On the reflex norm

Krzysztof Klosin

March 11, 2005

1 The reflex field

Let F denote a CM field, [F : Q] = 2g. Denote by F^{+} the maximal totally real subfield of F. Let C be a CM field which is Galois over Q and splits F. We denote by R the maximal totally real subfield of C. To ease notation, every time we write A ⊗ B, for two Q-algebras A and B, we mean A ⊗_{Q} B. There is a canonical map of F ⊗ C-algebras

F ⊗ C ⟶^{∼} ∏_{ϕ∈Hom_{Q}(F,C)} C_{ϕ}  (1)

given by c⊗f ↦ (cϕ(f))_{ϕ}, which is an isomorphism; the notation C_{ϕ} means C made into an F -algebra via ϕ. We have a similar decomposition for F ⊗ R. Indeed, F ⊗ R is canonically isomorphic to the (finite) product ∏_{𝔭∈Spec(F ⊗R)}(F ⊗ R)/𝔭, where each of the quotients is non-canonically R-isomorphic to C. We naturally have (F ⊗ R)/𝔭 ≅ (F ⊗ R)_{𝔭}. We let F_{𝔭} denote (F ⊗ R)_{𝔭} and P denote Spec(F ⊗ R). We also write F_{𝔭}^{+} for the corresponding factor field of F^{+} ⊗ R. We have a natural identification:

(Res_{F/Q} G_{m})_{R} ⟶^{∼} ∏_{𝔭∈P} Res_{F_{𝔭}/R} G_{m}

which for any R-algebra A is given by

(F ⊗ A)^{×} ⟶^{∼} (F ⊗ R ⊗_{R} A)^{×} ⟶^{∼} ∏_{𝔭∈P}(F_{𝔭} ⊗_{R} A)^{×}.

Let (F, Φ) be a CM type (over C, cf. Brian’s talk), i.e., Φ ⊂ Hom_{Q}(F, C) is a set of representatives for Hom_{Q}(F, C) mod Gal(F/F^{+}). Every ϕ ∈ Hom_{Q}(F, C) uniquely extends to an R-algebra map ϕ̃: F ⊗R → C (f ⊗ r ↦ ϕ(f)r) which must factor through exactly one of the factor fields F_{𝔭}, yielding an R-algebra isomorphism ϕ_{𝔭} : F_{𝔭} ⟶^{∼} C. Hence, we can think of Φ as the set {ϕ_{𝔭} : F_{𝔭} ⟶^{∼} C | 𝔭 ∈ P }. If ϕ_{𝔭} ∈ Φ we will denote the other R-algebra isomorphism F_{𝔭} ⟶^{∼} C by \overline{ϕ}_{𝔭} since it is related to ϕ_{𝔭} via Gal(C/R)..

Let S := Res_{C/R} G_{m}. We define a map h_{Φ} : S → (Res_{F/Q} G_{m})_{R} ≅ ∏_{𝔭∈P} Res_{F_{𝔭}/R} G_{m}, which for every

R-algebra A is given by h_{Φ}(A) = (∏_{ϕ_{𝔭}∈Φ} ϕ_{𝔭}^{−1}) ⊗ 1 : (C ⊗_{R} A)^{×} → (F ⊗ A)^{×} ≅ ∏_{𝔭∈P} F_{𝔭} ⊗_{R} A.

Example 1.1. On R-points, the map h_{Φ} is the map C^{×} → ∏_{𝔭∈P} F_{𝔭}^{×} given by z ↦ (ϕ_{𝔭}^{−1}(z))_{𝔭}. In fact, since S is a torus and R is infinite, S(R) is Zariski-dense in S and hence knowing h_{Φ} on R-points determines it completely.

Let T be the kernel of the composite Res_{F/Q} G_{m} → Res_{F^{+}/Q} G_{m} → (Res_{F^{+}/Q} G_{m}) /G_{m}, where the first arrow is the norm map N_{F/F^{+}} and the second is the natural projection. In Nick’s talk it was shown that the kernel of the first arrow is a torus, so connectedness of G_{m} implies connectedness of T, hence T ⊂ Res_{F/Q} G_{m} is a Q-torus.

Lemma 1.2. The map h_{Φ} factors through T_{R}.

