Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 38 (1997), No. 1, 135-143.

Commutativity of the Schur Algebra

Mina Teicher

Department of Mathematics and Computer Science Bar-Ilan University Ramat-Gan 52900, Israel

Abstract. In this paper we study the Schur algebra of the induced representation of the Weil representation of a certain subgroup of GL_{2}(C) and prove that it is commutative. It thus implies that the representation is multiplicity free.

Introduction

This paper is based on my talk given at the Schur-Gedenkkolloquium which was held at Humboldt Universität in Berlin.

Let G be an algebraic group, U a maximal unipotent subgroup of G.

U ≃ ( \begin{matrix}1 & ∗ & … & ∗ \\ & ... & 0 & ⋮ \\ & 0 & ... & ∗ \\ & & & 1 \end{matrix} )

For almost all G, every irreducible representation π of G is isomorphic to some component of the induced representation of a character ψ of U, (i.e., π ⊂ Ind_{U}^{G} ψ). Therefore, to know all the irreducible components of Ind_{U}^{G} ψ is like knowing all the irreducible representations of G. Thus, studying Ind_{U}^{G} ψ is very important.

An irreducible component of Ind_{U}^{G} ψ which is isomorphic to π is called the Whittaker Model of π. If we would know that all irreducible components appear only once in Ind_{U}^{G} ψ (i.e., Ind_{U}^{G} ψ is multiplicity free), then we could easily find all irreducible representations of G and conclude that the Whittaker Model is unique.

In 1962, Gelfand and Graev (see [1]) proved that for an algebraic group G and for U and ψ, as above, Ind_{U}^{G} ψ is multiplicity free. In 1974, Shalika proved it for GL(n, k) (see [5]) where k is a local field. In 1977, Piatetski-Shapiro proved it for GL(n, 𝔸) over a global field (see [3], [4]).

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix}1997 Heldermann Verlag

In this manuscript we shall discuss a generalized Whittaker Model, where U is replaced by D = M_{1}U and ψ is replaced by ω_{ψ}, the Weil representation of M_{1}U. We shall outline the proof that Ind_{D}^{G} ω_{ψ} is multiplicity free. We shall do it by proving that the Schur algebra of Ind_{D}^{G} ω_{ψ} is multiplicity free.

We shall start by recalling a few definitions. Then we shall explain how the Schur lemma connects to multiplicity problems. We shall discuss the Schur algebra of the induced representation. Then we shall present the group G which we are studying, and the group D = M_{1}U. We shall construct the Weil representation W_{ψ} of D and then outline the proof of “Ind_{D}^{G} ω_{ψ} is multiplicity free.”

The paper is divided as follows:

Introduction

1. Definitions and notations

2. The Schur algebra and multiplicity problems

3. The induced representation

4. Introducing the groups D < G

5. The Weil representation of D

6. The main theorem

1. Definitions and notations

In this paper G will denote a group. We recall basic definitions and notations:

T : (π_{1}, V_{1}) → (π_{2}, V_{2}) is a homomorphism of representations if: T : V_{1} → V_{2} is a homomorphism of vector spaces, and T π_{1}(g) = π_{2}(g)T ∀ g ∈ G.

(π_{1}, V_{1}) ≃ (π_{2}, V_{2}) are equivalent representations if: there exist T : (π_{1}, V_{1}) ⟶^{∼} (π_{2}, V_{2}) isomorphism of representations. (π_{1} ∼_{T} π_{2}).

V_{1} is an invariant subspace of V if: V_{1} is a subspace of V and π(g)v_{1} ∈ V_{1} ∀v_{1} ∈ V_{1}. We then have a subrepresentation, π |_{V_{1}} .

(π, V) is irreducible if: V has no invariant subspace.

(π, V) is decomposable if: V = ⊕ ∑ V_{i}, where V_{i} is an invariant subspace and π|_{V_{i}} is irreducible.

(π, V) is a finite dimensional representation if: dim V < ∞. A one-dimensional representation, a character, is denoted by χ.

(π, V) is multiplicity free if: V = ⊕ ∑_{i=1}^{m} V_{i}, V_{i} invariant and irreducible,

V_{i} ≁ V_{j} ∀ i ≠ j.

