Bourbaki

Proposition 11

Tag 0A7S, permanent. Cite it as [Bourbaki, Tag 0A7S].

Lie Groups and Lie Algebras, chapter III, § 1. Lie groups

Let XX be a Lie group and GG a Lie subgroup of XX.

(i) There exists on the homogeneous set X/GX/G one and only one analytic manifold structure such that the canonical projection π\pi of XX onto X/GX/G is a submersion. The law of operation of XX on X/GX/G is analytic. For all x∈Xx \in X, the kernel of Tx(π)T_x(\pi) is obtained from Te(G)T_e(G) by Te(γ(x))T_e(\gamma(x)).

(ii) If GG is normal in XX, X/GX/G is a Lie group with its group structure and the manifold structure defined in (i). The mapping π\pi is a Lie group morphism.

By General Topology, Chapter III, § 4, no. 1, Example 1, GG operates properly and freely on XX by right translation. Hence the first assertion of (i) follows from

Proposition 10 of no. 5. The second follows from the Remark of no. 5. Since π\pi is a submersion, the kernel of Tx(π)T_x(\pi) is the tangent space at xx to

π−1(π(x))=xG=γ(x)(G) \pi^{-1}(\pi(x)) = xG = \gamma(x)(G)
and hence is obtained from Te(G)T_e(G) by Te(γ(x))T_e(\gamma(x)).

Suppose that GG is normal. Let mm be the mapping (x,y)↦xy−1(x, y) \mapsto xy^{-1} of (X/G)×(X/G)(X/G) \times (X/G) into X/GX/G. Then (m∘(π×π))(x,y)=π(xy−1)(m \circ (\pi \times \pi))(x, y) = \pi(xy^{-1}) for all x,yx, y in XX. Hence m∘(π×π)m \circ (\pi \times \pi) is analytic. As π×π\pi \times \pi is a surjective submersion, mm is analytic (Differentiable and Analytic Manifolds, R. 5.9.5), whence (ii).

The homogeneous set X/GX/G with the manifold structure defined in (i) is called the quotient (left) Lie homogeneous space of XX by GG. The (right) Lie homogeneous space G∖XG \setminus X is defined analogously. When GG is normal, the Lie group X/GX/G defined in (ii) is called the quotient Lie group of XX by GG.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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