Proof. We want to show that h_{Φ}^{−1}(T_{R}) = S. As rational points are Zariski-dense in any torus over an infinite field, it is enough to check the equality holds on R-points, i.e., we want to show that h_{Φ}(S(R)) ⊂ T (R). By Example 1.1, this says that the map

C^{×} \underrightarrow{∏_{ϕ𝔭∈Φ} ϕ_{𝔭}^{−1}} ∏_{ϕ_{𝔭}∈Φ} F_{𝔭}^{×} \underrightarrow{∏ N_{F𝔭/F^{+}_{𝔭}} (·)} ∏_{𝔭∈P}(F_{𝔭}^{+})^{×}

lands in R^{×} diagonally embedded in ∏_{𝔭∈P} (F_{𝔭}^{+})^{×}. However this is clear as the canonical maps R^{×} → (F_{𝔭}^{+})^{×} are isomorphisms and for every 𝔭 ∈ P we have that ϕ_{𝔭} (N_{F_{𝔭}/(F^{+})_{𝔭}}(ϕ_{𝔭}^{−1}(z))) = N_{C/R}(z) ∈ R is independent of 𝔭.

Choose a totally negative element α ∈ R^{×} for which there exists δ ∈ C^{×} such that δ^{2} = α. (The existence of such α and δ was shown in Tong’s talk via weak approximation.) We define a map

μ_{Φ} : G_{m/C} ↪ S_{C} ⟶^{(h_{Φ})_{C}} T_{C},

where the first arrow j : G_{m/C} ↪ S_{C} is defined on A-points (for any C-algebra A) by

j(a) = \frac{1}{2} ⊗ (a + 1) + \frac{δ ⊗ δ(a − 1)}{2α}.

It is easy to see a priori (using that R is the maximal totally real subfield of the CM field C) that j is independent of α and δ and that j(aa′) = j(a)j(a′), so j really is unit-valued and is a group morphism. More concretely, if we identify S_{C} with G_{m} × c^{∗}(G_{m}) (with c denoting the non-trivial element of Gal(C/R)) then j is given (on Yoneda points) by z ↦ (z, 1).

Definition 1.3. The subfield E_{(F,Φ)} of C which is the field of definition of μ_{Φ} is called the reflex field of (F, Φ).

We will denote the descent of μ_{Φ} to E := E_{(F,Φ)} by μ_{Φ,E}, or simply by μ_{Φ} when it cannot cause confusion. We have Gal(C/E_{(F,Φ)}) = {σ ∈ Gal(C/Q) | σ^{∗}(μ_{Φ}) = μ_{Φ}}.

Remark 1.4. In Eiji’s talk the reflex field F^{∗} was defined to be the subfield of C generated by elements ∑_{ϕ∈Φ} ϕ(a), a ∈ F. He also proved that F^{∗} is the fixed field of the group {σ ∈ Gal(C/Q) | σΦ = Φ}. The next two lemmas imply that the notion of the reflex field in the sense of Definition 1.3 coincides with that defined in Eiji’s talk.

Lemma 1.5. σ^{∗}(μ_{Φ}) = μ_{σΦ}

Proof. Since μ_{Φ} is a map between connected algebraic groups in characteristic zero, it is uniquely determined by the induced map on the Lie algebras. For any affine group scheme G_{/k}, where k is a field, the Lie algebra Lie(G) is in canonical bijection with the points in G(k[ϵ]) mapping to 1 ∈ G(k) (here ϵ^{2} = 0). The map μ_{Φ} = (h_{Φ})_{C} ∘ j induces a map on C[ϵ]-points

C[ϵ]^{×} ⟶ (C ⊗_{R} C[ϵ])^{×} ⟶ (F ⊗ C[ϵ])^{×} = ∏_{𝔭∈P}(F_{𝔭} ⊗_{R} C[ϵ])^{×}  (2)

where the first arrow satisfies 1 + cϵ ↦ \frac{1}{2} ⊗ (2 + cϵ) + \frac{δ⊗δc}{2α}ϵ and the second arrow satisfies

1 + (c ⊗ c′)(1 ⊗ ϵ) ↦ (1 + (ϕ_{𝔭}^{−1}(c) ⊗ c′)(1 ⊗ ϵ))_{𝔭}.

