Proof. We may and will assume that $M$ is simple, and that there exist a projective smooth variety $Y$ of dimension $d$ over $K$, an integer $n$, and an André motivated cycle

$$\begin{eqnarray}\unicode[STIX]{x1D709}\in A_{\text{mot}}^{d}(Y\times _{K}Y)\end{eqnarray}$$

such that $\unicode[STIX]{x1D709}$ acts as the idempotent cutting out $M$ in $\mathfrak{h}^{n}(Y)$. Let $e_{p}$ (respectively, $e_{\unicode[STIX]{x1D70E}}$, $e_{\text{dR}}$) be the image of $\unicode[STIX]{x1D709}$ in the $p$-adic étale (respectively, $\unicode[STIX]{x1D70E}$-Betti, de Rham) realisation

$$\begin{eqnarray}e_{p}\in H^{2d}((Y\times Y)\otimes _{K}\overline{K},\mathbb{Q}_{p})(d),\quad e_{\unicode[STIX]{x1D70E}}\in H^{2d}(\unicode[STIX]{x1D70E}(Y\times Y),\mathbb{Q}(d)),\quad e_{\text{dR}}\in H_{\text{dR}}^{2d}(Y\times Y)(d).\end{eqnarray}$$

We have the diagram

Here we suppressed $Y\times Y$ in the argument for all cohomology theories as well as the degree $2d$ and the Tate twist $(d)$. Undecorated arrows are extensions of scalars and Art denotes Artin’s comparison isomorphism.

Main point. The diagram is commutative. In particular, the image of $\unicode[STIX]{x1D709}$ in any group in the diagram is the same, no matter which path emanating from $A_{\text{mot}}^{d}(Y\times Y)$ is followed.

This follows from the definition of André motivated cycles, together with the fact that the comparison isomorphisms on display are isomorphisms between Weil cohomology theories (and as such compatible with pullback, pushforward, cup product, cycle class, and Poincaré duality, that are involved in the definition of André motivated cycles). See André [Reference AndréAnd96, §§2.3 and 2.4].

(In contrast, it is not clear whether the similar diagram would be commutative if we replace the apex $A_{\text{mot}}$ with the larger space of the absolute Hodge cycles.)

Now applying the idempotents obtained from $\unicode[STIX]{x1D709}$ to the similar diagram without apex,

we get the diagram

Now, on the one hand, the Hodge–Tate weights of $M_{p}$ can be read off from the filtration on

$$\begin{eqnarray}(M_{p}\otimes _{\mathbb{Q}_{p}}B_{\text{dR}})^{\operatorname{Gal}(\overline{K}/K)}\simeq M_{\text{dR}}\quad \text{(filtered isomorphism)}.\end{eqnarray}$$

On the other hand, the Hodge numbers of $M_{\unicode[STIX]{x1D70E}}$ can also be read off from the (algebraic) Hodge filtration on $M_{\text{dR}}$ in a similar manner.◻

Proof. Let $\unicode[STIX]{x1D70B}:Y\longrightarrow X$ be any resolution of singularities. We use the main theorem of de Cataldo and Migliorini [Reference de Cataldo and MigliorinideCM15], strengthened in $K$-rationality by Patrikis [Reference PatrikisPat16, §8], to deduce the existence of an André motivated cycle

$$\begin{eqnarray}\unicode[STIX]{x1D709}:=\unicode[STIX]{x1D709}_{n}\in A_{\text{mot}}^{d}(Y\times Y)\end{eqnarray}$$

such that the Betti realisation

$$\begin{eqnarray}e_{\unicode[STIX]{x1D70E}}:=\text{cl}_{\unicode[STIX]{x1D70E}}(\unicode[STIX]{x1D709})\in H^{2d}(\unicode[STIX]{x1D70E}(Y\times Y),\mathbb{Q}(d))\end{eqnarray}$$

defines as a correspondence the idempotent for the direct summand

$$\begin{eqnarray}\text{IH}^{n}(\unicode[STIX]{x1D70E}(X),\mathbb{Q})\subseteq H^{n}(\unicode[STIX]{x1D70E}(Y),\mathbb{Q}),\end{eqnarray}$$

and similarly for the $p$-adic étale realisation

$$\begin{eqnarray}e_{p}:=\text{cl}_{\mathbb{Q}_{p}}(\unicode[STIX]{x1D709})\in H^{2d}((Y\times Y)\otimes _{K}\overline{K},\mathbb{Q}_{p}(d)).\end{eqnarray}$$

Using this, de Cataldo and Migliorini [Reference de Cataldo and MigliorinideCM15] and Patrikis [Reference PatrikisPat16] define the $K$-rational intersection de Rham cohomology of $X$,

$$\begin{eqnarray}\text{IH}_{\text{dR}}^{n}(X/K),\end{eqnarray}$$

as the image of the idempotent

$$\begin{eqnarray}e_{\text{dR}}:=\text{cl}_{\text{dR}}(\unicode[STIX]{x1D709})\in H^{2d}(Y\times Y)(d)\end{eqnarray}$$

acting on $H_{\text{dR}}^{n}(Y)$. Clearly, there is a comparison isomorphism of $\text{IH}_{\text{dR}}^{n}(X/K)\otimes _{K}\mathbb{C}$ with the (transcendental) Hodge structure on the Betti realisation $\text{IH}^{n}(\unicode[STIX]{x1D70E}(X),\mathbb{Q})\otimes \mathbb{C}$. Moreover, this last Hodge structure coincides with the Hodge structure constructed by Morihiko Saito; see de Cataldo [Reference de CataldodeC12, Theorem 4.3.5].

Apply Theorem 2.2.1 to this situation. ◻

Proof. This boils down to first finding an André (pure Nori) motive $\mathfrak{i}\mathfrak{h}(X)$ whose $\ell$-adic and Betti realisations give the $\ell$-adic and Betti intersection cohomology of $X$, and then lifting the action of the Hecke correspondences to one on $\mathfrak{i}\mathfrak{h}(X)$.

While the first step can be done as above in an ad hoc fashion, using a (non-canonical) resolution of singularities and using the theorems of de Cataldo, Migliorini and Patrikis, the second step is done most systematically (in our opinion) by using the theory of weight filtration. We learned the argument from Morel (cf. [Reference MorelMor08, §5]), which uses the more recent motivic constructions of Ivorra and Morel [Reference Ivorra and MorelIM19].

(Before giving the details, let us stress the main point and indicate where the innovations are. The intersection complex $\text{IC}$ (as a perverse sheaf in the derived category of constructible sheaves) was originally constructed as an iterated application of the two-step procedure: taking the direct image $Rj_{\ast }$ under open immersions $j$, and then truncating with respect to a topological stratification and a function called ‘perversity’; see the explicit formula in [Reference Beĭlinson, Bernstein and DeligneBBD82, Proposition 2.1.11].

One of the innovations in [Reference MorelMor08] was to realise $\text{IC}$ (for the middle perversity) over finite fields as the truncation with respect to the weight filtration; see [Reference MorelMor08, Theorem 3.1.4].

This renders the extension of the Hecke operators to $\text{IC}$ deceptively easy: one needs neither to worry about singularities in the boundary (which can be bad), to which one must pay heed if one uses the original (topological) definition, nor to rely on the toroidal compactifications, of which there is no canonical choice, and many are needed to extend the Hecke correspondences. This is why we adopt the idea.

In [Reference Ivorra and MorelIM19], Ivorra and Morel construct the four operations of Grothendieck (namely $f_{\mathscr{M}}^{\ast }$, $f_{\mathscr{M}}^{!}$, $f_{\ast }^{\mathscr{M}}$, $f_{!}^{\mathscr{M}}$ for morphisms $f$ between quasiprojective varieties) and the weight filtrations on the derived category of ‘perverse mixed motives’; this last abelian category is, moreover, shown to be equivalent to the Nori motives.

