Provided \({\mathbb {F}}\) is sufficiently large, irreducible \({\mathbb {F}}\)-representations of \(G_K\) and \(G_{K_\infty }\) are induced from characters, see Lemma 2.1.2. In this section and the next we investigate the extent with which this is true for objects of \({\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\). Throughout assume \(k \subset {\mathbb {F}}\).
Rank ones
Recall from Construction 4.3.4 how \({\mathfrak {S}} \otimes _{{\mathbb {Z}}_p} {\mathcal {O}}\) is made into an \({\mathcal {O}}[[u]]\)-algebra. Then \(k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}\) becomes an \({\mathbb {F}}[[u]]\)-algebra. Also let \(e_\tau \in k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}\) denote the image of the idempotent \({\widetilde{e}}_\tau \in {\mathfrak {S}} \otimes _{{\mathbb {Z}}_p} {\mathcal {O}}\) defined in Construction 4.3.4. Thus \(\varphi (e_{\tau \circ \varphi }) = e_\tau \).
The next lemma is proven by an easy change of basis argument (see [9, Lemma 6.2]).
Lemma 6.1.1
Fix \(\tau _0 \in {\text {Hom}}_{{\mathbb {F}}_p}(k,{\mathbb {F}})\). Let \(M \in {\text {Mod}}^{{\text {BK}}}_k({\mathcal {O}})\) be of rank one over \(k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}\). Then M is isomorphic to a Breuil–Kisin module
$$\begin{aligned} N = k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}, \qquad \varphi _N(1) = (x)\sum u^{r_\tau } e_\tau \end{aligned}$$
where \(r_\tau \in {\mathbb {Z}}\) and where \((x) = xe_{\tau _0} + \sum _{\tau \ne \tau _0} e_\tau \) for some \(x \in {\mathbb {F}}^\times \).
Remark 6.1.2
If N is as in Lemma 6.1.1 then \({\text {Weight}}_\tau (N) = \lbrace r_\tau \rbrace \). Note also that N satisfies the equivalent conditions of Lemma 5.3.4. Thus \(N \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) if and only if \(r_{\tau } \in [0,p]\).
Proposition 6.1.3
If N is as in Lemma 6.1.1 then the \(G_{K_\infty }\)-action on T(N) is through the restriction to \(G_{K_\infty }\) of the character
$$\begin{aligned} \psi _x\prod _{\tau } \omega _\tau ^{-r_\tau } \end{aligned}$$
Here \(\psi _x\) denotes the unramified character sending the geometric Frobenius to x, and the \(\omega _\tau \) are the characters defined in the paragraph after the proof of Lemma 2.1.1.
Proof
This is [9, Proposition 6.7]. However note that in loc. cit. they contravariantly associate a \(G_{K_\infty }\)-representation to Breuil–Kisin module; this is why the character appearing here is the inverse of that in loc. cit. \(\square \)
Induction and restriction
Notation 6.2.1
Let L / K be the unramified extension corresponding to a finite extension l / k, and let \(L_\infty = K_\infty L\). Set \({\mathfrak {S}}_L = W(l)[[u]]\). Extension of scalars along the inclusion \(f:{\mathfrak {S}} \rightarrow {\mathfrak {S}}_L\) describes a functor
$$\begin{aligned} f^*:{\text {Mod}}^{{\text {BK}}}_K \rightarrow {\text {Mod}}^{{\text {BK}}}_L \end{aligned}$$
For \(M \in {\text {Mod}}^{{\text {BK}}}_K\) the module \(f^*M = M \otimes _{{\mathfrak {S}}} {\mathfrak {S}}_L\) is made into a Breuil–Kisin module via the semilinear map \(m \otimes s \mapsto \varphi _M(m)\otimes \varphi (s)\); this map induces the isomorphism
$$\begin{aligned} (\varphi ^*f^*M)\Big [\tfrac{1}{E}\Big ] = (f^*\varphi ^*M)\Big [\tfrac{1}{E}\Big ] = f^*\Big (\varphi ^*M\Big [\tfrac{1}{E}\Big ]\Big ) \xrightarrow {f^*\varphi _M} f^*\Big (M\Big [\tfrac{1}{E}\Big ]\Big ) = (f^*M)\Big [\tfrac{1}{E}\Big ] \end{aligned}$$
where the first \(=\) comes from the fact that \(\varphi \circ f = f \circ \varphi \). The natural isomorphism
$$\begin{aligned} f^*M \otimes _{{\mathfrak {S}}_L} W(C^\flat ) \cong M \otimes _{{\mathfrak {S}}} W(C^\flat ) \end{aligned}$$
is clearly \(\varphi ,G_{L_\infty }\)-equivariant so \(T(f^*M) = T(M)|_{G_{L_\infty }}\).
Notation 6.2.2
With notation as in Notation 6.2.1, restriction of scalars along f induces a functor
$$\begin{aligned} f_*:{\text {Mod}}^{{\text {BK}}}_L \rightarrow {\text {Mod}}^{{\text {BK}}}_K \end{aligned}$$
If \(M \in {\text {Mod}}^{{\text {BK}}}_L\) we equip \(f_*M\) with the obvious semilinear map \(m \mapsto \varphi _M(m)\). Let us verify that this makes \(f_*M\) into a Breuil–Kisin module. The semilinear map induces the composite:
$$\begin{aligned} (\varphi ^*f_*M)\Big [\tfrac{1}{E}\Big ] \rightarrow (f_*\varphi ^*M)\Big [\tfrac{1}{E}\Big ] = f_*\Big (\varphi ^*M \Big [\tfrac{1}{E}\Big ]\Big ) \xrightarrow {f_*\varphi _M} f_*\Big (M\Big [\tfrac{1}{E}\Big ]\Big ) = (f_*M)\Big [\tfrac{1}{E}\Big ] \end{aligned}$$
which we claim is an isomorphism. It suffices to check the natural map \(\varphi ^*f_*M \rightarrow f_*\varphi ^*M\) is an isomorphism, and this follows because the commutative diagram
is a pushout.
Lemma 6.2.3
For all \(M \in {\text {Mod}}^{{\text {BK}}}_K\) and \(N \in {\text {Mod}}^{{\text {BK}}}_L\) there are functorial isomorphisms
$$\begin{aligned} {\text {Hom}}(M,f_*N) \cong f_*{\text {Hom}}(f^*M,N) \end{aligned}$$
in \({\text {Mod}}^{{\text {BK}}}_K\).
