Proof. By the exact sequence

\[ 0 \to \operatorname{soc}(\pi) \to \operatorname{soc}^{i+1}(\pi) \to \operatorname{soc}^i(\pi{/\!}\operatorname {soc}(\pi)) \to 0 \]

and induction on $i$ it suffices to prove that $\operatorname {soc}(\pi ) = \operatorname {soc}(\pi _1) \oplus \operatorname {soc}(\pi _2)$. For this, observe that if $X$ is a semisimple smooth representation of $k[H]$, and $\iota : X \to \pi$ is an $H$-linear map, then $\operatorname {im}(\mathrm {pr}_i \circ \iota ) \subseteq \operatorname {soc}(\pi _i)$ for $i = 1, 2$, where $\mathrm {pr}_i$ the projection $\pi \to \pi _i$, and so $\operatorname {soc}(\pi ) \subseteq \operatorname {soc}(\pi _1) \oplus \operatorname {soc}(\pi _2)$. The other direction is true by definition.

Proof. Every semisimple reducible $H$-representation contains two submodules incomparable by inclusion. Conversely, assume that all the graded factors of the socle filtration of $\pi$ are irreducible, and let $X \subset \pi$ be a proper submodule. Let $i$ be such that $\operatorname {soc}^i(\pi ) \subseteq X$ but $\operatorname {soc}^{i+1}(\pi ) \not \subseteq X$. Consider the submodule $X{/\!}\operatorname {soc}^i\pi \subset \pi {/\!}\operatorname{soc}^i \pi$. Since $\pi$ is smooth, if $X{/\!}\operatorname {soc}^i \pi \ne 0$, then it contains an irreducible $k[H]$-subspace. But since $\operatorname {soc}(\pi {/\!}\operatorname {soc}^i\pi )$ is irreducible, this implies that $\operatorname {soc}^{i+1}(\pi ) \subseteq X$, which we are assuming not to be the case. Thus, $X = \operatorname {soc}^i(\pi )$, hence the only submodules of $\pi$ are the $\operatorname {soc}^i(\pi )$, and the lattice of submodules of $\pi$ is totally ordered by inclusion.

Proof. Let $\pi _i' \subset \pi _i$ be a proper $H$-submodule. If $x \in \pi _i \setminus \pi _i'$, then the definition of a uniserial representation implies that $\pi _i' \subseteq \langle H \cdot x \rangle$, which is finite-dimensional by $H$-smoothness. The second claim follows from the first, counting dimensions on the exact sequence $0 \to \ker \lambda \to \pi _1 \to \operatorname {image} \lambda \to 0$.

Proof. Since $\operatorname {soc}_\mathrm {Iw}\sigma \cong \kappa ^+$ as $\mathrm {Iw}$-representations, the intersection $\sigma \cap \pi _\infty (\kappa )$ equals $0$. Since $\pi _\infty (\kappa )^{K_1}$ is one-dimensional, the projection of $\sigma$ to $\pi _\infty (\kappa )$ is at most one-dimensional, and so $\sigma \cap \pi _\infty ^+(\kappa )$ has codimension at most one in $\sigma$. We claim that the codimension is exactly one. For this, it suffices to note that

\[ \sigma \subset \operatorname{Ind}_\mathrm{Iw}^K(\kappa) = \pi_2(\kappa)^+ \oplus \kappa \subset \pi_\infty^+(\kappa) \oplus \pi_\infty(\kappa) \]

and so if $\sigma \subset \pi _\infty ^+(\kappa )$ then $\operatorname {Ind}_\mathrm {Iw}^K(\kappa )/\sigma$ has a two-dimensional space of $\mathrm {Iw}_1$-invariants, which is a contradiction (since this quotient is a Serre weight). It follows that $[1, x^{r-i}y^i] \in \pi _\infty ^+(\kappa )$ for all $0 \leq i < r$, and there exist nonzero $(HK_1, \kappa )$-eigenvectors $\varphi _r \in \pi _\infty (\kappa )^+$ and $\varphi \in \operatorname {soc}_\mathrm {Iw}\pi _\infty (\kappa )$ such that

(2.2.7)\begin{equation} [1, y^r] = \varphi_r + \varphi. \end{equation}

The lemma follows from this and the observation that if $x$ is an $(HK_1, \xi )$-eigenvector, then $\Pi x$ is a $(K_0(p^2), \xi ^+)$-eigenvector, as well as the computation $\kappa ^+ = \kappa \alpha ^r$.