This allows one to formally apply the algorithm described in [Reference MorelMor08, §5], but this time applied in the motivic derived category of [Reference Ivorra and MorelIM19], rather than in the derived category of constructible sheaves as in [Reference MorelMor08]. Then the Betti and $\ell$-adic realisations of these constructions give rise to the Hecke operators constructed previously on the level of complexes of sheaves.)

Now let us turn to a more detailed argument, and indicate which constructions in [Reference Ivorra and MorelIM19] replace those in the parallel argument from [Reference MorelMor08, §5].

Let $j:X^{\circ }{\hookrightarrow}X=X^{\text{BB}}$ be the open immersion of the Shimura variety and $c_{1},c_{2}:Y^{\circ }\longrightarrow X^{\circ }$ be the two finite étale maps that give rise to a given Hecke correspondence. With $j^{\prime }:Y^{\circ }{\hookrightarrow}Y$ denoting the open immersion into the Baily–Borel compactification, the $c_{i}$ extend canonically to $\overline{c_{i}}:Y\longrightarrow X$. We start with the identity correspondence

$$\begin{eqnarray}u=1:c_{1}^{\ast }\mathbb{Q}_{X^{\circ }}\longrightarrow c_{2}^{!}\mathbb{Q}_{X^{\circ }}\end{eqnarray}$$

arising from the natural isomorphisms $c_{1}^{\ast }\mathbb{Q}_{X^{\circ }}\simeq \mathbb{Q}_{Y^{\circ }}$ and $c_{2}^{!}\mathbb{Q}_{X^{\circ }}\simeq \mathbb{Q}_{Y^{\circ }}$.

Here and below, the functors $f^{\ast }$ and $f^{!}$, etc., refer to the ones in [Reference Ivorra and MorelIM19, Theorem 5.1], where the notation $f_{\mathscr{M}}^{\ast }$, $f_{\mathscr{M}}^{!}$, etc., is used.

Just as in [Reference MorelMor08, §5.1], but using the motivic constructions of the four functors and the base change morphisms, stated in [Reference Ivorra and MorelIM19, Theorem 5.1], we take $Rj_{\ast }^{\prime }={j^{\prime }}_{\ast }^{\mathscr{M}}$ and use the base change morphisms

We then use the fact that the lowest weight filtration of $Rj_{\ast }\mathbb{Q}_{X^{\circ }}$ is canonically isomorphic to the intersection complex $j_{!\ast }(\mathbb{Q}_{X^{\circ }}[d])[-d]$. For this, it is enough to show that the (motivic) weight filtration has $\ell$-adic realisation equal to the ($\ell$-adic) weight filtration. But this follows from the definitions and construction [Reference Ivorra and MorelIM19, Definitions 6.12, 6.13, and Proposition 6.16].

Then we proceed as in [Reference MorelMor08, Lemma 5.1.4] to obtain

$$\begin{eqnarray}\overline{u}:\overline{c_{1}}^{\ast }IC_{X}\longrightarrow \overline{c_{2}}^{!}IC_{X},\end{eqnarray}$$

lifting the cohomological correspondence in realisations. Again the point is that all the arrows in [Reference MorelMor08, Lemma 5.1.4] are constructed using (only) the functoriality of $Rj_{\ast }$, the base change morphisms, and the weight filtration. In our (motivic) context, we use the main theorem [Reference Ivorra and MorelIM19, Theorem 5.1] for the first two and the weight filtration constructed in [Reference Ivorra and MorelIM19, Proposition 6.16] for the last.◻

Proof. This amounts to showing that the (algebraic) Lie algebra $\mathfrak{g}$ of $G_{f,\ell }^{\circ }$ is equal to that of the right-hand side.

The containment $\subseteq$ follows from the fact that, if $F^{\prime }$ is any sufficiently large finite extension of $F$, then $\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}(\operatorname{Frob}_{\mathfrak{p}^{\prime }})$ has trace and determinant in $K_{f}^{\circ }$ for any prime $\mathfrak{p}^{\prime }$ of $F^{\prime }$ coprime to $\text{disc}(F)\mathfrak{n}\ell$, so that if $\unicode[STIX]{x1D706}$ and $\unicode[STIX]{x1D706}^{\prime }$ are any two primes of $K_{f}$ that lie over the same prime of $K_{f}^{\circ }$ over $(\ell )$, then $G_{f,\ell }^{\circ }$ is contained in the partial diagonal of $(\operatorname{Res}_{\mathbb{Q}}^{K_{f}}\text{GL}_{2})_{\mathbb{Q}_{\ell }}$ where the $\unicode[STIX]{x1D706}$- and the $\unicode[STIX]{x1D706}^{\prime }$-components are equal.

To prove the containment $\supseteq$, we first note that since $f$ is not of CM type, $\mathfrak{g}$ surjects onto each factor $\mathfrak{g}\mathfrak{l}_{2,\unicode[STIX]{x1D706}^{\circ }}$, which contains $\mathfrak{s}\mathfrak{l}_{2,\unicode[STIX]{x1D706}^{\circ }}$. If $\unicode[STIX]{x1D706}_{1}^{\circ }$ and $\unicode[STIX]{x1D706}_{2}^{\circ }$ are distinct primes of $K_{f}^{\circ }$ lying over $\ell$, then the representations of $\mathfrak{g}$ in the two factors are non-isomorphic, since the representations of (germs of) $\operatorname{Gal}(\overline{\mathbb{Q}}/F)$ have different traces, and the image of $\mathfrak{g}$ in $\mathfrak{g}\mathfrak{l}_{2,\unicode[STIX]{x1D706}_{1}^{\circ }}\times \mathfrak{g}\mathfrak{l}_{2,\unicode[STIX]{x1D706}_{2}^{\circ }}$ contains $\mathfrak{s}\mathfrak{l}_{2}\times \mathfrak{s}\mathfrak{l}_{2}$.

Then by Goursat’s lemma (in the middle of the proof of Ribet [Reference RibetRib76, Theorem 4.4.10]) the image of $\mathfrak{g}$ contains $\prod _{\unicode[STIX]{x1D706}^{\circ }|\ell }\mathfrak{s}\mathfrak{l}_{2,\unicode[STIX]{x1D706}^{\circ }}$, where $\unicode[STIX]{x1D706}^{\circ }$ ranges over the primes of $K_{f}^{\circ }$ lying over $(\ell )$. Since the determinant on $G_{f,\unicode[STIX]{x1D706}^{\circ }}^{\circ }$ is a dominant map onto $\mathbf{G}_{m}$, we get the desired equality.◻

Proof. (1) By induction on $|T|$, we are reduced to the case where $T$ consists of one element, say $T=\{t\}$. Twisting by $-t$ (i.e. taking $\otimes \{-t\}$) allows us to assume that $t=0$. Enumerate $S$ and $S^{\prime }$ in the order

$$\begin{eqnarray}s_{1}\leqslant s_{2}\leqslant \cdots \leqslant s_{m}\quad \text{and}\quad s_{1}^{\prime }\leqslant s_{2}^{\prime }\leqslant \cdots \leqslant s_{m}^{\prime },\end{eqnarray}$$

and let $a$ and $b$ be such that

$$\begin{eqnarray}s_{a}<0\leqslant s_{a+1}\quad \text{and}\quad s_{b}^{\prime }<0\leqslant s_{b+1}^{\prime };\end{eqnarray}$$

if all the $s_{i}$ are greater than or equal to $0$ (respectively, less than $0$), then we let $a:=0$ (respectively, $a:=m$), and similarly for $b$.

