in $ZGL_2(\mathbb{E})$, as desired.

If $\mathbb{F}_q, \lambda_0, \mu_0$ are as in the statement of the lemma, then $\bar{\rho}$ can be further conjugated such that $(\begin{smallmatrix} \lambda_0 & 0 \\ 0 & \mu_0 \end{smallmatrix}) \in \text{Im} \bar{\rho}$ while preserving the property that the image of $\bar{\rho}$ is in $ZGL_2(\mathbb{F}_q)$. $\square$

**Proposition 8.13.** Let $\bar{\rho}: \Pi \to \text{GL}_2(\mathbb{F})$ be regular and either exceptional or large. If $\bar{\rho}$ is octahedral, assume further that $\bar{\rho}$ is good. Assume that the order of $\det \bar{\rho}$ is a power of 2. Then there is a regular element $g_0 \in \Pi$ and a basis for $\bar{\rho}$ such that the following are simultaneously true:

(1) $\text{Im} \bar{\rho} \subseteq Z\text{GL}_2(\mathbb{E})$;

(2) $\bar{\rho}(g_0) = (\begin{smallmatrix} \lambda_0 & 0 \\ 0 & \mu_0 \end{smallmatrix})$ for some $\lambda_0, \mu_0 \in \mathbb{F}$;

(3) if $p \ge 7$ and $\bar{\rho}$ is large, then $\lambda_0, \mu_0 \in \mathbb{F}_p^\times$;

(4) there is a positive integer $n$ such that $g_0^n \in \Pi_0(\bar{\rho})$ and $\bar{\rho}(g_0^n)$ is not scalar.

**Proof.** By Lemma 8.12 we can always conjugate $\bar{\rho}$ so that $\text{Im} \bar{\rho} \subseteq Z\text{GL}_2(\mathbb{E})$. If $g_0 \in \Pi$ is a regular element and $\lambda_0$ and $\mu_0$ are the eigenvalues of $\bar{\rho}(g_0)$, then we may assume further that $\bar{\rho}(g_0) = (\begin{smallmatrix} \lambda_0 & 0 \\ 0 & \mu_0 \end{smallmatrix})$.

If $\bar{\rho}$ is large, then up to conjugation, $\text{Im} \bar{\rho} \supseteq \text{SL}_2(\mathbb{F}_p)$. Indeed, up to conjugation we may assume that the projective image of $\bar{\rho}$ contains $\text{PSL}_2(\mathbb{E})$. Therefore there is some $\lambda \in \mathbb{F}^\times$ such that $\lambda(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}) \in \text{Im} \bar{\rho}$. Note that the $n^{th}$ power of this matrix is $\lambda^n(\begin{smallmatrix} 1 & n \\ 0 & 1 \end{smallmatrix})$. Since $\lambda \in \mathbb{F}^\times$, its order $m$ is prime to $p$. Therefore we can write $1 = am + bp \equiv am \mod p$ for some $a, b \in \mathbb{Z}$. Thus

$$ (\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}) = \lambda^{am} (\begin{smallmatrix} 1 & am \\ 0 & 1 \end{smallmatrix}) = (\lambda(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}))^{am} \in \text{Im} \bar{\rho}. $$

Similarly, $(\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix}) \in \text{Im} \bar{\rho}$. Since $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $(\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix})$ generate $\text{SL}_2(\mathbb{F}_p)$, it follows that $\text{SL}_2(\mathbb{F}_p) \subseteq \text{Im} \bar{\rho}$.

If $p \ge 7$, then we can choose $\alpha \in \mathbb{F}_p^\times$ such that $\alpha^2 \ne \pm 1$. Then any $g_0 \in \Pi$ such that $\bar{\rho}(g_0)$ has eigenvalues $\alpha, \alpha^{-1}$ satisfies the first three conditions. Note that $\mathbb{P}\bar{\rho}(g_0) \in \text{PSL}_2(\mathbb{E})$. Recall that $\Pi_0(\bar{\rho})$ is a normal subgroup of 2-power index in $\Pi$ since the order of $\det \bar{\rho}$ is a power of 2. Furthermore, $\Pi/\Pi_0(\bar{\rho})$ is abelian. Therefore $\mathbb{P}\bar{\rho}(\Pi_0(\bar{\rho}))$ is either $\text{PGL}_2(\mathbb{E})$ or $\text{PSL}_2(\mathbb{E})$. In either case, we can find $g_0 \in \Pi_0(\bar{\rho})$ such that $\bar{\rho}(g_0)$ has eigenvalues $\alpha, \alpha^{-1}$. Thus all of the properties of the proposition are satisfied for this choice of $g_0$.

Next suppose that $\bar{\rho}$ is either tetrahedral or icosahedral. Once again, $\mathbb{P}\bar{\rho}(\Pi_0(\bar{\rho}))$ is a normal subgroup of $\mathbb{P}\bar{\rho}(\Pi)$ with 2-power index and abelian quotient. Since $\mathbb{P}\bar{\rho}(\Pi)$ is isomorphic to one of $A_4$ or $A_5$, it follows that $\mathbb{P}\bar{\rho}(\Pi_0(\bar{\rho})) = \mathbb{P}\bar{\rho}(\Pi)$. In particular, one can choose the regular element $g_0$ to be in $\Pi_0(\bar{\rho})$, and the resulting representation satisfies all of the desired conditions.

Finally, suppose that $\bar{\rho}$ is octahedral and good. If $p \equiv 1 \mod 3$ then any $g_0 \in \Pi$ such that $\mathbb{P}\bar{\rho}(g_0)$ has order 3 is a regular element. Since $\mathbb{P}\bar{\rho}(\Pi_0(\bar{\rho}))$ is a normal subgroup of $\mathbb{P}\bar{\rho}(\Pi)$ with 2-power index and abelian quotient, it follows that $\Pi_0(\bar{\rho})$ contains an element $g_0$ such that $\mathbb{P}\bar{\rho}(g_0)$ has order 3. Such a $g_0$ satisfies all of the necessary conditions.

Next suppose that $p \equiv 2 \mod 3$ and that $\bar{\rho}$ is strongly regular. Let $g_0 \in \Pi$ be a strongly regular element. Then $\bar{\rho}(g_0) = (\begin{smallmatrix} \lambda\zeta_4 & 0 \\ 0 & \lambda \end{smallmatrix})$ for some $\lambda \in \mathbb{E}^\times$. (Note that $\zeta_4 \in \mathbb{F}_p$ since $\bar{\rho}$ is regular and $p \equiv 2 \mod 3$.) We claim that $g_0 \in \Pi_0(\bar{\rho})$. Indeed, let $\sigma$ be a generator of $\text{Gal}(\mathbb{F}/\mathbb{E})$ and $\eta$ a character such that $(\sigma, \eta) \in \Sigma_{\bar{\rho}}$. Then $\Pi_0(\bar{\rho}) = \ker \eta$. Since $\lambda, \zeta_4 \in \mathbb{E}^\times$ we have

$$ \lambda(\zeta_4 + 1) = \sigma(\lambda(\zeta_4 + 1)) = \sigma_{\text{tr}} \bar{\rho}(g_0) = \eta(g_0) \text{tr} \bar{\rho}(g_0) = \eta(g_0)\lambda(\zeta_4 + 1). $$

As $\zeta_4 + 1 \ne 0$ it follows that $\eta(g_0) = 1$, and so $g_0 \in \Pi_0(\bar{\rho})$, as claimed. Therefore $g_0$ satisfies all of the necessary conditions.
Finally suppose that $p \equiv 2 \mod 3$ and there is a regular element $g_0 \in \Pi$ such that $g_0^2 \in \Pi_0(\bar{p})$. Note that $\mathbb{P}\bar{p}(g_0)$ has order 4 since $p \neq 1 \mod 3$. Therefore $\mathbb{P}\bar{p}(g_0^2)$ is nontrivial, so $\bar{p}(g_0^2)$ is not scalar. Therefore $g_0$ satisfies all of the conditions of the proposition. $\square$

Note that the $g_0$ chosen in Proposition 8.13 satisfies all of the conditions prior to Definition 2.21. In particular, if $(t, d): \Pi \to A$ is any admissible pseudodeformation of $\bar{p}$, then any $(t, d)$-representation that is adapted to the element $g_0$ from Proposition 8.13 is well adapted.

## 9. FULLNESS PEERS: $\mathcal{B}_{\rho}(\mathbb{E})$ AND $A^{\Sigma_{\rho}}$

Throughout Section 9 we fix a local pro-$p$ domain $A$ and an admissible pseudodeformation $(\Pi, \bar{p}, t, d)$ over $A$. We only consider $A$-valued conjugate self-twists throughout this section and thus write $\Sigma_t$ for $\Sigma_t(A)$ and similarly for $\tilde{\Sigma}_t$. The goal of Section 9 is to prove that $(t, d)$ is $A_0$-full whenever $(t, d)$ is not a priori small and regularity is satisfied. In view of Corollary 4.21 and Corollary 6.6, it suffices to show that $A^{\Sigma_t}$ and $\mathcal{B}_{\rho}(\mathbb{E})$ are fullness peers for some well chosen $(t, d)$-representation $\rho$. Let us point out an easy case when this is possible. If $\bar{p}$ has no conjugate self-twists, then $\mathbb{E} = \mathbb{F}$ by Corollary 4.13 and $\Sigma_t = 1$ by the diagram following Corollary 7.4. Furthermore, the assumption that $\tilde{\Sigma}_{\bar{p}} = 1$ implies that $\bar{p}$ is not dihedral and so $A = \mathcal{B}_{\rho}(\mathbb{F})$ by Theorem 2.23. Therefore we have

$$ \mathcal{B}_{\rho}(\mathbb{E}) = \mathcal{B}_{\rho}(\mathbb{F}) = A = A^{\Sigma_t}. $$

In general, the proof that $(t, d)$ is $A^{\Sigma_t}$-full, hence $A_0$-full, is structured as follows. The case when $\bar{p}$ is reducible is easily done in Proposition 9.1, so from Section 9.2 onwards we always assume that $\bar{p}$ is irreducible. In light of Corollary 6.6, the strategy is to prove that, under certain conditions on $\bar{p}$ and a good choice of a $(t, d)$-representation $\rho$, the two rings $\mathcal{B}_{\rho}(\mathbb{E})$ and $A^{\Sigma_t}$ have the same fields of fractions and $A^{\Sigma_t}$ is finitely generated as a $\mathcal{B}_{\rho}(\mathbb{E})$-module. This is done in Corollary 9.15, although key parts of it are proved in Corollary 9.9 and Proposition 9.14. Lemma 3.5 then implies that $A^{\Sigma_t}$ and $\mathcal{B}_{\rho}(\mathbb{E})$ are fullness peers. In Corollary 9.16, we combine Corollary 6.6, which established $\mathcal{B}_{\rho}(\mathbb{E})$-fullness, with Corollary 9.15 to show that $(t, d)$ is $A^{\Sigma_t}$-full under mild assumptions on $\bar{p}$. Since $A^{\Sigma_t}$ and $A_0$ are fullness peers in the constant-determinant setting, we conclude that our admissible pseudodeformation $(t, d)$ is $A_0$-full.

Let us now establish some assumptions on our fixed residual representation $\bar{p}: \Pi \to \mathrm{GL}_2(\mathbb{F})$. Assume that $\bar{p}$ is regular and, after Section 9.1, absolutely irreducible. Whenever $\bar{p}$ is absolutely irreducible, assume further that $\det \bar{p}$ is a power of 2, which can always be achieved by twisting $\bar{p}$ by a character by Lemma 8.4. Furthermore, the twisting operation does not change the field $\mathbb{E}$ by Proposition 4.8. Assume that $[\mathbb{F}: \mathbb{E}]$ is a power of 2, which is possible by Lemma 8.8. Note that we do not require $\mathbb{F}$ to be the trace ring of $\bar{p}$ since one may need to make a quadratic extension of the trace ring in order to make representations well-adapted in the dihedral case.

**9.1. The reducible case.** When $\bar{p}$ is reducible, we can use Corollary 6.6 to show that $(t, d)$ is $A^{\Sigma_t}$- and $A_0$-full.

**Proposition 9.1.** Suppose that $\bar{p} = \varepsilon \oplus \delta$ and that $\bar{p}$ is regular. If $(t, d)$ is not a priori small then $(t, d)$ is $A^{\Sigma_t}$- and $A_0$-full.

*Proof.* Let $(t', d') = (s(\delta^{-1})t, s(\delta^{-1})^2d)$, which is a pseudodeformation of $\bar{r} := \bar{p} \otimes \delta^{-1}$. Let $A_{t'}$ be the trace ring of $(t', d')$; its residue field $\mathbb{E}$ since $\bar{r}$ has no conjugate self-twists by Corollary 8.2. Then $(\Pi, \bar{r}, t', d')$ is an admissible pseudorepresentation over $A_{t'}$. Note that $(t', d')$ is not a priori small since $(t, d)$ is not. By Corollary 6.6, there is a well-adapted $(t', d')$-representation $r$ such that $(t', d')$ is $\mathcal{B}_r(\mathbb{E})$-full. But $A_{t'} = \mathcal{B}_r(\mathbb{E})$ by Theorem 2.23. Since $\bar{r}$ has no conjugate self-twists by Corollary 8.2, it follows that $\Sigma_{t'}$ is trivial. Thus $A_{t'}^{\Sigma_{t'}} = A_{t'} = \mathcal{B}_r(\mathbb{E})$.
By Corollary 3.12 it follows that $(t, d)$ is $A_{t'}^{\Sigma_{t'}}$-full. We know that $A^{\Sigma_t}$ and $A_{t'}^{\Sigma_{t'}}$ have the same fields of fractions, namely $K_0$, by Corollary 4.20. Furthermore $A$ is obtained by adjoining the values of $s(\delta)$ and $W(\mathbb{F})$ to $A_{t'}$. Therefore $A$, and hence $A^{\Sigma_t}$, is finitely generated over $A'$, so $(t, d)$ is $A^{\Sigma_{t'}}$-full by Lemma 3.5. Finally, $A_0$ and $A^{\Sigma_t}$ are fullness peers in this setting by Corollary 4.21. $\square$

**9.2. Choosing a good $(t, d)$-representation.** Throughout Section 9.2 through Section 9.4 we fix an absolutely irreducible regular representation $\bar{\rho}: \Pi \to \mathrm{GL}_2(\mathbb{F})$ such that the order of $\det \bar{\rho}$ is a power of 2. We assume that $[\mathbb{F}: \mathbb{E}]$ is a power of 2 by Lemma 8.8. If $\bar{\rho}$ is octahedral, we assume further that $\bar{\rho}$ is good. Furthermore, we fix a good basis for $\bar{\rho}$ as follows. If $\bar{\rho}$ is exceptional or large, choose a basis and a regular element $g_0 \in \Pi$ such that Proposition 8.13 holds. If $\bar{\rho} = \mathrm{Ind}_{\Pi_0}^{\Pi} \chi$ is dihedral, assume that $\bar{\rho}(\Pi_0)$ is diagonal and $\mathrm{Im} \bar{\rho}$ contains a matrix $\begin{pmatrix} 0 & b \\ c & 0 \end{pmatrix}$ such that $bc^{-1} \in \mathbb{F}_p$, which is possible by [Bel19, Proposition 6.3.2].

Recall that $(T, d): \Pi \to \mathcal{A}$ is the universal constant-determinant pseudorepresentation. Part of our arguments will require appealing to a universal $(T, d)$-representation. This requires choosing a good $(T, d)$-representation $\rho^{\mathrm{univ}}$ and also choosing our $(t, d)$-representation to be compatible with $\rho^{\mathrm{univ}}$. In particular, we want $I_1(\rho^{\mathrm{univ}})$ to be fixed by all conjugate self-twists of $(T, d)$. In Section 9.2 we make these choices and compatibilities precise. Since we need Proposition 9.2 and Corollary 9.3 for the universal ring $\mathcal{A}$, in Section 9.2 we do not require $\mathcal{A}$ to be a domain, only a local pro-$p$ ring.

Fix a generator $\sigma_1$ of $\Sigma_{\bar{\rho}} = \mathrm{Gal}(\mathbb{F}/\mathbb{E})$. We want to choose a character $\eta_1: \Pi \to \mathbb{F}^\times$ such that $(\sigma_1, \eta_1) \in \tilde{\Sigma}_{\bar{\rho}}$. There is a unique choice for $\eta_1$ when $\bar{\rho}$ is not dihedral. If $\bar{\rho}$ is dihedral and $\Sigma_t = 1$, choose $\eta_1$ to be the trivial character. Recall from the end of Section 7.2 that $\beta_t: \Sigma_t \to \Sigma_{\bar{\rho}}$ is given by reducing automorphisms of $\mathcal{A}$ modulo $\mathfrak{m}$. If $\bar{\rho}$ is dihedral and $\ker \beta_t = 1$ but $\Sigma_t \neq 1$, then there is a unique complement to $\Sigma_{\bar{\rho}}^\mathrm{di}$ in $\tilde{\Sigma}_{\bar{\rho}}$ that contains $\tilde{\beta}_t(\tilde{\Sigma}_t)$. Choose $\eta_1$ such that $(\sigma_1, \eta_1)$ generates that complement. Otherwise, when $\bar{\rho}$ is dihedral, we may take $\eta_1$ to be either of the two characters such that $(\sigma_1, \eta_1) \in \tilde{\Sigma}_{\bar{\rho}}$. Recall from (6) in Section 7.1 that $\Pi_0(\bar{\rho})$ is the intersection of the kernels of all twist characters of $\bar{\rho}$. Define

$$ \Pi_1 := \begin{cases} \ker \eta_1 & \text{if } \bar{\rho} \text{ is dihedral} \\ \Pi_0(\bar{\rho}) & \text{else.} \end{cases} $$

Let $A_1$ be the subring of $\mathcal{A}$ topologically generated by $t(\Pi_1)$. Note that $\bar{\rho}|_{\Pi_1}$ is absolutely irreducible by Lemma 8.5 and Corollary 8.7.

**Proposition 9.2.** There exists a well-adapted $(t, d)$-representation $\rho: \Pi \to \mathrm{GL}_2(A)$ such that $\rho|_{\Pi_1}$ takes values in $\mathrm{GL}_2(A_1)$ and such that $\rho$ is adapted to a regular element.

*Proof.* With the exception of the well-adaptedness statement, the proof is well known since $\bar{\rho}|_{\Pi_1}$ is absolutely irreducible. Indeed, a theorem of Rouquier [Rou96, Theorem 5.1] and Nyssen [Nys96] tells us that there are representations $\rho: \Pi \to \mathrm{GL}_2(A)$ and $\rho_1: \Pi_1 \to \mathrm{GL}_2(A_1)$ such that $\mathrm{tr} \rho = t$ and $\mathrm{tr} \rho_1 = t|_{\Pi_1}$. By a theorem of Carayol and Serre, $\rho|_{\Pi_1}$ and $\rho_1$ are conjugate by a matrix in $\mathrm{GL}_2(A)$ [Car94, Théorème 1].

For the well-adaptedness statement, let us first assume that $\bar{\rho}$ is not dihedral. Choose $\rho$ adapted to $g_0$ and $\rho_1$ adapted to $g_0^n$ with $g_0$ and $n$ as in Proposition 8.13. Then the matrix $M \in \mathrm{GL}_2(A)$ such that $M^{-1}\rho|_{\Pi_1}M = \rho_1$ commutes with $\rho(g_0^n) = \begin{pmatrix} s(\lambda_0^n) & 0 \\ 0 & s(\mu_0^n) \end{pmatrix} = \rho_1(g_0^n)$. Since $\lambda_0^n \neq \mu_0^n$ by Proposition 8.13, it follows that $M$ must be diagonal. In particular, $M$ commutes with $\rho(g_0)$. Hence $M^{-1}\rho M$ is still adapted to $g_0$ and satisfies the properties in the statement of the proposition.

The idea is similar when $\bar{\rho}$ is dihedral, except we can no longer assume that $\rho_1$ is adapted to the $g_0 \in \Pi_0$ such that $\bar{\rho}(g_0)$ generates the unique index-2 subgroup of the projective image of $\bar{\rho}$, because $g_0$ may not be in $\Pi_1$. Let $\rho$ be a well-adapted $(t, d)$-representation, say adapted to $g_0$ with $\rho(g_0) = \begin{pmatrix} s(\lambda_0) & 0 \\ 0 & s(\mu_0) \end{pmatrix}$.
Since $\bar{\rho}$ is regular, it follows that $\bar{\rho}|_{\Pi_0(\bar{\rho})}$ is multiplicity free over $\mathbb{E}$ by Proposition 8.6. Therefore, since $\rho$ is well adapted, the image of $\rho$ contains a matrix of the form $\begin{pmatrix} s^{(\lambda)} & 0 \\ 0 & s^{(\mu)} \end{pmatrix}$ with $\lambda \neq \mu$ and $\lambda, \mu \in \mathbb{E}^\times$. Let $h \in \Pi$ such that $\rho(h) = \begin{pmatrix} s^{(\lambda)} & 0 \\ 0 & s^{(\mu)} \end{pmatrix}$.

We claim that $h \in \Pi_1$. It suffices to prove that $h \in \Pi_0(\bar{\rho})$ since $\Pi_0(\bar{\rho}) = \Pi_0 \cap \ker \eta_1 \subset \ker \eta_1 = \Pi_1$. Note that $h \in \Pi_0$ as $\bar{\rho}(h)$ is diagonal. By Proposition 8.6, $h \in \Pi_0(\bar{\rho})$ if and only if the eigenvalues of $\bar{\rho}(h)$ are in $\mathbb{E}^\times$. But the eigenvalues $\lambda, \mu$ of $\bar{\rho}(h)$ were chosen to be in $\mathbb{E}^\times$. Therefore $h \in \Pi_1$.