Proof. Twisting by $\omega \circ \det ^s$ we can assume that $\sigma = \mathrm {Sym}^r(k^2)$. Since $v^+$ is a $(K_0(p), \kappa ^+)$-eigenvector, $\tau v^+$ is a $(K_0(p^2), \kappa ^+)$-eigenvector. It follows from Proposition 2.2.9 that $\pi _\infty (\kappa )^+$ is an injective envelope of the trivial representation of $U(\mathbf {Z}_p)$, and so its space of $U(\mathbf {Z}_p)$-invariants is one-dimensional. On the other hand, we know from Lemma 2.2.5 that the $(K_0(p^2), \kappa ^+)$-eigenspace in $\pi _\infty (\kappa )$ is one-dimensional. Hence, there exist $\mu, \nu \in k$ such that $\tau v^+ = \mu v^+ +\nu \psi _r$, and we have to prove that $\nu \ne 0$ and that $\mu = \lambda ^{-1}$. If $\nu = 0$, then $\tau v^+ \in \pi _\infty (\kappa )^+$ and so $\zeta (p)s v^+ = \Pi \tau v^+ \in \pi _\infty (\kappa )$, which contradicts (2.2.7), since $s v^+$ is a nonzero multiple of $[1, y^r]$. Now choose a $G$-linear isomorphism $\pi \cong \operatorname {Ind}_B^G(\mathrm {nr}_{\lambda ^{-1}} \otimes \mathrm {nr}_\lambda \omega ^{r+1})$. Since $\mu$ does not change if we replace $v^+$ by a nonzero scalar multiple, we can assume that $v^+$ is the restriction to $K$ of the unique $(\mathrm {Iw}, \kappa ^+)$-eigenvector $f: G \to \kappa$ supported in $B s \mathrm {Iw}$ and sending $s$ to $1$. Then we find that

\[ \tau v^+(s) = v^+ \begin{pmatrix} 0 & p \\ 1 & 0 \end{pmatrix} = v^+(ts) = \lambda^{-1}v^+(s) = \lambda^{-1} \]

and so $\mu = \lambda ^{-1}$ since $\psi _r(s) = 0$.

Proof. Since (2.2.9) follows from (2.2.8) applied to $z = \Pi ^{-1}\psi _0$, it suffices to prove (2.2.8). By Corollary 2.2.14 we know that $\alpha$ is an isomorphism. Fix $n \geq 0$. We claim that it suffices to prove that there exists $\mu \in k^\times$ such that $\alpha (\tau ^n \Pi z) = \mu \tau ^n \Pi \alpha (z)$. Indeed, by [Reference Barthel and LivnéBL94, (19)], the Hecke operator $T$ acts on $(\operatorname {soc}_K \pi _i)^{\mathrm {Iw}_1}$ by the formula

\[ z \mapsto \sum_{\lambda \in \mathbf{F}_p} \begin{pmatrix} 1 & [\lambda] \\ 0 & 1 \end{pmatrix} \begin{pmatrix} p & 0 \\ 0 & 1 \end{pmatrix} z. \]

Hence, the right-hand side equals $\lambda _i z$ in our setting. Iterating this, we deduce that there exists an element $k_{n+1}$ of the group algebra $k[K]$ such that $k_{n+1}\tau ^n\Pi z = \lambda _1^{n+1} z$ and $k_{n+1}\tau ^n \Pi \alpha (z) = \lambda _2^{n+1}\alpha (z)$. The $K$-linearity of $\alpha$ now implies that

\[ \lambda_1^{-(n+1)}\mu \cdot k_{n+1}(\tau^n \Pi\alpha(z)) = \alpha(z) = \lambda_2^{-(n+1)}\cdot k_{n+1}(\tau^n \Pi \alpha(z)) \]

and since $\alpha (z) \ne 0$ this implies that $\mu = \lambda _2^{-(n+1)}\lambda _1^{n+1}$, concluding the proof of the proposition. There remains to prove that $\alpha (\tau ^n \Pi z)$ and $\tau ^n \Pi \alpha (z)$ are collinear. This is true when $n = 0$ since $\alpha (\Pi z)$ and $\Pi \alpha (z)$ are both generators of $\operatorname {soc}_\mathrm {Iw} \pi _\infty (\kappa _2)^+$. In general, it suffices to prove that for all nonzero $(\mathrm {Iw}, \kappa _i)$-eigenvectors $\varphi _i \in \pi _i$ the support of $\tau ^n \varphi _i|_K$ is $K_0(p^{n+1})$. In fact, let

\[ \pi_i \xrightarrow{\sim} \operatorname{Ind}_{B(\mathbf{Z}_p)}^K(\kappa_i) \]

