Lifting of Characters for Nonlinear Simply Laced Groups

Jeffrey Adams and Rebecca Herb

March 20, 2007

1 Introduction

This is an alpha version of this paper. Sorry, the introduction gets written last.

2 Some Notation

Throughout this paper G will be the set of real points of a connected reductive complex Lie group G(ℂ). Let H(ℂ) be a Cartan subgroup of G(ℂ), and let Φ be the set of roots of H(ℂ) in G(ℂ). Let Φ^{+} be a set of positive roots, and ρ = \frac{1}{2} ∑_{α∈Φ^{+}} α as usual. We assume that the derived group G_{d}(ℂ) is acceptable, i.e. ρ exponentiates to a character of H(ℂ) ∩ G_{d}(ℂ). Let p : G̃ → G be a two-fold cover of G. We identify the kernel of p with ±1.

Let G_{ad}(ℂ) be the adjoint group, with real points G_{ad}. Write int(g) for the action of G_{ad}(ℂ) on G(ℂ), or G_{ad} on G.

The main results only apply if every factor of Φ has one root length or is of type G_{2}. Many of the auxiliary results of this paper hold in general so we will not assume this except when necessary. We plan to return to the general case in a future paper.

For any subgroup H of G we write H̃ = p^{−1}H. Writing Z(H) for the center, let Z_{0}(H) = p(Z(H̃)) ⊂ Z(H); then Z(H̃) = p^{−1}(Z_{0}(H)). In particular Z_{0}(G) ⊂ Z(G) plays an important role.

We denote real Lie algebras by Gothic letters 𝔥, 𝔤, 𝔱, …, and their complexifications by 𝔥(ℂ), 𝔤(ℂ), 𝔱(ℂ), . . .. We write σ for the action of the nontrivial element of the Galois group on 𝔤(ℂ) and G(ℂ), so 𝔤 = 𝔤(ℂ)^{σ} and G(ℝ) = G(ℂ)^{σ}.

Fix a Cartan involution θ of G(ℂ), so that K = G^{θ} is a maximal compact subgroup of G. Let H be a θ-stable Cartan subgroup of G. Then σ(α) = −θ(α) for all α ∈ Φ. Roots are classified as real, imaginary, complex, compact as in [10]. Write Φ = Φ_{r} ∪ Φ_{i} ∪ Φ_{cx} accordingly. We also have the decomposition of Φ_{i} = Φ_{i,c} ∪ Φ_{i,n} into compact and non-compact roots. If Φ^{+} is a set of positive roots, write Φ_{r}^{+} = Φ^{+} ∩ Φ_{r}, and Φ^{+}i, Φ_{cx}^{+} similarly. Write ρ = ρ(Φ^{+}) = \frac{1}{2} ∑_{α∈Φ^{+}} α as usual. Define ρ_{r}, ρ_{i} and ρ_{cx} similarly, so ρ = ρ_{r} + ρ_{i} + ρ_{cx}.

We say Φ^{+} is real if σΦ_{cx}^{+} = Φ_{cx}^{+}, i.e. if α is a positive non-imaginary root then σ(α) > 0. This is the set of roots of the nilpotent radical of a real parabolic of G.

Let P, R be the weight and root lattices, respectively, and denote the coweight and coroot lattices P^{∨} and R^{∨}.

Write 𝔥 = 𝔱 ⊕ 𝔞 as usual, and H = T A with T = H ∩ K and A = exp(𝔞). Typically H̃ is not abelian, and its center plays an important role.

A simple root system is said to be simply laced if all roots have the same length, and an arbitrary root system is simply laced if this holds for each simple factor. More succinctly a root system is simply laced if whenever α, β are non-proportional roots then ⟨α, β^{∨}⟩ = 0, ±1. We adopt the convention that in this case all roots are long.

We say a root system is oddly laced if whenever α, β are non-proportional roots then ⟨α, β^{∨}⟩ = 0 or is odd. Thus oddly laced as shorthand for each simple factor is simply laced or of type G_{2}. We also adopt the convention that in type G_{2} all roots are long. The reason for these conventions is Lemma 3.1. The main results in this paper hold for oddly laced groups, although (with the general case in mind) we will only make this assumption when necessary.

3 Admissible Triples

Fix G as in Section 2 and a two-fold cover p : G̃ → G. Let H be a Cartan subgroup of G. Fix a real or non-compact imaginary root α. Associated to α is the root subgroup M_{α}, which is locally isomorphic to SL(2, ℝ). As in [2, Definition 3.2] we say α is metaplectic if p^{−1}(M_{α}) is a non-linear group. It is easy to see SL(2, ℝ)/ ± 1 has no such cover, so we conclude M_{α} ≃ SL(2, ℝ) and p^{−1}(M_{α}) ≃ S̃L(2, ℝ), the unique non-trivial two-fold cover.

For the next Lemma see [3] or [2, Lemma 3.3].

Lemma 3.1 Assume G(ℂ) is simple and simply connected and that G̃ is non-linear. Fix a θ-stable Cartan subgroup H of G with roots Φ. Then every long real or non-compact imaginary root α ∈ Φ is metaplectic. Furthermore G admits a non-linear cover if and only if there is a Cartan subgroup H with a long real or long non-compact imaginary root. If this condition holds the non-linear two-fold cover is unique up to isomorphism.

It is enough to check this condition on the fundamental or the maximally split Cartan subgroup. If G is oddly laced it is enough to check this condition on any Cartan subgroup.

Definition 3.2 We say that p : G̃ → G is an admissible two-fold cover if for every Cartan subgroup H, every long real or long noncompact imaginary root is metaplectic.

Equivalently G̃ is admissible if and only if G̃_{i} is non-linear for every simple factor G_{i} of G which admits such a cover.

See [2, Definition 3.4]).

As discussed in the Introduction we are going to lift characters from a real form of a quotient of G(ℂ). We need to impose some conditions on this quotient. This will take up the remainder of this section.

Suppose C ⊂ Z(G(ℂ)) is a finite subgroup stable under the Galois action. Then \overline{G}(ℂ) = G(ℂ)/C is defined over ℝ, and let \overline{G} be its real points. Write \overline{p} : G(ℂ) → \overline{G}(ℂ) for the projection map. Let O(G) be the conjugacy classes of G. If g, g′ ∈ G are conjugate by G(ℂ) we say g, g′ are stably conjugate, and let O_{st}(G) be the set of stable conjugacy classes. (This is a naive definition, which agrees with the usual one for strongly regular semisimple elements.) Similar notation applies to other groups.

For g ∈ G let O(G, g) be the conjugacy class of g, and O_{st}(G, g) = {xgx^{−1} | x ∈ G(ℂ), xgx^{−1} ∈ G} the stable conjugacy class. Similar notation applies to other groups.

Definition 3.3 Assume C is a two-group. For h ∈ \overline{G} let ϕ(h) = s(h)^{2} where s : \overline{G}(ℂ) → G(ℂ) is any section.

Lemma 3.4 The map ϕ is well defined, and satisfies:

1. ϕ(\overline{G}) ⊂ G and ϕ(\overline{G}^{0}) ⊂ G^{0},

2. ϕ induces a map O(\overline{G}^{0}) → O(G^{0}),

3. ϕ induces a map O^{st}(\overline{G}) → O^{st}(G).

Proof. Since C is a two-group it is immediate that ϕ(h) is independent of the choice of s.

Suppose g ∈ \overline{G}, h ∈ G and p(h) = g. Then \overline{p}(σ(h)) = σ(\overline{p}(h)) = \overline{p}(h), so σ(h) = zh for some z ∈ C. Since C is a two-group σ(h)^{2} = h^{2}, so ϕ(g) ∈ G. Furthermore since p : G^{0} → \overline{G}^{0} is surjective the second assertion in (1) is clear.

Define

(3.5)  ϕ(O(\overline{G}^{0}, g)) = O(G^{0}, ϕ(g)) (g ∈ \overline{G}^{0}).

If x ∈ \overline{G}^{0} choose y ∈ G^{0} with \overline{p}(y) = x. Then ϕ(xgx^{−1}) = yϕ(g)y^{−1} so this is well defined, proving (2). Similarly define

(3.6)  ϕ(O_{st}(\overline{G}, g)) = O_{st}(G, ϕ(g)) (g ∈ \overline{G})

Suppose x ∈ \overline{G}(ℂ) and xgx^{−1} ∈ \overline{G}. Choose y ∈ G(ℂ) with \overline{p}(y) = x. Then ϕ(xgx^{−1}) = yϕ(g)y^{−1}, so again ϕ is well-defined. □

It is worthwhile noting that ϕ does not define a map O(\overline{G}) → O(G). Suppose we try to define ϕ(O(\overline{G}, g)) = O(G, ϕ(g)) as in (3.5). If x ∈ \overline{G} but x ∉ \overline{G}^{0} then there is no guarantee we can find y ∈ G such that ϕ(y) = x, so it may not hold that ϕ(g) is conjugate to ϕ(xgx^{−1}). (By (3) ϕ(g) is stably conjugate to ϕ(xgx^{−1})).

Example 3.7 Let G = SL(2, ℝ) and \overline{G} = P GL(2, ℝ). Let

t(θ) = (\begin{matrix}cos(θ) & sin(θ) \\ − sin(θ) & cos(θ) \end{matrix}) ∈ G.

Let \overline{t}(θ) be the image of t(θ) in \overline{G}. Then \overline{t}(θ) is conjugate to \overline{t}(θ), but ϕ(\overline{t}(θ)) = t(2θ) is not conjugate to ϕ(\overline{t}(−θ)) = t(−2θ) unless θ is an integral multiple of π/2.

We assume throughout this paper that O is a two-group.

Now fix a Cartan subgroup H of G, and let \overline{H}(ℂ) = H(ℂ)/C ⊂ \overline{G}(ℂ), with real points \overline{H}. We define ϕ : \overline{H}(ℂ) → H(ℂ) by the same formula as in Definition 3.3. Equivalently, if we write exp : 𝔥(ℂ) → H(ℂ), \overline{exp} : 𝔥(ℂ) → \overline{H}(ℂ), then for X ∈ 𝔥(ℂ), ϕ(\overline{exp}(X)) = exp(2X). It is immediate that

(3.8)  α(ϕ(h)) = α(h)^{2} (for all α ∈ Φ).

We make repeated use of some structure theory of Cartan subgroups which we now explain. Write 𝔥 = 𝔱 ⊕ 𝔞 and H = T A as in Section 2. Let

(3.9)  Γ(H) = exp(i𝔞) ∩ H = {exp(iX) | X ∈ 𝔞, exp(2iX) = 1}.

For α ∈ Φ_{R} (cf. Section 2) let m_{α} = α^{∨}(−1) = exp(πiα^{∨}) and define

(3.10)(a)  Γ_{R}(H) = ⟨m_{α} | α ∈ Φ_{R}⟩ ⊂ Γ(H) ∩ G_{d}^{0}.

It is well known that

(3.10)(b)  H = Γ(H)H^{0}, H ∩ G^{0} = Γ_{R}(H)H^{0}.

Up to conjugacy by G there are unique maximally split and fundamental Cartan subgroups H_{s} and H_{f}. Then

(3.10)(c)  G = H_{s}G^{0}, H_{f} ∩ G^{0} = H_{f}^{0}.

Let G_{d} be the derived group of G, and H_{d} = H ∩ G_{d}, with identity components G_{d}^{0} and H_{d}^{0}, respectively. Equations (a) and (b) imply

(3.10)(d)  G_{d} = G_{d}^{0}, H = Γ(H)Z(G)H_{d}^{0}.

We will also apply this to \overline{G} and \overline{H}.

Lemma 3.11 The homomorphism ϕ : \overline{H}(ℂ) → H(ℂ) has the following properties.

1. ϕ(wh) = wϕ(h) for all w ∈ W (Φ), h ∈ \overline{H}(ℂ),

2. ϕ(\overline{H}^{0}) = H^{0} ,

3. ϕ(h) = 1 for all h ∈ Γ_{R}(\overline{H}), 4. ϕ(\overline{H} ∩ \overline{G}^{0}) = H^{0} ,

5. ϕ(Γ(\overline{H})) = Γ(H) ∩ C,

6. ϕ(\overline{H}) = (Γ(H) ∩ C)H^{0}.

Proof. The first two are clear from the definition. For (3) note that ϕ(\overline{exp}(πiα^{∨})) = exp(2πiα^{∨}) = 1 for all α ∈ Φ_{R}. Then (4) follows since \overline{H} ∩ \overline{G}^{0} = Γ_{R}(\overline{H})\overline{H}^{0}.

Let 𝔥 = \overline{exp}(iX) ∈ Γ(\overline{H}) where X ∈ 𝔞 with \overline{exp}(2iX) = 1. Now ϕ(h) = exp(2iX) ∈ exp(i𝔞). But \overline{p}(ϕ(h)) = \overline{exp}(2iX) = 1 so that ϕ(h) ∈ C ∩ exp(i𝔞) ⊂ Γ(H) ∩ C. Conversely, let z_{G} = exp(iY) ∈ C ∩ Γ(H) where Y ∈ 𝔞 with exp(2iY) = 1. Since z_{G} ∈ C, \overline{p}(z_{G}) = \overline{exp}(iY) = 1. Let z = \overline{exp}(iY/2). Then z ∈ Γ(\overline{H}) and ϕ(z) = z_{G}. Finally (6) follows since \overline{H} = Γ(\overline{H})\overline{H}^{0}. □

It is a standard fact that the character of an irreducible genuine representation of G̃, considered as a function on the regular semisimple elements, vanishes off of Z(H̃) (for example see [1, §II]). Our character identities involve the image of ϕ in H, so we would like to know that ϕ(\overline{H}) ⊂ Z_{0}(H), with image as large as possible.

Let H_{s} be a maximally split Cartan subgroup of G, and assume ϕ(\overline{H}_{s}) ⊂ Z_{0}(H_{s}). Let C_{s} = Γ(H_{s}) ∩ C so that by Lemma 3.11 ϕ(\overline{H}_{s}) = C_{s}H_{s}^{0}. By assumption C_{s} ⊂ Z_{0}(H_{s}) ∩ Z(G), so (cf. (4.1)(g)) C̃_{s} ⊂ CentG̃(G̃^{0}) ∩ Z(H̃). Since H_{s} is maximally split G = H_{s}G^{0}, which implies CentG̃(G̃^{0}) ∩ Z(H̃_{s}) = Z(G̃). Therefore C_{s} ⊂ Z_{0}(G).

Now for a general θ-stable Cartan subgroup H = T A, there is a maximally split Cartan H_{s} = T_{s}A_{s} so that that A ⊂ A_{s}. Then Γ(H) ⊂ Γ(H_{s}), so that ϕ(\overline{H}) = (C ∩ Γ(H))H^{0} = (C_{s} ∩ Γ(H))H^{0}. That is, if we use C_{s} in place of C we have the same images ϕ(\overline{H}) for every Cartan subalgebra 𝔥. Thus we may as well assume that C ⊂ Z_{0}(G).

Lemma 3.12 Assume that C ⊂ Z_{0}(G). Then for every Cartan subgroup ϕ(Γ(\overline{H})) ⊂ Z_{0}(G) and ϕ(\overline{H}) ⊂ Z_{0}(G)H^{0} ⊂ Z_{0}(H). Furthermore ϕ(Z(\overline{G})) ⊂ Z_{0}(G).

Proof. The first part follows from the above discussion. Let H_{s} be maximally split with roots Φ. Let z ∈ Z(\overline{G}) ⊂ \overline{H}_{s}, z̃ ∈ Z(H̃_{s}) such that p(z̃) = ϕ(z). Then for every α ∈ Φ,

α(z̃) = α(ϕ(z)) = α(z)^{2} = 1

by (3.8). Thus z̃ ∈ CentG̃(G̃^{0}) ∩ Z(H̃_{s}) = Z(G̃) as above. □

To reiterate, so far we have assumed C is a σ-invariant subgroup of G(ℂ), C ⊂ Z_{0}(G), and z^{2} = 1 for all z ∈ C. For the definition of transfer factors we need to impose a further technical condition on C. Suppose χ̃ is a genuine character of Z(G̃). Then χ̃^{2} factors to a character of Z_{0}(G). Let

(3.13)(a)  Z_{2}(G) = {z ∈ Z_{0}(G) | z^{2} = 1}.

Then χ̃^{2} restricted to Z_{2}(G) is independent of the choice of χ̃: if p(z̃) = z then χ̃^{2}(z) = 1 if z̃^{2} = 1, or −1 otherwise.

