Journal of Lie Theory Volume 11 (2001) 23–55 \begin{matrix}c \\ ◯ \end{matrix} 2001 Heldermann Verlag Berlin

On some degenerate principal series representations of O(p, 2)

Takashi Fujimura

Communicated by G. Olshanski

Abstract. We consider representations of O(p, 2) (p > 4) induced from one-dimensional representations of a maximal parabolic subgroup. We first decompose them into K-types using Stiefel harmonics theory, then write down the actions of the noncompact part. Now the reducibility and the unitarizability of the irreducible constituents are deduced.

1. Introduction

The importance of parabolic induction in representation theory is widely acknowledged. Representations induced from one-dimensional representations of maximal parabolic subgroups are sometimes called “degenerate principal series”. This paper treats degenerate principal series of O(p, 2) in a manner analogous to those used in [9]. That is, we first decompose the representations into irreducibles of the maximal compact subgroup, and then calculate the infinitesimal action of the non-compact part. We get informations such as reducibility and unitarizability. The computations are done using Euler operators, thus avoiding the use of explicit formulae of the vectors.

For other settings (non p-adic), see for example [18] for SO(p, q), [13] for U(p, q), [14] and [26] for Sp(p, q), [24] for SU(2, 2), [15] for Mp(n, ℝ), [10] for Spin_{0}(n, n), SU(n, n) and Sp(n, n), [4] for SU(n, n, 𝔽), [1] for GL(2n, ℝ), [9] for SO(p, q), U(p, q) and Sp(p, q), [11], [22], [23], [20] and [27] for Hermitian symmetric spaces of tube type, [19] and [16] for U(n, n), [17] for Sp(2n, ℝ),

[8] for GL(n, ℝ) and GL(n, ℂ), and [3] for rank one case, suggesting further applications.

A special case of [21] gives a sufficient condition for our representations to be complementary series. Our result implies that the condition is also necessary.

Now we describe our result in more details. Let G = O(p, 2) where p > 4. We consider the maximal parabolic subgroup S of G which has Levi part isomorphic to GL(2, ℝ) × O(p − 2). G has only two maximal parabolic subgroups, namely, this one and the one treated by [9].

ISSN 0949–5932 / $2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag

In the Langlands decomposition S = MAN, A is one-dimensional, and its character is parametrized by a complex number c. (The precise definition will be given later.) We regard this character as a representation of S where M and N act trivially. We induce this representation to G, and this is what we will work on. We denote it by S^{c}(X^{+}). Its Harish-Chandra module is decomposed into the sum of blocks, each block invariant under the action of the maximal compact subgroup K = O(p)×O(2) of G. The blocks are parametrized by two non-negative integers l_{1}, l_{2} satisfying l_{1} ≥ l_{2}, which we identify with points (l_{1}, l_{2}) in a plane. Here is the main result.

Theorem. The Harish-Chandra module of S^{c}(X^{+}) is irreducible and unitary if and only if c′ = −2c − p + 1 ∈ iℝ or −p + 2 < c < −1.

[figure: -p+3; p-3; 0; Re c′; Im c′]

When it is reducible(i.e.c ∈ ℤ), its irreducible constituents are as follows.

• c ≤ −p : 1 non-unitary finite-dimensional quotient, 1 non-unitary subquotient, 1 unitary submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; −c − p + 2; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; −c − p + 3; unit.sub.]

• c = −p + 1 : 1 unitary 1-dimensional quotient, 1 non-unitary subquotient, 1 unitary submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; 1; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; 2; unit.quot.; unit.sub.]

• c = −p + 2 : 1 unitary quotient, 1 unitary submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; 0; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; 1; unit.sub.; unit.quot.]

• c = −1 : 1 unitary submodule, 1 unitary quotient.

• c ≥ 1 : 1 non-unitary finite-dimensional submodule, 1 non-unitary subquotient, 1 unitary quotient.

[figure: l_{1}; l_{2}; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; 0; unit.quot.; unit.sub.]

[figure: l_{1}; l_{2}; ⇐; ⇐; ⇐; ⇐; ⇐; ⇐; c; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; c + 1; unit.quot.]

• c = 0 : 1 unitary 1-dimensional submodule, 1 non-unitary subquotient, 1 unitary quotient.

[figure: l_{1}; l_{2}; ⇐; ⇐; ⇐; ⇐; ⇐; ⇐; 0; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; 1; unit.sub.; unit.quot.]

This completes the statement of the theorem.

In Section 2, we define the representation in more details, and realize it concretely on the space of functions satisfying some relations.

In Section 3, there is a quick review on the special case of Stiefel harmonics theory (which is generalized spherical harmonics theory).

In Section 4, the representation is identified with the space of functions on a Stiefel manifold. Using Section 3, we decompose the representation into the sum of irreducible K -modules, or “K -types”, where K is the maximal compact subgroup of G.

In Section 5, the action of 𝔭 is computed in terms of differential operators. Here, 𝔤 = 𝔨 + 𝔭 is the Cartan decomposition of the complexified Lie algebra. This is done essentially by writing explicitly a projection operator which extracts the harmonic polynomial part from an arbitrary polynomial, described in Section 6.

In Section 7, we deduce the reducibility. On points where it is reducible, the composition series is determined. We see that the representation is built up from “blocks”, forming a lattice in a plane. These blocks are separated by lines, each territory corresponding to an irreducible constituent.

Finally in Section 8, we investigate unitarizability. The complementary series is determined. We fix a standard inner product on each K -type. The ratios of an inner product restricted to K -types compared with the standard ones are related to each other by recursive relations. The existence of these ratios tells us whether or not the irreducible constituent is unitarizable.

I would like to thank my advisor Prof. H. Matumoto for many helpful suggestions. I was also informed that he had recently generalized my results.

2. Construction of the representation

Write I_{k} for the identity matrix of size k. Let

G = O(p, 2) = \{g ∈ GL(p + 2, ℝ) | ^{t}g (\begin{matrix}I_{p} & 0 \\ 0 & −I_{2} \end{matrix}) g = (\begin{matrix}I_{p} & 0 \\ 0 & −I_{2} \end{matrix})\} ,

K = \{(\begin{matrix}A & 0 \\ 0 & C \end{matrix}) | A ∈ O(p), C ∈ O(2)\} ,

𝔭_{0} = \{(\begin{matrix}0 & B \\ ^{t}B & 0 \end{matrix}) | B ∈ M(p, 2; ℝ)\} .

The Cartan decomposition is G = K exp 𝔭_{0}. We use German letters 𝔤_{0}, 𝔨_{0}, … for Lie algebras corresponding to G, K, …, and 𝔤, 𝔨, … for their complexifications.

G acts by left multiplication on ℝ^{p+2} ⊕ℝ^{p+2} (the standard representation), and we define:

v_{0} = (\begin{matrix}I_{2} \\ 0_{p−2,2} \\ I_{2} \end{matrix}) ∈ ℝ^{p+2} ⊕ ℝ^{p+2},
X = Gv_{0},
S = \{p ∈ G | ∃X ∈ GL(2, ℝ) s.t. pv_{0} = (\begin{matrix}X \\ 0 \\ X \end{matrix})\} .

S is a parabolic subgroup of G. (It is the stabilizer of a two-dimensional subspace in ℝ^{p+2}.)

We define a map χ : G → ℝ^{×} by

χ(g) = det C if gv_{0} = \binom{A_{1}}{C} {\}p \atop \}2}.

The restriction of χ to S is a group homomorphism.

For any complex number c, we define characters χ_{c}^{±} of S:

χ_{c}^{+}(p) = | det χ(p)|^{c},
χ_{c}^{−}(p) = sign(χ(p))| det χ(p)|^{c} (∀p ∈ S).

We study the Harish-Chandra module of:

S^{c±}(X) = {F ∈ C^{∞}(X) | F (gpv_{0}) = χ_{c}^{±}(p)F (gv_{0}) (∀g ∈ G, ∀p ∈ S)} .

These are G-modules by left translation. We can interpret these as (generalized) principal series representations as follows.

In the Langlands decomposition S = MAN, we can take as A:

A = \{a = exp H(h) | H(h) = ( \begin{matrix}& & h \\ & & & h \\ h \\ & h \end{matrix} ) , h ∈ ℝ, zeros elsewhere\} , which is one dimensional. For any complex number c′, we define a linear functional ν_{c′} on the Lie algebra of A:

ν_{c′}(H(h)) = c′h (∀h ∈ ℝ).

Since

χ_{c}^{+}(man) = exp(2ch),
χ_{c}^{−}(man) = sign(χ(m)) exp(2ch)
(∀m ∈ M, ∀a = exp H(h) ∈ A, ∀n ∈ N),

we get

S^{c+}(X) ≅ Ind_{MAN}^{G}(1_{M} ⊗ exp ν_{c′} ⊗ 1_{N}),
S^{c−}(X) ≅ Ind_{MAN}^{G}(sign χ|_{M} ⊗ exp ν_{c′} ⊗ 1_{N}).

Here we take 2c = −(c′ + p − 1) and 1_{M}, 1_{N} denote the trivial representations. The induction on the right hand side is shifted by ρ = ν_{p−1} to preserve unitarity.

Next, we pay attention to the connectivity. Let

G^{+} = O^{+}(p, 2)
= (O(p) × SO(2)) exp 𝔭_{0}
= {g ∈ G|χ(g) > 0} ,
k_{C} = (\begin{matrix}I_{p} \\ & 0 & 1 \\ & 1 & 0 \end{matrix}) ∈ O(p, 2) − O^{+}(p, 2),
X^{+} = O^{+}(p, 2)v_{0}
= {gv_{0} ∈ X|χ(g) > 0} ,
X^{−} = k_{C}O^{+}(p, 2)v_{0}
= {gv_{0} ∈ X|χ(g) < 0} .

O^{+}(p, 2) acts on X^{±}. Since X = X^{+} ⊔ X^{−} (disjoint union), we can define

sign_{X}(x) = \begin{cases}1 & (x ∈ X^{+}) \\ −1 & (x ∈ X^{−}) \end{cases}

which is an O^{+}(p, 2)-invariant function on X.

Elements in S^{c±}(X) are determined by their restrictions on X^{+}:

S^{c}(X^{+}) = \{F ∈ C^{∞}(X^{+}) | {F ((\begin{matrix}A_{1}X \\ CX \end{matrix})) = (det X)^{c}F ((\begin{matrix}A_{1} \\ C \end{matrix})) \atop ∀X ∈ GL^{+}(2, ℝ), ∀ (\begin{matrix}A_{1} \\ C \end{matrix}) ∈ X^{+}} \} .

F ∈ S^{c}(X^{+}) is extended to X^{−} by

F ((\begin{matrix}A_{1} \\ C \end{matrix}) ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )) = ±F ((\begin{matrix}A_{1} \\ C \end{matrix})) .

Thus, multiplication by sign_{X} is an O^{+}(p, 2)-module isomorphism between S^{c+}(X) and S^{c−}(X).

By O(p, 2) = O^{+}(p, 2) ∪ k_{C}O^{+}(p, 2), we first consider the decomposition of S^{c}(X^{+}) as an O^{+}(p, 2)-module, and then see the effects of k_{C}.

3. Stiefel harmonics theory

In Howe-Tan’s case [9], the module is identified with a space of functions on SO(p)/SO(p−1) ≃ S^{p−1}, and the theory of spherical harmonics is used for further analysis. In our case we have instead SO(p)/SO(p − 2), a Stiefel manifold. But we have at hand the Stiefel version of harmonics theory by Gelbart [6] and Ton-That [25]. We quote the results on SO(p, ℂ)/SO(p − 2, ℂ).

Theorem 3.1. We identify X_{p} = SO(p, ℂ) (\begin{matrix}I_{2} \\ 0_{p−2,2} \end{matrix}) with SO(p, ℂ)/SO(p −

2, ℂ). The space of functions on X_{p} is an SO(p, ℂ)-module by left translation.