By [Bel19, Proposition 2.4.2] we may assume that $\rho_1$ in the first paragraph of this proof is adapted to $h$. Therefore the matrix $M \in \text{GL}_2(A)$ such that $M^{-1}\rho|_{\Pi_1}M = \rho_1$ commutes with $\rho(h) = \begin{pmatrix} s^{(\lambda)} & 0 \\ 0 & s^{(\mu)} \end{pmatrix} = \rho_1(h)$. Since $\lambda \neq \mu$, it follows that $M$ is a diagonal matrix. Note that the second property in Definition 2.21 is unchanged by conjugation by a diagonal matrix. Therefore $M^{-1}\rho M$ is still well adapted and satisfies the statement of the proposition. $\square$

We recall that when $\bar{\rho}$ is dihedral, we may view elements in $B$ as elements of $A$ (Remark 2.12).

**Corollary 9.3.** There exists a well-adapted $(t, d)$-representation $\rho: \Pi \to \text{R}^\times$ such that $I_1(\rho) \subseteq A^{\Sigma_t}$. If $\bar{\rho}$ is dihedral and $\sigma \in \Sigma_t$ such that $\sigma$ and $\ker \beta_t$ generate $\Sigma_t$, then we may assume furthermore that $B_1(\rho)$ is pointwise fixed by $\sigma$.

*Proof.* Let $\rho$ be the $(t, d)$-representation from Proposition 9.2. Since the order of $\det \bar{\rho}$ is a power of 2, it follows that $[\Pi: \Pi_0(\bar{\rho})]$ is a power of 2. Since $\Gamma$ is pro-$p$ and $p \neq 2$, it follows that $\Gamma \subseteq \rho(\Pi_0(\bar{\rho})) \subseteq \text{GL}_2(A_1)$. Therefore $L_1(\rho) \subseteq \text{sl}_2(A_1)$, and so $I_1(\rho), B_1(\rho) \subseteq A_1$.

Let $(\sigma, \eta) \in \tilde{\Sigma}_t$ such that $\Pi_1 \subseteq \ker \eta$. Then for all $g \in \Pi_1$ we have

$$ \sigma_t(g) = \eta(g)t(g) = t(g), $$

and thus $A_1$ is contained in the subring of $A$ fixed by $\sigma$.

If $\ker \beta_t = 1$, then every $(\sigma, \eta) \in \tilde{\Sigma}_t$ satisfies $\Pi_1 \subseteq \ker \eta$ by definition of $\Pi_1$. Thus if $\ker \beta_t = 1$, then $A_1 \subseteq A^{\Sigma_t}$, and hence $I_1(\rho), B_1(\rho) \subseteq A^{\Sigma_t}$.

Now suppose that $\bar{\rho}$ is dihedral and $\ker \beta_t \neq 1$. Then half of the elements $(\sigma, \eta) \in \tilde{\Sigma}_t$ satisfy $\ker \eta \subseteq \ker \eta_1 = \Pi_1$, namely all those in the preimage under $\beta_t$ of the subgroup generated by $(\sigma_1, \eta_1)$ in $\tilde{\Sigma}_{\bar{\rho}}$. This proves the statement about $B_1(\rho)$ in the dihedral case. To see that $I_1(\rho)$ is fixed by all conjugate self-twists, it remains to show that $I_1(\rho)$ is fixed by the nontrivial element in $\ker \beta_t$. This follows from Proposition 7.5. $\square$

In light of Corollary 9.3, let us fix a well-adapted $(T, d)$-representation $\rho^{\text{univ}}: \Pi \to \text{GL}_2(\mathcal{A})$ such that $I_1(\rho^{\text{univ}}) \subseteq \mathcal{A}^{\Sigma_T}$. Assume furthermore in the case when the projective image of $\bar{\rho}$ is not dihedral that we have conjugated $\rho^{\text{univ}}$ by the relevant diagonal element so that Theorem 2.23 applies to $\rho^{\text{univ}}$, and thus to any quotient of $\rho^{\text{univ}}$. Recall from the commutative diagram following Corollary 7.4 that we have a natural reduction map $\beta_T: \Sigma_T \to \Sigma_{\bar{\rho}}$ that sends an automorphism $\sigma$ of $\mathcal{A}$ to the automorphism it induces on $\mathbb{F} = \mathcal{A}/\mathfrak{m}_{\mathcal{A}}$. In the case when $\bar{\rho}$ is dihedral, we need to choose a complement to $\ker \beta_T$ in $\Sigma_T$, whose generator we will denote by $\nu$. We choose $\nu$ such that $(\nu, \eta_1) \in \tilde{\Sigma}_T$, where $\eta_1$ is the character fixed prior to Proposition 9.2. By Corollary 9.3, we may and do assume that $B_1(\rho^{\text{univ}})$ is fixed by $\nu$.

The universal property of $(\mathcal{A}, (T, d))$ gives a surjective $W(\mathbb{F})$-algebra homomorphism $\alpha_t: \mathcal{A} \to \mathcal{A}$. Let $\rho_t := \alpha_t \circ \rho^{\text{univ}}: \Pi \to \text{GL}_2(A)$. It is a $(t, d)$-representation such that $I_1(\rho_t) \subseteq A^{\Sigma_t}$ by the diagram following Corollary 7.4. Furthermore, if $\bar{\rho}$ is dihedral and $\ker \beta_t = 1$, then $B_1(\rho_t) \subseteq A^{\Sigma_t}$ as well. By the functoriality of Pink-Lie algebras with respect to quotient maps, we have that $\alpha_t(I_1(\rho^{\text{univ}})) = I_1(\rho_t)$ and $\alpha_t(B_1(\rho^{\text{univ}})) = B_1(\rho_t)$. All of our theorems below will be specifically for this well-chosen representation $\rho_t$. To ease notation, write $\rho = \rho_t$.
Recall that by Theorem 2.23

$$A = \begin{cases} \mathcal{B}_{\rho}(\mathbb{F}) + W(\mathbb{F})B_{1}(\rho) & \text{if } \bar{\rho} \text{ is dihedral} \\ \mathcal{B}_{\rho}(\mathbb{F}) & \text{else.} \end{cases}$$

By Proposition 7.5 and the fact that $B_1(\rho) \subseteq A^{\Sigma_t}$ if $\bar{\rho}$ is dihedral and $\ker \beta_t = 1$, it follows that

$$A^{\Sigma_t} = \begin{cases} \mathcal{B}_{\rho}(\mathbb{F}^{\beta_t(\Sigma_t)}) + W(\mathbb{F}^{\beta_t(\Sigma_t)})B_1(\rho) & \text{if } \bar{\rho} \text{ dihedral and } \ker \beta_t = 1 \\ \mathcal{B}_{\rho}(\mathbb{F}^{\beta_t(\Sigma_t)}) & \text{else.} \end{cases}$$

We therefore define

$$J = J(\rho) := \begin{cases} W(\mathbb{E})I_1(\rho) + W(\mathbb{E})I_1(\rho)^2 + W(\mathbb{E})B_1(\rho) & \text{if } \bar{\rho} \text{ is dihedral and } \ker \beta_t = 1 \\ W(\mathbb{E})I_1(\rho) + W(\mathbb{E})I_1(\rho)^2 & \text{else.} \end{cases}$$

We claim that $J \subset \mathfrak{m}$ is a multiplicatively closed $W(\mathbb{E})$-module by Theorem 2.23. The key is to note that, since $\bar{\rho}$ is regular and $\rho$ is well adapted, it follows that Bellaïche's field $\mathbb{F}_q$ from p. 16 is contained in $\mathbb{E}$. Therefore it follows from Theorem 2.23 that $(W(\mathbb{E})I_1(\rho))^3 \subseteq W(\mathbb{E})I_1(\rho)$ and $W(\mathbb{E})I_1(\rho)B_1(\rho) \subseteq W(\mathbb{E})B_1(\rho)$ and $(W(\mathbb{E})B_1(\rho))^2 \subseteq W(\mathbb{E})I_1(\rho)$, which proves that $J$ is multiplicatively closed. Define

$$\mathfrak{A} := W(\mathbb{F}) + W(\mathbb{F})J.$$

We have $\mathfrak{A} = A$ unless $1 \neq \ker \beta_t$, in which case $\mathfrak{A} = A^+$ by Proposition 7.5.

Remark 9.4. The rings $W(\mathbb{E}) + J$ and $A^{\Sigma_t}$ differ only in their constants, $W(\mathbb{E})$ versus $W(\mathbb{F}^{\beta_t(\Sigma_t)})$. Furthermore, $W(\mathbb{E}) + J$ is often equal to $\mathcal{B}_{\rho}(\mathbb{E})$, and the goal of this section is to relate $\mathcal{B}_{\rho}(\mathbb{E})$ with $A^{\Sigma_t}$. Assume for a moment that $W(\mathbb{E}) + J = \mathcal{B}_{\rho}(\mathbb{E})$. Then the difference between $\mathcal{B}_{\rho}(\mathbb{E})$ and $A^{\Sigma_t}$ is entirely governed by understanding which elements of $\Sigma_{\bar{\rho}}$ lift to elements in $\Sigma_t$ under $\beta_t$. In particular, when there are elements in $\Sigma_{\bar{\rho}}$ that do not lift to $\Sigma_t$, we will be interested in writing the extra elements in $W(\mathbb{F}^{\beta_t(\Sigma_t)})$ as quotients of elements in $J$ to show that $Q(\mathcal{B}_{\rho}(\mathbb{E})) = Q(A^{\Sigma_t})$. $\triangle$

9.3. **Lifting conjugate self-twists to $\mathfrak{A}$**. In Section 9.3 we study a condition on $J$, called smallness (Definition 9.6), that dictates which conjugate self-twists of $\bar{\rho}$ lift to conjugate self-twists of $(t, d)$. This study culminates in Theorem 9.8. As a consequence, we prove in Corollary 9.9 that under such a smallness assumption, $A^{\Sigma_t} = \mathcal{B}_{\rho}(\mathbb{E})$. The reader is advised that, with the exception of the motivational remark following Definition 9.6, the assumption that $A$ is a domain is never used in Section 9.3.

Throughout Section 9.3, fix a subgroup $\Sigma \subseteq \Sigma_{\bar{\rho}}$, and let $\mathbb{F}' := \mathbb{F}^{\Sigma}$. Write $W := W(\mathbb{F})$ and $W' := W(\mathbb{F}')$. For an arbitrary ring $\mathcal{R}$ and a finite group $X$ of ring automorphisms of $\mathcal{R}$, for any $\varphi \in \operatorname{Hom}(X, \mathcal{R}^{\times})$, we write

$$\mathcal{R}^{\varphi} := \{s \in \mathcal{R} : \sigma s = \varphi(\sigma)s, \forall \sigma \in X\}.$$

As explained at the beginning of Section 9, we assume that $[\mathbb{F}:\mathbb{E}]$ is a power of 2. By Lemma 8.9 we may apply Lemma A.22 to conclude that $\mathbb{F} = \oplus_{\varphi \in \Sigma^*} \mathbb{F}^{\varphi}$, where $\Sigma^* := \operatorname{Hom}(\Sigma, \mathbb{F}^{\times})$. Note that since $\Sigma = \operatorname{Gal}(W/W')$, it follows that this decomposition lifts to $W$. More precisely, viewing elements of $\Sigma$ as automorphisms of $W$ and elements of $\Sigma^*$ as valued in $W^{\times}$ by composing with the Teichmüller map, we can define $W^{\varphi}$ for each $\varphi \in \Sigma^*$. Then Lemma A.22 gives $W = \oplus_{\varphi \in \Sigma^*} W^{\varphi}$.

For all $\varphi \in \Sigma$, define

$$\mathfrak{A}(\varphi) := W^{\varphi} + W^{\varphi}J,$$

where $W^{\varphi}J := \{\sum_i \alpha_i j_i | \alpha_i \in W^{\varphi}, j_i \in J\}$. Since $\mathfrak{A} = W + WJ$ it follows immediately that $\mathfrak{A} = \sum_{\varphi \in \Sigma^*} \mathfrak{A}(\varphi)$. We will be interested in understanding when this sum is direct, because in that case we will show that it is possible to find lifts of elements of $\Sigma$ in $\Sigma_t$. If $\mathfrak{a}$ is an ideal of $\mathfrak{A}$ and $\varphi \in \Sigma^*$, let $\mathfrak{a}(\varphi) := \mathfrak{A}(\varphi) \cap \mathfrak{a}$ and let $(\mathfrak{A}/\mathfrak{a})(\varphi) \subseteq \mathfrak{A}/\mathfrak{a}$ be the image of $\mathfrak{A}(\varphi)$ under the natural projection $\mathfrak{A} \to \mathfrak{A}/\mathfrak{a}$.
**Lemma 9.5.** The following are equivalent:

(1) $\mathfrak{A} = \oplus_{\varphi \in \Sigma^*} \mathfrak{A}(\varphi)$.

(2) For every $\mathfrak{A}$-ideal $\mathfrak{a}$ such that $\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} \mathfrak{a}(\varphi)$, we have $\mathfrak{A}/\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} (\mathfrak{A}/\mathfrak{a})(\varphi)$. Furthermore, there exists at least one such ideal $\mathfrak{a}$.

(3) There exists an $\mathfrak{A}$-ideal $\mathfrak{a}$ such that $\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} \mathfrak{a}(\varphi)$ and $\mathfrak{A}/\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} (\mathfrak{A}/\mathfrak{a})(\varphi)$.

*Proof.* First we show that (1) implies (2). We can take $\mathfrak{a} = 0$ for the existence statement in (2). Now suppose that $\mathfrak{a}$ is an $\mathfrak{A}$-ideal such that $\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} \mathfrak{a}(\varphi)$. If $\sum_{\varphi \in \Sigma^*} \bar{a}_{\varphi} = 0 \in \mathfrak{A}/\mathfrak{a}$ with each $\bar{a}_{\varphi} \in (\mathfrak{A}/\mathfrak{a})(\varphi)$, then letting $a_{\varphi} \in \mathfrak{A}(\varphi)$ be a lift of $\bar{a}_{\varphi}$, we see that $\sum_{\varphi \in \Sigma^*} a_{\varphi} \in \mathfrak{a} = \oplus_{\varphi \in \Sigma^*} \mathfrak{a}(\varphi)$. Thus, there are $\alpha_{\varphi} \in \mathfrak{a}(\varphi)$ such that $\sum_{\varphi \in \Sigma^*} a_{\varphi} = \sum_{\varphi \in \Sigma^*} \alpha_{\varphi}$. Since $\mathfrak{A} = \oplus_{\varphi \in \Sigma^*} \mathfrak{A}(\varphi)$, it follows that $a_{\varphi} = \alpha_{\varphi}$ for all $\varphi \in \Sigma^*$. Thus $\bar{a}_{\varphi} = 0 \in \mathfrak{A}/\mathfrak{a}$ for all $\varphi \in \Sigma^*$ and hence $\mathfrak{A}/\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} (\mathfrak{A}/\mathfrak{a})(\varphi)$.

The fact that (2) implies (3) is trivial.

To see that (3) implies (1), suppose that $\mathfrak{a}$ is an $\mathfrak{A}$-ideal such that $\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} \mathfrak{a}(\varphi)$ and $\mathfrak{A}/\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} (\mathfrak{A}/\mathfrak{a})(\varphi)$. For each $\varphi \in \Sigma^*$ fix a set $S_{\varphi} \subset \mathfrak{A}(\varphi)$ of representatives of $(\mathfrak{A}/\mathfrak{a})(\varphi)$ such that $0 \in S_{\varphi}$. Suppose that $\sum_{\varphi \in \Sigma^*} a_{\varphi} = 0$ with $a_{\varphi} \in \mathfrak{A}(\varphi)$. Then there is a unique way to write each $a_{\varphi}$ as

$$a_{\varphi} = s_{\varphi} + \alpha_{\varphi}$$

with $s_{\varphi} \in S_{\varphi}$ and $\alpha_{\varphi} \in \mathfrak{a}(\varphi)$. Modulo $\mathfrak{a}$, we see that

$$\sum_{\varphi \in \Sigma^*} \bar{s}_{\varphi} = 0.$$

Since $\mathfrak{A}/\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} (\mathfrak{A}/\mathfrak{a})(\varphi)$, it follows that $\bar{s}_{\varphi} = 0$ for all $\varphi$. As $0 \in S_{\varphi}$, it follows that $s_{\varphi} = 0$ for all $\varphi \in \Sigma^*$. Therefore $a_{\varphi} = \alpha_{\varphi} \in \mathfrak{a}(\varphi)$. As $\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} \mathfrak{a}(\varphi)$, it follows that each $a_{\varphi} = 0$. Thus $\mathfrak{A} = \oplus_{\varphi \in \Sigma^*} \mathfrak{A}(\varphi)$. $\square$

**Definition 9.6.** Let $\mathbb{L}_2 \subset \mathbb{L}_1$ be subfields of $\mathbb{F}$. We say that $J$ is small with respect to $\mathbb{L}_1/\mathbb{L}_2$ if

$$\ker(W(\mathbb{L}_1) \otimes_{W(\mathbb{L}_2)} W(\mathbb{L}_2)J \to W(\mathbb{L}_1)J) = 0,$$

where the map is multiplication inside $\mathfrak{A}$. Otherwise, we say that $J$ is big with respect to $\mathbb{L}_1/\mathbb{L}_2$.

To motivate Definition 9.6, recall from Remark 9.4 that we need to be able to write elements of $W(\mathbb{F}^{\beta_t(\Sigma_t)})$ as quotients of elements in $J$ whenever $\mathbb{F}^{\beta_t(\Sigma_t)} \neq \mathbb{E}$. Suppose that $\mathbb{L}_2 = \mathbb{E}$, $\mathbb{L}_1 = \mathbb{F}^{\beta_t(\Sigma_t)}$, and $[\mathbb{L}_1 : \mathbb{L}_2] = 2$. Write $\mathbb{L}_1 = \mathbb{L}_2(\alpha)$. Then $W(\mathbb{L}_1) = W(\mathbb{L}_2) \oplus s(\alpha)W(\mathbb{L}_2)$ and so

$$W(\mathbb{L}_1) \otimes_{W(\mathbb{L}_2)} W(\mathbb{L}_2)J = W(\mathbb{L}_2)J \oplus (s(\alpha)W(\mathbb{L}_2) \otimes_{W(\mathbb{L}_2)} W(\mathbb{L}_2)J).$$

If $J$ is big with respect to $\mathbb{L}_1/\mathbb{L}_2$, then we can find $x, y \in W(\mathbb{L}_2)J \setminus \{0\}$ such that $x + s(\alpha)y = 0$. Thus $s(\alpha) = x/y$, and hence $W(\mathbb{L}_1)$ is in the field of fractions of any domain containing $W(\mathbb{L}_2)J$. In contrast, the following proposition shows that when $J$ is small with respect to $\mathbb{F}/\mathbb{F}'$, elements of $\Sigma$ can be lifted to automorphisms of $\mathfrak{A}$.

**Proposition 9.7.** If $J$ is small with respect to $\mathbb{F}/\mathbb{F}'$, then $\mathfrak{A} = \oplus_{\varphi \in \Sigma^*} \mathfrak{A}(\varphi)$. In this case, every $\bar{\sigma} \in \Sigma$ can be lifted to an automorphism $\sigma$ of $\mathfrak{A}$ such that $\sigma$ acts trivially on $J$, and a lift with this property is unique.

*Proof.* Note that $\mathfrak{a} := WJ$ is an $\mathfrak{A}$-ideal since $\mathfrak{A} = W + WJ$ and $J$ is multiplicatively closed as discussed prior to Remark 9.4. The assumption that $J$ is small with respect to $\mathbb{F}/\mathbb{F}'$ implies that $WJ = \oplus_{\varphi \in \Sigma^*} W^{\varphi}J$. Indeed,

$$\bigoplus_{\varphi \in \Sigma^*} (W^{\varphi} \otimes_{W'} W'J) = \left( \bigoplus_{\varphi \in \Sigma^*} W^{\varphi} \right) \otimes_{W'} W'J = W \otimes_{W'} W'J \hookrightarrow WJ,$$

and the image of $W^{\varphi} \otimes_{W'} W'J$ is exactly $W^{\varphi}J$. Since $WJ = \sum_{\varphi \in \Sigma^*} W^{\varphi}J$, it follows that the multiplication map is an isomorphism and thus $WJ$ is graded by $\Sigma^*$. Note that $\mathfrak{a}(\varphi) = W^{\varphi}J$, so $\mathfrak{a} = \oplus_{\varphi \in \Sigma^*} \mathfrak{a}(\varphi)$.
By Lemma 9.5, for the first statement of the proposition it suffices to show that $\mathfrak{A}/WJ = \oplus_{\varphi \in \Sigma^*} (\mathfrak{A}/WJ)(\varphi)$. Note that

$$ \mathfrak{A}/WJ = (W + WJ)/WJ \cong W/(W \cap WJ) $$

and $W \cap WJ$ is a closed $W$-submodule of $pW$ since $J \subseteq \mathfrak{m}_\mathfrak{A}$. Thus we have $W \cap WJ = p^nW$ and $\mathfrak{A}/WJ \cong W/p^nW$ for some $1 \le n \le \infty$, where $p^\infty W := \{\infty\}$. Since $W$ is graded by $\Sigma^*$, it follows from Lemma 9.5 that $W/p^nW$ is graded by $\Sigma^*$ as well. Therefore $\mathfrak{A} = \oplus_{\varphi \in \Sigma^*} \mathfrak{A}(\varphi)$.