Since G is acceptable e^{ρ}(z) is well defined for z ∈ Z_{0}(G) ∩ G_{d}. Let

(3.13)(b)  ζ_{2}(z) = χ̃^{2}(z)e^{ρ}(z) z ∈ Z_{2}(G) ∩ G_{d}^{0} ⊂ Z(G_{d}^{0}).

This is independent of the choices of χ̃, H and Φ^{+}.

Definition 3.14 An admissible triple is a set (G̃, G, \overline{G}) where:

1. G is the set of real points of a connected reductive complex Lie group G(ℂ) such that the derived group G_{d}(ℂ) is acceptable, and every simple factor of the root system of G(ℂ) has one root length or is of type G_{2}.

2. p : G̃ → G is an admissible two-fold cover (Definition 3.2).

3. \overline{G} is the set of real points of \overline{G}(ℂ) = G(ℂ)/C where C is a finite central subgroup of G satisfying the following conditions:

(a) c^{2} = 1 for all c ∈ C,

(b) C ⊂ Z_{0}(G) = p(Z(G̃)),

(c) ζ_{2}(z) = 1 for all z ∈ C ∩ G_{d}^{0} (cf. (3.13)(b)).

Assume that (G̃, G, \overline{G}) is an admissible triple. Let H = T A be a Cartan subgroup of G. Let M = Cent_{G}(A) be a cuspidal Levi subgroup. Then M̃ ⊂ G̃ and \overline{M} = Cent\overline{G}(\overline{A}) ⊂ \overline{G} are also cuspidal Levi subgroups. We will show in Section ?? that (M̃, M, \overline{M}) is also an admissible triple. This means our constructions will be compatible with parabolic induction.

Suppose G̃ and G satisfy conditions (1) and (2). Then (G̃, G, \overline{G}) is an admissible triple for \overline{G}(ℂ) = G(ℂ)/C for any subgroup C satisfying conditions (3)(a-c). For example, C = {1} satisfies this condition, so that (G̃, G, G) is an admissible triple. By Lemma 3.11(6) taking C bigger makes the image of ϕ(\overline{H}) ⊂ Z_{0}(H) bigger, and in general, we get the best lifting results when we take C as large as possible. However, if we take C as large as possible for G, it may not be the maximal choice for Levi subgroups of G. Thus we do not specify C beyond the requirements of Definition 3.14.

Example 3.15 Let H = T A be a maximally split Cartan subgroup of G. We will show in Lemma 3.17 below that it is not always possible to pick C satisfying the conditions of Definition 3.14 such that ϕ(\overline{H}) = Z_{0}(H). Let M = C_{G}(A). We will show it is possible to pick a finite central subgroup C_{M} of M satisfying the conditions of Definition 3.14 for M such that ϕ(\overline{H}) = Z_{0}(H) where ϕ is defined using \overline{M}(ℂ) = M(ℂ)/C_{M}. Recall H = Γ(H)H^{0}. Thus H̃ = Γ̃(H)H̃^{0} where Γ̃(H) = p^{−1}Γ(H). Since H̃^{0} ⊂ Z(H̃), we have Z(H̃) = Z(Γ̃(H))H̃^{0}. Now Γ(H) ⊂ Z(M) so that Z(Γ̃(H)) ⊂ CM̃(M̃^{0}) ∩ Z(H̃) = Z(M̃). Thus Z_{0}(H) = Γ_{0}(H)H^{0} where Γ_{0}(H) = p(Z(Γ̃(H))) ⊂ Z_{0}(M) ∩ Γ(H). Since Γ_{0}(H) is a two-group, there is C_{M} ⊂ Γ_{0}(H) such that Z_{0}(H) = C_{M}H^{0} and C_{M} ∩ H^{0} = {1}. But C_{M} ∩ M_{d}^{0} ⊂ C_{M} ∩ H^{0} = {1}, so it is easy to see that C_{M} satisfies the conditions of Definition 3.14 for M and gives ϕ(\overline{H}) = C_{M}H^{0} = Z_{0}(H).

The following lemma, which is an immediate consequence of (5.12) is useful for producing admissible triples.

Lemma 3.16 Let H be a maximally split Cartan subgroup of G. Then ζ_{2}(c) = 1 for all c ∈ Z_{0}(G) ∩ Γ_{R}(H).

Lemma 3.17 Suppose that G(ℂ) is simple and simply connected, and let H be a maximally split Cartan subgroup of G. Then we can choose C so that ϕ(\overline{H}) = Z_{0}(H) except when G = SU(n, n) where n is even, G = Spin(p, q) where p and q are even with p > q ≥ 2 or when G = Spin^{∗}(2n) where n ≡ 0 mod 4.

Proof.

Clearly if Z_{0}(H) is connected we can take C = {1}. We know that Z_{0}(H) = Z_{0}(G)H^{0} = Z(G)H^{0} since G is connected. But since H = Γ_{R}(H)H^{0} where every element of Γ_{R}(H) is two group, it is easy to see that Z(G)H^{0} = Z_{e}(G)H^{0} where Z_{e}(G) is the group of elements in Z(G) with even order. Thus Z_{0}(H) is connected when Z_{e}(G) = {1}. This will be the case for all real forms when Φ is of type A_{2n}, E_{6}, E_{8}, or G_{2}. Thus we may as well assume that Φ is type A_{2n−1}, D_{n}, or E_{7}. In these cases Z_{e}(G) = Z(G). We may also assume that G is a real form such that Z_{0}(H) = Z(G)H^{0} is not connected. In particular, we can assume that G is not compact or complex. Suppose that Z(G) ⊂ Γ_{R}(H). Then Z(G) is a two-group, and by Lemma 3.16 we can take C = Z(G), giving us ϕ(\overline{H}) = Z(G)H^{0} = Z_{0}(G) as desired. This will always be the case if G is the split real form.

Suppose that Φ = A_{2n−1}. If G is type AI, then G is split. If G is type AII, then H is connected. If G is type AIII, H is connected unless G = SU(n, n). In this case Z_{0}(H) = H has two components. Now Z(G) =< z > is cyclic of order 2n and so Z_{2}(G) =< z_{0} > where z_{0} = z^{n} ∈ Γ_{R}(H) is the unique element of order two. Thus the biggest C we can take is C = {1, z_{0}}. When n is odd, z_{0} ∉ H^{0} so that ϕ(\overline{H}) = CH^{0} = H, but when n is even, z_{0} ∈ H^{0} so that ϕ(\overline{H}) = CH^{0} = H^{0} is a proper subgroup of Z_{0}(H).

Suppose that Φ = D_{n}. If G is type DI, then G = Spin(p, q), p ≥ q, p + q = 2n, and Φ_{R} is type D_{q}. H is connected unless q ≥ 2 and H is split if p = q. Thus we can assume that p > q ≥ 2. If p, q are odd, then Z(G) = Z_{2} ⊂ Γ_{R}(H). If p, q are even, then Z(G) = {1, z_{0}, z_{1}, z_{0}z_{1}} where z_{0} ∈ Γ_{R}(H), but z_{1} ∉ Γ(H). Thus the biggest C we can take is C = {1, z_{0}}. But z_{1} ∉ CH^{0} so that CH^{0} is a proper subgroup of Z_{0}(H). If G is type DIII, then G = Spin^{∗}(2n). H is connected if n is odd. Suppose that n = 2k is even. Then Z(G) = {1, z_{0}, z_{1}, z_{0}z_{1}} as above where z_{1} ∈ Γ_{R}(H), but z_{0} ∉ Γ(H) so the biggest C we can take is C = {1, z_{1}}. If k is even, then z_{0} ∉ CH^{0} while z_{0} ∈ CH^{0} when k is odd.

Finally, if Φ = E_{7}, then either G is split or Z_{0}(H) is connected. □

4 Structure Theory

Assume G_{d}(ℂ) is simply connected and τ is an automorphism of G. Then τ stabilizes G_{d} = G_{d}^{0}, and by Lemma 3.1 it lifts uniquely to an automorphism τ̃ of the (unique) admissible cover G̃_{d} of G_{d}. Suppose z ∈ Z(G_{d}) and τ(z) = z. Choose an inverse image z̃ of z and set

(4.1)(a)  {τ, z} = τ̃(z̃)z̃^{−1}.

Then p({τ, z}) = 1 so {τ, z} = ±1. It is independent of the choice of z̃.

Note that the real points G_{ad} of G_{ad}(ℂ) act by conjugation on G, and apply this discussion to τ = int(g) for g ∈ G_{ad}. Then int(g) fixes Z(G_{d}) pointwise, and we may therefore define

(4.1)(b)  {g, z} = {int(g), z} (g ∈ G_{ad}, z ∈ Z(G_{d})).

This defines a continuous pairing {, } : G_{ad} × Z(G_{d}) → ℤ/2ℤ. By continuity {g, z} = 1 for all g ∈ G_{ad}^{0}.

Let R(G) = G/Z(G)G^{0} and R_{ad}(G) = R(G_{ad}) = G_{ad}/G_{ad}^{0}. We obtain a canonical pairing

(4.1)(c)  {, } : R_{ad}(G) × Z(G_{d}) → ℤ/2ℤ.

By duality we obtain a canonical homomorphism

(4.1)(d)  γ_{ad} : R_{ad} → Hom(Z(G_{d}), ℤ/2ℤ).

We now drop the assumption that G_{d}(ℂ) is simply connected. Let G_{sc}(ℂ) be the simply connected cover of G_{d}(ℂ), with corresonding real form G_{sc}. The preceding discussion gives a canonical map:

(4.1)(e)  γ_{ad} : R_{ad} → Hom(Z(G_{sc}), ℤ/2ℤ).

Assume that we are given an admissible cover G̃ of G. Suppose g, h are commuting elements of G. Choose inverse images g̃, h̃ of g, h in G̃. Analogously to (4.1)(a) define

(4.1)(f)  \begin{aligned}\{g, h\} & = int(g̃)(h̃)h̃^{−1} \\ & = g̃h̃g̃^{−1}h̃^{−1}. \end{aligned}

As in (4.1)(a) {g, h} = ±1, independent of the choices, and we say {g, h} is the commutator of g, h (with respect to the cover G̃). It is continuous in each factor so

(4.1)(g)  {Z(G), G^{0}} = 1.

It follows that we obtain a pairing

(4.1)(h)  {, } : R(G) × Z(G_{d}^{0}) → ℤ/2ℤ

depending on the cover G̃. As in (4.1)(d) by duality this gives a homomorphism

(4.1)(i)  γ : R(G) → Hom(Z(G_{d}^{0}), ℤ/2ℤ).

For later use we note a simple consequence of (4.1)(g):

(4.2)  Z(G̃) = { \atop Z(G)} if G is connected.

In general we only have Z(G̃) ⊂ { \atop Z(G)}.

Remark 4.3 Unlike γ_{ad}, which is canonical, γ appears to depend on the choice of an admissible cover G̃ of G. In fact by (4.4) below γ is independent of this choice. However the existence of the map γ depends on the existence of an admissible cover.

The map G → G_{ad} factors to an inclusion R(G) ↪ R_{ad}(G). Also Z(G_{d}^{0}) is a quotient of Z(G_{sc}), and there is a natural inclusion Hom(Z(G_{d}^{0}), ℤ/2ℤ) ↪ Hom(Z(G_{sc}), ℤ/2ℤ). It is easy to see the following diagram is commutative:

(4.4)  R_{ad}(G)  Hom(Z(G_{sc}), ℤ/2ℤ) R(G)  Hom(Z(G_{d}^{0}), ℤ/2ℤ)^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} γ_{ad}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}  _{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} γ  _{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}

Proposition 4.5 Assume G is oddly laced. Then γ_{ad} and γ are injective.

Proof. It is enough to prove the first statement, so assume G_{d}(ℂ) is simply connected. It is easy to reduce to the case G(ℂ) simple, so we assume G(ℂ) is simple and simply connected. If G is of type G_{2} then G = G_{ad} and R_{ad}(G) = 1 so this case is obvious. Assume G is simply laced.

First assume G contains a compact Cartan subgroup T, which implies Z(G) = Z(G(ℂ)). The standard isomorphism P/R ≃ Hom(Z(G), ℂ^{×}) gives an isomorphism

(4.6)  P ∩ \frac{1}{2}R/R ≃ Hom(Z(G), ℤ/2ℤ).

Let Φ = Φ(G(ℂ), T (ℂ)). By [11, Proposition 9.5(a)]

(4.7)  R_{ad}(G) ≃ Norm_{W} (Φ_{I,c})/W (Φ_{I,c})

Let λ be the differential of a genuine character of T̃. It is not hard to see that using the isomorphism (4.7)

(4.8)  γ_{ad} : w → wλ − λ ∈ P ∩ \frac{1}{2}R/R ≃ Hom(Z(G), ℤ/2ℤ).

It is enough to show

(4.9)  wλ − λ ∈ R ⇒ w ∈ W (Φ_{I,c}).

This follows most easily from a straightforward case-by-case check. We sketch a more conceptual proof below.

Now suppose G does not contain a compact Cartan subgroup. We claim that if G is SL(2n+1, ℝ), SL(n, ℍ) or of type E_{6} then R_{ad}(G) = 1 and there is nothing to prove. Let Γ be the Galois group of ℂ/ℝ. It is well known (for example see [4]) there is an injection

(4.10)  α : R_{ad}(G) ↪ H^{1}(Γ, Z(G(ℂ))).

It is easy to compute the right hand side; it is trivial for SL(2n + 1, ℝ) or any real form of E_{6}. We leave to the reader verification that R_{ad}(G) = 1 when G is the adjoint form of SL(n, ℍ).

Suppose G = SL(2n, ℝ). It is easy to see R_{ad}(G) ≃ ℤ/2ℤ in this case. Write x, y for the elements of p^{−1}(−I) ⊂ Z(G̃). It is not hard to see that if g ∈ G_{ad}/G_{ad}^{0} then the lift τ̃ of int(g) to S̃L(2n, ℝ) satisfies τ̃(x) = y (cf. Example 4.13). This proves this case. The only remaining case is Spin(p, q) with p, q odd, in which case R_{ad}(G) = ℤ/2ℤ. This is similar to SL(2n, ℝ) and we leave the details to the reader. □

Remark 4.11 The injectivity of γ imposes a strong constraint on the existence of an admissible cover of G if G_{d}(ℂ) is not simply connected, or (essentially the same thing) if G_{d}(ℂ) is simply connected but R(G) is nontrivial.

Example 4.12 If G_{d}(ℂ) is adjoint then G has an admissible cover only if R(G) = 1.

Example 4.13 It is obvious that G = P GL(2n + 1, ℝ) ≃ SL(2n + 1, ℝ), and therefore G has an admissible cover. In this case R(G) = 1.

If G = P GL(2n, ℝ) then R(G) ≃ ℤ/2ℤ, so G does not have an admissible cover. We make this explicit.

Note that G^{0} ≃ SL(2n, ℝ)/ ± I. Let S̃L(2n, ℝ) → SL(2n, ℝ) be the admissible cover. Write Z(S̃L(2n, ℝ)) = {1, z, x, y} with p(z) = I, p(x) = p(y) = −I. It is easy to compute that Z(S̃L(2n, ℝ)) is isomorphic to ℤ/2ℤ× ℤ/2ℤ if n is even, or ℤ/4ℤ if n is odd.

Suppose p : G̃ → G is an admissible cover of G. Then G̃^{0} is a quotient of S̃L(2n, ℝ) by a central two element subgroup A. If A = {1, z} then G̃^{0} = SL(2n, ℝ) is not admissible. If n is odd then x, y have order 4, so there is no other such subgroup A: in this case G^{0} does not have an admissible two-fold cover.

Now suppose n is even, and A = {1, x} or {1, y}. Choose g ∈ G\G^{0} and g̃ ∈ p^{−1}(g). Then int(g̃) ∈ Aut(G̃^{0}) lies over int(g) ∈ Aut(G^{0}). However int(g) lifts uniquely to an automorphism τ̃ of S̃L(2n, ℝ), and τ̃ satisfies τ̃(x) = y. Therefore τ̃ does not factor to G̃^{0}. This contradiction proves that no such admissible cover exists.

Note that what is subtle is that if n is even then G^{0} has an admissible cover (take S̃L(2n, ℝ)/{1, x}), but this cover does not extend to G.

