PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Direct-sum, dual, Hom, and tensor representations

Statement

Let $(V_{i})_{i \in I}$ be representations of $g$ , and let $V, W$ be representations. Then the following formulas define representations:

$x (v_{i})_{i \in I} = (x v_{i})_{i \in I},$

$(x λ) (v) = - λ (x v),$

$(x T) (v) = x T (v) - T (x v),$

$x (v \otimes w) = x v \otimes w + v \otimes x w,$

on $⨁_{i} V_{i}$ , $V^{*}$ , ${Hom}_{k} (V, W)$ , and $V \otimes_{k} W$ , respectively.

Facts & Assumptions

Given: Representations of one Lie algebra $g$ on all displayed vector spaces.

[L1]

Their operators satisfy $[ρ (x), ρ (y)] = ρ ([x, y])$ (Representations of Lie algebras).

[L2]

A bilinear map induces a unique linear map from a tensor product (Universal property of the tensor product for balanced maps into abelian groups).

[L3]

Elements of an algebraic direct sum have finite support (The direct sum of an indexed family of modules).

Proof

technique · direct

1.1

The componentwise formula preserves finite support by [L3]. Its commutator is componentwise, so [L1] gives $[x, y] (v_{i}) = (x (y v_{i}) - y (x v_{i}))_{i}$ ; hence it is a representation, including for an empty family.

L1L3algebra

1.2

For the dual formula, $(x (y λ) - y (x λ)) (v) = λ (y x v - x y v) = - λ ([x, y] v) = ([x, y] λ) (v)$ by [L1]. Thus the minus sign gives the required bracket action.

L1algebra

1.3

On ${Hom}_{k} (V, W)$ , expand the commutator of $T \mapsto ρ_{W} (x) T - T ρ_{V} (x)$ and its $y$ -analogue. The two mixed terms cancel, leaving $ρ_{W} ([x, y]) T - T ρ_{V} ([x, y])$ , which is the prescribed $[x, y]$ -action by [L1].

L1algebra

1.4

For each fixed $x$ , the displayed tensor formula is bilinear in $(v, w)$ , so [L2] gives a linear endomorphism of $V \otimes W$ . Applying the $x$ - and $y$ -operators successively to a pure tensor produces two mixed terms in each order; they cancel in the commutator, leaving $[x, y] v \otimes w + v \otimes [x, y] w$ . Since pure tensors span, this is the representation identity everywhere.

L1L2algebra

2.1

Each construction is linear in $x$ and satisfies the bracket identity, so all four displayed actions are Lie-algebra representations without finite-dimensional assumptions.

step 1.1step 1.2step 1.3step 1.4∎

Depends on

Used by

Tensor, dual, and Hom representation formulas Example
The dual action needs a minus sign False statement
Symmetric and exterior powers are representations Proposition

Dependency tree · two levels

15 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

Etingof, MIT 18.745 notes, §11.2, printed pp. 62–63 (standard reference, not scraped)
Kirillov, An Introduction to Lie Groups and Lie Algebras, §4.2, printed pp. 50–52 (standard reference, not scraped)