Notes on the Knapp-Zuckerman theory

The point of these notes is to redefine some of their concepts in terms of the L-group. I observe, however, that it is best and indeed essential for further applications that their results be formulated for reductive groups rather than just for simply-connected semi-simple groups. I use the notation of CRRAG (On the classification of representations of real algebraic groups) modified sometimes according to Borel’s suggestions.

Since we are dealing with tempered representations we start from φ : W_{C/R} → ^{L}G with image which is essentially compact. We suppose φ defines an element of Φ(G). Choose a parabolic ^{L}P in ^{L}G which is minimal with respect to the property that φ(W_{C/R}) ⊆ ^{L}P. ^{L}P defines P and M. Let ρ (with character Θ) be one of the representations of M associated to φ. Thus ρϵΠ_{φ}, if φ is regarded as taking W_{C/R} to ^{L}M. It is

Ind(G, P, ρ)

that Knapp-Zuckerman study.

They define W on p. 3, formula [2]. We want another definition. For this we observe that Ω(^{L}T^{0},^{L}G^{0}). Here T is a CSG (Cartan subgroup) of M. We want to regard W as a subgroup of the latter group. We may assume, along the lines of CRRAG that φ(C^{X}) ⊆ ^{L}T, that φ(W_{C/R}) normalizes ^{L}T, and that ^{L}T ⊆ ^{L}M, a chosen Levi factor of ^{L}P.

LEMMA 1. W is the quotient Norm (^{L}T) ∩ Cent φ(W_{C/R})/^{L}T^{0} ∩ Cent φ(W_{C/R}), the normalizer and centralizer being taken in ^{L}G^{0}.

Let {1, σ} be 𝔊(C/R) so that W_{C/R} is generated by C^{×} and σ with σ^{2} = −1. As on pages 48 and 49 of CRRAG with M replacing G the homomorphism φ is defined by μ, ν with ν = φ(σ)μ and by λ_{0}. If ω in Ω_{R}(T, G) normalizes M then

ωϵW ⟺ ωρ ∼ ρ ⟺ ωμ = ω_{1}μ, ωλ_{0} ≡ ω_{1}λ_{0} mod (^{L}X_{∗} + (1 − φ(σ)) (^{L}X_{∗} ⊗ C))

with ω_{1} ϵ Ω_{R}(T, M) and ^{L}X_{∗} = Hom (GL(1)), ^{L}T. Replace ω by ω_{1}^{−1}ω. Since ω normalizes M,

φ(σ)ω = ωφ(σ)

on ^{L}X_{∗} and

ωμ = μ ⟺ ωμ = μ, ω v = v ⟺ w φ(z) w for z ϵ C^{×}

if w ϵ^{L}G^{0} represents ω. We write

LM = ^{L}M^{0} × W_{C/R}

and let

φ(σ) = a × σ

with

λ^{∨}(a) = e^{2πi⟨λ_{0},λ^{∨}⟩}.

By the first paragraph on p. 37 of Problems in the theory of automorphic forms we may choose w so that wσ = σw. But this is the wrong choice.

ω(a) = σ(b)b^{−1}a.

Replace w by bw then

wφ(σ)w^{−1} = σ(b)b^{−1}abσ(b)^{−1} × σ = a × σ = φ(σ).

In other words this new choice of w satisfies

wφ(v)w^{−1} = φ(v) ∀vϵW_{C/R}.

Since ωϵΩ_{R}(T, M) and ωμ = μ imply that ω = 1 we have found

W ↪ Norm (^{L}T^{0}) ∩ Cent φ(W_{C/R})/^{L}T^{0} ∩ Cent φ(W_{C/R}).