Corollary 4.14 Assume G is oddly laced. Then

(4.15)  Z({ \atop Z(G)})G̃^{0} = Z(G̃)G̃^{0}

Proof. It is obvious that Z(G̃) ⊂ Z({ \atop Z(G)}) so one inclusion is obvious, and it is enough to show Z({ \atop Z(G)}) ⊂ Z(G̃)G̃^{0}. Suppose z̃ ∈ Z({ \atop Z(G)}) and let z = p(z̃) ∈ Z(G). Then {z, g} = 1 for g ∈ Z(G) and (by (4.1)(g)) G^{0}. So τ_{z} : g → {z, g} is an element of Hom(R(G), ℤ/2ℤ). By Proposition 4.5 the Pontrijagin dual of γ (4.1)(i) is a surjection Z(G_{d}^{0}) → Hom(R(G), ℤ/2ℤ). Choose an element z′ ∈ Z(G_{d}^{0}) with image τ_{z}. Then {zz′, g} = 1 for all g ∈ R(G). That is, letting z̃′ ∈ G̃^{0} be an inverse image of z′, z̃z̃′ ∈ Z(G̃), and z̃ ∈ Z(G̃)G̃^{0}. □

We conclude this section with another application of Proposition 4.5. Suppose G̃ is an admissible cover of G, and G contains a relatively compact Cartan subgroup H. Let

(4.16)

W_{0} = {w ∈ W | w : 𝔥 → 𝔥 exponentiates to H̃^{0}}
W_{2} = {w ∈ W_{0} | wz̃ = z̃ for all z̃ ∈ Z(G̃) ∩ H̃^{0}}
W_{∗} = Norm_{W} (Φ_{I,c}).

We have obvious inclusions

(4.17)  W (Φ_{I,c}) ⊂ W (G, H) ⊂ W_{2} ⊂ W_{0} ⊂ W_{∗}.

By [11, Proposition 9.5(a)] we have

(4.18)(a)  R_{ad}(G) ≃ W_{∗}/W (Φ_{I,c})
(4.18)(b)  R(G) ≃ W (G, H)/W (Φ_{I,c})

Lemma 4.19 W_{2} = W (G, H).

Proof. Let

(4.20)  \begin{aligned}A & = Z_{0}(G) ∩ H^{0} ∩ G_{d}^{0} \\ & = Z_{0}(G) ∩ G_{d}^{0} \end{aligned}

since (cf. 3.10)(c) H ∩ G^{0} = H^{0}. Let w ∈ W_{0}. It is clear that

(4.21)  w ∈ W_{2} ⇔ wz̃ = z̃ for all z ∈ Ã.

If w ∈ W_{0} corresponds to g ∈ G_{ad} via (4.18)(a), we have

(4.22)  wz̃ = {g, p(z̃)}z̃.

In other words it is enough to show

(4.23)  {g, z} = 1 for all z ∈ A ⇔ g ∈ G.

Assume that G_{d}(ℂ) is simply connected. Then using the pairing on R_{ad}(G)× Z(G_{d}) (4.1)(c) we need to show

(4.24)  A^{⊥} = R(G).

It is easy to see that A = R(G)^{⊥} so we want to show

(4.25)  (R(G)^{⊥})^{⊥} = R(G).

This follows from the injectivity of γ_{ad}.

The argument if G_{d}(ℂ) is not simply connected is a slight generalization of this. Write Z(G_{d}^{0}) = Z(G_{sc})/B for some B. Then R(G) ⊂ B^{⊥} ⊂ R_{ad}(G). The pairing R_{ad}(G) × Z(G_{sc}) → ℤ/2ℤ descends to a pairing B^{⊥} × Z(G_{d}^{0}) → ℤ/2ℤ, and B^{⊥} ↪ Hom(Z(G_{d}^{0}), ℤ/2ℤ). The preceding argument now applies with B^{⊥} in place of R_{ad}(G).

By Lemma 5.34

(4.26)(a)  e^{ρ(Φ_{1}^{+})−ρ(Φ_{2}^{+})}(ta)Γ(H, Φ_{1}^{+})(h, g̃) = Γ(H, Φ_{2}^{+})(h, g̃).
□

4.1 Cartan Subgroups

We need some structural facts about Cartan subgroups. Fix an admissible cover G̃ of G, and a Cartan subgroup H of G.

Suppose τ is an automorphism of G_{d}^{0} and α ∈ Φ(G, H) is a long real root. We assume τ(H ∩ G_{d}^{0}) = H ∩ G_{d}^{0} and τ(α) = α. Let M_{α} be as in Section 3; we have τ(M_{α}) = M_{α}. Let T_{α} ≃ S^{1} be a τ-stable compact Cartan subgroup of M_{α}. Then for all t ∈ T_{α}, τ(t) = t or t^{−1}. Let m_{α} = α^{∨}(−1) ∈ Z(M_{α}).

Proposition 4.27 Suppose α is a long real root. Then

(4.28)(a)  \{τ, m_{α}\} = \begin{cases}1 & τ(t) = t for all t ∈ T_{α} \\ −1 & τ(t) = t^{−1} for all t ∈ T_{α} \end{cases}

For all h ∈ H,

(4.28)(b)  {h, m_{α}} = sgn(α(h)).

If β ∈ Φ_{r}(G, H) then

(4.28)(c)  {m_{α}, m_{β}} = (−1)^{⟨α,β^{∨}⟩}.

Proof. We can compute τ(m_{α}) by working in T_{α}. If τ acts trivially on T_{α} then the same holds for the action of τ̃ on T̃_{α}. On the other hand if τ(t) = t^{−1} then τ̃(t̃) = t̃^{−1} for all t̃∈ T̃_{α}. Since G̃ is admissible, m_{α} has order 4, and we conclude τ̃(m_{α}) = −m_{α}.

It is a standard fact (essentially a calculation in GL(2, ℝ)) that for all h ∈ H, t ∈ T_{α} hth^{−1} = t^{sgn(α(h)}. This proves (4.28)(b), and (4.28)(c) follows from this and the identity α(m_{β}) = (−1)^{⟨α,β^{∨}⟩}. □

Proposition 4.29 Assume G is oddly laced. Then

(4.30)  Z(H̃) = Z(G̃)H̃^{0}

We first prove

Lemma 4.31 In the setting of the Proposition we have

(4.32)  Z(H̃) ⊂ { \atop Z(G)}H̃^{0}.

Proof. It is enough to show if h ∈ H, {h, m_{α}} = 1 for all real roots α then h ∈ Z(G)H^{0}. (This is where the oddly laced condition appears: otherwise we only have this identity for the long roots.) By (4.28)(b) it is enough to show

(4.33)  α(h) > 0 for all α ∈ Φ_{r}(G, H) implies h ∈ Z(G)H^{0}

This is a straightforward calculation using roots and weights. Choose a basis of the root lattice of the form

α_{1}, …, α_{m}, α_{m+1}, …, α_{m+n}, α_{m+n+1}, … α_{m+n+2r}

where α_{i} is real for 1 ≤ i ≤ m, imaginary for m + 1 ≤ i ≤ m + n, and θα_{m+n+2i−1} = −α_{m+n+2i} for all 1 ≤ i ≤ r. It is a basic fact about lattices that such a basis exists. Let λ_{1}^{∨}, …, λ_{m+n+2r}^{∨} be the dual basis of coweights.

For each i choose x_{i} ∈ ℂ so that e^{x_{i}} = α_{i}(h). If i ≤ m we may assume x_{i} ∈ ℝ since α_{i}(h) > 0. We may also assume x_{m+n+2i} = \overline{x}_{m+n+2i−i} for all 1 ≤ i ≤ r. Let X = ∑_{i} x_{i}λ_{i}^{∨}, h_{1} = exp(X). It follows easily that X ∈ 𝔥, h_{1} ∈ H^{0}, and α(h) = α(h_{1}) for all roots α. Let z = hh_{1}^{−1}; this is contained in G, and also Z(G(ℂ)), since α(z) = 1 for all roots α. Therefore z ∈ Z(G), and therefore h = zh_{1} ∈ Z(G)H^{0}. □

Proof of the Proposition. The statement is equivalent to

(4.34)  Z_{0}(H) = Z_{0}(G)H^{0}.

It is enough to show Z_{0}(H) ⊂ Z_{0}(G)H^{0} (the reverse inclusion is obvious). We first prove this for a maximally split Cartan subgroup H_{s} of G. Suppose h ∈ Z_{0}(H). By the Lemma h = zy with z ∈ Z(G), y ∈ H^{0}. Then y ∈ Z_{0}(H), so z = hy^{−1} ∈ Z_{0}(H). It is enough to show z ∈ Z_{0}(G). We have {z, g} = 1 for all g ∈ H, and also for g ∈ G^{0} by (4.1)(g) Since H is maximally split G = HG^{0}, so z ∈ Z_{0}(G).

The general case will be proved after Lemma 4.45, using Cayley transforms. □

4.2 Cayley Transforms

Suppose H is a Cartan subgroup and α is a real or non-compact imaginary root. We define the Cayley transform H_{α} of H of H with respect to α as in [10]. Then H_{α} has a non-compact imaginary or real root β, and the Cayley transform of H_{α} with respect to β is H.

In order to emphasize the symmetry of the situation we change notation and let H_{α} = T_{α}A_{α} be a Cartan subgroup with a real root α. Then we let H_{β} = T_{β}A_{β} be its Cayley transform, with non-compact imaginary root β.

We have elements Z_{α} ∈ 𝔭 and Z_{β} ∈ 𝔨, and let B_{α} = exp(ℝZ_{α}) ≃ ℝ^{+} ⊂ H_{α}, and B_{β} = exp(ℝZ_{β}) ≃ S^{1} ⊂ H_{β}.

It follows easily from [10, 8.3.4 and 8.3.13] that we have

(4.35)  \begin{aligned}(H_{α} ∩ H_{β})B_{α} & ⊂ H_{α} index 1 or 2 \\ (H_{α} ∩ H_{β})B_{β} & = H_{β} \end{aligned}

The root α takes positive real values on (H_{α} ∩ H_{β})B_{α}. We say α is type I if the first inclusion in (4.35) is an equality. Otherwise α is of type II, in which case H_{α} is generated by the left hand side and an element t satisfying α(t) = −1.

We say β is type I if s_{β} ∉ W (G, H_{β}), and type II otherwise. Then α, β are both of type I or both of type II.

Definition 4.36 Suppose χ_{α} is a character of H_{α}, and χ_{β} is a character of H_{β}. We say χ_{α} is a Cayley transform of χ_{β}, and vice-versa, if

(4.37)(a)  χ_{α}(h) = χ_{β}(h) (h ∈ H_{α} ∩ H_{β})
(4.37)(b)  Ad^{∗}(g)(dχ_{α}) = dχ_{β} for some g ∈ Aut(𝔤).

To be precise, in (b) the element g satisfies Ad(g)(𝔥_{β}(ℂ)) = 𝔥_{α}(ℂ), and the identity is between elements of 𝔥_{β}^{∗}.

Here is a convenient alternative characterization of Cayley transforms.

Lemma 4.38 In the setting of the Definition, χ_{α} is a Cayley transform of χ_{β}, and vice versa, if and only if (4.37)(a) holds, and

(4.39)  ⟨dχ_{α}, α^{∨}⟩ = ±⟨dχ_{β}, β^{∨}⟩.

Lemma 4.40

(1) Fix a character χ of H_{α}. There is a Cayley transform χ_{β} of χ if and only if

(4.41)  ⟨dχ, α^{∨}⟩ ∈ ℤ

and

(4.42)  χ(m_{α}) = (−1)^{⟨dχ,α^{∨}⟩}.

Assume these hold. Then there are two choices of Cayley transform, given by (4.37)(a) and

(4.43)  χ_{β}^{±}(exp(xZ_{β})) = e^{±i⟨dχ,α^{∨}⟩x}

The characters χ_{β}^{±} are conjugate by W (G, H_{β}) if and only if α is type I.

(2) Fix a character χ of H_{β}. Define χ_{α} restricted to (H_{α}∩H_{β})B_{α} by (4.37)(a), and

(4.44)  χ_{α}(exp(xZ_{α})) = e^{⟨dχ,β^{∨}⟩x}.

If β is of type I this defines a character of H_{α}. If β is of type II define χ_{α}^{±} to be the two extensions of χ_{α} to H_{α}.

Sketch of proof. This is elementary from the identities [10, 8.3.4 and 8.3.13]. See [10][Lemma 8.3.7 and 8.3.15], where the setting is a regular character and the construction differs from this one by a ρ-shift. For this reason proof of the Lemma is in fact much easier than those in [10]. □

We now consider Cayley transforms for G̃. The analogue of (4.35) is:

Lemma 4.45 Assume G is oddly laced and G̃ is an admissible cover of G. Then

(4.46)(a)  (Z(H̃_{α}) ∩ Z(H̃_{β}))B̃_{β} = Z(H̃_{β})
(4.46)(b)  (Z(H̃_{α}) ∩ Z(H̃_{β}))B̃_{α} = Z(H̃_{α})

Proof. We first prove (a). We have to show

(4.47)  (Z_{0}(H_{α}) ∩ Z_{0}(H_{β}))B_{β} = Z_{0}(H_{β}).

Since B_{β} is connected, B_{β} = Z_{0}(B_{β}) ⊂ Z_{0}(H_{β}), and the inclusion ⊂ is clear. For the opposite inclusion suppose g ∈ Z_{0}(H_{β}). By (4.35) write g = hb with h ∈ H_{α} ∩ H_{β}, b ∈ B_{β}. Then h = gb^{−1} ∈ Z_{0}(H_{β}) since both g and b are in Z_{0}(H_{β}). It is enough to show h ∈ Z_{0}(H_{α}).

Note that {h, x} = 1 for x ∈ H_{α} ∩ H_{β} (since h ∈ Z_{0}(H_{β})). Since B_{α} is connected B_{α} = Z_{0}(B_{α}) ⊂ Z_{0}(H_{α}), and therefore {h, x} = 1 for x ∈ B_{α}.

Therefore {h, x} = 1 for x ∈ (H_{α} ∩ H_{β})B_{α}. If α is type I the right hand side is H_{α} and we are done. Otherwise choose t satisfying α(t) = −1, so that H_{α} is generated by (H_{α} ∩ H_{β})B_{α} and t.

It is enough to show {h, t} = 1. If this is not the case replace h with hm_{α} and b with bm_{α} (note that m_{α} ∈ H_{α} ∩ H_{β} and B_{β}). By (4.28)(b) {m_{α}, t} = −1, so {hm_{α}, t} = 1.

For (b) the inclusion ⊂ is immediate since B_{α} is connected. On the other hand suppose g ∈ H_{α} but g ∉ (H_{α}∩H_{β})B_{α}. Then α(g) < 0, and by (4.28)(b) {m_{α}, g} = −1, so g ∉ Z_{0}(H_{α}). This proves the reverse inclusion. □

We may now complete the proof of Proposition 4.29. Proof. We have already shown this for the most split Cartan subgroup. By repeated use of the Cayley transform it is enough to show that in the previous setting

(4.48)  Z(H̃_{α}) = Z(G̃)H̃_{α}^{0} ⇒ Z(H̃_{β}) = Z(G̃)H̃_{β}^{0}

By the Lemma we have

(4.49)  \begin{aligned}Z(H̃_{β}) & = (Z(H̃_{α}) ∩ Z(H̃_{β}))B̃_{β} \\ & = (Z(G̃)H̃_{α}^{0} ∩ Z(H̃_{β}))B̃_{β} \\ & = Z(G̃)(H̃_{α}^{0} ∩ Z(H̃_{β}))B̃_{β} \\ & = Z(G̃)H̃_{β}^{0} \end{aligned}

For the last equality we have used that (H_{α}^{0} ∩ H_{β})B_{β} ⊂ H_{β}^{0}, which implies (H̃_{α}^{0} ∩ Z(H̃_{β}))B̃_{β} ⊂ H̃_{β}^{0}. □

If G̃ is an admissible cover of a oddly laced group G, we define Cayley transforms for genuine characters as in Definition 4.36, using Z(H̃_{α}), Z(H̃_{β}) in place of H_{α}, H_{β}, and Lemma 4.45 in place of (4.35). By 4.29 we can restate the definition of Cayley transform in this setting.

Lemma 4.50 Assume G is oddly laced and G̃ is an admissible cover of G. In the setting of Lemma 4.45, χ̃_{α} and χ̃_{β} are each others Cayley transforms if and only if

(4.51)(a)  χ̃_{α}(z) = χ̃_{β}(z) for all z ∈ Z(G̃)
(4.51)(b)  Ad^{∗}(g)(dχ_{α}) = dχ_{β} for some g ∈ Aut(𝔤).

Lemma 4.52 Assume G is oddly laced and G̃ is an admissible cover of G.

(1) Suppose χ̃_{α} is a genuine character of Z(H̃_{α}). A Cayley transform χ̃_{β} of χ̃_{α} exists if and only if

(4.53)  ⟨dχ̃_{α}, α^{∨}⟩ ∈ ℤ + \frac{1}{2}.