We recall the following classical theorem:

Theorem 1.1 If G is compact and π is a finite dimensional representation, then (π, V) is decomposable.

2. The Schur algebra and multiplicity problems

The basic theorem of representation theory is the so-called “Schur Lemma”.

Schur Lemma 2.1 Let (π_{1}, V_{1}), (π_{2}, V_{2}) be two finite dimensional irreducible representations of G. Let f : V_{1} → V_{2} be a homomorphism of the representations, fπ_{1}(g) = π_{2}(g)f ∀ g ∈ G. Then:

π_{1} ≁ π_{2} ⇒ f ≡ 0

π_{1} ∼_{T} π_{2} ⇒ f = λT λ ∈ ℂ.

By the Schur lemma, we transfer multiplicity problems of representations to commutativity properties of algebras. We recall:

Definition. (Schur algebra, Aut_{G}(V))

Aut_{G}(V) = automorphism ring of a representation π : G → Aut(V)
= {f : V → V; f homomorphism, fπ(g) = π(g)f ∀ g ∈ G}.

The following is an immediate result of the Schur lemma, which we shall use in the proof of the main result:

Theorem 2.2 Let (π, V) be a finite dimensional decomposable representation of G, V =

∑_{i=1}^{m} n_{i}V_{i}. Then π is multiplicity free ⇔ Aut_{G}(V) is commutative.

The proof follows from the fact that Aut_{G} (V) = ∑ M_{n_{i}} (ℂ). Thus Aut_{G}(V) is commutative ⇔ ∑ M_{n_{i}} (ℂ) is commutative ⇔ n_{i} = 1 ∀i ⇔ (π, V) is multiplicity free.

3. The induced representation

The result in this paper deals with induced representation. We recall:

Definition. (Induced Representation.) Let D be a subgroup of G, let (π, W) be a representation of D, dim W < ∞. The induced representation of π from D to G, denoted Ind_{D}^{G} π is: (V, τ) where

V = {f : G → W |f(δg) = π(δ)f(g) V δ ∈ D, ∀g ∈ G}

τ(g_{0})f(g) = f(gg_{0}).

The Schur algebra of an induced representation has a model which is more convenient for us. We prove:

Theorem 3.1 Let G be a finite group D < G, (π, W) a representation of D. (V, τ) = Ind_{D}^{G} π. Then

Aut_{g} V ≃^{S} {K : G → End_{C} W |K(δ_{1}gδ_{2}) = π(δ_{1})K(g)π(δ_{2})
∀ δ_{1}, δ_{2} ∈ D ∀ g ∈ G}

K_{1} ∗ K_{2}(g) = ∑_{X∈G} K_{1}(gX^{−1})K_{2}(x).

Proof. By the definition of induced representation V = {f : G → W |f(δg) = π(δ)f(g)}. For K : G → End_{C} W as above we associate S(K) : V → V which is defined by S(K)(f)(g) = ∑_{h∈G/D} K(gh^{−1})(f(h)). S turns out to be an isomorphism.

4. Introducing the groups D < G

In the sequel k is a finite field, char(k) ≠ 2, and ⟨, ⟩ is an antisymmetric form on k^{∗}. Let G = {g ∈ GL_{2n}(ℂ)|⟨gx, gy⟩ = ⟨x, y⟩ · λ(g) ∀ x, y ∈ k^{2n}}. Take e_{1}, an isotropic vector, i.e., ⟨e_{1}, e_{1}⟩ = 0. Let Q = {g ∈ G|g(e_{1}) = αe_{1} for some

α ∈ k^{∗}}. The group Q is decomposed to a unipotent part and a reductive part (classically known as a Levi decomposition), i.e., Q = MU where:

M ≃ S_{p}(2n − 2) × k^{∗}

U ≃ Heisenberg group of order 2n − 2.

Let M_{1} = {m ∈ M|mu = um ∀ u ∈ Center (U)}. We define: D = M_{1}U.

We shall compute G, D for the special case where 2n = 4 and

⟨x, y⟩ = ^{t}xJy  J = ( \begin{matrix}& & & 1 \\ & & 1 \\ & −1 \\ −1 \end{matrix} )  x, y ∈ k^{4}.