To obtain the full lemma we have only to show that if w lies in Norm (^{L}T^{0}) ∩ Cent φ(W_{C/R}) then the corresponding element of the Weyl group stabilizes M and lies in Ω_{R}(T, G). It stabilizes ^{L}M because α^{∨} is a root of ^{L}M if and only if φ(σ)α^{∨} = −α^{∨}. Hence it stabilizes M. By Lemma 5.2 of Shelstad’s thesis

ω = ω_{1}ω_{2}

with ω_{1} ϵ Ω_{C}(T, M), ω_{2} ϵ Ω_{R}(T, G). Then

w φ = φ ⇒ ω_{1}^{−1} μ = ω_{2} μ, ω_{1}^{−1} v = ω_{2}v, ω_{1}^{−1}λ_{0} ≡ ω_{2}λ_{0}.

Another lemma of Shelstad implies that ω_{1} ϵ Ω_{R}(T, M). Hence

ω ϵΩ_{R}(T, G).

The advantage of introducing the L-group appears immediately when Knapp’s R-group is discussed. Let S be the centralizer of φ(W_{C/R}) in ^{L}G^{0} and S^{0} the connected component.

LEMMA 2. If G is semi-simple and simply-connected then the R-group is S/S^{0}.

Let ^{L}𝔱 be the Lie algebra of ^{L}T and set

L𝔱 = ^{L}𝔱_{+} +^{L}𝔱.

where ^{L}𝔱_{+} and ^{L}𝔱_{−} are the +1 and -1 eigenspaces for φ(σ). I claim that ^{L}𝔱_{+} which certainly lies in 𝔰, the Lie algebra of S^{0}, is in fact a Cartan subalgebra of S^{0}. Indeed

𝔰 ⊆ ^{L}𝔱_{+} + Σ_{⟨μ,α^{∨}⟩=⟨ν,α^{∨}⟩=0}CX_{α^{∨}}.

If ⟨μ, α^{∨}⟩ = ⟨ν, α^{∨}⟩ = 0 then α^{∨} cannot be a root of ^{L}T in ^{L}M. Hence

φ(σ)α^{∨} ≠ −α^{∨}

and α^{∨} is not 0 on ^{L}𝔱_{+}. The assertion follows.

We may indentify Hom(^{L}𝔱, C) with 𝔱 ⊗ C as a 𝔊(C/R)-module if 𝔱 is the Lie algebra of T. If α^{∨} is a root of ^{L}T^{0} in ^{L}G^{0} with φ(σ)α^{∨} ≠ −α^{∨} set

𝔞_{α^{∨}} = (^{L}𝔱_{−} + Cα^{∨})^{⊥}.

Then G_{α^{∨}} the centralizer of 𝔞_{α^{∨}} in G is defined over R and M is the Levi factor of a maximal P SG of G_{α^{∨}}. Let μ(ρ, α^{∨}) be the value of the Plancherel measure for

Ind (G_{α^{∨}}(R), M(R), ρ).

Let

𝔛_{α^{∨}} = {β^{∨}|φ(σ)β^{∨} ≠ −β^{∨}, G_{β^{∨}} = G_{α^{∨}}}.

The centralizer of ^{L}𝔱_{+} is

L𝔱_{+} + Σ_{φ(σ)α^{∨}=−α^{∨}}CX_{α^{∨}}

and this is the Lie algebra of ^{L}M. Moreover

S/S^{0} ≃ Norm_{S} (^{L}𝔱_{+})/ Norm_{S^{0}}(^{L}𝔱_{+}).

If wϵ Norm_{S}(^{L}𝔱_{+}) then w normalizes ^{L}𝔱 and we have

Norm_{S} (^{L}𝔱_{+})/^{L}T_{+} ≃ W.

The lemma and indeed more will be established once the following facts are proved. They will be proved for any G.

(i)  dim 𝔰_{α^{∨}} = dim ((Σ_{β^{∨}ϵ𝔛_{α^{∨}}} CX_{β^{∨}}) ∩ 𝔰) ≤ 1.

(ii) It is equal to 1 if and only if μ(ρ, α^{∨}) = 0. (iii) If it is one then 𝔰_{α^{∨}} defines a root space of ^{L}𝔱_{+} in 𝔱. The corresponding reflection in ^{L}𝔱_{+} is the same

as that defined by the real root of T in G_{α^{∨}}. There are a number of possibilities to consider.

