20 DAN BARBASCH AND PAVLE

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This should be equal to τ up to W , and this happens precisely when k = n. (Recall

that W consists of permutations and arbitrary sign changes.)

So we see that for even n, HD(πeven) consists of E(τ), for τ as in (5.18), without

multiplicity other than the global multiplicity [Spin : E(ρ)], while HD(πodd) = 0.

For odd n, the situation is reversed: HD(πeven) = 0, while HD(πodd) consists

of E(τ), with multiplicity [Spin : E(ρ)].

5.6. Type D. Let G = SO(2n, C). We use the usual coordinates.

Since 2λ must be regular integral, in this case the WF-sets of the nontrivial

unipotent representations can only be nilpotents with columns 2b, 2a − 1, 1, where

a + b = n.

By [B], there are two unipotent (so also unitary) representations with 2λ W -

conjugate to (2a−1, 2a−3,... , 1; 2b−2,..., 0); the spherical one, and the one with

lowest K-type (1, 0,..., 0) and parameter

(a − 1/2,..., 3/2, −1/2,b − 1,..., 1, 0) × (a − 1/2,..., 3/2, 1/2,a − 1,..., 1, 0).

Made dominant for the standard positive system,

(5.20) 2λ = (2b − 2, 2b,..., 2a + 2, 2a, 2a − 1, 2a − 2,..., 1, 0).

(When b = a, the parameter is (2a − 1, 2a − 2,..., 1, 0).) Since

(5.21) ρ = (a + b − 1,a + b − 2,..., 1, 0),

we see that

(5.22) τ = 2λ − ρ = (b − a − 1,..., 1, 0, 0,..., 0)

2a

The K-structure of our unipotent representations is given by

(5.23)

μ = (α1,...,α2a, 0,..., 0), αj ∈ N, αj ∈ 2N,

μ = (α1,...,α2a, 0,..., 0), αj ∈ N, αj ∈ 2N + 1.

The argument is similar to type B, but more involved because π is not trivial.

The ﬁrst case is for the spherical representation, the second for the other one.

Therefore,

(5.24) μ − ρ = (α1 − (a + b − 1),...,α2a − (b − a), −(b − a − 1),..., −1, 0).

Since τ has 2a + 1 zeros, the only way μ − ρ can be conjugate to τ is to have

(5.25) α1 = a + b − 1, α2 = a + b − 2,...,α2a = b − a.

Using (5.23), we conclude that for even a, the spherical unipotent representation

has HD equal to E(τ), with multiplicity [Spin : E(ρ)], while the nonspherical

representation has HD = 0. For odd a the situation is reversed: the spherical

representation has HD = 0, while the nonspherical one has HD equal to E(τ),

with multiplicity [Spin : E(ρ)]. (Recall that for type D the Weyl group consists of

permutations combined with an even number of sign changes, but we can use all

sign changes because we are wroking with the full orthogonal group.)

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PANDˇ

ZI

´

C

This should be equal to τ up to W , and this happens precisely when k = n. (Recall

that W consists of permutations and arbitrary sign changes.)

So we see that for even n, HD(πeven) consists of E(τ), for τ as in (5.18), without

multiplicity other than the global multiplicity [Spin : E(ρ)], while HD(πodd) = 0.

For odd n, the situation is reversed: HD(πeven) = 0, while HD(πodd) consists

of E(τ), with multiplicity [Spin : E(ρ)].

5.6. Type D. Let G = SO(2n, C). We use the usual coordinates.

Since 2λ must be regular integral, in this case the WF-sets of the nontrivial

unipotent representations can only be nilpotents with columns 2b, 2a − 1, 1, where

a + b = n.

By [B], there are two unipotent (so also unitary) representations with 2λ W -

conjugate to (2a−1, 2a−3,... , 1; 2b−2,..., 0); the spherical one, and the one with

lowest K-type (1, 0,..., 0) and parameter

(a − 1/2,..., 3/2, −1/2,b − 1,..., 1, 0) × (a − 1/2,..., 3/2, 1/2,a − 1,..., 1, 0).

Made dominant for the standard positive system,

(5.20) 2λ = (2b − 2, 2b,..., 2a + 2, 2a, 2a − 1, 2a − 2,..., 1, 0).

(When b = a, the parameter is (2a − 1, 2a − 2,..., 1, 0).) Since

(5.21) ρ = (a + b − 1,a + b − 2,..., 1, 0),

we see that

(5.22) τ = 2λ − ρ = (b − a − 1,..., 1, 0, 0,..., 0)

2a

The K-structure of our unipotent representations is given by

(5.23)

μ = (α1,...,α2a, 0,..., 0), αj ∈ N, αj ∈ 2N,

μ = (α1,...,α2a, 0,..., 0), αj ∈ N, αj ∈ 2N + 1.

The argument is similar to type B, but more involved because π is not trivial.

The ﬁrst case is for the spherical representation, the second for the other one.

Therefore,

(5.24) μ − ρ = (α1 − (a + b − 1),...,α2a − (b − a), −(b − a − 1),..., −1, 0).

Since τ has 2a + 1 zeros, the only way μ − ρ can be conjugate to τ is to have

(5.25) α1 = a + b − 1, α2 = a + b − 2,...,α2a = b − a.

Using (5.23), we conclude that for even a, the spherical unipotent representation

has HD equal to E(τ), with multiplicity [Spin : E(ρ)], while the nonspherical

representation has HD = 0. For odd a the situation is reversed: the spherical

representation has HD = 0, while the nonspherical one has HD equal to E(τ),

with multiplicity [Spin : E(ρ)]. (Recall that for type D the Weyl group consists of

permutations combined with an even number of sign changes, but we can use all

sign changes because we are wroking with the full orthogonal group.)

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