be the restriction map to $K$. Since by hypothesis $\kappa _1 = \kappa _2$, by Corollary 2.2.14 the map $\alpha$ corresponds to a nonzero scalar multiple of the identity under these isomorphisms. It follows that $\alpha (\tau ^n \Pi z)$ and $\tau ^n \Pi \alpha (z)$ are $(K_0(p^{n+1}), \kappa )$-eigenvectors in $\operatorname {Ind}_{B(\mathbf {Z}_p)}^K(\kappa )$ supported in $K_0(p^{n+1})$, and the space of such eigenvectors is one-dimensional. Thus, let $\varphi _i \in \pi _i$ be a nonzero $(\mathrm {Iw}, \kappa _i)$-eigenvector. Since $\pi _i^{\mathrm {Iw}_1}$ is two-dimensional and $G = Bs\mathrm {Iw} \coprod B\mathrm {Iw}$, we know that $\varphi _i$ is supported in $B\mathrm {Iw}$. Hence, $\tau ^n \varphi _i$ is supported in $B\mathrm {Iw} \tau ^{-n}$, and we have to prove that $B\mathrm {Iw} \tau ^{-n} \cap K = K_0(p^{n+1})$. Taking inverses, it suffices to prove that $\tau ^n \mathrm {Iw} B \cap K = K_0(p^{n+1})$. The Iwahori decomposition implies that $\tau ^n \mathrm {Iw} B = \overline {U}(p^{n+1}\mathbf {Z}_p)B$. It follows that $\tau ^n \mathrm {Iw} B \cap K = \overline {U}(p^{n+1}\mathbf {Z}_p)B(\mathbf {Z}_p) = K_0(p^{n+1})$ using the Iwahori decomposition for $K_0(p^{n+1})$.

Proof. When $\psi$ is a generator of $\operatorname {soc}_\mathrm {Iw} \pi _\infty (\kappa )$ this is true by Proposition 2.2.18. Now let $\psi$ be a $K_0(p^2)$-eigenvector but not a $K_0(p)$-eigenvector, and fix a generator $\psi _0 \in \operatorname {soc}_\mathrm {Iw} \pi _\infty (\kappa )$. We have seen in the proof of Proposition 2.2.18 that $\tau \psi _0$ is a $(K_0(p^2), \kappa )$-eigenvector supported in $K_0(p^2)$. Now Lemma 2.2.1 implies that $\tau \psi _0$ generates the $p$-dimensional $\mathrm {Iw}$-submodule $\pi _2(\kappa ) \subset \pi _\infty (\kappa )$. By Proposition 2.2.9 we know that $\pi _2(\kappa ) = \pi _\infty (\kappa )^{\overline {U}(p^2\mathbf {Z}_p)}$ is $\overline {U}(p\mathbf {Z}_p)$-uniserial, and so there exists $h\in k[\overline {U}(p\mathbf {Z}_p)]$ such that $\psi = h \tau \psi _0$. Since $\tau$ normalizes $\overline {U}(p\mathbf {Z}_p)$, for all $k\geq 1$ there exists $h_k \in k[\overline {U}(p\mathbf {Z}_p)]$ such that $\tau ^k h \tau ^{-k} = h_k$ as endomorphisms of any $k[G]$-module. It follows that

\[ \alpha(\tau^k \psi) = \alpha(h_k \tau^{k+1} \psi_0) = h_k\lambda_2^{-(k+1)}\lambda_1^{k+1} \tau^{k+1}(\alpha(\psi_0)) = \lambda_2^{-(k+1)}\lambda_1^{k+1}\tau^k h\tau\alpha(\psi_0) = \lambda_2^{-k}\lambda_1^k\tau^k\alpha(\psi), \]

where the final equality follows from $\alpha (\psi ) = h\alpha (\tau \psi _0) = h\lambda _2^{-1}\lambda _1\tau \alpha (\psi _0)$.

Proof. This is similar to Proposition 2.2.11. It suffices to prove the result over $\overline {k}$. By [Reference PaškūnasPaš10, Proposition 4.7], the injection $\chi \to M_\sigma$ is an injective envelope in the category of smooth representations of $HU(\mathbf {Z}_p)$. By [Reference PaškūnasPaš10, Proposition 5.9], the representation $M_\sigma |_{HU(\mathbf {Z}_p)}$ is uniserial and the layers of its socle filtration satisfy

(2.3.2)\begin{equation} \operatorname{soc}^{k+1}(M_\sigma) {/\!}\operatorname{soc}^{k}(M_\sigma) \cong \alpha^{-1} \otimes \operatorname{soc}^{k}(M_\sigma) {/\!}\operatorname{soc}^{k-1}(M_\sigma). \end{equation}

By the following lemma, we see that (2.3.2) is also true for the $\mathrm {Iw}$-socle filtration of $M_\sigma$.

Lemma 2.3.3 The representation $M_\sigma |_{\mathrm {Iw}}$ is uniserial. The socle filtration of $M_\sigma |_{HU(\mathbf {Z}_p)}$ is a filtration by $\mathrm {Iw}$-stable subspaces, and it coincides with the socle filtration of $M_\sigma |_{\mathrm {Iw}}$.

