Endoscopy on SL₂-eigenvarieties

Let w ∣ p be a place of F ̃ , and let ϖ w be a uniformizer of F ̃ w which we also view as the element of A F ̃ × with ϖ w in the 𝑤-component and 1’s elsewhere. Then

where e w is the ramification index of the extension F ̃ / Q p .

Proof

Let p w ⊆ O F ̃ denote the prime ideal corresponding to 𝑤 and choose n ∈ Z > 0 so that p w n is principal. Let a ∈ O F ̃ generate p w n . We can write the idèle ϖ w n as

where a ∈ F ̃ × (embedded diagonally), a ∞ − 1 has 1 at all finite components and a − 1 at all infinite components, and 𝑢 is simply ϖ w n ⁢ a − 1 ⁢ a ∞ . Since 𝑎 generates p w n , one sees that u ∈ O ̂ F ̃ × . Since 𝜓 has open kernel (and F ̃ is totally complex), we have ψ ⁢ ( a ∞ ) = 1 and ψ ⁢ ( u ) is a root of unity. In particular, v p ⁢ ( ψ ⁢ ( a ∞ ) ) = v p ⁢ ( ψ ⁢ ( u ) ) = 0 and we see that

By definition (and since 𝑎 generates p w n ), v p ⁢ ( σ ⁢ ( a ) ) = 0 if σ ∉ Σ w and v p ⁢ ( σ ⁢ ( a ) ) = e w − 1 ⁢ n if σ ∈ Σ w . Thus v p ⁢ ( ψ ⁢ ( ϖ w n ) ) = e w − 1 ⁢ n ⁢ ∑ σ ∈ Σ w k σ , and dividing by 𝑛 gives the result. ∎

Let v ∣ p be a place of 𝐹, which we assume splits as v = w ⁢ w ̄ in F ̃ . Let ϖ w and ϖ w ̄ be uniformizers of F ̃ w and F ̃ w ̄ , respectively. Assume that ψ ⁢ ( ϖ w ) = − ψ ⁢ ( ϖ w ̄ ) . Then we must have

Proof

If ψ ⁢ ( ϖ w ) = − ψ ⁢ ( ϖ w ̄ ) , then we must have v p ⁢ ( ψ ⁢ ( ϖ w ) ) = v p ⁢ ( ψ ⁢ ( ϖ w ̄ ) ) . Since e w = e w ̄ , the result follows directly from Lemma A.1.1. ∎

Let a , b ∈ Z with a + b = w ∈ 2 ⁢ Z . Set k 0 = k 3 = a and k 1 = k 2 = b . Write α ∈ A L × as α = γ ⁢ u ⁢ x as above. Then

is an algebraic character of 𝐿 of weight ( k i ) i = 0 3 . Moreover, ψ a , b ⁢ ( ϖ + ) = ψ a , b ⁢ ( ϖ − ) .

Proof

The first statement is clear by definition as long as ψ a , b is well defined (note that the kernel contains O ̂ L × ⁢ L ∞ × ). If we write 𝛼 in a different way as α = δ ⁢ v ⁢ y with δ ∈ L × , v ∈ O ̂ L × and y ∈ L ∞ × , then γ = δ ⁢ ω for some ω ∈ O L × , so we need to check that ∏ i = 0 3 σ i ⁢ ( ω ) k i = 1 for all ω ∈ O L × . This can be checked on generators, so it is enough to check it for ω = ( 1 + i ) / 2 and ω = 1 + 2 , which is a short straightforward calculation (using that 𝑤 is even). Similarly, a straightforward computation gives that ψ a , b ⁢ ( ϖ + ) = ψ a , b ⁢ ( ϖ − ) = 5 w . ∎

Let a , b ∈ Z with a + b = w ∈ 2 ⁢ Z and set

Then there exists an algebraic character 𝜓 of 𝐿 of weight ( k i ) i = 0 3 which is unramified at 2 + i and 2 − i and such that ψ ⁢ ( ϖ + ) = − ψ ⁢ ( ϖ − ) .

Proof

Let L ′ / L be a quadratic extension such that 2 + i splits and 2 − i is inert in L ′ , which exists by e.g. [2, §10.2, Theorem 5]^[15]. If 𝜒 is the quadratic character corresponding to L ′ / L , then we can view it as an algebraic character of 𝐿 which is trivial on L × ⁢ L ∞ × and satisfies χ ⁢ ( 2 + i ) = 1 and χ ⁢ ( 2 − i ) = − 1 . Then, if ψ a , b is as in Lemma A.1.3, ψ = χ ⋅ ψ a , b satisfies the requirements of the corollary. ∎

Let 𝜓 be an algebraic character as above with weights ( k σ ) σ of 𝜓 satisfying k σ ≠ k σ ̄ for all 𝜎. Let v ∣ p and assume that 𝑣 splits as v = w ⁢ w ̄ in F ̃ , and that 𝜓 is unramified at 𝑤 and w ̄ .

1. If ∑ σ ∈ Σ w k σ ≠ ∑ σ ∈ Σ w ̄ k σ , then a v ⁢ ( π ) ≠ 0 . In particular, if F v = Q p , then a v ⁢ ( π ) ≠ 0 .

2. It can happen that a v ⁢ ( π ) = 0 .

Proof

We have a v ⁢ ( π ) = 0 if and only if ψ ⁢ ( ϖ w ) = − ψ ⁢ ( ϖ w ̄ ) , so the first part follows from Corollary A.1.2 and the second part follows from the example in Corollary A.1.4, choosing a ≠ b . ∎

𝐵 is normal and Cohen–Macaulay. Moreover,

1. if n = 1 , then 𝐵 is regular.