If we define $\unicode[STIX]{x1D6F4}_{S}(i):=s_{1}+\cdots +s_{i}$ for $i\in [0,m]$, the condition $S\leqslant S^{\prime }$ becomes

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{S}(i)\leqslant \unicode[STIX]{x1D6F4}_{S^{\prime }}(i)\quad \text{for all }i\in [0,m].\end{eqnarray}$$

We need to prove

(3.4.3.1)$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{S\oplus \{0\}}(i)\leqslant \unicode[STIX]{x1D6F4}_{S^{\prime }\oplus \{0\}}(i)\quad \text{for all }i\in [0,m+1].\end{eqnarray}$$

(1a) Suppose that $a<b$. Then for $i\in [0,a]\cup [b+1,m+1]$, (3.4.3.1) is clearly satisfied. For $i\in [a+1,b]$, since $0$ is inserted into $S^{\prime }$ in $(b+1)$th place, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{S^{\prime }\oplus \{0\}}(i) & = & \displaystyle \unicode[STIX]{x1D6F4}_{S^{\prime }\oplus \{0\}}(b+1)-(s_{i+1}^{\prime }+\cdots +s_{b}^{\prime }+0)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6F4}_{S^{\prime }}(b)-(s_{i+1}^{\prime }+\cdots +s_{b}^{\prime })\geqslant \unicode[STIX]{x1D6F4}_{S^{\prime }}(b)\nonumber\end{eqnarray}$$

while, since $0$ is inserted into $S$ in $(a+1)$th place, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{S\oplus \{0\}}(i)=\unicode[STIX]{x1D6F4}_{S}(i-1)=\unicode[STIX]{x1D6F4}_{S}(b)-(s_{i}+\cdots +s_{b})\leqslant \unicode[STIX]{x1D6F4}_{S}(b), & & \displaystyle \nonumber\end{eqnarray}$$

and we get (3.4.3.1).

(1b) Suppose that $a\geqslant b$. Then again for $i\in [0,b]\cup [a+1,m+1]$, (3.4.3.1) is trivially satisfied. For $i\in [b+1,a]$, we have this time:

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{S^{\prime }\oplus \{0\}}(i) & = & \displaystyle \unicode[STIX]{x1D6F4}_{S^{\prime }}(i-1)=\unicode[STIX]{x1D6F4}_{S^{\prime }}(b)+(s_{b+1}^{\prime }+\cdots +s_{i-1}^{\prime })\geqslant \unicode[STIX]{x1D6F4}_{S^{\prime }}(b),\nonumber\\ \displaystyle \unicode[STIX]{x1D6F4}_{S\oplus \{0\}}(i) & = & \displaystyle \unicode[STIX]{x1D6F4}_{S}(i)=\unicode[STIX]{x1D6F4}_{S}(b)+(s_{b+1}+\cdots +s_{i})\leqslant \unicode[STIX]{x1D6F4}_{S^{\prime }}(b).\nonumber\end{eqnarray}$$

This completes the proof of (1).

(2) By decomposing $T$ into singletons and using the distributive law, we deduce (2) from (1).

(3) The duals $S^{\vee }$ and ${S^{\prime }}^{\vee }$ are enumerated:

$$\begin{eqnarray}-\!s_{m}\leqslant -s_{m-1}\leqslant \cdots \leqslant -s_{1}\quad \text{and}\quad -\!s_{m}^{\prime }\leqslant -s_{m-1}^{\prime }\leqslant \cdots \leqslant -s_{1}^{\prime }.\end{eqnarray}$$

The assumption that $S$ and $S^{\prime }$ end at the same point means that $\unicode[STIX]{x1D6F4}_{S}(m)=\unicode[STIX]{x1D6F4}_{S^{\prime }}(m)$. Thus

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{S^{\vee }}(i)=\unicode[STIX]{x1D6F4}_{S}(m-i)-\unicode[STIX]{x1D6F4}_{S}(m)\leqslant \unicode[STIX]{x1D6F4}_{S^{\prime }}(m-i)-\unicode[STIX]{x1D6F4}_{S^{\prime }}(m)=\unicode[STIX]{x1D6F4}_{{S^{\prime }}^{\vee }}(i)\end{eqnarray}$$

for all $i\in [1,m]$, and this completes the proof of the proposition.◻

Proof. Let $X=\operatorname{Hom}(K,F^{s})$ and $X^{\prime }:=\operatorname{Hom}(K^{\prime },F^{s})$. Then we have a surjective map of $G$-sets,

$$\begin{eqnarray}X\longrightarrow X^{\prime },\end{eqnarray}$$

obtained by restriction. Since $K/K^{\prime }$ is separable, each fiber has exactly $n$ elements.

The first inequality follows from inspecting the images of the $g$-orbits in $X$, and the second follows from the first by definition.◻

Proof. (1) Recall that $\operatorname{Gal}(\widetilde{K}/\mathbb{Q})$ acts transitively on $\operatorname{Hom}(K,\mathbb{Q})$, and therefore has order divisible by $p=[K:\mathbb{Q}]$. Since the image of $\operatorname{Gal}(\overline{\mathbb{Q}}/F)$ in $\operatorname{Gal}(\widetilde{K}/\mathbb{Q})$ (via $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$) has index dividing $[F:\mathbb{Q}]$, the image also has order divisible by $p$. Therefore the image contains an element $g$ of exact order $p$, which has $\unicode[STIX]{x1D706}(g,\operatorname{Hom}(K,\mathbb{Q}))=p=[K:\mathbb{Q}]$.

(2) Use the fact that a symmetric group has no proper normal subgroup of odd index to deduce that the image of $\operatorname{Gal}(\overline{\mathbb{Q}}/\widetilde{F})$ in $\operatorname{Gal}(\widetilde{K}/\mathbb{Q})$ is the full symmetric group.

(3) By assumption, the natural map

$$\begin{eqnarray}(\unicode[STIX]{x1D719}_{1},\unicode[STIX]{x1D719}_{2}):\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\longrightarrow \operatorname{Gal}(\widetilde{F}/\mathbb{Q})\times \operatorname{Gal}(\widetilde{K}/\mathbb{Q})\end{eqnarray}$$

is surjective and by definition $\operatorname{Gal}(\overline{\mathbb{Q}}/F)\supseteq \operatorname{Gal}(\overline{\mathbb{Q}}/\widetilde{F})=\ker (\unicode[STIX]{x1D719}_{1})$. Therefore $\operatorname{Gal}(\overline{\mathbb{Q}}/F)$ surjects onto $\operatorname{Gal}(\widetilde{K}/\mathbb{Q})$. The statements about the slope and bisecting elements follow from this.◻

Proof. This follows from Serre [Reference SerreSer12, Proposition 5.12] applied to $Z:=\unicode[STIX]{x1D711}^{-1}(S)$, which is a proper algebraic subset of $G$ by the assumptions and has Zariski density $0$ by definition.◻

Proof. We have already fixed a rational prime $\ell$ that splits completely in $K_{f}$. Now for all rational primes $p\neq \ell$, we fix once and for all an isomorphism  and pull back the $p$-adic valuation on the target to get a rank-$1$ (discontinuous) valuation $v_{p}$ on $\overline{\mathbb{Q}}_{\ell }$, normalised by$v_{p}(p)=1$.

To prove (1), we may pass to $\widetilde{F}$, and consider $\operatorname{Frob}_{\mathfrak{p}}$ for any prime $\mathfrak{p}$ lying over $p$, because doing so does not change the Newton polygon.