We denote by P the space of polynomials on ℂ^{p} ⊕ ℂ^{p}, with coordinates x_{i}, y_{i}(1 ≤ i ≤ p). Then,

(a) The space of SO(p, ℂ)-finite vectors in C^{∞}(X_{p})is

A(X_{p}) = {restriction to X_{p} of a polynomial ∈ P } .

(b) A(X_{p}) ≅ H by restriction. Here,

H = {h ∈ P | Δ_{x^{2}}h = 0, Δ_{y^{2}}h = 0, Δ_{xy}h = 0} ,

Δ_{x^{2}} = ∑_{i=1}^{p} \frac{∂^{2}}{∂x_{i}^{2}} , Δ_{y^{2}} = ∑_{i=1}^{p} \frac{∂^{2}}{∂y_{i}^{2}} , Δ_{xy} = ∑_{i=1}^{p} \frac{∂^{2}}{∂x_{i}∂y_{i}} .

(c) From now on, assume p > 4. If we denote by J the space of SO(p, ℂ) invariant polynomials, then

P ≅ J ⊗ H.

(d) J is generated by ξ_{x^{2}}, ξ_{y^{2}}, ξ_{xy}. Here,

ξ_{x^{2}} = ∑_{i=1}^{p} x_{i}^{2}, ξ_{y^{2}} = ∑_{i=1}^{p} y_{i}^{2}, ξ_{xy} = ∑_{i=1}^{p} x_{i}y_{i}.

(e) H has also a GL(2, ℂ)-action by right translation. Then, we have the decomposition into irreducibles as SO(p, ℂ) × GL(2, ℂ)-modules (multiplicity free):

H ≅ ⨁_{{l_{1},l_{2}∈ℤ \atop l_{1}≥l_{2}≥0}} H_{(l_{1},l_{2})},

H_{(l_{1},l_{2})} ≅ σ_{(l_{1},l_{2},0,... ,0)}^{p} ⊗ ρ_{(l_{1},l_{2})}^{2},

where σ_{(l_{1},l_{2},0,... ,0)}^{p} is the irreducible representation of SO(p, ℂ) with highest weight

(\overbrace{l_{1}, l_{2}, 0, …, 0}^{[p/2]}) and ρ_{(l_{1},l_{2})}^{2} is the irreducible representation of GL(2, ℂ) with highest weight (l_{1}, l_{2}).

In fact, their results are stated more generally for SO(p)/SO(p − n) (n < [p/2]). Part (a) is due to Helgason [7] and Part (e) to Kashiwara and Vergne [12]. For general n, see [5].

4. Decomposition into K -types

Let K_{0} = SO(p) × SO(2), K_{1} = O(p) × SO(2). We first decompose S^{c}(X^{+}) into the sum of irreducible K_{0}-modules.

Lemma 4.1. We have an isomorphism of K_{0}-modules:

S^{c}(X^{+}) ≅ C^{∞}(SO(p)/SO(p − 2)).

The action of K_{0} on the right hand side is given by

((k_{1}, k_{2}) h) (x) = h(k_{1}^{−1}xk_{2})

for

\begin{matrix}∀(k_{1}, k_{2}) ∈ SO(p) × SO(2), ∀h ∈ C^{∞}(SO(p)/SO(p − 2)), \\ ∀x ∈ SO(p)/SO(p − 2) = SO(p) ( \begin{matrix}I_{2} \\ 0 \end{matrix} ) . \end{matrix}

In particular, the Harish-Chandra module of S^{c}(X^{+}) is

S^{c}(X^{+})_{K−finite} ≅ A(SO(p)/SO(p − 2))

≅ H

= ⨁_{{l_{1},l_{2}∈ℤ \atop l_{1}≥l_{2}≥0}} H_{(l_{1},l_{2})}.

We denote this isomorphism by j : H ⟶^{≅} S^{c}(X^{+}).

Proof. Since G^{+} = K_{1}MAN, the elements of S^{c}(X^{+}) are determined by their values on K_{1}v_{0}. Thus,

S^{c}(X^{+}) = {F ∈ C^{∞}(G^{+}v_{0}) | F (gpv_{0}) = χ_{c}(p)F (gv_{0}) (∀g ∈ G^{+}, ∀p ∈ S ∩ G^{+})}

= {F ∈ C^{∞}(Kv_{0}) | F (kpv_{0}) = F (kv_{0}) (∀k ∈ K_{1}, ∀p ∈ S ∩ K_{1})}

≅ \{F ∈ C^{∞}(O(p) ( \begin{matrix}I_{2} \\ 0 \end{matrix} ) × SO(2)) | \begin{matrix}F ((A ( \begin{matrix}X \\ 0 \end{matrix} ) , CX)) = F ((A ( \begin{matrix}I_{2} \\ 0 \end{matrix} ) , C)) \\ (∀A ∈ O(p), ∀C ∈ SO(2), ∀X ∈ SO(2)) \end{matrix}\}

= \{F ∈ C^{∞}(O(p) ( \begin{matrix}I_{2} \\ 0 \end{matrix} ) × SO(2)) | \begin{matrix}F ((A ( \begin{matrix}I_{2} \\ 0 \end{matrix} ) , C)) = F ((AC^{−1}, I_{2})) \\ (∀A ∈ O(p), ∀C ∈ SO(2)) \end{matrix}\}

≅ C^{∞}(O(p) ( \begin{matrix}I_{2} \\ 0 \end{matrix} )).

Here, we used the fact that χ_{c} is trivial on

S ∩ K_{1} = \{( \begin{matrix}X \\ & Y \\ & & X \end{matrix} ) | X ∈ SO(2), Y ∈ O(p − 2)\} .

The rest is Thm.3.1.

We want to decompose the irreducible SO(p) × GL(2, ℝ)-module H_{(l_{1},l_{2})} into the sum of irreducible SO(p) × SO(2)-modules. For this purpose, we complexify the actions and apply the Caley transform.

Lemma 4.2. Let J_{C} = \frac{1}{\sqrt{2}} (\begin{matrix}1 & i \\ i & 1 \end{matrix}) . The map ϵ : P → P defined by

(ϵ(f))(x) = f(xJ_{C}^{−1})

restricts to an isomorphism of SO(p, ℂ) × GL(2, ℂ)-modules:

H_{(l_{1},l_{2})} ⟶_{≅}^{ϵ} H_{(l_{1},l_{2})}

if we define the GL(2, ℂ)-action on the image ϵ(H) by

\begin{matrix}(gf)(x) = f(xJ_{C}^{−1}gJ_{C}) \\ (g ∈ SO(2, ℂ), f ∈ ϵ(H), x ∈ ℂ^{p} ⊕ ℂ^{p}). \end{matrix}

Proof. That ϵ preserves the space of harmonic functions follows from the chain rule in differentiation. Since left translations and right translations commute, the rest is obvious.

The set of diagonal matrices T ⊆ GL(2, ℂ) is a Cartan subgroup. We decompose H_{(l_{1},l_{2})} into weight spaces:

H_{(l_{1},l_{2})} = ⨁_{m=0}^{l_{1}−l_{2}} H_{(l_{1},l_{2}),m}.

H_{(l_{1},l_{2}),m} has weight(l_{1} − m, l_{2} + m) with respect to T. Note that H_{(l_{1},l_{2}),m} consists of harmonic polynomials which are

homogeneous of degree l_{1} − m with respect to x_{1}, …, x_{p}, homogeneous of degree l_{2} + m with respect to y_{1}, …, y_{p}.

Lemma 4.3.

S^{c}(X^{+}) ≅ ϵ(H) = ⨁_{{l_{1},l_{2}∈ℤ \atop l_{1}≥l_{2}≥0}} ⨁_{m=0}^{l_{1}−l_{2}} H_{(l_{1},l_{2}),m}

is the irreducible decomposition of S^{c}(X^{+}) as an SO(p, ℂ) × SO(2, ℂ)-module. In particular, it is multiplicity free.

Proof. Since J_{C}^{−1}SO(2, ℂ)J_{C} = \{(\begin{matrix}a & 0 \\ 0 & a^{−1} \end{matrix}) | a ∈ ℂ\}, the action of SO(2, ℂ)

on ϵ(H) is reduced to the action of T, which is determined by weights. The lemma follows by Lemma4.1 and Lemma 4.2.

Lemma 4.4.

S^{c}(X^{+}) ≅ϵ(H)
= ⨁_{\begin{matrix}l_{1},l_{2}∈ℤ \\ l_{1}≥l_{2}≥0 \\ l_{1}−l_{2}odd \end{matrix}} ( ⨁_{m=0}^{\frac{l_{1}−l_{2}−1}{2}} (H_{(l_{1},l_{2}),m} ⊕ H_{(l_{1},l_{2}),l_{1}−l_{2}−m}))

⊕ ⨁_{\begin{matrix}l_{1},l_{2}∈ℤ \\ l_{1}≥l_{2}≥0 \\ l_{1}−l_{2}even \end{matrix}} ( ⨁_{m=0}^{\frac{l_{1}−l_{2}}{2} −1} (H_{(l_{1},l_{2}),m} ⊕ H_{(l_{1},l_{2}),l_{1}−l_{2}−m}) ⊕ H_{(l_{1},l_{2}), \frac{l_{1}−l_{2}}{2}} )

is the irreducible decomposition of S^{c}(X^{+}) as an O(p, ℂ) × O(2, ℂ)-module. In particular, it is multiplicity free.

Proof. This follows from the fact that the action of k_{C} (∈ G − G^{+}) is substitution of the coordinates between the x_{i}’s and y_{i}’s.

5. 𝔭-action

The next thing to do is to calculate the action of the non compact part. We transfer harmonic polynomials to S^{c}(X^{+}) by the K -isomorphism Φ = j ∘ ϵ^{−1}, apply the Lie algebra actions, and pull-back again to ϵ(H).

Proposition 5.1. For (s_{1}, s_{2}, s_{3}) ∈ {±(1, 0, 0), ±(1, 0, 1), ±(0, 1, −1), ±(0, 1, 0)}, there exist linear maps

T_{(l_{1},l_{2}),m}^{s_{1},s_{2},s_{3}} : 𝔭 ⊗ H_{(l_{1},l_{2}),m} → H_{(l_{1}+s_{1},l_{2}+s_{2}),m+s_{3}},

satisfying the followings:

• They do not depend on c,

• They are not zero maps (when the target space is not zero), and

• For ∀X ∈ 𝔭, ∀h ∈ H_{(l_{1},l_{2}),m},

X(Φ(h)) =(c − l_{1}) (Φ(T_{(l_{1},l_{2}),m}^{(1,0,0)}(X ⊗ h)) + Φ(T_{(l_{1},l_{2}),m}^{(1,0,1)}(X ⊗ h)))
+(c − l_{2} + 1) (Φ(T_{(l_{1},l_{2}),m}^{(0,1,−1)}(X ⊗ h)) + Φ(T_{(l_{1},l_{2}),m}^{(0,1,0)}(X ⊗ h)))
+(c + l_{1} + p − 2) (Φ(T_{(l_{1},l_{2}),m}^{(−1,0,−1)}(X ⊗ h)) + Φ(T_{(l_{1},l_{2}),m}^{(−1,0,0)}(X ⊗ h)))
+(c + l_{2} + p − 3) (Φ(T_{(l_{1},l_{2}),m}^{(0,−1,0)}(X ⊗ h)) + Φ(T_{(l_{1},l_{2}),m}^{(0,−1,1)}(X ⊗ h))) .

We need several lemmas to prove this. Anyone who is willing to admit this proposition may safely skip the rest of this section and the next one. The proof itself is given in the end of the next section.

We fix some notations:

E_{x}^{x} = ∑_{i=1}^{p} x_{i} \frac{∂}{∂x_{i}} ,  E_{y}^{y} = ∑_{i=1}^{p} y_{i} \frac{∂}{∂y_{i}} ,

E_{y}^{x} = ∑_{i=1}^{p} x_{i} \frac{∂}{∂y_{i}} ,  E_{x}^{y} = ∑_{i=1}^{p} y_{i} \frac{∂}{∂x_{i}} .

