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

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Every orbit is an injectively immersed homogeneous space

Statement

Assume ${A C}_{ω}$ . For a smooth action of $G$ on $M$ and $x \in M$ , the map

${\overline{Φ}}_{x} : G / G_{x} ⟶ M, g G_{x} ⟼ g \cdot x,$

is a $G$ -equivariant injective immersion with image $G \cdot x$ . Transporting the quotient structure through this map gives the orbit its canonical immersed homogeneous-space structure. The immersion need not be an embedding.

Facts & Assumptions

Given: A smooth left action of $G$ on $M$ and a point $x \in M$ .

[F1]

The stabilizer is a closed embedded Lie subgroup. Stabilizers are closed embedded Lie subgroups.

[F2]

For a closed subgroup, $G / G_{x}$ is a smooth homogeneous manifold and the quotient map is a submersion. Quotient manifold by a closed Lie subgroup.

[F3]

The kernel of the orbit-map differential at the identity is $T_{e} G_{x}$ , and its tangent image is the infinitesimal orbit. Kernel of the infinitesimal orbit map.

[F4]

Constant-rank normal forms describe immersed images locally. The constant-rank theorem for manifolds.

[F5]

The preceding closed-subgroup and quotient results carry countable choice. The Axiom of Countable Choice ( ${A C}_{ω}$ ).

Proof

technique · factor the orbit map through its stabilizer cosets

1.1F1algebra

If $g_{1} G_{x} = g_{2} G_{x}$ , then $g_{2}^{- 1} g_{1} \in G_{x}$ and $g_{1} \cdot x = g_{2} \cdot x$ , so the formula is well defined. Conversely, equality of the two orbit points puts $g_{2}^{- 1} g_{1}$ in $G_{x}$ , proving injectivity. Every orbit point is $g \cdot x$ , so the image is exactly $G \cdot x$ .

2.1step 1.1algebra

For $a, g \in G$ , ${\overline{Φ}}_{x} (a \cdot g G_{x}) = a g \cdot x = a \cdot {\overline{Φ}}_{x} (g G_{x})$ , so the map is $G$ -equivariant.

3.1F2F3step 1.1step 2.1

By [F2], the quotient map $q : G \to G / G_{x}$ is a submersion and has smooth local sections. Since $Φ_{x} = {\overline{Φ}}_{x} \circ q$ , on the domain of such a section $s$ one has ${\overline{Φ}}_{x} = Φ_{x} \circ s$ , proving smoothness. If $v \in T_{e G_{x}} (G / G_{x})$ and $d {\overline{Φ}}_{x} (v) = 0$ , choose $X \in g$ with $d q_{e} X = v$ . Then $d Φ_{x} (X) = 0$ , so [F3] gives $X \in T_{e} G_{x} = \ker d q_{e}$ and hence $v = 0$ . Equivariance from step 2.1 transports this injectivity to every coset. Thus the factor is an injective immersion.

4.1F1F2F3F4F5step 1.1step 2.1step 3.1∎

The local form [F4] now makes the image locally an immersed coordinate plane, and transport along the bijection in step 1.1 gives the canonical intrinsic orbit structure. The induced $G$ -action is smooth and transitive by step 2.1. If $G_{x} = G$ , the orbit is a zero-dimensional point; if $G_{x} = {e}$ , its dimension is $\dim G$ . Self-accumulating immersed orbits are allowed, which is why embeddedness is not asserted. ${A C}_{ω}$ is used through [F1]--[F3], and no further choice is made.

Every orbit is an injectively immersed homogeneous space

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources