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Tensor products of primitive vectors

Statement

Assume the Axiom of Choice inherited from the named suppliers. Let $(V, r)$ and $(V^{'}, r^{'})$ be rational representations of a split reductive group $(G, T)$ with primitive vectors $v, v^{'}$ of weights $λ, λ^{'}$ (Primitive vectors for a Borel pair). Then $v \otimes v^{'}$ is a primitive vector of $V \otimes V^{'}$ of weight $λ + λ^{'}$ . Consequently tensor powers and tensor products of primitive vectors are primitive, with the summed weights.

Facts & Assumptions

Given: Rational representations $(V, r)$ , $(V^{'}, r^{'})$ of the split reductive group $(G, T)$ with primitive vectors $v \in V$ , $v^{'} \in V^{'}$ and weights $λ, λ^{'}$ , and the unipotent radical $U = B_{u}$ of a Borel subgroup $B \supseteq T$ .

[F1]

Tensor coactions. The representation/comodule dictionary applies to arbitrary vector spaces. For finite-dimensional pairs, Tensor products, exterior powers and Hom spaces of finite-dimensional rational representations are rational gives the tensor coaction explicitly. The same formula from the two finite coaction expansions on each elementary tensor is defined in arbitrary dimension by the tensor universal property; its counit and coassociativity identities follow from those of the two factors and multiplicativity of the Hopf-algebra counit and coproduct. This is verified directly in step 1.1. (Rational representations and comodules of an affine group scheme, Universal property of the tensor product for balanced maps into abelian groups)

[F2]

Primitive vectors. A nonzero vector is primitive of weight $χ$ exactly when it is fixed by $U$ and is a $T$ -eigenvector with character $χ$ ; in particular $u \cdot v = v$ and $t \cdot v = λ (t) v$ for all $R$ -points $u \in U (R)$ , $t \in T (R)$ , and likewise for $v^{'}$ (Primitive vectors for a Borel pair, Rational representations and comodules of an affine group scheme).

[F3]

Nonzero elementary tensors. Under the existing AC premise every vector space has a basis (Every vector space has a basis). For a nonzero vector choose a nonzero coordinate in that basis and rescale its coordinate functional to value $1$ . Thus there are $ϕ \in V^{*}$ and $ϕ^{'} \in (V^{'})^{*}$ with $ϕ (v) = ϕ^{'} (v^{'}) = 1$ . Their bilinear product defines a linear functional on $V \otimes V^{'}$ taking $v \otimes v^{'}$ to $1$ , by the tensor universal property. (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , Universal property of the tensor product for balanced maps into abelian groups)

Proof

technique · direct

1.1F1F3givenalgebra

For finite-dimensional $V, V^{'}$ , [F1] supplies the tensor representation. In arbitrary dimension, write $ρ (v) = \sum_{i} v_{i} \otimes a_{i}$ and $ρ^{'} (v^{'}) = \sum_{j} v_{j}^{'} \otimes b_{j}$ . Each expansion is finite, even for arbitrary-dimensional representations. The formula $c (v \otimes v^{'}) = \sum_{i, j} v_{i} \otimes v_{j}^{'} \otimes a_{i} b_{j}$ is induced by a bilinear map, hence defines a linear coaction candidate by [F1]. Its counit sends this expression to $v \otimes v^{'}$ because $ε (a_{i} b_{j}) = ε (a_{i}) ε (b_{j})$ . Its two iterated coactions agree because the coactions of both factors are coassociative and $Δ (a_{i} b_{j}) = Δ (a_{i}) Δ (b_{j})$ . Thus the representation/comodule dictionary gives a rational representation on $V \otimes V^{'}$ ; evaluating at any $R$ -point $g$ gives $g (v \otimes v^{'}) = g v \otimes g v^{'}$ . The two coordinate functionals of [F3] evaluate $v \otimes v^{'}$ to $1$ , so this vector is nonzero. This proves all tensor inputs needed below in arbitrary dimension.

1.2F1F2

For every $R$ -point $t \in T (R)$ one has $t \cdot (v \otimes v^{'}) = t \cdot v \otimes t \cdot v^{'} = λ (t) λ^{'} (t) (v \otimes v^{'}) = (λ + λ^{'}) (t) (v \otimes v^{'})$ , so $v \otimes v^{'}$ is a $T$ -eigenvector of weight $λ + λ^{'}$ .

2.1F1F2step 1.1

For every $R$ -point $u \in U (R)$ one has $u \cdot (v \otimes v^{'}) = u \cdot v \otimes u \cdot v^{'} = v \otimes v^{'}$ , so $v \otimes v^{'}$ is fixed by $U$ .

3.1F2step 1.1step 2.1step 1.2

By [F2] a nonzero $U$ -fixed $T$ -eigenvector of weight $λ + λ^{'}$ is primitive of that weight, so $v \otimes v^{'}$ is primitive of weight $λ + λ^{'}$ .

4.1step 3.1algebra∎

Iterating the construction gives that tensor powers $v^{\otimes m}$ are primitive of weight $m λ$ and tensor products of finitely many primitive vectors are primitive with the sum of the weights. For $m = 0$ , the empty tensor is $1$ in the trivial module $k$ , primitive of weight $0$ .

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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)