Proof. Since every $\mathrm {Iw}$-subspace of $M_\sigma$ is $HU(\mathbf {Z}_p)$-stable, the $\mathrm {Iw}$-subspaces of $M_\sigma$ are totally ordered by inclusion, and so $M_\sigma |_{\mathrm {Iw}}$ is uniserial. The irreducible $\overline {k}$-representations of $HU(\mathbf {Z}_p)$ and $\mathrm {Iw}$ are both inflated from characters of $H$, hence the rest of the lemma follows from Lemma 2.1.4.

Twisting by $\Pi$, we find that

\[ \operatorname{soc}_\mathrm{Iw}^{k+1}(M_{\sigma}^+) {/\!}\operatorname{soc}_\mathrm{Iw}^{k}(M_{\sigma}^+) \cong \alpha \otimes \operatorname{soc}_\mathrm{Iw}^{k}(M^+_\sigma) {/\!}\operatorname{soc}_\mathrm{Iw}^{k-1}(M^+_\sigma), \]

because $\alpha ^+ = \alpha ^{-1}$. Now the proof proceeds as for Proposition 2.2.11.

Proof. Since both $M_\sigma |_{\mathrm {Iw}}$ and $M_\sigma |_{HU(\mathbf {Z}_p)}$ are uniserial, the module $M_{\sigma, n}$ is $\mathrm {Iw}$-stable and $M_{\sigma, n} = \langle \mathrm {Iw} \cdot t^{2n}v_\sigma \rangle$. Since $v_{\sigma _i}$ is an $\mathrm {Iw}$-eigenvector, the vector $t^{2n} v_{\sigma _i}$ is an eigenvector for

\[ t^{2n} \cdot \mathrm{Iw} \cdot t^{-2n} = \begin{pmatrix} \mathbf{Z}_p^\times & p^{2n} \mathbf{Z}_p \\ p^{-2n+1}\mathbf{Z}_p & \mathbf{Z}_p^\times \end{pmatrix}, \]

hence, it is an eigenvector for

\[ K_0^+(p^{2n+1}) = \Pi \cdot K_0(p^{2n+1}) \cdot \Pi^{-1} = \begin{pmatrix} \mathbf{Z}_p^\times & p^{2n}\mathbf{Z}_p \\ p\mathbf{Z}_p & \mathbf{Z}_p^\times \end{pmatrix} \]

with the same $H$-eigencharacter as $v_{\sigma _i}$, namely $\chi _i \cong \sigma _i^{\mathrm {Iw}_1}$. The isomorphism $\operatorname {Ind}_{K_0^+(p^{2n+1})}^{K_0(p)}(\chi _i)\cong (\operatorname {Ind}_{K_0(p^{2n+1})}^{K_0(p)}(\chi _i^+))^+$ is a consequence of Lemma 2.3.9 to follow.

Proof. It suffices to prove the same result over $\overline {k}$. To do so, it suffices to prove that if there exists an $\mathrm {Iw}$-linear isomorphism $M_{\sigma _1}/X_1 \cong M_{\sigma _2}/X_2$, then $\sigma _1 = \sigma _2$ and $X_1 = X_2$. In fact, assume that this is true, and let

\[ \lambda: M_{\sigma_1}/X_1 \to M_{\sigma_2} / X_2 \]

be a nonzero $\mathrm {Iw}$-linear morphism. Then $\lambda$ induces an injection

\[ \lambda: M_{\sigma_1}{/\!}\ker(\lambda) \to M_{\sigma_2}/X_2, \]

which is an isomorphism by Lemma 2.1.5. By our assumption, this implies that $\sigma _1 = \sigma _2$ and $\ker (\lambda ) = X_2$, hence $X_1 \subseteq X_2$. There remains to prove that $\operatorname {End}_{k[\mathrm {Iw}]}(M_\sigma /X, M_\sigma /X)$ is one-dimensional. To see this, note that for all $\mathrm {Iw}$-linear $\lambda : M_\sigma /X \to M_\sigma /X$ there exists $x \in \overline {k}^\times$ such that $\lambda -x|_{\operatorname {soc}_\mathrm {Iw}(M_\sigma /X)} = 0$. If $\lambda - x \ne 0$, then $\lambda - x$ factors through an isomorphism

\[ M_\sigma{/\!}\ker(\lambda-x) \to M_\sigma/X, \]

which we are assuming not to exist, since $X$ is strictly contained in $\ker (\lambda -x)$.