We now investigate the effect of the σ^{∗}-operation on the C-linear map Lie μ_{Φ}. We consider the following situation. Let G, H be two affine group schemes of finite type over Q and let V and W be the tangent spaces at identity of G and H respectively. A C-group map G_{C} ⟶^{ψ} H_{C} induces a C-linear transformation C ⊗ V ⟶^{T_{ψ}} C ⊗ W that is Lie(ψ). On the other hand, for σ ∈ Gal(C/Q) the map σ^{∗}ψ induces the Lie algebra map C ⊗ V ⟶^{σ^{∗}T_{ψ}} C ⊗ W that is the scalar extension of T_{ψ} via σ, so we get a commutative diagram of C-vector spaces

\xymatrix{C_{σ} ⊗_{C} C ⊗ V \ar[r]^{id ⊗T_{ψ}} & C_{σ} ⊗_{C} C ⊗ W \\ C ⊗ V \ar[u]_{≀}^{c⊗v↦c⊗1⊗v} \ar[r]^{σ^{∗}T_{ψ}} & C ⊗ W \ar[u]^{≀}_{c⊗w↦c⊗1⊗w}}  (3)

Hence, if v ∈ V and T_{ψ}(v) = ∑ c_{i} ⊗ w_{i} ∈ C ⊗ W, then (σ^{∗}T_{ψ})(v) = ∑ σ(c_{i}) ⊗ w_{i} = (σ ⊗ 1) ∘ T_{ψ}(v). So in our case, σ^{∗}(μ_{Φ})(1 + ϵ) = (1 ⊗ σ)(μ_{Φ}(1 + ϵ)) for v ∈ Q[ϵ]^{×} ⊂ C[ϵ]^{×}, where 1 ⊗ σ : (F ⊗ C)^{×} → (F ⊗ C)^{×} is induced from 1 ⊗ σ : F ⊗ C → F ⊗ C. We have the following commutative diagram of rings:

\xymatrix{F ⊗ C \ar[d]_{≀}^{f⊗z↦(f_{𝔭}⊗z)_{v}} \ar[r]_{1⊗σ} & F ⊗ C \ar[d]_{≀} \\ ∏_{𝔭∈P} F_{𝔭} ⊗_{R} C \ar[d]_{≀}^{(f_{𝔭}⊗z_{𝔭})_{𝔭}↦(ϕ_{𝔭}(f_{𝔭})z_{𝔭},\overline{ϕ_{𝔭}}(f_{𝔭})z_{𝔭})} \ar[r]^{(f_{𝔭}⊗z_{𝔭})_{𝔭}↦(f_{𝔭}⊗σ(z_{𝔭}))_{𝔭}} & ∏_{𝔭∈P} F_{𝔭} ⊗_{R} C \ar[dd]_{≀}^{(f_{𝔭}⊗z_{𝔭})_{𝔭}↦(τ(f_{τ})z_{τ})_{τ}} \\ ∏_{𝔭∈P} C_{ϕ_{𝔭}} × C_{\overline{ϕ}_{𝔭}} \ar[d]^{=} \\ ∏_{τ:F ↪C} C_{τ} \ar[r]^{(z_{τ})_{τ} ↦(σ(z_{σ^{−1}τ}))_{τ}} & ∏_{τ:F ↪C} C_{τ}}  (4)

where f_{τ} := f_{𝔭} and z_{τ} := z_{𝔭} if τ and \overline{τ} are the two ways to identify F_{𝔭} with C as R-algebras. Let v = 1 + ϵ ∈ Q[ϵ]^{×}. We first calculate the image of μ_{Φ}(1 + ϵ) ∈ (F ⊗ C[ϵ])^{×} inside (∏_{τ:F ↪C} C_{τ} [ϵ])^{×}.

By the definition of maps (2), we have μ_{Φ}(1 + ϵ) = (1 + \frac{1}{2} (1 + \frac{ϕ_{𝔭}^{−1}(δ)⊗δ}{α} ) (1 ⊗ ϵ))_{𝔭} ∈ ∏_{𝔭∈P} (F_{𝔭} ⊗ C)_{𝔭}^{×} which

gives the element (1 + ϵ, 1) ∈ ∏_{𝔭∈P} C_{ϕ_{𝔭}}^{×} × C_{\overline{ϕ}_{𝔭}}^{×}, i.e., μ_{Φ}(1 + ϵ) = (β_{τ})_{τ}, where β_{τ} = 1 + ϵ if τ ∈ Φ and β_{τ} = 1 otherwise.