The key fact (from Brylinski and Labesse [Reference Brylinski and LabesseBL84]) that we use from the constructions (see Deligne [Reference DeligneDel71], Ohta [Reference OhtaOht83], Carayol [Reference CarayolCar86], Wiles [Reference WilesWil88], Taylor [Reference TaylorTay89] and Blasius and Rogawski [Reference Blasius and RogawskiBR93] and the references therein) of the Galois representations associated with the $\{\unicode[STIX]{x1D70E}(f)\}$ is the following: the $\ell$-adic étale realisation $M(f)_{\ell }$ of $M(f)$ is the direct sum of the tensor inductions (see Curtis and Reiner [Reference Curtis and ReinerCR87, §80C]):

(4.1.1.1)

This implies that for transversals (coset representatives) $g_{1},\ldots ,g_{d}\in \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ modulo $\operatorname{Gal}(\overline{\mathbb{Q}}/F)$, we have that for any element $\unicode[STIX]{x1D6FE}$ in the finite-index normal subgroup $\operatorname{Gal}(\overline{\mathbb{Q}}/\widetilde{F})$ of $\operatorname{Gal}(\overline{\mathbb{Q}}/F)$, the action of $\unicode[STIX]{x1D6FE}$ on $M(f)_{\ell }$ is given by

(4.1.1.2)$$\begin{eqnarray}\bigoplus _{\unicode[STIX]{x1D70E}}(\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}(f),\unicode[STIX]{x1D706}}(g_{1}\unicode[STIX]{x1D6FE}g_{1}^{-1})\otimes \unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}(f),\unicode[STIX]{x1D706}}(g_{2}\unicode[STIX]{x1D6FE}g_{2}^{-1})\otimes \cdots \otimes \unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D70E}(f),\unicode[STIX]{x1D706}}(g_{d}\unicode[STIX]{x1D6FE}g_{d}^{-1})).\end{eqnarray}$$

Let $K^{\prime }:=K_{f}(\operatorname{Frob}_{\mathfrak{p}})$ be the splitting field over $K_{f}$ of the polynomial

(4.1.1.3)$$\begin{eqnarray}X^{2}-\text{Tr}(\unicode[STIX]{x1D70C}(\operatorname{Frob}_{\mathfrak{ p}}))X+\det (\unicode[STIX]{x1D70C}(\operatorname{Frob}_{\mathfrak{p}})),\end{eqnarray}$$

and let $R_{\mathfrak{p}}=\{\unicode[STIX]{x1D6FC}_{\mathfrak{p}},\unicode[STIX]{x1D6FD}_{\mathfrak{p}}\}\subset K^{\prime }$ be the roots. Each embedding $\unicode[STIX]{x1D70E}:K_{f}\longrightarrow \mathbb{Q}_{\ell }$ extends to (at most two) embeddings $\unicode[STIX]{x1D70E}^{\prime }:K^{\prime }\longrightarrow \overline{\mathbb{Q}}_{\ell }$, and the image $\unicode[STIX]{x1D70E}^{\prime }(R_{\mathfrak{p}})\subset \overline{\mathbb{Q}}_{\ell }$ is independent of the choice $\unicode[STIX]{x1D70E}^{\prime }$. Therefore we can unambiguously form the multiset of slopes:Footnote ⁵

$$\begin{eqnarray}S_{\mathfrak{p},\unicode[STIX]{x1D70E}}=\bigg\{\frac{v_{p}(\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D6FC}_{\mathfrak{p}}))}{v_{p}(\mathbb{N}\mathfrak{p})},\frac{v_{p}(\unicode[STIX]{x1D70E}^{\prime }(\unicode[STIX]{x1D6FD}_{\mathfrak{p}}))}{v_{p}(\mathbb{N}\mathfrak{p})}\bigg\}.\end{eqnarray}$$

Since $\text{Tr}(\unicode[STIX]{x1D70C}(\operatorname{Frob}_{\mathfrak{p}}))$ is an algebraic integer, its $p$-adic valuation is greater than or equal to $0$. Also, $\det (\unicode[STIX]{x1D70C}(\operatorname{Frob}_{\mathfrak{p}}))$ is $\mathbb{N}\mathfrak{p}$ times a root of unity, so its $p$-adic valuation is equal to $v_{p}(\mathbb{N}\mathfrak{p})$. These two facts imply the inequalities on the $\unicode[STIX]{x1D70E}$-slopes of (4.1.1.3) for all $\unicode[STIX]{x1D70E}$:

(4.1.1.4)$$\begin{eqnarray}\{0,1\}\leqslant S_{\mathfrak{p},\unicode[STIX]{x1D70E}}\leqslant \{1/2,1/2\}\end{eqnarray}$$

(see § 3.4 for the partial order by Newton polygon). Moreover, if $\mathbb{N}\mathfrak{p}=p$ (in particular, if $p$ splits completely in $F$) and $p$ is unramified in $K_{f}$, then one of the two inequalities must be an equality.

In view of the description of cohomology in terms of tensor induction (4.1.1.1) and (4.1.1.2), we have

$$\begin{eqnarray}\text{NP}(\operatorname{Frob}_{\mathfrak{p}|_{M(f)}})=\bigoplus _{\unicode[STIX]{x1D70E}\in \operatorname{Hom}(K_{f},\mathbb{Q}_{\ell })}S_{\mathfrak{p},\unicode[STIX]{x1D70E}}^{\otimes d},\end{eqnarray}$$

and since

$$\begin{eqnarray}\text{HTP}(M(f))=HP(M(f))=P(d;k_{f},0)=(\{0,1\}^{\otimes d})^{\oplus k_{f}},\end{eqnarray}$$

we have proven (1).

In order to proceed further, we first recall the following known fact (generalised Ramanujan–Peterson conjecture; see Taylor [Reference TaylorTay95b] and Blasius[Reference BlasiusBla06]),

From this point on, we assume that $\mathfrak{p}$ is a prime of absolute degree $1$ over $(p)$ (in addition to being coprime to $\text{disc}(F)\cdot \mathfrak{n}\cdot \ell$). We also assume that $p$ is unramified in $K_{f}$.

(2) We now look more closely at

$$\begin{eqnarray}a_{\mathfrak{p}}:=\text{Tr}(\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}(\operatorname{Frob}_{\mathfrak{p}})).\end{eqnarray}$$

By the assumption that $f$ is not of CM type, the image of $\operatorname{Gal}(\mathbb{Q}/\widetilde{F})$ under $\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}$ is Zariski dense in $\text{GL}_{2}$ over $K_{f,\unicode[STIX]{x1D706}}$. Since $\text{Tr}$ is a regular morphism of the algebraic variety $\text{GL}_{2}$ into $\mathbb{A}^{1}$ and takes value $2\neq 0$ at the identity element $I_{2}$, the set of primes $\mathfrak{p}$ of $\widetilde{F}$ such that $a_{\mathfrak{p}}=0$ has density zero by Lemma 3.6.1. We exclude them from this point on.

Let ${\wp}_{1},\ldots ,{\wp}_{m}$ be the primes of $K_{f}$ lying over $(p)=\mathfrak{p}\cap \mathbb{Z}$, and write the ideal factorisation

(4.1.3.1)$$\begin{eqnarray}a_{\mathfrak{p}}\cdot \mathscr{O}_{K_{f}}={\wp}_{1}^{e_{1}}\cdots {\wp}_{k}^{e_{m}}\cdot I,\end{eqnarray}$$

where $I$ is an integral ideal coprime to $p$, and we carry on with the argument preceding the lemma.