Proof
The standard adjunction between \(f^*\) and \(f_*\) provides functorial \({\mathfrak {S}}\)-linear isomorphisms \({\text {Hom}}_{{\mathfrak {S}}}(M,f_*N) \rightarrow {\text {Hom}}_{{\mathfrak {S}}_L}(f^*M,N)\). Explicitly, this map sends \(\alpha \) onto the homomorphism \(m \otimes s \mapsto s \alpha (m)\). As this is \(\varphi \)-equivariant we get isomorphisms as claimed. \(\square \)
Lemma 6.2.4
Let \(N \in {\text {Mod}}^{{\text {BK}}}_L\). Then there are functorial identifications \(\iota _N:T(f_*N) \rightarrow {\text {Ind}}_{L_\infty }^{K_\infty } T(N)\) such that the diagram
commutes for all \(M \in {\text {Mod}}^{{\text {BK}}}_K\). The top horizontal arrow is obtained from the identification in Lemma 6.2.3 by taking \(\varphi \)-invariants, and the lower horizontal arrow is given by Frobenius reciprocity.
Proof
Let \({\mathcal {O}}_{{\mathcal {E}},L}\) be the p-adic completion of \({\mathfrak {S}}_L[\frac{1}{u}]\). The map \(f: {\mathfrak {S}} \rightarrow {\mathfrak {S}}_L\) extends to a map \(f: {\mathcal {O}}_{{\mathcal {E}}} \rightarrow {\mathcal {O}}_{{\mathcal {E}},L}\) and so we can make sense of the operations \(f^*\) and \(f_*\) on etale \(\varphi \)-modules. Write \(M^{{\text {et}}} = M \otimes _{{\mathfrak {S}}} {\mathcal {O}}_{{\mathcal {E}}}\) and \(N^{{\text {et}}} = N \otimes _{{\mathfrak {S}}_L} {\mathcal {O}}_{{\mathcal {E}},L}\). Then clearly \(f^* (M^{{\text {et}}}) = (f^*M)^{{\text {et}}}\) and, because \({\mathcal {O}}_{{\mathcal {E}},L} = {\mathcal {O}}_{{\mathcal {E}}} \otimes _{{\mathfrak {S}}} {\mathfrak {S}}_L\), we also have that \(f_*(N^{{\text {et}}}) = (f_*N)^{{\text {et}}}\). We obtain maps
$$\begin{aligned}&{\text {Hom}}_{{\text {BK}}}(M,f_*N) \rightarrow {\text {Hom}}_{{\text {et}}}(M^{{\text {et}}},f_*N^{{\text {et}}}), \qquad \\&{\text {Hom}}_{{\text {BK}}}(f^*M,N) \rightarrow {\text {Hom}}_{{\text {et}}}(f^*M^{{\text {et}}},N^{{\text {et}}}) \end{aligned}$$
which commute with T. The analogue of Lemma 6.2.3 in the setting of etale \(\varphi \)-modules is proved in exactly the same way, and the obtained identification is compatible with the maps above. Thus, to prove the lemma we may replace \({\text {Hom}}_{{\text {BK}}}\) with \({\text {Hom}}_{{\text {et}}}\) (homsets in the category of etale \(\varphi \)-modules) and M and N with \(M^{{\text {et}}}\) and \(N^{{\text {et}}}\) in the diagram of the lemma.
Since \(M^{{\text {et}}} \mapsto T(M^{{\text {et}}})\) is an equivalence of categories, the map \( ({\text {Frob}}) \circ T \circ (6.2.3) \circ T^{-1}\) describes an identification
$$\begin{aligned} {\text {Hom}}_{G_{K_\infty }}(V,T(f_*N)) \rightarrow {\text {Hom}}_{G_{K_\infty }}(V,{\text {Ind}}_{L_\infty }^{K_\infty } T(N)) \end{aligned}$$
(6.2.5)
for any continuous \(G_{K_\infty }\)-representation V on a finitely generated \({\mathbb {Z}}_p\)-module. As (6.2.5) is functorial in V, Yoneda’s lemma provides the isomorphism \(\iota _N\). As (6.2.5) is functorial in N we see that \(\iota _N\) is functorial. \(\square \)
Lemma 6.2.6
Assume \(k \subset l \subset {\mathbb {F}}\).
- 1.
If \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) then \(f^*M \in {\text {Mod}}^{{\text {SD}}}_l({\mathcal {O}})\) and for each \(\theta \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) we have
$$\begin{aligned} {\text {Weight}}_\theta (f^*M) = {\text {Weight}}_{\theta |_k}(M) \end{aligned}$$
- 2.
If \(N \in {\text {Mod}}^{{\text {SD}}}_l({\mathcal {O}})\) then \(f_*N \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) and
$$\begin{aligned} {\text {Weight}}_\tau (f_*N) = \bigcup _{\theta |_k = \tau } {\text {Weight}}_\theta (N) \end{aligned}$$
Proof
By functoriality both \(f^*\) and \(f_*\) preserve \({\mathcal {O}}\)-actions. Note that the inclusion \(k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}} \rightarrow l[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}\) sends \(e_\tau \mapsto \sum _{\theta |_k = \tau } e_\theta \). Thus \((f^*M)_\theta = M_{\theta |_k}\) and \((f_*N)_\tau = \prod _{\theta |_k = \tau } N_\theta \). Both (1) and (2) then follow by verifying the second condition of Lemma 5.3.4. \(\square \)
Approximation by induced Breuil–Kisin modules
We consider the situation from Notation 6.2.1. Thus L / K is a finite unramified extension, corresponding to an extension l / k of residue fields, and \(L_\infty = L(\pi ^{1/p^\infty })\). We also have the map \(f:{\mathfrak {S}} \rightarrow {\mathfrak {S}}_L\).
Lemma 6.3.1
Suppose \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) and assume that \(T(M) \cong {\text {Ind}}_{L_\infty }^{K_\infty } T'\). Then there exists an \(N \in {\text {Mod}}^{{\text {SD}}}_l({\mathcal {O}})\) with \(T(N) = T'\), together with a \(\varphi \)-equivariant inclusion
$$\begin{aligned} M \hookrightarrow f_*N \end{aligned}$$
of \(k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}\)-modules which becomes an isomorphism after inverting u.
Proof
There is a non-zero map \(T(M)|_{G_{L_\infty }} \rightarrow T'\) corresponding under Frobenius reciprocity to the isomorphism \(T(M) \cong {\text {Ind}}^{K_\infty }_{L_\infty } T'\). Lemma 5.1.3 produces a surjection \(f^*M \rightarrow N\) where \(N \in {\text {Mod}}^{{\text {BK}}}_l({\mathcal {O}})\) is of rank one with \(T(N) = T'\). Applying Lemma 6.2.4 to \(f^*M \rightarrow N\) we obtain a map
$$\begin{aligned} M \rightarrow f_*N \end{aligned}$$
which, after applying T, induces the identification \(T(N) = T'\). Thus \(M \rightarrow f_*N\) becomes an isomorphism after inverting u and is, in particular, injective. Lemma 6.2.6 implies \(f^*M \in {\text {Mod}}^{{\text {SD}}}_l({\mathcal {O}})\), since \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\). Therefore \(N \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) by Proposition 5.4.7. \(\square \)
When T(M) is irreducible and \({\mathbb {F}}\) is sufficiently large T(M) is induced from a character. Thus, Lemma 6.3.1 produces an inclusion \(M \hookrightarrow f_*N\) with N of rank one. Lemma 6.1.1 allows us to describe N explicitly. In this case we would like to know which submodules of \(f_*N\) arise in this way. The following example shows that there are non-trivial (i.e. \(M \ne f_*N\)) possibilities.
