DefinitionDefinition: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The left action on feature maps induced by an action on sites

Definition

Let $G$ be a group acting on the left on a set $S$ , and let $V$ be a set of feature values. A feature map is a function $f : S \to V$ . For $u \in G$ , the induced transformation of feature maps is

(L_{u} f) (x) = f (u^{- 1} \cdot x) (x \in S) .

It is a left action on the set of feature maps: the group acts by reindexing sites and leaves the feature values unchanged. In particular, for maps $F : G \to V$ on the group itself with its left multiplication action, $(L_{u} F) (g) = F (u^{- 1} g)$ .

Definition references

left action of a group on a set

functions and declared domains/codomains

Cohen and Welling, “Group Equivariant Convolutional Networks,” §4.4, Eqs. (4)–(5)

Goodfellow, Bengio, and Courville, Deep Learning, Chapter 9 §9.2

Facts & Assumptions

Given: A group $G$ acts on the left on a set $S$ , and $f : S \to V$ is a feature map. Verify that the displayed transformations form a left action.

[F1]

Left group actions, transitive actions, and faithful actions: A left action satisfies $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ for every $g, h \in G$ and $x \in S$ .

[F2]

Group and abelian group: A group operation is associative, has an identity, and every element has a two-sided inverse.

Proof

1.1givenF1algebra

For each $x \in S$ , $(L_{e} f) (x) = f (e^{- 1} \cdot x) = f (x)$ by the identity law in [F1], so $L_{e} f = f$ .

2.1givenF1F2algebra∎

For $u, v \in G$ , the element $v^{- 1} u^{- 1}$ is a two-sided inverse of $u v$ , since $(u v) (v^{- 1} u^{- 1}) = u (v v^{- 1}) u^{- 1} = e$ and $(v^{- 1} u^{- 1}) (u v) = v^{- 1} (u^{- 1} u) v = e$ . Two two-sided inverses $a, b$ of an element $z$ coincide because $a = a (z b) = (a z) b = b$ ; hence $(u v)^{- 1} = v^{- 1} u^{- 1}$ . Therefore, for each $x \in S$ , $(L_{u} (L_{v} f)) (x) = f (v^{- 1} \cdot (u^{- 1} \cdot x)) = f ((v^{- 1} u^{- 1}) \cdot x) = f ((u v)^{- 1} \cdot x) = (L_{u v} f) (x)$ by [F1]. Thus $L_{u} L_{v} = L_{u v}$ .

Depends on

Used by

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

Taco S. Cohen and Max Welling, Group Equivariant Convolutional Networks (standard reference, not scraped)
Ian Goodfellow, Yoshua Bengio, and Aaron Courville, Deep Learning (standard reference, not scraped)