These are infinitesimal translations corresponding to the GL(2, ℂ)-action on H. Under these notations,

H_{(l_{1},l_{2}),0} = {h_{0} ∈ H | E_{y}^{x}h_{0} = 0} ,
H_{(l_{1},l_{2}),m} = {(E_{x}^{y})^{m}h_{0} | h_{0} ∈ H_{(l_{1},l_{2}),0}} .
For a basis of 𝔭 = \{(\begin{matrix}0 & B \\ ^{t}B & 0 \end{matrix})\} , we may choose
{X_{i,k} = −E_{i,k} − E_{k,i}|i = 1, 2, k = 1, 2, …, p}

where E_{i,k} are matrix units.

Lemma 5.2. The explicit formula for j is

j(h) ((\begin{matrix}A_{1} \\ C \end{matrix})) = (det C)^{c}h(A_{1}C^{−1})  (∗)

for h ∈ H, (\begin{matrix}A_{1} \\ C \end{matrix}) ∈ O^{+}(p, q) ( \begin{matrix}I_{2} \\ 0 \\ I_{2} \end{matrix} ) . We extend j to a map:

j : P → C^{∞} (\{( \begin{matrix}A_{1} \\ C \end{matrix} ) ∈ ℝ^{p+2} ⊕ ℝ^{p+2}| det C ≠ 0\})

by the same formula (∗). Then, for 1 ≤ k ≤ p, h ∈ H,

X_{1,k}j(h) = j (\frac{∂h}{∂x_{k}} + cx_{k}h − x_{k}E_{x}^{x}h − y_{k}E_{y}^{x}h) ,

X_{2,k}j(h) = j (\frac{∂h}{∂y_{k}} + cy_{k}h − y_{k}E_{y}^{y}h − x_{k}E_{x}^{y}h) .

(In general, the polynomials in the parentheses of j in the right hand sides are no longer harmonic.)

Proof. Let (\begin{matrix}x_{1} & y_{1} \\ ⋮ & ⋮ \\ x_{p} & y_{p} \\ z_{1} & w_{1} \\ z_{2} & w_{2} \end{matrix}) be the coordinates for ℝ^{p+2}⊕ℝ^{p+2}. Then, for a function

F on ℝ^{p+2} ⊕ ℝ^{p+2}, the action of X_{i,k} in terms of differential operators is

X_{i,k}F = z_{i} \frac{∂F}{∂x_{k}} + x_{k} \frac{∂F}{∂z_{i}} + w_{i} \frac{∂F}{∂y_{k}} + y_{k} \frac{∂F}{∂w_{i}} .

Fujimura 33

We want to compute this for F = j(h). In terms of coordinates,

F ((\begin{matrix}x_{1} & y_{1} \\ ⋮ & ⋮ \\ x_{p} & y_{p} \\ z_{1} & w_{1} \\ z_{2} & w_{2} \end{matrix})) = D^{c}h (( ⋮  ⋮_{\frac{x_{p}w_{2}−y_{p}z_{2}}{D} \frac{−x_{p}w_{1}+y_{p}z_{1}}{D}}^{\frac{x_{1}w_{2}−y_{1}z_{2}}{D} \frac{−x_{1}w_{1}+y_{1}z_{1}}{D}} )) ,

where D = det (\begin{matrix}z_{1} & w_{1} \\ z_{2} & w_{2} \end{matrix}) = z_{1}w_{2} − z_{2}w_{1}.

Using the chain rule,

\frac{∂F}{∂x_{k}} ((\begin{matrix}A_{1} \\ C \end{matrix})) =D^{c} (\frac{w_{2}}{D} \frac{∂h}{∂x_{k}} (A_{1}C^{−1}) − \frac{w_{1}}{D} \frac{∂h}{∂y_{k}} (A_{1}C^{−1})) ,
\frac{∂F}{∂y_{k}} ((\begin{matrix}A_{1} \\ C \end{matrix})) =D^{c} (−\frac{z_{2}}{D} \frac{∂h}{∂x_{k}} (A_{1}C^{−1}) + \frac{z_{1}}{D} \frac{∂h}{∂y_{k}} (A_{1}C^{−1})) ,
\frac{∂F}{∂z_{1}} ((\begin{matrix}A_{1} \\ C \end{matrix})) =cw_{2}D^{c−1}h(A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{−(x_{i}w_{2} − y_{i}z_{2})w_{2}}{D^{2}} \frac{∂h}{∂x_{i}} (A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{(x_{i}w_{2} − y_{i}z_{2})w_{2}}{D^{2}} \frac{∂h}{∂y_{i}} (A_{1}C^{−1}),

\frac{∂F}{∂w_{1}} ((\begin{matrix}A_{1} \\ C \end{matrix})) =c(−z_{2})D^{c−1}h(A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{−(x_{i}w_{2} − y_{i}z_{2})(−z_{2})}{D^{2}} \frac{∂h}{∂x_{i}} (A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{−(x_{i}w_{2} − y_{i}z_{2})z_{1}}{D^{2}} \frac{∂h}{∂y_{i}} (A_{1}C^{−1}),

\frac{∂F}{∂z_{2}} ((\begin{matrix}A_{1} \\ C \end{matrix})) =c(−w_{1})D^{c−1}h(A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{(x_{i}w_{1} − y_{i}z_{1})w_{2}}{D^{2}} \frac{∂h}{∂x_{i}} (A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{−(−x_{i}w_{1} + y_{i}z_{1})(−w_{1})}{D^{2}} \frac{∂h}{∂y_{i}} (A_{1}C^{−1}),

\frac{∂F}{∂w_{2}} ((\begin{matrix}A_{1} \\ C \end{matrix})) =cz_{1}D^{c−1}h(A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{−(x_{i}w_{1} − y_{i}z_{1})z_{2}}{D^{2}} \frac{∂h}{∂x_{i}} (A_{1}C^{−1})

+ ∑_{i=1}^{p} \frac{−(−x_{i}w_{1} + y_{i}z_{1})z_{1}}{D^{2}} \frac{∂h}{∂y_{i}} (A_{1}C^{−1}).

Thus,

(X_{1,k}F) ((\begin{matrix}A_{1} \\ C \end{matrix}))
= (z_{1} \frac{∂F}{∂x_{k}} + x_{k} \frac{∂F}{∂z_{1}} + w_{1} \frac{∂F}{∂y_{k}} + y_{k} \frac{∂F}{∂w_{1}} ) ((\begin{matrix}A_{1} \\ C \end{matrix}))
=D^{c} \frac{z_{1}w_{2} − z_{2}w_{1}}{D} \frac{∂h}{∂x_{k}} (A_{1}C^{−1}) + cD^{c−1}(x_{k}w_{2} − y_{k}z_{2})h(A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{−(x_{i}w_{2} − y_{i}z_{2})(x_{k}w_{2} − y_{k}z_{2})}{D^{2}} \frac{∂h}{∂x_{i}} (A_{1}C^{−1})

+ D^{c} ∑_{i=1}^{p} \frac{(x_{i}w_{2} − y_{i}z_{2})(x_{k}w_{1} − y_{k}z_{1})}{D^{2}} \frac{∂h}{∂y_{i}} (A_{1}C^{−1})

=D^{c}(\frac{∂h}{∂x_{k}} (A_{1}C^{−1}) + c\frac{x_{k}w_{2} − y_{k}z_{2}}{D} (h(A_{1}C^{−1}))

− ∑_{i=1}^{p} \frac{x_{i}w_{2} − y_{i}z_{2}}{D} \frac{x_{k}w_{2} − y_{k}z_{2}}{D} (\frac{∂h}{∂x_{i}} (A_{1}C^{−1}))

+ ∑_{i=1}^{p} \frac{x_{i}w_{2} − y_{i}z_{2}}{D} \frac{x_{k}w_{1} − y_{k}z_{1}}{D} (\frac{∂h}{∂y_{i}} (A_{1}C^{−1})) )

=D^{c}(\frac{∂h}{∂x_{k}} (A_{1}C^{−1}) + c (x_{k}h) (A_{1}C^{−1})

− ∑_{i=1}^{p} (x_{i}x_{k} \frac{∂h}{∂x_{i}} ) (A_{1}C^{−1}) + ∑_{i=1}^{p} (x_{i}(−y_{k})\frac{∂h}{∂y_{i}} ) (A_{1}C^{−1}))

=j (\frac{∂h}{∂x_{k}} + cx_{k}h − x_{k}E_{x}^{x}h − y_{k}E_{y}^{x}h) .

X_{2,k}F is similar.

Lemma 5.3. We extend Φ = j ∘ ϵ^{−1} to a map from P. For h ∈ H_{(l_{1},l_{2}),m} ⊆ ϵ(H),1 ≤ k ≤ p,

\begin{aligned}X_{1,k}Φ(h) = Φ(( & \frac{1}{\sqrt{2}} \frac{∂}{∂x_{k}} + \frac{i}{\sqrt{2}} \frac{∂}{∂y_{k}} + \frac{1}{\sqrt{2}}(c − l_{1} + m)x_{k} \\ & − \frac{i}{\sqrt{2}}(c − l_{2} − m)y_{k} + \frac{i}{\sqrt{2}}x_{k}E_{x}^{y} − \frac{1}{\sqrt{2}}y_{k}E_{y}^{x})h), \end{aligned}

\begin{aligned}X_{2,k}Φ(h) = Φ(( & \frac{i}{\sqrt{2}} \frac{∂}{∂x_{k}} + \frac{1}{\sqrt{2}} \frac{∂}{∂y_{k}} − \frac{i}{\sqrt{2}}(c − l_{1} + m)x_{k} \\ & + \frac{1}{\sqrt{2}}(c − l_{2} − m)y_{k} − \frac{1}{\sqrt{2}}x_{k}E_{x}^{y} + \frac{i}{\sqrt{2}}y_{k}E_{y}^{x})h). \end{aligned}

(In general, the polynomials in the parentheses of Φ in the right hand sides are no longer harmonic.)

Fujimura 35

Proof. By the definition of ϵ, for ∀f ∈ P,

\frac{∂}{∂x_{k}} ϵ^{−1}f = ϵ^{−1} ((\frac{1}{\sqrt{2}} \frac{∂}{∂x_{k}} + \frac{i}{\sqrt{2}} \frac{∂}{∂y_{k}} ) f) ,

\frac{∂}{∂y_{k}} ϵ^{−1}f = ϵ^{−1} ((\frac{i}{\sqrt{2}} \frac{∂}{∂x_{k}} + \frac{1}{\sqrt{2}} \frac{∂}{∂y_{k}} ) f) ,

x_{k}ϵ^{−1}f = ϵ^{−1} ((\frac{1}{\sqrt{2}}x_{k} − \frac{i}{\sqrt{2}}y_{k}) f) ,

y_{k}ϵ^{−1}f = ϵ^{−1} ((−\frac{i}{\sqrt{2}}x_{k} + \frac{1}{\sqrt{2}}y_{k}) f) ,

E_{x}^{x}ϵ^{−1}f = ϵ^{−1} ((\frac{1}{2}E_{x}^{x} + \frac{1}{2}E_{y}^{y} + \frac{i}{2}E_{y}^{x} − \frac{i}{2}E_{x}^{y}) f) ,

E_{y}^{y}ϵ^{−1}f = ϵ^{−1} ((\frac{1}{2}E_{x}^{x} + \frac{1}{2}E_{y}^{y} − \frac{i}{2}E_{y}^{x} + \frac{i}{2}E_{x}^{y}) f) ,

E_{y}^{x}ϵ^{−1}f = ϵ^{−1} ((\frac{i}{2}E_{x}^{x} − \frac{i}{2}E_{y}^{y} + \frac{1}{2}E_{y}^{x} + \frac{1}{2}E_{x}^{y}) f) ,

E_{x}^{y}ϵ^{−1}f = ϵ^{−1} ((−\frac{i}{2}E_{x}^{x} + \frac{i}{2}E_{y}^{y} + \frac{1}{2}E_{y}^{x} + \frac{1}{2}E_{x}^{y}) f) .

