TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Lie-group homomorphisms have constant rank

Statement

If $F : G \to H$ is a smooth Lie-group homomorphism, then

$rank (d F_{g}) = rank (d F_{e})$

for every $g \in G$ . In particular, $F$ is a constant-rank smooth map, with no connectedness assumption on either group.

Facts & Assumptions

Given: A smooth Lie-group homomorphism $F : G \to H$ and $g \in G$ .

[F1]

Left translations are diffeomorphisms, so their differentials are linear isomorphisms. Left and right translations on a Lie group.

[F2]

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

Proof

technique · direct

1.1

The homomorphism identity is $F \circ L_{g} = L_{F (g)} \circ F$ . Differentiating it at $e$ and using [F2] gives $d F_{g} \circ d (L_{g})_{e} = d (L_{F (g)})_{e} \circ d F_{e}$ .

F2algebra

2.1

Both translation differentials in step 1.1 are isomorphisms by [F1]. Therefore $d F_{g} = d (L_{F (g)})_{e} \circ d F_{e} \circ d (L_{g^{- 1}})_{g}$ , so composing with the two isomorphisms does not change rank. This proves the displayed equality. Dimension zero, disconnected groups, and the zero differential require no separate argument, and no choice principle is used.

F1step 1.1algebra∎

Depends on

Used by

Images are immersed Lie subgroups Theorem
Kernels are closed embedded normal Lie subgroups Theorem

Dependency tree · two levels

7 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I (standard reference, not scraped)