To conclude the proof of the theorem, let

\[ \lambda : M_{\sigma_1}/X_1 \to M_{\sigma_2}/X_2 \]

be an $\mathrm {Iw}$-linear isomorphism. By Lemma 2.3.10 we deduce that $\dim (\sigma _1) = \dim (\sigma _2)$, which implies that $\dim (M_{\sigma _1, n}) = \dim (M_{\sigma _2, n})$ for all $n$. We have to prove that $\dim X_1 = \dim X_2$ and $\sigma _1 = \sigma _2$ (or, equivalently, $\chi _1 = \chi _2$). Assume this is false. After possibly replacing $\lambda$ by $\lambda ^{-1}$, we can assume that $\dim (X_1) \geq \dim (X_2)$, and so $\lambda$ factors through a morphism

\[ \lambda_n: M_{\sigma_1, n}/X_1 \to M_{\sigma_2, n}/X_2, \]

since the dimension of the left-hand side is smaller than the dimension of the right-hand side, and $M_{\sigma _2}/X_2$ is a uniserial representation (compare the first paragraph of the proof of Theorem 2.2.12). We are going to prove that the dimension of $\mathrm {coker}(\lambda _n)$ tends to infinity with $n$, which would contradict the fact that the dimension of $\ker (\lambda _n)$ is independent of $n$ for $n$ large enough (since it coincides with $\ker (\lambda )$ for $n$ large enough, by uniseriality). To do so, fix $n > 0$. We are going to prove that $\mathrm {coker}(\lambda _m) \geq p^{n-1}$ if $m$ is large enough: by the above discussion, this suffices to conclude the proof of the theorem. By Lemma 2.3.5 we know that $M_{\sigma _i, m}$ is a quotient of $\operatorname {Ind}_{K_0^+(p^{2m+1})}^{K_0(p)}(\chi _i)$. Since the dimension of $M_{\sigma _i ,m}$ tends to infinity with $m$, for all $m$ large enough the module $M_{\sigma _i, m}$ will have an $\mathrm {Iw}$-quotient $Y_{\sigma _i, m}$ of dimension $p^{n}$. Since $\operatorname {Ind}_{K_0^+(p^{2m+1})}^{K_0(p)}(\chi _i)$ is uniserial, necessarily $Y_{\sigma _i, m } \cong \operatorname {Ind}_{K_0^+(p^{n+1})}^{K_0(p)}(\chi _i)$. Furthermore, if $\dim M_{\sigma _1, m}/X_1 \geq p^n$, the map $\lambda _m: M_{\sigma _1, m}/X_1 \to M_{\sigma _2, m}/X_2$ induces an $\mathrm {Iw}$-linear map $\lambda _{Y, m}: Y_{\sigma _1, m} \to Y_{\sigma _2, m}$, and $\mathrm {coker}(\lambda _m)$ surjects onto $\mathrm {coker}(\lambda _{Y, m})$.

If $\dim (X_1) > \dim (X_2)$, then $\lambda _m$ cannot be surjective: in fact, in this case its domain has smaller dimension than its codomain, because $\dim (M_{\sigma _1, n}) = \dim (M_{\sigma _2, n})$. Hence, $\lambda _{Y, m}$ is not surjective, since if $\lambda _{Y, m}$ is surjective, then $\lambda _m$ is surjective on $\mathrm {Iw}$-cosocles, hence $\lambda _m$ is surjective. On the other hand, if $\chi _1 \ne \chi _2$, then $\lambda _{Y, m}$ is not a surjection either, since it cannot be an isomorphism when $\chi _1 \ne \chi _2$. In each of these cases we deduce by Lemmas 2.2.6 and 2.3.9 that $\dim (\mathrm {coker} \lambda _m) \geq \dim \mathrm {coker} (\lambda _{Y, m}) \geq p^{n-1}$, and the claim follows.

Proof. By Proposition 2.3.2, an $\mathrm {Iw}$-linear map $\pi _{\sigma _1} \to \pi _{\sigma _2}$ has to send $M_{\sigma _1}$ to $M_{\sigma _2}$ and $\Pi M_{\sigma ^{[s]}_1}$ to $\Pi M_{\sigma ^{[s]}_2}$. By Theorem 2.3.6 (and after twisting by $\Pi$), these restrictions are zero if $\sigma _1 \not = \sigma _2$, and otherwise they are scalar endomorphisms. The corollary follows as these have to agree when restricted to $\operatorname {soc}_{\mathrm {Iw}}(\pi _{\sigma })$.