Since E_{x}^{x} and E_{y}^{y} are Euler operators which tell us the degree of homogeneity,

E_{x}^{x}h = (l_{1} − m)h, E_{y}^{y}h = (l_{2} + m)h

(h ∈ H_{(l_{1},l_{2}),m}.)

By Lemma 5.2,

X_{1,k}Φ(h) =X_{1,k}j(ϵ^{−1}(h))
=j ((\frac{∂}{∂x_{k}} + cx_{k} − x_{k}E_{x}^{x} − y_{k}E_{y}^{x}) (ϵ^{−1}(h)))
=j ∘ ϵ^{−1}(((\frac{1}{\sqrt{2}} \frac{∂}{∂x_{k}} + \frac{i}{\sqrt{2}} \frac{∂}{∂y_{k}} ) + c (\frac{1}{\sqrt{2}}x_{k} − \frac{i}{\sqrt{2}}y_{k})

− (\frac{1}{\sqrt{2}}x_{k} − \frac{i}{\sqrt{2}}y_{k}) (\frac{1}{2}(l_{1} − m) + \frac{1}{2}(l_{2} + m) + \frac{i}{2}E_{y}^{x} − \frac{i}{2}E_{x}^{y})

− (−\frac{i}{\sqrt{2}}x_{k} + \frac{1}{\sqrt{2}}y_{k}) (\frac{i}{2}(l_{1} − m) − \frac{i}{2}(l_{2} + m) + \frac{1}{2}E_{y}^{x} + \frac{1}{2}E_{x}^{y}))h)

=Φ((\frac{1}{\sqrt{2}} \frac{∂}{∂x_{k}} + \frac{i}{\sqrt{2}} \frac{∂}{∂y_{k}} + \frac{1}{\sqrt{2}}(c − l_{1} + m)x_{k}

− \frac{i}{\sqrt{2}}(c − l_{2} − m)y_{k} + \frac{i}{\sqrt{2}}x_{k}E_{x}^{y} − \frac{1}{\sqrt{2}}y_{k}E_{y}^{x})h).

X_{2,k}Φ(h) is similar.

In order to pull-back the 𝔭-action by Φ, we have to modify the polynomials occuring in this lemma to harmonic ones. As will be shown, it is possible to decompose them according to P ≅ J ⊗ H (c.f.Thm.3.1), and furthermore to determine to which H_{(l_{1},l_{2}),m} they belong. This is done in the next section.

6. Decomposition of polynomials

We fix k (1 ≤ k ≤ p) during this section.

Lemma 6.1. Let P ∈ P be homogeneous of degree m_{1} in x_{1}, …, x_{p}, homogeneous of degree m_{2} in y_{1}, …, y_{p}. Suppose (E_{y}^{x})^{2}P = 0. Then, in the decomposition

P = (P − \frac{1}{m_{1} − m_{2} + 2}E_{x}^{y}E_{y}^{x}P ) + \frac{1}{m_{1} − m_{2} + 2}E_{x}^{y}(E_{y}^{x}P),

we have

E_{y}^{x} (P − \frac{1}{m_{1} − m_{2} + 2}E_{x}^{y}E_{y}^{x}P ) = 0.

Proof.

E_{y}^{x}E_{x}^{y}E_{y}^{x}P = E_{x}^{y}E_{y}^{x}E_{y}^{x}P + (E_{x}^{x} − E_{y}^{y})E_{y}^{x}P
= E_{x}^{y}0 + ((m_{1} + 1) − (m_{2} − 1))E_{y}^{x}P = (m_{1} − m_{2} + 2)E_{y}^{x}P

gives us

E_{y}^{x} (P − \frac{1}{m_{1} − m_{2} + 2}E_{x}^{y}E_{y}^{x}P ) = 0.

Lemma 6.2. For h_{0} ∈ H_{(l_{1},l_{2}),0},

\frac{∂h_{0}}{∂y_{k}} ∈ H_{(l_{1},l_{2}−1),0},

\frac{∂h_{0}}{∂x_{k}} = (\frac{∂h_{0}}{∂x_{k}} + \frac{1}{l_{1} − l_{2} + 1}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ) − \frac{1}{l_{1} − l_{2} + 1}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ,

with

\frac{1}{l_{1} − l_{2} + 1}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ∈ H_{(l_{1}−1,l_{2}),0}.

Proof. Recall that h_{0} ∈ H_{(l_{1},l_{2}),0} implies E_{y}^{x}h_{0} = 0.

For \frac{∂h_{0}}{∂y_{k}},

Δ\frac{∂h_{0}}{∂y_{k}} = \frac{∂}{∂y_{k}} Δh_{0} = 0 (for Δ = Δ_{x^{2}}, Δ_{xy}, Δ_{y^{2}}),

that is, \frac{∂h_{0}}{∂y_{k}} ∈ H. It is homogeneous of degree l_{1} w.r.t. x_{1}, …, x_{p} and of degree l_{2} − 1 w.r.t. y_{1}, …, y_{p}.

E_{y}^{x} \frac{∂h_{0}}{∂y_{k}} = \frac{∂}{∂y_{k}} (E_{y}^{x}h_{0}) = 0

Fujimura 37

shows that

\frac{∂h_{0}}{∂y_{k}} ∈ H_{(l_{1},l_{2}−1),0}.

For \frac{∂h_{0}}{∂x_{k}},

Δ\frac{∂h_{0}}{∂x_{k}} = \frac{∂}{∂x_{k}} Δh_{0} = 0 (for Δ = Δ_{x^{2}}, Δ_{xy}, Δ_{y^{2}}),

that is, \frac{∂h_{0}}{∂x_{k}} ∈ H. Since

(E_{y}^{x})^{2} \frac{∂h_{0}}{∂x_{k}} = E_{y}^{x} \frac{∂}{∂x_{k}} (E_{y}^{x}h_{0}) − E_{y}^{x} \frac{∂h_{0}}{∂y_{k}} = 0,

we can apply Lemma 6.1 with P = \frac{∂h_{0}}{∂x_{k}}, m_{1} = l_{1} − 1, m_{2} = l_{2}. Lemma 6.3. For h_{0} ∈ H_{(l_{1},l_{2}),0}, define

(x_{k}h_{0})^{}˜ =x_{k}h_{0} − ξ_{xy} (\frac{1}{l_{1} + l_{2} + p − 3} \frac{∂h_{0}}{∂y_{k}} )

− ξ_{x^{2}}(\frac{1}{2l_{1} + p − 2} (\frac{∂h_{0}}{∂x_{k}} + \frac{1}{l_{1} − l_{2} + 1}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} )

− \frac{1}{(l_{1} − l_{2} + 1)(l_{1} + l_{2} + p − 3)}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ),

(y_{k}h_{0})^{}˜ =y_{k}h_{0} − ξ_{xy}(\frac{2l_{1} + p − 4}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)} \frac{∂h_{0}}{∂x_{k}}

+ \frac{−2(2l_{1} + p − 4)}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(2l_{2} + p − 4)}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} )

− ξ_{x^{2}}(\frac{−1}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)}E_{x}^{y} \frac{∂h_{0}}{∂x_{k}}

+ \frac{2}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(2l_{2} + p − 4)}(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}} )

− ξ_{y^{2}} \frac{l_{1} + l_{2} + p − 4}{(l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)} \frac{∂h_{0}}{∂y_{k}} .

Then,

(x_{k}h_{0})^{}˜ ∈ H, (y_{k}h_{0})^{}˜ ∈ H.

A fortiori, E_{y}^{x}(y_{k}h_{0})^{}˜ = (x_{k}h_{0})^{}˜ . Proof. We first deal with

(x_{k}h_{0})^{}˜ =x_{k}h_{0} − \frac{1}{l_{1} + l_{2} + p − 3}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} − \frac{1}{2l_{1} + p − 2}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}}

+ \frac{1}{(2l_{1} + p − 2)(l_{1} + l_{2} + p − 3)}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} .

Δ_{x^{2}}x_{k}h_{0} = x_{k}(Δ_{x^{2}}h_{0}) + 2\frac{∂h_{0}}{∂x_{k}} = 2\frac{∂h_{0}}{∂x_{k}} ,

Δ_{x^{2}}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}} = ξ_{x^{2}}(Δ_{x^{2}} \frac{∂h_{0}}{∂x_{k}}) + (2p + 4E_{x}^{x})\frac{∂h_{0}}{∂x_{k}} = (2p + 4l_{1} − 4)\frac{∂h_{0}}{∂x_{k}} ,
Δ_{x^{2}}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = ξ_{x^{2}}(Δ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}) + (2p + 4E_{x}^{x})E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

= (2p + 4l_{1} − 4)E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ,

and

Δ_{x^{2}}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} = ξ_{xy}(Δ_{x^{2}} \frac{∂h_{0}}{∂y_{k}}) + 2E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = 2E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

give

Δ_{x^{2}}(x_{k}h_{0})^{}˜ =Δ_{x^{2}}x_{k}h_{0} − \frac{1}{l_{1} + l_{2} + p − 3}Δ_{x^{2}}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} − \frac{1}{2l_{1} + p − 2}Δ_{x^{2}}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}}

+ \frac{1}{(2l_{1} + p − 2)(l_{1} + l_{2} + p − 3)}Δ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

=2\frac{∂h_{0}}{∂x_{k}} − \frac{1}{l_{1} + l_{2} + p − 3}2E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} − \frac{1}{2l_{1} + p − 2}(2p + 4l_{1} − 4)\frac{∂h_{0}}{∂x_{k}}

+ \frac{1}{(2l_{1} + p − 2)(l_{1} + l_{2} + p − 3)}(2p + 4l_{1} − 4)E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

=0.

Δ_{y^{2}}x_{k}h_{0} = x_{k}(Δ_{y^{2}}h_{0}) = 0,
Δ_{y^{2}}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}} = ξ_{x^{2}}(Δ_{y^{2}} \frac{∂h_{0}}{∂x_{k}}) = 0,
Δ_{y^{2}}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = ξ_{x^{2}}(Δ_{y^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}) = 0,

and

Δ_{y^{2}}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} = ξ_{xy}(Δ_{y^{2}} \frac{∂h_{0}}{∂y_{k}}) + 2E_{y}^{x} \frac{∂h_{0}}{∂y_{k}} = 2\frac{∂}{∂y_{k}} (E_{y}^{x}h_{0}) = 0

give

Δ_{y^{2}}(x_{k}h_{0})^{}˜ = 0.