An example
Take \(K = {\mathbb {Q}}_p\) and let L / K be of degree 5 with residue extension l / k. Let \(N \in {\text {Mod}}^{{\text {SD}}}_l({\mathcal {O}})\) be the rank one object defined by
$$\begin{aligned} N = l[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}, \quad \varphi _N(1) = u^x e_{\theta \circ \varphi ^4} + u^n e_{\theta \circ \varphi ^3} + e_{\theta \circ \varphi ^2} + u^n e_{\theta \circ \varphi } + e_{\theta } \end{aligned}$$
Here we have fixed \(\theta \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) and \(1 \le n \le p, 0 \le x \le p\). Let \(M \subset f_*N\) be the sub-module generated over \({\mathbb {F}}[[u]]\) by \(e_{\theta \circ \varphi ^4},e_{\theta \circ \varphi ^3} + e_{\theta \circ \varphi }, e_{\theta \circ \varphi ^2}, ue_{\theta \circ \varphi },e_{\theta }\). One computes that
$$\begin{aligned} \varphi (e_{\theta \circ \varphi ^4}, e_{\theta \circ \varphi ^3} + e_{\theta \circ \varphi }, e_{\theta \circ \varphi ^2}, ue_{\theta \circ \varphi }, e_\theta ) = (e_{\theta \circ \varphi ^4}, e_{\theta \circ \varphi ^3} + e_{\theta \circ \varphi }, e_{\theta \circ \varphi ^2}, ue_{\theta \circ \varphi }, e_\theta ) X \end{aligned}$$
where
$$\begin{aligned} X = \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad -1 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \end{pmatrix} \begin{pmatrix} u^n &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad u^{n-1} &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad u^{p} &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad u^x \end{pmatrix} \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ -1 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \end{pmatrix} \end{aligned}$$
This shows that \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\). One checks that \(M \ne f_*N'\) for any rank one \(N' \subset N\).
Irreducibility and strong divisibility
Let L / K, l / k and \(L_\infty /K_\infty \) be as in Notation 6.2.1; we obtain \(f:{\mathfrak {S}} \rightarrow {\mathfrak {S}}_L\). Let \(N \in {\text {Mod}}_l^{{\text {SD}}}({\mathcal {O}})\) be the rank one object given by
$$\begin{aligned} N = l[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}, \quad \varphi _N(1) = \sum _{\theta \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})} u^{r_\theta } e_\theta \end{aligned}$$
Since \(N \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) each \(r_\theta \in [0,p]\). Note this N is as in Lemma 6.1.1, except we’ve fixed \(x = 1\). This is to simplify notation (it will be easy to reduce from the general case to this one). The following proposition describes which Breuil–Kisin modules embed into \(f_*N\) as in Lemma 6.3.1.
Proposition 6.5.1
Assume \(T(f_*N)\) is irreducible. Let \(M \subset f_*N\) be a finite free \(k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}\)-sub-module with \(M[\frac{1}{u}] = (f_*N)[\frac{1}{u}]\). Then \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) if and only if the following conditions are satisfied.
- 1.
If \(m \in M\) then \(\varphi (m) \in M\) and if \(m \in f_*N\) and \(\varphi (m) \in uM\) then \(m \in M\).
- 2.
If \(m \in f_*N\) then \(um \in M\).
- 3.
If \(\sum \alpha _\theta e_\theta \in M\) with \(\alpha _{\theta } \in {\mathbb {F}}\) then
$$\begin{aligned} \sum _{r_\theta \equiv r~{\text { mod}}p} \alpha _{\theta } e_{\theta } \in M \end{aligned}$$
for every \(0 \le r \le p\).
Proof (that SD implies (1), (2) and (3))
If \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) then \(F^0 M_k = M_k\) and \(F^{p+1}M_k = 0\). The first condition implies \(\varphi (m) \in M\) whenever \(m \in M\). The second condition implies
Let us show this implies (2). As \(M[\frac{1}{u}] = (f_*N)[\frac{1}{u}]\) there is, for each \(\theta \), a smallest integer \(\delta _{\theta } \ge 0\) with \(u^{\delta _{\theta }} e_\theta \in M\). Since \(\varphi (u^{\delta _{\theta \circ \varphi }}e_{\theta \circ \varphi }) = u^{\delta _{\theta \circ \varphi }p - \delta _{\theta } + r_\theta } u^{\delta _{\theta }}e_\theta \) and \(u^{\delta _{\theta }}e_\theta \not \in uM\) we see \(\delta _{\theta \circ \varphi }p - \delta _{\theta } + r_\theta \in [0,p]\). Therefore \(\delta _{\theta \circ \varphi }p - \delta _{\theta } \le p\) and
$$\begin{aligned} (p^{[l:{\mathbb {F}}_p]}-1)\delta _\theta = \sum _{i=0}^{[l:{\mathbb {F}}_p]-1} p^i(p\delta _{\theta \circ \varphi ^{i+1}} - \delta _{\theta \circ \varphi ^i}) \le p(p^{[l:{\mathbb {F}}_p]}-1)/(p-1) \end{aligned}$$
This implies \(\delta _\theta \in [0,1]\) if \(p>2\), and \(\delta _\theta \in [0,2]\) if \(p=2\). If \(p=2\) and \(\delta _{\theta \circ \varphi } =2\) then, as \(r_\theta + p\delta _{\theta \circ \varphi } - \delta _\theta \in [0,p]\), we must have \(\delta _\theta = 2\) and \(r_\theta = 0\). Thus \(r_\theta =0\) for all \(\theta \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) and so T(N) is the trivial character. In this case \(T(f_*N)\) is not irreducible.
Now we can deduce the second part of (1). If \(m \in f_*N\) and \(\varphi (m) \in uM\) then \(\varphi (um) \in u^{p+1}M\); as (2) holds we have \(um \in M\) and so the above bullet point give \(um \in uM\). Hence \(m \in M\).