Assume this holds. Define χ̃_{β} by (4.37)(a) and

(4.54)(a)  χ̃_{β}(ẽxp(xZ_{β})) = e^{±⟨dχ̃_{α},α^{∨}⟩ix}.

If α is type II either sign is allowed. If α is type I let m̃_{β} = ẽxp(πZ_{β}) ∈ Z(H̃_{α}) ∩ Z(H̃_{β}). Note that χ̃_{β}(m_{β}) = ±i. Then there is a unique choice of sign, satisfying

(4.54)(b)  e^{±iπ⟨dχ̃_{α},α^{∨}⟩} = χ̃(m_{β}).

(2) Suppose χ̃_{β} is a genuine character of Z(H̃_{β}). Then there are two choices of Cayley transform χ_{α} of χ_{β}, given by (4.37)(a) and

(4.54)(c)  χ̃_{α}(ẽxp(xZ_{α})) = e^{⟨dχ̃_{β},β^{∨}⟩x.}

Proof. For (1) it is immediate from the definition that χ̃_{β} is defined by (4.37)(a) and (4.54)(a). If α is type II then Z(H̃_{α}) ∩ Z(H̃_{β}) ∩ B̃_{β} = 1, and there is no further condition. If α is type I then m̃_{β} ∈ Z(H̃_{α}) ∩ Z(H̃_{β}) ∩ B̃_{β}, which gives condition (4.54)(b). Case (2) is similar, and easier. □

Remark 4.55 It is easy to see that in the cases when there are two Cayley transform χ̃_{±} in Lemma 4.52, χ̃_{±} are conjugate by G̃. In [10] and elsewhere Cayley transforms are used to construct “pseudocharacters”, in which one is concerned only about G-conjugacy classes. In this sense Cayley transforms for G̃ are single-valued.

5 Characters of Cartan Subgroups.

Let (G̃, G, \overline{G}) be an admissible triple. The definition of transfer factors involves certain factors defined for each Cartan. In this section we define these and establish their properties. We apply this to the definition of transfer factors in Section 6.

Let H be a Cartan subgroup of G and let Φ = Φ(G, H). Define the character ζ_{cx} = ζ_{cx}(G, H) of Γ(\overline{H}) as in [2, Section 15]. That is ζ_{cx} = ∏_{S} α where S is a set of complex roots such that the set of all complex roots is {±α, ±σα | α ∈ S}. This is independent of the choice of S, factors to G_{ad}, and satisfies

(5.1)(a)  ζ_{cx}(g) = ζ_{cx}(wg) (w ∈ W (\overline{G}, \overline{H}), g ∈ Γ(\overline{H}))
(5.1)(b)  ζ_{cx}(h) = e^{ρ_{cx}}(h) = e^{ρ−ρ_{r}}(h) (h ∈ Γ(\overline{H}))

for any real set of positive roots (Section 2).

Suppose Φ^{+} is a set of positive roots and χ̃ is a genuine character of Z(G̃). Since G_{d}(ℂ) is acceptable ρ exponentiates to a character of (H ∩ G_{d})^{0}, and χ̃^{2}e^{ρ} is defined on (H ∩ G_{d})^{0}. By Definition 3.14(3c) this character factors to \overline{H}_{d}^{0}, where \overline{H}_{d} = \overline{H} ∩ \overline{G}_{d}.

Definition 5.2 Let H be a Cartan subgroup of G. Choose

1. a real set of positive roots Φ^{+} (see Section 2),

2. a genuine character χ̃ of Z(H̃),

3. a character χ of \overline{H}.

Assume these satisfy:

(5.3)(a)  χ(h) = (χ̃^{2}e^{ρ})(h), (h ∈ \overline{H}_{d}^{0})
(5.3)(b)  χ(h) = ζ_{cx}(h) (h ∈ Γ_{R}(\overline{H})).

Fix (Φ^{+}, χ̃, χ) satisfying these conditions. Suppose g̃ ∈ H̃, h ∈ \overline{H} satisfy p(g̃) = ϕ(h). Then we define

(5.4)  Γ(χ̃, χ)(g̃, h) = \frac{χ̃(g̃)}{χ(h)}

Let

(5.5)(a)  S(H, Φ^{+}) = {(χ̃, χ)} satisfying (5.3)(a) and (b),
(5.5)(b)  T (H, Φ^{+}) = {Γ(χ̃, χ) | (χ̃, χ) ∈ S(H, Φ^{+})},
(5.5)(c)  S(H) = ∪_{Φ^{+}}S(H, Φ^{+}),
(5.5)(d)  T (H) = ∪_{Φ^{+}}T (H, Φ^{+})

(the unions are over real sets of positive roots).

We begin with some elementary properties.

Lemma 5.6 Fix real Φ^{+} and (χ̃, χ) ∈ S(H, Φ^{+}).

(1) Suppose μ is a character of Z_{0}(H). Then (χ̃μ, χ(μ ∘ ϕ)) ∈ S(H, Φ^{+}) and

(5.7)  Γ(χ̃μ, χ(μ ∘ ϕ)) = Γ(χ̃, χ).

(2) Suppose Φ_{1}^{+} is another real set of positive roots. There exists a character τ = τ(Φ^{+}, Φ_{1}^{+}) such that (χ̃, χτ) ∈ S(H, Φ_{1}^{+}), and

(5.8)  Γ(χ̃, χ)(g̃, h) = Γ(χ̃, χτ)(g̃, h)

for all h ∈ Γ(\overline{H})Z(\overline{G}), g̃ ∈ H̃ satisfying p(g̃) = ϕ(h).

Proof. The first part is straightforward. For the second define τ(h) = e^{ρ(Φ^{+})−ρ(Φ_{1}^{+})}(h) for h ∈ \overline{H}_{d}^{0}, and τ(h) = 1 for h ∈ Γ(\overline{H})Z(\overline{G}). It is easy to see that the fact that both Φ^{+} and Φ_{1}^{+} are real implies e^{ρ(Φ^{+})−ρ(Φ_{1}^{+})}(h) = 1 for all h ∈ Γ(\overline{H})Z(\overline{G}) ∩ \overline{H}_{d}^{0}, so τ is well defined. It follows easily that it has the desired properties. □

If ψ is a one-dimensional representation of \overline{G} define

(5.9)  ψ · Γ(χ̃, χ) = Γ(χ̃, χψ|\overline{H})

This defines an action of the characters of \overline{G} on T (H, Φ^{+}).

Lemma 5.10 The space T (H, Φ^{+}) is non-empty, and the action of the group of characters of \overline{G} on T (H, Φ^{+}) is transitive. If H is maximally split it is simply transitive.

Proof. Fix any genuine character χ̃ of Z(H̃). For existence it is enough to show

(5.11)  ζ_{cx}(h) = e^{ρ}χ̃^{2}(h) h ∈ Γ_{R}(\overline{H}) ∩ \overline{H}_{d}^{0}.

Note that p^{−1}(\overline{H}_{d}^{0}) ⊂ p^{−1}(\overline{H}^{0}) = H^{0}C ⊂ H^{0}Z_{0}(G) = Z_{0}(H) by 3.14(3b)) and Proposition 4.29. Pulling back to G it is therefore enough to show

(5.12)  ζ_{cx}(h) = e^{ρ}χ̃^{2}(h) h ∈ Γ_{R}(H) ∩ Z_{0}(H).

Any element h ∈ Γ_{R}(H) may be written h = ∏_{i=1}^{n} m_{α_{i}} where {α_{1}, …, α_{n}} are simple roots for Φ_{R}^{+}. If h ∈ Z_{0}(H) then as in [2, Section 14] χ̃^{2}(h) = (−1)^{n}. Therefore

(5.13)  χ̃^{2}(h) = e^{ρ_{r}}(h) h ∈ Γ_{R}(H) ∩ Z_{0}(H).

We conclude that

(5.14)  \begin{aligned}χ̃^{2}e^{ρ}(h) & = e^{ρ_{r}}(h)e^{ρ_{r}+ρ_{cz}+ρ_{i}}(h) \\ & = e^{ρ_{cx}}(h) \end{aligned}

since e^{2ρ_{r}}(h) = e^{ρ_{i}}(h) = 1. Then (5.12) follows from (5.1)(b)

Suppose (χ̃_{1}, χ_{1}), (χ̃_{2}, χ_{2}) ∈ S(H, Φ^{+}). Define a character of Γ(\overline{H})Z(\overline{G}) by

(5.15)  ψ_{0}(h) = \frac{χ_{1}}{χ_{2}} (h)\frac{χ̃_{1}}{χ̃_{2}} (ϕ(h)).

If h ∈ Γ_{R}(\overline{H}) then by Lemmma 3.11(3) ϕ(h) = 1, and by (5.3)(b) χ_{1}(h) = χ_{2}(h), so ψ_{0}(h) = 1. If h ∈ Γ(\overline{H})Z(\overline{G}) ∩ \overline{H}_{d}^{0} then

(5.16)  ψ_{0}(h) = \frac{(χ̃_{1}^{2}e^{ρ})(h)}{(χ̃_{2}^{2}e^{ρ})(h)} \frac{χ̃_{2}(ϕ(h))}{χ̃_{1}(ϕ(h))} = 1

Therefore ψ_{0} is trivial on Γ_{R}(\overline{H})(Γ(\overline{H})Z(\overline{G})∩\overline{H}_{d}^{0}). By (3.10)(b) Γ_{R}(\overline{H})\overline{H}_{d}^{0} = \overline{H}_{d}, and Γ_{R}(\overline{H})(Γ(\overline{H})Z(\overline{G}) ∩ \overline{H}_{d}^{0}) = Γ(\overline{H})Z(\overline{G}) ∩ \overline{H}_{d}. Therefore \overline{H}/\overline{H}_{d} = Γ(\overline{H})Z(\overline{G})\overline{H}_{d}/\overline{H}_{d} ≃ Γ(\overline{H})Z(\overline{G})/Γ(\overline{H})Z(\overline{G}) ∩ \overline{H}_{d}, so ψ_{0} defines a character of \overline{H}/\overline{H}_{d}. We have a surjection

(5.17)  Hom(\overline{G}/\overline{G}_{d}, ℂ) → Hom(\overline{H}/\overline{H}_{d}, ℂ)

dual to the injection \overline{H}/\overline{H}_{d} ↪ \overline{G}/\overline{G}_{d}, which is an isomorphism if H is maximally split. If we choose any preimage ψ of ψ_{0} in (5.17) then it is easy to see that Γ(χ̃_{2}, χ_{2}) = Γ(χ̃_{1}, χ_{1})ψ. It follows that characters of \overline{G} act transitively on T (H, Φ^{+}), and this action is simply transitive if H is maximally split. □

We now need to choose elements of T (H, Φ^{+}) consistently for all H. We do this by reducing to the most split Cartan subgroup.

Lemma 5.18 Suppose (χ̃, χ) ∈ S(H). Then

(5.19)  χ(h) = χ(wh) (w ∈ W (\overline{G}, \overline{H}), h ∈ Γ(\overline{H})Z(\overline{G}))

We defer the proof to the end of this section.

Fix a maximally split Cartan subgroup H_{s} = T_{s}A_{s} of G and (χ̃_{s}, χ_{s}) ∈ S(H_{s}). Let \overline{H}_{s} = \overline{T}_{x}\overline{A}_{s} be the corresponding Cartan subgroup of \overline{G}. If \overline{H} = \overline{T} \overline{A} is any Cartan subgroup of \overline{G} there exists x ∈ \overline{G} so that x\overline{A}x^{−1} ⊂ \overline{A}_{s}, and therefore xΓ(\overline{H})x^{−1} ⊂ Γ(\overline{H}_{s}). Suppose h ∈ Γ(\overline{H})Z(\overline{G}), g̃ ∈ H̃ satisfy p(g̃) = ϕ(h). By Lemmas 3.11(5) and 3.12 ϕ(h) ∈ Z_{0}(G), so g̃ ∈ Z(G̃) ⊂ Z(H̃_{s}). Define

(5.20)  Γ(χ̃_{s}, χ_{s})(h, g̃) = \frac{χ̃_{s}(g̃)}{χ_{s}(xhx^{−1})}.

By the Lemma this is independent of the choice of x.

Proposition 5.21 Fix a maximally split Cartan subgroup H_{s} and (χ̃_{s}, χ_{s}) ∈ S(H_{s}). Suppose H is a Cartan subgroup, χ̃ ∈ X_{g}(H̃) and Φ^{+} is a real set of positive roots for H. Then there exists unique χ ∈ X(\overline{H}) such that (χ̃, χ) ∈ S(H, Φ^{+}) and

(5.22)  Γ(χ̃_{s}, χ_{s})(g̃, h) = Γ(χ̃, χ)(g̃, h)

for all h ∈ Γ(\overline{H})Z(\overline{G}), g̃ ∈ H̃ satisfying p(g̃) = ϕ(h).

We say (χ̃, χ) ∈ S(H, Φ^{+}) are compatible with (χ̃_{s}, χ_{s}) if they satisfy the conditions of the Proposition, in particular

(5.23)(a)  \frac{χ̃(g̃)}{χ(h)} = \frac{χ̃_{s}(g̃)}{χ_{s}(xhx^{−1})}

for all h, g̃ as in the Proposition, and x as in the preceding discussion. Using the fact that ϕ(h) = g̃ and that χ̃_{s} and χ̃ are both genuine this condition can be expressed

(5.23)(b)  χ(h) = (χ̃/χ̃_{s})(ϕ(h))χ_{s}(xhx^{−1}).

Using Cayley transforms (Section 4.2) we will reduce to the following Lemma. Suppose H_{α} is a Cartan subgroup, with real root α, and H_{β} is the Cayley transform of H_{α}, with non-compact imaginary root β.

Lemma 5.24 Suppose Φ_{α}^{+} is a real set of positive roots for H_{α} and (χ̃_{α}, χ_{α}) ∈ S(H_{α}, Φ_{α}^{+}). Suppose Φ_{β}^{+} is a real set of positive roots for H_{β}, and χ̃_{β} is a genuine character of Z(H̃_{β}). Then there exists unique χ_{β} such that (χ̃_{β}, χ_{β}) ∈ X(H_{β}, Φ_{β}^{+}) and

(5.25)  Γ(χ̃_{α}, χ_{α})(g̃, h) = Γ(χ̃_{β}, χ_{β})(g̃, h)

for all h ∈ Γ(\overline{H}_{β})Z(\overline{G}), g̃ ∈ H̃_{β} satisfying p(g̃) = ϕ(h).

In addition assume c_{α}(Φ_{α}^{+}) = Φ_{β}^{+}. Then (5.25) holds for all h ∈ \overline{H}_{α} ∩\overline{H}_{β}.

Proof. To avoid runaway notation let \overline{H} = \overline{H}_{β} for a moment. The character χ_{β} is determined on Γ(\overline{H})Z(\overline{G}) by (5.25) and on \overline{H}_{d}^{0} by (5.3)(a). Since \overline{H}^{0} ⊂ Z(\overline{G})\overline{H}_{d}^{0}, by (3.10)(b) \overline{H} = Γ(\overline{H})Z(\overline{G})\overline{H}_{d}^{0}, and uniqueness is immediate.

For existence, by Lemma 5.6 it is enough to prove this for a single choice of Φ_{α}^{+} and Φ_{β}^{+}. It is not hard to see we can choose Φ_{α}^{+} real so that Φ_{β}^{+} = c_{α}(Φ_{α}^{+}) is also real (this implies α is simple for Φ_{r}^{+}).

By 5.6(1) again we may choose χ̃ arbitrarily. We have ⟨dχ̃, α^{∨}⟩ ∈ ℤ + \frac{1}{2} and ⟨dχ, α^{∨}⟩ ∈ ℤ. By Lemma 4.52 choose χ̃_{β} to be a Cayley transform of χ̃_{α}. Choose g ∈ Aut(𝔤) so that Ad^{∗}(g)(dχ̃_{α}) = dχ̃_{β}. By Lemma 4.40 let χ_{β} be the (unique) Cayley transform of χ_{α} satisfying Ad^{∗}(g)(dχ_{α}) = dχ_{β}.

It is enough to show (χ̃_{β}, χ_{β}) ∈ T (H_{β}, Φ_{β}^{+}), for it is then obvious from the definition of Cayley transforms that (5.25) holds.

We have

(5.26)  \begin{aligned}dχ_{β} & = Ad^{∗}(g)(dχ_{α}) \\ & = Ad^{∗}(g)(2dχ̃_{α} + ρ(Φ_{α}^{+})) \\ & = 2dχ̃_{β} + ρ(Φ_{β}^{+}) \end{aligned}

by the choice of Φ_{α}^{+} and Φ_{β}^{+}. This verifies (5.3)(a).