Δ_{xy}x_{k}h_{0} = x_{k}(Δ_{xy}h_{0}) + \frac{∂}{∂y_{k}} = \frac{∂h_{0}}{∂y_{k}} ,
Δ_{xy}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}} = ξ_{x^{2}}(Δ_{xy} \frac{∂h_{0}}{∂x_{k}}) + 2E_{y}^{x} \frac{∂h_{0}}{∂x_{k}}

= 2 (\frac{∂}{∂x_{k}} (E_{y}^{x}h_{0}) − \frac{∂}{∂y_{k}} h_{0}) = −2\frac{∂h_{0}}{∂y_{k}} ,

Δ_{xy}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = ξ_{x^{2}}(Δ_{xy}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}) + 2E_{y}^{x}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

= 2 (E_{x}^{y}(E_{y}^{x} \frac{∂h_{0}}{∂y_{k}}) + (E_{x}^{x} − E_{y}^{y})\frac{∂h_{0}}{∂y_{k}} ) = 2(l_{1} − l_{2} + 1)\frac{∂h_{0}}{∂y_{k}} ,

Fujimura 39

and

Δ_{xy}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} = ξ_{xy}(Δ_{xy} \frac{∂h_{0}}{∂y_{k}}) + (p + E_{x}^{x} + E_{y}^{y})\frac{∂h_{0}}{∂y_{k}} = (p + l_{1} + l_{2} − 1)\frac{∂h_{0}}{∂y_{k}}

give

Δ_{xy}(x_{k}h_{0})^{}˜ =Δ_{xy}x_{k}h_{0} − \frac{1}{l_{1} + l_{2} + p − 3}Δ_{xy}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} − \frac{1}{2l_{1} + p − 2}Δ_{xy}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}}

+ \frac{1}{(2l_{1} + p − 2)(l_{1} + l_{2} + p − 3)}Δ_{xy}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

=\frac{∂h_{0}}{∂y_{k}} − \frac{1}{l_{1} + l_{2} + p − 3}(p + l_{1} + l_{2} − 1)\frac{∂h_{0}}{∂y_{k}} − \frac{1}{2l_{1} + p − 2} (−2\frac{∂h_{0}}{∂y_{k}} )

+ \frac{1}{(2l_{1} + p − 2)(l_{1} + l_{2} + p − 3)}2(l_{1} − l_{2} + 1)\frac{∂h_{0}}{∂y_{k}}

=0.

Thus,

Δ(x_{k}h_{0})^{}˜ = 0 (forΔ = Δ_{x^{2}}, Δ_{xy}, Δ_{y^{2}}),

that is, (x_{k}h_{0})^{}˜ ∈ H. We turn to

(y_{k}h_{0})^{}˜ =y_{k}h_{0} − \frac{2l_{1} + p − 4}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)}ξ_{xy} \frac{∂h_{0}}{∂x_{k}}

+ \frac{2(2l_{1} + p − 4)}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(2l_{2} + p − 4)}ξ_{xy}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

+ \frac{1}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂x_{k}}

− \frac{2}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(2l_{2} + p − 4)}ξ_{x^{2}}(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}}

− \frac{l_{1} + l_{2} + p − 4}{(l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)}ξ_{y^{2}} \frac{∂h_{0}}{∂y_{k}} .

Δ_{x^{2}}y_{k}h_{0} = 0,
Δ_{x^{2}}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂x_{k}} = 2(2l_{1} + p − 4)E_{x}^{y} \frac{∂h_{0}}{∂x_{k}} ,
Δ_{x^{2}}ξ_{x^{2}}(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}} = 2(2l_{1} + p − 4)(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}} ,
Δ_{x^{2}}ξ_{y^{2}} \frac{∂h_{0}}{∂y_{k}} = 0,
Δ_{x^{2}}ξ_{xy} \frac{∂h_{0}}{∂x_{k}} = 2E_{x}^{y} \frac{∂h_{0}}{∂x_{k}} ,

and

Δ_{x^{2}}ξ_{xy}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = 2(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}}

give Δ_{x^{2}}(y_{k}h_{0})^{}˜ = 0.

Δ_{y^{2}}y_{k}h_{0} = 2\frac{∂h_{0}}{∂y_{k}} ,

Δ_{y^{2}}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂x_{k}} = 0,
Δ_{y^{2}}ξ_{x^{2}}(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}} = 0,
Δ_{y^{2}}ξ_{y^{2}} \frac{∂h_{0}}{∂y_{k}} = 2(2l_{2} + p − 2)\frac{∂h_{0}}{∂y_{k}} ,
Δ_{y^{2}}ξ_{xy} \frac{∂h_{0}}{∂x_{k}} = −2\frac{∂h_{0}}{∂y_{k}} ,

and

Δ_{y^{2}}ξ_{xy}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = 2(l_{1} − l_{2} + 1)\frac{∂h_{0}}{∂y_{k}}

give Δ_{y^{2}}(y_{k}h_{0})^{}˜ = 0.

Δ_{xy}y_{k}h_{0} = \frac{∂h_{0}}{∂x_{k}} ,
Δ_{xy}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂x_{k}} = −2E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} + 2(l_{1} − l_{2} − 1)\frac{∂h_{0}}{∂x_{k}} ,
Δ_{xy}ξ_{x^{2}}(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}} = 0,
Δ_{xy}ξ_{y^{2}} \frac{∂h_{0}}{∂y_{k}} = 4(l_{1} − l_{2})E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ,
Δ_{xy}ξ_{xy} \frac{∂h_{0}}{∂x_{k}} = 2E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ,

and

Δ_{xy}ξ_{xy}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = (l_{1} + l_{2} + p − 1)E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

give Δ_{xy}(y_{k}h_{0})^{}˜ = 0. Thus, (y_{k}h_{0})^{}˜ ∈ H.

E_{y}^{x}y_{k}h_{0} = x_{k}h_{0},
E_{y}^{x}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂x_{k}} = −ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} + (l_{1} − l_{2} − 1)ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}} ,
E_{y}^{x}ξ_{x^{2}}(E_{x}^{y})^{2} \frac{∂h_{0}}{∂y_{k}} = 2(l_{1} − l_{2})ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ,
E_{y}^{x}ξ_{y^{2}} \frac{∂h_{0}}{∂y_{k}} = 2ξ_{xy} \frac{∂h_{0}}{∂y_{k}} ,
E_{y}^{x}ξ_{xy} \frac{∂h_{0}}{∂x_{k}} = −ξ_{xy} \frac{∂h_{0}}{∂y_{k}} + ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}} ,

and

E_{y}^{x}ξ_{xy}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = (l_{1} − l_{2} + 1)ξ_{xy} \frac{∂h_{0}}{∂y_{k}} + ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}

Fujimura 41

give E_{y}^{x}(y_{k}h_{0})^{}˜

=x_{k}h_{0}
−\frac{2l_{1} + p − 4}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)} (−ξ_{xy} \frac{∂h_{0}}{∂y_{k}} + ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}} )
+\frac{2(2l_{1} + p − 4)}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(2l_{2} + p − 4)} ((l_{1} − l_{2} + 1)ξ_{xy} \frac{∂h_{0}}{∂y_{k}} + ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} )
+\frac{1}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)} (−ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} + (l_{1} − l_{2} − 1)ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}} )
−\frac{2}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(2l_{2} + p − 4)}2(l_{1} − l_{2})ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}
−\frac{l_{1} + l_{2} + p − 4}{(l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)}2ξ_{xy} \frac{∂h_{0}}{∂y_{k}}
=x_{k}h_{0} − \frac{1}{l_{1} + l_{2} + p − 3}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} − \frac{1}{2l_{1} + p − 2}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}}
+\frac{1}{(2l_{1} + p − 2)(l_{1} + l_{2} + p − 3)}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}}
=(x_{k}h_{0})^{}˜ .

Thus, E_{y}^{x}(y_{k}h_{0})^{}˜ = (x_{k}h_{0})^{}˜ .

Lemma 6.4. For h_{0} ∈ H_{(l_{1},l_{2}),0}, let

h_{0,k}^{∂x} = \frac{1}{(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)} (\frac{∂h_{0}}{∂x_{k}} + \frac{1}{l_{1} − l_{2} + 1}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ) ,
h_{0,k}^{∂y} = \frac{1}{(l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(l_{1} − l_{2} + 1)} \frac{∂h_{0}}{∂y_{k}} ,
h_{0,k}^{x} = \frac{1}{l_{1} − l_{2} + 1}(x_{k}h_{0})^{}˜ ,
h_{0,k}^{y} = (y_{k}h_{0})^{}˜ − \frac{1}{l_{1} − l_{2} + 1}E_{x}^{y}(x_{k}h_{0})^{}˜ .

Then,

h_{0,k}^{∂x} ∈ H_{(l_{1}−1,l_{2}),0}, h_{0,k}^{∂y} ∈ H_{(l_{1},l_{2}−1),0},
h_{0,k}^{x} ∈ H_{(l_{1}+1,l_{2}),0}, h_{0,k}^{y} ∈ H_{(l_{1},l_{2}+1),0},

42 Fujimura and

\frac{∂h_{0}}{∂x_{k}} = (l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)h_{0,k}^{∂x}
− (l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)E_{x}^{y}h_{0,k}^{∂y},
\frac{∂h_{0}}{∂y_{k}} = (l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(l_{1} − l_{2} + 1)h_{0,k}^{∂y},
x_{k}h_{0} =(l_{1} − l_{2} + 1)h_{0,k}^{x} + ξ_{xy} ((2l_{2} + p − 4)(l_{1} − l_{2} + 1)h_{0,k}^{∂y})
+ ξ_{x^{2}} ((l_{1} + l_{2} + p − 3)h_{0,k}^{∂x} − (2l_{2} + p − 4)E_{x}^{y}h_{0,k}^{∂y}) ,
y_{k}h_{0} =h_{0,k}^{y} + E_{x}^{y}h_{0,k}^{x} + ξ_{xy} ((2l_{1} + p − 4)h_{0,k}^{∂x} − (2l_{1} + p − 4)E_{x}^{y}h_{0,k}^{∂y})
+ ξ_{x^{2}} (−E_{x}^{y}h_{0,k}^{∂x} + (E_{x}^{y})^{2}h_{0,k}^{∂y}) + ξ_{y^{2}}(l_{1} − l_{2} + 1)(l_{1} + l_{2} + p − 4)h_{0,k}^{∂y}.

Proof. The latter part is a reformulation of Lemma 6.2 and Lemma 6.3.

h_{0,k}^{∂x} ∈ H_{(l_{1}−1,l_{2}),0} and h_{0,k}^{∂y} ∈ H_{(l_{1},l_{2}−1),0} are also in Lemma 6.2. h_{0,k}^{x} ∈ H_{(l_{1}+1,l_{2}),0} follows from E_{y}^{x}(x_{k}h_{0})^{}˜ = 0:

E_{y}^{x}x_{k}h_{0} = x_{k}(E_{y}^{x}h) = 0,
E_{y}^{x}ξ_{x^{2}} \frac{∂h_{0}}{∂x_{k}} = ξ_{x^{2}}E_{y}^{x} \frac{∂h_{0}}{∂x_{k}} = ξ_{x^{2}} \frac{∂}{∂x_{k}} (E_{y}^{x}h_{0}) − ξ_{x^{2}} \frac{∂}{∂y_{k}} h = −ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}} ,
E_{y}^{x}ξ_{x^{2}}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = ξ_{x^{2}}E_{y}^{x}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} = ξ_{x^{2}}E_{x}^{y}(E_{y}^{x} \frac{∂h_{0}}{∂y_{k}}) + ξ_{x^{2}}(E_{x}^{x} − E_{y}^{y})\frac{∂h_{0}}{∂y_{k}}

= (l_{1} − l_{2} + 1)ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}} ,

and

E_{y}^{x}ξ_{xy} \frac{∂h_{0}}{∂y_{k}} = ξ_{xy}(E_{y}^{x} \frac{∂h_{0}}{∂y_{k}}) + ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}} = ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}}

give

E_{y}^{x}h_{0,k}^{x} = \frac{1}{l_{1} − l_{2} + 1}((E_{y}^{x}x_{k}h_{0}) − E_{y}^{x}ξ_{xy} (\frac{1}{l_{1} + l_{2} + p − 3} \frac{∂h_{0}}{∂y_{k}} )

− E_{y}^{x}ξ_{x^{2}}(\frac{1}{2l_{1} + p − 2} (\frac{∂h_{0}}{∂x_{k}} + \frac{1}{l_{1} − l_{2} + 1}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} )

− \frac{1}{(l_{1} − l_{2} + 1)(l_{1} + l_{2} + p − 3)}E_{x}^{y} \frac{∂h_{0}}{∂y_{k}} ))

= \frac{1}{l_{1} − l_{2} + 1}(−\frac{1}{l_{1} + l_{2} + p − 3}ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}} − \frac{1}{2l_{1} + p − 2} (−ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}} )

− \frac{1}{(2l_{1} + p − 2)(l_{1} − l_{2} + 1)}(l_{1} − l_{2} + 1)ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}}

+ \frac{1}{(l_{1} − l_{2} + 1)(l_{1} + l_{2} + p − 3)}(l_{1} − l_{2} + 1)ξ_{x^{2}} \frac{∂h_{0}}{∂y_{k}} )

= 0.