Now, σ^{∗}(μ_{Φ})(1 + ϵ) = (1 ⊗ σ)(μ_{Φ}(1 + ϵ)) viewed as an element of (∏_{F ↪C} C_{τ} [ϵ])^{×} is identified with Ψ_{σ}((β_{τ})_{τ}) by commutativity of diagram (4), where Ψ_{σ} is the right vertical arrow in diagram (4). We have Ψ_{σ}((β_{τ})_{τ}) = (σ(β_{σ^{−1}τ}))_{τ}, so by the definition of β_{τ} we have σ(β_{σ^{−1}τ}) = 1 + ϵ if σ^{−1}τ ∈ Φ and σ(β_{σ^{−1}τ}) = 1 otherwise. Hence σ^{∗}(μ_{Φ}) = μ_{σΦ}.

Lemma 1.6. μ_{Φ} = μ_{Φ′} if and only if Φ = Φ′.

Proof. One implication is trvial. The other implication follows immediately from the description of μ_{Φ} in the proof of Lemma 1.5. Indeed, we calculated there that μ_{Φ}(1 + ϵ) = (β_{τ})_{τ} ∈ (∏_{τ:F ↪C} C_{τ})^{×}, where β_{τ} = 1 + ϵ if τ ∈ Φ and β_{τ} = 1 otherwise. Hence if μ_{Φ} = μ_{Φ′}, we must have Φ = Φ′.

2 The reflex norm

We will now define the notion of the reflex norm. Let L′/L be a finite separable field extension (not necessarily Galois), and X a commutative affine L-group of finite type. Fix a separable closure L^{sep}/L. There is a canonical “trace” map

Res_{L′/L} X_{L′} → X,  (5)

which for an L-algebra A is given by

X_{L′}(A ⊗_{L} L′) = X(A ⊗_{L} L′) → X(A ⊗_{L} L^{sep})^{Gal(L^{sep}/L)} = X(A)

a′ ↦ ∏_{σ∈Hom_{L}(L′,L^{sep})} σ(a′).

This is independent of the choice of L^{sep} and is natural in X. Setting L = Q, L′ = E = E_{(F,Φ)}, and X = T, we obtain a canonical map

ψ : Res_{E/Q} T_{E} → T.

Remark 2.1. Such a map can also be obtained with Res_{F/Q} G_{m} replacing T, and the two are compatible via the inclusion ι : T ↪ Res_{F/Q} G_{m} since (5) is natural in X. We will use this fact in the proof of Proposition 2.5. On C-points with X = Res_{F/Q} G_{m}, the “trace” Res_{L′/Q} X_{L′} → X is the map (F ⊗L′⊗C)^{×} → (F ⊗C)^{×} induced by the product of F ⊗ C-algebra maps x ⊗ y ⊗ z ↦ x ⊗ σ(y)z over all σ ∈ Hom_{Q}(L′, C), i.e., this is the usual ring-theoretic norm on units.

Definition 2.2. The map r_{Φ} = ψ ∘ Res_{E/Q} μ_{Φ} : Res_{E/Q} G_{m} → T is the reflex norm.

We will now give an alternative definition of the reflex norm and show that it is equivalent to the one given above.

Let A be an abelian variety over C, with i : F ↪ End^{∘}(A) of type (F, Φ). Let E be the reflex field of (F, Φ). The abelian variety (A, i) typically does not descend to E. However the tangent space at identity t_{A/C} is isomorphic ∏_{ϕ∈Φ} C_{ϕ} as an F ⊗ C-module and this concrete model ∏_{ϕ∈Φ} C_{ϕ} has a natural C-semilinear action of Gal(C/E) via σ((a_{ϕ})_{ϕ}) = (σ(a_{σ^{−1}ϕ}))_{ϕ}, so it descends to E in the following sense.

Lemma 2.3. Up to isomorphism, there exists a unique F ⊗ E-module t_{Φ} such that t_{Φ} ⊗_{E} C ≅ ∏_{ϕ∈Φ} C_{ϕ} as F ⊗ C-modules compatible with the natural Gal(C/E)-actions on both sides.

Proof. See Eiji’s talk. We only note here that explicitly t_{Φ} is defined to be the F ⊗E-submodule of Gal(C/E)- invariants in ∏_{ϕ∈Φ} C_{ϕ}.