To prove (3) we first make the following claim. Suppose that \(\sum \alpha _{\theta } e_\theta \in M\) with \(\alpha _{\theta } \in {\mathbb {F}}[[u]]\) (so this sum is more general than that in (3)) and that \(u^r \sum \alpha _{\theta } e_\theta \in M^\varphi \) for \(0 \le r \le p\). Then:
There exist \(\widetilde{\alpha _{\theta ,r}} \in {\mathbb {F}}[[u]]\) such that \(\sum \widetilde{\alpha _{\theta ,r}} e_\theta \in M\), \(u^{r-1} \sum \widetilde{\alpha _{\theta ,r}} e_\theta \in M^\varphi \) and
$$\begin{aligned} \widetilde{\alpha _{\theta ,r}} \equiv {\left\{ \begin{array}{ll} \alpha _{\theta }~{\text {mod}} u &{} \text {if }r_\theta \ne r, \text { except possibly if }r_\theta = 0\text { and }r = p \\ 0~{\text {mod}} u &{} \text {if }r_\theta = r \end{array}\right. } \end{aligned}$$
To verify the claim we use that, since M is strongly divisible, the map \(M_k \rightarrow M^\varphi _k\) is an isomorphism of filtered modules. As \(u^r\sum \alpha _{\theta } e_\theta \in F^rM^\varphi \) it follows that there exists an element \(\beta \in F^r M\) such that \(\varphi (\beta ) - u^r\sum \alpha _{\theta } e_\theta \in uM^\varphi \). If \(\beta = \sum \beta _\theta e_{\theta \circ \varphi }\) then
$$\begin{aligned} \sum \varphi (\beta _{\theta }) u^{r_\theta } e_\theta - u^r \sum \alpha _{\theta } e_\theta = \sum \left( \varphi (\beta _{\theta }) u^{r_\theta } - u^r\alpha _{\theta } \right) e_\theta \in uM^\varphi \cap u^r M\nonumber \\ \end{aligned}$$
(6.5.2)
As \(u^r M \subset u^rN\) and \(uM^\varphi \subset uN^\varphi \) we deduce that
$$\begin{aligned} v_u \left( \varphi (\beta _{\theta }) u^{r_\theta } - u^r\alpha _{\theta } \right) > {\text {max}} \lbrace r_\theta ,r-1\rbrace \end{aligned}$$
Here \(v_u\) denotes the u-adic valuation. If \(r_\theta = r\) this inequality implies \(\alpha _{\theta } \equiv \varphi (\beta _{\theta })\) modulo u, and so we can write \(\varphi (\beta _{\theta }) = \alpha _{\theta } + u \gamma _{\theta }\) for some \(\gamma _{\theta } \in {\mathbb {F}}[[u]]\). If \(r > r_\theta \) the inequality implies \(\varphi (\beta _{\theta }) \equiv 0\) modulo u, and so we can write \(\varphi (\beta _{\theta }) = u^p\gamma _{\theta }\) for some \(\gamma _{\theta } \in {\mathbb {F}}[[u]]\). If \(r_\theta > r\) then we simply write \(\varphi (\beta _{\theta }) = \gamma _{\theta }\). Dividing (6.5.2) by \(u^r\) we therefore see that
$$\begin{aligned} \sum _{r_\theta \ne r} \alpha _{\theta } e_\theta - \sum _{r_\theta = r} u \gamma _{\theta } e_\theta - \sum _{r> r_\theta } u^{p-r+r_\theta } \gamma _{\theta } e_\theta - \sum _{r_\theta > r} u^{r_\theta - r} \gamma _{\theta } e_\theta \in M \end{aligned}$$
and that \(u^{r-1}\) times this element is contained in \(M^\varphi \). As such, taking
$$\begin{aligned} \widetilde{\alpha _{\theta ,r}} ={\left\{ \begin{array}{ll} -u\gamma _{\theta } &{} \text {if }r_\theta = r \\ \alpha _{\theta } - u^{p-r+r_\theta } \gamma _{\theta } &{} \text {if }r> r_\theta \\ \alpha _{\theta } - u^{r_\theta -r} \gamma _{\theta } &{} \text {if } r_\theta > r \end{array}\right. } \end{aligned}$$
gives the claim.
We now use the claim to verify (3). Suppose \(\sum \alpha _\theta e_\theta \in M\), now with \(\alpha _{\theta } \in {\mathbb {F}}\). As already remarked, the fact that \({\text {Weight}}(M) \subset [0,p]\) implies \(u^pM \subset M^\varphi \). In particular \(u^p\sum \alpha _{\theta } e_\theta \in M^\varphi \) so the claim applies, and produces \(\sum \widetilde{\alpha _{\theta ,p}} e_\theta \in M\). Using that \(ue_\theta \in M\) for every \(\theta \) we deduce that there are \(\gamma _\theta \in {\mathbb {F}}\) such that \(\sum _{r_\theta \ne p} \alpha _{\theta } e_\theta + \sum _{r_\theta = 0} \gamma _{\theta } e_\theta \in M\). Hence
$$\begin{aligned} \sum _{r_\theta = p} \alpha _{\theta } e_\theta - \sum _{r_\theta = 0} \gamma _\theta e_\theta \in M \end{aligned}$$
As \(u^{p-1} \sum \widetilde{\alpha _{\theta ,p}} e_\theta \in M^\varphi \) we can then apply the claim to \( \sum \widetilde{\alpha _{\theta ,p}} e_\theta \); this yield \(\sum \widetilde{\alpha _{\theta ,p-1}} e_\theta \in M\). Again using that \(ue_\theta \in M\) for each \(\theta \), we deduce that \(\sum _{r_\theta \ne p,p-1} \alpha _{\theta } e_\theta + \sum _{r_\theta = 0} \gamma _{\theta } e_\theta \in M\), and hence
$$\begin{aligned} \sum _{r_\theta = p-1} \alpha _{\theta } e_\theta \in M \end{aligned}$$
Repeatedly applying the claim in this fashion we deduce that \(\sum _{r_\theta = r} \alpha _{\theta } e_\theta \) for \(0< r< p\) and that \(\sum _{r_\theta = 0} \alpha _{\theta } e_\theta + \sum _{r_\theta = 0} \gamma _{\theta } e_\theta \in M\). In particular we find
$$\begin{aligned} \sum _{r_\theta = p} \alpha _{\theta } e_\theta + \sum _{r_\theta = 0 } \alpha _{\theta } e_\theta = \sum _{r_\theta \equiv 0 \mod p} \alpha _{\theta } e_\theta \in M \end{aligned}$$
which finishes the proof that \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) implies (1), (2) and (3) hold. \(\square \)
Finishing the proof of Proposition 6.5.1
Let N be as in the previous subsection and suppose that \(M \subset f_*N\) is a free \(k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}\)-module with \(M[\frac{1}{u}] = (f_*N)[\frac{1}{u}]\). Assume that M satisfies conditions (1), (2) and (3) from Proposition 6.5.1. We are going to prove that \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\). Along the way we shall describe the weights of M in terms of the \(r_\theta \).