Fujimura 43

Finally, h_{0,k}^{y} ∈ H_{(l_{1},l_{2}+1),0} is obtained by Lemma 6.1 with P = E_{y}^{x}(y_{k}h_{0})^{}˜ , m_{1} = l_{1} and m_{2} = l_{2} + 1, using (E_{y}^{x})^{2}(y_{k}h_{0})^{}˜ = E_{y}^{x}(x_{k}h_{0})^{}˜ = 0. Lemma 6.5.

y_{k}(E_{x}^{y})^{m} = (E_{x}^{y})^{m}y_{k},
x_{k}(E_{x}^{y})^{m} = (E_{x}^{y})^{m}x_{k} − m(E_{x}^{y})^{m−1}y_{k},
(E_{x}^{y})^{m}ξ_{xy} = ξ_{xy}(E_{x}^{y})^{m} + mξ_{y^{2}}(E_{x}^{y})^{m−1},
(E_{x}^{y})^{m}ξ_{y^{2}} = ξ_{y^{2}}(E_{x}^{y})^{m},
(E_{x}^{y})^{m}ξ_{x^{2}} = ξ_{x^{2}}(E_{x}^{y})^{m} + 2mξ_{xy}(E_{x}^{y})^{m−1} + m(m − 1)ξ_{y^{2}}(E_{x}^{y})^{m−2},
\frac{∂}{∂x_{k}} (E_{x}^{y})^{m} = (E_{x}^{y})^{m} \frac{∂}{∂x_{k}} ,
\frac{∂}{∂y_{k}} (E_{x}^{y})^{m} = (E_{x}^{y})^{m} \frac{∂}{∂y_{k}} + m\frac{∂}{∂x_{k}} (E_{x}^{y})^{m−1},
E_{y}^{x}(E_{x}^{y})^{m} = (E_{x}^{y})^{m}E_{y}^{x} + (E_{x}^{y})^{m−1}m(E_{x}^{x} − E_{y}^{y} − m + 1).

Proof. The case m = 1 being easy, this is proved by induction on m. Proposition 6.6. For h_{0} ∈ H_{(l_{1},l_{2}),0}, put h = (E_{x}^{y})^{m}h_{0} ∈ H_{(l_{1},l_{2}),m}. (Recall that any element of H_{(l_{1},l_{2}),m} is of this form.) Then,

\frac{∂h}{∂x_{k}} =(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(E_{x}^{y})^{m}h_{0,k}^{∂x}
− (l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y},
\frac{∂h}{∂y_{k}} =m(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}
+ (l_{1} + l_{2} − m + 1)(l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y},
x_{k}h =(l_{1} − l_{2} − m + 1)(E_{x}^{y})^{m}h_{0,k}^{x} − m(E_{x}^{y})^{m−1}h_{0,k}^{y}
+ξ_{xy}(m(2l_{2} + 2m + p − 4)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}
+ (l_{1} − l_{2} − m + 1)(2l_{2} + 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y})
+ξ_{x^{2}} ((l_{1} + l_{2} + m + p − 3)(E_{x}^{y})^{m}h_{0,k}^{∂x} − (2l_{2} + m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y}) ,
y_{k}h =(E_{x}^{y})^{m+1}h_{0,k}^{x} + (E_{x}^{y})^{m}h_{0,k}^{y}
+ξ_{xy} ((2l_{1} − 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂x} − (2l_{1} − 2m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y})
+ξ_{x^{2}} (−(E_{x}^{y})^{m+1}h_{0,k}^{∂x} + (E_{x}^{y})^{m+2}h_{0,k}^{∂y})
+ξ_{y^{2}}(m(2l_{1} − m + p − 3)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}
+ (l_{1} − l_{2} − m + 1)(l_{1} + l_{2} − m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y}).

Proof. This follows from Lemma 6.4 and Lemma 6.5.

Now the proof of Prop. 5.1.

44 Fujimura Proof. For P ∈ P,

j(ξ_{x^{2}}P) = j(ξ_{y^{2}}P) = j(P), j(ξ_{xy}P) = 0

by the definition of j. Since

ϵ^{−1}ξ_{xy} = (\frac{i}{2}ξ_{x^{2}} + \frac{i}{2}ξ_{y^{2}}) ϵ^{−1}(P),
ϵ^{−1}ξ_{x^{2}} = (iξ_{xy} + \frac{1}{2}ξ_{x^{2}} − \frac{1}{2}ξ_{y^{2}}) ϵ^{−1}(P),
ϵ^{−1}ξ_{y^{2}} = (iξ_{xy} − \frac{1}{2}ξ_{x^{2}} + \frac{1}{2}ξ_{y^{2}}) ϵ^{−1}(P),

we have

Φ(ξ_{x^{2}}P) = Φ(ξ_{y^{2}}P) = 0, Φ(ξ_{xy}P) = Φ(iP).  (∗∗)

Recall Lemma 5.3 for h = (E_{x}^{y})^{m}h_{0} ∈ H_{(l_{1},l_{2}),m}:

X_{1,k}Φ(h) = Φ((\frac{1}{\sqrt{2}} \frac{∂}{∂x_{k}} + \frac{i}{\sqrt{2}} \frac{∂}{∂y_{k}} + \frac{1}{\sqrt{2}}(c − l_{1} + m)x_{k}

− \frac{i}{\sqrt{2}}(c − l_{2} − m)y_{k} + \frac{i}{\sqrt{2}}x_{k}E_{x}^{y} − \frac{1}{\sqrt{2}}y_{k}E_{y}^{x})h).

By Prop. 6.6,

X_{1,k}Φ(h) = Φ(\frac{1}{\sqrt{2}}((l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(E_{x}^{y})^{m}h_{0,k}^{∂x}

− (l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y})

+ \frac{i}{\sqrt{2}}(m(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}

+ (l_{1} + l_{2} − m + 1)(l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y})

+ \frac{1}{\sqrt{2}}(c − l_{1} + m)((l_{1} − l_{2} − m + 1)(E_{x}^{y})^{m}h_{0,k}^{x} − m(E_{x}^{y})^{m−1}h_{0,k}^{y}

+ ξ_{xy}(m(2l_{2} + 2m + p − 4)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}

+ (l_{1} − l_{2} − m + 1)(2l_{2} + 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y})

+ ξ_{x^{2}} (⋯))

− \frac{i}{\sqrt{2}}(c − l_{2} − m)((E_{x}^{y})^{m+1}h_{0,k}^{x} + (E_{x}^{y})^{m}h_{0,k}^{y}

+ ξ_{xy}((2l_{1} − 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂x}

− (2l_{1} − 2m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y})

+ ξ_{x^{2}} (⋯) + ξ_{y^{2}} (⋯))

+ \frac{i}{\sqrt{2}}((l_{1} − l_{2} − m)(E_{x}^{y})^{m+1}h_{0,k}^{x} − (m + 1)(E_{x}^{y})^{m}h_{0,k}^{y}

+ ξ_{xy}((m + 1)(2l_{2} + 2m + p − 2)(E_{x}^{y})^{m}h_{0,k}^{∂x}

+ (l_{1} − l_{2} − m)(2l_{2} + 2m + p − 2)(E_{x}^{y})^{m+1}h_{0,k}^{∂y})

+ ξ_{x^{2}} (⋯))

− \frac{1}{\sqrt{2}}m(l_{1} − l_{2} − m + 1)((E_{x}^{y})^{m}h_{0,k}^{x} + (E_{x}^{y})^{m−1}h_{0,k}^{y}

+ ξ_{xy}((2l_{1} − 2m + p − 2)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}

− (2l_{1} − 2m + p − 2)(E_{x}^{y})^{m}h_{0,k}^{∂y})

+ ξ_{x^{2}} (⋯) + ξ_{y^{2}} (⋯))).

Here we used

E_{y}^{x}(E_{x}^{y})^{m}h_{0} = (E_{x}^{y})^{m}E_{y}^{x}h_{0} + (E_{x}^{y})^{m−1}m(E_{x}^{x} − E_{y}^{y} − m + 1)h_{0}

= m(l_{1} − l_{2} − m + 1)(E_{x}^{y})^{m−1}h_{0}.

By (∗∗),

X_{1,k}Φ(h) =Φ(\frac{1}{\sqrt{2}}((l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(E_{x}^{y})^{m}h_{0,k}^{∂x}

− (l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y})

+ \frac{i}{\sqrt{2}}(m(l_{1} + l_{2} + p − 3)(2l_{1} + p − 2)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}

+ (l_{1} + l_{2} − m + 1)(l_{1} + l_{2} + p − 3)(2l_{2} + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y})

+ \frac{1}{\sqrt{2}}(c − l_{1} + m)((l_{1} − l_{2} − m + 1)(E_{x}^{y})^{m}h_{0,k}^{x} − m(E_{x}^{y})^{m−1}h_{0,k}^{y}

+ i(m(2l_{2} + 2m + p − 4)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}

+ (l_{1} − l_{2} − m + 1)(2l_{2} + 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y}))

− \frac{i}{\sqrt{2}}(c − l_{2} − m)((E_{x}^{y})^{m+1}h_{0,k}^{x} + (E_{x}^{y})^{m}h_{0,k}^{y}

+ i((2l_{1} − 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂x}

− (2l_{1} − 2m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y})

+ (l_{1} − l_{2} − m + 1)(l_{1} + l_{2} − m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y}))

+ \frac{i}{\sqrt{2}}((l_{1} − l_{2} − m)(E_{x}^{y})^{m+1}h_{0,k}^{x} − (m + 1)(E_{x}^{y})^{m}h_{0,k}^{y}

+ i((m + 1)(2l_{2} + 2m + p − 2)(E_{x}^{y})^{m}h_{0,k}^{∂x}

+ (l_{1} − l_{2} − m)(2l_{2} + 2m + p − 2)(E_{x}^{y})^{m+1}h_{0,k}^{∂y}))

− \frac{1}{\sqrt{2}}m(l_{1} − l_{2} − m + 1)((E_{x}^{y})^{m}h_{0,k}^{x} + (E_{x}^{y})^{m−1}h_{0,k}^{y}

+ i((2l_{1} − 2m + p − 2)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}

− (2l_{1} − 2m + p − 2)(E_{x}^{y})^{m}h_{0,k}^{∂y})))

= Φ(\frac{1}{\sqrt{2}}(c − l_{1})(l_{1} − l_{2} − m + 1)(E_{x}^{y})^{m}h_{0,k}^{x}

− \frac{i}{\sqrt{2}}(c − l_{1})(E_{x}^{y})^{m+1}h_{0,k}^{x}

− \frac{1}{\sqrt{2}}(c − l_{2} + 1)m(E_{x}^{y})^{m−1}h_{0,k}^{y}

− \frac{i}{\sqrt{2}}(c − l_{2} + 1)(E_{x}^{y})^{m}h_{0,k}^{y}

+ \frac{i}{\sqrt{2}}(c + l_{1} + p − 2)m(2m + 2l_{2} + p − 4)(E_{x}^{y})^{m−1}h_{0,k}^{∂x}

+ \frac{1}{\sqrt{2}}(c + l_{1} + p − 2)(2l_{1} − 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂x}

+ \frac{i}{\sqrt{2}}(c + l_{2} + p − 3)(l_{1} − l_{2} − m + 1)(2m + 2l_{2} + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y}

− \frac{1}{\sqrt{2}}(c + l_{2} + p − 3)(2l_{1} − 2m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y}).