In this case G = {g ∈ GL_{n}(ℂ)| ^{t}gJg = λ(g)J for some λ(g) ∈ k^{∗}} and is denoted by GS_{p}(4, k).

Let {e_{i}|i = 1, …, 4} be the standard basis of k^{4}. Then Q = {g ∈ G|g(e_{1}) = αe_{1}} is equal to

Q = \{( \begin{matrix}α \\ 0 & & ∗ \\ 0 \\ 0 & 0 & 0 & β \end{matrix} )\}

Let E = span{e_{2}, e_{3}}. Let M = {g ∈ Q|g(E) = E and g(e_{1}) = βe_{4} for some β ∈ k^{∗}}. Thus:

M = {( α ; h ; β )| h ∈ GS_{p}(2, k) αβ = λ(h) = λ(g), }

Clearly, M = GS_{p}(2, k) × k^{∗}.

Let U = {g ∈ G|g(e_{1}) = e_{1}, g(e) = e + αe_{1} for some α ∈ k ∀ e ∈ E}. Thus

U = \{( \begin{matrix}1 & x & y & z \\ & 1 & 0 & y \\ & & 1 & −x \\ & & & 1 \end{matrix} )| x, y, z ∈ k\} .

Let us denote (x, y, z) = ( \begin{matrix}1 & x & y & z \\ & 1 & 0 & y \\ & & 1 & −x \\ & & & 1 \end{matrix} ) . By matrix multiplication we get:

(x, y, z) · (x_{1}, y_{1}, z_{1}) = (x + x_{1}, y + y, xy_{1} − x_{1}y + z + z_{1}).

Evidently: Z(U) = {(0, 0, k)}. Let M_{1} = {m ∈ M|mu = um ∀ u ∈ Z(U)}. Then:

M_{1} = \{( \begin{matrix}α \\ & h \\ & & α \end{matrix} )| h ∈ GS_{p}(2, k), λ(h) = α^{2}\}

= \{α ( \begin{matrix}1 \\ & h′ \\ & & 1 \end{matrix} )| h′ ∈ S_{p}(2, k), α ∈ k^{∗}\} .

Clearly, M_{1} ∼ S_{p}(2, k) × k^{∗}. Since M_{1}U = UM_{1}, M_{1}U is actually a subgroup and we denote D = M_{1}U.

5. The Weil representation of D

We construct the Weil representation of D in several steps:

a) The Weil representation of U Let ψ be a nontrivial additive character of k, ψ ≢ 1. Let X = {(x, 0, z)} ⊆ U. Clearly, X is commutative. We extend ψ to X by ψ(x, 0, z) = ψ(z). Its induced representation to U, ω_{ψ} = Ind_{X}^{U} ψ is called the Weil representation of U.

It is well-known that ω_{ψ} is the only irreducible representation of U whose restriction to Z(U) is ψ. Let us denote its vector space by V.

b) The Weil representation of S_{p}(2n − 2, k) The group S_{p}(2n − 2, k) acts on U/Z(U) ≃ k^{2n−2} in a natural way. This action defines ∀ g ∈ S_{p}(2n − 2, k) a new representation of U : u → ω_{ψ}(gu). By the above theorem, ω_{ψ}(·) and ω_{ψ}(g·) are equivalent. Then ∀ g ∈ S_{p}(2n − 2, k) ∃ A(g) equivalence operator A(g) : V → V, s.t. ω_{ψ}(gu) = A(g)^{−1}ω_{ψ}(u)A(g).

It turns out that generally A(g_{1}g_{2}) = C(g_{1}g_{2})A(g_{1})A(g_{2}) (i.e., A(g) is a projective representation). But for k finite, C(g_{1}g_{2}) ≡ 1 and thus, A(g) is a representation. We call A : S_{p}(2n − 2, k) → Aut(V) the Weil representation of S_{p}(2n − 2, k).

c) The Weil representation of M_{1} Let μ be a multiplicative character of k^{∗}. Recall that M_{1} = S_{p}(2n − 2, k) × k^{∗}. We define ω_{ψ} : M_{1} → Aut(V) by: ω_{ψ}(m, α) = μ(α)A(m).

d) The Weil representation of D The Weil representations of M_{1} and U are compatible (see [2]), so they glue to a representation of D, ω_{ψ} : D → Aut(V).