Construction 6.6.1
For a fixed \(\lambda \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) define an ordering on \({\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) by asserting that
$$\begin{aligned} \lambda \circ \varphi<_\lambda \lambda \circ \varphi ^2<_\lambda \cdots<_{\lambda } \lambda \circ \varphi ^{[l:{\mathbb {F}}_p] - 1} <_\lambda \lambda \end{aligned}$$
Using this ordering we define \(X \subset {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) by
$$\begin{aligned} \theta \not \in X \Leftrightarrow \text {there exists }\alpha _{\kappa } \in {\mathbb {F}}\text { such that }e_\theta +\sum _{\kappa <_\lambda \theta } \alpha _{\kappa } e_{\kappa } \in M \end{aligned}$$
(6.6.2)
Clearly X depends upon the choice of \(\lambda \).
Lemma 6.6.3
-
1.
If \(\sum _{\kappa \in X} \alpha _{\kappa }e_\kappa \) is an \({\mathbb {F}}\)-linear combination contained in M then \(\sum _{\kappa \in X} \alpha _{\kappa } e_\kappa = 0\).
-
2.
If \(\theta \not \in X\) there exists a unique \({\mathbb {F}}\)-linear combination
$$\begin{aligned} e_{\theta } + \sum \alpha _{\kappa } e_{\kappa } \in M, \quad \alpha _{\kappa } \in {\mathbb {F}} \end{aligned}$$
in which the sum runs over \(\kappa \in X\) satisfying (i) \(\kappa <_\lambda \theta \) (ii) \(r_{\kappa } \equiv r_\theta \) modulo p and (iii) \(\kappa |_k = \theta |_k\). In particular, the element lies in \(M_{\theta |_k}\).
Proof
(1) If \(\sum _{\kappa \in X} \alpha _{\kappa } e_{\kappa } \ne 0\) then there exists a maximal \(\kappa \) (with respect to \(<_{\lambda }\)) with \(\alpha _{\kappa } \ne 0\). From \(\sum _{\kappa \in X} \alpha _{\kappa } e_\kappa \in M\) it follows this maximal \(\kappa \) is not contained in X, a contradiction.
(2) As \(\theta \not \in X\), there exists \(e_\theta + \sum \alpha _{\kappa } e_{\kappa } \in M\) with the sum running over \(\kappa <_\lambda \theta \). Arguing inductively one shows there exists such a sum running only over those \(\kappa <_\lambda \theta \) with \(\kappa \in X\). There can be at most one sum of this form; indeed their difference would be a sum as in (1) and so would be zero. Condition (3) of Proposition 6.5.1 therefore implies the sum may be taken to run over \(\kappa \) additionally satisfying (ii). As \(M = \prod _{\tau \in {\text {Hom}}_{{\mathbb {F}}_p}(k,{\mathbb {F}})} M_{\tau }\) we also have (iii). \(\square \)
Definition 6.6.4
Consider \({\mathbb {F}}\)-linear combinations of the form
$$\begin{aligned} e_{\iota } + \sum _{0 < j \le I} \alpha _j e_{\iota \circ \varphi ^j} \in M \end{aligned}$$
(6.6.5)
with \(0 \le I < [l:{\mathbb {F}}_p]\) and \(\iota \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\). We say (6.6.5) is minimal if there exists no \(\iota ' \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) together with an \({\mathbb {F}}\)-linear combination \(e_{\iota '} + \sum _{0 < j \le J} \alpha _j e_{\iota ' \circ \varphi ^j} \in M\) such that \(J<I\). Note that for a fixed \(\iota \) there can exist at most one minimal sum as in (6.6.5); if there were two their difference would have shorter length.
Note that when there exists a \(\theta \) such that \(e_\theta \in M\) then the minimal elements are simply scalar multiples of \(e_\theta \) for any \(\theta \) with \(e_\theta \in M\).
Lemma 6.6.6
If (6.6.5) is a minimal sum then \(r_{\iota \circ \varphi ^j} = r_\iota \) whenever \(\alpha _j \ne 0\) and \(j \le I\).
Proof
Uniqueness of minimal elements and condition (3) of Proposition 6.5.1 implies \(r_{\iota } \equiv r_{\iota \circ \varphi ^i}\) modulo p. Since each \(r_{\iota \circ \varphi ^j} \in [0,p]\) this will be an equality, except possibly if \(r_\iota = 0\) or p. In this case set
$$\begin{aligned} z = u^{\gamma _0} e_{\iota \circ \varphi } + \sum _{0 <j \le I} u^{\gamma _{j}} \alpha _{j} e_{\iota \circ \varphi ^{j+1}} \end{aligned}$$
where \(\gamma _{j} = 0\) if \(r_{\iota \circ \varphi ^j} = p\) and \(\gamma _{j} = 1\) if \(r_{\iota \circ \varphi ^j} = 0\). Then \(\varphi (z)\) equals \(u^p\) times (6.6.5) and so condition (1) of Proposition 6.5.1 implies \(z \in M\). Thus, either all \(\gamma _i =0\) or all equal 1, otherwise we would obtain an element of M contradicting the minimality of (6.6.5). \(\square \)
The next proposition is where we use that \(T(M)= T(f_*N)\) is irreducible.
Proposition 6.6.7
There exists \(\lambda \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) such that
- 1.
If \(\theta \in X\) and \(\theta \circ \varphi \not \in X\) then \(r_\theta >0\).
- 2.
If \(\theta \not \in X\) and \(\theta \circ \varphi \in X\) then \(r_{\theta } = 0\).
- 3.
If \(\theta \in X\) and \(e_{\theta \circ \varphi } \not \in M\) then \(0 \le r_\theta \le 1\). In particular this holds if \(\theta \circ \varphi \in X\).
Proof
First observe (3) holds for any choice of \(\lambda \). Indeed, if \(e_{\theta \circ \varphi } \not \in M\) then condition (1) of Proposition 6.5.1 would imply \(\varphi (e_{\theta \circ \varphi }) = u^{r_\theta } e_\theta \not \in uM\). If \(r_\theta \ge 2\) then \(u^{r_\theta } e_\theta \in u^2f_*N\), which is contained in uM by (2) of Proposition 6.5.1.
