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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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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Symmetric and exterior powers are representations

Statement

If $V$ is a representation of $g$ , the diagonal tensor action on $V^{\otimes n}$ preserves the symmetric and repeated-vector relation subspaces. It therefore descends to representations on $S^{n} (V)$ and $Λ^{n} (V)$ for every $n \geq 0$ and over every field.

Facts & Assumptions

Given: A Lie-algebra representation $V$ and an integer $n \geq 0$ .

[L1]

Iterating the tensor-product construction gives the diagonal action $x (v_{1} \otimes \dots \otimes v_{n}) = \sum_{i} v_{1} \otimes \dots \otimes x v_{i} \otimes \dots \otimes v_{n}$ (Direct-sum, dual, Hom, and tensor representations).

[L2]

The two quotient relation subspaces are those in Symmetric and exterior powers over an arbitrary field.

[L3]

A linear map killing a quotient relation subspace factors uniquely through the quotient (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

Proof

technique · direct

1.1

Every diagonal operator $D_{x}$ commutes with every permutation of tensor positions, because permuting after applying $x$ in one position gives the same summand as applying $x$ in the permuted position. Hence $D_{x} (t - π t) = D_{x} t - π D_{x} t$ lies in the symmetric relation subspace.

L1L2algebra

1.2

Consider a pure tensor with equal entries $v$ in positions $p \neq q$ . Terms of $D_{x}$ differentiating another position still have equal entries in positions $p, q$ . The sum of the two remaining terms, with $x v$ in position $p$ or $q$ , equals the tensor having $v + x v$ in both positions minus the tensors having $v$ in both and $x v$ in both; it therefore belongs to the span of repeated-vector tensors in every characteristic. Thus the exterior relation subspace is stable.

L1L2algebra

1.3

Stability makes $D_{x}$ induce an endomorphism on each quotient by [L3]. Since $[D_{x}, D_{y}] = D_{[x, y]}$ on $V^{\otimes n}$ , the same equality holds after passing to either quotient, and the induced actions are representations.

L1L3algebra

2.1

For $n = 0$ the resulting action on $k$ is zero, and for $n = 1$ it is the original action on $V$ ; the same construction proves the assertion for every $n$ without averaging or dividing by $n!$ .

step 1.1step 1.2step 1.3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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)