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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Cartan decomposition identifies p with the noncompact symmetric space

Statement

Assume the Axiom of Choice. Let (G,K) be a Riemannian symmetric pair of noncompact type with Cartan decomposition g0=k0⊕p0 (Riemannian symmetric pair of noncompact type). Then the map Φ:p0⟶G/K,X⟼exp⁡(X)K, is a diffeomorphism; here G/K carries the manifold structure of the quotient by the closed subgroup K (Quotient manifold by a closed Lie subgroup).

Facts & Assumptions

Given: The Axiom of Choice; a connected real semisimple Lie group G with finite center, a global Cartan involution Θ, K=GΘ, the Cartan decomposition g0=k0⊕p0, and the quotient map q:G→G/K.

[A1]

The Axiom of Choice is assumed (The Axiom of Choice), supplying L1 and the countable-choice assumptions of L2 and L3.

[L1]

The map K×p0→G, (k,X)↦kexp⁡X, is a diffeomorphism; K is closed with Lie algebra k0 (Global Cartan decomposition for a connected finite center semisimple Lie group).

[L2]

The quotient map q:G→G/K is a surjective smooth submersion (Quotient manifold by a closed Lie subgroup). Local submersion coordinates have the form (u,v)↦u (Local normal form for submersions).

[L3]

For real s,t, exp⁡((s+t)X)=exp⁡(sX)exp⁡(tX), so exp⁡(−X)=exp⁡(X)−1 (Exponential scales one-parameter subgroups).

Proof

technique · direct
1.1L1L3algebra

Let D:K×p0→G be D(k,X)=kexp⁡X. Define DR:p0×K→G by DR(X,k)=exp⁡X k. It is a diffeomorphism: explicitly DR(X,k)=D(k−1,−X)−1 by [L3], a composition of D with the product diffeomorphism (X,k)↦(k−1,−X) and the smooth inversion diffeomorphism of G. Thus every g is uniquely exp⁡X k, and its first coordinate F(g)=X is smooth.

2.1L2step 1.1algebra

For h∈K, gh=exp⁡X (kh), so uniqueness gives F(gh)=F(g). Hence F factors as F‾∘q for a unique set map F‾:G/K→p0. It is smooth: near any quotient point, fix the v-coordinate in a local submersion chart of [L2] to obtain a smooth section s of q; on that neighborhood F‾=F∘s. Smoothness is local, so no global section choice is needed.

3.1L1L2step 1.1step 2.1algebraA1∎

The map Φ(X)=q(exp⁡X) is smooth by [L1] and [L2]. Uniqueness in step 1.1 gives F(exp⁡X)=X, hence F‾∘Φ=id⁡. Conversely, if g=exp⁡X k, then Φ(F‾(gK))=exp⁡XK=gK. These smooth maps are mutually inverse, proving the assertion. The zero-dimensional case is included: if p0=0, [L1] gives G=K and both sides are singletons.

Depends on

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Sources