We now define a map N_{Φ} : Res_{E/Q} G_{m} → Res_{F/Q} G_{m} which for any Q-algebra A is given by

N_{Φ}(A) : (E ⊗ A)^{×} → (F ⊗ A)^{×},

via s ↦ det_{F ⊗_{Q}A} (s : t_{Φ} ⊗_{Q} A → t_{Φ} ⊗_{Q} A).

Example 2.4. Let F be a quadratic imaginary field. Then Φ = {ϕ : F ↪ C} is a one-element set, hence E = ϕ(F). In this case t_{Φ} is a one-dimensional F -vector space and N_{Φ}(Q) : E^{×} → F^{×} is given by ϕ^{−1}.

The following proposition shows that N_{Φ} coincides with the reflex norm in the sense of Definition 2.2.

Proposition 2.5. Let ι : T ↪ Res_{F/Q} G_{m} be the inclusion map. Then ι ∘ r_{Φ} = N_{Φ}.

Remark 2.6. It is important that the target group of the reflex norm is the torus T, and not merely the full group Res_{F/Q} G_{m}. The reason which makes this fact significant for us is (as will be proved later, using results in Nick’s talk) that the Q-points of T are discrete in T (A_{f}), where A_{f} denotes the finite adeles of Q. This makes the quotient space T (Q)\T (A_{f}) Hausdorff. This is not the case if we replace T with Res_{F/Q} G_{m} unless the group of intgral units of F is finite, which is to say F is imaginary quadratic.

Proof of Proposition 2.5. First note that E ⊂ C by definition. To ease notation we will denote by E the set Hom_{Q}(E, C). Also for each σ ∈ E we choose σ̃ ∈ Gal(C/Q) satisfying σ̃|_{E} = σ. Then σ̃ϕ makes sense for every ϕ ∈ Φ.

We will show that the following diagram commutes:

\xymatrix{(E ⊗ C)^{×} \ar[r]^{Res_{E/Q} μ_{Φ,E}} \ar[d]_{E⊗C-action} & (F ⊗ E ⊗ C)^{×} \ar[d]^{∏_{σ∈E}[σ](·)} \\ Aut_{F ⊗C}(t_{Φ} ⊗ C) \ar[r]^{_{det_{F ⊗C}}} & (F ⊗ C)^{×} = ∏_{τ:↪C} C_{τ}^{×}}  (6)

where the top map is on C-points and [σ](f ⊗ e ⊗ z) = f ⊗ σ(e)z as an F ⊗ C-algebra map from F ⊗ E ⊗ C to F ⊗ C. By Remark 2.1 the composite of the right vertical arrow with the top arrow is ι ∘ r_{Φ}(C), while the composite of the other two arrows is N_{Φ}(C). We will first describe the map N_{Φ}(C) = det_{F ⊗C} ∘(E⊗C)-action. We have an F ⊗ E ⊗ C-linear decomposition

t_{Φ} ⊗ C = t_{Φ} ⊗_{E} E ⊗ C = ∏_{σ∈E} t_{Φ} ⊗_{E} C_{σ} = ∏_{σ∈E} t_{Φ} ⊗_{E} C ⊗_{C} C_{σ} = ∏_{σ∈E} ∏_{ϕ∈Φ} C_{ϕ} ⊗_{C} C_{σ} = ∏_{σ∈E} ∏_{ϕ∈Φ} C_{ϕ,σ}

where the second equality from the right follows from the definition of t_{Φ} in the proof of Lemma 2.3; the notation C_{ϕ,σ} means C made into an E-algebra and F -algebra via σ and σ̃ϕ respectively. Moreover E ⊗ C = ∏_{σ∈E} C_{σ} acts on t_{Φ} ⊗ C = ∏_{σ∈E} ∏_{ϕ∈Φ} C_{ϕ,σ} via the diagonal action of C_{σ} on ∏_{ϕ∈Φ} C_{ϕ,σ}. To compute det_{F ⊗C} we need to rearrange the factors in ∏_{σ∈E} ∏_{ϕ∈Φ} C_{ϕ,σ} according to the action of F ⊗ C.

As F ⊗ C-modules we have t_{Φ} ⊗ C = ∏_{τ:F ↪C} V_{τ}, where V_{τ} = ∏_{(ϕ,σ)∈Σ(Φ,τ)} C_{ϕ,σ}, with Σ(Φ, τ) := {(ϕ, σ) ∈ Φ × E | σ̃ϕ = τ}. For σ ∈ E, let σ′ : E ⊗ C → C be the C-algebra map induced by e ⊗ z ↦ σ(e)z. On each V_{τ} an element ξ ∈ E ⊗ C acts by multiplication by the diagonal matrix diag (σ′(ξ))_{(ϕ,σ)∈Σ(Φ,τ)}, so the C_{τ} -determinant of this C-linear action equals ∏_{(ϕ,σ)∈Σ(Φ,τ)} σ′(ξ) ∈ C_{τ} for each τ : F ↪ C. After we prove that diagram (6) commutes, this computation will provide a practical way to calculate the reflex norm (cf. Remark 2.7).

We will now describe the composite of the top and right maps in diagram (6). We first consider the top map; the difficulty is that μ_{Φ} over C is concrete but μ_{Φ,E} is “abstract”. Note that given two Q-schemes X

and Y, and an E-map X_{E} ⟶^{f} Y_{E}, with E a subfield of C, we have a commutative diagram:

\xymatrix{(Res_{E/Q} X_{E}) (C) = X_{E}(E ⊗ C) \ar[r]^{Res_{E/Q} f} \ar[d]^{≀} & (Res_{E/Q} Y_{E}) (C) = Y_{E}(E ⊗ C) \ar[d]^{≀} \\ (Res_{E⊗C/C} X_{E⊗C/C}) (C) \ar[r]^{Res_{E⊗C/C} f_{C}} \ar[d]^{≀} & (Res_{E⊗C/C} Y_{E⊗C/C}) (C) \ar[d]^{≀} \\ ∏_{σ:E↪C} Res_{C_{σ}/C} X_{C_{σ}}(C) \ar[r]^{∏_{σ} Res_{Cσ /C} f_{Cσ}} \ar[d]^{≀} & ∏_{σ:E↪C} Res_{C_{σ}/C} Y_{C_{σ}}(C_{σ}) \ar[d]^{≀} \\ ∏_{σ:E↪C} X_{C_{σ}}(C_{σ}) \ar[r]^{∏_{σ} f_{Cσ}} & ∏_{σ:E↪C} Y_{C_{σ}}(C_{σ})}  (7)

where f_{C_{σ}} = σ̃^{∗}(f_{C}) : X_{C_{σ}} → Y_{C_{σ}}. Hence if X = G_{m}, Y = Res_{F/Q} G_{m}, and f = μ_{Φ,E} then since μ_{Φ/C_{σ}} = σ̃^{∗}(μ_{Φ}) = μ_{σ̃Φ} (by Lemma 1.5), we have that Res_{E/Q} μ_{Φ,E} on C-points is identified with the map

∏_{σ:E↪C} C_{σ}^{×} \underrightarrow{∏_{σ} μ_{σ̃Φ}} ∏_{σ:E↪C} (Res_{F/Q} G_{m})_{C_{σ}} (C_{σ}) = ∏_{σ:E↪C} (F ⊗ C_{σ})^{×} .  (8)

For any σ : E ↪ C, we have the following commutative diagram of rings

\xymatrix{C ⊗_{R} C_{σ} \ar[r]^{∏_{𝔭∈P} (σ̃ϕ_{𝔭})^{−1}⊗1} \ar[d]^{≀} & F ⊗ C_{σ} = ∏_{𝔭∈P} F_{𝔭} ⊗_{R} C_{σ} \ar[d]^{(f_{𝔭}⊗z_{σ})_{𝔭,σ}↦(σ̃ϕ_{𝔭}(f_{σ̃ϕ𝔭})z_{σ},σ̃\overline{ϕ_{𝔭}(f_{σ̃ϕ𝔭})}z_{σ})_{𝔭}} \\ C_{σ} × C\overline{σ} \ar[r]^{(z_{σ},z\overline{σ})↦(z_{σ},z\overline{σ})_{𝔭}} & ∏_{𝔭∈P} C_{ϕ_{𝔭},σ} × C_{\overline{ϕ}_{𝔭},σ}}  (9)

and a commutative diagram of unit groups

\xymatrix{C_{σ}^{×} \ar[r]^{j_{Cσ}} \ar[dr]_{z_{σ}↦(z_{σ},1)} & (C ⊗_{R} C_{σ})^{×} \ar[d]_{≀} \\  & C_{σ}^{×} × C× ; \overline{σ}}  (10)