For the second statement, we let $\sigma$ act by $W(\bar{\sigma})$ on $W$ and trivially on $J$. The only question is to verify that this is well defined. Since $\mathfrak{A} = \oplus_{\varphi \in \Sigma^*} \mathfrak{A}(\varphi)$, it suffices to show that $\sigma$ is well defined on each $\mathfrak{A}(\varphi)$. That is, we must show

$$ \sum_{i=1}^{n} \alpha_i j_i = 0 \implies \sum_{i=1}^{n} W(\bar{\sigma})(\alpha_i) j_i = 0, $$

where $\alpha_i \in W^\varphi$, $j_i \in J$. Since $J$ is small with respect to $\mathbb{F}/\mathbb{F}'$, $\sum_{i=1}^n \alpha_i j_i = 0$ implies that $\sum_{i=1}^n \alpha_i \otimes j_i = 0 \in W \otimes_{W'} W'J$. Since $\alpha_i \in W^\varphi$, we know that $W(\bar{\sigma})(\alpha_i) = s(\varphi(\bar{\sigma}))\alpha_i$ for all $i$. Hence

$$ 0 = \sum_{i=1}^{n} \alpha_i \otimes j_i \implies 0 = s(\varphi(\bar{\sigma})) \sum_{i=1}^{n} \alpha_i \otimes j_i. $$

Therefore $\sum_{i=1}^n W(\bar{\sigma})(\alpha_i)j_i = 0$, since it is the image of $s(\varphi(\bar{\sigma})) \sum_{i=1}^n \alpha_i \otimes j_i$ under $W \otimes_{W'} W'J \to WJ$. $\square$

Now that we have lifted elements of $\Sigma$ to automorphisms of $\mathfrak{A}$ under the smallness assumption, we would like to verify that the lifts are conjugate self-twists of $(t, d)$ when $\mathfrak{A} = A$ and that they come from conjugate self-twists when $\mathfrak{A} = A^+$. (Recall that $A^+$ is only defined when $\ker \beta_t$ is nontrivial; see Proposition 7.5.)

**Theorem 9.8.** If $J$ is small with respect to $\mathbb{F}/\mathbb{F}'$ then $\Sigma$ is contained in the image of $\beta_t: \Sigma_t \to \Sigma_{\bar{\rho}}$. Furthermore, every lift of $\bar{\sigma} \in \Sigma$ to $\Sigma_t$ acts trivially on $J$.

*Proof.* Fix $\bar{\sigma} \in \Sigma$. By Proposition 9.7, there is a unique $\sigma \in \mathrm{Aut}\,\mathfrak{A}$ that acts as $W(\bar{\sigma})$ on $W$ and fixes $J$. If $\ker \beta_t = 1$, then $\mathfrak{A} = A$. If $\ker \beta_t \ne 1$, then $\mathfrak{A} = A^+$ and we need to extend $\sigma$ to $A = A^+ \oplus A^-$. We do this by declaring that $\sigma$ fixes $A^-$; we will still denote the automorphism of $A$ by $\sigma$. We already know that $\sigma$ acts trivially on $J$, so it is enough to prove that $\sigma \in \Sigma_t$.

Our strategy is to show that $\sigma$ comes from an appropriate element of $\Sigma_T$. More precisely, we claim that there is some $(\tilde{\sigma}, \eta) \in \tilde{\Sigma}_T$ such that $\sigma \circ \alpha_t = \alpha_t \circ \tilde{\sigma}$, where $\alpha_t: A \to A$ is the $W$-algebra homomorphism given by universality. If this is true, then for all $g \in \Pi$ we have

$$ \sigma_t(g) = \sigma \circ \alpha_t(T(g)) = \alpha_t \circ \tilde{\sigma}(T(g)) = \alpha_t(\eta(g)T(g)) = \eta(g)\alpha_t(T(g)) = \eta(g)t(g) $$

since $\alpha_t$ is a $W$-algebra homomorphism and $\eta(g) \in W$. Thus $\sigma \in \Sigma_t$.

First suppose that $\bar{\rho}$ is not dihedral. Then there is a unique lift $\tilde{\sigma}$ of $\bar{\sigma}$ in $\Sigma_T$ by Lemma 7.1 and Proposition 7.3. By Proposition 7.3, we know that $\tilde{\sigma}$ acts as $W(\bar{\sigma})$ on the image of $W$ in $\mathfrak{A}$. Furthermore, $\tilde{\sigma}$ acts trivially on $I_1(\rho^{\mathrm{univ}})$ by our fixed choice of $\rho^{\mathrm{univ}}$ after Corollary 9.3. Since $\bar{\rho}$ is not dihedral, it follows that $\mathfrak{A} = W + WJ(\rho^{\mathrm{univ}})$. By the construction of $\rho^{\mathrm{univ}}$ and $\rho_t$, we have $\alpha_t(I_1(\rho^{\mathrm{univ}})) = I_1(\rho)$ and thus $\alpha_t(J(\rho^{\mathrm{univ}})) = J(\rho) = J$. Recall that $\sigma$ acts trivially on $J$. Both $\tilde{\sigma}$ and $\sigma$ act on $W$ by $W(\bar{\sigma})$. Thus for any $\sum_{i=1}^n a_i x_i$ with $a_i \in W, x_i \in J(\rho^{\mathrm{univ}}) \cup \{1\}$, we have

$$ \sigma \circ \alpha_t \left( \sum_{i=1}^{n} a_i x_i \right) = \sum_{i=1}^{n} W(\bar{\sigma})(a_i) \alpha_t(x_i) = \alpha_t \circ \tilde{\sigma} \left( \sum_{i=1}^{n} a_i x_i \right). $$

If $\bar{\rho}$ is dihedral, then there are two lifts of $\bar{\sigma}$ in $\Sigma_T$ by Lemma 7.1 and Proposition 7.3. One acts on $\mathfrak{A}^-$ by +1 and the other acts by -1 by Proposition 7.5 and since we chose $\rho^{\mathrm{univ}}$ such that $B_1(\rho^{\mathrm{univ}})$ is fixed by $\nu$, which generates a complement of $\ker \beta_T$. Let $\tilde{\sigma} \in \Sigma_T$ be the lift of $\bar{\sigma}$ that
is in $\langle \nu \rangle$. Thus $\tilde{\sigma}$ acts trivially on $J(\rho^{\text{univ}})$ and $B_1(\rho^{\text{univ}})$. Then an argument similar to that in the previous paragraph shows that $\sigma \circ \alpha_t = \alpha_t \circ \tilde{\sigma}$. $\square$

**Corollary 9.9.** If $J$ is small with respect to $\mathbb{F}/\mathbb{E}$ then $A^{\Sigma_t} = W(\mathbb{E}) + J$. Suppose furthermore that either $\bar{\rho}$ is not dihedral or $\ker \beta_t \neq 1$. Then $A^{\Sigma_t} = \mathcal{B}_{\rho}(\mathbb{E})$.

*Proof.* By Theorem 9.8 applied to $\Sigma = \Sigma_t$, the map $\beta_t$ is a surjection and $\Sigma_t$ acts trivially on $J$. If $\ker \beta_t = 1$, then $A = \mathfrak{A} = W + WJ$, so $A^{\Sigma_t} = W(\mathbb{E}) + J$.

If $\ker \beta_t \neq 1$, then $A = W + W I_1(\rho) + W I_1(\rho)^2 + W B_1(\rho)$ and $J = W(\mathbb{E})I_1(\rho) + W(\mathbb{E})I_1(\rho)^2$. Note that $A^{\Sigma_t} \subseteq A^+ = W + WJ$ since the nontrivial element in $\ker \beta_t$ acts by $-1$ on $B_1(\rho)$ by Proposition 7.5. As above, we have that

$$A^{\Sigma_t} = (W + WJ)^{\Sigma_t} = W(\mathbb{E}) + J.$$

The last sentence in the statement of the corollary follows from the definition of $J$. $\square$

*Remark 9.10.* Note that none of the arguments in Section 9.3 require that $A$ is a domain. In particular, when $J$ is small with respect to $\mathbb{F}/\mathbb{E}$ and either $\bar{\rho}$ is not dihedral or $\ker \beta_t \neq 1$, Corollary 9.9 gives a conceptual interpretation of the ring $\mathcal{B}_{\rho}(\mathbb{E})$. $\triangle$

**9.4. When $J$ is big with respect to $\mathbb{F}/\mathbb{E}$.** Corollary 9.9 requires the assumption that $J$ is small with respect to $\mathbb{F}/\mathbb{E}$. We do not always expect this to be true. The purpose of Section 9.4 is to show that $A^{\Sigma_t}$ and $\mathcal{B}_{\rho}(\mathbb{E})$ have the same fraction field and $A^{\Sigma_t}$ is a finite type $\mathcal{B}_{\rho}(\mathbb{E})$-module even without the smallness assumption. This is done in Corollary 9.15, although the two key inputs to that theorem are Proposition 9.13 and Proposition 9.14. Then we can apply Lemma 3.5 and Corollary 6.6 to conclude that $\rho_t$, and thus $(t, d)$, is $A^{\Sigma_t}$-full in Corollary 9.16.

The discussion following Definition 9.6 shows why one may expect to get $Q(A^{\Sigma_t}) = Q(\mathcal{B}_{\rho}(\mathbb{E}))$ when smallness fails and $[\mathbb{F}^{\beta_t(\Sigma_t)} : \mathbb{E}] = 2$. Unfortunately, the assumption that $[\mathbb{F}^{\beta_t(\Sigma_t)} : \mathbb{E}] = 2$ is rather critical to that argument. This is the primary reason we insist that $[\mathbb{F} : \mathbb{E}]$ be a power of 2 throughout this section. It allows us to split up the extension $\mathbb{F}^{\beta_t(\Sigma_t)}/\mathbb{E}$ into a series of quadratic extensions, and thus we can apply the argument following Definition 9.6 inductively. This is the essential idea of the argument; we now prepare some notation to formalize it.

Write $[\mathbb{F}^{\beta_t(\Sigma_t)} : \mathbb{E}] = 2^n$ for some $n \geq 0$. For integers $0 \leq i \leq n$, let $\mathbb{E}_i$ be the unique extension of $\mathbb{E}$ of degree $2^i$. In particular, $\mathbb{E}_0 = \mathbb{E}$ and $\mathbb{E}_n = \mathbb{F}^{\beta_t(\Sigma_t)}$, and $[\mathbb{E}_i : \mathbb{E}_{i-1}] = 2$ for all $1 \leq i \leq n$. For $0 \leq i \leq n$, let $W_i$ denote the image of $W(\mathbb{E}_i)$ in $\mathfrak{A}$. Define

$$\mathfrak{A}_i := W_i + W_i J \subseteq \mathfrak{A}.$$

In particular, $\mathfrak{A}_0 = W(\mathbb{E}) + J$ and $\mathfrak{A}_n = W(\mathbb{F}^{\beta_t(\Sigma_t)}) + W(\mathbb{F}^{\beta_t(\Sigma_t)})J$. Since $A$ is a domain, so are all of the $\mathfrak{A}_i$, and we write $Q(\mathfrak{A}_i)$ for the field of fractions of $\mathfrak{A}_i$.

In the case when $\mathfrak{A} = A^+$, there is a 2-to-1 group homomorphism $\Sigma_t \to \text{Aut } A^+$ given by restricting elements of $\Sigma_t$ to $A^+$. Let $\Sigma_t(\mathfrak{A})$ denote the image of this map when $\mathfrak{A} = A^+$, and otherwise (that is, whenever $\ker \beta_t = 1$) let $\Sigma_t(\mathfrak{A}) = \Sigma_t$. In either case we can identify $\Sigma_t(\mathfrak{A})$ with a subgroup of $\Sigma_{\bar{\rho}}$ via $\beta_t$, and we have $\mathbb{E}_n = \mathbb{F}^{\beta_t(\Sigma_t(\mathfrak{A}))}$. We write $\Sigma_t(\mathfrak{A})^* := \text{Hom}(\Sigma_t(\mathfrak{A}), \mathfrak{A}^*)$.

We begin with two preliminary lemmas about the relationship between smallness and the $\mathbb{E}_i$.

**Lemma 9.11.** We have that $J$ is small with respect to $\mathbb{F}/\mathbb{E}_n$; that is, $\ker(W \otimes_{W_n} W_n J \to WJ) = 0$.

*Proof.* Recall that $WJ$ is an $\mathfrak{A}$-ideal that is stable under the action of $\Sigma_t(\mathfrak{A})$ since $\Sigma_t$ fixes $J$ by the construction of $\rho_t$. By Lemma 8.9, we can apply Lemma A.22 with $X = \Sigma_t(\mathfrak{A})$. Therefore

$$WJ = \bigoplus_{\varphi \in \Sigma_t(\mathfrak{A})^*} (WJ)^{\varphi},$$

where $(WJ)^{\varphi} := \{x \in WJ : \sigma x = \varphi(\sigma)x, \forall \sigma \in \Sigma_t(\mathfrak{A})\}$. Recall that $W^{\varphi}J$ was defined prior to Lemma 9.5. We claim that

$$(WJ)^{\varphi} = W^{\varphi}J.$$
Clearly $(WJ)^\varphi \supseteq W^\varphi J$ since $\Sigma_t(\mathfrak{A})$ acts trivially on $J$. On the other hand,

$$WJ = \bigoplus_{\varphi \in \Sigma_t(\mathfrak{A})^*} (WJ)^\varphi = \sum_{\varphi \in \Sigma_t(\mathfrak{A})^*} (WJ)^\varphi \supseteq \sum_{\varphi \in \Sigma_t(\mathfrak{A})^*} W^\varphi J = WJ,$$

so we must have equality.

For each $\varphi \in \Sigma_t(\mathfrak{A})^*$, choose $x_\varphi \in \mathbb{F}^\varphi \setminus \{0\}$. Then $\{s(x_\varphi) : \varphi \in \Sigma_t(\mathfrak{A})^*\}$ is a $W_n$-basis for $W$. Thus we have

$$W \otimes_{W_n} W_n J = \bigoplus_{\varphi \in \Sigma_t(\mathfrak{A})^*} W_n s(x_\varphi) \otimes_{W_n} W_n J.$$

If $x \in \ker(W \otimes_{W_n} W_n J \to WJ)$, then we can write

$$x = \sum_{\varphi \in \Sigma_t(\mathfrak{A})^*} s(x_\varphi) \otimes y_\varphi$$

for some $y_\varphi \in W_n J$. Then we have

$$0 = \sum_{\varphi \in \Sigma_t(\mathfrak{A})^*} s(x_\varphi) y_\varphi$$

and $s(x_\varphi)y_\varphi \in W^\varphi J$. Since $WJ = \bigoplus_{\varphi \in \Sigma_t(\mathfrak{A})^*} W^\varphi J$, it follows that each $s(x_\varphi)y_\varphi = 0$. As $\mathfrak{A}$ is a domain and $s(x_\varphi) \neq 0$, it follows that $y_\varphi = 0$ for all $\varphi \in \Sigma_t(\mathfrak{A})^*$. $\square$

**Lemma 9.12.** We have

$$\ker(W \otimes_{W_{n-1}} W_{n-1} J \to WJ) = W \otimes_{W_n} \ker(W_n \otimes_{W_{n-1}} W_{n-1} J \to W_n J).$$

*Proof.* Let $K := \ker(W_n \otimes_{W_{n-1}} W_{n-1} J \to W_n J)$. We have an exact sequence of $W_n$-modules

$$0 \to K \to W_n \otimes_{W_{n-1}} W_{n-1} J \to W_n J \to 0.$$

Since $W$ is free over $W_n$, tensoring with $W$ over $W_n$ gives an exact sequence

$$0 \to W \otimes_{W_n} K \to W \otimes_{W_{n-1}} W_{n-1} J \to W \otimes_{W_n} W_n J \to 0.$$

We can identify the last nonzero term in this sequence with $WJ$ by Lemma 9.11.

Thus $W \otimes_{W_n} K = \ker(W \otimes_{W_{n-1}} W_{n-1} J \to WJ)$.

<figure><img src="image_5.png" /></figure>

**Proposition 9.13.** If $J$ is big with respect to $\mathbb{F}/\mathbb{E}$ then $Q(\mathfrak{A}_n) = Q(\mathfrak{A}_{n-1})$.

*Proof.* We claim that $J$ is big with respect to $\mathbb{E}_n/\mathbb{E}_{n-1}$. Indeed, if $J$ were small with respect to $\mathbb{E}_n/\mathbb{E}_{n-1}$, then $J$ would be small with respect to $\mathbb{F}/\mathbb{E}_{n-1}$ by Lemma 9.12. Therefore we could apply Theorem 9.8 with $\Sigma = \text{Gal}(\mathbb{F}/\mathbb{E}_{n-1})$, which implies that $\mathbb{E}_n \subseteq \mathbb{E}_{n-1}$, a contradiction.

Let $\{1, \alpha\}$ be an $\mathbb{E}_{n-1}$-basis for $\mathbb{E}_n$. Then $\{1, s(\alpha)\}$ is a $W_{n-1}$-basis for $W_n$ and so

$$W_n \otimes_{W_{n-1}} W_{n-1} J = (W_{n-1} \otimes_{W_{n-1}} W_{n-1} J) \oplus (W_{n-1} s(\alpha) \otimes_{W_{n-1}} W_{n-1} J).$$

Since $J$ is big with respect to $\mathbb{E}_n/\mathbb{E}_{n-1}$, there exist $x, y \in W_{n-1} J \setminus \{0\}$ such that

$$x + s(\alpha)y = 0.$$

Thus, $s(\alpha) = -x/y \in Q(\mathfrak{A}_{n-1})$. It follows that $W_n \subset Q(\mathfrak{A}_{n-1})$ and hence $Q(\mathfrak{A}_n) = Q(\mathfrak{A}_{n-1})$. $\square$

<figure><img src="image_6.png" /></figure>

Finally, we descend from $Q(\mathfrak{A}_n)$ to $Q(\mathfrak{A}_0)$ by induction on $n$.

**Proposition 9.14.** For all $2 \le k \le n$, if $Q(\mathfrak{A}_k) = Q(\mathfrak{A}_{k-1})$ then $Q(\mathfrak{A}_{k-1}) = Q(\mathfrak{A}_{k-2})$. In particular, if $J$ is big with respect to $\mathbb{F}/\mathbb{E}$, then $Q(A^{\Sigma_t}) = Q(W(\mathbb{E}) + J)$.
Proof. Note that for any $k \ge 1$ we have $Q(\mathfrak{A}_k) = Q(\mathfrak{A}_{k-1})$ if and only if $W_k \subseteq Q(\mathfrak{A}_{k-1})$. Assume that $Q(\mathfrak{A}_k) = Q(\mathfrak{A}_{k-1})$ for some $k, 2 \le k \le n$. Choose $\bar{\alpha} \in \mathbb{E}_{k-2}, \bar{\beta} \in \mathbb{E}_{k-1}$ such that $\mathbb{E}_{k-1} = \mathbb{E}_{k-2}(\sqrt{\bar{\alpha}})$ and $\mathbb{E}_k = \mathbb{E}_{k-1}(\sqrt{\bar{\beta}})$. Define $\alpha := s(\bar{\alpha})$ and $\beta := s(\bar{\beta})$, so $W_{k-1} = W_{k-2}(\sqrt{\alpha})$ and $W_k = W_{k-1}(\sqrt{\beta})$. It suffices to show that $\sqrt{\alpha} \in Q(\mathfrak{A}_{k-2})$.

Since $Q(\mathfrak{A}_k) = Q(\mathfrak{A}_{k-1})$, we can write $\sqrt{\beta} = x/y$ with $x, y \in \mathfrak{A}_{k-1} \setminus \{0\}$. By multiplying $x$ and $y$ by any nonzero element of $W_{k-1}J$, we may assume that $x, y \in W_{k-1}J \setminus \{0\}$.

Note that we can write $y = i_1 + \sqrt{\alpha}i_2$ with $i_1, i_2 \in W_{k-2}J$. If $i_1 - \sqrt{\alpha}i_2 \ne 0$, then by multiplying $x$ and $y$ by $i_1 - \sqrt{\alpha}i_2$, we may assume that $y \in W_{k-2}J \setminus \{0\}$. If $i_1 - \sqrt{\alpha}i_2 = 0$ and $y \notin W_{k-2}J$, then we must have $i_2 \ne 0$ since $y \ne 0$ and $\sqrt{\alpha} = i_1/i_2 \in Q(\mathfrak{A}_{k-2})$, as desired. We assume henceforth that $y \in W_{k-2}J$.

