We had $F$ a totally real field of even degree, $D/F$ a definite quaternion algebra split at all finite places, a coefficient ring $L \supseteq \mathcal{O} \twoheadrightarrow \mathcal{O}/\lambda = \mathbb{F}$, and a representation $$ r \colon G_F \to \GL_2(\mathcal{O}) $$ ramified only at $\lbrace v \mid \ell \rbrace \amalg R$ that we wanted to show is automorphic, where for $v \in R$
- $q_v \equiv 1 \pmod{\ell}$,
- $\bar{r} \vert_{G_{F_v}} = 1$,
- $r \vert_{I_{F_v}}$ is unipotent.
We also had a character $\chi = \prod_{v \in R} \chi_v \colon \prod k(v)^\times \to \mathcal{O}^\times$ of $\ell$-power order. We chose a set $Q$ of primes disjoint from $R$ such that for all $v \in Q$
- $q_v \equiv 1 \pmod{\ell}$,
- $\bar{r}(\mathrm{Frob}_ v)$ has distinct eigenvalues $\alpha_v, \beta_v$.
We looked at a universal deformation ring $R_{\chi,Q}^\mathrm{univ}$ wit hthe property that for all $v \in R$ we have $$ \operatorname{char}_ {r(\sigma)}(T) = (T - \chi_v^\mathrm{gal}(\sigma)) (T - \chi_v^\mathrm{gal}(\sigma)^{-1}) $$ for all $\sigma \in I_{F_v}$, and no condition at $v \in Q$. We showed that $r_{\chi,Q}^\mathrm{univ} \vert_{G_{F_v}} = \chi_{\alpha_v} \oplus \chi_{\beta_v}$ for $v \in Q$. Letting $H_v$ be the maximal $\ell$-power order quotient of $k(v)^\times$, we could then define an $H_Q$-action on $R_{\chi,Q}^\mathrm{univ}$ via $\chi_{\alpha_v}$. There was also a corresponding space of automorphic forms $S(\chi, Q)$ with an action of $R_{\chi,Q}^\mathrm{univ}$.
For $v \in R$, we considered the framed deformation ring $R_{\bar{r} \vert_{G_{F_v}}, \chi_v}^\square$ and similarly $R_{\chi,Q}^\square$ which parametrizes $(r, \lbrace \alpha_v \rbrace_{v \in R})$, where this is equivalent to $(\beta r \beta^{-1}, \lbrace \beta \alpha_v \rbrace_{v \in R})$ where $\alpha_v, \beta \in \ker(\GL_2(R^\mathrm{univ}) \to \GL_2(\mathbb{F}))$. Then we had a non-canonical isomorphism $$ R_{\chi,Q}^\square \cong R_{\chi,Q}^\mathrm{univ}[[A_{v,i,j}]]_ {v \in R, 1 \le i, j \le 2} / (A_{v_0,1,1}). $$ We then looked at the base ring $$ R_\chi^\mathrm{loc} = \bigotimes_{v \in R}^\wedge R_{\bar{r} \vert_{G_{F_v}}, \chi_v}^\square. $$ We also had a space of automorphic forms $$ S(\chi, Q)^\square = S(\chi, Q) \otimes_{R_{\chi,Q}^\mathrm{univ}} R_{\chi,Q}^\square. $$
For every positive integer $N \gt 0$, there exists a set $Q_N$ such that
- $\lvert Q_N \rvert = s$,
- $q_v \equiv 1 \pmod{\ell^N}$ for all $v \in Q_N$,
- $H_{\mathscr{L}_ {Q_N}^\perp}^1 = 0$.
