PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Maurer--Cartan form is a pointwise isomorphism and left invariant

Statement

Assume ${A C}_{ω}$ . Let $θ$ be the left Maurer--Cartan form of a Lie group $G$ . Every fibre map $θ_{g} : T_{g} G \to g$ is a linear isomorphism, with inverse $d (L_{g})_{e}$ . For $h, g \in G$ and $V \in T_{g} G$ , define the pullback here by

$(L_{h}^{*} θ)_{g} (V) = θ_{h g} (d (L_{h})_{g} V) .$

Then $L_{h}^{*} θ = θ$ for every $h \in G$ . Moreover, for every $X \in g$ and its left-invariant extension $X^{L}$ ,

$θ_{g} (X_{g}^{L}) = X$

at every $g \in G$ . The countable-choice assumption is used exactly through the supplied Maurer--Cartan definition and invariant-extension theorem.

Facts & Assumptions

Given: ${A C}_{ω}$ , a Lie group $G$ with identity $e$ , elements $g, h \in G$ , a vector $V \in T_{g} G$ , and $X \in g = T_{e} G$ .

[F1]

${A C}_{ω}$ is countable choice. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F2]

The left Maurer--Cartan form is $θ_{g} = d (L_{g^{- 1}})_{g}$ , and its associated bundle map is the inverse of smooth left trivialization. Left Maurer--Cartan form.

[F3]

Left translations are $L_{a} (b) = a b$ , with inverse $L_{a^{- 1}}$ . Left and right translations on a Lie group.

[F4]

Differentials obey the chain rule. The chain rule for differentials of smooth maps.

[F5]

The left-invariant extension of $X \in g$ is $X_{g}^{L} = d (L_{g})_{e} X$ . Left-invariant vector fields evaluate isomorphically at the identity.

Proof

technique · direct

1.1

By [F2], the bundle map $V_{g} \mapsto (g, θ_{g} V_{g})$ is inverse to $Φ_{L} (g, X) = d (L_{g})_{e} X$ . Consequently each $θ_{g}$ is a linear isomorphism with inverse $d (L_{g})_{e}$ .

F2algebra

1.2

The group law in [F3] gives $L_{(h g)^{- 1}} \circ L_{h} = L_{g^{- 1}}$ . Therefore [F2] and the chain rule [F4] give $(L_{h}^{*} θ)_{g} (V) = d (L_{(h g)^{- 1}})_{h g} d (L_{h})_{g} V = d (L_{g^{- 1}})_{g} V = θ_{g} (V) .$ Since $g$ and $V$ were arbitrary, $L_{h}^{*} θ = θ$ .

F2F3F4algebra

2.1

By [F5], [F2], and step 1.1, $θ_{g} (X_{g}^{L}) = θ_{g} (d (L_{g})_{e} X) = X .$ Thus the $g$ -valued function $θ (X^{L})$ is the constant function with value $X$ .

F2F5step 1.1

3.1

A Lie group is nonempty. In dimension zero all tangent spaces are zero and the unique fibre maps give every asserted identity; in dimension one the same proof applies. Lie groups are boundaryless by convention, and no metric, nondegeneracy, or endpoint occurs. The stated ${A C}_{ω}$ is inherited through [F2] and [F5]; the chain-rule identities are pointwise and add no selection. The proposition asserts equalities and an explicit inverse, not a biconditional.

F1F2F3F4F5step 1.1step 1.2step 2.1∎

Depends on

Used by

Maurer--Cartan structure equation Theorem

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

Robert L. Bryant, An Introduction to Lie Groups and Symplectic Geometry (standard reference, not scraped)