For each embedding $\unicode[STIX]{x1D70E}:K_{f}\longrightarrow \mathbb{Q}_{\ell }$, let ${\wp}_{i(\unicode[STIX]{x1D70E})}$ be the inverse image in $\mathscr{O}_{K_{f}}$ of the maximal ideal of the integral closure $\overline{\mathbb{Z}_{p}}\subset \overline{\mathbb{Q}}_{p}$ of $\mathbb{Z}_{p}$ under the composite of $\unicode[STIX]{x1D70E}$ with the fixed $\overline{\mathbb{Q}}_{\ell }\simeq \overline{\mathbb{Q}}_{p}$:

Then $S_{\mathfrak{p},\unicode[STIX]{x1D70E}}$ is the Newton polygon of the polynomial (obtained by applying $\unicode[STIX]{x1D70E}$ to (4.1.1.3)) with respect to $v_{p}$ on $\mathbb{Q}_{\ell }$,

$$\begin{eqnarray}X^{2}-\unicode[STIX]{x1D70E}(a_{\mathfrak{ p}})X+\unicode[STIX]{x1D70E}(\det (\unicode[STIX]{x1D70C}(\operatorname{Frob}_{\mathfrak{p}}))),\end{eqnarray}$$

and as such is equal to

$$\begin{eqnarray}S_{\mathfrak{p},\unicode[STIX]{x1D70E}}=\left\{\begin{array}{@{}ll@{}}\{0,1\}\quad & \text{if }e_{i(\unicode[STIX]{x1D70E})}=0,\\ \{1/2,1/2\}\quad & \text{if }e_{i(\unicode[STIX]{x1D70E})}>0.\end{array}\right.\end{eqnarray}$$

Let $f_{1},\ldots ,f_{m}$ be the degrees of the residue class extensions:

$$\begin{eqnarray}f_{i}:=\dim _{\mathbb{F}_{p}}\mathscr{O}_{K_{f}}/{\wp}_{i}.\end{eqnarray}$$

Then we have

$$\begin{eqnarray}\text{NP}(\operatorname{Frob}_{\mathfrak{p}}\!|_{M(f)})=P(d;k_{f},k(p)),\end{eqnarray}$$

where $k(p)$ is the sum of those $f_{i}$ for which $e_{i}>0$.

Since $a_{\mathfrak{p}}\neq 0$, we may apply the product formula. By the lemma, we have

$$\begin{eqnarray}\mathop{\prod }_{v|\infty }\Vert a_{\mathfrak{p}}\Vert _{v}\leqslant (2\sqrt{p})^{k_{f}},\end{eqnarray}$$

while by the factorisation (4.1.3.1),

$$\begin{eqnarray}\mathop{\prod }_{v|p}\Vert a_{\mathfrak{p}}\Vert _{v}=p^{-\!\mathop{\sum }_{i=1}^{m}e_{i}f_{i}}\end{eqnarray}$$

and

$$\begin{eqnarray}\mathop{\prod }_{v\nmid p,\infty }\Vert a_{\mathfrak{p}}\Vert _{v}=\mathop{\prod }_{v\nmid p,\infty }\Vert I\Vert _{v}\leqslant 1.\end{eqnarray}$$

Therefore

$$\begin{eqnarray}1=\mathop{\prod }_{v}\Vert a_{\mathfrak{p}}\Vert \leqslant 2^{k_{f}}p^{k_{f}/2-\mathop{\sum }_{i}e_{i}f_{i}},\end{eqnarray}$$

which implies, for all $p>2^{2k_{f}}$,

$$\begin{eqnarray}\frac{k_{f}}{2}\geqslant \mathop{\sum }_{i=1}^{m}e_{i}f_{i}\geqslant \mathop{\sum }_{i:e_{i}>0}f_{i}=k(p).\end{eqnarray}$$

This proves (2).

For (3), we assume in addition that $p$ splits completely in $\widetilde{F^{\circ }}$, and choose a prime $\mathfrak{p}$ of $\widetilde{F^{\circ }}$ lying over $p$, so that by definition

$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{f,\ell }(\operatorname{Frob}_{\mathfrak{p}})\in G_{f,\ell }^{\circ }(\mathbb{Q}_{\ell })\quad \text{and}\quad a_{\mathfrak{p}}\in \mathscr{O}_{K_{f}^{\circ }}.\end{eqnarray}$$

If $k_{f}^{\circ }=1$, that is, if $K_{f}^{\circ }=\mathbb{Q}$, then $a_{\mathfrak{p}}\in \mathbb{Z}$ and $|a_{\mathfrak{p}}|_{\infty }<2\sqrt{p}$. As soon as $p\geqslant 5$, the only way $p|a_{\mathfrak{p}}$ is then $a_{\mathfrak{p}}=0$, which we have excluded above.

Suppose therefore that $k_{f}^{\circ }=2$, and consider the homomorphisms

and the regular map of algebraic varieties

(4.1.3.2)$$\begin{eqnarray}\text{Tr}(\wedge ^{2}(\unicode[STIX]{x1D704})\otimes \det ^{-1}):((\operatorname{Res}_{\mathbb{ Q}}^{K_{f}^{\circ }}\text{GL}_{2})^{\det \subseteq \mathbb{Q}^{\times }})\otimes \mathbb{Q}_{\ell }\longrightarrow \mathbb{A}_{\mathbb{Q}_{\ell }}^{1},\end{eqnarray}$$

where $\wedge ^{2}(\unicode[STIX]{x1D704})$ takes values in $\text{GL}_{6}$ and $\det$ takes values in $\mathbf{G}_{m}$.

In order to prove (3), we find a set of primes $\mathfrak{p}$ of $\widetilde{F^{\circ }}$ of density equal to $1$ such that $a_{\mathfrak{p}}$ is not divisible by any prime of $K_{f}^{\circ }$ lying over $(p)$. If $p$ is inert in $K_{f}^{\circ }$, the bound (2) suffices, and we exclude the finitely many primes that are ramified in $K_{f}^{\circ }$, so we assume that $p$ splits:

(4.1.3.3)$$\begin{eqnarray}p\cdot \mathscr{O}_{K_{f}^{\circ }}={\wp}_{1}{\wp}_{2}\quad \text{where }{\wp}_{1}\neq {\wp}_{2}\text{ are primes}.\end{eqnarray}$$

By (2), we may assume that at most one of the two primes can divide $a_{\mathfrak{p}}$, say $a_{\mathfrak{p}}\in {\wp}_{1}$ but $a_{\mathfrak{p}}\not \in {\wp}_{2}$. Let $\unicode[STIX]{x1D716}$ be the non-trivial field automorphism of $K_{f}^{\circ }$, let $\unicode[STIX]{x1D6FC}_{\mathfrak{p}}$ be a root of the polynomial $X^{2}-a_{\mathfrak{p}}X+p=0$, and let $\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime }$ be a root of $X^{2}-\unicode[STIX]{x1D716}(a_{\mathfrak{p}})X+p=0.$