We now verify (5.3)(b). Suppose γ is a real root of \overline{H}_{β}. Then m_{γ} ∈ \overline{H}_{α} ∩ \overline{H}_{β}, and χ_{β}(m_{γ}) = χ_{α}(m_{γ}) = ζ_{cx}(\overline{G}, \overline{H}_{α})(m_{γ}), by (5.3)(b) for (χ̃_{α}, χ_{α}). We want to show this equals ζ_{cx}(\overline{G}, \overline{H}_{β})(m_{γ}), i.e.

(5.27)  ζ_{cx}(\overline{G}, \overline{H}_{α})(m_{γ}) = ζ_{cx}(\overline{G}, \overline{H}_{β})(m_{γ}).

If the simple factor containing α, β is of type G_{2} an explicit calculation shows that both sides are 1, so assume G is simply laced.

The root γ̃ of \overline{H}_{α} corresponding to γ is also real, and ⟨γ̃, α^{∨}⟩ = 0. Suppose δ is a complex root of \overline{H}_{α}. The corresponding root of \overline{H}_{β} is also complex unless s_{α}δ = −σδ (s_{α}δ = σδ does not occur), in which case it is imaginary. We claim sγ̃δ ≠ ±σδ, i.e. sγ̃δ ≠ ±s_{α}δ̃. Since δ ≠ ±α the only possibility is with the minus sign, which would give

(5.28)  2δ = ⟨δ, γ̃^{∨}⟩γ̃ + ⟨δ, α^{∨}⟩α̃.

By considering rank two root systems it is easy to see this cannot happen in the simply laced case.

Therefore both δ and sγ̃(δ) contribute to ζ_{cx}(\overline{G}, \overline{H}_{α})(m_{γ}), and their total contribution is δ(m_{γ})(sγ̃δ)(m_{γ}) = δ(m_{γ})δ(m_{γ}) = 1.

A similar argument holds for complex roots δ of \overline{H}_{β} for which s_{β}δ = σδ. It follows that the terms in ζ_{cx}(\overline{G}, \overline{H}_{α}) and ζ_{cx}(\overline{G}, \overline{H}_{β}) are the same, with the exception of those just discussed, which are 1. □

For later use we note a consequence of the last part of the proof.

Lemma 5.29 Suppose H_{1} = T_{1}A_{1} and H_{2} = T_{2}A_{2} are Cartan subgroups with A_{1} ⊂ A_{2}. Suppose α ∈ Φ_{r}(G, H_{1}). (1)

ζ_{cx}(G, H_{1})(m_{α}) = ζ_{cx}(G, H_{2})(m_{α}).

(2) Let M_{1} = Cent_{G}(A_{1}) and suppose α ∈ Φ_{r}(M_{1}, H_{2}). Then

ζ_{cx}(G, H_{2})(m_{α}) = ζ_{cx}(M_{1}, H_{2})(m_{α}).

Proof. The first equality follows from a repeated application of (5.27). The second is similar. Choose a set S of complex roots such that

(5.30) {β ∈ Φ_{cx}(G, H_{1})\Φ_{cx}(M, H_{1}) | ⟨β, α^{∨}⟩ ≠ 0} = {±β, ±θβ | β ∈ S}.

If β ∈ S then s_{α}β is also complex, is not contained in Φ(M_{2}, H_{2}), and is not equal to ±β, ±σβ Then S is can be written as a union over pairs {β, s_{α}β} and the result follows as in the proof of (5.27). □

Proof of the Proposition. This is now straightforward. If H is maximally split there is nothing to prove. If H is obtained from H_{s} by a series of Cayley transforms we conclude the result by a repeated application of Lemma 5.24. It is easy to check that if the conditions of the Proposition hold for a Cartan subgroup H, they hold for every G-conjugate of H. Up to conjugacy every Cartan subgroup is conjugate to one obtained by a series of Cayley transforms from H_{s}, and the result follows. □

Definition 5.31 Fix (χ̃_{s}, χ_{s}) ∈ S(H_{s}). Suppose H is a Cartan subgroup and Φ^{+} is a real set of positive roots. Choose χ̃ ∈ X_{g}(H̃). Let χ ∈ X(\overline{H}) be the character given by Proposition 5.21. Define

(5.32)  Γ(H, Φ^{+}) = Γ(χ̃, χ).

Lemma 5.33 In the setting of the Definition Γ(H, Φ^{+}) is independent of the choice of χ̃.

This follows easily from Lemma 5.6.

We note some properties of Γ(H, Φ^{+}) which will be useful later. First of all its dependence on Φ^{+} follows easily from Lemma 5.6(2) and its proof.

Lemma 5.34 Suppose Φ_{1}^{+}, Φ_{2}^{+} are real sets of positive roots for H. Suppose g̃ ∈ H̃, h ∈ \overline{H} satisfy p(g̃) = ϕ(h), and write h = γh_{0} with γ ∈ Γ(\overline{H}) and h_{0} ∈ \overline{H}^{0}. Then

(5.35)  Γ(H, Φ_{1}^{+})(h, g̃) = Γ(H, Φ_{2}^{+})(h, g̃)e^{ρ(Φ_{2}^{+})−ρ(Φ_{1}^{+})}(h_{0}).

We obtain a character of the center of 𝔤:

Lemma 5.36 Given χ̃ ∈ X_{g}(H̃), let χ ∈ X(\overline{H}) be the character given by Proposition 5.21 and set

(5.37)  μ = dχ − 2dχ̃ − ρ(Φ^{+})

Then μ|_{𝔥_{d}} = 0, and the restriction of μ to the center of 𝔤 is independent of H, Φ^{+} and χ̃.

Proof. The fact that μ|_{𝔥_{d}} = 0 is (5.3)(a). On the other hand ρ(Φ^{+})|_{𝔷} = ρ(Φ_{s}^{+})|_{𝔷} = −, and by (5.22) 2dχ̃ − χ and 2dχ̃_{s} − χ_{s} have the same restriction to 𝔷. □

We conclude this section with the proof of Lemma 5.18. Proof. We need to show χ(g(wg)^{−1}) = 1 for g ∈ Γ(\overline{H}). This is obvious if w ∈ W (Φ_{i}). If w ∈ Φ_{r} then g(wg)^{−1} ∈ Γ_{R}(\overline{H}). Then χ(g(wg)^{−1}) = ζ_{cx}(g(wg)^{−1}) = 1 by (5.3)(b) and (5.1)(a).

By [11, Proposition 3.12] W (\overline{G}(ℂ), \overline{H}(ℂ))^{θ}, which contains W (\overline{G}, \overline{H}), is generated by W (Φ_{i}), W (Φ_{r}) and elements of the form s_{α}s_{θα} where ⟨α, θα^{∨}⟩ = 0. So it is enough to show χ(g(wg)^{−1}) = 1 for w of this form.

Write g = \overline{exp}πiX with X ∈ 𝔞. Since g^{2} = 1, α(X) ∈ ℤ. Then

(5.38)  g(wg)^{−1} = e^{πiα(X)(α^{∨}+θα^{∨})} ∈ Γ(\overline{H}) ∩ \overline{H}_{d}^{0}

Let ρ = ρ(Φ^{+}). Then

(5.39)  \begin{aligned}χ((wg)g^{−1}) & = (χ̃^{2}e^{ρ})(g(wg)^{−1}) (by (5.3)(a)) \\ & = (−1)^{α(X)⟨2dχ̃,α^{∨}+θα^{∨}⟩}e^{ρ−w^{−1}ρ}(g) \end{aligned}

By [2, Corollary 8.5] ⟨dχ̃, α^{∨} + θα^{∨}⟩ ∈ ℤ. The fact that Φ^{+} is real implies ρ − w^{−1}ρ is a sum of imaginary roots and terms β − θβ with β complex. Therefore e^{w^{−1}ρ−ρ}(g) = 1. □

Lemma 5.40 Let Φ^{+} be a σ-stable choice of positive roots. Let χ̃_{i} ∈ X_{g}(H̃), i = 1, 2, and χ_{i} = χ(χ̃_{i}, Φ^{+}). Let h ∈ \overline{H}, g̃ ∈ Z(H̃) with p(g̃) = ϕ(h). Then

\frac{χ̃_{1}(g̃)}{χ_{1}(h)} = \frac{χ̃_{2}(g̃)}{χ_{2}(h)}.

6 Transfer Factors

In this section we define transfer factors for an admissible triple (G̃, G, \overline{G}). There are some choices involved; we make these choices consistently for different Cartan subgroups by fixing (χ̃_{s}, χ_{s}) ∈ S(H̃_{s}) where H_{s} is a maximally split Cartan subgroup of G.

Suppose 𝔥 = 𝔱 ⊕ 𝔞 is a θ-stable Cartan subalgebra, with corresponding Cartan subgroups H̃, H and \overline{H} of G̃, G and \overline{G}, respectively. Let H̃′, H′, \overline{H}′ be the regular elements of of these groups.

Let Φ^{+} be a real set of positive roots of H in G. For h ∈ H̃′, H′ or \overline{H}′ define

(6.1)(a)  Δ^{0}(Φ^{+}, h) = ∏_{α∈Φ^{+}}(1 − e^{−α}(h))
(6.1)(b)  ϵ_{r}(Φ^{+}, h) = sign ∏_{α∈Φ_{r}^{+}}(1 − e^{−α}(h))
(6.1)(c)  Δ^{1}(Φ^{+}, h) = ϵ_{r}(Φ^{+}, h)Δ^{0}(Φ^{+}, h)
(6.1)(d)  |Δ(h)| = |e^{ρ}(h)Δ^{0}(Φ^{+}, h)|

In the last definition we may write |Δ_{G}|, |ΔG̃| or |Δ\overline{G}| to indicate the group involved.

Here is the definition of transfer factors.

Definition 6.2 Fix a maximally split Cartan subgroup H_{s} of G, and (χ̃_{s}, χ_{s}) ∈ S(H_{s}). Let H be a Cartan subgroup of G. Choose a real set positive roots Φ^{+} for H. Recall Γ(H, Φ^{+})(h, g̃) is given by Definition 5.31. Suppose h ∈ \overline{H}′, g̃ ∈ H̃′ satisfy p(g̃) = ϕ(h). Define

(6.3)  ΔG̃ ; \overline{G}(h, g̃) = \frac{Δ^{1}(Φ^{+}, h)}{Δ^{1}(Φ^{+}, g̃)}Γ(H, Φ^{+})(h, g̃).

Recall that by (5.32)

(6.4)  ΔG̃ ; \overline{G}(h, g̃) = \frac{Δ^{1}(Φ^{+}, h)}{Δ^{1}(Φ^{+}, g̃)} \frac{χ̃(g̃)}{χ(h)}

for any (χ̃, χ) ∈ S(H, Φ^{+}), compatible with (χ̃_{s}, χ_{s}) in the sense of Proposition 5.21.

It is clear that ΔG̃ ; \overline{G}(h, g̃) is a genuine function of g̃.

Proposition 6.5 ΔG̃ ; \overline{G} is independent of the choice of Φ^{+}, and depends only on the choice of (χ̃_{s}, χ_{s}) ∈ S(H_{s}).

Proof. We have to show that \frac{Δ^{1}(Φ^{+},h)}{Δ^{1}(Φ^{+},g̃)} and Γ(H, Φ^{+}) satisfy inverse transformation properties with respect to Φ^{+}.

Fix g̃ ∈ H̃′ and h ∈ \overline{H}′ such that ϕ(h) = p(g̃). By (3.10)(b) write h = γh_{0} with γ ∈ Γ(\overline{H}) and h_{0} ∈ \overline{H}^{0}. Suppose Φ_{1}^{+}, Φ_{2}^{+} are two choices of real positive roots. By Lemma 5.34

(6.6)(a)  Γ(H, Φ_{1}^{+})(h, g̃) = Γ(H, Φ_{2}^{+})(h, g̃)e^{ρ(Φ_{2}^{+})−ρ(Φ_{1}^{+})}(h_{0}).

By the definition of ΔG̃ ; \overline{G} it is enough to show

(6.6)(b)  \frac{Δ^{1}(Φ_{1}^{+}, h)}{Δ^{1}(Φ_{1}^{+}, g̃)} = \frac{Δ^{1}(Φ_{2}^{+}, h)}{Δ^{1}(Φ_{2}^{+}, g̃)}e^{ρ(Φ_{1}^{+})−ρ(Φ_{2}^{+})}(h_{0}).

Multiplying (a) and (b) gives the desired equality.

We first consider the ϵ_{r} term of (6.1)(c). By (3.10)(b) write h = γh_{0} with γ ∈ Γ(\overline{H}) and h_{0} ∈ \overline{H}^{0}. Write h_{0} = t_{0}exp(X) with t_{0} ∈ \overline{T}^{0} and X ∈ 𝔞. Since h is regular Φ_{r}^{+}(X) = {α ∈ Φ_{r} : α(X) > 0} is a choice of positive roots for Φ_{r}. For α ∈ Φ_{r} e^{−α}(h) = e^{−α}(γ)e^{−α(X)}. Since γ^{2} = 1, e^{−α}(γ) = e^{α}(γ) = ±1, so for any choice of Φ^{+} we have

(6.7)(a)  ϵ_{r}(Φ^{+}, h) = ∏_{α∈Φ_{r}^{+}} sgn(1 − e^{α}(γ)e^{−α(X)}).

Note that sgn(1−e^{α}(γ)e^{−α(X)}) = 1 unless e^{α}(γ) = 1 and α(X) < 0. Therefore

(6.7)(b)  \begin{aligned}ϵ_{r}(h, Φ^{+}) & = ∏_{α∈Φ_{r}^{+}∩Φ_{r}^{−}(X)} −e^{α}(γ) \\ & = (−1)^{|Φ_{r}^{+}∩Φ_{r}^{−}(X)|}e^{ρ(Φ_{r}^{+})−ρ(Φ_{r}^{+}(X))}(γ). \end{aligned}

Now write g̃ = γ̃g̃_{0} with p(γ̃) = ϕ(γ) and p(g̃_{0}) = ϕ(h_{0}). By Lemma 3.11(5) γ̃ ∈ Z(G̃), so e^{α}(γ̃) = 1. By a similar argument to the preceding we conclude ϵ_{r}(g̃, Φ^{+}) = (−1)^{|Φ_{r}^{+}∩Φ_{r}^{−}(X)|}. Thus

(6.7)(c)  \frac{ϵ_{r}(h, Φ^{+})}{ϵ_{r}(g̃, Φ^{+})} = e^{ρ(Φ_{r}^{+})−ρ(Φ_{r}^{+}(X))}(γ).

Now suppose Φ_{1}^{+}, Φ_{2}^{+} are two choices of real positive roots. Letting Φ_{r,j}^{+} = Φ_{j}^{+} ∩ Φ_{r} we have

(6.7)(d)  \frac{ϵ_{r}(h, Φ_{1}^{+})}{ϵ_{r}(g̃, Φ_{1}^{+})} = e^{ρ(Φ_{r,1}^{+})−ρ(Φ_{r,2}^{+})}(γ)\frac{ϵ_{r}(h, Φ_{2}^{+})}{ϵ_{r}(g̃, Φ_{2}^{+})}

Now consider the Δ^{0} term of (6.1)(c). Write Φ_{1}^{+} = wΦ_{2}^{+} for some w ∈ W (Φ), so that

(6.7)(e)  \frac{Δ^{0}(Φ_{1}^{+}, h)}{Δ^{0}(Φ_{1}^{+}, g̃)} = \frac{ϵ(w)e^{ρ(Φ_{2}^{+})−ρ(Φ_{1}^{+})}(h)Δ^{0}(Φ_{2}^{+}, h)}{ϵ(w)e^{ρ(Φ_{2}^{+})−ρ(Φ_{1}^{+})}(g̃)Δ^{0}(Φ_{2}^{+}, g̃)}.

Using the fact that ρ(Φ_{2}^{+}) − ρ(Φ_{1}^{+}) is a sum of roots, p(g̃) = ϕ(h) and (3.8) we conclude

e^{ρ(Φ_{2}^{+})−ρ(Φ_{1}^{+})}(g̃) = e^{ρ(Φ_{2}^{+})−ρ(Φ_{1}^{+})}(ϕ(h)) = e^{2ρ(Φ_{2}^{+})−2ρ(Φ_{1}^{+})}(h).

Therefore

(6.7)(f)  \frac{Δ^{0}(Φ_{1}^{+}, h)}{Δ^{0}(Φ_{1}^{+}, g̃)} = e^{ρ(−Φ_{2}^{+})+ρ(Φ_{1}^{+})}(h)\frac{Δ^{0}(Φ_{2}^{+}, h)}{Δ^{0}(Φ_{2}^{+}, g̃)}.