Next we show (2) holds whenever \(r_\lambda =0\). Suppose \(\theta \not \in X\) and \(r_\theta > 0\) (we’re assuming that \(r_\lambda = 0\) so \(\theta \ne \lambda \)). We show \(\theta \circ \varphi \not \in X\). Choose \(e_\theta + \sum _{\kappa <_\lambda \theta } \alpha _{\kappa }e_\kappa \in M\) as in Lemma 6.6.3. Set \(z = e_{\theta \circ \varphi } + \sum _{r_\kappa \ne 0} \alpha _{\kappa } e_{\kappa \circ \varphi } + u\sum _{r_\kappa = 0} \alpha _{\kappa } e_{\kappa \circ \varphi }\). Using that \(r_\theta \equiv r_\kappa \) modulo p and \(r_\theta >0\) we see that \(\varphi (z) = u^{r_\theta }( e_\theta + \sum \alpha _{\kappa }e_\kappa )\). Condition (1) of Proposition 6.5.1 implies \(z \in M\). Since \(\theta \ne \lambda \), if \(\kappa <_\lambda \theta \) then \(\kappa \circ \varphi <_\lambda \theta \circ \varphi \). Therefore z shows \(\theta \circ \varphi \not \in X\).
Now choose a minimal sum as in (6.6.5) (if none exists then we must have \(M = u(f_*N)\) and so \(X = {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\), in which case conditions (1), (2), and (3) hold vacuously). We are going to show that either \(\lambda = \iota \) satisfies the conditions of the proposition or \(e_{\iota \circ \varphi } + \sum \alpha _{j} e_{\iota \circ \varphi ^{j+1}} \in M\). Let us explain why this implies the proposition. If there exists no \(\lambda \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) satisfying conditions (1)–(3) then it would follow that \(e_{\iota \circ \varphi ^i} + \sum \alpha _{j} e_{\iota \circ \varphi ^{j+i}} \in M\) for every \(i \ge 0\). Lemma 6.6.6 would then imply \(r_{\iota '} = r_{\iota ' \circ \varphi ^I}\) for every \(\iota ' \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\). If \(\omega \) is the character through which \(G_{K}\) acts on T(N) we would then have \(\omega = \prod _\iota \omega _{\iota }^{-r_\iota } = \prod \omega _{\iota \circ \varphi ^I}^{-r_\iota } = \omega ^{p^I}\). If \(I >0\) this contradicts the irreducibility of \(T(M) = {\text {Ind}}_L^K T(N)\). If \(I = 0\) it would follow that every \(e_{\iota '} \in M\), so \(X = \emptyset \) and conditions (1)–(3) hold vacuously.
Set \(z = e_{\iota \circ \varphi } + \sum \alpha _{j} e_{\iota \circ \varphi ^{j+1}}\). If \(r_{\iota } >0\) then we always have \(z \in M\) for the following reason: Lemma 6.6.6 implies \(r_{\iota \circ \varphi ^j} = r_\iota \) whenever \(\alpha _{\iota \circ \varphi ^j} \ne 0\) and so \(\varphi (z) \in u^{r_\iota }M\) from which we deduce \(z \in M\) using condition (1) of Proposition 6.5.1. If instead \(r_{\iota } = 0\) set \(\lambda = \iota \). The first and second paragraph of this proof shows that (2) and (3) hold. If (1) holds also then we are done, so assume it does not. There must then exist \(\theta \in X\) with \(\theta \circ \varphi \not \in X\) and \(r_{\theta } = 0\). We use this to show \(z \in M\). If \(\theta = \lambda \) then \(\lambda \circ \varphi \not \in X\) which means \(e_{\lambda \circ \varphi } \in M\); by minimality \(z = e_{\lambda \circ \varphi }\) and we are done. Let us therefore assume \(\theta \ne \lambda \). Consider the unique
$$\begin{aligned} f_{\theta \circ \varphi } = e_{\theta \circ \varphi } + \sum _{\kappa <_{\lambda } \theta \circ \varphi } \alpha _{\kappa }e_\kappa \end{aligned}$$
from Lemma 6.6.3. As \(r_\theta = 0\) and \(\kappa <_{\lambda } \theta \circ \varphi \) implies \(\kappa \circ \varphi ^{-1} <_{\lambda } \theta \), except if \(\kappa = \lambda \circ \varphi \), we obtain
$$\begin{aligned} \varphi (f_{\theta \circ \varphi }) = e_\theta + \alpha _{\lambda \circ \varphi } e_\lambda + \sum _{\kappa \circ \varphi ^{-1}<_{\lambda } \theta } \alpha _{\kappa \circ \varphi } u^{r_{\kappa \circ \varphi ^{-1}}} e_{\kappa \circ \varphi ^{-1}} \in M \end{aligned}$$
Removing those terms with \(r_{\kappa \circ \varphi ^{-1}} > 0\) and re-indexing, we obtain
$$\begin{aligned} e_\theta + \alpha e_{\lambda } + \sum _{\kappa <_{\lambda } \theta } \beta _{\kappa } e_{\kappa } \in M \end{aligned}$$
(6.6.8)
for some \(\alpha , \beta _{\kappa } \in {\mathbb {F}}\). If \(\alpha = 0\) then (6.6.8) contradicts the assumption that \(\theta \in X\). If we write \(\theta = \lambda \circ \varphi ^J\) and \(J<I\) then (6.6.8) contradicts the assumption that (6.6.5) is minimal. If \(I < J\) then the difference between (6.6.8) and the product of \(\alpha \) and (6.6.5) again contradicts the assumption that \(\theta \in X\). Thus \(I = J\). The uniqueness of minimal elements then implies (6.6.8) equals \(\alpha \) times (6.6.5). Thus \( z = \frac{f_{\theta \circ \varphi }}{\alpha } \in M\) which completes the proof. \(\square \)
Proof (End of the proof of Proposition 6.5.1)
We have to show M is strongly divisible. Fix \(\lambda \) as in Proposition 6.6.7 and for \(\theta \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) set
$$\begin{aligned} f_\theta = {\left\{ \begin{array}{ll} e_{\theta } + \sum \alpha _{\kappa } e_{\kappa } \text { as in Lemma}~6.6.3 &{} \text {if }\theta \not \in X \\ ue_{\theta } &{} \text {if }\theta \in X \end{array}\right. } \end{aligned}$$
For \(\tau \in {\text {Hom}}_{{\mathbb {F}}_p}(k,{\mathbb {F}})\) the \(f_\theta \) with \(\theta |_k = \tau \) form an \({\mathbb {F}}[[u]]\)-basis of \(M_\tau \). To see this let \(W \subset M_\tau \) be the subspace they span. It is easy to see that if \(\theta |_k = \tau \) then \(ue_\theta \in W\). It therefore suffices to show any \(\sum \alpha _\theta e_{\theta } \in M_\tau \) with \(\alpha _{\theta } \in {\mathbb {F}}\) is in W. We see that \(\sum \alpha _{\theta }e_\theta - \sum _{\theta \not \in X} \alpha _{\theta }f_\theta \) is an \({\mathbb {F}}\)-linear combination of \(e_\theta \) with \(\theta \in X\), and is contained in M. Such a linear combination must be zero (cf. the proof of Lemma 6.6.3) so \(W = M_\tau \), as claimed.