(a) 𝔛_{α^{∨}} consists of a single element. Then φ(σ)α^{∨} = α^{∨} and α, the corresponding root of T, is real. Since σμ = ν, ⟨μ, α^{∨}⟩ = ⟨ν, α^{∨}⟩ and dim 𝔰_{α^{∨}} = 1 if and only if ⟨μ, α^{∨}⟩ = 0 and

φ(σ)X_{α^{∨}} = X_{α^{∨}}.

Certainly T (R) is not fundamental. According to the formula on p. 141 of Harish–Chandra’s preprint Harmonic analysis III, μ(ρ, α^{∗}) is 0 if and only if

ν_{α} = 0 and \frac{(−1)^{ρ_{α}}}{2} (σ_{a^{∗}}(γ) + σ_{a^{∗}}(γ^{−1})) ≠ 1.

Now

ν_{α} = ⟨μ, α^{∨}⟩.

Also 𝔰_{a^{∗}} is now of dimension one and

σ_{a^{∗}}(γ) = σ_{a^{∗}}(γ^{−1}) = χ(α^{∨}(−1)).

Here χ is associated to φ : W_{C/R} → ^{L}M as on p. 50 of CRRAG and if the definition of a coroot is taken into account

γ = α^{∨}(−1).

Thus (cf. p. 51 of CRRAG)

χ(α^{∨}(−1)) = e^{2πi⟨λ_{0},α^{∨}⟩}.

Apologies are necessary for this phase of the discussion but the transition from Harish-Chandra’s notation to that used in CRRAG is clumsy.

On the other hand

φ(σ) = a × σ

and

φ(σ)X_{α^{∨}} = e^{2πi⟨λ_{0},α^{∨}⟩}φ′(σ)(X_{α^{∨}})

if φ′(σ) = a′ × σ, a′ϵ^{L}M_{der}, a^{−1}a′ϵ^{L}T^{0}. The assertion (ii) will be verified if we show that

φ′(σ)(X_{α^{∨}}) = −(−1)^{ρ_{α}}X_{α^{∨}}.

Now, by p. 122 of Harmonic Analysis III

ρ_{α} = ⟨ρ_{α^{∨}}, α^{∨}⟩

if ρ_{α^{∨}} is one-half the sum of the positive roots of G_{α^{∨}}. But in the present circumstances the derived algebra of 𝔤_{α^{∨}} is a direct sum because α^{∨} is perpendicular to all roots of G_{α^{∨}} except ±α^{∨}. Thus

⟨ρ_{α}, α^{∨}⟩ = \frac{1}{2}⟨α, α^{∨}⟩ = 1.

Moreover α^{∨} must be a simple root and so by the definition of ^{L}M

φ′(σ)(X_{α^{∨}}) = σ(X_{α^{∨}}) = 1.

The assertion (ii) follows. Since the reflections corresponding to α and α^{∨} are the same, the assertion (iii) does also.

(b) Suppose φ(σ)α^{∨} = α^{∨} and β^{∨} different from α^{∨} lies in 𝔛_{α^{∨}}.

(i) Suppose

⟨μ, β^{∨}⟩ = ⟨ν, β^{∨}⟩ = 0.

Then

⟨μ, φ(σ)β^{∨}⟩ = ⟨ν, φ(σ)β^{∨}⟩ = 0.

Since φ(σ)β^{∨} lies in the span of {α^{∨}, β^{∨}} and is different from β^{∨}, both μ and ν vanish on this twodimensional space. As a consequence there are no roots γ^{∨} on it orthogonal to α^{∨}. For then φ(σ)γ^{∨} would be −γ^{∨} and as a consequence

⟨μ, γ^{∨}⟩ ≠ 0.

This leaves only

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of type A_{2}.

