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Primitive vectors of the induced coordinate module

Statement

Assume the Axiom of Choice inherited from the named suppliers. Let $E (λ)$ be the induced coordinate module of a split reductive group $(G, T)$ with Borel $B$ and unipotent radical $U = B_{u}$ (The induced coordinate module E(lambda)). If $E (λ) \neq 0$ , then the space $E (λ)^{U}$ of $U$ -fixed elements is one-dimensional, its nonzero elements are primitive vectors of weight $λ$ (Primitive vectors for a Borel pair), and evaluation at the identity $f \mapsto f (1)$ is an isomorphism $E (λ)^{U} \to k$ . In particular $E (λ) \neq 0$ if and only if $E (λ)$ contains a primitive vector of weight $λ$ , unique up to scalar.

Facts & Assumptions

Given: A split reductive group $(G, T)$ with Borel $B \supseteq T$ , unipotent radical $U = B_{u}$ , opposite Borel $B^{0} = B^{-}$ , and an element $λ \in X (T)$ with $E (λ) \subseteq O (G)$ as in the cited definition.

[F1]

Big cell. $U \times B^{0} \to G$ , $(u, b) \mapsto u b$ , is an open immersion onto a dense open subscheme of $G$ , and $G$ is smooth, hence reduced (Bruhat decomposition for a split reductive group).

[F2]

Determination on the big cell. Two elements of $E (λ)$ agreeing on the image of $U \times B^{0}$ agree on $G$ : morphisms from the reduced scheme $G$ equal on a dense open are equal (Agreement on a schematically dense open).

[F3]

The action on $U$ . For $f \in E (λ)$ put $f_{U} (u) = f (u^{- 1})$ . For $u_{0} \in U (R)$ one has $(u_{0} f)_{U} (u) = f ((u u_{0})^{- 1}) = f_{U} (u u_{0})$ ; in particular $f \in E (λ)^{U}$ is fixed by every $U (R)$ exactly when $f_{U}$ is invariant under right translation by $U (R)$ (The induced coordinate module E(lambda), Rational representations of an affine group scheme are comodules of its coordinate Hopf algebra).

[F4]

Unipotent fixed vectors. A nonzero rational representation of the unipotent group $U$ has a nonzero $U$ -fixed vector; and a $U$ -invariant regular function on $U$ is constant, since right-translation invariance gives $f_{U} (u) = f_{U} (e)$ by translating the identity by $u$ (Unipotent algebraic groups and unipotent representations).

[F5]

Translation law. $f (u b) = f (u) λ (b^{- 1})$ for $u \in U (R)$ , $b \in B^{0} (R)$ , and $f (t) = f (1) λ (t^{- 1})$ for $t \in T (R)$ (The induced coordinate module E(lambda)).

Proof

technique · direct

1.1F1F2F5given

If $f \in E (λ)$ vanishes on $U$ , then $f = 0$ : by [F5] $f$ vanishes on the whole big cell $U \cdot B^{0}$ [F1], and [F2] applies.

1.2F4given

If $E (λ) \neq 0$ , then $E (λ)^{U} \neq 0$ : $E (λ)$ is a nonzero rational representation of the unipotent group $U$ , so it has a nonzero fixed vector by [F4].

2.1F3F4step 1.1step 1.2

For $f \in E (λ)^{U}$ the function $f_{U}$ is invariant under right translation by $U (R)$ for every $R$ by [F3], hence constant by [F4]; its constant value is $f_{U} (1) = f (1)$ , so $f$ is determined by the scalar $f (1)$ . Consequently the $k$ -linear evaluation map $E (λ)^{U} \to k$ , $f \mapsto f (1)$ , is injective; by step 1.2 it is nonzero (a nonzero fixed vector has $f (1) \neq 0$ by step 1.1), so $E (λ)^{U}$ is one-dimensional and evaluation is an isomorphism onto $k$ .

3.1F5step 2.1

Let $0 \neq f \in E (λ)^{U}$ . For $t \in T (R)$ one has $(t f) (1) = f (t^{- 1}) = f (1) λ (t)$ by [F5], so $t \cdot f = λ (t) f$ and $f$ is a $T$ -eigenvector of weight $λ$ . Since $f$ is $U$ -fixed, it is primitive of weight $λ$ by the definition.

4.1step 1.2step 2.1step 3.1∎

Conversely, a primitive vector of weight $λ$ in $E (λ)$ is $U$ -fixed, hence lies in the one-dimensional space $E (λ)^{U}$ of step 2.1; so it is unique up to a nonzero scalar, and its existence forces $E (λ) \neq 0$ . Together with steps 1.2, 2.1 and 3.1 this proves all assertions.

Remarks

The statement is the source's Proposition 22.22 in the conventions of this page: the fixed space is $E (λ)^{U}$ for the unipotent radical of the Borel $B$ used to define primitive vectors, and the big cell is $U \cdot B^{0} = U \cdot B^{-}$ . The earlier scaffold's formulation with $E (λ)^{U^{-}}$ was corrected here; the computations for ${SL}_{2}$ in the example show that $E (λ)^{U^{-}}$ is not the one-dimensional space of primitive vectors.
The one-dimensionality of $E (λ)^{U}$ is what makes the primitive vector "unique up to scalar" and underlies the classification.

Depends on

Used by

Every dominant weight of a split semisimple group is a primitive weight Lemma

Dependency tree · two levels

51 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Groups (corrected 2022 printing, Cambridge University Press) (standard reference, not scraped)
Robert Steinberg, Lectures on Chevalley Groups (Yale University, 1967; notes prepared by J. Faulkner and R. Wilson) (standard reference, not scraped)