2. If n = 2 , then 𝐵 is a complete intersection but not regular.

3. If 𝑛 is odd and at least 3, then 𝐵 is 2-Gorenstein (but not Gorenstein).

4. If 𝑛 is even and at least 4, then 𝐵 is Gorenstein but not a complete intersection.

Proof

We may equivalently work with the affine form R = E ⁢ [ x 1 , … , x n ] ⊆ A and S = R H instead. That 𝑆 is Cohen–Macaulay and normal is a general property of quotient singularities (indeed of rational singularities). When n = 1 , S = E ⁢ [ x 1 2 ] and part (1) follows. Part (2) follows from the description

when n = 2 .

From now on, assume n ≥ 3 . Let U ⊆ Spec ⁡ S = X be the complement of the origin and write 𝑗 for the inclusion. Then ω X = j ∗ ⁢ ( ω U ) . In particular, H 0 ⁢ ( X , ω X ) = H 0 ⁢ ( U , ω U ) . Set U i = X \ { x i = 0 } ; then 𝑈 is the union of the U i , and H 0 ⁢ ( U i , ω U ) consists of all top forms

where the degrees of all terms in f ∈ R have the same parity 𝑑, and d − b ≡ n modulo 2. Gluing, we see that H 0 ⁢ ( X , ω X ) consists of the top forms

where all terms in g ∈ R have the same parity as 𝑛. In particular, ω X is free of rank 1 when 𝑛 is even, so 𝑆 is Gorenstein, but when 𝑛 is odd, the top forms x i ⁢ d ⁢ x 1 ∧ ⋯ ∧ d ⁢ x n , i = 1 , … , n , form a minimal set of generators of ω X in any neighbourhood of 0, so 𝑆 is then not Gorenstein. A similar calculation replacing ω X by ω X ⊗ 2 shows that 𝑆 is 2-Gorenstein when 𝑛 is odd.

It remains to show that 𝐵 is not a complete intersection when 𝑛 is even. Let 𝑇 be the polynomial ring T = E ⁢ [ u i , v j ⁢ k ] with i = 1 , … , n and 1 ≤ j < k ≤ n . The ring 𝑇 surjects onto 𝑆 by sending u i to x i 2 and v j ⁢ k to x j ⁢ x k . Let 𝐼 be the kernel of this surjection. Consider the elements f i ⁢ j = u i ⁢ u j − v i ⁢ j 2 ∈ I . We claim that the f i ⁢ j form a regular sequence, but that they do not generate 𝐼. Comparing the dimensions of 𝑆 and 𝑇 and the length of that sequence, this would imply that 𝑆 is not a complete intersection (by [59, Tag 09PZ]). Set

To show that the f i ⁢ j form a regular sequence, it suffices to show, for each i < j , that no irreducible component of

is contained in Z i ⁢ j . Let x ∈ Z i ⁢ j ∩ Y i ⁢ j be any point. By deforming the v i ⁢ j -coordinate of 𝑥 but not the other coordinates, we see that we stay in Y i ⁢ j but move out of Z i ⁢ j . This shows that no component of Y i ⁢ j is contained in Z i ⁢ j , as desired, and finishes the proof that the f i ⁢ j form a regular sequence. It remains to show that they do not generate 𝐼. To see this, consider, for example, the element v 12 ⁢ v 23 − u 2 ⁢ v 13 ∈ I . If

then we would obtain a contradiction by setting u i = 1 for all 𝑖 and v j ⁢ k = − 1 for all j < k . This finishes the proof. ∎

Let X = Spa ⁡ ( A ) be an affinoid rigid space and let x ∈ X be a point corresponding to a maximal ideal 𝔪. Then the natural map A → O X , x induces an isomorphism A ̂ m → O ̂ X , x .

Let f : X → Y be a morphism of rigid analytic varieties over a non-archimedean field 𝐾, and let x ∈ X be a 𝐾-point with image f ⁢ ( x ) = y ∈ Y . Assume that 𝑓 is locally quasi-finite at 𝑥. Then the following are equivalent:

1. 𝑓 is a local isomorphism at 𝑥;

2. 𝑓 is étale at 𝑥;

3. 𝑓 induces an isomorphism O ̂ Y , y ≅ O ̂ X , x .

Proof

That (1) implies (2) is trivial, and that (2) implies (3) is clear since 𝑓 induces an isomorphism on residue fields (since 𝑥 is defined over the base field 𝐾). It remains to prove that (3) implies (1).

First, note that O Y , y → O X , x is flat since it induces an isomorphism on completions and the local rings O X , x and O Y , y are Noetherian (using that completion of Noetherian local rings is faithfully flat). In other words, 𝑓 is flat at 𝑥. Since flatness is an open property, we may shrink 𝑋 to ensure that 𝑓 is flat. Second, using [34, Proposition 1.5.4], we may assume that 𝑋 and 𝑌 are affinoid and 𝑓 is finite. Since the subset of rank 1 points in 𝑋 is Hausdorff and f − 1 ⁢ ( y ) is finite, a standard topology argument allows us to shrink 𝑋 and 𝑌 so that f − 1 ⁢ ( y ) = { x } . Let 𝐴 and 𝐵 be the affinoid 𝐾-algebras so that X = Spa ⁡ B and Y = Spa ⁡ A , and consider f ∗ : A → B . Then

so f ∗ has rank 1 (since it has so after tensoring with O ̂ Y , y ) and is therefore an isomorphism, as desired. This finishes the proof. ∎