Write $x = a + b\sqrt{\alpha}$ for some $a, b \in W_{k-2}J$. Then we have $y\sqrt{\beta} = a + b\sqrt{\alpha}$ and thus

$$ (7) \qquad y^2\beta = a^2 + \alpha b^2 + 2ab\sqrt{\alpha}. $$

Since $\beta \in W_{k-1}$, we may write $\beta = e + f\sqrt{\alpha}$ for some $e, f \in W_{k-2}$. Note that $f \ne 0 \bmod p$ since $[\mathbb{E}_k : \mathbb{E}_{k-1}] = 2$ and $\mathbb{E}_k = \mathbb{E}_{k-1}(\sqrt{\beta})$. Substituting this into equation (7), we see that

$$ (y^2f - 2ab)\sqrt{\alpha} = a^2 + \alpha b^2 - y^2e \in W_{k-2}J. $$

Note that $y^2f - 2ab \in W_{k-2}J$ since all of $y, f, a, b \in W_{k-2}J$. If $y^2f - 2ab \ne 0$, then we can conclude that $\sqrt{\alpha} \in Q(\mathfrak{A}_{k-2})$ as desired.

Henceforth, assume that $y^2f = 2ab$. Then we also have $y^2e = a^2 + \alpha b^2$. Thus $2f^{-1}eab = a^2 + \alpha b^2$. Note that $a, b \ne 0$ since $2ab = y^2f$ and we know $y, f \ne 0$. Then we have

$$ 2f^{-1}e = \frac{a}{b} + \alpha \frac{b}{a}. $$

Therefore $\frac{a}{b}$ is a root of $t^2 - 2f^{-1}et + \alpha \in W_{k-2}[t]$. The discriminant of this polynomial is $4(f^{-2}e^2 - \alpha)$. We claim that $W_{k-1} \supseteq W_{k-2}(\sqrt{f^{-2}e^2 - \alpha})$. Note that since $\beta = s(\bar{\beta})$ and $s$ is multiplicative we have that $e^2 - f^2\alpha = s(N_{\mathbb{E}_{k-1}/\mathbb{E}_{k-2}}(\bar{\beta}))$, and therefore $\sqrt{e^2 - f^2\alpha} = s(\sqrt{N_{\mathbb{E}_{k-1}/\mathbb{E}_{k-2}}(\bar{\beta})}) \in s(\mathbb{E}_{k-1}^\times)$. Therefore $\frac{a}{b} \in W_{k-1}$. Thus we can write $\sqrt{\beta} = \frac{x/b}{y/b} = \frac{\frac{a}{b} + \sqrt{\alpha}}{y/b}$, and so

$$ \frac{y}{b} = \left( \frac{a}{b} + \sqrt{\alpha} \right) \sqrt{\beta}^{-1}. $$

Note that $\frac{a}{b} + \sqrt{\alpha} \ne 0$ since $y \ne 0$. It follows that $(\frac{a}{b} + \sqrt{\alpha})\sqrt{\beta}^{-1}$ generates $W_k$ over $W_{k-1}$ since $\frac{a}{b} + \sqrt{\alpha} \in W_{k-1}$ and $\sqrt{\beta}$ generates $W_k$ over $W_{k-1}$. Thus $(\frac{a}{b} + \sqrt{\alpha})\sqrt{\beta}^{-1} = \frac{y}{b} \in Q(\mathfrak{A}_{k-2})$ and so $W_k \subset Q(\mathfrak{A}_{k-2})$. Therefore $Q(\mathfrak{A}_k) = Q(\mathfrak{A}_{k-2})$.

For the second statement of the proposition, note that by Proposition 9.13 we have $Q(\mathfrak{A}_n) = Q(\mathfrak{A}_0)$ for all $0 \le k \le n$. We have $\mathfrak{A}_0 = W(\mathbb{E}) + J$ by definition. Since $\Sigma_t$ acts trivially on $J$, it follows that

$$ A^{\Sigma_t} = \mathfrak{A}^{\Sigma_t}(\mathfrak{A}) = W_n + W_nJ = \mathfrak{A}_n. \quad \square $$

**Corollary 9.15.** We have

(1) $A^{\Sigma_t} \supseteq \mathcal{B}_\rho(\mathbb{E})$;

(2) $A^{\Sigma_t}$ is a finitely generated $\mathcal{B}_\rho(\mathbb{E})$-module;

(3) $A^{\Sigma_t}$ has the same field of fractions as $\mathcal{B}_\rho(\mathbb{E})$.

In particular, $A^{\Sigma_t}$ and $\mathcal{B}_\rho(\mathbb{E})$ are fullness peers.

**Proof.** The definition of $\mathbb{E}$ and Corollary 9.3 imply (1).

For (2), if either $\bar{\rho}$ is not dihedral or ker $\beta_t \ne 1$, then $A^{\Sigma_t} = \mathcal{B}_\rho(\mathbb{F}^{\beta_t(\Sigma_t)})$, which is finitely generated over $\mathcal{B}_\rho(\mathbb{E})$ since $W(\mathbb{F})^{\Sigma_t}$ is finitely generated over $W(\mathbb{E})$. In the case when $\bar{\rho}$ is dihedral and ker $\beta_t = 1$, recall that $\mathcal{B}_\rho(\mathbb{E})$ is a noetherian ring (Proposition 7.6(1)) and $A = \mathcal{B}_\rho(\mathbb{F}) + W(\mathbb{F})B_1(\rho)$
is a noetherian $\mathcal{B}_\rho(\mathbb{E})$-module by Proposition 7.6(2) and the fact that $W(\mathbb{F})$ is noetherian over $W(\mathbb{E})$. Thus the $\mathcal{B}_\rho(\mathbb{E})$-submodule $A^{\Sigma_t}$ of $A$ is necessarily noetherian and hence finitely generated.

The third point has largely been established already. When $J$ is small with respect to $\mathbb{F}/\mathbb{E}$, it follows from Corollary 9.9. When $J$ is big with respect to $\mathbb{F}/\mathbb{E}$ and either $\bar{\rho}$ is not dihedral or $\ker \beta_t \neq 1$, this follows from Proposition 9.14 since in those cases $W(\mathbb{E}) + J = \mathcal{B}_\rho(\mathbb{E})$. Finally, when $\bar{\rho}$ is dihedral, $\ker \beta_t = 1$, and $J$ is big with respect to $\mathbb{F}/\mathbb{E}$ we have

$$Q(A^{\Sigma_t}) = Q(W(\mathbb{E}) + J) = Q(\mathcal{B}_\rho(\mathbb{E})),$$

where the first equality follows from Proposition 9.14 and they second from Proposition 7.6(3).

The final statement now follows from Lemma 3.5. $\square$

We now have the following corollary, which summarizes the most general theorem we have for images of admissible pseudodeformations with 2-power determinant.

**Corollary 9.16.** Let $\bar{\rho}: \Pi \to \mathrm{GL}_2(\mathbb{F})$ be a regular representation such that the order of $\det \bar{\rho}$ is a power of 2. If $\bar{\rho}$ is octahedral, assume furthermore that $\bar{\rho}$ is good. Let $A$ be a domain and $(t, d): \Pi \to A$ an admissible pseudodeformation of $\bar{\rho}$. If $(t, d)$ is not a priori small, then $(t, d)$ is $A^{\Sigma_t}$-full, hence $A_0$-full.

*Proof.* By Corollary 6.6, $\rho_t$ is $\mathcal{B}_\rho(\mathbb{E})$-full. Corollary 9.15 implies that $\rho_t$ is $A^{\Sigma_t}$-full, hence $A_0$-full by Corollary 4.21. $\square$

# 10. MAIN FULLNESS RESULTS

In Section 10 we draw conclusions from Corollary 9.16 that are useful in applications and when comparing our work with previous results in the literature. Although the constant determinant assumption is important to be able to use Bellaïche's work in Section 9, in practice one rarely works in a constant-determinant setting. Here we give the most general fullness result we can prove; in particular, we remove the constant-determinant assumption present in Corollary 9.16. We then recast our main theorem and other highlights of the theory of fullness in the language of representation theory rather than pseudorepresentations.

To ensure that our main result can be read independent of much of the rest of the paper, we briefly recall our notation and terminology. Let $p$ be an odd prime, $A$ a local pro-$p$ domain with maximal ideal $\mathfrak{m}$ and residue field $\mathbb{F}$, and $\Pi$ a $p$-finite profinite group (Definition 2.6). We are interested in a continuous pseudodeformation $(t, d): \Pi \to A$ of a semisimple representation $\bar{\rho}: \Pi \to \mathrm{GL}_2(\mathbb{F})$. Unlike much of the paper, in Section 10 we never require $A$ to be the trace algebra of $(t, d)$. We say $(t, d)$ is not a priori small if it is not reducible, dihedral, or equal to a twist of its Teichmüller lift (Definition 2.3), and it is $A_0$-full if the image of some representation carrying $(t, d)$ contains, up to conjugation, an $A_0$-congruence subgroup (Definition 3.1).

Recall that $A_0$ is the adjoint trace ring of $(t, d)$ (Definition 4.6); its residue field is the trace field of $\bar{\rho}$ and is denoted $\mathbb{E}$. We say that $\bar{\rho}$ is regular if its image contains a matrix whose eigenvalue ratio is in $\mathbb{E}^\times \setminus \{\pm 1\}$: see Definition 2.19 and Remark 2.20 immediately following. When $\bar{\rho}$ is octahedral (that is, projective image $S_4$), see Definition 8.10 for the definition of goodness.

**Theorem 10.1.** Let $p > 2$ be prime, $A$ a local pro-$p$ domain with residue field $\mathbb{F}$, and $\Pi$ be a $p$-finite profinite group. Let $\bar{\rho}: \Pi \to \mathrm{GL}_2(\mathbb{F})$ be a regular semisimple representation that is good if $\bar{\rho}$ is octahedral. If $(t, d): \Pi \to A$ is a pseudodeformation of $\bar{\rho}$ that is not a priori small, then $(t, d)$ is $A_0$-full.

*Proof.* Let $\chi: \Pi \to A^\times$ be a character such that $(t', d') := (\chi t, \chi^2 d)$ is a constant-determinant pseudorepresentation, and write $\bar{\rho}' := \bar{\chi} \otimes \bar{\rho}$, where $\bar{\chi}: \Pi \to \mathbb{F}^\times$ is the reduction of $\chi$ modulo $\mathfrak{m}$. Assume that $\chi$ is chosen such that $\bar{\rho}'$ has no conjugate self-twists if $\bar{\rho}$ is reducible and the order of $\det \bar{\rho}'$ is a power of 2 if $\bar{\rho}$ is absolutely irreducible. This is possible by Corollary 8.2 in the reducible
case and Lemma 8.4 in the absolutely irreducible case. Furthermore, note that if $\bar{\rho}$ is octahedral and good, then so is $\bar{\rho}'$ by Lemma 8.11. Let $A'$ be the subring of $A$ topologically generated by $t'(\Pi)$. We have seen in Proposition 9.1 and Corollary 9.16 that if $\bar{\rho}'$ is regular (and under the further assumption that $\bar{\rho}$ is good when $\bar{\rho}$ is octahedral) and $(t', d')$ is not a priori small, then $(t', d')$ is $A_0$-full. This is sufficient by Corollary 3.12. $\square$

**Remark 10.2.** Let $(t, d)$ be as in Theorem 10.1. Then its constant-determinant twist $(t', d') : \Pi \to A$ satisfies the conditions of Corollary 9.16, so that it is $A_{t'}^{\Sigma_{t'}(A_{t'})}$-full. By Corollary 3.12, $(t, d)$ itself is also $A_t^{\Sigma_t(A)}$-full. However it does not follow that that $(t, d)$ is $A_t^{\Sigma_t(A)}$-full. Indeed, $(t, d)$ may be affected by the pathologies pointed out in Examples 4.3 and 4.10, or even worse, the pro-$p$-part of $d$ may be transcendental over $A_0$. Such obstacles are not faced by well-behaved $A_0$. $\triangle$

We end this section by recasting all our headline results, including our main result (Theorem 10.1) in the language of representation theory, convenient for comparing applications to results in the literature in Section 12.

Building on the notation recalled above, let $\rho : \Pi \to \text{GL}_2(Q(A))$ be a representation whose trace lands in $A$ and is continuous, and let $(t, d) = (\text{tr}\,\rho, \text{det}\,\rho)$. Recall that $\rho$ is not a priori small if $\rho$ is strongly absolutely irreducible (Proposition 2.4). Write $A_\rho$ for the trace algebra of $\rho$ — the $W(\mathbb{F})$-subalgebra of $A$ topologically generated by $t(\Pi)$ — and $K$ for its field of fractions. Let $K_0$ denote the field of fractions of $A_0$, the adjoint trace ring of $\rho$. There is a unique semisimple representation $\bar{\rho}$ whose trace is equal to the reduction of $t$ modulo $m$. Finally, recall that we say that the determinant of $\rho$ is $A_0$-constant if the pro-$p$ part of $\text{det}\,\rho$ is $A_0$-valued. In this case $K/K_0$ is Galois (Theorem 4.19).

**Theorem 10.3.** Let $\rho$ be not a priori small. Suppose $\bar{\rho}$ is regular, and good if octahedral.

(1) **$A_0$-fullness:** $\rho$ is $A_0$-full.

(2) **Optimality:** If $\rho$ is $A'$-full for a subring $A' \subseteq A$, then $A_0$ contains a fullness peer of $A'$.

(3) **All CSTs fix $A_0$:** For any extension $B$ of $A$ we have $\Sigma_\rho(B) = \Sigma_\rho(B/A_0)$.

If further $K$ is a separable extension of $K_0$, then:

(4) **CSTs carve out $K_0$:** $(K^{\text{sep}})^{\Sigma_\rho(K^{\text{sep}})} = K_0$.

If further still $\rho$ has $A_0$-constant determinant, then:

(5) **All CSTs are simple:** Every $K^{\text{sep}}$-valued conjugate self-twist restricts to a simple $A_\rho$-valued one: $\Sigma_\rho(K^{\text{sep}}) \to \Sigma_\rho(A_\rho)$. Moreover, $K^{\Sigma_\rho(A_\rho)} = K_0$.

(6) **CST-invariants fullness:** $\rho$ is $A_\rho^{\Sigma_\rho(A_\rho)}$-full.

*Proof.* Recall that $\rho$ is $A'$-full if and only if $(\text{tr}\,\rho, \text{det}\,\rho)$ is $A'$-full by Lemma 3.2(2). Thus the first statement follows from Theorem 10.1 and the second from Theorem 5.3. The third statement follows from (1) by Corollary 4.24. The fourth statement follows from Corollary 4.12(2) with $L = K^{\text{sep}}$, $E = K_0$, $F = K$ and $F_0$ the (nontopological) adjoint trace field of $(t, d)$ viewed as valued in $K$. For the fifth statement, use Corollary 4.24 to obtain the restriction map $\Sigma_t(K^{\text{sep}}/A_0) \to \Sigma_t(A_\rho)$, surjective since $K^{\text{sep}}$ is normal over $K$. Alternatively, $K$ is Galois over $K_0$ with $\Sigma_t(A_\rho) = \text{Gal}(K/K_0)$ by Theorem 4.19 and then $\Sigma_t(K^{\text{sep}}) = \text{Gal}(K^{\text{sep}}/K_0)$ by Theorem 4.11. The last statement follows from (1) and Corollary 4.21. $\square$

## 11. RESIDUALLY LARGE REPRESENTATIONS

Here we show that by imposing stronger conditions on the residual image when $\bar{\rho}$ is large, we obtain a more precise description of the image of $\rho$. That is, we assume that $\rho: \Pi \to \text{GL}_2(A)$ is a continuous representation such that $\text{Im}\,\bar{\rho} \supseteq \text{SL}_2(\mathbb{E})$ and $\bar{\rho}$ is large. Under this assumption we have a more precise understanding of the image of $\rho$ than simply fullness. Unlike our main fullness result, in this section $A$ is any local pro-$p$ ring; it need not be a domain.
Historically, this is the case that has been studied the most, starting with the work of Boston in [MW86, Appendix]. Boston shows that if $A$ is a complete local noetherian ring and $H$ is a closed subgroup of $SL_2(A)$ that projects onto $SL_2(A/m^2)$, then $H = SL_2(A)$ [MW86, Appendix Proposition 2]. Bellaïche has pointed out that Boston's result follows from his work [Bel19, Remark 6.8.4]. As an application of the description of the image found in Proposition 11.1, we show in Theorem 11.3 that one can replace the hypothesis that $H$ projects onto $SL_2(A/m^2)$ with the hypothesis that $H$ projects onto $SL_2(A/m)$ and $A$ is the trace ring of $H$ to obtain the same conclusion. Theorems of this form have been obtained in special cases, for instance for the Galois representation attached to the mod-$p$ Hecke algebra [Amo21, Theorem, Introduction]. Note that assuming $H$ projects onto $SL_2(A/m)$, the hypothesis about $A$ being the trace algebra can always be arranged while Boston's hypothesis about projecting onto $SL_2(A/m^2)$ may fail. Moreover, our description of the image in Proposition 11.1 does not require that the residual image contain all of $SL_2(F)$, rather only $SL_2(E)$, and is thus more general that the setting of Boston's work.

Let $\rho: \Pi \to GL_2(A)$ be a continuous representation such that $\mathrm{Im} \, \bar{\rho} \supseteq SL_2(E)$. Note that such a $\bar{\rho}$ is large if and only if $E \neq \mathbb{F}_3, \mathbb{F}_5$. Let $\chi: \Pi \to A^\times$ be the character described in the proof of Theorem 10.1, and let $r := \chi \otimes \rho$. Assume that $\rho$ is conjugated in such a way that Theorem 2.23 applies to $r$. Recall that $\mathcal{B}_r(E) = W(E)[I_1(r)]$.

**Proposition 11.1.** Assume $\#E \ge 7$. Then

(1) $\mathrm{Im} \, r \supseteq SL_2(\mathcal{B}_r(E))$ as a finite index subgroup;

(2) $\mathrm{Im} \, \rho \supseteq SL_2(\mathcal{B}_r(E))$;

(3) $\mathcal{B}_r(E)$ is the largest subring $B$ of $A$ for which $\mathrm{Im} \, \rho \supseteq SL_2(B)$.

*Proof.* For ease of notation, write $\mathfrak{m}_r$ for the maximal ideal of $\mathcal{B}_r(E)$.

Since $\mathrm{Im} \, \bar{r} \supseteq SL_2(E)$, it follows that $\mathrm{Im} \, r \supseteq SL_2(W(E))$ by [Man15, Main Theorem]. In particular, $p \in I_1(r)$ and so $\mathfrak{m}_r = I_1(r)$ by Theorem 2.23. Let $H$ be the subgroup of $G := \mathrm{Im} \, r$ generated by $\Gamma = \Gamma(r)$ and $SL_2(W(E))$. Then $H$ is a finite index subgroup of $G$ since $\Gamma$ is.

We claim that $H = SL_2(\mathcal{B}_r(E))$. Indeed, note that $\Gamma = \Gamma_{\mathcal{B}_r(E)}(\mathfrak{m}_r)$ by [Bel19, Corollary 6.8.3] and the fact that $\mathfrak{m}_r = I_1(r)$. In particular, this shows that $H \subseteq SL_2(\mathcal{B}_r(E))$. In fact, $H$ is a subgroup of $SL_2(\mathcal{B}_r(E))$ such that $H/\Gamma = SL_2(E) = SL_2(\mathcal{B}_r(E))/\Gamma$. Thus we must have equality.

Now suppose that $\mathrm{Im} \, r \supseteq SL_2(B)$ for some subring $B$ of $A$. Without loss of generality, we may assume $B$ is closed, hence local. Then $\Gamma \supseteq \Gamma_B(\mathfrak{m}_B)$, which implies that $I_1(r) \supseteq \mathfrak{m}_B$. On the other hand, if $\mathrm{Im} \, r \supseteq SL_2(B)$ then $\mathrm{Im} \, \bar{r} \supseteq SL_2(B/\mathfrak{m}_B)$. By definition of $E$, we know that $E$ is the largest subfield of $F$ such that $\mathrm{Im} \, \bar{r} \supseteq SL_2(E)$. Thus we must have $B/\mathfrak{m}_B \subseteq E$. It follows that $B \subseteq \mathcal{B}_r(E)$.

As for $\rho$, note that there is a character $\tilde{\chi}: \mathrm{Im} \, \rho \to A^\times$ such that $\mathrm{Im} \, r = \{x\tilde{\chi}(x): x \in \mathrm{Im} \, \rho\}$. Now $\tilde{\chi}$ must be trivial on $SL_2(B)$ for any ring $B$ whose residue field has more than three elements by Corollary 3.10. Therefore $\mathrm{Im} \, \rho$ and $\mathrm{Im} \, r$ contain the same copies of $SL_2$. $\square$

**Corollary 11.2.** If $\#E \ge 7$ and $\mathrm{Im} \, \bar{\rho} \supseteq SL_2(E)$, then $\mathrm{Im} \, \rho$ contains $SL_2(A_0)$ up to conjugation.