I claim next that if γ^{∨} lies in X_{α^{∨}} and is defferent from α^{∨}, β^{∨}, and φ(σ)β^{∨} then either ⟨μ, γ^{∨}⟩ ≠ 0 or ⟨ν, γ^{∨}⟩ ≠ 0. If not, consider all roots in the span of {α^{∨}, β^{∨}, γ^{∨}}. They form a root system of rank 3 on which φ(σ) acts. If δ^{∨} lies in this system then ⟨μ, δ^{∨}⟩ = ⟨ν, δ^{∨}⟩ = 0 so φ(σ)δ^{∨} ≠ −δ^{∨}. As a consequence

δ^{∨} + φ(σ)δ^{∨} = aα^{∨} a ≠ 0

and

{δ^{∨}|⟨α, δ^{∨}⟩ ≥ 0}

defines a system of positive roots stable under φ(σ). Let α_{1}^{∨}, α_{2}^{∨}, α_{3}^{∨} be the simple roots. They are permuted amongst themselves by φ(σ). Thus by a suitable numbering

a_{1}^{∨} = α^{∨} a_{3}^{∨} = φ(σ)a_{2}^{∨}.

Then

aα^{∨} = α_{2}^{∨} + a_{3}^{∨}.

This is a contradiction.

Also we may take

X_{α^{∨}} = [X_{β^{∨}}, φ(σ)X_{β^{∨}}]

and

φ(σ)X_{α^{∨}} = −X_{α^{∨}}.

Thus

𝔰_{α^{∨}} = C(X_{β^{∨}} + φ(σ)X_{β^{∨}})

has dimension 1. Since

⟨μ, β′⟩ = (λ + iν)(H_{β}),

the right side conforming to Harish-Chandra’s notation, the measure μ(ρ, α^{∨}) is certainly zero. The reflection defined by 𝔰_{α^{∨}} is clearly correct on F_{𝔱_{+}}.

(ii) Suppose that for every β^{∨} different from α^{∨} in 𝔛_{α^{∨}}

⟨μ, β^{∨}⟩ ≠ 0 or ⟨ν, β^{∨}⟩ ≠ 0.

Then dim𝔰_{α^{∨}} = 1 if and only if

⟨μ, α^{∨}⟩ = 0, φ(σ)X_{α^{∨}} = X_{α^{∨}}.

Again the first condition is equivalent to ν_{α} = 0. We have to show that when this is so then the second is equivalent to

\frac{(−1)^{ρ_{α}}}{2} (σ_{a^{∗}}(γ) + σ_{a^{∗}}(γ^{−1})) ≠ 1.

Let

φ(σ)X_{α^{∨}} = λX_{α^{∨}}.

We show that

\frac{(−1)^{ρ_{α}}}{2} (σ_{a^{∗}}(γ) + σ_{a^{∗}}(γ^{−1})) = −λ.

This is enough, for λ = ±1. As before

σ_{a^{∗}}(γ) = σ_{a^{∗}}(γ^{−1}) = e^{2πi⟨λ_{0},α^{∨}⟩}

and

φ(σ)X_{α^{∨}} = −(−1)^{⟨ρ_{α^{∨}} ,α^{∨}⟩}X_{α^{∨}}.

if φ′(σ) is defined as before. What we must do is show that

φ′(σ)(X_{α^{∨}}) = −(−1)^{⟨ρ_{α^{∨}} ,α^{∨}⟩}X_{α^{∨}}.

This is a statement about a reductive group G_{α^{∨}} and a Levi factor M of a maximal parabolic, M and G both having compact CSG′s. It is not bound to the present situation and may be proved by induction on the rank of G_{α^{∨}}. Let β^{∨} be the largest root of one of the simple factors of ^{L}M_{der} and introduce a_{2}, a_{1} as on p. 46 of CRRAG. We may take a′ = a_{2}a_{1}. If ρ′ is the analogue of ρ_{α^{∨}} for the roots perpendicular to β^{∨} then by induction

a_{1} × σ(X_{α^{∨}}) = −(−1)^{⟨ρ′,α^{∨}⟩}X_{α^{∨}}

What we have to do is show that

a_{2}(X_{α^{∨}}) = (−1)^{ℓ}X_{α^{∨}}, ℓ = \frac{1}{2} ∑_{{⟨γ,β^{∨}⟩≠0 \atop γ>0}} ⟨γ, α^{∨}⟩.

