Exercise set 2.2

Throughout, K denotes a field and V a vector space over K of finite dimension d ≥ 1. We fix an integer m ≥ 1 and denote by G the group GLK(m). We denote by AK(m) the ring of polynomial functions on G and by SK(m) and SK(m,n) the appropriate Schur algebras (as defined in the lecture).

(1)
(Difficulty level 1) Let U and W be finite dimensional K-vector spaces. Convince yourself that the following are equivalent for a (set) map φ : U → W:
(a)
There exists a basis {wi} of W such that the K-valued functions φi on U defined by φ(u) = ∑ iφi(u)wi are all polynomial.
(b)
For any basis {wi} of W, the K-valued functions φi on U defined by φ(u) = ∑ iφi(u)wi are all polynomial.
(c)
For any linear functional ζ on W, the K-valued function ζ ∘ φ on U is polynomial.
(d)
For any polynomial function f on W, the K-valued function f ∘ φ on U is polynomial.

If any of these holds, then φ is said to be a polynomial map from U to W.

(2)
(Difficulty level 2) Let ρ : G → GL(V ) be a polynomial representation and ρ′ : G → GL(V *) the contragredient representation of ρ. Observe that g↦→ρ′(g)-1 is a polynomial map from G to EndKV * (although not a group homomorphism, but only an anti-homomorphism).
(3)
(Difficulty level 1) Verify that SK(m,1) is isomorphic to the matrix algebra Mm(K).
(4)
(Difficulty level 2) For α in SK(m), let αn denote its projection to SK(m). (Recall that this projection is induced from the inclusion of homogeneous polynomials of degree n in the ring AK(m).) We think of α↦→αn as a map from SK(m) to SK(m), considering αn to be an element of SK(m) under the inclusion of SK(m,n) ⊆ SK(m). (Recall that this inclusion is induced by the projection of AK(m) onto its homogeneous component of degree n.) Show that δI,n (where I stands for the identity element of G, the m×m identity matrix), as n varies over the non-negative integers, are pairwise orthogonal central idempotents. (Caution: they are not primitive central idempotents, except in very special cases, as we will see.)
(5)
(Difficulty level 3) We now outline a proof of the fact that a (finite dimensional) polynomial SK(m)-module arises from a polynomial representation of G. More precisely, we show that any polynomial SK(m)-module ˜ρ : SK(m) → EndK(V ) that is homogeneous of degree n arises from a homogeneous polynomial representation ρ : G → GL(V ) of G of degree n. The candidate for ρ is clear: ρ(x) := ˜ρ (δx) for x in G. It is also clear that ρ is a representation since δxy = δx ⋆ δy). It remains only to prove that ⟨ξ,ρ(g)v⟩ is a polynomial function of g, as g varies over G (for any fixed v in V and ξ in V *).

Since SK(m,n) is the dual of the finite dimensional K-vector space AK(m,n), it is clear that there exists a homogeneous polynomial cξ,v in AK(m,n) such that

           ∫
⟨ξ,ρ˜(αn)v⟩ =   cξ,v(x)dαn(x)  for all αn in SK (m, n).

Since v = ˜ρ (δI,n)v, it follows that

˜ρ(α)v = ˜ρ(α )(˜ρ(δI,n)v) = ˜ρ(α⋆ δI,n)v = ˜ρ(αn)v for all α in SK (m).

On the other hand, since cξ,v is homogeneous of degree n, we have

∫            ∫

  cξ,vdα (x) =   cξ,vdαn(x)  for all α in SK(m ).

From the equations in the last three displays, we conclude that

           ∫
⟨ξ,˜ρ(α)v⟩ =  cξ,v(x)dα(x)  for all α in SK(m ).

Putting α = δg, we obtain ⟨ξ,ρ(g)v⟩ = cξ,v(g). □