Proof. Since $\pi _1 \cong \operatorname {c-Ind}_{G^+}^G(\pi _{\sigma _1})$, we have $\pi _1|_N \cong \operatorname {c-Ind}_{\mathrm {Iw} Z}^N \pi _{\sigma _1}$. Now by Frobenius reciprocity we have

\[ \mathrm{Hom}_N(\pi_1, \pi_2) = \mathrm{Hom}_{\mathrm{Iw} Z}(\pi_{\sigma_1}, \pi_{\sigma_2}) \oplus \mathrm{Hom}_{\mathrm{Iw} Z}(\pi_{\sigma_1}, \pi_{\sigma_2^{[s]}}). \]

Corollary 2.3.7 together with the fact that $\sigma _2$ and $\sigma _2^{[s]}$ are not isomorphic implies that the right-hand side is at most one-dimensional, and does not vanish if and only if $\pi _1$ and $\pi _2$ are $G$-isomorphic.

Proof. We can assume that $\theta _\alpha$ has the same representation space of $\theta$ with the action $\theta _\alpha (\alpha (h)) t =\theta (h) t$. Given a smooth function $f: G \to \theta$ with $f(hg) = \theta (h) f(g)$, let $f_\alpha (g) = f (\alpha ^{-1}(g))$. Then $f_\alpha (\alpha (h) g) = f(h \alpha ^{-1}(g)) = \theta (h) f_\alpha (g) = \theta _\alpha (\alpha (h))f_\alpha (g)$, hence $f_\alpha \in \operatorname {Ind}_{\alpha H}^G(\theta _\alpha )$. The map we are looking for is $f \mapsto f_\alpha$.

Proof. It suffices to prove the same result over $\overline {k}$. Let

\[ \lambda: M_{\sigma_1}/X_1 \to M_{\sigma_2} / X_2 \]

be a nonzero $\mathrm {Iw}$-linear morphism. We will prove that $\dim (\sigma _1) = \dim (\sigma _2)$ by a numerical argument based on the computation of $\dim M_{\sigma _i, n}$ from [Reference PaškūnasPaš10, Proposition 4.7]. More precisely, if we define the integers $0< r_i < p-1, 0 \leq a_i \leq p-2$ by the formula

\[ \sigma_i \cong \mathrm{Sym}^{r_i}(k^2) \otimes \operatorname{det}^{a_i}, \]

then

(2.3.3)\begin{equation} d_{i, n} = \dim M_{\sigma_i, n} - 1 = r_i + (p-1-r_i)p + \cdots + r_i p^{2n -2} + (p-1-r_i)p^{2n-1} \end{equation}

and we need to prove that $r_1 = r_2$. We will do so by assuming $r_1 \ne r_2$ and deriving a contradiction from the fact that the $j$th $p$-adic digits of $d_{1, n}$ and $d_{2, n}$ are different for all values of $j$.

Replacing $X_1$ by $\ker (\lambda )$, we can assume without loss of generality that $\lambda$ is injective. It follows from Lemma 2.1.5 that $\lambda$ is an isomorphism, and replacing $\lambda$ by $\lambda ^{-1}$ we can assume without loss of generality that $r_2< r_1$. Since $d_{i, n}$ tends to infinity with $n$, and $X_i$ is finite-dimensional, there exists $n>0$ such that $X_i \subset M_{\sigma _i, n-2}$ for $i = 1,2$. Fix such an integer $n$. Then

\[ \lambda(M_{\sigma_1, n}/X_1) \subset M_{\sigma_2, n}/X_2 \]

since the dimension of the left-hand side is smaller than the dimension of the right-hand side and $M_{\sigma _2}/X_2$ is uniserial (in fact, $\dim M_{\sigma _2, n} - \dim X_2 > (p-1-r_2)p^{2n-1} > \dim M_{\sigma _1, n}$). Since $\lambda$ is injective, the dimension of the left-hand side is equal to

(2.3.4)\begin{equation} (\dim M_{\sigma_1, n-2} - \dim X_1) + r_1 p^{2n -4} + (p-1-r_1)p^{2n-3} + r_1 p^{2n -2} + (p-1-r_1)p^{2n-1}, \end{equation}

and since $M_{\sigma _2, n}/X_2$ is uniserial we find that $\lambda (M_{\sigma _1, n}/X_1)$ contains every $\mathrm {Iw}$-submodule of $M_{\sigma _2, n}/X_2$ whose dimension is smaller than $\dim M_{\sigma _1, n}/X_1$. It follows that $\lambda (M_{\sigma _1, n}/X_1)$ contains the unique $\mathrm {Iw}$-submodule $Z_{\sigma _2, n} \subset M_{\sigma _2, n}/X_2$ of dimension

\[ (\dim M_{\sigma_2, n-2} - \dim X_2) + r_2 p^{2n -4} + (p-1-r_2)p^{2n-3} + r_2 p^{2n -2}. \]

We are going to conclude by studying the dimension of the image of $\lambda (M_{\sigma _1, n}/X_1)$ in $M_{\sigma _2, n}/Z_{\sigma _2, n}$. Since $M_{\sigma _2, n}/Z_{\sigma _2, n}$ has dimension $(p-1-r_2)p^{2n-1}$, it follows from Lemmas 2.3.5 and 2.2.2 that there is an $\mathrm {Iw}$-stable filtration

\[ 0 = \operatorname{Fil}^{p-1-r_2} \subset \cdots \subset \operatorname{Fil}^1 \subset \operatorname{Fil}^0 = M_{\sigma_2, n}/Z_{\sigma_2, n} \]

such that $\operatorname {Fil}^i{\!\!/\!}\operatorname {Fil}^{i+1} \cong \pi _{2n}(\chi _2\alpha ^{-i})^+$.

