SOME SMALL UNIPOTENT REPRESENTATIONS OF INDEFINITE ORTHOGONAL GROUPS AND THE THETA CORRESPONDENCE

ANNEGRET PAUL AND PETER TRAPA

Abstract.

1. Introduction

2. Preliminaries and notation

Recall that each K-type μ which occurs in F has associated to it a degree (the minimum degree of polymomials in the μ-isotypic subspace), and that if π and π′ are representations of G̃ and G̃′, respectively, which correspond to each other, then each K-type for G̃ which is of minimal degree in π will occur in H and correspond to a K-type for G̃′ of minimal degree in π′. Since for given choice of n, Sp(2n, ℝ) is a member of many dual pairs, we refer to the degree of a K-type σ for S̃p(2n, ℝ) for the dual pair (Sp(2n, ℝ), O(p, q)) as the (p, q)-degree of σ. As we will see below in Proposition 2.1, The degree of a K-type for O(p, q) depends on the difference p − q only, and the degree of a K-type for O(p, q) is independent of n.

Proposition 2.1. Let p, q, and n be non-negative integers, p_{0} = [\frac{p}{2}], and q_{0} = [\frac{q}{2}]. The correspondence of K-types in the space of joint harmonics H for the dual pair (Sp(2n, ℝ), O(p, q)) is given as follows.

(1) Let

(2.2)  μ = (a_{1}, a_{2}, …, a_{x}, 0, …, 0; ϵ) ⊗ (b_{1}, b_{2}, …, b_{y}, 0, …, 0; η)

be a K-type for O(p, q), with a_{x} > 0 and b_{y} > 0. Then μ occurs in H if and only if n ≥ x + \frac{1−ϵ}{2}(p − 2x) + \frac{1−η}{2} (q − 2y) + y. In that case, μ corresponds to

(2.3)  (\frac{p−q}{2} , \frac{p−q}{2} , …, \frac{p−q}{2} ) + (a_{1}, …, a_{x}, \underbrace{1, …, 1}_{\frac{1−ϵ}{2} (p−2x)}, 0, …, 0, \underbrace{−1, …, −1}_{\frac{1−η}{2} (q−2y)}, −b_{y}, …, −b_{1}).

(2) If a K-type μ for O(p, q) as in (2.2) occurs in the Fock space, then the degree of μ is

(2.4)  ∑_{i=1}^{x} a_{i} + ∑_{i=1}^{y} b_{i} + \frac{1 − ϵ}{2}(p − 2x) + \frac{1 − η}{2} (q − 2y).

For a K-type for S̃p(2n, ℝ) ξ which occurs in F, write

(2.5)  ξ = (\frac{p−q}{2} , \frac{p−q}{2} , …, \frac{p−q}{2} ) + (a_{1}, a_{2}, …, a_{n}).

Then the (p, q)-degree of ξ is ∑_{i=1}^{n} |a_{i}|.

Proposition 2.6. Let p, q, and n be non-negative integers, and suppose π and π′ are genuine irreducible admissible representations of S̃p(2n, ℝ) and O(p, q) which correspond to each other in the correspondence for the dual pair (Sp(2n, ℝ), O(p, q)). If p + q ≤ 2n + 1 then every LKT of π is of minimal (p, q)-degree in π, and if 2n ≤ p + q then every LKT of π′ is of minimal degree in π′.

Proof.

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Proposition 2.7. Let p, q, m, and n be non-negative integers such that p + q ≤ m ≤ \frac{n}{2}. Let π = θ_{m}(11_{p,q}), and π′ = θ_{2m,n}(π).

(1) The K-types of π are precisely those of the form

(2.8)  (\frac{p − q}{2} , …, \frac{p − q}{2} ) + (a_{1}, a_{2}, …, a_{p}, 0, …, 0, −b_{q}, …, −b_{1}),

where a_{i} and b_{i} are non-negative even integers for all i, and each K-type occurs with multiplicity one.