Suppose γ > 0, ⟨γ, β^{∨}⟩ ≠ 0, ⟨γ, α^{∨}⟩ ≠ 0 and γ^{∨} is not in plane spanned by α^{∨}, β^{∨}. Then:

1) γ^{∨} = a_{2}γ^{∨} ⇒ γ = a_{2}γ ⇒ ⟨γ, β^{∨}⟩ = 0 impossible

2) γ^{∨} = φ(σ)γ^{∨} ⇒ γ^{∨} = ±α^{∨} impossible

3) γ^{∨} = a_{2}φ(σ)γ^{∨} ⇒ γ^{∨} in plane of α^{∨}, β^{∨} because (α^{∨}, β^{∨}) = 0. Thus γ, a_{2}γ, φ(σ)γ, a_{2}φ(σ)γ are distinct and positive. Since

⟨γ, α^{∨}⟩ = ⟨a_{2}γ, α^{∨}⟩ = ⟨φ(σ)γ, α^{∨}⟩ = ⟨a_{2}φ(σ)γ, α^{∨}⟩

the sum of the four of them even after division by 2 is even and may be dropped from the exponent. So may those ⟨γ, α^{∨}⟩ which are 0. We confine ourselves to γ with γ^{∨} in the plane of α^{∨}, β^{∨}.

The possibilities are as follows. A) No roots except ±α^{∨}, ±β^{∨} in the plane. Then exponent is 0 and

a_{2}(X_{α^{∨}}) = X_{α^{∨}}.

Notes on the Knapp-Zuckerman theory 8 B)

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\frac{1}{2}Σ⟨γ, α^{∨}⟩ = \frac{1}{2}⟨α, α^{∨}⟩ = 1, a_{2}(X_{α^{∨}}) = −X_{α^{∨}}

C)

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 ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.} β α α^{∨}  β^{∨}

\frac{1}{2}Σ⟨γ, α^{∨}⟩ = ⟨α, α^{∨}⟩ = 2, a_{2}(X_{α^{∨}}) = X_{α^{∨}}

D)

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\frac{1}{2}Σ⟨γ, α^{∨}⟩ = 2⟨α, α^{∨}⟩ = 4, a_{2}(X_{α^{∨}}) = X_{α^{∨}}

E) The roles of α, α^{∨} and β, β^{∨} are reversed

\frac{1}{2}Σ⟨γ, α^{∨}⟩ = ⟨α, α^{∨}⟩ = 2, a_{2}(X_{α^{∨}}) = X_{α^{∨}}.

All that is claimed in A) through E) is easy to check. Finally it is clear that the reflection defined by 𝔰_{α^{∨}} is that defined by α or α^{∨}.

i) Suppose that φ(σ)β^{∨} ≠ β^{∨} for all β^{∨} in 𝔛_{α^{∨}}. Then β^{∨} + φ(σ)β^{∨} is not a root, nor is

\frac{β^{∨} + φ(σ)β^{∨}}{2} .

(1) Suppose that ⟨μ, α^{∨}⟩ = ⟨ν, α^{∨}⟩ = 0. Then α^{∨} − φ(σ)α^{∨} is not a root and ⟨α^{∨}, φ(σ)α^{∨}⟩ = 0. Since α^{∨} and φ(σ)α^{∨} have the same length, the root diagram of the plane spanned by α^{∨}, φ(σ)α^{∨} is

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I claim that if β^{∨} lies in 𝔛_{α^{∨}} but not in this plane then either ⟨μ, β^{∨}⟩ = 0 or ⟨ν, β^{∨}⟩ = 0. Otherwise in the three-dimensional plane spanned by α^{∨}, φ(σ)α^{∨}, β^{∨}, φ(σ)β^{∨} we have a root system and