Note that the composite of the top map in diagram (9) with j_{C_{σ}} is the map μ_{σ̃Φ}. Hence if z_{σ} ∈ C_{σ}^{×}, then μ_{σ̃Φ}(z_{σ}) = (z_{σ}, 1)_{𝔭} if we view it as an element of ∏_{𝔭∈P} C_{ϕ_{𝔭},σ}^{×} × C_{\overline{ϕ}_{𝔭},σ}^{×} via the right vertical arrow

in diagram (9). If we identify ∏_{𝔭∈P} C_{ϕ_{𝔭},σ} × C_{\overline{ϕ}_{𝔭},σ} with ∏_{τ:F ↪C} C_{τ,σ}, then (z_{σ}, 1)_{𝔭} is identified with the element (z_{τ,σ})_{τ,σ}, where z_{τ,σ} := z_{σ} if τ ∈ Φ and z_{τ,σ} := 1 otherwise.

Now, for every σ_{0} : E ↪ C we have the following commuative diagram of rings:

\xymatrix{E ⊗ F ⊗ C \ar[r]^{[σ_{0}](·)} \ar[d]^{≀} & F ⊗ C \ar[dd]^{≀} \\ ∏_{σ:E↪C} F ⊗ C_{σ} \ar[d]^{≀} \\ ∏_{σ:E↪C} ∏_{𝔭∈P} F_{𝔭} ⊗_{R} C_{σ} \ar[r]^{(f_{𝔭,σ}⊗z_{𝔭,σ})_{𝔭,σ}↦(f_{𝔭,σ_{0}} ⊗z_{𝔭,σ_{0}})_{𝔭}} \ar[d]^{≀} & ∏_{𝔭∈P} F_{𝔭} ⊗_{R} C \ar[d]^{≀} \\ ∏_{σ:E↪C} ∏_{ϕ_{𝔭}∈Φ} C_{ϕ_{𝔭},σ} × C_{\overline{ϕ}_{𝔭},σ} \ar[d]^{≀} & ∏_{𝔭∈P} C_{ϕ_{𝔭}} × C_{\overline{ϕ}_{𝔭}} \ar[d]^{≀} \\ ∏_{σ:E↪C} ∏_{τ:F ↪C} C_{τ,σ} \ar[r]^{(z_{τ,σ})_{τ,σ}↦(z_{τ_{0},σ_{0}})_{σ̃_{0}τ_{0}=τ}} & ∏_{τ:F ↪C} C_{τ}}  (11)

Our goal is to compute ∏_{σ∈A}[σ] (Res_{E/Q} μ_{Φ,E}) (ξ), where ξ ∈ (Res_{E/Q} G_{m}) (C) = (E ⊗ C)^{×}, but for z_{σ} := σ′(ξ) by using diagrams (7) and (11) this is the same as first computing ∏_{σ:E↪C} μ_{σ̃Φ}(z_{σ}), which is

(z_{τ,σ})_{τ,σ} ∈ ∏_{σ:E↪C} ∏_{τ:F ↪C} C_{τ,σ}^{×},

then composing it with the bottom arrow in diagram (11) giving (z_{τ_{0},σ_{0}})_{σ̃_{0}τ_{0}=τ} ∈ ∏_{τ:F ↪C} C_{τ}^{×}, and finally taking the product over all σ_{0} : E ↪ C, yielding

( ∏_{\begin{matrix}τ_{0},σ_{0} \\ σ̃_{0}τ_{0}=τ \end{matrix}} z_{τ_{0},σ_{0}} )_{τ} = ( ∏_{σ∈Σ(Φ,τ)} z_{σ})_{τ} ∈ ∏_{τ:F ↪C} C_{τ}^{×}.

This is the same element we obtained when we calculated the composite of the other two maps in diagram (6).

Remark 2.7. The proof of Proposition 2.5 gives a concrete expression for the reflex norm of an element e ∈ E^{×}. Indeed, fix τ : F ↪ C. Then the reflex norm r_{Φ}(e) is τ^{−1} (∏_{σ∈Σ(Φ,τ)} σ(e)). Note that this is in

fact independent of the choice of τ, as it should be. Moreover, we have r_{Φ}(e)\overline{r_{Φ}(e)} = N_{E/Q}(e).