Then the four eigenvalues of $(\unicode[STIX]{x1D704}\circ \unicode[STIX]{x1D70C}^{\circ })(\operatorname{Frob}_{\mathfrak{p}})$ are $\{\unicode[STIX]{x1D6FC}_{\mathfrak{p}},p/\unicode[STIX]{x1D6FC}_{\mathfrak{p}},\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime },p/\unicode[STIX]{x1D6FC}_{\mathfrak{p}}\}$, and the six eigenvalues of $(\wedge ^{2}\unicode[STIX]{x1D704}\circ \unicode[STIX]{x1D70C}^{\circ })(\operatorname{Frob}_{\mathfrak{p}})$ are $\{p,\unicode[STIX]{x1D6FC}_{\mathfrak{p}}\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime },\unicode[STIX]{x1D6FC}_{\mathfrak{p}}(p/\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime }),(p/\unicode[STIX]{x1D6FC}_{\mathfrak{p}})\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime },(p/\unicode[STIX]{x1D6FC}_{\mathfrak{p}})(p/\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime }),p\}$. Therefore the value of (4.1.3.2) at $\operatorname{Frob}_{\mathfrak{p}}$ is

$$\begin{eqnarray}\frac{1}{p}\biggl(2p+\biggl(\unicode[STIX]{x1D6FC}_{\mathfrak{p}}+\frac{p}{\unicode[STIX]{x1D6FC}_{\mathfrak{p}}}\biggr)\biggl(\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime }+\frac{p}{\unicode[STIX]{x1D6FC}_{\mathfrak{p}}^{\prime }}\biggr)\biggr)=2+\frac{a_{\mathfrak{p}}\unicode[STIX]{x1D716}(a_{\mathfrak{p}})}{p}\in \mathbb{Z}.\end{eqnarray}$$

Now by Lemma 4.1.3, which implies

$$\begin{eqnarray}|a_{\mathfrak{p}}|_{\infty },|\unicode[STIX]{x1D716}(a_{\mathfrak{p}})|_{\infty }\leqslant 2\sqrt{p},\end{eqnarray}$$

and by our assumption that $p$ is unramified in $K_{f}^{\circ }$, which implies that the inequalities are strict, we have

(4.1.3.4)$$\begin{eqnarray}\text{Tr}(\wedge ^{2}(\unicode[STIX]{x1D704})\otimes \det ^{-1})(\operatorname{Frob}_{\mathfrak{ p}})\in [-1,5]\cap \mathbb{Z}.\end{eqnarray}$$

Therefore $a_{\mathfrak{p}}$ does not belong to any prime of $K_{f}^{\circ }$ lying over $(p)$, as soon as we avoid (4.1.3.4). But since

$$\begin{eqnarray}\text{Tr}(\wedge ^{2}(\unicode[STIX]{x1D704})\otimes \det ^{-1})(1)=6\end{eqnarray}$$

(here $1$ denotes the identity element in the group $G_{f,\ell }^{\circ }$), the set of $\mathfrak{p}$ for which (4.1.3.4) holds has density $0$ by Lemma 3.6.1. Now if $\mathfrak{p}$ avoids (4.1.3.4), then

$$\begin{eqnarray}a_{\mathfrak{p}}\mathscr{O}_{K_{f}}=(a_{\mathfrak{p}}\mathscr{O}_{K_{f}^{\circ }})\mathscr{O}_{K_{f}}\end{eqnarray}$$

is coprime to $(p)$, and we have $k(p)=0$. This completes the proof of (3).

For the sake of continuity in exposition, we treat the conditional (5) before the unconditional (4). By an argument similar to that for (2), but applied to the restriction $\unicode[STIX]{x1D70C}^{\circ }:\operatorname{Gal}(\overline{\mathbb{Q}}/\widetilde{F^{\circ }})\longrightarrow \text{GL}_{2}(K_{f,\unicode[STIX]{x1D706}^{\circ }}^{\circ })$ (where $\unicode[STIX]{x1D706}^{\circ }=\unicode[STIX]{x1D706}\cap K_{f}^{\circ }$), for a prime $\mathfrak{p}$ of density $1$ in $\widetilde{F^{\circ }}$, if we write

$$\begin{eqnarray}a_{\mathfrak{p}}\mathscr{O}_{K_{f}^{\circ }}={\wp}_{1}^{e_{1}}\cdots {\wp}_{m^{\prime }}^{e_{m^{\prime }}}I,\end{eqnarray}$$

where ${\wp}_{1},\ldots ,{\wp}_{m^{\prime }}$ are the primes of $K_{f}^{\circ }$ lying over $p$ and $I$ is coprime to $p$, then we have

(4.1.3.5)$$\begin{eqnarray}\mathop{\sum }_{i:e_{i}>0}\dim _{\mathbb{F}_{p}}(\mathscr{O}_{K_{f}^{\circ }}/{\wp}_{i})\leqslant \frac{k_{f}^{\circ }}{2},\end{eqnarray}$$

and

$$\begin{eqnarray}k(p)=\frac{k_{f}}{k_{f}^{\circ }}\mathop{\sum }_{i:e_{i}>0}\dim _{\mathbb{F}_{p}}(\mathscr{O}_{K_{f}^{\circ }}/{\wp}_{i}).\end{eqnarray}$$

If $k_{f}^{\circ }$ is odd, then (4.1.3.5) trivially implies (5), so we assume that $k_{f}^{\circ }$ is even.

Now if the equality holds in (4.1.3.5), then we necessarily have $e_{i}=1$ whenever $e_{i}>0$, and by Lemma 4.1.3 and the product formula,

$$\begin{eqnarray}(2\sqrt{p})^{k_{f}^{\circ }}\geqslant \mathop{\prod }_{v|\infty }\Vert a_{\mathfrak{p}}\Vert _{v}=\biggl(\mathop{\prod }_{v|p}\Vert a_{\mathfrak{p}}\Vert _{v}\cdot \mathop{\prod }_{v\nmid p\infty }\Vert a_{\mathfrak{p}}\Vert _{v}\biggr)^{-1}\in p^{k_{f}^{\circ }/2}\mathbb{Z}.\end{eqnarray}$$

In other words, if $i_{1},\ldots ,i_{k_{f}^{\circ }}$ are the real embeddings of $K_{f}^{\circ }$, then we have

$$\begin{eqnarray}\mathop{\prod }_{j=1}^{k_{f}^{\circ }}\frac{i_{j}(a_{\mathfrak{p}})}{\sqrt{p}}\in \mathbb{Z}\cap [-2^{k_{f}^{\circ }},2^{k_{f}^{\circ }}].\end{eqnarray}$$

In $\mathbb{R}^{k_{f}^{\circ }}$, consider the nowhere dense real analytic subsets

$$\begin{eqnarray}B_{j}=\bigg\{(x_{1},\ldots ,x_{k_{f}^{\circ }}):\mathop{\prod }_{a=1}^{k_{f}^{\circ }}x_{a}=j\bigg\}\end{eqnarray}$$

for non-zero $j\in \mathbb{Z}\cap [-2^{k_{f}^{\circ }},2^{k_{f}^{\circ }}]$, $B_{0}=\{(0,\ldots ,0)\}$, and let $B$ be their union.

If $f$ satisfies (RST), then the set $A(f)$ in § 3.3 is equidistributed in $\unicode[STIX]{x1D711}(\mathbf{x})d\unicode[STIX]{x1D707}_{L}(\mathbf{x})$, where $\unicode[STIX]{x1D711}:[-2,2]^{k_{f}^{\circ }}\longrightarrow \mathbb{R}_{{\geqslant}0}$ is a continuous function and $d\unicode[STIX]{x1D707}_{L}$ is the Lebesgue measure. Therefore the set of primes $\mathfrak{p}$ of $\widetilde{F^{\circ }}$ such that

$$\begin{eqnarray}\biggl(\frac{i_{1}(a_{\mathfrak{p}})}{\sqrt{\mathbb{N}\mathfrak{ p}}},\ldots ,\frac{i_{k_{f}^{\circ }}(a_{\mathfrak{p}})}{\sqrt{\mathbb{N}\mathfrak{ p}}}\biggr)\in B\end{eqnarray}$$

has density $0$, since $\int _{B}\unicode[STIX]{x1D711}\,d\unicode[STIX]{x1D707}_{L}=0$.