*Proof.* By Proposition 11.1 it suffices to show that $\mathcal{B}_r(E) \supseteq A_0$, and this only needs to be shown when $r = \rho$ is the constant-determinant universal representation. Without loss of generality we may twist to assume that the order of $\det \rho$ is a power of 2. By Theorem 2.23 we have $A = \mathcal{B}_r(F)$ since $\bar{\rho}$ is large. By Corollary 9.3 we may conjugate $\rho$ to assume that $I_1(\rho)$ is fixed by all (A-valued) conjugate-self twists of $\rho$, and by Corollary 7.4 all conjugate-self twists of $\bar{\rho}$ lift to conjugate self-twists of $\rho$ since $\rho$ is universal. Therefore

$$ \mathcal{B}_r(E) = W(F)^{\Sigma_{\rho}}[I_1(\rho)] = A^{\Sigma_{\rho}} \supseteq A_0. \quad \square $$

**Theorem 11.3.** Let $A$ be a pro-$p$ local noetherian ring with $p \ne 2$ and residue field $F$. Let $H$ be a closed subgroup of $SL_2(A)$ that projects onto $SL_2(F)$. If $A$ is the trace ring of $H$, that is, $A$ is topologically generated by $\mathrm{tr} H$, then $H = SL_2(A)$.
*Proof.* Let $\rho: H \to SL_2(A)$ be the natural inclusion. Note that $\rho$ has constant determinant. Since $A$ is the trace ring of $\rho$, Theorem 2.23 implies that $A = B_\rho(\mathbb{F})$. Thus by Proposition 11.1, it suffices to show that $\mathbb{E} = \mathbb{F}$. Since $\mathrm{Im}\,\bar{\rho} = SL_2(\mathbb{F})$, it follows that $\mathbb{E}$ is the subfield of $\mathbb{F}$ generated by the squares of traces of $SL_2(\mathbb{F})$. A straightforward matrix calculation shows that $\mathbb{E} = \mathbb{F}$. $\square$

*Remark* 11.4. In the preprint [AB], Aryas-de-Reina and Böckle prove a large image result for a residually full representation $\Pi \to G(A)$, where $G$ is an adjoint group and $A$ is the ring of definition of the representation. It seems possible to recover Theorem 11.3 by applying their result to the projective representation $\mathbb{P}\rho: \Pi \to PGL_2(A)$ attached to $\rho$ and using the fact that the ring of definition of $\mathbb{P}\rho$ is the ring fixed by the conjugate self-twists of $\rho$. $\triangle$

## 12. APPLICATIONS TO GALOIS REPRESENTATIONS

In this section we specialize Theorem 10.1 to some arithmetic settings, more specifically to representations coming from elliptic, Hilbert, and Bianchi cuspidal eigenforms (Section 12.1 through Section 12.3) and cuspidal $p$-adic families of elliptic and Hilbert eigenforms (Section 12.4 through Section 12.6). We explain how to recover, and in some cases improve, the results already present in the literature. In particular, since our methods are entirely agnostic about the group $\Pi$, they reveal that many of the classical big-image results are fundamentally *algebraic* in nature: they do not rely on the arithmetic input, such as local information at the places where a Galois representation is ramified, that went into the original proof.

**12.1. Classical modular eigenforms.** Let $f$ be a non-CM cuspidal modular eigenform of some level and some weight $k \ge 2$ defined over a number field $K$. Fix an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ and let $G_{\overline{\mathbb{Q}}} := \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. For any prime $p$ of $K$ lying over a rational prime $p$, let $K_p$ be the completion of $K$ at $p$ and $O_p$ its ring of integers. A construction of Deligne attaches to this data an irreducible continuous representation $\rho_{f,p}: G_{\overline{\mathbb{Q}}} \to GL_2(K_p)$, unramified almost everywhere and hence factoring through a $p$-finite extension, whose traces of Frobenius elements at unramified primes correspond to Hecke eigenvalues of $f$. Because $G_{\overline{\mathbb{Q}}}$ is compact we may view $\rho_{f,p}$ as taking values in $GL_2(O_p)$.

The following result about the image of $\rho_{f,p}$ was proved by Ribet and Momose in the 1980s, generalizing an earlier theorem of Serre about Tate modules of elliptic curves. Let $K_{p,0}$ be the subfield of $K_p$ fixed by all the generalized conjugate self-twists of $\rho_{f,p}$, and $O_{p,0}$ its ring of integers.

**Theorem 12.1** (Ribet, Momose at $p$: first version [Rib85, Mom81]).

*For all but finitely many primes $p$ of $K$, the representation $\rho_{f,p}$ is $O_{p,0}$-full.*

To show the extent to which our work recovers the result of this theorem at $p$, we first make more explicit Ribet and Momose's condition on $p$. Let $K_0$ be the subfield of $K$ fixed by the conjugate self-twists of $f$; the $p$-adic field $K_{p,0}$ defined above is the completion of $K_0$ at the prime under $p$. Let $H \subseteq G_{\overline{\mathbb{Q}}}$ be the intersection of the kernels of all the conjugate self-twist characters of $f$:

$$ (8) \qquad H := \bigcap_{(\sigma, \eta) \in \tilde{\Sigma}_f} \ker \eta. $$

Then $H$ is a finite-index normal subgroup of $G_{\overline{\mathbb{Q}}}$. Because all the conjugate self-twist characters of $\rho_{f,p}$ are trivial on $H$, the trace of $\rho_{f,p}|_H$ lands in $K_{p,0}$. As described just after Definition A.11, there is therefore a $K_{p,0}$-quaternion algebra $D_p$ splitting over $K_p$ with $\rho_{f,p}(H) \subset D_p^\times \subset GL_2(K_p)$.

Ribet and Momose describe a global analogue to this picture. They define a global $K_0$-quaternion algebra $D$ split over $K$ with $D(K_{p,0}) \cong D_p$: that is, for each prime $p$, the restriction to $H$ of $\rho_{f,p}$ can be viewed as taking values in $D(K_{p,0})^\times$. By compactness again we may view $\rho_{f,p}(H)$ as a subgroup of the units of a maximal order $O_{D,p}$ of $D(K_{p,0})$. Ribet and Momose's adelic open-image theorems say that the image of $\rho_{f,p}$ always contains an open subgroup of the norm-1 units of $O_{D,p}^\times$, and for
all but finitely many $\mathfrak{p}$ it contains all of those norm-1 units. In particular, if $D(K_{\mathfrak{p},0})$ is split, then up to conjugation $O_{D,\mathfrak{p}}^{\times} = \text{GL}_2(O_{\mathfrak{p},0})$; and we can therefore make Theorem 12.1 more precise.

**Theorem 12.2** (Ribet, Momose at $\mathfrak{p}$: second version [Rib85, Mom81]).
If $D(K_{\mathfrak{p},0})$ is split, then the representation $\rho_{f,\mathfrak{p}}$ is $O_{\mathfrak{p},0}$-full.

*Remark 12.3.* In fact, the statement in Theorem 12.2 is an if-and-only-if. Indeed, no element of $D(K_{\mathfrak{p},0})$ can have distinct $\text{GL}_2$-eigenvalues in $K_{\mathfrak{p},0}$. A matrix $g$ with eigenvalues $\alpha, \beta$ satisfies $(g - \alpha)(g - \beta) = 0$. If $\alpha, \beta$ are in the center of a division algebra containing $g$, and at the same time are eigenvalues of $g$ in any matrix setting, then the Cayley-Hamilton equation $(g - \alpha)(g - \beta) = 0$ means that either $g = \alpha$ or $g = \beta$. Thus no embedding of $D(K_{\mathfrak{p},0})$ into $\text{GL}_2(K_{\mathfrak{p}})$ can contain any congruence subgroup of $\text{SL}_2(O_{\mathfrak{p},0})$, or even of $\text{SL}_2(\mathbb{Z}_{\mathfrak{p}})$. The same is true for all of $\rho(\Pi)$: see Theorem 4.19 and Proposition 3.11 or Proposition 12.5 below. In other words, no nonsplit $\rho_{f,\mathfrak{p}}$ can ever satisfy our present definition of fullness. As we've defined it, fullness is fundamentally a $\text{GL}_2$ property; a fitting notion for more general algebraic groups generalizing Ribet and Momose's openness beyond Krull dimension 1 is outside the scope of this investigation. $\triangle$

We now show that our results recover Ribet and Momose's theorem at $\mathfrak{p}$ in most cases. Let $\mathbb{F}$ be the residue field of $O_{\mathfrak{p}}$, a finite extension of $\mathbb{F}_{\mathfrak{p}}$, and let $\bar{\rho}: G_{\mathbb{Q}} \to \text{GL}_2(\mathbb{F})$ be the semisimplification $\rho_{f,\mathfrak{p}}$ modulo the maximal ideal of $O_{\mathfrak{p}}$. Recall that $\bar{\rho}$ is regular if its image contains a matrix whose eigenvalue ratio is not $\pm 1$ but is contained in the trace algebra $\mathbb{E}$ of ad $\bar{\rho}$ (which is a subfield of the residue field of $O_{\mathfrak{p},0}$): see Definition 2.19 and Remark 2.20 immediately following. If $\bar{\rho}$ is octahedral (that is, projective image $S_4$), see Definition 8.10 for the notion of *goodness*.

**Theorem 12.4** (Our results recovering Ribet and Momose at $\mathfrak{p}$). Assume that $\mathfrak{p}$ is odd and that $\bar{\rho}$ is regular; if $\bar{\rho}$ is octahedral, assume further that $\bar{\rho}$ is good. Then $\rho_{f,\mathfrak{p}}$ is $O_{\mathfrak{p},0}$-full.

*Proof.* Since $f$ is cuspidal, non-CM, and has weight $k \ge 2$, its associated representation $\rho_{f,\mathfrak{p}}$ is strongly absolutely irreducible ([Rib77, Proposition 4.4]) and hence not a priori small. By Theorem 10.3, $\rho_{f,\mathfrak{p}}$ is $A_0$-full, where as usual $A_0 \subseteq O_{\mathfrak{p}}$ is the adjoint trace ring. By Corollary 4.21, $A_0$ and $O_{\mathfrak{p},0}$ are fullness peers, and $\rho_{f,\mathfrak{p}}$ is $O_{\mathfrak{p},0}$-full. $\square$

The regularity assumption in Theorem 12.4 a posteriori forces $D(K_{\mathfrak{p},0})$ to split (Remark 12.3). We can also see that a nonsplit $D(K_{\mathfrak{p},0})$ means an irregular $\bar{\rho}$ directly:

**Proposition 12.5.** If $D(K_{\mathfrak{p},0})$ is a division algebra, then $\bar{\rho}$ is reducible and not regular.

*Proof.* We first show that $\bar{\rho}|_H$ is reducible and not regular. Let $L/K_{\mathfrak{p},0}$ be the unique quadratic unramified extension, $\pi$ is a uniformizer of either, and $\sigma$ the nontrivial element of $\text{Gal}(L/K_{\mathfrak{p},0})$. Note that we do not assume that $L$ is a subfield of $K_{\mathfrak{p}}$. Write $\ell, k$ for the residue fields of $L, K_{\mathfrak{p},0}$, respectively; then $[\ell : k] = 2$ and $\mathbb{E} \subseteq k$. By compactness $\rho(H)$ can be viewed as a subgroup of

$$ \left\{ \begin{pmatrix} \alpha & \pi\beta \\ \sigma(\beta) & \sigma(\alpha) \end{pmatrix} : \alpha \in O_L^\times, \beta \in O_L \right\} \subset \text{GL}_2(O_L), $$

the maximal order of $D(K_{\mathfrak{p},0})$ as viewed inside $\text{GL}_2(L)$, which gives a representation $\rho'$ of $H$ over $L$. By inspection, it is clear that its residual representation $\bar{\rho}'$ is, up to semisimplification, a sum of two characters to $\ell$ conjugate over $k$. This means that the eigenvalue ratio $r$ of any element in $\bar{\rho}'(H)$ is in the form $r = a^{\#k-1}$ for some $a \in \ell^\times$. Such an element is in $k$ if and only if $r^{\#k-1} = 1$; in other words, if and only if

$$ 1 = a^{(\#k)^2 - 2(\#k) + 1} = a^{-2(\#k) - 2} = r^{-2}. $$

But this last is only possible if $r = \pm 1$. In other words, $\rho'$ is residually neither absolutely irreducible nor regular. Since $\rho'$ is isomorphic to $\rho|_H$ over $\overline{\mathbb{Q}}_p$, the same is true for $\rho|_H$ as well.

We now follow [Nek12, B.4.8(1)] to claim that the same is true for $\rho$ on all of $\Pi$. Change notation to let $L$ be any subextension of $K_{\mathfrak{p}}$ that is quadratic over $K_{\mathfrak{p},0}$ and hence splits $D(K_{\mathfrak{p},0})$, with $\sigma$ a
generator of Gal($L/K_{p,0}$) and $\pi$ a uniformizer of $K_{p,0}$ that is not a norm from $L$. Since $H$ is normal in $\Pi$ and $\rho(H)$ spans $D(K_{p,0})$ over $K_{p,0}$, the image of $\rho$, up to conjugation, will be contained in the normalizer of the subgroup $Q = \left\{ \begin{pmatrix} \alpha & \pi\beta \\ \sigma(\beta) & \sigma(\alpha) \end{pmatrix} : (\alpha, \beta) \in L^2 - \{(0,0)\} \right\}$, isomorphic to $D(K_{p,0})^\times$, in $\mathrm{GL}_2(K_p)$. One can show that this normalizer is just $K_p^\times Q^{(\mathrm{viii})}$, so that passing from $H$ to $\Pi$ does not affect the projective image of $\bar{\rho}$. Therefore $\bar{\rho}$ on all of $\Pi$ remains reducible and not regular. $\square$

In other words, the regularity assumption in Theorem 12.4 eliminates the division algebra case. One might hope for a converse, so that our methods could recover all of Theorem 12.2. But alas this is not so: there are certainly cases where $D(K_{p,0})$ is a matrix algebra but regularity is not satisfied, so that our methods do not apply. In addition to $p=2$, we do not conclude fullness if $\mathbb{F}=\mathbb{F}_3$, even if $f$ has no conjugate self-twists and hence $D$ is globally split, as is the case for a non-CM elliptic curve over $\mathbb{Q}$. If the image of $\bar{\rho}$ is too small to accommodate the regularity assumption, our methods cannot handle it.

**12.2. Hilbert modular eigenforms.** Everything in Section 12.1 has been generalized to Hilbert modular forms. In particular, our results recover, in much the same manner and to much the same extent, the big-image results of Nekovář generalizing Ribet and Momose’s work over $\mathbb{Q}$. We summarize the situation very briefly.

Let $F$ be a totally real field and $f$ a non-CM cuspidal Hilbert modular eigenform over $F$ all of whose weights are at least 2. Fix an algebraic closure $\bar{F}$ of $F$, and let $G_F := \mathrm{Gal}(\bar{F}/F)$. Fix a prime $p$ and an embedding $\iota_p: \bar{\mathbb{Q}} \hookrightarrow \bar{\mathbb{Q}}_p$. Let $\rho_{f,\iota_p}: G_F \to \mathrm{GL}_2(\bar{\mathbb{Q}}_p)$ be the Galois representation attached to $f$ in the usual way, which we may view as having coefficients in the ring of integers $\mathcal{O}$ of some finite extension of $\mathbb{Q}_p$. Let $\mathcal{O}_0 \subseteq \mathcal{O}$ be the ring of integers of the fixed field of all the conjugate self-twists of $\rho_{f,\iota_p}$. Like Ribet and Momose, Nekovář constructs a division algebra $D$ over the fixed field $K_0$ of $\Sigma_f$ and proves an adelic open-image result, which implies $\mathcal{O}_0$-fullness when $D$ splits.

**Theorem 12.6 ([Nek12]).** For all but finitely many $\iota_p$, the representation $\rho_{f,\iota_p}$ is $\mathcal{O}_0$-full.

Our results depend on hypotheses on the reduction $\bar{\rho}_{f,\iota_p}$ of $\rho_{f,\iota_p}$ modulo the maximal ideal of $\mathcal{O}$.

**Theorem 12.7 (Our results recovering Nekovář at $\iota_p$).** Suppose that $p$ is odd and $\bar{\rho}_{f,\iota_p}$ is regular; if $\bar{\rho}_{f,\iota_p}$ is octahedral suppose further that it is good. Then $\rho_{f,\iota_p}$ is $\mathcal{O}_0$-full.

The proof is analogous to that of Theorem 12.4. In particular, the fact that the weight of $f$ is at least 2 at each infinite place means that $\rho_{f,\iota_p}$ has distinct Hodge-Tate weights, which implies that no twist of it has finite image (see Proposition 2.4 for context). Or apply [CEG, Lemma 3.2.12].

**12.3. Bianchi modular forms and generalizations.** Unlike Hilbert modular forms, which are automorphic forms on $\mathrm{GL}_2$ over totally real fields, Galois representations associated to automorphic forms of $\mathrm{GL}_2$ over CM fields have only been constructed relatively recently, and even then only under some technical assumptions. We briefly summarize how our results can be applied to that context.

Let $E$ be a CM field with maximal totally real subfield $F$. Fix an algebraic closure $\bar{E}$, and let $G_E := \mathrm{Gal}(\bar{E}/E)$. Let $\pi$ be a cuspidal automorphic representation of $\mathrm{GL}_2(\mathbb{A}_E)$, where $\mathbb{A}_E$ denotes

<footnote>(viii) Certainly, $K_p^\times Q$ normalizes. Conversely, if $(a \ b) \in \mathrm{GL}_2(K_p)$ normalizes $(\begin{smallmatrix} 0 & \pi \\ \pi & 0 \end{smallmatrix})$, then the off-diagonal relation gives us $\sigma((\pi a^2 - \pi^2 c^2)\delta^{-1}) = (\pi d^2 - b^2)\delta^{-1}$, where $\delta = ad - bc$; if it normalizes $(\begin{smallmatrix} 0 & \pi t \\ -t & 0 \end{smallmatrix})$, where $t \in L$ with $\sigma(t) = -t$, then the off-diagonal gives $\sigma((\pi a^2 + \pi^2 c^2)\delta^{-1}) = (\pi d^2 + b^2)\delta^{-1}$; if it normalizes $(\begin{smallmatrix} t & 0 \\ -t & 0 \end{smallmatrix})$ then the off-diagonal is $\sigma(a(\pi c)\delta^{-1}) = db\delta^{-1}$. Combining the first two, we obtain $\sigma(a^2\delta^{-1}) = d^2\delta^{-1}$ and $\sigma((\pi c)^2\delta^{-1}) = b^2\delta^{-1}$; adding the third gives us, for example if $a$ is invertible, $\sigma(\pi c/a) = b/d$. By considering the diagonal relations, we also get $\sigma(b/a) = \pi c/d$ and $\sigma(a/d) = d/a$. In other words any normalizing element with a nonzero entry in the upper left looks like $(\begin{smallmatrix} a & ua\sigma(\pi c/a) \\ c & ua \end{smallmatrix})$, with $a \in K_p^\times$ arbitrary, $c \in aL$, and $u := d/a \in L$ of norm 1. From Hilbert 90, any norm-1 $u$ is $\sigma(\alpha)/\alpha$ for some $\alpha \in L$; letting $x := a/\alpha$ and $\beta := c\alpha/a$ puts our matrix in the desired form $x(\begin{smallmatrix} \alpha & \pi\sigma(\beta) \\ \beta & \sigma(\alpha) \end{smallmatrix}) \in K_p^\times Q$. Or see [Nek12, B.1.6] for a more conceptual argument.</footnote>
the adeles of $E$; when $E$ is imaginary quadratic, $\pi$ is called a Bianchi modular form. Assume that $\pi$ is of cohomological type with central character $\omega$. Following Mok [Mok14], assume moreover that $\omega$ arises from an algebraic idele class character $\tilde{\omega}$ on $\mathbb{A}_F^\times$ via the norm map and that $\tilde{\omega} = \otimes_v \tilde{\omega}_v$ such that $\tilde{\omega}_v(-1)$ takes the same value for all archimedean places of $F$. (When $F = \mathbb{Q}$, this is simply the condition that $\omega$ is invariant under complex conjugation.) Suppose there is no nontrivial quadratic character $\delta$ of $E$ such that $\pi \cong \pi \otimes \delta$; this is analogous to the non-CM assumption present in Section 12.1 and Section 12.2.

For each rational prime $p$ and fixed embedding $\iota_p: \overline{\mathbb{Q}} \hookrightarrow \overline{\mathbb{Q}}_p$, associated to $\pi$ there is a continuous irreducible representation $\rho_{\pi, \iota_p}: G_E \to \text{GL}_2(\overline{\mathbb{Q}}_p)$, which we may view as having coefficients in the ring of integers $\mathcal{O}$ of some finite extension of $\mathbb{Q}_p$. In this generality, the existence of $\rho_{\pi, \iota_p}$ is due to Mok [Mok14], who generalized the construction of Taylor in the imaginary quadratic case [Tay94]. Mok also shows, building on work of Berger and Harcos in the imaginary quadratic case [BH07], that $\rho_{\pi, \iota_p}$ is unramified outside a finite set of places and hence factors through a $p$-finite group $\Pi$.

Let $\mathcal{O}_0$ be the ring of integers of the fixed field of all the conjugate self-twists of $\rho_{\pi, \iota_p}$.

**Theorem 12.8.** Suppose $p$ is odd and $\bar{\rho}_{\pi, \iota_p}$ is regular and good if octahedral. Then $\rho_{\pi, \iota_p}$ is $\mathcal{O}_0$-full.

The proof is analogous to that of Theorem 12.4 and Theorem 12.7. The Hodge-Tate weights of $\rho_{\pi, \iota_p}$ are distinct by [Mok14, Theorem 5.17].

To our knowledge, Theorem 12.8 is the most general fullness result in the literature in this context, though Taylor proves the weaker theorem that the image of $\rho_{\pi, \iota_p}$ is Zariski dense in the imaginary quadratic case [Tay94, Corollary 2]. Our Theorem 12.8 may be well known to experts.

