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Tangent space of a homogeneous quotient

Statement

Assume ${A C}_{ω}$ . For a closed subgroup $H \leq G$ , the differential of $q : G \to G / H$ at the identity induces a canonical linear isomorphism

$\overline{d q_{e}} : g / h \overset{≅}{\to} T_{e H} (G / H) .$

At $g H$ this identification is transported by left translation and satisfies

$d q_{g} \circ d (L_{g})_{e} = d (L_{g}^{G / H})_{e H} \circ d q_{e} .$

Facts & Assumptions

Given: ${A C}_{ω}$ , a finite-dimensional real Lie group $G$ , a closed subgroup $H$ , and the quotient map $q : G \to G / H$ .

[A1]

The quotient manifold exists and $q$ is a surjective submersion. The Axiom of Countable Choice ( ${A C}_{ω}$ ), Quotient manifold by a closed Lie subgroup.

[F1]

A surjective linear map factors through the quotient by its kernel. A module homomorphism vanishing on $N$ factors uniquely through $M / N$ .

[F2]

The tangent space of a regular fibre is the kernel of the differential. The tangent space of a regular level set is the kernel.

Proof

technique · quotient the differential by its kernel

1.1A1F2algebra

Since $q$ is a submersion by [A1], $e H$ is a regular value. Its fibre is $q^{- 1} (e H) = H$ , so [F2] gives $\ker d q_{e} = T_{e} H = h$ . Also $d q_{e}$ is surjective.

2.1F1step 1.1

By [F1], $d q_{e}$ factors uniquely through a linear map $\overline{d q_{e}} : g / h \to T_{e H} (G / H)$ . It is injective because its kernel would lift to $\ker d q_{e} = h$ , and it is surjective because $d q_{e}$ is. Hence it is the claimed canonical isomorphism.

3.1A1step 2.1algebra∎

Equivariance of the quotient map says $q \circ L_{g} = L_{g}^{G / H} \circ q$ . Differentiating at $e$ gives the displayed identity. Both translation differentials are isomorphisms, so it transports the identity-coset description to every $g H$ . The formulas include $H = G$ , $H = {e}$ , and disconnected groups. Choice is used only through [A1].

Depends on

Dependency tree · 2 levels

Sources

Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I