Taking the product of (b) and (d) gives

(6.7)(g)  \frac{Δ^{1}(Φ_{1}^{+}, h)}{Δ^{1}(Φ_{1}^{+}, g̃)} = e^{ρ(Φ_{r,1}^{+})−ρ(Φ_{r,2}^{+})}(γ)e^{−ρ(Φ_{2}^{+})+ρ(Φ_{1}^{+})}(h)\frac{Δ^{1}(Φ_{2}^{+}, h)}{Δ^{1}(Φ_{2}^{+}, g̃)}.

Writing h = γh_{0}, and since γ^{2} = 1 we have (6.7)(h)

e^{ρ(Φ_{r,1}^{+})−ρ(Φ_{r,2}^{+})}(γ)e^{−ρ(Φ_{2}^{+})+ρ(Φ_{1}^{+})}(h)
= e^{ρ(Φ_{r,1}^{+})−ρ(Φ_{r,2}^{+})−ρ(Φ_{2}^{+})+ρ(Φ_{1}^{+})}(γ)e^{−ρ(Φ_{2}^{+})+ρ(Φ_{1}^{+})}(h_{0})
= e^{ρ(Φ_{1}^{+})−ρ(Φ_{r,1}^{+})}(γ)e^{ρ(Φ_{2}^{+})−ρ(Φ_{r,2}^{+})}(γ)e^{−ρ(Φ_{2}^{+})+ρ(Φ_{1}^{+})}(h_{0})
= e^{ρ(Φ_{1}^{+})−ρ(Φ_{2}^{+})}(h_{0})

since (by (5.1)(b))

(6.7)(i)  e^{ρ(Φ_{1}^{+})−ρ(Φ_{r,1}^{+})}(γ) = ζ_{cx}(\overline{G}, \overline{H})(γ) = e^{ρ(Φ_{2}^{+})−ρ(Φ_{r,2}^{+})}(γ).

Therefore

(6.7)(j)  \frac{Δ^{1}(Φ_{1}^{+}, h)}{Δ^{1}(Φ_{1}^{+}, g̃)} = e^{ρ(Φ_{1}^{+})−ρ(Φ_{2}^{+})}(h_{0})\frac{Δ^{1}(Φ_{2}^{+}, h)}{Δ^{1}(Φ_{2}^{+}, g̃)}.

This completes the proof. □

Lemma 6.8 Assume μ ∈ i𝔷^{∗} where μ is defined as in (??). With notation as in the Definition,

(6.9)  |ΔG̃ ; \overline{G}(h, g̃)| = \frac{|Δ\overline{G}(h)|}{|ΔG̃(g̃)|}.

Proof. Let h ∈ \overline{H}′, g̃ ∈ H̃′ with ϕ(h) = p(g̃). Then

|ΔG̃ ; \overline{G}(h, g̃)| = \frac{|Δ\overline{G}(h)|e^{ρ(Φ^{+})}(g̃)χ̃(g̃)|}{|ΔG̃(g̃)||e^{ρ(Φ^{+})}(h)χ(Φ^{+}, h)|}.

Thus it suffices to prove that

|e^{ρ(Φ^{+})}(g̃)χ̃(g̃)| = |e^{ρ(Φ^{+})}(h)χ(Φ^{+}, h)|.

As in Lemma 5.40 we can write h = z\overline{exp}X where z ∈ \overline{Γ}(\overline{H})Z(\overline{G}) and X ∈ 𝔥_{d} and g̃ = z̃ẽxp2X where z̃ ∈ Z(G̃) with p(z̃) = ϕ(z). Now |e^{ρ(Φ^{+})}(h)| = |e^{<ρ(Φ^{+}),X>}| and |e^{ρ(Φ^{+})}(g̃)| = |e^{<ρ(Φ^{+}),2X>}| Thus by Lemma 5.40,

|e^{ρ(Φ^{+})}(g̃)χ̃(g̃)||e^{−ρ(Φ^{+})}(h)χ^{−1}(Φ^{+}, h)| = |χ̃_{s}(z̃)χ_{s}^{−1}(z)|.

Write z = t\overline{exp}Z where k ∈ \overline{Γ}(\overline{H})Z(\overline{G}) ∩ \overline{T} and Z ∈ 𝔷 ∩ 𝔭. Then z̃ = t̃ẽxp2Z where t̃∈ T̃, so that |χ̃_{s}(t̃)| = 1 = |χ_{s}(t)|. But

|χ̃_{s}(ẽxp2Z)χ_{s}^{−1}(\overline{exp}Z)| = |e^{<2dχ̃_{s}−dχ_{s},Z>}| = |e^{−<μ,Z>}| = 1

since < μ, Z >∈ iℝ. □

The transfer factors satisfy the following invariance property with respect to the Weyl group.

Lemma 6.10 Suppose g̃ ∈ H̃′, h ∈ \overline{H}′ satisfy p(g̃) = ϕ(h). Fix x̃ ∈ G̃ and let x = \overline{p}(p(x̃)) ∈ \overline{G}. Then

(6.11)  ΔG̃ ; \overline{G}(xhx^{−1}, x̃g̃x̃^{−1}) = ΔG̃ ; \overline{G}(h, g̃).

Proof. Fix a real set of positive roots Φ^{+} and (χ, χ̃) ∈ S(H, Φ^{+}). Let 𝔥_{1} = Ad(x)𝔥, H_{1} = xHx^{−1}, χ_{1} = xχx^{−1}, and χ̃_{1} = x̃χ̃x̃^{−1}. We claim that (χ_{1}, χ̃_{1}) ∈ S(H_{1}), and that (χ̃_{1}, χ_{1}) is compatible with (χ̃_{s}, χ_{s}) in the sense of Proposition 5.21.

Condition 5.3(b) is immediate. Further, for all t ∈ \overline{H}_{1,d}^{0},

χ_{1}(t) = χ_{1}(x^{−1}tx) = χ̃^{2}e^{ρ(Φ^{+})}(x^{−1}tx) = χ̃_{1}^{2}e^{ρ(Φ_{1}^{+})}(t)

so (a) holds as well.

Assume γ ∈ Γ(\overline{H})Z(\overline{G}). By (5.23)(b)

(6.12)  \begin{aligned}χ_{1}(γ) = χ(x^{−1}γx) & = (χ̃/χ̃_{s})(ϕ(x^{−1}γx))χ_{s}(yx^{−1}γxy^{−1}) \\ & = (χ̃_{1}/χ̃_{s})(ϕ(γ))χ_{s}((yx^{−1})γ(yx^{−1})^{−1}). \end{aligned}

This proves (5.23)(b) holds for (χ̃_{1}, χ_{1}).

Therefore we can use (χ̃_{1}, χ_{1}) and Φ_{1}^{+} to define the transfer factor for \overline{H}_{1}. But

χ̃_{1}(x̃g̃x̃^{−1}) = χ̃(g̃), χ_{1}(xhx^{−1}) = χ(h);
Δ^{1}(Φ_{1}^{+}, xhx^{−1}) = Δ^{1}(Φ^{+}, h), Δ^{1}(Φ_{1}^{+}, x̃g̃x̃^{−1}) = Δ^{1}(Φ^{+}, g̃).

Therefore

ΔG̃ ; \overline{G}(xhx^{−1}, x̃g̃x̃^{−1}) = ΔG̃ ; \overline{G}(h, g̃).

□

We next show see how the transfer factor depends on the choice of (χ̃_{s}, χ_{s}). For now we include this data in the notation and write ΔG̃ ; \overline{G}(χ̃_{s}, χ_{s}).

Lemma 6.13 Suppose (χ̃_{s}, χ_{s}), (χ̃′_{s}, χ′_{s}) ∈ S(H_{s}). Then there is a character ψ : \overline{G} → ℂ^{×} so that for every Cartan subgroup H and h ∈ \overline{H}′, g̃ ∈ H̃′ with ϕ(h) = p(g̃) we have

(6.14)  ΔG̃ ; \overline{G}(χ̃_{s}, χ_{s}, h, g̃) = ψ(h)ΔG̃ ; \overline{G}(χ̃′_{s}, χ′_{s}, h, g̃).

Conversely, suppose that ψ : \overline{G} → ℂ^{×} is a character and (χ̃_{s}, χ_{s}) ∈ S(H_{s}). Then there exists (χ̃′_{s}, χ′_{s}) ∈ S(H_{s}) such that (6.14) is satisfied.

Proof. This follows readily from Lemma 5.10. □

Proof. Since χ̃_{s}, χ̃′_{s} are both genuine characters of Z(H̃_{s}), χ̃_{s}χ̃_{s}^{′−1} factors to Z_{0}(H_{s}). Thus we can define a character ψ_{0} by

(6.15)  ψ_{0}(γ) = (χ′/χ)(γ)(χ̃/χ̃′)(ϕ(γ)) (γ ∈ Γ(\overline{H}_{s})Z(\overline{G})).

We claim ψ_{0} is trivial on Γ(\overline{H}_{s})Z(\overline{G}) ∩ \overline{G}_{d}^{0}.

If γ ∈ \overline{Γ}(Φ_{R,s}), then ϕ(γ) = 1 by Lemma 3.11 and χ_{s}(γ) = χ′_{s}(γ) = ζ_{cpx}(\overline{G}, \overline{H}_{s})(z), so ψ_{0}(γ) = 1. If z ∈ Γ(\overline{H}_{s})Z(\overline{G}) ∩ \overline{H}_{s,d}^{0}, then

ψ_{0}(z) = χ̃_{2,s}^{2}χ̃_{1,s}^{−2}e^{ρ(Φ_{2,s}^{+})−ρ(Φ_{1,s}^{+})}(z)χ̃_{1,s}χ̃_{2,s}^{−1}(ϕ(z)) = e^{ρ(Φ_{2,s}^{+})−ρ(Φ_{1,s}^{+})}(z) = 1

by Lemma ??. Thus ψ_{0}(z) = 1 for all z ∈ \overline{Γ}(Φ_{R,s})(Γ(\overline{H}_{s})Z(\overline{G}) ∩ \overline{H}_{s,d}^{0}) = Γ(\overline{H}_{s})Z(\overline{G}) ∩ \overline{G}_{d}^{0}. Now since \overline{G} = Γ(\overline{H}_{s})Z(\overline{G})\overline{G}_{d}^{0}, we can extend ψ_{0} to \overline{G} by ψ(zh_{0}) = ψ_{0}(z), z ∈ Γ(\overline{H}_{s})Z(\overline{G}), h_{0} ∈ \overline{G}_{d}^{0}.

Now let 𝔥 = 𝔱 ⊕ 𝔞 be an arbitrary Cartan subalgebra of 𝔤. We may as well assume that 𝔞 ⊂ 𝔞_{s}. Let χ̃_{i} ∈ X_{g}(H̃) and let Φ^{+} be a σ-stable choice of positive roots. Let χ_{i} be the character of \overline{H} corresponding to χ̃_{i} using Φ^{+} and (χ̃_{i,s}, χ_{i,s}).

Fix h ∈ \overline{H}, g̃ ∈ Z(H̃) with p(g̃) = ϕ(h). Then

ΔG̃ ; \overline{G}(χ̃_{1,s}, χ_{1,s}, h, g̃) = c(h, g̃)ΔG̃ ; \overline{G}(χ̃_{2,s}, χ_{2,s}, h, g̃)

where

c(h, g̃) = \frac{χ̃_{1}χ̃_{2}^{−1}(g̃)}{χ_{1}χ_{2}^{−1}(h)}.

Write h = z\overline{exp}X, g̃ = z̃ẽxp2X as in Lemma 5.40 where z ∈ Γ(\overline{H})Z(\overline{G}), X ∈ 𝔥_{d}, z̃ ∈ Z(G̃), with p(z̃) = ϕ(z). Now by Lemma 5.40

c(h, g̃) = \frac{χ̃_{1,s}χ̃_{2,s}^{−1}(z̃)}{χ_{1,s}(z)χ_{2,s}^{−1}(z)} = \frac{χ̃_{1,s}χ̃_{2,s}^{−1}(ϕ(z))}{χ_{1,s}(z)χ_{2,s}^{−1}(z)} = ψ_{0}(z) = ψ(h).

Conversely, fix a character ψ of \overline{G} and (χ̃_{1,s}, χ_{1,s}) ∈ X(H̃_{s}, \overline{H}_{s}). Define χ̃_{2,s} = χ̃_{1,s} and χ_{2,s} = ψχ_{1,s}. By Lemma ?? (χ̃_{2,s}, χ_{2,s}) ∈ X(H̃_{s}, \overline{H}_{s}), and it is easy to see that it satisfies (6.14). □

7 Lifting: Basic Propertires

Assume that (G̃, G, \overline{G}) is an admissible triple. In this section we define lifting from \overline{G} to G̃ and derive some of the basic properties of lifting. This depends on the choice of a set of transfer factors. Thus we fix a maximally split Cartan subgroup H_{s} of G and (χ̃_{s}, χ_{s}) ∈ S(H_{s}), and obtain a set of transfer factors by Definition 6.2.

Recall (Section 3) O(G, g) is the conjugacy class of g ∈ G, and

(7.1)  O_{st}(G, g) = {xgx^{−1} | x ∈ G(ℂ)} ∩ G

is the stable orbit. We also have O(\overline{G}, g) and O_{st}(\overline{G}, g), and the map (Lemma 3.4(3)) ϕ : O_{st}(\overline{G}) → O_{st}(G).

Definition 7.2 Suppose Õ ∈ O(G̃) is a strongly regular semisimple orbit for G̃, and and \overline{O}_{st} ∈ O_{st}(\overline{G}) is a strongly regular semisimple stable orbit for \overline{G}. Let p(Õ)_{st} be the stable orbit for G containing p(Õ).

Define ΔG̃ ; \overline{G}(\overline{O}_{st}, Õ) = 0 unless ϕ(\overline{O}_{st}) = p(Õ)_{st}.

Suppose ϕ(\overline{O}) = p(Õ)_{st}. Choose h ∈ \overline{O} and g̃ ∈ Õ such that ϕ(h) = p(g̃). Define

(7.3)  ΔG̃ ; \overline{G}(\overline{O}_{st}, Õ) = ΔG̃ ; \overline{G}(h, g̃).

By Lemma 6.10 this is independent of the choices of h and g̃.

Definition 7.4 Suppose \overline{θ} is a stable virtual character of \overline{G}. Then θ(\overline{O}_{st}) is defined for any strongly regular semisimple orbit \overline{O}_{st}. Suppose Õ is a strongly regular semisimple orbit for G̃. Define

(7.5)  LiftG̃ ; \overline{G}(\overline{θ})(Õ) = ∑_{\overline{O}_{st}∈O_{st}(\overline{G})} ΔG̃ ; \overline{G}(\overline{O}_{st}, Õ)\overline{θ}(\overline{O}_{st}).

This is a finite sum, over

(7.6)  \{\overline{O}_{st} | ϕ(\overline{O}_{st}) = p(Õ)_{st}\}.

Thus LiftG̃ ; \overline{G}Θ is a genuine class function defined on the set of strongly regular semisimple orbits in G̃.

Lemma 7.7 Suppose g̃ is a strongly regular semisimple element. Let H = Cent_{G}(p(g̃)) and \overline{H} = \overline{p}(H). Then

(7.8)  LiftG̃ ; \overline{G}(\overline{θ})(g̃) = ∑_{\{h∈\overline{H} | ϕ(h)=p(g̃)\}} ΔG̃ ; \overline{G}(h, g̃)\overline{θ}(h).

In particular LiftG̃ ; \overline{G}(\overline{θ})(g̃) = 0 unless p(g̃) is in the image of ϕ. Assume this holds, and choose h satisfying ϕ(h) = p(g̃). Then

(7.9)  LiftG̃ ; \overline{G}(\overline{θ})(g̃) = ∑_{\{t∈\overline{H} | ϕ(t)=1\}} ΔG̃ ; \overline{G}(th, g̃)\overline{θ}(th).

Proof. Let g = p(g̃). The assertion is clear unless g is in the image of ϕ, so assume this holds.

We claim

(7.10)  \{\overline{O}_{st} | ϕ(\overline{O}_{st}) = p(Õ)_{st}\} = \{O_{st}(\overline{G}, h) | ϕ(h) = g\}.

Suppose ϕ(\overline{O}_{st}) = p(Õ)_{st}. Then ϕ(h′) = g′ for some h′ ∈ \overline{O}_{st} and where g′ is G(ℂ) conjugate to g. Since ϕ takes stable orbits to stable orbits, this implies ϕ(h) = g for some h ∈ \overline{O}_{st}. Therefore the left hand side is contained in the right.