**Remark 12.9.** In contrast to the case of elliptic or Hilbert modular forms that can be $p$-adically interpolated in families with dense classical points and thus have associated “big” Galois pseudorepresentations (see Sections 12.4 to 12.6), a $p$-adic family of Bianchi modular forms often has only finitely many classical points [CM09, Theorem 8.9], [Ser19, Theorem 1.1]. Therefore no Galois pseudorepresentations have been attached to Bianchi families by conventional methods. $\Delta$

**12.4. Hida $p$-adic families of modular forms.** In this section, we explore the extent to which our methods recover known big-image results for Galois representations attached to ordinary $p$-adic families of modular forms, often called Hida families.

Fix $p > 2$, and let $A$ be the ring corresponding to a primitive non-CM irreducible component of Hida's cuspidal shallow Hecke algebra parametrizing $p$-ordinary cuspforms of some fixed tame level and weight $k$ such that $k - 1 \equiv i \pmod{p-1}$ for some fixed $0 \le i \le p-1$ (for $i = 0$ this is the ring $\mathbb{I}'$ on [Lan16, p. 158]). Then $A$ is a finite extension of the Iwasawa algebra $\Lambda := \mathbb{Z}_p[1 + p\mathbb{Z}_p] \cong \mathbb{Z}_p[T]$, which parametrizes the corresponding component of weight space. Let $\mathbb{F}$ be the residue field of $A$ and $K$ the fraction field of $A$.

Let $(t, d): G_{\mathbb{Q}} \to A$ be the pseudorepresentation obtained from gluing together those attached to the classical cuspforms in the family, all of which have the same semisimplifed residual representation $\bar{\rho}: G_{\mathbb{Q}} \to \text{GL}_2(\mathbb{F})$. Let $\varepsilon_p: G_{\mathbb{Q}} \to \mathbb{Z}_p^\times$ be the $p$-adic cyclotomic character and $\langle \cdot \rangle: \mathbb{Z}_p^\times \to 1 + p\mathbb{Z}_p$ the projection onto the pro-$p$ part of $\mathbb{Z}_p^\times$. The weight character $\kappa: G_{\mathbb{Q}} \to \Lambda^\times$ is given by $\kappa(g) = (1 + T)^{\langle \varepsilon_p(g) \rangle}$. Let $\chi$ be the tame Dirichlet character associated to the family. The determinant of $\rho$ is given by

$$ (9) \qquad d = \kappa \chi s(\bar{\varepsilon}_p)^i, $$

Both $\bar{\rho}$ and $(t, d)$ factor through $\Pi$, the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside $p$ and the level, a $p$-finite profinite group. Let $\Pi_p \subset \Pi$ be a decomposition group at $p$ and $I_p \subset \Pi_p$ its inertia subgroup. The ordinary condition guarantees that there exists a $(t, d)$-representation $\rho$, which we view as $\text{GL}_2(K)$-valued by Lemma 2.10, with $\rho|_{\Pi_p} = (\epsilon_0^* \delta)$, where $\delta$ an unramified character and $\epsilon$ coincides with $\kappa$ on wild inertia and therefore surjects onto $(1 + T)^{1+p\mathbb{Z}_p}$ [Hid12, Theorem 4.3.2].
The image of $\rho$ has been studied by Boston [MW86, Appendix], Fischman [Fis02], Hida [Hid15], and Lang [Lan16]. The latter two are the more recent and most general results, so we focus there.

**Theorem 12.10** (Λ-fullness for Hida families, Hida [Hid15]). If $\bar{\rho}$ restricted to $\Pi_p$ is multiplicity free and $\rho$ is realizable by a representation over $A$ then $\rho$ is Λ-full.

Hida's Λ-fullness strongly suggested that every conjugate self-twist of $(t, d)$ should fix Λ (see Theorem 5.4 here for a proof of this fact). Following Hida, Lang analyzes how the image of $\rho$ is constrained by $\Sigma_t(A/\Lambda)$, the conjugate self-twists that fix Λ pointwise. To state her result, we let $H_Λ \subseteq \Pi$ be the intersection of all the $\ker \eta$ for $(\sigma, \eta)$ in $\Sigma_t(A/\Lambda)$.

**Theorem 12.11** (Big image for Hida families, [Lan16, Theorem 2.4]). Suppose that $\mathbb{F} \neq \mathbb{F}_3$, that $\bar{\rho}$ is absolutely irreducible, and that there is an element in $\bar{\rho}(H_Λ \cap \Pi_p)$ whose eigenvalue ratio is in $\mathbb{E}^\times \setminus \{1\}$.$^{(ix)}$ Then $\rho$ is $A^{\Sigma_\rho(A/\Lambda)}$-full.

A posteriori Lang's fullness result by itself justifies considering only those conjugate self-twists that fix $\Lambda$, even without Hida's $\Lambda$-fullness: see Theorem 5.4. Our work both recovers virtually all of Lang's result (exception: [Lan16] is able to handle some cases where $\mathbb{P}\bar{\rho}$ is the Klein-4 group) and extends it to include residually reducible $\bar{\rho}$. Let $A_0$ be the adjoint trace ring of $\rho$; see Definition 4.6. For the notion of $A_0$-constant determinant, see Definition 4.14; for good octahedral $\bar{\rho}$ see Definition 8.10.

**Theorem 12.12** (Our work recovering [Lan16]).

(1) If $\bar{\rho}$ is regular and good if octahedral, then $\rho$ is $A_0$-full.

(2) If $\Pi_p$ contains a regular element for $\bar{\rho}$, then $A_0$ contains $\Lambda$.

Consequently, if $\Pi_p$ contains a regular element for $\bar{\rho}$ and $\bar{\rho}$ is further good if octahedral, then

(3) $\rho$ is $\Lambda$-full;

(4) $\rho$ is $A^{\Sigma_\rho(A)}$-full;

(5) every conjugate self-twist of $\rho$ fixes $\Lambda$, so that $\Sigma_\rho(A) = \Sigma_\rho(A/\Lambda)$ and $\rho$ is $A^{\Sigma_\rho(A/\Lambda)}$-full.

*Proof.* For (1), the representation $\rho$ is not reducible since the Hida family is cuspidal, and it is not dihedral since the Hida family is not CM. The fact that $\rho(\Pi) \not\cong \bar{\rho}(\Pi)$ follows from the fact that a Hida family has classical specializations of weight at least 2. Therefore we know that $\rho$ is $A_0$-full by Theorem 10.3.

For (2), let $d_1: \Pi \to A^\times$ be the pro-$p$ part of $d = \det \rho$, and let $\rho' = d_1^{-1/2} \otimes \rho$ with $(t', d') = (\mathrm{tr} \, \rho', \det \rho')$ the constant-determinant (pseudo)representation of $\rho$. Note that $\rho'|_{\Pi_p}$ is still upper triangular since $\rho|_{\Pi_p}$ is. Let $g_0 \in \Pi_p$ be a regular element with residual eigenvalues $\lambda_0, \mu_0$, and let $r$ be a $(t', d')$-representation adapted to $(g_0, \lambda_0, \mu_0)$. By the proof of [Bel19, Theorem 6.2.1], we see that, up to replacing $g_0$ with the limit of a sequence of its powers, we may assume that $r(g_0) = \begin{pmatrix} s(\lambda_0) & 0 \\ 0 & s(\mu_0) \end{pmatrix}$. Viewing both $\rho'$ and $r$ as $\mathrm{GL}_2(K)$-valued by Lemma 2.10, we see that they are isomorphic since they have the same trace and are irreducible. In particular, $\rho'(\Pi_p)$ contains an element with eigenvalues $s(\lambda_0), s(\mu_0)$, which (up to swapping $\lambda_0$ and $\mu_0$) is necessarily of the form $M := \begin{pmatrix} s(\lambda_0) & * \\ 0 & s(\mu_0) \end{pmatrix}$. On the other hand, using the description of $\epsilon$ and $\delta$ above, we see that $\rho'(\Pi_p)$ contains $J := \begin{pmatrix} (1+T)^{1/2} & * \\ 0 & (1+T)^{-1/2} \end{pmatrix}$.

<footnote>$^{(ix)}$There is a small error in [Lan16], which we correct here. Theorem 2.4 as stated *loc. cit.* requires merely that $\bar{\rho}$ restricted to $H_\Lambda \cap \Pi_p$ be multiplicity free, but in fact the result relies on the stronger regularity condition given here. Indeed, on [Lan16, p. 174] the definition of $\overline{L}[\lambda]$ only makes sense if one knows $\overline{L}$ is closed under multiplication by $\lambda$, where $\lambda$ is an adjoint eigenvalue of the regular element.</footnote>
We compute adjoint-trace elements. Both $a := \frac{(\mathrm{tr} M)^2}{\det M} = \frac{(s(\lambda_0)+s(\mu_0))^2}{s(\lambda_0)s(\mu_0)} = 2 + \frac{s(\lambda_0)}{s(\mu_0)} + \frac{s(\mu_0)}{s(\lambda_0)}$ and

$$b := \frac{(\mathrm{tr} MJ)^2}{\det MJ} = \frac{(s(\lambda_0)(1+T) + s(\mu_0))^2}{s(\lambda_0)s(\mu_0)(1+T)} = \left( a + \left( 2 + 2\frac{s(\lambda_0)}{s(\mu_0)} \right) T + \frac{s(\lambda_0)}{s(\mu_0)} T^2 \right) (1+T)^{-1}.$$

are in $A_0$ by construction. The last expression shows that $b$ is in $W(\mathbb{E})[[T]]$, since $\lambda_0\mu_0^{-1} \in \mathbb{E}$ by the regularity assumption. Moreover, the $T$-coefficient of $b$ is $2 + 2\frac{s(\lambda_0)}{s(\mu_0)} - a = \frac{s(\lambda_0)}{s(\mu_0)} - \frac{s(\mu_0)}{s(\lambda_0)}$, which is in $W(\mathbb{E})^\times$ since $\lambda_0\mu_0^{-1} \neq \pm 1$. It follows that the closed $W(\mathbb{E})$-algebra generated by $b$ in $A_0$ is all of $W(\mathbb{E})[[T]]$. In other words $\Lambda \subseteq W(\mathbb{E})[[T]] \subseteq A_0$, as claimed.

For (3), combine (1) and (2). For (4), from (2) and the expression for $d$ in (9) $(t, d)$ has $A_0$-constant determinant. Now use Corollary 4.21. For (5), combine (3), Theorem 5.4, and (4). $\square$

**Remark 12.13.** The regularity-on-$\Pi_p$ hypothesis in Theorem 12.12(2) can easily be checked in terms of the data of the tame Nebentypus character $\chi$ and the mod-$p$ eigenvalue $a_p$ of $U_p$. Indeed, since $\delta$ sends Frobenius to $U_p$, to verify regularity we need to check whether $\bar{\epsilon}\bar{\delta}^{-1}$ takes on a value in $\mathbb{E}^\times \setminus \{\pm 1\}$. Writing $\chi|_{\Pi_p} = \chi^{\mathrm{unr}}\chi^{\mathrm{tame}}$ with $\chi^{\mathrm{unr}}$ unramified and $\chi^{\mathrm{tame}}$ a character on $\mu_{p-1}$ by local class field theory, we see that the tame part of $\bar{\epsilon}\bar{\delta}^{-1}$ is $\chi^{\mathrm{tame}}\bar{\epsilon}_p^i$ and the unramified part is $\chi^{\mathrm{unr}}\bar{\delta}^{-2}$. Note that the tame part necessarily takes values in $\mathbb{F}_p^\times$, so if $\chi^{\mathrm{tame}}\bar{\epsilon}_p^i$ has order greater than 2, then the regularity-on-$\Pi_p$ hypothesis is automatically satisfied. Otherwise, one must look to the unramified part and check whether some power of $\chi^{\mathrm{unr}}(p)a_p^{-2}$ lies in $\mathbb{E}^\times \setminus \{\pm 1\}$. $\triangle$

**Remark 12.14.** **Do all conjugate self-twists of $p$-adic families fix weight space?** In an abstract algebraic setting, given a representation of a profinite group $\Pi$ over a ring $A$ that is finite over $\Lambda = \mathbb{Z}_p[[T]]$, we cannot expect to prove that $\rho$ is $\Lambda$-full — equivalently, that every conjugate self-twist of $\rho$ fixes $\Lambda$ — because it is simply not true.

On the other hand, one intuitively expects $\Lambda$, which parametrizes weight space, to be preserved by any conjugate self-twist of a $p$-adic family. Indeed, if $(\sigma, \chi)$ is a conjugate self-twist that doesn't fix $\Lambda$, then modulo any prime ideal of $\Lambda$, the character $\chi$ will relate pairs of eigenforms of different $p$-adic weights — an implausible scenario. Therefore, although our $A_0$-fullness result in Theorem 12.12(1) is purely algebraic, in order to recover the full strength of [Lan16], and to match our intuition of how conjugate self-twists in $p$-adic families behave, we necessarily need to use some modular-form-theoretic input. For Hida families the ordinary condition suffices: see the proof of Theorem 12.12 above, especially part (2). We have not extended this result to Coleman families, which is a drawback on our big-image result in Theorem 12.15. Ultimately one hopes to formalize the geometric intuition alluded to above. $\triangle$

**12.5. Coleman $p$-adic families of classical modular forms.** We now relax the ordinary assumption present in Section 12.4 and derive the consequences of our main theorem in the context of pseudorepresentations arising from the Coleman-Mazur eigencurve, comparing with known results in the literature.

Let $X$ be a cuspidal irreducible component of the $p$-adic Coleman-Mazur eigencurve of some fixed tame level ([Buz07], [CM98]); having dealt with the ordinary case in Section 12.4, we assume that $X$ is nonordinary. Let $A$ be the ring of analytic functions on $X$ bounded by 1. It is a compact $\mathbb{Z}_p$-algebra (hence pro-$p$) since $X$ is nested [BC09, Lemma 7.2.11(ii), Corollary 7.2.12]. In fact, the map from $X$ to weight space endows $A$ with a $\Lambda$-algebra structure. As usual, $A$ is a local domain since $X$ is irreducible. As in the Hida family setting, one obtains a 2-dimensional pseud representation $(t, d): G_{\mathbb{Q}} \to A$ by gluing together those attached to classical cuspforms in the family, all of which have the same semisimplified residual representation $\bar{\rho}: G_{\mathbb{Q}} \to \mathrm{GL}_2(\mathbb{F})$. Then $(t, d)$ is unramified outside of a finite set of primes, namely $p$ and the tame level, and thus factors through a $p$-finite quotient $\Pi$ of $G_{\mathbb{Q}}$. Unlike the theorems presented in Section 12.1 through Section 12.4, there are no previous fullness results known for $(t, d)$ (though see Remark 12.16 below), so we proceed directly to a statement of our result in this setting.
**Theorem 12.15.** If $\bar{\rho}$ is regular and good if octahedral, then $(t, d)$ is $A_0$-full.

*Proof.* By Theorem 10.1 it suffices to show that $(t, d)$ is not a priori small. It is not reducible since it is cuspidal. Any $(t, d)$-representation does not have finite image since $X$ admits classical specializations of weight at least 2. Finally, since $X$ is nonordinary its CM points are isolated and hence $X$ necessarily admits a classical non-CM positive slope specialization: see [CIT16, Corollary 3.6]. Thus $(t, d)$ is not dihedral. □

*Remark 12.16.* Besides Bellaïche's work [Bel19], the only previous work on images of Galois representations of finite slope $p$-adic families of modular forms was done in [CIT16]. We briefly compare Theorem 12.15 to their main result [CIT16, Theorem 1.3].

* **Setup:** Rather than working with $A$ as above, they restrict to an irreducible component $\mathbb{I}^\circ$ of what they call the “adapted slope $\le h$ Hecke algebra” — essentially a bounded-slope piece of $X$ as above: see [CIT16, §3.1]. Note that one can replace $A$ above by $\mathbb{I}^\circ$ and retain the veracity of Theorem 12.15.

* **Assumptions on $\bar{\rho}$:** In [CIT16, Theorem 1.3] the authors assume that $\bar{\rho}$ is absolutely irreducible, even when restricted to the intersection of the kernels of twist characters. Moreover, their regularity assumption is stronger than ours in that it requires the mod-$p$ eigenvalues of the regular element to be in $\mathbb{F}_p^\times$ rather than requiring their ratio to lie in $\mathbb{E}^\times$.

* **Conclusion:** As mentioned above, [CIT16, Theorem 1.3] is not a true fullness result. Rather, it shows “rigid-Lie fullness” — a certain rigid analytic Lie algebra attached to the image of a $(t, d)$-representation contains the rigid analytic Lie algebra of a congruence subgroup. While highly suggestive, one does not know how to recover an actual congruence subgroup in the image from this result. Following [Lan16], Conti and his coauthors show rigid-Lie fullness with respect to the ring fixed by $\Sigma_t(\mathbb{I}^\circ/\Lambda)$. $\Delta$

*Remark 12.17.* Although the determinant $d$ has a form similar to (9) — the universal character $\kappa$ times a finite-order character — in this setting we have not proved that $d$ is $A_0$-constant: we do not know whether the image of $\Lambda$ is contained in $A_0$. See Remark 12.14 for why one expects this to be true nonetheless. $\Delta$

## 12.6. $p$-Adic families of Hilbert modular forms.
Since our methods are agnostic about the group $\Pi$, one can proceed with a similar analysis in the context of $p$-adic families of Hilbert modular forms, which we briefly outline here. We believe these are the first big image results in this context.

As in Section 12.2, fix a totally real field $F$. Let $X$ be a cuspidal irreducible component of a $p$-adic eigenvariety interpolating classical Hilbert modular forms over $F$ of a fixed tame level; there are several possible constructions, for instance [Urb11] or [AIP16]. Let $A$ be the ring of analytic functions on $X$ bounded by 1, which is again a pro-$p$ local domain. Gluing together the pseudorepresentations attached to the classical cusp forms parametrized by $X$, all of which have the same semisimplified residual representation $\bar{\rho}: G_F \to \text{GL}_2(\mathbb{F})$, yields a 2-dimensional pseudorepresentation $(t, d): G_F \to A$. It is unramified outside the tame level and $p$ and hence factors through a $p$-finite quotient $\Pi$ of $G_F$.

**Theorem 12.18.** Suppose that $\bar{\rho}$ is regular and good if octahedral. If $X$ admits a non-CM classical specialization, then $(t, d)$ is $A_0$-full.

*Proof.* By Theorem 10.1 it suffices to check that $(t, d)$ is not a priori small, which follows from the fact that $X$ admits classical specializations that are cuspidal, not CM, and whose weights are at least 2. □

*Remark 12.19.* As in Section 12.5, we do not know whether $(t, d)$ has $A_0$-constant determinant in this case and hence lack an $A_0^{\Sigma_t}$-fullness result. As in Remark 12.14, we expect the image in $A$ of the ring $\Lambda_F$ of analytic functions on weight space to be contained in $A_0$. Note that in this case, $\Lambda_F$ is a power series ring over $\mathbb{Z}_p$; the number of variables depends on the totally real field $F$. $\Delta$
**A.1. Representations with isomorphic adjoint differ by a character.** Throughout Appendix A.1, let $G$ be a group and $F$ a separably closed field of odd characteristic. All representations are assumed to be finite-dimensional. Let $sl_n(F)$ denote the $F$-vector space of $n \times n$-matrices of trace 0 and $ad^0: GL_n(F) \to GL_{n^2-1}(F)$ the representation obtained by letting $GL_n(F)$ act on $sl_n(F)$ by conjugation. The primary goal of this section is to prove that if $\rho_1, \rho_2: G \to GL_2(F)$ are semisimple representations such that $ad^0 \rho_1 \cong ad^0 \rho_2$, then $\rho_1 \cong \rho_2 \otimes \eta$ for some character $\eta: G \to F^\times$. This is done in Theorem A.10. The easier case when the $\rho_i$ are not dihedral is treated first in Appendix A.1.1. Appendix A.1.2 is an analysis of dihedral representations that allows us to conclude Theorem A.10 in full generality. The results of this section are probably well known to experts, but we give proofs for lack of a reference in the generality we need. We were guided by Venkatarama's answer to MathOverflow question 297746. In the nondihedral case, this result can be found in [KMP00, Lemma 2.9]. When the representations $\rho_1$ and $\rho_2$ arise from classical modular forms, the result can be found in [DK00, Appendix].

**A.1.1. The nondihedral case.** Given a representation $\rho: G \to GL_n(F)$, we write $\rho^*$ for its dual representation. That is, if $V$ is the representation space of $\rho$, then $V^* := \text{Hom}(V, F)$ is the representation space of $\rho^*$ with $G$-action given by $(g\varphi)(v) := g\varphi(g^{-1}v)$. In terms of matrices, if we fix a basis for $V$ and take the dual basis for $V^*$, then $\rho^*(g)$ is the inverse transpose of $\rho(g)$.

If $\rho$ is 2-dimensional, then an explicit calculation shows that $\rho^* \cong \rho \otimes \Lambda^2 \rho^*$, where $\Lambda^2$ denotes the second exterior power of $\rho$. (The conjugating matrix can be taken to be $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.) We have that

$$1 \oplus ad^0 \rho \cong \rho \otimes \rho^*.$$

In particular, $ad^0 \rho$ is self dual. Furthermore,

$$1 \oplus ad^0 \rho \cong \rho \otimes \rho^* \cong \rho \otimes \rho \otimes \Lambda^2 \rho^* \cong 1 \oplus (\text{Sym}^2 \rho \otimes \Lambda^2 \rho^*),$$

$$\text{and so } ad^0 \rho \cong \text{Sym}^2 \rho \otimes \Lambda^2 \rho^* = \text{Sym}^2 \rho \otimes \det \rho^{-1}.$$

The following lemma is essentially a version of Schur's lemma that will be useful in what follows.