In what follows we need precise formulas for ω_{ψ} and thus we compute:

Formulas of ω_{ψ} : U → Aut(V) for 2n = 4 and J = ( \begin{matrix}& & & 1 \\ & & 1 \\ & −1 \\ −1 \end{matrix} )

V = {f : U → ℂ|f((x, 0, z)u) = ψ(z)f(V)}
V ∼ {f : {(0, y, 0)} → ℂ} ∼ {f : k → ℂ}.

The action:

ω_{ψ}(0, y, 0)f(w) = f(w + y)
ω_{ψ}(x, 0, 0)f(w) = f((0, w, 0) · (x, 0, 0))
= f(x, w, −xw)
= f(x, 0 − 2xw) · (0, w, 0)
= ψ(−2xw)f(w)
ω_{ψ}(0, 0, z)f(w) = f((0, w, 0) · (0, 0, z))
= f(0, w, z))
= f((0, 0, z) · (0, w, 0))
= ψ(z)f(w).

Formulas of ω_{ψ} on S_{p}(2, k)

ω_{ψ} ( \begin{matrix}1 & 0 \\ c & 1 \end{matrix} ) f(w) = ψ (\frac{1}{2}cw^{2}) f(ω)
ω_{ψ} ( \begin{matrix}a \\ & a^{−1} \end{matrix} ) f(w) = f(wa^{−1})
ω_{ψ} ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) f(w) = ∑_{c∈k}(cw)f(c)

In [6], A. Weil, who constructed this representation, proved that it is irreducible. Thus we have:

Theorem 5.1 (1964, Weil) The representation ω_{ψ} of D = M_{1}U is irreducible.

We are ready now to formulate and prove our main result.

6. The main theorem

We know (see §I) that the induced representation of the Weil representation of D, Ind_{D}^{G} ω_{ψ}, is decomposable. We prove here that it is multiplicity free.

Theorem. Ind_{D}^{G} ω_{ψ} is multiplicity free.

Proof. Let us denote the space of ω_{ψ} by V, V = {f : k → c}. By Theorem 2.2 it is enough to prove that the Schur algebra of Ind_{D}^{G} ω_{ψ} is commutative. By Theorem 3.1, we have another presentation of the Schur algebra of an induced representation, i.e., the Schur algebra ≃ {K : G → Aut(V) | K(δ_{1}gδ_{2}) = ω_{ψ}(δ_{1})K(g)ω_{ψ}(δ_{2}) ∀ δ_{1}, δ_{2} ∈ D ∀ g ∈ G}. We denote this algebra by K, and we shall prove that K is commutative.

To prove that K is commutative we introduce an involution on K, i.e., a function τ, τ : K → K such that: ττ(K) = K ∀ K ∈ K and τ(K_{1}K_{2}) = τ(K_{2})τ(K_{1}) ∀K_{1}, K_{2} ∈ K and we prove that it is actually the identity. Denote ^{τ} K = τ(K). We shall prove τ K = K ∀ K ∈ K. We do it in the following steps:

1) Introduce an involution on G that stabilizes D and stabilizes Z(U) elementwise.

2) Choose a base B of V s. t. ω_{ψ}(^{τ} δ) = ^{t}(ω_{ψ}(δ)) for t transposes w.r.t. B.

3) Define ^{τ} K(g) = ^{t}K(^{τ} g) and prove that ^{τ} K ∈ K, (thus τ is an involution on K).

4) Find representatives for Q\G/Q.

5) Find representatives for M_{1}\M and M/M_{1}.

6) Find representatives for D\G/D.

7) Prove that ^{τ} K(η) = K(η) for every representative from step 6.

8) Deduce that ^{τ} K = K ∀K ∈ K and thus K is commutative.