If $f$ satisfies ($t$-ST$^{\prime }$), then there exists a sequence $1\leqslant j_{1}<\cdots <j_{t}\leqslant k_{f}^{\circ }$ such that $\text{pr}_{j_{1},\ldots ,j_{t}}(A(f))$ is equidistributed in the $t$-fold product of the Sato–Tate measure. Then the set

$$\begin{eqnarray}C=\bigg\{(x_{1},\ldots ,x_{k_{f}^{\circ }}):\mathop{\prod }_{a=1}^{t}|x_{j_{a}}|<2^{-(k_{f}^{\circ }-t)}\bigg\}\end{eqnarray}$$

is disjoint from $B_{j}$ for all non-zero $j\in \mathbb{Z}\cap [-2^{k_{f}^{\circ }},2^{k_{f}^{\circ }}]$. Since the set of $\mathfrak{p}$ such that $a_{\mathfrak{p}}=0$ has density $0$ as noted above, the lower natural density of the set of primes $\mathfrak{p}$ such that

$$\begin{eqnarray}\biggl(\frac{i_{1}(a_{\mathfrak{p}})}{\sqrt{\mathbb{N}\mathfrak{ p}}},\ldots ,\frac{i_{k_{f}^{\circ }}(a_{\mathfrak{p}})}{\sqrt{\mathbb{N}\mathfrak{ p}}}\biggr)\not \in B\end{eqnarray}$$

is bounded from below by

(4.1.3.6)$$\begin{eqnarray}c(k_{f}^{\circ },t):=\int _{|y_{1}|\cdots |y_{t}|<2^{t-k_{f}^{\circ }}}d\unicode[STIX]{x1D707}_{ST}(y_{1})\cdots \,d\unicode[STIX]{x1D707}_{ST}(y_{t})>0\quad \text{for }1\leqslant t<k_{f}^{\circ },\end{eqnarray}$$

where $d\unicode[STIX]{x1D707}_{ST}(y)=(1/2\unicode[STIX]{x1D70B})\sqrt{4-y^{2}}\,dy$ concentrated in $[-2,2]$.

(4) There is nothing to prove if $\unicode[STIX]{x1D70E}_{\widetilde{F}}(K_{f})\geqslant 1/2$, so we assume that

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{0}:=\unicode[STIX]{x1D706}_{\widetilde{F}}(K_{f})>k_{f}/2.\end{eqnarray}$$

We return to the primes $\mathfrak{p}$ of $\widetilde{F}$ of absolute degree $1$ over $\mathbb{Q}$. The conjugacy class of $\operatorname{Frob}_{\mathfrak{p}}$ in $\operatorname{Gal}(\overline{\mathbb{Q}}_{S}/\widetilde{F})$ maps into the conjugacy class of $\operatorname{Frob}_{p}$ in $\operatorname{Gal}(\overline{\mathbb{Q}}_{S}/\mathbb{Q})$. (Here $\overline{\mathbb{Q}}_{S}$ is the maximal subfield of $\overline{\mathbb{Q}}$ unramified outside $\text{disc}(F)\cdot \text{disc}(K_{f})\cdot \ell \cdot \mathbb{N}(\mathfrak{n})$.) The following diagram exhibits the interaction of $\widetilde{F}$ and $K_{f}$ (§ 3.5) and the Galois representation at hand:

Let $\unicode[STIX]{x1D6E4}$ be the image of $\operatorname{Gal}(\overline{\mathbb{Q}}_{S}/\widetilde{F})$ in $\operatorname{Aut}(\operatorname{Hom}(K_{f},\overline{\mathbb{Q}}))$, and let $H^{\sharp }\subseteq \unicode[STIX]{x1D6E4}$ be the non-empty subset of elements $h$ such that $\unicode[STIX]{x1D706}(h,\operatorname{Hom}(K_{f},\overline{\mathbb{Q}}))=\unicode[STIX]{x1D706}_{0}$; one sees easily that $H^{\sharp }$ is stable under conjugation by $\unicode[STIX]{x1D6E4}$. Then if $\operatorname{Frob}_{\mathfrak{p}}$ maps into $H^{\sharp }$, there exists a prime ideal ${\wp}$ of $K_{f}$ lying over $(p)$ with

$$\begin{eqnarray}\dim _{\mathbb{F}_{p}}\mathscr{O}_{K_{f}}/{\wp}=\unicode[STIX]{x1D706}_{0}.\end{eqnarray}$$

Now the bound in (2) prevents ${\wp}$ from occurring in the ideal factorisation of $a_{\mathfrak{p}}$ (4.1.3.1) with multiplicity greater than $0$. Therefore $k(p)$ is at most the sum of the degrees of the residue class field extensions at the other primes of $K_{f}$ lying over $(p)$, and

$$\begin{eqnarray}k(p)\leqslant k_{f}-\unicode[STIX]{x1D706}_{0}=k_{f}\cdot \unicode[STIX]{x1D70E}_{\widetilde{F}}(K_{f}).\end{eqnarray}$$

The density of such primes $\mathfrak{p}$ (i.e. not dividing $\text{disc}(F)\cdot \text{disc}(K_{f})\cdot \ell \cdot \mathbb{N}(\mathfrak{n})$, having $a_{\mathfrak{p}}\neq 0$, and whose $\operatorname{Frob}_{\mathfrak{p}}$ mapping into $H^{\sharp }$) in $\widetilde{F}$ is, by the Chebotarev density theorem, equal to $|H^{\sharp }|/|\unicode[STIX]{x1D6E4}|>0$, and we get (4).

The proof of (4$^{\prime }$) is parallel to that of (4), except that we consider the degree-$1$ primes of $\widetilde{F^{\circ }}$ and $a_{\mathfrak{p}}\in \mathscr{O}_{K_{f}^{\circ }}$, and we omit it.

(6) Consider a similar diagram (with $\widetilde{F}$ replaced with $\widetilde{F^{\circ }}$) to the one in (4), and let $\unicode[STIX]{x1D6E4}^{\circ }$ be the image of $\operatorname{Gal}(\overline{\mathbb{Q}}_{S}/\widetilde{F^{\circ }})$ in $\operatorname{Aut}(\operatorname{Hom}(K_{f}^{\circ },\overline{\mathbb{Q}}))$. This time, choose $H^{\sharp }$ to consist of those elements of $\unicode[STIX]{x1D6E4}^{\circ }$ that bisect $\operatorname{Hom}(K_{f}^{\circ },\overline{\mathbb{Q}})$. Again, $H^{\sharp }$ is stable under conjugation by $\unicode[STIX]{x1D6E4}^{\circ }$.

Then for any prime $\mathfrak{p}$ of $\widetilde{F^{\circ }}$ such that $\operatorname{Frob}_{\mathfrak{p}}$ maps into $H^{\sharp }$ and $p$ is unramified in $K_{f}^{\circ }$, there are exactly two primes of $K_{f}^{\circ }$ lying over $(p)$ with the same degree of residue class extension (namely, $k_{f}^{\circ }/2$). The density of such primes $\mathfrak{p}$ in $\widetilde{F^{\circ }}$ is $|H^{\sharp }|/|\unicode[STIX]{x1D6E4}^{\circ }|>0$ by the Chebotarev density theorem.

