Formula for y

November 6, 2009

More details can be found in Some Notes on Parametrizing Representations on the atlas web site. In particular this has formulas for computing Cayley transforms and cross actions in various coordinates.

The main point is the formula for y. See Marc’s email in the Appendix. We have fixed G, G^{∨}, and T, T^{∨} as usual. Let X^{∗} = X^{∗}(T), which is

canonically identified with X_{∗}(T^{∨}).

0.1 Maps of the Weil group

Give (y, λ) satisfying:

1. λ ∈ X^{∗} ⊗ ℂ,

2. y ∈ Norm_{G^{∨Γ}−G^{∨}}(T^{∨}),

3. y^{2} = exp(2πiλ)

Define ϕ by

1. ϕ(z) = z^{λ}\overline{z}^{yλ}

2. ϕ(j) = exp(−πiλ)y. The first line is shorthand for ϕ(e^{w}) = exp(wλ + y\overline{w}λ).

0.2 Character of covers of T (ℝ)

Fix γ ∈ \frac{1}{2}Z^{∗}. Then the two-fold cover T_{γ} of the complex torus T is defined. It is an abelian algebraic group; its identity component is a torus of index 1 or 2. In our applications we often have γ = ρ.

If θ is a Cartan involution of T satisfying θγ − γ ∈ X^{∗} then the inverse image T (ℝ)_{γ} of the real form T (ℝ) of T defined by θ is defined.

Fix θ and (λ, κ) satisfying:

1. λ ∈ X^{∗} ⊗ ℂ,

2. κ ∈ (γ + X^{∗})/(1 − θ)X^{∗},

3. λ + θλ = κ + θκ

In (2) we identify κ with a representative in γ + X^{∗}; any formula involving κ + θκ is well defined.

Associated to (λ, κ) is a genuine character Λ(λ, κ) of T (ℝ)_{γ}. See [1, Proposition 5.8].

0.3 Characters of tori defined by (y, λ)

Fix γ ∈ \frac{1}{2}X^{∗} as in Section 0.2. Suppose we are given (y, λ, y_{0}) with (y, λ) as in Section 0.1, y_{0} in the same fiber as y, and y_{0}^{2} = exp(2πiγ). That is:

1. y, y_{0} ∈ Norm_{G^{∨Γ}−G^{∨}}(T^{∨}),

2. yy_{0}^{−1} ∈ T^{∨},

3. λ ∈ X^{∗} ⊗ ℂ

4. y^{2} = exp(2πiλ),

5. y_{0}^{2} = exp(2πiγ).

Assume θ is a Cartan involution of T satisfying θ^{t} = −Ad(y). This defines a character Λ[y, λ, y_{0}] of T_{γ}(ℝ) as follows. By (2) choose τ ∈ X^{∗}⊗ℂ satisfying y = exp(2πiτ)y_{0}, and let

(1)  κ = λ − (τ − θτ).

We check that conditions (1-3) in Section 0.2 hold. It is obvious that another choice of τ modifies κ by an element of (1 − θ)X^{∗}, and λ + θλ = κ + θκ. Also

(2)

exp(2πiκ) = exp(2πi(λ − τ + θτ)
= y^{2}exp(2πi(−τ + θτ))
= exp(2πi(τ + θ^{∨}τ))y_{0}^{2}exp(2πi(−τ + θτ))
= y_{0}^{2} = exp(2πiγ).

We can therefore define

(3)  Λ[y, λ, y_{0}] = Λ(λ, κ)

0.4 Formula for y

The main point is now a formula inverting (3): given (λ, κ) and y_{0}, give a formula for y, so that Λ[y, λ, y_{0}] = Λ(λ, κ).

To spell things out, suppose we are given (θ, γ, λ, κ) satisfying

1. γ ∈ \frac{1}{2}X^{∗},

2. γ − θγ ∈ X^{∗},

3. λ ∈ X^{∗} ⊗ ℂ,

4. κ ∈ γ + X^{∗}/(1 − θ)X^{∗}

5. λ + θλ = κ + θκ.

Then the genuine character Λ(λ, κ) of T_{γ}(ℝ) is defined as in Section 0.2. Fix y_{0} ∈ Norm_{G^{∨Γ}−G^{∨}}(T^{∨}) so that θ^{t} = −Ad(y) satisfying

(4)  y_{0}^{2} = exp(2πiγ).

Then there exists y satisfying 1. y ∈ Norm_{G^{∨Γ}−G^{∨}}(T^{∨}),

2. y^{2} = exp(2πiλ),

3. yy_{0}^{−1} ∈ T^{∨} so that Λ[y, λ, y_{0}] is defined, and

(5)  Λ[y, λ, y_{0}] = Λ(λ, κ).

We want a formula for y.

Set y = exp(2πiμ)y_{0}, and solve for μ. By (1) we need to choose μ so that

(6)  κ = λ − (μ − θμ).

This is equivalent to κ − λ = μ − θμ, and this is possible since by (5)

(7)  λ − κ = −θ(λ − κ).

In fact we can set

(8)  \begin{aligned}μ & = \frac{1}{4}((λ − κ) − θ(λ − κ)) \\ & = \frac{1}{2}(λ − κ). \end{aligned}

We check this:

(9)  \begin{aligned}λ − (μ − θμ) & = λ − \frac{1}{2}((λ − κ) − θ(λ − κ)) \\ & = λ − \frac{1}{2}(2(λ − κ)) by (7) \\ & = λ − (λ − κ) = κ \end{aligned}

as desired.