1) Let T = T_{4} = ( \begin{matrix}& & & 1 \\ & & 1 \\ & 1 \\ 1 \end{matrix} ). Clearly T is symmetric of order 2. We define an

involution ^{τ} g on G as follows: ^{τ} g = ^{t}T gT. Since ^{t}T = T, T^{2} = T and t is an involution, we get that ^{τ} (^{τ} g) = g, ^{τ} (g_{1}g_{2}) = ^{τ} g_{2} ^{τ} g_{1}, and τ is an involution. Clearly, ^{τ} (x, y, z) =

(−x, y, z) and ^{τ} ( \begin{matrix}α \\ & αh \\ & & α \end{matrix} ) = ( \begin{matrix}α \\ & αT_{2}^{t}hT_{2} \\ & & α \end{matrix} ). Thus, τ stabilizes D and stabilizes Z(U)

elementwise.

2) There exists the following theorem:

Theorem. Let G be a group, (π, V) an irreducible representation of G, τ an involution on G. Then there exists a base B of V such that the corresponding transpose t satisfies π(^{τ} g) = ^{t}π(g).

As quoted in Theorem 5.1, ω_{ψ} is irreducible and thus we can apply it to the above theorem to get a transpose t such that: ω_{ψ}(^{τ} δ) = ^{t}ω_{ψ}(δ) ∀ δ ∈ D.

3) We define ^{τ} K by ^{τ} K(g) = ^{t}(K(^{τ} g)). Now: ^{τ} K(δ_{1}gδ_{2}) = ^{t} (K (^{τ} δ_{2} ^{τ} g ^{τ} δ_{1})) = t (ω_{ψ}(^{τ} δ_{2}) ^{τ} K(g)ω_{ψ}(^{τ} δ_{1})) = ^{t} (^{t}ω_{ψ}(δ_{2}) ^{t}K(^{τ} g)) ^{t}ω_{ψ}(δ_{1})) = ω_{ψ}(δ_{1}) ^{τ} K(g)ω_{ψ}(δ_{2}). Thus, τ K ∈ K. It is easy to check that τ is an involution on K.

4) Consider ℙ^{3} = ℙ(k^{4}) = {ℓ(x)|x ∈ k^{4} − {0}} . G acts transitively on ℙ^{3} by g(ℓ(x)) = ℓ(g(x)). Let Q be the stabilizer of ℓ(e_{1}). Q also acts on ℙ^{3} and there are 3 orbits to this action:

(1) {ℓ(e_{1})}.

(2) {ℓ(e)|⟨e, e_{4}⟩ ≠ 0 ℓ(e) ≠ ℓ(e_{1})} = orbit (ℓ(e_{3})).

(3) {ℓ(e)|⟨e, e_{4}⟩ = 0} = orbit (ℓ(e_{4})).

Since G acts transitively on ℙ^{3}, ℙ^{3} ≃ G/Q. Since there are 3 orbits to the action of Q on ℙ^{3} we have Q\G/Q ≃ {3 orbits}. In this correspondence an orbit of ℓ(e) corresponds to \overline{g} ∈ Q\G/Q such that g(e_{1}) = e. Thus, Q\G/Q has 3 cosets whose representatives are:

( \begin{matrix}I_{2} \\ & I_{2} \end{matrix} ) , ( \begin{matrix}& I_{2} \\ I_{2} \end{matrix} ) , ( \begin{matrix}& T_{2} \\ T_{2} \end{matrix} ).

5) We shall prove that \{( \begin{matrix}I_{2} \\ & αI_{2} \end{matrix} ) α ∈ k^{∗}\} are representatives of M/M_{1}. The proof that they are also representatives for M_{1}\M is similar.

( \begin{matrix}α \\ & g \\ & & β \end{matrix} ) = ( \begin{matrix}\frac{α}{β}I_{2} \\ & I_{2} \end{matrix} ) ( \begin{matrix}β \\ & g^{1} \\ & & β \end{matrix} ) .