Now the bound in (5), which is in effect because we assume (RST), keeps either of the two primes from appearing in the ideal decomposition of $a_{\mathfrak{p}}\mathscr{O}_{K_{f}^{\circ }}$ with multiplicity greater than $0$, except for a set of primes $\mathfrak{p}$ with density $0$.◻

Proof. Let $\unicode[STIX]{x1D706}$ be a prime of $K_{f}$ lying over $(\ell )$ such that the connected component $G^{\circ }$ of the Zariski closure $G=G_{f,\unicode[STIX]{x1D706}}$ of the image of $\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}$ is a torus. The argument employed in proving part (1) of Theorem 4.1.1 goes through without change: the non-CM condition was not used. This way we get the inequality (1) for primes $p$ coprime to $\text{disc}(F)\cdot \mathfrak{n}\cdot \ell$, and for those splitting completely in $F$ and unramified in $K_{f}$ in addition, an integer $k(p)\in [0,k_{f}]$ such that

$$\begin{eqnarray}\text{NP}(\operatorname{Frob}_{p}\!|_{M(f)})=P(d;k_{f},k(p)).\end{eqnarray}$$

Let $F^{\circ }$ be the Galois extension of $F$ cut out by the two representations with finite image,

$$\begin{eqnarray}\operatorname{Gal}(\overline{\mathbb{Q}}/F^{\circ })=\ker (\operatorname{Gal}(\overline{\mathbb{Q}}/F)\longrightarrow G(K_{f,\unicode[STIX]{x1D706}})/G^{\circ }(K_{f,\unicode[STIX]{x1D706}}))\cap \ker (\det (\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}})(1)),\end{eqnarray}$$

and let $\widetilde{F^{\circ }}$ be the compositum of $F^{\circ }$ and $\widetilde{F}$.

Since the restriction of $\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}$ to $\widetilde{F^{\circ }}$ is then abelian and $K_{f}$-rational, by a theorem of Serre [Reference SerreSer98, ch. III, §3], augmented with a transcendence result of Waldschmidt [Reference WaldschmidtWal81] (see Henniart [Reference HenniartHen82]), this restriction is locally algebraic (and semisimple by assumption). Then by a theorem of Ribet [Reference RibetRib76, §1.6] (which extends that of Serre [Reference SerreSer98, § III.2.3]), there exist (i) a two-dimensional $K_{f}$-rational vector subspace $V_{0}$ of $K_{f,\unicode[STIX]{x1D706}}^{\oplus 2}$, (ii) a modulus $\mathfrak{m}$ of $\widetilde{F^{\circ }}$, and (iii) a rational representation

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{0}:S_{\mathfrak{m}}\otimes _{\mathbb{Q}}K_{f}\longrightarrow \text{GL}_{V_{0}}\end{eqnarray}$$

such that $\unicode[STIX]{x1D70C}|_{\widetilde{F^{\circ }}}$ is isomorphic to the $\unicode[STIX]{x1D706}$-adic representation associated with $\unicode[STIX]{x1D719}_{0}$.

The image of $\unicode[STIX]{x1D719}_{0}$ is a maximal algebraic torus of $\text{GL}_{V_{0}}$, since $\det \unicode[STIX]{x1D719}_{0}$ gives the Tate structure $\mathbb{Q}_{\ell }(-1)$ on $\operatorname{Gal}(\overline{\mathbb{Q}}/F^{\circ })$, and the cyclotomic character is not divisible by $2$ as the character of any number field.

Let $K^{\prime }$ be the splitting field over $K_{f}$ of this algebraic torus, so that $[K^{\prime }:K_{f}]\leqslant 2$ and $[G_{f,\unicode[STIX]{x1D706}}:G_{f,\unicode[STIX]{x1D706}}^{\circ }]\leqslant 2$. (It is worth clarifying that, unlike the $K^{\prime }$ introduced in the proof of Theorem 4.1.1 in a similar context, this $K^{\prime }$ depends only on $\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}$ and is independent of $\mathfrak{p}$.)

For every prime $\mathfrak{p}$ of $\widetilde{F^{\circ }}$ of absolute degree $1$ and coprime to $\text{disc}(F)\cdot \mathfrak{n}\cdot \ell$, let $\{\unicode[STIX]{x1D6FC}_{\mathfrak{p}},\unicode[STIX]{x1D6FD}_{\mathfrak{p}}\}\subset K^{\prime }$ be the two eigenvalues of $\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}(\operatorname{Frob}_{\mathfrak{p}})$. Since they are Weil $p$-integers by Lemma 4.1.3 and $\det (\unicode[STIX]{x1D70C}(\operatorname{Frob}_{\mathfrak{p}}))=p$, we have

(4.2.1.1)$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{\mathfrak{p}}=\frac{p}{\unicode[STIX]{x1D6FC}_{\mathfrak{p}}}=\overline{\unicode[STIX]{x1D6FC}_{\mathfrak{p}}},\end{eqnarray}$$

where the bar denotes the complex conjugation on $\mathbb{Q}(\unicode[STIX]{x1D6FC}_{\mathfrak{p}})=\mathbb{Q}(\unicode[STIX]{x1D6FD}_{\mathfrak{p}})$.

Now consider only those $\mathfrak{p}$ such that, in addition, $(p):=\mathfrak{p}\,\cap \,\mathbb{Z}$ splits completely in $K^{\prime }$, a fortiori also in $\mathbb{Q}(\unicode[STIX]{x1D6FC}_{\mathfrak{p}})$; the resulting set is clearly principally abundant. Then, since $(p)$ is unramified in $\mathbb{Q}(\unicode[STIX]{x1D6FC}_{\mathfrak{p}})$, $\unicode[STIX]{x1D6FC}_{\mathfrak{p}}$ cannot be totally real, and generates a CM field. Let $\{{\wp}_{1},\ldots ,{\wp}_{m},\overline{{\wp}_{1}},\ldots ,\overline{{\wp}_{m}}\}$ be the set of primes of $\mathbb{Q}(\unicode[STIX]{x1D6FC}_{\mathfrak{p}})$ lying over $(p)$, where $2m=[\mathbb{Q}(\unicode[STIX]{x1D6FC}_{\mathfrak{p}}):\mathbb{Q}]$. Equation (4.2.1.1) further shows that, perhaps after renaming the primes, we get

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{\mathfrak{p}}\cdot \mathscr{O}_{\mathbb{Q}(\unicode[STIX]{x1D6FC}_{\mathfrak{p}})}={\wp}_{1}\cdots {\wp}_{m}\quad \text{and}\quad \unicode[STIX]{x1D6FD}_{\mathfrak{p}}\cdot \mathscr{O}_{\mathbb{Q}(\unicode[STIX]{x1D6FC}_{\mathfrak{p}})}=\overline{{\wp}_{1}}\cdots \overline{{\wp}_{m}}.\end{eqnarray}$$

It follows that

$$\begin{eqnarray}\text{Tr}(\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}(\operatorname{Frob}_{\mathfrak{p}}))=\unicode[STIX]{x1D6FC}_{\mathfrak{p}}+\unicode[STIX]{x1D6FD}_{\mathfrak{p}}\end{eqnarray}$$

does not belong to any prime ideal of $\mathscr{O}_{K^{\prime }}$ lying over $(p)$. Since $\text{Tr}(\unicode[STIX]{x1D70C}_{f,\unicode[STIX]{x1D706}}(\operatorname{Frob}_{\mathfrak{p}}))\in \mathscr{O}_{K_{f}}$, it belongs to no prime of $\mathscr{O}_{K_{f}}$ lying over $(p)$ either. This proves that $k(p)=0$ and completes the proof of Theorem 4.2.1.◻

Proof. (1) The key point is that under the assumptions, we can use the crystalline realisation to compute the two polygons in part (c). Namely, for almost all $\mathfrak{p}$, we have the free $W(k(\mathfrak{p}))$-module $\mathfrak{R}_{\text{cris},\mathfrak{p}}$, equipped with the Hodge filtration and the crystalline Frobenius $\unicode[STIX]{x1D719}_{\text{cris},\mathfrak{p}}$ (induced from those on $\mathfrak{M}$).