**Lemma A.1.** If $\rho: G \to GL_n(F)$ is a semisimple representation such that $ad^0 \rho$ contains a copy of the trivial representation, then $\rho$ is reducible.

*Proof.* Let $V$ be the $F$-vector space on which $G$ acts via $\rho$. Then $\text{End} V$ is the representation space for $1 \oplus ad^0 \rho$, where 1 is the trivial representation, which corresponds to scalar endomorphisms of $V$. If $ad^0 \rho$ contains a copy of the trivial representation, then there is a nonscalar $\varphi \in \text{End} V$ that commutes with the action of $G$. By Schur's lemma, $\rho$ must be reducible. $\square$

**Lemma A.2.** Let $\rho_1, \rho_2: G \to GL_2(F)$ be semisimple reducible representations. If $ad^0 \rho_1 \cong ad^0 \rho_2$, then there exists a character $\eta: G \to F^\times$ such that $\rho_1 \cong \eta \otimes \rho_2$.

*Proof.* By assumption there exist for $i=1,2$ characters $\lambda_i, \mu_i: G \to F^\times$ such that $\rho_i \cong \lambda_i \oplus \mu_i$. It is straightforward to calculate

$$\lambda_1 \mu_1^{-1} \oplus 1 \oplus \lambda_1^{-1} \mu_1 \cong ad^0 \rho_1 \cong ad^0 \rho_2 \cong \lambda_2 \mu_2^{-1} \oplus 1 \oplus \lambda_2^{-1} \mu_2.$$

Thus, up to switching $\lambda_2$ and $\mu_2$, we must have $\lambda_1 \mu_1^{-1} = \lambda_2 \mu_2^{-1}$. Let $\eta = \mu_1 \mu_2^{-1}$. Then

$$\rho_1 \cong \lambda_1 \oplus \mu_1 = \lambda_2 \mu_1 \mu_2^{-1} \oplus \mu_1 = (\mu_1 \mu_2^{-1}) \otimes (\lambda_2 \oplus \mu_2) \cong \eta \otimes \rho_2.$$

$\square$

**Lemma A.3.** Let $\rho_1, \rho_2: G \to GL_2(F)$ be semisimple representations such that both $ad^0 \rho_i$ are irreducible. If $ad^0 \rho_1 \cong ad^0 \rho_2$, then there exists a character $\eta: G \to F^\times$ such that $\rho_1 \cong \eta \otimes \rho_2$.
*Proof.* We begin by showing that $\rho_1 \otimes \rho_2$ must be reducible (which does not make use of the assumption that $\mathrm{ad}^0 \rho_i$ is irreducible). Indeed, by Lemma A.1 if $\rho_1 \otimes \rho_2$ were irreducible then its endomorphism ring would contain a single copy of the trivial representation. But

$$ \begin{aligned} \mathrm{End}(\rho_1 \otimes \rho_2) &= (\rho_1 \otimes \rho_2) \otimes (\rho_1 \otimes \rho_2)^* \cong (\rho_1 \otimes \rho_1^*) \otimes (\rho_2 \otimes \rho_2^*) \\ &\cong (1 \oplus \mathrm{ad}^0 \rho_1) \otimes (1 \oplus \mathrm{ad}^0 \rho_2) \\ &\cong 1 \oplus \mathrm{ad}^0 \rho_1 \oplus \mathrm{ad}^0 \rho_1 \oplus (\mathrm{ad}^0 \rho_1 \otimes \mathrm{ad}^0 \rho_1) \\ &\cong 1 \oplus \mathrm{ad}^0 \rho_1 \oplus \mathrm{ad}^0 \rho_1 \oplus (\mathrm{ad}^0 \rho_1 \otimes (\mathrm{ad}^0 \rho_1)^*), \end{aligned} $$

and $\mathrm{ad}^0 \rho_1 \otimes (\mathrm{ad}^0 \rho_1)^* \cong \mathrm{End}(\mathrm{ad}^0 \rho_1)$ contains a copy of the trivial representation, a contradiction.

Next we show that $\rho_1 \otimes \rho_2$ cannot be the sum of two 2-dimensional representations. Indeed, suppose that $\rho_1 \otimes \rho_2 \cong r_1 \oplus r_2$, where $r_1, r_2: G \to \mathrm{GL}_2(F)$ are representations. Take the second exterior product on both sides. We have

$$ \Lambda^2(\rho_1 \otimes \rho_2) \cong (\Lambda^2 \rho_1 \otimes \mathrm{Sym}^2 \rho_2) \oplus (\mathrm{Sym}^2 \rho_1 \otimes \Lambda^2 \rho_2) $$

and

$$ \Lambda^2(r_1 \oplus r_2) \cong \Lambda^2 r_1 \oplus \Lambda^2 r_2 \oplus (r_1 \otimes r_2). $$

Since $\mathrm{ad}^0 \rho_i \cong \mathrm{Sym}^2 \rho_i \otimes \Lambda^2 \rho_i^*$, we have $\mathrm{Sym}^2 \rho_1 \otimes \Lambda^2 \rho_2 \cong \mathrm{Sym}^2 \rho_2 \otimes \Lambda^2 \rho_1$. But if

$$ (\Lambda^2 \rho_1 \otimes \mathrm{Sym}^2 \rho_2)^{\oplus 2} \cong \Lambda^2 r_1 \oplus \Lambda^2 r_2 \oplus (r_1 \otimes r_2), $$

then this contradicts irreducibility of $\mathrm{ad}^0 \rho_i$. Thus $\rho_1 \otimes \rho_2$ must contain a 1-dimensional representation; call it $\chi$. Then we claim that $\rho_2 \cong \rho_1^* \otimes \chi \cong \rho_1 \otimes \det \rho_1^{-1} \otimes \chi$, and so $\rho_1$ and $\rho_2$ differ by a twist.

To see that $\rho_2 \cong \rho_1^* \otimes \chi$, recall that $\rho_1 \otimes \rho_2 \cong \mathrm{Hom}(\rho_1^*, \rho_2)$. Thus having a 1-dimensional $G$-stable subspace corresponds to a nonzero linear map $\varphi: \rho_1^* \to \rho_2$ such that $g\varphi = \lambda(g)\varphi$ for some $\lambda(g) \in F^\times$ for all $g \in G$. Define $f: \rho_1^* \to \rho_2 \otimes \chi^{-1}$ by $v \mapsto \varphi(v) \otimes e$, where $e$ is a basis for the 1-dimensional vector space on which $G$ acts by $\chi$. Note that $f \neq 0$ since $\varphi \neq 0$. It is straightforward to check that $f(gv) = gf(v)$ for all $g \in G$. Therefore $\mathrm{Hom}(\rho_1^*, \rho_2 \otimes \chi^{-1}) \neq 0$. Since $\rho_1^*$ and $\rho_2 \otimes \chi^{-1}$ are irreducible, it follows that they must be isomorphic. $\square$

The following observation can be checked easily via a direct calculation on $2 \times 2$ matrices.

**Lemma A.4.** For any $g \in \mathrm{GL}_2(F)$ with (not necessarily distinct) eigenvalues $\lambda, \mu$, the eigenvalues of $\mathrm{ad}^0 g$ are $1, \lambda\mu^{-1}, \lambda^{-1}\mu$. In particular, we have

$$ \mathrm{tr} \, \mathrm{ad}^0 g = \frac{\mathrm{tr}(g)^2}{\det(g)} - 1. $$

A.1.2. The dihedral case. In Appendix A.1.2 we assume for simplicity that the characteristic of $F$ is not equal to 2. The goal of Appendix A.1.2 is to remove the assumption that both $\rho_i$ are reducible or both $\mathrm{ad}^0 \rho_i$ are irreducible from Lemmas A.2 and A.3. We begin with a lemma that shows that, in light of Lemmas A.2 and A.3, we only need to consider the case when both $\rho_1$ and $\rho_2$ are dihedral representations.

**Lemma A.5.** If $\rho: G \to \mathrm{GL}_2(F)$ is irreducible but $\mathrm{ad}^0 \rho$ is reducible, then $\rho$ is dihedral.

*Proof.* If $\mathrm{ad}^0 \rho$ is reducible, then so is $\mathrm{Sym}^2 \rho$ and $\mathrm{Sym}^2 \rho^*$ since $\mathrm{ad}^0 \rho \cong \mathrm{Sym}^2 \rho \otimes \det \rho^{-1}$. But $\mathrm{Sym}^2 \rho^*$ can be identified with the action of $G$ on the $F$-vector space of quadratic forms on $F^2$. Thus, there is a quadratic form $Q$ on which $G$ acts by a scalar. Since $F$ is separably closed and $\mathrm{char}F \neq 2$, all quadratic forms are equivalent. In particular, we may assume that $Q(x, y) = xy$. But one checks immediately that the only matrices that preserve $Q$ up to scalars are diagonal and antidiagonal. Thus $\rho$ must be dihedral. $\square$

The rest of this section is devoted to an analysis of dihedral representations.
**Lemma A.6.** Assume that $\rho: G \to \text{GL}_2(F)$ is a semisimple representation. If $\rho \cong \eta \otimes \rho$ for some nontrivial character $\eta: G \to F^\times$, then the image of $\rho|_{\ker \eta}$ is abelian.

*Proof.* This argument essentially comes from [Rib77, Proposition 4.4]. Note that $\det \rho = \eta^2 \det \rho$ and so $\eta^2 = 1$. Set $H := \ker \eta$. Thus $[G: H] = 2$ since $\eta$ is nontrivial. By assumption, there is a matrix $M \in \text{GL}_2(F)$ such that $M\rho(g)M^{-1} = \eta(g)\rho(g)$ for all $g \in G$. In particular, $\rho(H)$ is contained in the commutant of $M$.

We claim that $M$ is semisimple. It suffices to show that $M$ has distinct eigenvalues. Up to a change of basis for $\rho$, we may assume that $M$ is upper triangular, say $M = (a_0^b)^{-1}$. The eigenvalues of $M$ acting on $M_2(F)$ by conjugation are $1, 1, ac^{-1}, a^{-1}c$ by Lemma A.4. Note that for any $g \in G \setminus H$, we have

$$M\rho(g)M^{-1} = -\rho(g).$$

Thus $-1 = ac^{-1}$, which implies that $a \neq c$ and thus $M$ has distinct eigenvalues, as claimed. Therefore $M$ is semisimple and so its commutant, and hence $\rho(H)$, is abelian. $\square$

If $H$ is a subgroup of $G$ of index 2, then we use $c$ to denote a fixed element in $G \setminus H$. For a character $\chi: H \to F^\times$ and $g \in G$, we write $\chi^g: H \to F^\times$ for the character defined by $\chi^g(h) := \chi(g^{-1}hg)$. It is not difficult to check that $\chi^g$ depends only on the coset of $g$ in $G/H$. Set $\chi^- := \chi/\chi^c$. We will write $\eta_H: G \to G/H \cong \{\pm 1\}$ for the canonical projection map. With this notation, we recall an explicit description of $\text{Ind}_H^G \chi$. Namely, $\text{Ind}_H^G \chi$ is isomorphic to the representation

$$(10) \qquad g \mapsto \begin{cases} \begin{pmatrix} \chi(g) & 0 \\ 0 & \chi^c(g) \end{pmatrix} & \text{if } g \in H \\ \begin{pmatrix} 0 & \chi(gc) \\ \chi^c(gc^{-1}) & 0 \end{pmatrix} & \text{otherwise.} \end{cases}$$

Using Frobenius reciprocity it is easy to see that $\text{Ind}_H^G \chi$ is irreducible if and only if $\chi \neq \chi^c$.

**Lemma A.7.**

(1) If $\rho = \text{Ind}_H^G \chi$ for a character $\chi: H \to F^\times$ and $[G: H] = 2$, then $\rho \cong \rho \otimes \eta_H$.

(2) Conversely, if $\rho: G \to \text{GL}_2(F)$ is a dihedral representation, then there is a subgroup $H$ of $G$ of index 2 and a character $\chi: H \to F^\times$ such that $\rho \cong \text{Ind}_H^G \chi$ and $\chi \neq \chi^c$.

(3) Furthermore, $H$ as in (2) is unique unless $\chi^2 = (\chi^c)^2$.

(4) If $\chi^2 = (\chi^c)^2$ then there are exactly three index 2 subgroups $H_i$ of $G$ for $i = 1, 2, 3$ for which there exist characters $\chi_i: H_i \to F^\times$ such that $\rho \cong \text{Ind}_{H_i}^G \chi_i$.

*Proof.* For the first point, note that $\chi$ is a constituent of $(\rho \otimes \eta_H)|_H = \rho|_H$. By Frobenius reciprocity and dimension counting, it follows that $\text{Ind}_H^G \chi \cong \rho \otimes \eta_H$.

If $\rho$ is dihedral, then there is a nontrivial character $\eta: G \to F^\times$ such that $\text{tr} \rho = \eta \text{tr} \rho$ and $\det \rho = \eta^2 \det \rho$. In particular, $\eta^2 = 1$ and so $\eta$ is a quadratic character. Let $H := \ker \eta$. Then $H$ is a subgroup of $G$ of index 2 and $\rho|_H$ is reducible by Lemma A.6. Let $\chi: H \to F^\times$ be one of the constituents of $\rho|_H$. By Frobenius reciprocity, $\text{Ind}_H^G \chi$ is a constituent of $\rho$ and we deduce equality for dimension reasons. Thus we have $\rho|_H = \chi \oplus \chi^c$. Since $\rho$ is irreducible by the definition of being dihedral, it follows by Frobenius reciprocity that $\chi \neq \chi^c$. This finishes the proof of the second point.

For the third point, suppose that $\rho = \text{Ind}_{H'}^G \chi'$ for some character $\chi': H' \to F^\times$ and $[G: H'] = 2$. Let $c' \in G \setminus H'$. Then by restricting to $H$ we have $\chi \oplus \chi^c = (\eta_{H'})|_H \cdot \chi \oplus (\eta_{H'})|_H \cdot \chi^c$. Thus we either have $\chi = (\eta_{H'})|_H \cdot \chi$ or $\chi = (\eta_{H'})|_H \cdot \chi^c$. In the first case, we see that $H = \ker \eta_{H'} = H'$. In the second case we conclude that $\chi^2 = (\chi^c)^2$ since $\eta_{H'}$ is quadratic.
Finally, suppose that $\chi^2 = (\chi^c)^2$. Then $H_0 := \ker(\chi/\chi^c)$ is a subgroup of index 2 in $H$. We claim that $H_0$ is normal in $G$. Recall that $\chi^c$ is independent of the choice of $c \in G \setminus H$. If $h \in H_0$ and $g \in G \setminus H$ then

$$ \chi(g^{-1}hg)/\chi^c(g^{-1}hg) = \chi(g^{-1}hg)/\chi^g(g^{-1}hg) = \chi^g(h)/\chi(h) \\ = (\chi/\chi^g)(h)^{-1} = (\chi/\chi^c)(h)^{-1} = 1. $$

Furthermore, the above calculation shows that the class of $c$ generates a subgroup of $G/H_0$ of order 2 distinct from $H$. Thus $G/H_0$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$. We claim that if $H'$ is any of the three subgroups of $G$ of index 2 containing $H_0$, then there is a character $\chi': H' \to F^\times$ such that $\rho \cong \mathrm{Ind}_{H'}^G \chi'$. By Frobenius reciprocity, it suffices to show that $\rho|_{H'}$ is reducible. Since $\rho|_{H_0} = \chi|_{H_0} \oplus \chi^c|_{H_0}$, it follows from Frobenius reciprocity that $\rho|_{H'} = \mathrm{Ind}_{H_0}^{H'} \chi|_{H_0}$. But $\chi|_{H_0} = \chi^c|_{H_0}$ and so it follows (again by Frobenius reciprocity) that $\rho|_{H'}$ is reducible. $\square$

Combining the following lemma with Frobenius reciprocity, we see that the irreducibility of $\mathrm{Ind}_H^G \chi$ is related to the question of whether the character $\chi: H \to F^\times$ extends to a character of $G$.

**Lemma A.8.** Let $H$ be a subgroup of $G$ of index 2 and $\chi: H \to F^\times$ a character. Then there exists an extension of $\chi$ to a character $G \to F^\times$ if and only if $\chi = \chi^c$. If $\chi$ extends to a character of $G$, then there are exactly two different extensions, and they differ by $\eta_H$.

*Proof.* If such an $L$ and extension of $\chi$ exist, then certainly $\chi = \chi^c$. On the other hand, since $c^2 \in H$, we know that $\chi(c^2)$ is well defined. Since $F$ is algebraically closed, we may choose a square root $r$ of $\chi(c^2)$ in $F$. Define a new character $\tilde{\chi}: G \to L^\times$ by

$$ \tilde{\chi}(g) := \begin{cases} \chi(g) & \text{if } g \in H \\ r\chi(c^{-1}g) & \text{if } g \notin H. \end{cases} $$

To see that $\tilde{\chi}$ is a character, it suffices to verify that it is multiplicative. That is, one must check that $\tilde{\chi}(h)\tilde{\chi}(ch') = \tilde{\chi}(hch')$ and $\tilde{\chi}(ch)\tilde{\chi}(ch') = \tilde{\chi}(chch')$ for $h, h' \in H$. It is easy to see by direct computation that these are satisfied if $\chi = \chi^c$. $\square$

**Lemma A.9.** Let $\rho = \mathrm{Ind}_H^G \chi$ be a dihedral representation. Then $\mathrm{ad}^0 \rho \cong \eta_H \oplus \mathrm{Ind}_H^G \chi^-$. If $\mathrm{ad}^0 \rho$ is the sum of three characters, then $\chi^2 = (\chi^c)^2$ and $\mathrm{ad}^0 \rho \cong \eta_{H_1} \oplus \eta_{H_2} \oplus \eta_{H_3}$, where the $H_i$ are the index 2 subgroups of $G$ given in Lemma A.7.

*Proof.* The first claim is an explicit calculation. Let $e_1 := \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, $e_2 := \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, $e_3 := \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. Assume that $\rho$ is given by (10). Then with respect to the basis $e_1, e_2, e_3$ we see that

$$ \mathrm{ad}^0(g) = \begin{cases} \begin{pmatrix} 1 & 0 & 0 \\ 0 & \chi^-(g) & 0 \\ 0 & 0 & \chi^-(g)^{-1} \end{pmatrix} & \text{if } g \in H \\ \begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & \chi^-(gc) \\ 0 & \chi^-(gc^{-1})^{-1} & 0 \end{pmatrix} & \text{otherwise.} \end{cases} $$

We observe that $\eta_H$ appears in the upper left corner. Furthermore, $(\chi^-)^c = (\chi^-)^{-1}$. Therefore the lower right 2 × 2-matrix in $\mathrm{ad}^0 \rho$ is isomorphic to $\mathrm{Ind}_H^G \chi^-$ by (10). Thus $\mathrm{ad}^0 \rho \cong \eta_H \oplus \mathrm{Ind}_H^G \chi^-$. If $\mathrm{ad}^0 \rho$ is the sum of three characters, then $\mathrm{Ind}_H^G \chi^-$ is reducible and thus $\chi^- = (\chi^-)^c$. That is, $\chi^2 = (\chi^c)^2$. By Lemma A.7, it follows that there are exactly three subgroups $H_i$ of $G$ of index 2 for which $\rho \cong \mathrm{Ind}_{H_i}^G \chi_i$. By the above calculation, each $\eta_{H_i}$ must be a constituent of $\mathrm{ad}^0 \rho$. By counting dimensions, we find that $\mathrm{ad}^0 \rho \cong \eta_{H_1} \oplus \eta_{H_2} \oplus \eta_{H_3}$. $\square$

**Theorem A.10.** Let $F$ be a field whose characteristic is not 2. Let $\rho_1, \rho_2: G \to \mathrm{GL}_2(F)$ be semisimple representations. If $\mathrm{ad}^0 \rho_1 \cong \mathrm{ad}^0 \rho_2$ then there is a character $\eta: G \to L^\times$ such that $\rho_1 \cong \eta \otimes \rho_2$.