λ ( \begin{matrix}α \\ & g \\ & & β \end{matrix} ) = αβ, λ ( \begin{matrix}\frac{α}{β} I_{2} \\ & I_{2} \end{matrix} ) = \frac{α}{β}. Thus, ( \begin{matrix}β \\ & g^{1} \\ & & β \end{matrix} ) is also an element of M and λ ( \begin{matrix}β \\ & g^{1} \\ & & β \end{matrix} )

= β^{2}. Thus, ( \begin{matrix}β \\ & g^{1} \\ & & β \end{matrix} ) ∈ M_{1}. Thus, \{( \begin{matrix}\frac{α}{β} I_{2} \\ & I_{2} \end{matrix} )\} are representatives for M/M_{1}. We replace

\frac{α}{β} by α and we get that \{( \begin{matrix}αI_{2} \\ & I_{2} \end{matrix} ) | α ∈ k^{∗}\} is a set of representatives for M/M_{1}.

6) From 4) and 5) we get that we have 3 types of representatives for Q\G/Q.

type I: ( \begin{matrix}αI_{2} \\ & I_{2} \end{matrix} ) α ∈ k^{∗}.

type II: ( \begin{matrix}0 & αI_{2} \\ I_{2} & 0 \end{matrix} ) α ∈ k^{∗}.

type III: ( \begin{matrix}0 & αT_{2} \\ T_{2} & 0 \end{matrix} ) α ∈ k^{∗}.

7) We shall only prove here that ^{τ} K(η) = ^{τ} K for every η of type I. Let η = ( \begin{matrix}αI_{2} & 0 \\ 0 & I_{2} \end{matrix} ) α ∈ k^{∗}. We want to prove that ^{τ} K(η) = K(η) for every K ∈ K.

Case I: α ≠ 1 Since α ≠ 1 and ψ is not trivial, there exists c ∈ k such that ψ(c) ≠ ψ (\frac{c}{α}) . Let K ∈ K.

Since ( \begin{matrix}1 & & & c \\ & 1 \\ & & 1 \\ & & & 1 \end{matrix} ) η = η ( \begin{matrix}1 & & & \frac{c}{α} \\ & 1 \\ & & 1 \\ & & & 1 \end{matrix} ), we have ω_{ψ} ( \begin{matrix}1 & & & c \\ & 1 \\ & & 1 \\ & & & 1 \end{matrix} ) K(η) = K(η)ω_{ψ} ( \begin{matrix}1 & & & \frac{c}{α} \\ & 1 \\ & & 1 \\ & & & 1 \end{matrix} ) .

Thus ψ(c)K(η) = K(η)ψ (\frac{c}{α}) . Since ψ(c) ≠ ψ (\frac{c}{α}) , we get K(η) = 0. This is true for every K ∈ K and thus also ^{τ} K(η) = 0. In particular, ^{τ} K(η) = K(η).

Case II: α = 1 In this case η = I and δη = ηδ ∀ δ ∈ D. Let K ∈ K. We have ω_{ψ}(δ)K(η) = K(η)ω_{ψ}(δ) ∀ δ ∈ D, and thus by the Schur lemma, 2.1, K(η) = c · I. Since ^{τ} η = η we get ^{τ} K(η) = ^{t}K(η) = ^{t}(cI) = cI = K(η).

8) Take K ∈ K. Since ^{τ} K(η) = K(η) for every representative of Q\G/Q by 7), it is true for every element of G thus ^{τ} K = K. Thus, the involution τ on K is the identity, therefore, K is commutative.

References

[1] Gelfand, I. M.; Graev, M. I: Construction of irreducible representations of simple algebraic groups over finite field. Acad. Sci. USSR 147 (1962), no. 3, 529. Russian.

[2] Gerardin, P.: Weil’s representations associated to finite fields. J. Algebra 46 (1977), 54–101.

[3] Piatetski-Shapiro, I. I.: Euler Subgroups in Lie Groups and their Representations. (I. M. Gelfand ed.) John Wiley, New York-Toronto 1971, 597–620.

[4] Piatetski-Shapiro, I. I.: Multiplicity one theorems. Proc. Symp. Pure Math. 33 (1979), 209–212.

[5] Shalika, J. A.: The multiplicity one theorem for GL(n). Ann. Math. 100 (1974), 171–193.

[6] Weil, A.: Sur certain groupe d’opérateurs unitaires. Acta Math. (Upsala) 11 (1964), no. 1–4, 143–211.

Received January 3, 1996