{γ^{∨}|⟨γ, α^{∨} + φ(σ)α^{∨}⟩ ≥ 0}

is a set of positive roots, for

{γ, α^{∨} + φ(σ)α^{∨}⟩

is never 0, because if it were then φ(σ)γ^{∨} = −γ^{∨}. Since ⟨μ, γ^{∨}⟩ = ⟨ν, γ^{∨}⟩ = 0 this is impossible. Then φ(σ) permutes the three simple roots amongst themselves, and leaves one fixed. This is a contradiction. Thus

𝔰_{α} = C(X_{α^{∨}} + φ(σ)X_{α^{∨}})

has dimension one. Since T is fundamental in G_{α^{∨}}, the formula on p. 97 of Harmonic analysis III shows that μ(ρ, α^{∨}) = 0. The three assertions follow again.

(ii) Suppose that for any β^{∨} in 𝔛_{α^{∨}} either ⟨μ, β^{∨}⟩ ≠ 0 or ⟨ν, β^{∨}⟩ ≠ 0. Then 𝔰_{α^{∨}} = 0. By the same formula in Harmonic analysis III,

μ(ρ, α^{∨}) ≠ 0.

Lemma 2 is now completely proved. I should observe, for it will remove a confusion that could otherwise arise, that

\overline{−⟨μ, α^{∨}⟩} = ⟨ν, α^{∨}⟩

for any α^{∨}.

It is also possible to give Zuckerman’s proof that the R-group is a sum of Z_{2}’s in the above context. Let Norm_{S}^{+}(^{L}𝔱_{+}) bet the set of elements of Norm_{S}(^{L}𝔱_{+}) that take positive roots of S_{0} to positive roots. Then

R = S/S^{0} ≃ Norm_{S}^{+}(^{L}𝔱_{+})/^{L}𝔱_{+}.

Let

𝔰_{1} = ^{L}𝔱 + Σ_{⟨μ,α^{∨}⟩=⟨ν,α^{∨}⟩=0}CX_{α^{∨}}

The elements of Norm_{S}^{+}(^{L}𝔱_{+}) take 𝔰_{1} to itself. Let Q be the operator

\frac{1}{|R|}Σ_{R}r

on 𝔱 ⊗ C. Since the centralizer of φ(C^{×}) is connected, S lies in the connected group S_{1} with Lie algebra 𝔰_{1}. Thus by Chevalley’s theorem R is contained in the group generated by the reflections associated to the roots α^{∨} of 𝔰_{1} for which Qα^{∨} = 0.

If α^{∨} is a root of 𝔰_{1} then φ(σ)α^{∨} ≠ −α^{∨}. Suppose φ(σ)α^{∨} ≠ α^{∨}. Then

X_{α^{∨}} + φ(σ)X_{α^{∨}} ≠ 0

and lies in 𝔰. Thus α^{∨} restricted to ^{L}𝔱_{+} defines a root of 𝔰. Since the elements of r stabilize ^{L}𝔱_{+} and each r takes positive roots of ^{L}𝔱_{+} in 𝔰 to positive roots,

Qα^{∨} ≠ 0.

Thus if α^{∨} is a root of 𝔰_{1} then

Qα^{∨} = 0 ⇒ φ(σ)α^{∨} = α^{∨}.

Moreover α^{∨} cannot be a root of 𝔰 and therefore

φ(σ)X_{α^{∨}} = −X_{α^{∨}}.

Finally if Qα^{∨} = 0, Qβ^{∨} = 0 then α^{∨} ± β^{∨} is not a root because φ(σ)X_{α^{∨}+β^{∨}} = φ(σ)[X_{α^{∨}}, X_{β^{∨}}] = [−X_{α^{∨}}, −X_{β^{∨}}] = X_{α^{∨}+β^{∨}} and α^{∨} + β^{∨} would have to be a root of 𝔰. This is inconsistent with

Q(α^{∨} + β^{∨}) = 0.

The set of positive α^{∨} for which ⟨μ, α^{∨}⟩ = ⟨ν, α^{∨}⟩ = 0 and Qα^{∨} = 0 is the strongly orthogonal system needed for Zuckerman’s argument.