Define

T_{(l_{1},l_{2}),m}^{(1,0,0)}(X_{1,k} ⊗ h) = \frac{1}{\sqrt{2}}(l_{1} − l_{2} − m + 1)(E_{x}^{y})^{m}h_{0,k}^{x},
T_{(l_{1},l_{2}),m}^{(1,0,1)}(X_{1,k} ⊗ h) = −\frac{i}{\sqrt{2}}(E_{x}^{y})^{m+1}h_{0,k}^{x},
T_{(l_{1},l_{2}),m}^{(0,1,−1)}(X_{1,k} ⊗ h) = −\frac{1}{\sqrt{2}}m(E_{x}^{y})^{m−1}h_{0,k}^{y},
T_{(l_{1},l_{2}),m}^{(0,1,0)}(X_{1,k} ⊗ h) = −\frac{i}{\sqrt{2}}(E_{x}^{y})^{m}h_{0,k}^{y},
T_{(l_{1},l_{2}),m}^{(−1,0,−1)}(X_{1,k} ⊗ h) = \frac{i}{\sqrt{2}}m(2m + 2l_{2} + p − 4)(E_{x}^{y})^{m−1}h_{0,k}^{∂x},
T_{(l_{1},l_{2}),m}^{(−1,0,0)}(X_{1,k} ⊗ h) = \frac{1}{\sqrt{2}}(2l_{1} − 2m + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂x},
T_{(l_{1},l_{2}),m}^{(0,−1,0)}(X_{1,k} ⊗ h) = \frac{i}{\sqrt{2}}(l_{1} − l_{2} − m + 1)(2m + 2l_{2} + p − 4)(E_{x}^{y})^{m}h_{0,k}^{∂y},
T_{(l_{1},l_{2}),m}^{(0,−1,1)}(X_{1,k} ⊗ h) = −\frac{1}{\sqrt{2}}(2l_{1} − 2m + p − 4)(E_{x}^{y})^{m+1}h_{0,k}^{∂y}.

X_{2,k} is similar.

7. Reducibility

Since the decomposition of S^{c}(X^{+}) into irreducible K -modules is multiplicity free, Prop. 5.1 is a criterion to see whether or not we can pass from a K_{0}-type (which is of the form H_{(l_{1},l_{2}),m}) to another. That is, starting from a K_{0}-type, we can go to an adjacent K_{0}-type with an action of 𝔭 if and only if the coefficients in Prop. 5.1 for that direction is not zero.

The first observation is:

Lemma 7.1. If an irreducible constituent of S^{c}(X^{+}) contains H_{(l_{1},l_{2}),m_{0}} for some l_{1}, l_{2} and m_{0}, then it contains all of H_{(l_{1},l_{2})} = ⨁_{m=0}^{l_{1}−l_{2}} H_{(l_{1},l_{2}),m}.

Proof. If l_{1} = 0, then l_{2} = m_{0} = 0, H_{(0,0),0} = H_{(0,0)} and we have nothing to prove.

We may assume l_{1} ≥ 1. To go from H_{(l_{1},l_{2}),m_{0}} to H_{(l_{1},l_{2}),m_{0}+1}, consider two different paths:

H_{(l_{1},l_{2}),m_{0}} → H_{(l_{1}+1,l_{2}),m_{0}+1} → H_{(l_{1},l_{2}),m_{0}+1},
H_{(l_{1},l_{2}),m_{0}} → H_{(l_{1}−1,l_{2}),m_{0}} → H_{(l_{1},l_{2}),m_{0}+1}.

The first path is available iff the coefficients of T_{(l_{1},l_{2}),m}^{(1,0,1)} and of T_{(l_{1}+1,l_{2}),m+1}^{(−1,0,0)} are both non-zero, namely,

c − l_{1} ≠ 0, c + l_{1} + p − 1 ≠ 0.

The second path is available iff

c + l_{1} + p − 2 ≠ 0, c − l_{1} + 1 ≠ 0.

Since these do not occur at the same time, we conclude that we can always go from H_{(l_{1},l_{2}),m_{0}} to H_{(l_{1},l_{2}),m_{0}+1}. Conversely, we can also prove that we can always go from H_{(l_{1},l_{2}),m_{0}} to H_{(l_{1},l_{2}),m_{0}−1}, proving the assertion of the lemma.

This lemma says that each H_{(l_{1},l_{2})} is a sort of “building block” for irreducible constituents, and in consequence that we can forget the parameter m. The next thing to do is to determine which of these blocks are tied together.

The formula in Prop. 5.1 is rewritten as:

X(Φ(h)) =A^{+0}(l_{1}, l_{2})Φ(h_{X}^{+0}) + A^{−0}(l_{1}, l_{2})Φ(h_{X}^{−0})
+ A^{0+}(l_{1}, l_{2})Φ(h_{X}^{0+}) + A^{0−}(l_{1}, l_{2})Φ(h_{X}^{0−})

with

A^{+0}(l_{1}, l_{2}) = c − l_{1},  A^{−0}(l_{1}, l_{2}) = c + l_{1} + p − 2,
A^{0+}(l_{1}, l_{2}) = c − l_{2} + 1,  A^{0−}(l_{1}, l_{2}) = c + l_{2} + p − 3,

h_{X}^{+0} ∈ H_{(l_{1}+1,l_{2})},  h_{X}^{−0} ∈ H_{(l_{1}−1,l_{2})},
h_{X}^{0+} ∈ H_{(l_{1},l_{2}+1)},  h_{X}^{0−} ∈ H_{(l_{1},l_{2}−1)}.

To understand better what is going on, we translate the ingredients into diagrams.

The block H_{(l_{1},l_{2})} is identified with the point (l_{1}, l_{2}) ∈ ℝ^{2}. The blocks occuring in H are {(l_{1}, l_{2}) ∈ ℤ^{2}|l_{1} ≥ l_{2} ≥ 0}, with multiplicity one.

[figure: l_{1}; l_{2}]

The four lines A^{±0}(l_{1}, l_{2}) = 0 and A^{0±}(l_{1}, l_{2}) = 0 are “potential barriers” which are obstructions to the transitions.

[figure: l_{1}; l_{2}; ⇐; ⇐; ⇐; ⇐; ⇐; ⇐; c; A^{+0} = 0; l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; −c − p + 2; A^{−0} = 0; l_{1}; l_{2}; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; c + 1; A^{0+} = 0; l_{1}; l_{2}; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; −c − p + 3; A^{0−} = 0]

In general, starting from a point (l_{1}, l_{2}) corresponding to a block, we can go to four directions ((l_{1} ± 1, l_{2}), (l_{1}, l_{2} ± 1)). If a barrier meets a block, the transition crossing the barrier is not permitted, and this yields a submodule and a quotient module.

All that remains is to determine the positions of the barriers according to the value of c.

Theorem 7.2. The decomposition of the Harish-Chandra module of S^{c}(X^{+}) is

(0) c ∉ ℤ : The barriers never cross the K -types, so it is irreducible. From now on, assume c ∈ ℤ.

(1) c ≤ −p + 1 : 1 finite-dimensional quotient, 1 subquotient, 1 submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; −c − p + 2; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; \begin{matrix}−c \\ −p + 3 \end{matrix}; quot.; subquot.; sub.]

(2) c = −p + 2 : 1 quotient, 1 submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; 0; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; 1; sub.; quot.]

(3) −p + 3 ≤ c ≤ −2 : irreducible.

(4) c = −1 : 1 submodule, 1 quotient.

[figure: l_{1}; l_{2}; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; 0; quot.; sub.]

(5) c ≥ 0 : 1 finite-dimensional submodule, 1 subquotient, 1 quotient.

[figure: l_{1}; l_{2}; ⇐; ⇐; ⇐; ⇐; ⇐; ⇐; c; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; c + 1; sub; subquot.; quot.]

8. Unitarizability

We determine which of the irreducible constituents are unitarizable, that is, admit a G-invariant positive definite Hermitian form.

We define an inner product ⟨·, ·⟩_{(l_{1},l_{2}),m} on H_{(l_{1},l_{2}),m} by

⟨h_{1}, h_{2}⟩_{(l_{1},l_{2}),m} = ∫_{O(p)×SO(2)} Φ(h_{1})(k)\overline{Φ(h_{2})(k)}dk
(h_{1}, h_{2} ∈ H_{(l_{1},l_{2}),m})

where dk is the normalized Haar measure on K_{1}. Then, this is a K_{1}-invariant hermitian inner product, called “standard inner product”.

Lemma 8.1. On each K_{1}-type H_{(l_{1},l_{2}),m}, K_{1}-invariant hermitian inner products are unique up to scalar multiple.

Proof. Suppose we have ⟨·, ·⟩ and ⟨·, ·⟩′. The map A defined by ⟨h_{1}, h_{2}⟩ = ⟨A(h_{1}), h_{2}⟩′ intertwines the K_{1}-action, and hence is a scalar operator by Schur’s Lemma.

If an irreducible constituent of S^{c}(X^{+}) admits a K_{1}-invariant hermitian inner product ⟨·, ·⟩, its restriction to Φ(H_{(l_{1},l_{2}),m}) must be a constant multiple of the standard one:

⟨Φ(h_{1}), Φ(h_{2})⟩ = c_{(l_{1},l_{2}),m}⟨h_{1}, h_{2}⟩_{(l_{1},l_{2}),m} h_{1}, h_{2} ∈ H_{(l_{1},l_{2}),m}.

Then, c_{(l_{1},l_{2}),m} ∈ ℝ. Furthermore, if ⟨·, ·⟩ is positive definite, c_{(l_{1},l_{2}),m} > 0.

For example, since we shifted the parameter of the parabolic induction to preserve unitarity, the whole S^{c}(X^{+}) is unitarizable for c′ ∈ iℝ. We call this the “unitary axis”. It corresponds to c_{(l_{1},l_{2}),m} = 1 for all the K_{1}-types.

Conversely, since the K_{1}-types are mutually orthogonal (this follows from the fact that the decomposition into K_{1}-types is multiplicity free), the constants {c_{(l_{1},l_{2}),m}} completely determine the inner product.

The condition for the positive definite K_{1}-invariant inner product ⟨·, ·⟩ to be G^{+}-invariant is rephrased as

⟨Xf_{1}, f_{2}⟩ + ⟨f_{1}, Xf_{2}⟩ = 0 ∀X ∈ 𝔭, ∀f_{1}, f_{2} ∈ S^{c}(X^{+}).  (♯)

Lemma 8.2. (♯) is equivalent to:

(c − l_{1})c_{(l_{1}−1,l_{2}),m} + (\overline{c} + l_{1} + p − 1)c_{(l_{1},l_{2}),m} = 0,
(c − l_{1})c_{(l_{1}+1,l_{2}),m+1} + (\overline{c} + l_{1} + p − 1)c_{(l_{1},l_{2}),m} = 0,
(c − l_{2} + 1)c_{(l_{1},l_{2}−1),m−1} + (\overline{c} + l_{2} + p − 2)c_{(l_{1},l_{2}),m} = 0,
(c − l_{2} + 1)c_{(l_{1},l_{2}+1),m} + (\overline{c} + l_{2} + p − 2)c_{(l_{1},l_{2}),m} = 0,

whenever c_{(·,·),·} is well-defined on the irreducible constituent of S^{c}(X^{+}).

Proof. Writing f_{1} = Φ(h_{1}) (h_{1} ∈ H_{(l_{1,1},l_{1,2}),m_{1}} ) and f_{2} = Φ(h_{2}) (h_{2} ∈ H_{(l_{2,1},l_{2,2}),m_{2}} ), we use Prop. 5.1 for XΦ(h_{1}) and XΦ(h_{2}).