Then we have a surjection $$ R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s+\lvert R \rvert-1}]] \twoheadrightarrow R_{\chi,Q}^\square. $$ So we get a diagram $$ \begin{CD} \mathcal{O}[[s_1, \dotsc, s_{s + 4 \lvert R \rvert - 1}]] @>>> \mathcal{O}[[A_{v,i,j}]]/(A_{v_0,1,1})[H_{Q_N}] \br @. @VVV \br R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s + \lvert R \rvert - 1}]] @>>> R_{\chi,Q_N}^\square @>>> \End(S(\chi, Q_N)^\square), \end{CD} $$ where $S(\chi, Q_N)$ is finite free over $\mathcal{O}[[A_{v,i,j}]] / (A_{v_0,1,1})[H_{Q_N}]$. Here, both rings on the left side has the same dimension. Also if we quotient by the augmentation ideal $\mathfrak{a}_ Q$, we get $$ R_{\xi,Q_N}^\square / \mathfrak{a}_ Q \cong R_{\chi,\emptyset}^\mathrm{univ}, \quad S(\chi, \emptyset) = S(\chi, Q_N)^\square / \mathfrak{a}_ Q \neq (0), $$ where this space is nonzero because $r$ is residually automorphic.
Here is where the flexibility of $\chi$ becomes useful. Fix $\chi$ with $\chi_v \neq 1$ for all $v \in R$. Then the diagram between $\chi$ and $1$ are isomorphic modulo $\lambda$ because $\chi$ is $\ell$-power torsion.
There was a way of taking an inverse limit as $N \to \infty$. Then we get $$ \mathcal{O}[[s_1, \dotsc, s_{s + 4\lvert R \rvert - 1}]] \to R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s+\lvert R \rvert-1}]] \twoheadrightarrow R_\chi^\mathrm{loc}[[x_1, \dotsc, x_{s+\lvert R \rvert-1}]] / \mathfrak{a}_ \infty \twoheadrightarrow R_{\chi,\emptyset}^\mathrm{univ}, $$ acting on $$ S(\chi, \infty)^\square \twoheadrightarrow S(\chi, \infty)^\square / \mathfrak{a}_ \infty = S(\chi, \emptyset). $$
Fact 1. The space $\Spec R_\chi^\mathrm{loc}$ is irreducible (as long as $\chi_v \neq 1$ for all $v$).
Now we have the same inequality $$ s + 4\lvert R \rvert = \dim R_\chi[[\underline{x}]] \ge \dim \operatorname{supp}_ {R_\chi[[\underline{x}]]} S(\chi,\infty)^\square \ge \operatorname{depth}_ {R_\chi[[\underline{x}]]} S(\chi,\infty)^\square \ge s + 4\lvert R \rvert. $$ So we have equality everywhere, so $$ \operatorname{supp}_ {R_\chi[[\underline{x}]]} S(x, \infty)^\square = \Spec R_\chi^\mathrm{loc}[[\underline{x}]]. $$ Taking everything modulo $\lambda$, we get $$ \operatorname{supp}_ {R_\chi^\mathrm{loc}[[\underline{x}]]/\lambda} S(\chi,\infty)^\square/\lambda = \Spec R_\chi^\mathrm{loc}[[\underline{x}]]/\lambda, $$ and so $$ \operatorname{supp}_ {R_1^\mathrm{loc}[[\underline{x}]]/\lambda} S(1,\infty)^\square/\lambda = \Spec R_1^\mathrm{loc}[[\underline{x}]]/\lambda. $$
Fact 2. There is a bijection between components of $\Spec R_1^\mathrm{loc}$ and components of $R_1^\mathrm{loc}/\lambda$. More precisely, every component of $\Spec R_1^\mathrm{loc}$ contains a unique component of $R_1^\mathrm{loc}/\lambda$.
This implies that we have $$ \operatorname{supp}_ {R_1^\mathrm{loc}[[\underline{x}]]} S(1,\infty)^\square = \Spec R_1^\mathrm{loc}[[\underline{x}]]. $$ Taking the quotient by $\mathfrak{a}_ \infty$, we then get $$ \operatorname{supp}_ {R_1^\mathrm{loc}[[\underline{x}]] / \mathfrak{a}_ \infty} S(1, \emptyset) = \Spec R_1^\mathrm{loc}[[\underline{x}]] / \mathfrak{a}_ \infty, $$ where the left hand side is contained in $R_{1,\emptyset}^\mathrm{univ}$ and the right hand side contains $\Spec R_{1,\emptyset}^\mathrm{univ}$. This implies that $$ \operatorname{supp}_ {R_{1,\emptyset}^\mathrm{univ}} S(1,\emptyset) = \Spec R_{1,\emptyset}^\mathrm{univ}. $$