This proves:

Lemma 10 Fix θ, γ, and suppose (λ, κ) are given, so Λ(λ, κ) is a genuine character of T_{γ}(ℝ). Fix y_{0} satisfying y_{0}^{2} = exp(2πiγ) and θ^{t} = Ad(y). Let

(11)  y = exp(πi(λ − κ))y_{0}.

Then Λ(λ, κ) = Λ[y, λ, y_{0}].

In particular:

Lemma 12 Suppose we are in the usual atlas setting, with integral infinitesimal character. Fix (x, y) ∈ Z. Let τ^{∨} = p(y), the twisted involution defined by y. Fix y[τ^{∨}] also lying over τ^{∨}, satisfying y[τ^{∨}]^{2} = exp(2πiρ).

Let y = exp(πi(λ − κ))y[τ^{∨}]. Then

(13)  Λ(λ, κ) = Λ(y, λ, y_{0}).

Note this holds for any y[τ^{∨}] with the correct square. While this gives a well defined genuine character of T_{γ}(ℝ), we need to choose y[τ^{∨}] more carefully; this is the basepoint issue.

In this setting λ is always dominant, and this formulation gives a dominant character of T_{γ}(ℝ). This may get moved by an element of the Weyl group when making an actual standard representation.

I’ve used my terminology to avoid going crazy. Marc uses different convention:

my (λ, κ) are Marc’s (γ, λ), and he chooses y[τ^{∨}] = σ_{w}δ_{1}, so in his terms we have

y = exp(πi(γ − λ))σ_{w}δ_{1}

which agrees with what he wrote in his email (see the appendix).

0.5 Basepoint

To be added later…

0.6 Appendix

Subject:    Re: minor matters
From:    "Marc van Leeuwen" <Marc.van-Leeuwen@math.univ-poitiers.fr>
Date:    Thu, November 5, 2009 7:51 am
...
First the identification of the parameters (what is a bit confusing is that all
of them have several "components", some of which overlap with those of others).
There is a strong involution (actually given as KGB element) x, which has as
component an involution theta of the root datum (the other part, the torus part
of x, will play no role for y). Then there is the discrete parameter lambda,
which is a character of the rho-cover of the real Cartan at theta (please
correct me if the wording is wrong) and lies in rho+X^* and is defined modulo
(1-theta)X^*; one can distinguish the free part of lambda (determined by
(1+theta)lambda) and the torsion part of lambda, although the decomposition may
not be mathematically very meaningful. Finally there is the parameter gamma,
taken as a rational weight, which is a representative of the infinitesimal
character; it has a discrete part (projection on +1 eigenspace of theta) that
is bound to the free part of lambda by the relation (1+theta)(gamma-lambda)=0,

and a continuous part (projection on -1 eigenspace of theta) that is also known
as nu. I suppose that the Weyl group can act on everything to give an
equivalent set of parameters, but this aspect confuses me because (1) the
formulas seem to depend on choosing definite representatives and (2) if we are
going to determine a block element (x,y) then certainly we cannot operate by
cross actions on x (and y) and hope to get an equivalent block element;
therefore I will ignore any W action and suppose (x,lambda,gamma) is fixed.
What I need is a formula to find y associated to these parameters, in the form
 y = t.sigma_w.delta where t is a dual torus element whose square is central in
G^\vee(\gamma), the dual group whose root system is the subsystem of the coroot
system of G integral on gamma (this means that all coroots integral on gamma
should have values plus or minus 1 on t); sigma_w is the canonical lift of a
twisted involution in the dual Tits group (in fact w is determined by theta),
and \delta is some fixed basic involution of G^\vee. In order to have a formula
with some chance of being valid I take \delta_1=exp(i\pi\rho)\delta_0 with
delta_0 the pinning-preserving involution of G^\vee (never mind whether that
makes sense). This choice makes sigma_w.delta_1 the base-point for the dual
fiber. The formula I propose is
   y = exp(i\pi(gamma-lambda)) . sigma_w.delta_1
There are a number of sanity checks that this formula seems to pass. The square
of y should be exp(2i\pi(\gamma)); I think I checked that by computation in the
Tits group of G^\vee (more or less). Then y should encompass the torsion
information of lambda (because that is what lives in the dual fiber group that
atlas uses to compute with y); note that taking the difference gamma-lambda
precisely ignores the free part of lambda, and the discrete part of gamma. The
value of y is only meaningful modulo the image of the +1 \Q-eigenspace of
theta^t on \Q\tensor X_* (or the -1 eigenspace of -theta^t if you prefer);
changing just the free part of lambda gives a change precisely in that
direction (this is not really a check, but I find it reassuring nonetheless).
Then we should have coincidence with the formula (4.3) in the sp4forms paper
given at theta=-1 (which implies w=e); that formula has an unspecified binary
parameter epsilon, and setting epsilon = rho-lambda (mod 2X^*) gives my formula
above (this was indeed the starting point for my formula). Finally y should
transform correctly under cross actions and Cayley transforms; since I do not
know how to do these in general at the level of (x,lambda,gamma) I could only
check that for complex cross actions, where things seem to work well. So I

cross my fingers and hope this is the right formula. (By the way, I must have
mentioned this formula at least twice in previous emails, but with perhaps not
enough explanation).
-- Marc

References

[1] Jeffrey Adams and David A. Vogan, Jr. L-groups, projective representations, and the Langlands classification. Amer. J. Math., 114(1):45–138, 1992.