Let $b$ be the smallest integer such that $\operatorname {Fil}^{b} \subseteq \lambda (M_{\sigma _1, n}/X_1)/Z_{\sigma _2, n}$, so that $1 \leq b \leq p-1-r_2$. Since $M_{\sigma _2, n}/Z_{\sigma _2, n}$ is uniserial, it follows that $\lambda (M_{\sigma _1, n}/X_1)/Z_{\sigma _2, n}$ is strictly contained in $\operatorname {Fil}^{b-1}$, and so the induced map

\[ \lambda(M_{\sigma_1, n}/X_1)/Z_{\sigma_2, n} \to \operatorname{Fil}^{b-1}{\!\!/\!}\operatorname{Fil}^{b} \]

is not surjective. By Lemmas 2.3.5 and 2.2.6 it follows that there exist integers $0 \leq c \leq p-1, 0 \leq d \leq 2n-2,$ such that

\[ \dim \lambda(M_{\sigma_1, n}/X_1)/Z_{\sigma_2, n} = (p-1-r_2-b)p^{2n-1} + c p^d. \]

In all, we find that the dimension of $\lambda (M_{\sigma _1, n}/X_1)$ equals

(2.3.5)\begin{equation} (\dim M_{\sigma_2, n-2} - \dim X_2) + r_2 p^{2n -4} + (p-1-r_2)p^{2n-3} + r_2 p^{2n -2} + (p-1-r_2-b)p^{2n-1} + c p^d \end{equation}

and so this number has to equal (2.3.4). Since $0 < r_i < p-1$ and $\dim M_{\sigma _i, n-2} < p^{2n-4}$, adding $c p^d$ can modify at most two of the coefficients of $p^{2n-4}, p^{2n-3}, p^{2n-2}$ in (2.3.5). Hence, (2.3.4) and (2.3.5) have different $p$-adic expansions, and so cannot be equal. This concludes the proof that $r_1 = r_2$.

Proof. Note that any element in the kernel of this map is contained in $\mathrm {Ext}^1_{k[G], \zeta }(\pi _1, \pi _2)$, i.e. it has central character $\zeta$. Hence, by [Reference PaškūnasPaš13, Lemma 5.7], it suffices to prove the theorem over $\overline {k}$. By [Reference PaškūnasPaš10, Theorem 1.1], the theorem is true if $\sigma _1$ and $\sigma _2$ are not conjugate, since then the representations $\pi _i$ are not $G$-isomorphic and the space of $G$-extensions vanishes. Thus, it suffices to prove that if $\sigma$ is a generic Serre weight, $\pi = (\operatorname {c-Ind}_{KZ}^G \sigma )/T$, and

(2.3.6)\begin{equation} 0 \to \pi \to X \to \pi \to 0 \end{equation}

is an exact sequence of $G$-representations with central character $\zeta$ that splits over $\mathrm {Iw}$, then the sequence is split over $G$.

For this, choose $v_\sigma$ in the $K$-socle of the quotient, invariant under $\mathrm {Iw}_1$ and such that $\langle K \cdot v_\sigma \rangle \cong \sigma$. Then $\langle K \cdot \Pi v_\sigma \rangle \cong \sigma ^{[s]}$, because $\pi ^{\mathrm {Iw}_1}$ is two-dimensional. Let $w_\sigma$ be an $\mathrm {Iw}_1$-invariant lift of $v_\sigma$ to $X$ with $\mathrm {Iw}$-eigencharacter $\chi$, which exists since we assume that the extension is $\mathrm {Iw}$-split.