For each \(\theta \) we now construct elements \(g_{\theta \circ \varphi } \in M_{\theta \circ \varphi |_k}, h_\theta \in M_{\theta |_k}\) so that \(\varphi (g_{\theta \circ \varphi }) = u^{r_\theta + ps_{\theta \circ \varphi } - s_\theta } h_\theta \) where
$$\begin{aligned} s_\theta ={\left\{ \begin{array}{ll} 1 &{} \text {if }\theta \in X \\ 0 &{} \text {if }\theta \notin X \end{array}\right. } \end{aligned}$$
(6.6.9)
We do this on a case-by-case basis. We note each \(f_\theta \in M_{\theta |_k}\) by (iii) of Lemma 6.6.3(2).
Suppose \(\theta \not \in X\) and \(\theta \circ \varphi \in X\). Set \(h_\theta := f_\theta = e_\theta + \sum _{\kappa <_\lambda \theta , \kappa \in X} \alpha _{\kappa } e_{\kappa } \in M_{\theta |_k}\). (2) of Proposition 6.6.7 implies \(r_\theta = 0\), so each \(r_\kappa \), being congruent to \(r_\theta \) modulo p, equals 0 or p. If \(r_\kappa = p\) then (3) of Proposition 6.6.7 implies \(e_{\kappa \circ \varphi } \in M\), and so \(\kappa \circ \varphi \not \in X\) and \(f_{\kappa \circ \varphi } = e_{\kappa \circ \varphi }\). If \(r_\kappa = 0\) then (1) of Proposition 6.6.7 implies \(\kappa \circ \varphi \in X\). Thus
$$\begin{aligned} g_{\theta \circ \varphi } := f_{\theta \circ \varphi } + \sum _{\kappa <_\lambda \theta , \kappa \in X} \alpha _{\kappa } f_{\kappa \circ \varphi } \in M \end{aligned}$$
is such that \(\varphi (g_{\theta \circ \varphi }) = u^ph_\theta \). Since \(h_\theta \in M_{\theta }\) we must have \(g_{\theta \circ \varphi } \in M_{\theta \circ \varphi |_k}\).
Suppose \(\theta \not \in X\), \(\theta \circ \varphi \not \in X\) and \(r_\theta =0\). Set \(g_{\theta \circ \varphi } := f_{\theta \circ \varphi } = e_{\theta \circ \varphi } + \sum _{\kappa <_\lambda \theta \circ \varphi ,\kappa \in X} \alpha _{\kappa } e_{\kappa } \in M_{\theta \circ \varphi |_k}\). Since \(\kappa \in X\), if \(\kappa \circ \varphi ^{-1} \not \in X\) then \(r_{\kappa \circ \varphi ^{-1}} = 0\) by (2) of Proposition 6.6.7. By (3) of Proposition 6.6.7, if \(\kappa \circ \varphi ^{-1} \in X\) then \(r_{\kappa \circ \varphi ^{-1}} \in [0,1]\). Therefore the difference between \(\varphi (g_{\theta \circ \varphi })\) and
$$\begin{aligned} h_\theta := f_{\theta } + \sum _{\kappa<_\lambda \theta \circ \varphi ,\kappa \circ \varphi ^{-1} \not \in X} \alpha _{\kappa } f_{\kappa \circ \varphi ^{-1}} + \sum _{\kappa <_\lambda \theta \circ \varphi ,\kappa \circ \varphi \in X,r_{\kappa \circ \varphi ^{-1}} = 1} \alpha _{\kappa } f_{\kappa \circ \varphi ^{-1}} \end{aligned}$$
is an \({\mathbb {F}}\)-linear combination of \(e_\kappa \) with \(\kappa \in X\). Since this \({\mathbb {F}}\)-linear combination is contained in M it must be zero by (1) of Lemma 6.6.3. Therefore \(\varphi (g_{\theta \circ \varphi }) = h_\theta \). Since \(g_{\theta \circ \varphi |_k}\) we must have \(h_\theta \in M_{\theta |_k}\).
Suppose \(\theta \not \in X\), \(\theta \circ \varphi \not \in X\) and \(r_\theta > 0\). Set \(h_\theta := f_\theta = e_\theta + \sum _{\kappa <_\lambda \theta ,\kappa \in X} \alpha _{\kappa }e_\kappa \in M_{\theta |_k}\). Each \(r_\kappa \equiv r_\theta \) modulo p and so \(\varphi \) sends
$$\begin{aligned} e_{\theta \circ \varphi } + \sum _{r_\kappa > 0} \alpha _{\kappa }e_{\kappa \circ \varphi } + u\sum _{r_\kappa = 0} \alpha _{\kappa } e_{\kappa \circ \varphi } \end{aligned}$$
onto \(u^{r_\theta } h_\theta \) (note the term \(u\sum _{r_\kappa = 0} \alpha _{\kappa } e_{\kappa \circ \varphi }\) appears only if \(r_\theta = p\)). As \(r_\theta >0\) this displayed sum is contained in M by condition (1) of Proposition 6.5.1. We claim this displayed sum is equal to
$$\begin{aligned} g_{\theta \circ \varphi } := f_{\theta \circ \varphi } + \sum _{\kappa<_\lambda \theta , \kappa \circ \varphi \not \in X} \alpha _{\kappa } f_{\kappa \circ \varphi } + \sum _{\kappa <_\lambda \theta , \kappa \circ \varphi \in X, r_\kappa =0} \alpha _{\kappa } f_{\kappa \circ \varphi } \end{aligned}$$
To see this note that, by (1) of Proposition 6.6.7, if \(r_\kappa = 0\) then \(\kappa \circ \varphi \in X\) and if \(r_{\kappa \circ \varphi } \not \in X\) then \(r_\kappa > 0\). From this it follows that the difference between these two sums, which is an element of M, is an \({\mathbb {F}}\)-linear combination of \(e_\kappa \) with \(\kappa \in X\). This difference is therefore zero, and so \(\varphi (g_{\theta \circ \varphi }) = u^{r_\theta } h_\theta \). As \(h_\theta \in M_{\theta |_k}\) we have \(g_{\theta \circ \varphi } \in M_{\theta \circ \varphi |_k}\).