We just need to show that if ϕ(h) = ϕ(h′) = g then O_{st}(\overline{G}, h) ≠ O_{st}(\overline{G}, h′). Suppose not, so there exists x ∈ \overline{G}(ℂ) such that xhx^{−1} = h′. Choose a preimage y of x in G(ℂ). Then g = ϕ(h′) = ϕ(xhx^{−1}) = yϕ(h)y^{−1} = ygy^{−1}. Since g is strongly regular this implies y ∈ H, so x ∈ \overline{H} and h′ = h.

Since ϕ restricted to \overline{H} is a homomorphism the final assertion is clear. □

We derive some elementary properties of lifting.

Lemma 7.11 LiftG̃ ; \overline{G}(Θ) is supported on Z(G̃)G̃_{d}^{0}.

Proof. Let \overline{H} be a Cartan subgroup of \overline{G}. Then \overline{H} = Z(\overline{G})Γ(\overline{H})\overline{H}_{d}^{0}, so that ϕ(\overline{H}) ⊂ Z_{0}(G)H_{d}^{0}. Thus p^{−1}ϕ(\overline{H}) ⊂ Z(G̃)G̃_{d}^{0}. □

Define S̃ = p^{−1}ϕ(Z(\overline{G})). Then

(7.12)  Z(G̃)^{0} ⊂ S̃ ⊂ Z(G̃).

Lemma 7.13 Let H be a Cartan subgroup of G. Suppose g̃ ∈ H̃′, h ∈ \overline{H}′ satisfy ϕ(h) = p(g̃). Suppose we are also given z̃ ∈ S̃, z ∈ Z(\overline{G}) satisfying ϕ(z) = p(z̃). Then

(7.14)  ΔG̃ ; \overline{G}(z̃g̃, zh) = χ̃_{s}(z̃)χ_{s}(z^{−1})ΔG̃ ; \overline{G}(g̃, h).

Proof. Let Φ^{+} be a real set of positive roots and choose (χ̃, χ) ∈ S(H, Φ^{+}). Since z ∈ Z(\overline{G}) and z̃ ∈ S̃ ⊂ Z(G̃), we have

Δ^{1}(Φ^{+}, zh) = Δ^{1}(Φ^{+}, h), Δ^{1}(Φ^{+}, z̃g̃) = Δ^{1}(Φ^{+}, g̃).

Furthermore

(7.15)  \frac{χ̃(z̃g̃)}{χ(zh)} = \frac{χ̃(z̃)}{χ(z)} \frac{χ̃(g̃)}{χ(h)} = \frac{χ̃_{s}(z̃)}{χ_{s}(z)} \frac{χ̃(g̃)}{χ(h)}

since \frac{χ̃(z̃)}{χ(z)} = \frac{χ̃_{s}(z̃)}{χ_{s}(z)} by (5.23)(a). Inserting this into the defintion of ΔG̃ ; \overline{G} gives the result. □

Define Z_{1}(\overline{G}) = \{z ∈ Z(\overline{G}) : ϕ(z) = 1\}.

Lemma 7.16 Let Θ be a stable character of \overline{G} with central character ζ. Then LiftG̃ ; \overline{G}Θ ≡ 0 unless ζ(z) = χ_{s}(z) for all z ∈ Z_{1}(\overline{G}).

Proof. Fix a Cartan subgropu H, g̃ ∈ H̃′, and z ∈ Z_{1}(\overline{G}). Then

\{h ∈ \overline{H} : ϕ(h) = p(g̃)\} = \{zh : h ∈ \overline{H}, ϕ(h) = p(g̃)\}.

Further, for all h ∈ \overline{H} such that ϕ(h) = p̃(g̃),

ΔG̃ ; \overline{G}(g̃, zh) = χ_{s}(z^{−1})ΔG̃ ; \overline{G}(g̃, h)

by Lemma 7.13 applied to h, g̃, z, and z̃ = 1. Thus

LiftG̃ ; \overline{G}(Θ)(g̃) = ∑_{\{h∈\overline{H}:ϕ(h)=p(g̃)\}} ΔG̃ ; \overline{G}(g̃, zh)Θ(zh) = ζ(z)χ_{s}(z^{−1})LiftG̃ ; \overline{G}(Θ)(g̃).

□

Fix a character ζ of Z(\overline{G}). Assume that ζ(z) = χ_{s}(z) for all z ∈ Z_{1}(\overline{G}). Then ζχ_{s}^{−1} defines a character of Z(\overline{G})/Z_{1}(\overline{G}), and ϕ induces an isomorphism of this group and ϕ(Z(\overline{G})) = p(S̃). We can pull this character back to a character of S̃. Tensoring with χ̃_{s} we obtain a genuine character of S̃, which we denote ζ̃.

In other words for z̃ ∈ S̃ choose z ∈ Z(\overline{G}) so that ϕ(z) = p(z̃) and define

(7.17)  ζ̃(z̃) = χ̃_{s}(z̃)(ζχ_{s}^{−1})(z).

This is independent of the choice of z, and is a genuine character of S̃.

Lemma 7.18 Let Θ be a stable character of \overline{G} with central character ζ such that ζ = χ_{s} on Z_{1}(\overline{G}). Then for all g̃ ∈ G̃, z̃ ∈ S̃,

LiftG̃ ; \overline{G}(Θ)(z̃g̃) = ζ̃(z̃)LiftG̃ ; \overline{G}(Θ)(g̃).

Proof. Fix a Cartan subgroup H of G, g̃ ∈ H̃′, z̃ ∈ S̃, and z ∈ Z(\overline{G}) such that ϕ(z) = p(z̃). Then

\{h ∈ \overline{H} : ϕ(h) = p(z̃g̃)\} = \{zh : h ∈ \overline{H}, ϕ(h) = p(g̃)\}.

Thus by Lemma 7.13,

LiftG̃ ; \overline{G}(Θ)(z̃g̃) = ∑_{\{h∈\overline{H}:ϕ(h)=p(g̃)\}} ΔG̃ ; \overline{G}(z̃g̃, zh)(Θ)(zh)

= (ζχ_{s}^{−1})(z)χ̃_{s}(z̃)LiftG̃ ; \overline{G}(Θ)(g̃) = ζ̃(z̃)LiftG̃ ; \overline{G}(Θ)(g̃).

□

8 Invariant Eigendistributions

In this section we review results of Hirai and Harish-Chandra on invariant eigendistributions. In the next section we apply these results to study lifting from G to G̃.

Let G be a reductive Lie group in Harish-Chandra’s class, and let H be a Cartan subgroup of G. Suppose F : H′ → ℂ is differentiable. Corresponding to X ∈ 𝔥 is the differential operator

(8.1)  D_{X}F (h) = \frac{d}{dt}|_{t=0}F (hexp(tX)) (h ∈ H′).

Fix a set of positive roots Φ^{+}, with corresponding ρ = ρ(Φ^{+}). Then we also define

(8.2)  D_{X}^{ρ}F (h) = \frac{d}{dt}|_{t=0}e^{<ρ,tX>}F (hexp(tX)) (h ∈ H′).

The map D_{X} → D_{X}^{ρ} =< ρ, X > +D_{X} can be extended to an automorphism D → D^{ρ} of S(𝔥(ℂ)).

Let ν be a character of the center Z of the universal enveloping algebra of 𝔤(ℂ). [Let λ be the corresponding 𝔥(ℂ)^{∗} be the character of I(𝔥(ℂ)) corresponding to ν by the Harish-Chandra homomorphism.]

We say F : H′ → ℂ satisfies condition (C1, Φ^{+}, ν ) if F is real analytic on H′ and

(C1,Φ^{+}, ν)  D^{ρ}F (h) = λ(D)F (h) (h ∈ H′, D ∈ I(𝔥(ℂ))).

Let H′(R) = {h ∈ H | e^{α}(h) ≠ 1, α ∈ Φ_{R}}. We say F satisfies condition (C2) if

(C2)  F extends to a real analytic function on H′(R).

Now let Θ be a class function on G′ and define

(8.3)  Ψ(H, Φ^{+}, h) = Δ^{1}(Φ^{+}, h)Θ(h) (h ∈ H′).

Assume Ψ(H, Φ^{+}) satisfies (C1, Φ^{+}, ν ) and (C2). Suppose α is a simple root for Φ_{r}^{+}. Let J be the Cartan subgroup obtained from H via a Cayley transform c defined by α. Let Φ_{J}^{+} = cΦ^{+} and ρ_{J} = ρ(Φ_{J}^{+}). Let H(α) be the set of h ∈ H such that e^{α}(h) = 1, but e^{β}(h) ≠ 1 for all β ≠ ±α.

Fix h ∈ H(α). Then hexp(tα̌) ∈ H′ for 0 ≠ t sufficiently small, and there are well-defined one-sided limits

(8.4)  Dρ ; α̌Ψ^{±}(H, Φ^{+}, h) = lim_{t→0^{±}} Dρ ; α̌Ψ(H, Φ^{+}, hexp(tα̌)).

Let β be the imaginary root of J corresponding to α, so iβ^{∨} ∈ 𝔧 and h ∈ J′(R). Thus Dβ̌ = −iD_{iβ̌} is defined as in (8.1), and Dρ_{J} ; β̌ Ψ(J, Φ_{J}^{+}, h) is defined. We say Ψ(H, Φ^{+}) satisfies condition (C3) if

(C3) Dρ ; α̌Ψ^{+}(H, Φ^{+}, h) − Dρ ; α̌Ψ^{−}(H, Φ^{+}, h) = 2Dρ_{J} ; β̌ Ψ(J, Φ_{J}^{+}, h) (h ∈ H(α)).

The following theorem is a generalization of results of Hirai [8], [9] in the case that G is acceptable and satisfies an extra condition.

Theorem 8.5 [7] Let Θ be a class function on G′ and let ν be a character of Z. Then Θ is an invariant eigendistribution with infinitesimal character ν if and only for every Cartan subgroup H of G and every real root β of H, there is a choice of positive roots Φ^{+} such that β is simple for Φ_{R}^{+} and the function Ψ(H, Φ^{+}) satisfies conditions (C1, Φ^{+}, ν), (C2) , and (C3).

Moreover, suppose Θ is an invariant eigendistribution. Then Ψ(H, Φ^{+}) satisfies (C1, Φ^{+}, ν) and (C2) for every choice of Φ^{+}, and satisfies (C3) for every choice of Φ^{+} such that β is simple for Φ_{R}^{+}.

Write G = Kexp(𝔭) where K is a maximal compact subgroup and 𝔤 = 𝔨 ⊕ 𝔭 is the Cartan decomposition. Let ∥ · ∥_{G} be a Euclidean norm on 𝔭 and define

(8.6)  τ_{G}(kexpX) = ∥X∥, k ∈ K, X ∈ 𝔭.

Let Θ be an invariant eigendistribution on G. Then by §12 of [5], Θ is tempered if and only for each Cartan subgroup H of G there are numbers C, r ≥ 0, so that

(8.7)  |Δ_{G}(h)||Θ(h)| ≤ C(1 + τ_{G}(h))^{r}, g ∈ H′,

where |Δ_{G}(h)| is defined as in (6.1)(d).

Let 𝔷 be the center of 𝔤. As in [5] we can decompose G = ^{0}GZ_{p} where 0G has Lie algebra [𝔤, 𝔤] ⊕ (𝔷 ∩ 𝔨) and Z_{p} = exp(𝔷 ∩ 𝔭). Let Θ be an invariant eigendistribution on G. We say Θ is relatively tempered (relatively supertempered) on G if its restriction to ^{0}G is tempered (supertempered) on 0G. By §4 of [6], Θ is supertempered on ^{0}G if and only if for every Cartan subgroup ^{0}H of ^{0}G and every r ≥ 0,

(8.8)  sup_{g∈^{0}H′} |Δ_{G}(h)||Θ(h)|(1 + τ_{G}(h))^{r} < ∞.

We will also use the following result from §4 of [6].

Theorem 8.9 [6] Assume that G has a relatively compact Cartan subgroup B and let Θ be a relatively supertempered invariant eigendistribution such that Θ(b) = 0, b ∈ B′. Then Θ = 0.

9 Lifting

We now return to the context of lifting. That is, we assume that (G̃, G, \overline{G}) is an admissible triple and fix (χ̃_{s}, χ_{s}) ∈ X(H̃_{s}, \overline{H}_{s}) where 𝔥_{s} is a maximally split Cartan subalgebra of 𝔤. All three of the groups G̃, G, and \overline{G} are in the Harish-Chandra class.

Let Θ be a stable invariant eigendistribution on \overline{G} with infinitesimal character ν. Define Θ̃ = LiftG̃ ; \overline{G}Θ. We know that Θ̃ is a class function on G̃′.

Let 𝔥 be a Cartan subalgebra of 𝔤 and let Φ^{+} be a choice of positive roots for 𝔥. Define

(9.1)  Ψ(H̃, Φ^{+}, g̃) = Δ^{1}(Φ^{+}, g̃)Θ̃(g̃), g̃ ∈ H̃′;
(9.2)  Ψ(\overline{H}, Φ^{+}, h) = Δ^{1}(Φ^{+}, h)Θ(h), h ∈ \overline{H}′.

The following lemma follows from Theorem 8.5.

Lemma 9.3 Ψ(\overline{H}, Φ^{+}) satisfies conditions (C1, Φ^{+}, ν) and (C2) for every choice of Φ^{+}. Let β ∈ Φ_{R}^{+}. Then Ψ(\overline{H}, Φ^{+}) satisfies condition (C3) for any choice of Φ^{+} such that β is a simple root for Φ_{R}^{+}.

Fix χ̃ ∈ X_{g}(H̃) and a σ-stable choice of Φ^{+}. Write ρ = ρ(Φ^{+}). Let χ = χ(χ̃, Φ^{+}) and write Ψ(H̃) = Ψ(H̃, Φ^{+}) and Ψ(\overline{H}) = Ψ(\overline{H}, Φ^{+}).

Lemma 9.4 For all g̃ ∈ H̃, X ∈ 𝔥 such that g̃ẽxpX ∈ H̃′,

Ψ(H̃, g̃ẽxpX) = ∑_{h∈X(g̃)} d(h, g̃)e^{<−ρ−μ,X/2>}Ψ(\overline{H}, h\overline{exp}(X/2)),

where

X(g̃) = \{h ∈ \overline{H} : ϕ(h) = p(g̃)\},

and for h ∈ X(g̃),

d(h, g̃) = χ̃(g̃)χ^{−1}(h).

Proof. Let g̃ ∈ H̃′. ¿From the definition of lifting and transfer factor it is clear that

Ψ(H̃, g̃) = ∑_{h∈X(g̃)} χ̃(g̃)χ^{−1}(h)Ψ(\overline{H}, h).

Fix g̃ ∈ H̃, X ∈ 𝔥 such that g̃ẽxpX ∈ H̃′. Then

X(g̃ẽxpX) = \{h\overline{exp}(X/2) : h ∈ X(g̃)\}.

Thus Ψ(H̃, g̃ẽxpX) =

∑_{h∈X(g̃)} d(h, g̃)χ̃(ẽxpX)χ^{−1}(\overline{exp}(X/2))Ψ(\overline{H}, h\overline{exp}(X/2)).

But

χ̃(ẽxpX)χ(\overline{exp}(X/2))^{−1} = e^{<2dχ̃−dχ,X/2>} = e^{<−ρ−μ,X/2>}.
□

Let ν be the infinitesimal character of Θ and let λ ∈ 𝔥(ℂ)^{∗} correspond to ν by the Harish-Chandra homomorphism.

Lemma 9.5 Ψ(H̃) is real analytic on H′. Let g̃ ∈ H̃′. Then for all X ∈ 𝔥_{d},

D_{X}^{ρ}Ψ(H̃, g̃) = (1/2) ∑_{h∈X(g̃)} d(h, g̃)D_{X}^{ρ}Ψ(\overline{H}, h).

For all D ∈ S(𝔷(ℂ)),

D^{ρ}Ψ(H̃, g̃) = ((λ − μ)/2)(D)Ψ(H̃, g̃).

Proof. The fact that Ψ(H̃) is real analytic on H̃′ follows from Lemmas 9.3 and 9.4. Let X ∈ 𝔥, t ∈ ℝ. Then using Lemma 9.4 we have

e^{<ρ,tX>}Ψ(H̃, g̃ẽxptX) = sum_{h∈X(g̃)}d(h, g̃)e^{<ρ−μ,tX/2>}Ψ(\overline{H}, h\overline{exp}(tX/2)).