We are going to study the representations $\langle K \cdot w_\sigma \rangle$ and $\langle K \cdot \Pi w_\sigma \rangle$, which are quotients of finite principal series $\operatorname {Ind}_{\mathrm {Iw}}^K(\chi )$ and $\operatorname {Ind}_{\mathrm {Iw}}^K(\chi ^s)$, respectively. Recall (see, for instance, [Reference Breuil and PaškūnasBP12, § 2]) that the representation $\operatorname {Ind}_{\mathrm {Iw}}^K(\chi )$ has a two-dimensional space of $\mathrm {Iw}_1$-invariants, spanned by the function $\varphi \in \operatorname {Ind}_{\mathrm {Iw}}^K(\chi )$ supported in $\mathrm {Iw}$ and satisfying $\varphi (1) = 1$, and the function $f_0 = S_0 \varphi$, where we have written $S_0 = \sum _{\lambda \in \mathbf {F}_p}\bigl (\begin{smallmatrix} [\lambda ] & 1 \\ 1 & 0 \end{smallmatrix}\bigr )$. The functions $\varphi$ and $f_0$ are $H$-eigenvectors, with eigencharacter $\chi$ and $\chi ^s$, respectively. We have an exact sequence

\[ 0 \to \operatorname{Ind}_{\mathrm{Iw}}^K(\chi)_0 \to \operatorname{Ind}_{\mathrm{Iw}}^K(\chi) \xrightarrow{\mathrm{res}} \chi \to 0 \]

defined by restricting functions to $\mathrm {Iw}$, which is $\mathrm {Iw}$-linearly split. Looking at the $H$-eigencharacter, it follows that $f_0 \in \operatorname {Ind}_{\mathrm {Iw}}^K(\chi )_0$ and generates its $\mathrm {Iw}$-socle.

Proposition 2.3.13 The representations $\langle K \cdot w_\sigma \rangle$ and $\langle K \cdot \Pi w_\sigma \rangle$ are irreducible.

Proof. Assume that

\[ \alpha: \operatorname{Ind}_{\mathrm{Iw}}^K(\chi) \to \langle K \cdot w_\sigma \rangle,\quad \varphi \mapsto w_\sigma \]

is injective. Since $\langle K \cdot v_\sigma \rangle$ is irreducible, we have $S_0 \varphi \in \pi \subset X$. Since $S_0 \varphi$ is an $\mathrm {Iw}$-eigenvector with eigencharacter $\chi ^s$, we deduce furthermore that $S_0 \varphi \in \pi _{\sigma ^{[s]}}$.

By assumption, there is an $\mathrm {Iw}$-linear retraction $r: X \to \pi$ of the inclusion of $\pi$ in $X$. Consider the composition

\[ \operatorname{Ind}_{\mathrm{Iw}}^K(\chi)_0 \xrightarrow{\alpha} X \xrightarrow{r} \pi \to \pi_{\sigma^{[s]}}. \]

Since $S_0 \varphi$ generates the $\mathrm {Iw}$-socle of $\operatorname {Ind}_{\mathrm {Iw}}^K(\chi )_0$, this composition is injective. But the dimension of $\operatorname {Ind}_{\mathrm {Iw}}^K(\chi )_0$ is equal to $p$. Since the first congruence subgroup $K_1$ acts trivially on $\operatorname {Ind}_{\mathrm {Iw}}^K(\chi )_0$, this contradicts the fact that $\dim (\pi _{\sigma ^{[s]}}^{K_1}) = p-1$ (see [Reference Breuil and PaškūnasBP12, § 20] or [Reference MorraMor13, Theorem 1.4]).

The same proof works for $\operatorname {Ind}_{\mathrm {Iw}}^K(\chi ^s) \to \langle K \cdot \Pi w_{\sigma } \rangle$, or argue by symmetry, using that

\[ (\operatorname{c-Ind}_{KZ}^G \sigma)/T \cong (\operatorname{c-Ind}_{KZ}^G \sigma^{[s]})/T \]

to conclude.

Proof. Let $0 \to \pi _{\sigma _1} \to X \to \pi _{\sigma _2} \to 0$ be a short exact sequence of $G^+$-representations that is split over $\mathrm {Iw} Z$. The exact sequence

\[ 0 \to \pi_{\sigma_1} \oplus \pi_{\sigma_1}^+ \to X \oplus X^+ \to \pi_{\sigma_2} \oplus \pi_{\sigma_2}^+ \to 0 \]

is $\mathrm {Iw} Z$-split, because $(-)^+ = {\operatorname {ad}}(\Pi )^*(-)$ and $\Pi$ normalizes $\mathrm {Iw} Z$. It is isomorphic to the restriction to $G^+$ of

\[ 0 \to \operatorname{Ind}_{G^+}^G(\pi_{\sigma_1}) \to \operatorname{Ind}_{G^+}^G(X) \to \operatorname{Ind}_{G^+}^G(\pi_{\sigma_2}) \to 0. \]

Since $\operatorname {Ind}_{G^+}^G(\pi _{\sigma _i}) \cong (\operatorname {c-Ind}_{KZ}^G \sigma _i)/T$, this sequence is $G$-split by Theorem 2.3.12. But then the inclusion $\pi _{\sigma _1} \to X \oplus X^+$ has a $G^+$-linear retraction, which we can restrict to $X$ to prove that the original exact sequence was already $G^+$-split.