**Theorem A.10.** Let $F$ be a field whose characteristic is not 2. Let $\rho_1, \rho_2: G \to \mathrm{GL}_2(F)$ be semisimple representations. If $\mathrm{ad}^0 \rho_1 \cong \mathrm{ad}^0 \rho_2$ then there is a character $\eta: G \to L^\times$ such that $\rho_1 \cong \eta \otimes \rho_2$.
**Proof.** The case when either of $\rho_1$ or $\rho_2$ is not dihedral is settled by Lemmas A.2, A.3 and A.5. Therefore we may assume that both $\rho_1$ and $\rho_2$ are dihedral. By Lemma A.9 there are index-2 subgroups $H_i$ of $G$ and characters $\chi_i: H_i \to F^\times$ such that $\rho_i \cong \mathrm{Ind}_{H_i}^G \chi_i$. Note that the set of possible such $H_i$ can be read off from $\mathrm{ad}^0 \rho_i$ since $\eta_{H_i}$ is a constituent of $\mathrm{ad}^0 \rho_i$ by Lemma A.9 and $H_i = \ker \eta_{H_i}$. In particular, since $\mathrm{ad}^0 \rho_1 \cong \mathrm{ad}^0 \rho_2$, we may assume that $H := H_1 = H_2$. By Lemma A.9 we have $\mathrm{Ind}_H^G \chi_1^- \cong \mathrm{Ind}_H^G \chi_2^-$. By restricting to $H$ it follows that $\chi_1^- \oplus (\chi_1^-)^c \cong \chi_2^- \oplus (\chi_2^-)^c$, and so up to replacing $\chi_2$ with $\chi_2^c$ (which is okay since $\mathrm{Ind}_H^G \chi_2 \cong \mathrm{Ind}_H^G \chi_2^c$), it follows that $\chi_1^- = \chi_2^-$. That is, $\chi_1\chi_2^{-1} = (\chi_1\chi_2^{-1})^c$. By Lemma A.8 there is a character $\eta: G \to L^\times$ such that $\eta|_H = \chi_1\chi_2^{-1}$. We claim that $\rho_1 \cong \eta \otimes \rho_2$. Indeed, this is true upon restriction to $H$ since

$$ \rho_1|_H = \chi_1 \oplus \chi_1^c = \eta|_H \otimes (\chi_2 \oplus \chi_2^c) = (\eta \otimes \rho_2)|_H. $$

Therefore $\rho_1 \cong \eta \otimes \rho_2$ by Frobenius reciprocity since $\rho_1$ is irreducible and thus $\chi_1 \neq \chi_1^c$. $\square$

**A.2. Trace field extensions.** In this brief subsection, let $G$ be an abstract group, $F$ an abstract field, and $(t, d) : G \to F$ a (2-dimensional) pseudorepresentation.

**Definition A.11.** The trace field of $(t, d)$ is the subfield of $F$ generated by the image $t(G)$ over the prime subfield of $F$. A pseudorepresentation $(t, d) : G \to F$ is realizable over an extension $L$ of $F$ if there exists a semisimple representation $\rho : G \to \mathrm{GL}_2(L)$ that carries $(t, d)$ — that is, with $t = \mathrm{tr} \rho$ and $d = \det \rho$.

**Lemma A.12.** Let $(t, d) : G \to F$ is a pseudorepresentation with trace field $F$. If the characteristic of $F$ is not 2, then $(t, d)$ realizable over an at-most quadratic extension of $F$.

**Proof.** To start with, $(t, d)$ is always realizable by a semisimple representation $V$ over $\bar{F}$ [Che14, Theorem 2.12]. If we suppose that $V$ is irreducible, then the image of the associated $F$-algebra map $\rho : F[G] \to \mathrm{End}_{\bar{F}}(V)$ surjects onto the full matrix algebra $\mathrm{End}_{\bar{F}}(V)$ after extending scalars to $\bar{F}$, and is therefore a quaternion algebra $D$ over $F$. There are now two possibilities. Either $D$ is split, in which case $D^\times \cong \mathrm{GL}_2(F)$ is a realization of $(t, d)$ over $F$. Or $D$ is an $F$-division algebra, in which case $\rho|_G : G \to D^\times$ carries $(t, d)$, in the sense that $\rho(g)$ has reduced trace $t(g)$ and reduced norm $d(g)$, and any quadratic extension $L$ of $F$ that splits $D$ carries a realization of $(t, d)$ as an irreducible representation $G \to \mathrm{GL}_2(L)$.

On the other hand, suppose $(t, d)$ splits into a sum of two characters $\chi, \chi' : G \to \bar{F}^\times$. The image of $\chi$ is a subgroup of $\bar{F}^\times$ whose every element is contained in an at-most-quadratic extension of $F$. Suppose that $\alpha = \chi(a)$ and $\beta = \chi(b)$ for $a, b \in G$ generate different quadratic extensions of $F$. Then on one hand, $\alpha\beta = \chi(ab)$ must generate the third quadratic subextension of $F(\alpha, \beta)$. But on the other hand, we claim that $\chi'(ab) = \chi'(a)\chi'(b)$ is equal to $\alpha\beta$: indeed, let $c_\alpha$ be the generator of $\mathrm{Gal}(F(\alpha, \beta)/F)$ viewed as an element of $\mathrm{Gal}(F(\alpha, \beta)/F)$, so that $\chi'(a) = c_\alpha(\alpha)$; define $c_\beta$ similarly. Then $c_\alpha c_\beta$ generates $\mathrm{Gal}(F(\alpha, \beta)/F(\alpha\beta))$ and hence fixes $\alpha\beta$. Therefore $\chi(ab) + \chi'(ab) = 2\alpha\beta$, which is not in $F^{(x)}$ — a contradiction. $\square$

The following proposition is true in any dimension, but we state it here for dimension 2.

**Proposition A.13.** Let $(t, d) : G \to F$ be a pseudorepresentation and $H \subseteq G$ a finite-index normal subgroup. Then the trace field of $(t, d)$ is a finite extension of the trace field of $(t|_H, d|_H)$.

**Proof.** Replace $F$ by the trace field of $(t, d)$, and let $E \subseteq F$ be the trace field of $(t|_H, d|_H)$. Let $\rho$ be a semisimple representation carrying $(t, d)$ over an extension of $F$. Note that $F$ is algebraic over $E$: every $g \in G$ satisfies $g^{[G:H]} \in H$, so that every eigenvalue of $\rho(g)$, and hence $t(g)$, is algebraic over the finite extension of $E$ containing the eigenvalues of $g^{[G:H]}$. Since $F$ is contained in the field

<footnote>(x) The constraint on the characteristic is necessary: consider $G = \mathbb{Z}^2$ and $F = \mathbb{F}_2(x, y)$, with $(t, d)$ the pseudocharacter corresponding to the scalar 2-dimensional representation $(1, 0)$ to $\sqrt{x}$ and $(0, 1)$ to $\sqrt{y}$.</footnote>
generated by all these eigenvalues, $F$ is algebraic over $E$, and in particular $\rho$ is realizable over an algebraic extension of $E$.

By Clifford's theorem [Cra19, Theorem 7.1.1], $\rho|_H$ is still semisimple, so it is carried by a representation $V$ over a finite extension $E'$ of $E$. Since $F/E$ is algebraic, $V_E = V \otimes_{E'} \bar{E}$ carries all of $\rho$. Let $g_1, \dots, g_n$ be coset representatives for $H$ in $G$; write each $\rho(g_i)$ as a matrix in a fixed $E'$-basis of $V$ extended to $V_E$. Let $M$ be the subfield of $\bar{E}$ generated over $E'$ by the matrix coefficients of all the $\rho(g_i)$. Then $M$ is a finite extension of $E'$, and hence of $E$, containing the values of tr $\rho$. Therefore $F/E$ is finite. $\square$

**A.3. Rings with involution.** Throughout Appendix A.3, let $A$ be a commutative noetherian ring equipped with an involution $*$. Note that we will need to apply the results in this section to the universal constant-determinant pseudodeformation ring $\mathcal{A}$, so we cannot assume that $A$ is a domain. Let $A^\epsilon = \{a \in A: a^* = \epsilon a\}$ for $\epsilon \in \{+, -\}$. We will assume throughout that $*$ is not the identity on $A$ so that $A^- \neq 0$. It is easy to see that $A^+$ is a subring of $A$ and $A^-$ is an $A^+$-module. The following results have been adapted from [Lan75] and [CL77], where they are presented in the context when $A$ may be noncommutative.

**Definition A.14.** We say that an $A$-ideal $\mathfrak{a}$ is a $*-\textit{ideal}$ if $\mathfrak{a}^* = \mathfrak{a}$. We say that $A$ is $*-\textit{prime}$ if whenever $\mathfrak{a}$ and $\mathfrak{b}$ are $*-\textit{ideals}$ such that $\mathfrak{a}\mathfrak{b} = 0$ then either $\mathfrak{a} = 0$ or $\mathfrak{b} = 0$.

**Lemma A.15.** If $A$ is $*-\textit{prime}$ then $A$ is reduced.

*Proof.* Let $0 \neq a \in A$ be nilpotent. Then there is a smallest integer $n > 1$ such that $a^n = 0$. Let $\mathfrak{a} = aA$ and $\mathfrak{b} = a^{n-1}A$. Note that $\mathfrak{a} \neq 0$ and $\mathfrak{b} \neq 0$ by the minimality of $n$. If $\mathfrak{a}$ and $\mathfrak{b}$ are $*-\textit{ideals}$ then we have reached a contradiction since $\mathfrak{a}\mathfrak{b} = a^n A = 0$. In particular, if $a + a^* = 0$ then $a^* = -a \in aA$ and so $\mathfrak{a}, \mathfrak{b}$ are $*-\textit{ideals}$.

If $a + a^* \neq 0$, then $a + a^*$ is still nilpotent since $A$ is commutative. By replacing $a$ with $a + a^*$ in the above argument, we find that $\mathfrak{a}$ and $\mathfrak{b}$ are $*-\textit{ideals}$ and thus we reach a contradiction. $\square$

**Lemma A.16.** If $A$ is a noetherian commutative ring with $2 \in A^\times$, then $A^+$ is a noetherian ring.

*Proof.* The following argument comes from [CL77, Lemma]. Let $I_1 \subseteq I_2 \subseteq \dots$ be an ascending chain of ideals in $A^+$. Then $I_1 A \subseteq I_2 A \subseteq \dots$ is an ascending chain of ideals in $A$. Since $A$ is noetherian, there is some $n$ such that $I_n A = I_m A$ for all $m \geq n$.

Fix $m \geq n$ and $a \in I_m \subseteq A^+$. Since $a \in I_m A = I_n A$ we may write

$$a = \sum_{i} b_{i}x_{i}$$

with $b_i \in I_n$ and $x_i \in A$. Applying the involution $*$ yields

$$a = a^* = \sum_{i} b_{i}x_{i}^{*}.$$

Thus

$$2a = \sum_{i} b_{i}(x_{i} + x_{i}^{*}).$$

Since $x_i+x_i^* \in A^+$ and $2 \in A^\times$ it follows that $a = \frac{1}{2} \sum_i b_i(x_i+x_i^*) \in I_n$. In particular, $I_m = I_n$. $\square$

We would like to show that $A$ is finitely generated as an $A^+$-module, which is equivalent to $A$ being a noetherian $A^+$-module since $A^+$ is a noetherian ring by Lemma A.16. The following lemma follows the proof of [Lan75, Lemma 6].

**Lemma A.17.** If there is an element $d \in A^-$ that is not a zero divisor in $A$, then $A$ is noetherian as an $A^+$-module.
*Proof.* Since $d$ is not a zero divisor, it follows that $A$ is isomorphic to $dA$ as an $A^+$-module. On the other hand, for any $a \in A$ we can write

$$da = \frac{1}{2}(d(a - a^*)) + \frac{1}{2}(d(a + a^*)) \in A^+ + dA^+.$$

Thus $dA$ is a submodule of the finitely generated $A^+$-module $A^+ + dA^+$. Since $A^+$ is noetherian by Lemma A.16, it follows that $dA$, and hence $A$, is a finitely generated (and hence noetherian) $A^+$-module. $\square$

**Proposition A.18.** If $A$ is a commutative noetherian ring with $2 \in A^\times$, then $A$ is a noetherian $A^+$-module.

*Proof.* This proof combines elements of the proofs of [CL77, Theorem] and [Lan75, Theorem 7].

Suppose not. Let $a_0$ be the largest $*$-ideal of $A$ such that $A/a_0$ is not a noetherian $A^+$-module, which exists since $A$ is a noetherian ring and is not noetherian as an $A^+$-module. Thus, by replacing $A$ with $A/a_0$, we may assume that $A/a$ is a noetherian $A^+$-module for any $*$-ideal $a \neq 0$.

We claim that, under this assumption, $A$ is reduced. It suffices to show that $A$ is $*$-prime by Lemma A.15. Suppose that $a$ and $b$ are nonzero $*$-ideals of $A$ such that $ab = 0$. Note that we can view $a$ as an $A/b$-module since $ab = 0$. We know that $a$ is noetherian as an $A/b$-module since $a$ is noetherian as an $A$-module. Furthermore, $A/b$ is a noetherian $A^+$-module since $b \neq 0$. Thus $a$ is noetherian as an $A^+$-module. We also know that $A/a$ is a noetherian $A^+$-module since $a \neq 0$. Therefore $A$ is a noetherian $A^+$-module, a contradiction. Thus $A$ is $*$-prime and hence reduced.

Since $A$ is a noetherian ring, it has only finitely many minimal prime ideals; call them $\mathfrak{p}_1, \dots, \mathfrak{p}_n$. Since $A$ is reduced, we have that

$$\bigcap_{i=1}^{n} \mathfrak{p}_i = 0.$$

Note that $n = 1$ corresponds to the case when $A$ is a domain, and in that case we have already seen that $A$ is a noetherian $A^+$-module by Lemma A.17. Thus we assume henceforth that $n > 1$ and thus each $\mathfrak{p}_i \neq 0$.

If $\mathfrak{p}_i^* \cap \mathfrak{p}_i \neq 0$, then $\mathfrak{p}_i \cap \mathfrak{p}_i^*$ is a $*$-ideal and so $A/(\mathfrak{p}_i \cap \mathfrak{p}_i^*)$ is a noetherian $A^+$-module. If every $\mathfrak{p}_i$ satisfies $\mathfrak{p}_i \cap \mathfrak{p}_i^* \neq 0$ then we can view $A$ as a subring of

$$\bigoplus_{i=1}^{n} A/(\mathfrak{p}_i \cap \mathfrak{p}_i^*),$$

which is noetherian as an $A^+$-module. In particular, $A$ is a noetherian $A^+$-module, a contradiction, which proves the proposition.

Suppose there is some $k$ such that $\mathfrak{p}_k \cap \mathfrak{p}_k^* = 0$. It is easy to check that $\mathfrak{p}_k^*$ is another minimal prime ideal of $A$. We claim that $n = 2$ in this case. Indeed, if $\mathfrak{p}$ is any minimal prime ideal of $A$, then we have $\mathfrak{p}_k \mathfrak{p}_k^* \subseteq \mathfrak{p}_k \cap \mathfrak{p}_k^* = 0$ and thus $\mathfrak{p}_k \mathfrak{p}_k^* = 0 \in \mathfrak{p}$. Thus $\mathfrak{p} = \mathfrak{p}_k$ or $\mathfrak{p} = \mathfrak{p}_k^*$.

Let us write $\mathfrak{p} = \mathfrak{p}_k$ henceforth. We can embed $A$ into $A/\mathfrak{p} \times A/\mathfrak{p}^*$ by identifying $a \in A$ with $(a + \mathfrak{p}, a + \mathfrak{p}^*)$. Note that $A^+ \cap \mathfrak{p} = 0$ since if $a \in A^+ \cap \mathfrak{p}$ then $a = a^* \in \mathfrak{p}^* \cap \mathfrak{p} = 0$. Similarly, $A^+ \cap \mathfrak{p}^* = 0$. In particular, $A^+$ injects into $A/\mathfrak{p}$ and is therefore a domain.

Note that by Lemma A.17, we may assume that every element of $A^-$ is a zero divisor in $A$. However, both $A/\mathfrak{p}$ and $A/\mathfrak{p}^*$ are domains, so the only zero divisors in $A/\mathfrak{p} \times A/\mathfrak{p}^*$ are elements of the form $(a + \mathfrak{p}, \mathfrak{p}^*)$ or $(\mathfrak{p}, a + \mathfrak{p}^*)$. Recall that $(A^-)^2 \subseteq A^+$. In particular, if $(a + \mathfrak{p}, \mathfrak{p}^*) \in A^-$, then $(a^2 + \mathfrak{p}, \mathfrak{p}^*) \in A^+$. That is, there is some $a_+ \in A^+$ such that $a_+ - a^2 \in \mathfrak{p}$ and $a_+ \in \mathfrak{p}^*$. But we have already seen that $A^+ \cap \mathfrak{p}^* = 0$. Similarly, any $(\mathfrak{p}, a + \mathfrak{p}^*) \in A^-$ must be trivial. In other words, $A^- = 0$, a contradiction. Therefore $A$ must be noetherian as an $A^+$-module. $\square$

Given any ideal $a$ of $A$, we define $a^\epsilon := a \cap A^\epsilon$. We call $a$ a graded ideal if $a = a^+ \oplus a^-$. 68
**Proposition A.19.** Let $A$ be a commutative local noetherian ring such that $A$ and $A^+$ have the same residue field. Assume that $2 \in A^\times$. If $A'$ is the quotient of $A$ by a nongraded prime ideal, then $A'$ has the same field of fractions as the image of $A^+$ in $A'$.

*Proof.* Write $f: A \to A'$ for the quotient map. It suffices to show that every element of $f(A^-)$ can be written as a quotient of elements in $f(A^+)$. Since the prime ideal $\mathfrak{p} = \ker f$ is assumed to be nongraded, it follows that there is some $a \in \mathfrak{p}$ such that, if we decompose $a = a^+ + a^-$ with $a^+ \in A^+$ and $a^- \in A^-$, then neither $a^+$ nor $a^-$ is in $\mathfrak{p}$. It follows that $f(a^-) = -f(a^+)$, and so $f(a^-) \in f(A^+)$. Note that $f(a^-) \neq 0$ since $a^- \notin \mathfrak{p}$. For any $x \in A^-$ we have that $xa^- \in A^+$ since $(A^-)^2 \subseteq A^+$. Thus $f(x) = f(xa^-)/f(a^-) \in Q(f(A^+))$, as desired. $\square$

**A.4. Automorphisms and gradings.** We recall how ring automorphisms give rise to gradings.

Let $A$ be a complete local ring and $X$ a finite abelian subgroup of the group of ring automorphisms of $A$. We write $\mu_n(A) := \{a \in A^\times : a^n = 1\}$. Given a character $\varphi: X \to A^\times$, we define

$$A^{\varphi} := \{a \in A : \sigma a = \varphi(\sigma)a, \forall \sigma \in X\}.$$

The following lemma is standard, so we leave the proof to the reader.

**Lemma A.20.** Let $\mathbb{F}$ be a finite field of characteristic $p$ and $A$ a pro-$p$ local ring with residue field $\mathbb{F}$. If $p \nmid n$, then $\mu_n(A) = s(\mu_n(\mathbb{F}))$.

Assume the following:

(*) for every positive integer $n$, if $X$ contains an element of order $n$, then $\#\mu_n(A) = n$.

Then one has $\#X = \#\operatorname{Hom}(X, A^\times)$. (It is easily checked when $X$ is cyclic, and then for general $X$ one applies the structure theorem of finite abelian groups.)

**Corollary A.21.** Assume $(*)$. If $p \nmid \#X$, then for any $1 \neq \sigma \in X$ we have

$$\sum_{\varphi \in \operatorname{Hom}(X, A^\times)} \varphi(\sigma) = 0.$$

*Proof.* First suppose that $X$ is cyclic of order $n$ and $\sigma$ is a generator for $X$. Then $\operatorname{Hom}(X, A^\times)$ is cyclic, generated by any $\varphi_0$ such that $\varphi_0(\sigma)$ is a primitive $n^{\text{th}}$ root of unity. Let $H := \langle \varphi_0^k \rangle$ be a nontrivial subgroup of $\operatorname{Hom}(X, A^\times)$. Then by Lemma A.20 we have

$$\sum_{\varphi \in H} \varphi(\sigma) = \sum_{i=0}^{n/k} \varphi_0^{ki}(\sigma) = \sum_{\omega \in \mu_{n/k}(A)} \omega = \sum_{\omega \in \mu_{n/k}(\mathbb{F})} s(\omega) = 0.$$

Now we allow $X$ to be any finite abelian group such that $p \nmid \#X$ and $\sigma$ any nontrivial element of $X$. Then we have an exact sequence

$$0 \to \operatorname{Hom}(X/\langle\sigma\rangle, A^\times) \to \operatorname{Hom}(X, A^\times) \to \operatorname{Hom}(\langle\sigma\rangle, A^\times).$$

Thus $\sum_{\varphi \in \operatorname{Hom}(X, A^\times)} \varphi(\sigma)$ is an integral multiple of $\sum_{\varphi \in H} \varphi(\sigma)$, where $H$ is the image of $\operatorname{Hom}(X, A^\times)$ in $\operatorname{Hom}(\langle\sigma\rangle, A^\times)$. This sum is 0 by the first paragraph. $\square$

**Lemma A.22.** Let $A$ and $X$ be as above. Assume that $\#X \in A^\times$ and that condition $(*)$ holds. Then $A$ admits a grading given by $A = \bigoplus_{\varphi \in \operatorname{Hom}(X, A^\times)} A^{\varphi}$. Furthermore, for any $\mathbb{Z}[1/\#X][X]$-submodule $M \subseteq A$, letting $M^{\varphi} := M \cap A^{\varphi}$, there is a decomposition

$$M = \bigoplus_{\varphi \in \operatorname{Hom}(X, A^\times)} M^{\varphi}.$$

*Proof.* For $\varphi \in \operatorname{Hom}(X, A^\times)$, define $e_\varphi := \frac{1}{\#X} \sum_{\sigma \in X} \varphi(\sigma)\sigma^{-1} \in \mathbb{Z}[1/\#X][X]$. A straightforward computation shows that $\{e_\varphi : \varphi \in \operatorname{Hom}(X, A^\times)\}$ is an orthogonal system of idempotents in $\mathbb{Z}[1/\#X][X]$. (Note that Corollary A.21 is needed to show that $\sum_{\varphi} e_\varphi = 1$.) There is a natural ring homomorphism $\mathbb{Z}[1/\#X][X] \to \operatorname{End} A$; pushing forward the $e_\varphi$ to $\operatorname{End} A$ gives the result. $\square$
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