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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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

1.1L1L3algebra

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.

1.2L1algebra

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.

1.3L1algebra

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].

1.4L1L2algebra

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.

2.1step 1.1step 1.2step 1.3step 1.4∎

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

Depends on

Used by

Tensor-product multiplicities for finite-dimensional simple modules Definition
Clebsch–Gordan decomposition for sl2 Example
Tensor, dual, and Hom representation formulas Example
Three times three for sl3 Example
The dual action needs a minus sign False statement
The lowest weight space is the nilradical-invariant line Lemma
Formal characters are additive and multiplicative Proposition
Highest weight of the dual representation Proposition
Symmetric and exterior powers are representations Proposition
Top summand in a tensor product 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)