⟨X(Φ(h_{1})), Φ(h_{2})⟩ + ⟨Φ(h_{1}), X(Φ(h_{2}))⟩

=⟨(c − l_{1,1}) (Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(1,0,0)}(X ⊗ h_{1})) + Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(1,0,1)}(X ⊗ h_{1})))
+(c − l_{1,2} + 1) (Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,1,−1)}(X ⊗ h_{1})) + Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,1,0)}(X ⊗ h_{1})))
+(c + l_{1,1} + p − 2) (Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(−1,0,−1)}(X ⊗ h_{1})) + Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(−1,0,0)}(X ⊗ h_{1})))
+(c + l_{1,2} + p − 3) (Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,−1,0)}(X ⊗ h_{1})) + Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,−1,1)}(X ⊗ h_{1}))) , Φ(h_{2})⟩

+⟨Φ(h_{1}), (c − l_{2,1}) (Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(1,0,0)}(X ⊗ h_{2})) + Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(1,0,1)}(X ⊗ h_{2})))
+(c − l_{2,2} + 1) (Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,1,−1)}(X ⊗ h_{2})) + Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,1,0)}(X ⊗ h_{2})))

+(c + l_{2,1} + p − 2) (Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(−1,0,−1)}(X ⊗ h_{2})) + Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(−1,0,0)}(X ⊗ h_{2})))
+(c + l_{2,2} + p − 3) (Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,−1,0)}(X ⊗ h_{2})) + Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,−1,1)}(X ⊗ h_{2})))⟩.

By the orthogonality property of the K_{1}-types,

⟨X(Φ(h_{1})), Φ(h_{2})⟩ + ⟨Φ(h_{1}), X(Φ(h_{2}))⟩

=δ_{l_{1,1}+1,l_{2,1}}δ_{l_{1,2},l_{2,2}}δ_{m_{1},m_{2}}(c − l_{1,1})⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(1,0,0)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1}+1,l_{2,1}}δ_{l_{1,2},l_{2,2}}δ_{m_{1}+1,m_{2}}(c − l_{1,1})⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(1,0,1)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2}+1,l_{2,2}}δ_{m_{1}−1,m_{2}}(c − l_{1,2} + 1)⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,1,−1)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2}+1,l_{2,2}}δ_{m_{1},m_{2}}(c − l_{1,2} + 1)⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,1,0)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1}−1,l_{2,1}}δ_{l_{1,2},l_{2,2}}δ_{m_{1}−1,m_{2}}(c + l_{1,1} + p − 2)⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(−1,0,−1)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1}−1,l_{2,1}}δ_{l_{1,2},l_{2,2}}δ_{m_{1},m_{2}}(c + l_{1,1} + p − 2)⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(−1,0,0)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2}−1,l_{2,2}}δ_{m_{1},m_{2}}(c + l_{1,2} + p − 3)⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,−1,0)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2}−1,l_{2,2}}δ_{m_{1}+1,m_{2}}(c + l_{1,2} + p − 3)⟨Φ(T_{(l_{1,1},l_{1,2}),m_{1}}^{(0,−1,1)}(X ⊗ h_{1})), Φ(h_{2})⟩
+δ_{l_{1,1},l_{2,1}+1}δ_{l_{1,2},l_{2,2}}δ_{m_{1},m_{2}}(\overline{c − l_{2,1}})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(1,0,0)}(X ⊗ h_{2}))⟩
+δ_{l_{1,1},l_{2,1}+1}δ_{l_{1,2},l_{2,2}}δ_{m_{1},m_{2}+1}(\overline{c − l_{2,1}})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(1,0,1)}(X ⊗ h_{2}))⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2},l_{2,2}+1}δ_{m_{1},m_{2}−1}(\overline{c − l_{2,2} + 1})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,1,−1)}(X ⊗ h_{2}))⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2},l_{2,2}+1}δ_{m_{1},m_{2}}(\overline{c − l_{2,2} + 1})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,1,0)}(X ⊗ h_{2}))⟩
+δ_{l_{1,1},l_{2,1}−1}δ_{l_{1,2},l_{2,2}}δ_{m_{1},m_{2}−1}(\overline{c + l_{2,1} + p − 2})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(−1,0,−1)}(X ⊗ h_{2}))⟩
+δ_{l_{1,1},l_{2,1}−1}δ_{l_{1,2},l_{2,2}}δ_{m_{1},m_{2}}(\overline{c + l_{2,1} + p − 2})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(−1,0,0)}(X ⊗ h_{2}))⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2},l_{2,2}−1}δ_{m_{1},m_{2}}(\overline{c + l_{2,2} + p − 3})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,−1,0)}(X ⊗ h_{2}))⟩
+δ_{l_{1,1},l_{2,1}}δ_{l_{1,2},l_{2,2}−1}δ_{m_{1},m_{2}+1}(\overline{c + l_{2,2} + p − 3})⟨Φ(h_{1}), Φ(T_{(l_{2,1},l_{2,2}),m_{2}}^{(0,−1,1)}(X ⊗ h_{2}))⟩.

Here, δ_{i,j} is 1 if i = j, and 0 otherwise. Because of these δ’s, most terms vanish. There are some remaining terms only when (l_{1,1}, l_{1,2}), m_{1} and (l_{2,1}, l_{2,2}), m_{2} are “adjacent”. (♯) becomes:

(c − l_{1})c_{(l_{1}−1,l_{2}),m}⟨T_{(l_{1},l_{2}),m}^{(1,0,0)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1}−1,l_{2}),m}
+(\overline{c} + l_{1} + p − 1)c_{(l_{1},l_{2}),m}⟨h_{1}, T_{(l_{1}+1,l_{2}),m}^{(−1,0,0)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m} = 0
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1}+1,l_{2}),m}),
(c − l_{1})c_{(l_{1}+1,l_{2}),m+1}⟨T_{(l_{1},l_{2}),m}^{(1,0,1)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1}+1,l_{2}),m+1}
+(\overline{c} + l_{1} + p − 1)c_{(l_{1},l_{2}),m}⟨h_{1}, T_{(l_{1}+1,l_{2}),m+1}^{(−1,0,−1)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m} = 0
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1}+1,l_{2}),m+1}),

(c − l_{2} + 1)c_{(l_{1},l_{2}+1),m−1}⟨T_{(l_{1},l_{2}),m}^{(0,1,−1)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1},l_{2}+1),m−1}
+(\overline{c} + l_{2} + p − 2)c_{(l_{1},l_{2}),m}⟨h1, T_{(l_{1},l_{2}+1),m−1}^{(0,−1,1)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m} = 0
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1},l_{2}+1),m−1}),
(c − l_{2} + 1)c_{(l_{1},l_{2}+1),m}⟨T_{(l_{1},l_{2}),m}^{(0,1,0)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1},l_{2}+1),m}
+(\overline{c} + l_{2} + p − 2)c_{(l_{1},l_{2}),m}⟨h_{1}, T_{(l_{1},l_{2}+1),m}^{(0,−1,0)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m} = 0
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1},l_{2}+1),m}).

By the case of unitary axis, the above relations hold for ∀c′ = −2c−p+1 ∈ iℝ with c_{(l_{1},l_{2}),m} = 1 for all pairs (l_{1}, l_{2}, m). Substituting, we get

⟨T_{(l_{1},l_{2}),m}^{(1,0,0)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1}+1,l_{2}),m} = ⟨h_{1}, T_{(l_{1}+1,l_{2}),m}^{(−1,0,0)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m}
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1}+1,l_{2}),m}),
⟨T_{(l_{1},l_{2}),m}^{(1,0,1)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1}+1,l_{2}),m+1}) = ⟨h_{1}, T_{(l_{1}+1,l_{2}),m+1}^{(−1,0,−1)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m}
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1}+1,l_{2}),m+1}),
⟨T_{(l_{1},l_{2}),m}^{(0,1,−1)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1},l_{2}+1),m−1}) = ⟨h1, T_{(l_{1},l_{2}+1),m−1}^{(0,−1,1)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m}
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1},l_{2}+1),m−1}),
⟨T_{(l_{1},l_{2}),m}^{(0,1,0)}(X ⊗ h_{1}), h_{2}⟩_{(l_{1},l_{2}+1),m}) = ⟨h_{1}, T_{(l_{1},l_{2}+1),m}^{(0,−1,0)}(X ⊗ h_{2})⟩_{(l_{1},l_{2}),m}
(∀h_{1} ∈ H_{(l_{1},l_{2}),m}, ∀h_{2} ∈ H_{(l_{1},l_{2}+1),m}).

Thus, we get the assertions.

From this Lemma, we get c_{(l_{1},l_{2}),m} = c_{(l_{1},l_{2}),m±1}. So, we can again ignore the parameter m, and this settles the difference between G^{+} and G. From now on, we write c_{(l_{1},l_{2})} instead of c_{(l_{1},l_{2}),m}.

The recursive relations become

A^{+0}(l_{1}, l_{2})c_{(l_{1}+1,l_{2})} + \overline{A^{−0}(l_{1} + 1, l_{2})c_{(l_{1},l_{2})}} = 0,
A^{−+}(l_{1}, l_{2})c_{(l_{1},l_{2}+1)} + \overline{A^{+−}(l_{1}, l_{2} + 1)c_{(l_{1},l_{2})}} = 0.

The condition c_{(l_{1},l_{2})} ∈ ℝ is equivalent to c′ ∈ ℝ or c′ ∈ iℝ. The latter case is the unitary axis, the module is irreducible and unitarizable.

In case c′ ∈ ℝ, we get

c_{(l_{1}+1,l_{2})} = \frac{A^{−0}(l_{1} + 1, l_{2})}{A^{+0}(l_{1}, l_{2})}c_{(l_{1},l_{2})},

c_{(l_{1},l_{2}+1)} = \frac{A^{+−}(l_{1}, l_{2} + 1)}{A^{−+}(l_{1}, l_{2})}c_{(l_{1},l_{2})}.

Once we fix a c_{(l_{1},l_{2})} in an irreducible constituent, the other c_{(·,·)}’s are automatically determined. To make all the c_{(·,·)}’s positive, we have to look at the signature of A^{··}(l_{1}, l_{2}). A careful examination of these formulae and Thm.7.2 yields the following. The numbers correspond to those of Thm.7.2, except that we have to single out the case of trivial representation in case (1) and (5).

Theorem 8.3. The Harish-Chandra module of S^{c}(X^{+}) is irreducible and unitary if and only if c′ = −2c−p+1 ∈ iℝ(unitary axis) or (0), (3) −p+2 < c < −1.

[figure: -p+3; p-3; 0; Re c′; Im c′]

When it is reducible (i.e.c ∈ ℤ),

(1) c ≤ −p : 1 non-unitary finite-dimensional quotient, 1 non-unitary subquotient, 1 unitary submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; −c − p + 2; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; \begin{matrix}−c \\ −p + 3 \end{matrix}; unit.sub.]

(1bis) c = −p + 1 : 1 unitary 1-dimensional quotient, 1 non-unitary subquotient, 1 unitary submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; 1; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; 2; unit.quot.; unit.sub.]

(2) c = −p + 2 : 1 unitary quotient, 1 unitary submodule.

[figure: l_{1}; l_{2}; ⇒; ⇒; ⇒; ⇒; ⇒; ⇒; 0; ⇑; ⇑; ⇑; ⇑; ⇑; ⇑; 1; unit.sub.; unit.quot.]

(4) c = −1 : 1 unitary submodule, 1 unitary quotient.

[figure: l_{1}; l_{2}; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; 0; unit.quot.; unit.sub.]

(5)c ≥ 1 : 1 non-unitary finite-dimensional submodule, 1 non-unitary subquotient, 1 unitary quotient.

(5bis) c = 0 : 1 unitary 1-dimensional submodule, 1 non-unitary subquotient, 1 unitary quotient.

[figure: l_{1}; l_{2}; ⇐; ⇐; ⇐; ⇐; ⇐; ⇐; c; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; c + 1; unit.quot.]

[figure: l_{1}; l_{2}; ⇐; ⇐; ⇐; ⇐; ⇐; ⇐; 0; ⇓; ⇓; ⇓; ⇓; ⇓; ⇓; 1; unit.sub.; unit.quot.]

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Takashi Fujimura 154-0011 1-6-5 Kamiuma Setagaya-ku Tokyo Japan

Received February 26, 1998 and in final form April 20, 2000