(2) Let a = \frac{p−q+n}{2} − m. Then π′ has a unique LKT Λ, given by

(2.9)  Λ = \{ \begin{matrix}(a, a, …, a, \underbrace{0, …, 0}_{q}; 0, …, 0) & if a ≥ 0 and even; \\ (a, a, …, a, \underbrace{1, …, 1}_{q}; 0, …, 0) & if a > 0 and odd; \\ (0, …, 0; \underbrace{−a, …, −a}_{m−p}, 0, …, 0) & if a ≤ 0 and even; \\ (0, …, 0; \underbrace{−a, …, −a}_{m−p}, \underbrace{1, …, 1}_{p}, 0, …, 0) & if a < 0 and odd. \end{matrix}

Proof. The first part of the proposition is Proposition 2.1, together with Corollary 2.7(c), of [22]. Let

(2.10)  σ = (\frac{p−q}{2} , …, \frac{p−q}{2} ) + (a_{1}, a_{2}, …, a_{p}, 0, …, 0, −b_{q}, …, −b_{1}),

be a K-type of π. Then

(2.11)  σ = (\frac{2m−n}{2}, …, \frac{2m−n}{2}) + (a + a_{1}, …, a + a_{p}, a, …, a, a − b_{q}, …, a − b_{1}),

and the (2m, n)-degree of σ is

(2.12)  d = ∑_{i=1}^{p} |a + a_{i}| + ∑_{i=1}^{q} |a − b_{q}| + (m − p − q)|a|.

Suppose a is even. Then the degree of σ is minimized if a_{i} = 0 and b_{i} = a for all i if a ≥ 0, and by choosing a_{i} = −a and b_{i} = 0 for all i if a is negative. So π has a unique K-type σ_{0} of minimal degree. The K-type Λ corresponding to σ_{0} in the space of joint harmonics (see Proposition 2.1) must then be the unique LKT of π′ by Proposition 2.6.

If a is odd then π has more than one K-type of minimal (p, q)-degree: for a < 0 choose b_{i} = 0 for all i, and a_{i} = −a + 1 or −a − 1 in such a way that the resulting weight is dominant. The corresponding K-types for O(p, q) are then those of the form

(2.13)  Λ_{k,l} = (\underbrace{1, …, 1}_{k}, 0, …, 0; \underbrace{−a, …, −a}_{m−p}, \underbrace{1, …, 1}_{l}, 0, …, 0)

with k +l = p. By Proposition 2.6, the LKT’s of π′ are precisely those Λ_{k,l} for which the Vogan-norm ([46] Definition 5.4.18) of Λ_{k,l},

(2.14)  ||Λ_{k,l}|| =< Λ_{k,l} + 2ρ_{c}, Λ_{k,l} + 2ρ_{c} >

is minimal. Since the quantity < Λ_{k,l}, Λ_{k,l} > only depends on k + l = p, we can minimize the quantity n_{k,l} = \frac{1}{2}(||Λ_{k,l}||− < 2ρ_{c}, 2ρ_{c} > − < Λ_{k,l}, Λ_{k,l} >) instead. If r = [\frac{n}{2}] then 2ρ_{c} = (2m − 2, 2m − 4, …, 2, 0; n − 2, n − 4, …, n − 2r). So

(2.15)  n_{k,l} = ∑_{i=1}^{k}(2m − 2i) + ∑_{i=1}^{m−p}(−a)(n − 2i) + ∑_{i=1}^{l}(n − 2m + 2p − 2i).

The second sum is independent of k and l, and it is clear that n_{k,l} will be minimized by k = p, l = 0 if 2m ≤ n − 2m + 2p, and by k = 0, l = p if 2m ≥ n − 2m + 2p. It remains to show that we must have 4m > n + 2p. We know that a = \frac{p−q+n}{2} − m < 0, so p − q + n < 2m, Moreover, we have assumed that p + q ≤ m, so we get 2p + n < 3m ≤ 4m, and we are done.

The case a > 0 is much easier and left to the reader.

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Department of Mathematics, Western Michigan University, Kalamazoo, MI 49008

E-mail address: annegret.paul@wmich.edu

Department of Mathematics, University of Utah, Salt Lake City, UT 84112

E-mail address: ptrapa@math.utah.edu