Thus

D_{X}^{ρ}Ψ(H̃, g̃) = ∑_{h∈X(g̃)} d(h, g̃)(D_{X/2}^{ρ}− < μ, X/2 >)Ψ(\overline{H}, h).

Now if X ∈ 𝔥_{d}, then < μ, X >= 0. If X = Z ∈ 𝔷, then D_{Z/2} ∈ I(𝔥(ℂ)), so that D_{Z/2}^{ρ}Ψ(\overline{H}, h) = λ(Z/2)Ψ(\overline{H}, h) by Lemma 9.3. Thus

D_{Z}^{ρ}Ψ(H̃, g̃) = ∑_{h∈X(g̃)} d(h, g̃) < λ−μ, Z/2 > Ψ(\overline{H}, h) = ((λ−μ)/2)(D_{Z})Ψ(H̃, g̃).

By induction we see that D^{ρ}Ψ(H̃, g̃) = ((λ − μ)/2)(D)Ψ(H̃, g̃) for all D ∈ S(𝔷(ℂ)). □

Lemma 9.6 For all D ∈ I(𝔥(ℂ)) and g̃ ∈ H̃′,

D^{ρ}Ψ(H̃, g̃) = ((λ − μ)/2)(D)Ψ(H̃, g̃).

Proof. We have I(𝔥(ℂ)) = I(𝔥_{d}(ℂ))S(𝔷(ℂ)). By Lemma 9.5 the result is true for D ∈ S(𝔷(ℂ)).

Let D ∈ I(𝔥_{d}(ℂ)) be homogeneous of degree k ≥ 0. Since Ψ(\overline{H}) satisfies (C1,ν), using Lemma 9.5 and an inductive argument, for g̃ ∈ H̃′,

D^{ρ}Ψ(H̃, g̃) = 2^{−k} ∑_{h∈X(g̃)} d(h, g̃)D^{ρ}Ψ(\overline{H}, h)

= 2^{−k}λ(D) ∑_{h∈X(g̃)} d(h, g̃)Ψ(\overline{H}, h) = (λ/2)(D)Ψ(H̃, g̃).

But μ(D) = 0, D ∈ I(𝔥_{d}(ℂ)), so the formula is also valid for D ∈ I(𝔥_{d}(ℂ)). □

Since (ν − μ)/2 corresponds to (λ − μ)/2 by the Harish-Chandra homomorphism, we have proven that Ψ(H̃) satisfies condition (C1,(ν − μ)/2).

Define

(9.7)  \overline{H}′(R) = \{h ∈ \overline{H} : e^{α}(h) ≠ 1, α ∈ Φ_{R}\};

(9.8)  H̃′(R) = {g̃ ∈ H̃ : e^{α}(g̃) ≠ 1, α ∈ Φ_{R}}.

The next lemma shows that Ψ(H̃) satisfies (C2).

Lemma 9.9 Ψ(H̃) extends to a real analytic function on H̃′(R).

Proof. Since the restriction of Θ̃ to H̃ is supported on Z(G̃)H̃^{0}, Ψ(H̃) is also supported on Z(G̃)H̃^{0}. Fix t̃∈ Z(G̃)T̃^{0}. Then connected components of t̃H̃^{0} ∩ H̃′(R) have the form t̃T̃^{0}ẽxpC(Φ_{R}^{+}) where C(Φ_{R}^{+}) = {X ∈ 𝔞 : α(X) > 0 ∀α ∈ Φ_{R}^{+}} and Φ_{R}^{+} is a choice of positive roots for Φ_{R}.

By Lemma 9.4, for all X ∈ 𝔥 such that t̃ẽxpX ∈ H̃′, Ψ(H̃, t̃ẽxpX) =

∑_{h∈X(t̃)} d(h, t̃)e^{<−ρ−μ,X/2>}Ψ(\overline{H}, h\overline{exp}(X/2)).

Fix h ∈ X(t̃). Then for all α ∈ Φ_{R}, e^{2α}(h) = e^{α}(t̃) = 1, so that e^{α}(h) = ±1. Define Φ_{R}(h) = {α ∈ Φ_{R} : e^{α}(h) = 1}. Then connected components of h\overline{H}^{0} ∩ \overline{H}′(R) have the form h\overline{T}^{0}\overline{exp}C(Φ_{R}^{+}(h)) where C(Φ_{R}^{+}(h)) = \{X ∈ 𝔞 : α(X) > 0 ∀α ∈ Φ_{R}^{+}(h)} and Φ_{R}^{+}(h) is a choice of positive roots for Φ_{R}(h). Let Φ_{R}^{+} be a choice of positive roots for Φ_{R}. Then Φ_{R}^{+}(h) = Φ_{R}^{+}∩Φ_{R}(h) is a choice of positive roots for Φ_{R}(h) and X/2 ∈ 𝔱 + C(Φ_{R}^{+}(h)) for all X ∈ 𝔱 + C(Φ_{R}^{+}). Thus X → e^{<−ρ−μ,X/2>}Ψ(\overline{H}, h\overline{exp}(X/2)) extends to a real analytic function on 𝔱 + C(Φ_{R}^{+}). Thus

X → ∑_{h∈X(t̃)} d(h, t̃)e^{<−ρ−μ,X/2>}Ψ(\overline{H}, h\overline{exp}(X/2))

extends to a real analytic function on 𝔱 + C(Φ_{R}^{+}). Thus there is an open neighborhood U of t̃ in t̃T̃^{0} such that Ψ(H̃) extends to be real analytic on Uẽxp(C(Φ_{R}^{+})). □

Fix β ∈ Φ_{R} and let 𝔧 be the Cartan subalgebra of 𝔤 obtained from 𝔥 by the Cayley transform c in the root β.

Lemma 9.10 Assume that Φ^{+} is σ-stable, β ∈ Φ^{+}, and that Φ_{J}^{+} = cΦ^{+} is also σ-stable. Then β is a simple root for Φ_{R}^{+}.

Proof. Let α ∈ Φ_{R}^{+}, α ≠ β. If < α, β̌ >= 0, then s_{β}α ∈ Φ_{R}^{+}. Suppose that < α, β̌ >≠ 0. Then cα ∈ Φ_{J,CP X}^{+}. Now since Φ_{J}^{+} = cΦ^{+} is also σstable, (cα)^{σ} = c(s_{β}α) ∈ cΦ^{+}, so that s_{β}α ∈ Φ_{R}^{+}. Thus s_{β}α ∈ Φ_{R}^{+} for all α ∈ Φ_{R}^{+}, α ≠ β, and hence β is simple. □

Lemma 9.11 Let Φ^{+} be a σ-stable choice of positive roots for Φ such that Φ_{J}^{+} = cΦ^{+} is also σ-stable. Let g̃ ∈ H̃(β). Then

(Dρ ; β̌Ψ)^{+}(H̃, Φ^{+}, g̃) − (Dρ ; β̌Ψ)^{−}(H̃, Φ^{+}, g̃) = 2D_{cβ̌}^{ρ_{J}}Ψ(J̃, Φ_{J}^{+}, g̃).

Proof. Fix Φ^{+} and Φ_{J}^{+} = cΦ^{+} as in the lemma and drop it from the notation in the various Ψ functions. Fix χ̃ ∈ X_{g}(H̃) and χ̃_{J} ∈ X_{g}(J̃), and let χ = χ(χ̃, Φ^{+}) and χ_{J} = χ(χ̃_{J}, Φ_{J}^{+}) be the corresponding characters of \overline{H} and \overline{J}.

Fix g̃ ∈ H̃(β) and h ∈ X(g̃). For all α ∈ Φ we have e^{2α}(h) = e^{α}(g̃). Thus for α ≠ ±β we have e^{α}(h) ≠ 1, while e^{β}(h) = ±1. Thus h is either regular or semiregular with respect to β.

Suppose that e^{β}(h) = −1. Then h ∈ \overline{H}′ is regular so that

(Dρ ; β̌Ψ)^{+}(\overline{H}, h) = (Dρ ; β̌Ψ)^{−}(\overline{H}, h).

Thus

(Dρ ; β̌Ψ)^{+}(H̃, g̃) − (Dρ ; β̌Ψ)^{−}(H̃, g̃)

= (1/2) ∑_{h∈X(g̃,β)} d(h, g̃)[(Dρ ; β̌Ψ)^{+}(\overline{H}, h) − (Dρ ; β̌Ψ)^{−}(\overline{H}, h)],

where

X(g̃, β) = {h ∈ X(g̃) : e^{β}(h) = 1}.

Let h ∈ X(g̃, β) = H(β) ∩ X(g̃). Then since Ψ(\overline{H}) satisfies (C3),

(Dρ ; β̌Ψ)^{+}(\overline{H}, h) − (Dρ ; β̌Ψ)^{−}(\overline{H}, h) = 2D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, h).

Thus

(Dρ ; β̌Ψ)^{+}(H̃, g̃) − (Dρ ; β̌Ψ)^{−}(H̃, g̃) = ∑_{h∈X(g̃,β)} d(h, g̃)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, h).

Let X_{J}(g̃) = \{b ∈ \overline{J} : ϕ(b) = p(g̃)\}. Then X_{J}(g̃) ⊂ J′(R). Thus applying Lemma 9.5 to J̃,

D_{cβ̌}^{ρ_{J}}Ψ(J̃, g̃) = ∑_{b∈X_{J} (g̃)} d_{J}(b, g̃)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, b),

where for b ∈ X_{J}(g̃),

d_{J}(b, g̃) = χ̃_{J}(g̃) χ_{J}(b)^{−1}.

Now X_{J}(g̃, cβ) = {b ∈ X_{J}(g̃) : e^{cβ}(b) = 1} = X(g̃, β). Let h ∈ X_{J}(g̃, cβ). Then by Lemma ?? we have

χ̃(g̃)χ^{−1}(h) = χ̃_{J}(g̃)χ_{J}^{−1}(h),

so that d(h, g̃) = d_{J}(h, g̃). Thus to complete the proof of the lemma we must show that

∑_{b∈X_{J} (g̃,cβ)^{c}} d_{J}(b, g̃)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, b) = 0

where X_{J}(g̃, cβ)^{c} = {b ∈ X_{J}(g̃) : e^{cβ}(b) = −1}.

Write β′ = cβ, and let w denote the reflection in β′. Since β′ is an imaginary root and Θ is a stable class function on \overline{G}′, Θ(wb) = Θ(b) and ϵ_{R}(Φ_{J}^{+}, wb) = ϵ_{R}(Φ_{J}^{+}, b) for all b ∈ \overline{J}′. Further, since ϵ(w) = −1,

Δ^{0}(Φ_{J}^{+}, wb) = −e^{wρ_{J} −ρ_{J}} (b)Δ^{0}(Φ_{J}^{+}, b), b ∈ \overline{J}′.

Thus

Ψ(\overline{J}, wb) = −e^{ρ_{J} −wρ_{J}} (b)Ψ(\overline{J}, b), b ∈ \overline{J}′.

Fix b ∈ X_{J}(g̃, β′)^{c}, so that e^{β′}(b) = −1. Then wb = b\overline{exp}(πiβ̌′), ϕ(wb) = ϕ(b) = p(g̃), and e^{β′}(wb) = e^{β′}(b) = −1. Thus wb ∈ X_{J}(g̃, β′)^{c}, and since wb\overline{exp}(itβ̌′) = w(b\overline{exp}(−itβ̌′)) for all t ∈ ℝ,

e^{<ρ_{J} ,itβ̌′>}Ψ(\overline{J}, (wb)\overline{exp}(itβ̌′)) = −e^{ρ_{J} −wρ_{J}} (b)e^{<ρ_{J} ,−itβ̌′>}Ψ(\overline{J}, b\overline{exp}(−itβ̌′)).

Thus

D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, wb) = e^{ρ_{J} −wρ_{J}} (b)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, b).

Since (wb)b^{−1} = \overline{exp}(πiβ̌′), e^{ρ_{J} −wρ_{J}} (b) = (−1)^{<ρ_{J} ,β̌′>} and χ_{J}((wb)b^{−1}) = (−1)^{<dχ_{J} ,β̌′>}. Thus

d_{J}(wb, g̃)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, wb) = (−1)^{<dχ_{J} +ρ_{J} ,β̌′>}d_{J}(b, g̃)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, b).

But since β′ is noncompact,

< dχ_{J} + ρ_{J}, β̌′ >=< 2dχ̃_{J} + 2ρ_{J}, β̌′ >=≡< 2dχ̃_{J}, β̌′ >≡ 1

by Lemma ??. Thus

d_{J}(wb, g̃)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, wb) = −d_{J}(b, g̃)D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, b).

Thus if \overline{exp}(πiβ̌′) ≠ 1 so that wb ≠ b, the terms for b and wb cancel. If \overline{exp}(πiβ̌′) = 1 so that wb = b, this forces

D_{cβ̌}^{ρ_{J}}Ψ(\overline{J}, b) = 0.
□

We have shown that the functions Ψ(H̃, Φ^{+}) satisfy conditions (C1,(ν − μ)/2), (C2), and (C3). Thus by Theorem 8.5 we have the following.

Theorem 9.12 Let Θ be a stable invariant eigendistribution on \overline{G} with infinitesimal character ν. Then LiftG̃ ; \overline{G}Θ is an invariant eigendistribution on G̃ with infinitesimal character (ν − μ)/2 where μ ∈ 𝔷(ℂ)^{∗} depends on the choice of (χ̃_{s}, χ_{s}) used to define the transfer factor and is defined as in (??).

Theorem 9.13 Let Θ be a tempered stable invariant eigendistribution on \overline{G}. If (χ̃_{s}, χ_{s}) is chosen so that μ ∈ i𝔷^{∗}, then LiftG̃ ; \overline{G}Θ is tempered on G̃.

Proof. Since \overline{G} and G̃ have the same Lie algebra, we can use the same Euclidean norm ∥ · ∥ on 𝔭 to define both τ_{H} and τG̃.

Let 𝔥 be a Cartan subalgebra of 𝔤. Fix g̃ ∈ H̃′ and write g̃ = t̃ẽxpX where t̃ ∈ T̃′, X ∈ 𝔞. Then τG̃(g̃) = ∥X∥. Further, for any h ∈ \overline{H} such that ϕ(h) = p(g̃), we have h = t\overline{X}/2 where t ∈ \overline{T} with ϕ(t) = p(t̃). Thus τ_{H}(h) = ∥X/2∥ = τG̃(g̃)/2.

Now, using Lemma 6.8 and (8.8) applied to \overline{G},

|ΔG̃(g̃)||Θ̃(g̃)| ≤ ∑_{\{h∈\overline{H}:ϕ(h)=p(g̃)\}} |ΔG̃(g̃)||ΔG̃ ; \overline{G}(h, g̃)||Θ(h)|

= ∑_{\{h∈\overline{H}:ϕ(h)=p(g̃)\}} |Δ\overline{G}(h)||Θ(h)| ≤ [\overline{H}_{1}]C(1 + τG̃(g̃)/2)^{r},

where \overline{H}_{1} = \{h ∈ \overline{H} : ϕ(h) = 1\}. Thus there is a constant C′ such that

|ΔG̃(g̃)||Θ̃(g̃)| ≤ C′(1 + τG̃(g̃))^{r}

for all g̃ ∈ H̃′. Now Θ̃ is tempered using (8.8) applied to G̃. □

Theorem 9.14 Let Θ be a stable invariant eigendistribution on \overline{G}. If Θ is relatively tempered, then LiftG̃ ; \overline{G}Θ is relatively tempered. If Θ is relatively supertempered, then LiftG̃ ; \overline{G}Θ is relatively supertempered.

Proof. Let 𝔥 be a Cartan subalgebra of 𝔤. Then^{0}H̃ = H̃ ∩^{0}G̃ = T̃ẽxp𝔞_{d} and 0\overline{H} = \overline{H} ∩^{0}\overline{G} = \overline{T} \overline{exp}𝔞_{d}. Fix g̃ ∈ ^{0}H̃′. Then g̃ = t̃ẽxpX where t̃∈ T̃′, X ∈ 𝔞_{d}. Then for any h ∈ \overline{H} such that ϕ(h) = p(g̃), we have h = t\overline{X}/2 where t ∈ \overline{T} with ϕ(t) = p(t̃). Thus h ∈ ^{0}\overline{H}. Further, as in the proof of Lemma 6.8, it is easy to see that

|ΔG̃(g̃)||ΔG̃ ; \overline{G}(h, g̃)| = |Δ\overline{G}(h)|

for any choice of (χ̃_{s}, χ_{s}). The first statement now follows easily by the same argument as that of Theorem 9.13. For the second statement we use as similar argument along with (8.8) applied to both \overline{G} and G̃. □

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