Suppose \(\theta \in X\) and \(\theta \circ \varphi \not \in X\). Set \(g_{\theta \circ \varphi } := f_{\theta \circ \varphi } = e_{\theta \circ \varphi } + \sum _{\kappa <_\lambda \theta \circ \varphi ,\kappa \in X} \alpha _{\kappa }e_{\kappa } \in M_{\theta \circ \varphi |_k}\), and set
$$\begin{aligned} h_\theta := f_\theta + \sum _{\kappa<_\lambda \theta \circ \varphi ,\kappa \circ \varphi ^{-1} \not \in X} \alpha _{\kappa } f_{\kappa \circ \varphi ^{-1}} + \sum _{\kappa <_\lambda \theta \circ \varphi ,\kappa \circ \varphi ^{-1} \in X, r_{\kappa \circ \varphi ^{-1}} = 1} \alpha _{\kappa } f_{\kappa \circ \varphi ^{-1}} \end{aligned}$$
We claim \(\varphi (g_{\theta \circ \varphi }) = u^{r_\theta -1}h_\theta \). If \(e_{\theta \circ \varphi } \in M\) then this is clear since \(g_{\theta \circ \varphi } = e_{\theta \circ \varphi }\) and \(h_\theta = ue_\theta \). If \(e_{\theta \circ \varphi } \not \in M\) then (1) and (3) of Proposition 6.6.7 implies \(r_\theta = 1\), so we have to show \(\varphi (g_{\theta \circ \varphi }) = h_\theta \). Proposition 6.6.7 tells us \(\kappa \in X\) and \(\kappa \circ \varphi ^{-1} \not \in X\) implies \(r_{\kappa \circ \varphi ^{-1}} = 0\), while if \(\kappa \in X\) and \(\kappa \circ \varphi ^{-1} \in X\) then \(r_{\kappa \circ \varphi ^{-1}} \in [0,1]\). Using these two facts we see that the difference between \(\varphi (g_{\theta \circ \varphi })\) and \(h_\theta \) is an \({\mathbb {F}}\)-linear combination of \(e_\kappa \) with \(\kappa \in X\). Since this difference is contained in M it must be zero. As \(g_{\theta \circ \varphi } \in M_{\theta \circ \varphi |_k}\) we have \(h_\theta \in M_{\theta |_k}\).
Finally, if \(\theta \in X\) and \(\theta \circ \varphi \in X\) set \(g_{\theta \circ \varphi } := f_{\theta \circ \varphi } \in M_{\theta \circ \varphi |_k}\) and \(h_{\theta } := f_\theta \in M_{\theta |_k}\). Then \(\varphi (g_{\theta \circ \varphi }) = u^{r_\theta + p-1} h_\theta \).
To finish the proof it suffices to show that for \(\theta \) with \(\theta |_k = \tau \), the \(g_{\theta \circ \varphi }\) form an \({\mathbb {F}}[[u]]\)-basis of \(M_{\tau \circ \varphi }\), and the \(h_\theta \) form an \({\mathbb {F}}[[u]]\)-basis of \(M_\tau \). If H is the \({\mathbb {F}}[[u]]\)-linear endomorphism of \(M_\tau \) sending \(f_\theta \) onto \(h_\theta \) then \(H - {\text {Id}}\) sends \(f_\theta \) onto \({\mathbb {F}}\)-linear combinations of \(f_{\kappa \circ \varphi ^{-1}}\) with \(\kappa <_\lambda \theta \circ \varphi \). Hence \(H- {\text {Id}}\) is nilpotent, H is an automorphism, and the \(h_\theta \) form an \({\mathbb {F}}[[u]]\)-basis as claimed. A similar observation shows the \(g_{\theta \circ \varphi }\) also form an \({\mathbb {F}}[[u]]\)-basis. \(\square \)
Using Remark 5.2.6 we deduce:
Corollary 6.6.10
With \(s_\theta \) as in (6.6.9)
$$\begin{aligned} {\text {Weight}}_\tau (M) = \lbrace r_\theta + ps_{\theta \circ \varphi } - s_\theta \mid \theta |_k = \tau \rbrace \end{aligned}$$
Putting everything together
Applying what we’ve shown so far in this subsection gives:
Proposition 6.7.1
Let \(M \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) with T(M) irreducible. Then there exist integers \(\widetilde{r_{\theta }}\) indexed over \(\theta \in {\text {Hom}}_{{\mathbb {F}}_p}(l,{\mathbb {F}})\) such that (i):
$$\begin{aligned} T(M) \otimes _{{\mathbb {F}}} \overline{{\mathbb {F}}}_p = \psi \otimes {\text {Ind}}_{L_\infty }^{K_\infty } \Bigg (\prod _{\theta } \omega _{\theta }^{-\widetilde{r_{\theta }}}\Bigg ) \end{aligned}$$
for some unramified character \(\psi \) and for \(L_\infty = L(\pi ^{1/p^\infty })\) with L an unramified extension K, and such that (ii):
$$\begin{aligned} {\text {Weight}}_\tau (M) = \lbrace \widetilde{r_\theta } \mid \theta |_k = \tau \rbrace \end{aligned}$$
Proof
Lemma 6.3.1 produces a rank one \(N \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\), which we assume is as in Lemma 6.1.1, together with an embedding \(M \hookrightarrow f_*N\). We want to apply the results of Sects. 6.5 and 6.6, so we require the \(x \in {\mathbb {F}}^\times \) appearing in the definition of N to equal 1. Let us explain how to reduce to this case. Let \({\text {ur}}_x \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) be the rank one object given by
$$\begin{aligned} {\text {ur}}_x = k[[u]] \otimes _{{\mathbb {F}}_p} {\mathbb {F}}, \quad \varphi _{{\text {ur}}_x}(1) = x e_{\tau _0} + \sum _{\tau \ne \tau _0} e_\tau \end{aligned}$$
Set \({\widetilde{M}} = {\text {Hom}}({\text {ur}}_x,M)^{{\mathcal {O}}}\) (recall Construction 4.3.3). One easily checks that \({\widetilde{M}} \in {\text {Mod}}^{{\text {SD}}}_k({\mathcal {O}})\) and that \({\text {Weight}}_\tau ({\widetilde{M}}) = {\text {Weight}}_\tau (M)\) for each \(\tau \) by verifying that condition (2) of Lemma 5.3.4 holds. The last sentence of Construction 4.3.3 implies
$$\begin{aligned} T({\widetilde{M}}) = {\text {Hom}}(\psi _x,{\text {Ind}}_{L_\infty }^{K_\infty } \chi ) = {\text {Ind}}_{L_\infty }^{K_\infty } (\psi _x^{-1} \chi ) \end{aligned}$$
Thus, if the proposition holds for \({\widetilde{M}}\) it holds for M, and so we may assume \(x= 1\). Applying Corollary 6.6.10, \({\text {Weight}}_\tau (M) = \lbrace r_\theta + ps_{\theta \circ \varphi } - s_\theta \mid \theta |_k = \tau \rbrace \). On the other hand, \(\chi = T(N)\) and this equals \(\prod _\theta \omega _{\theta }^{r_\theta +ps_{\theta \circ \varphi } - s_\theta }\) by Proposition 6.1.3. Therefore, we can take \(\widetilde{r_{\theta }} = r_\theta + ps_{\theta \circ \varphi } - s_\theta \). \(\square \)