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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Connected Lie groups are central quotients of simply connected integrations

Statement

Assume countable choice. Every connected real Lie group G is isomorphic to G~/Γ, where G~ is its simply connected covering Lie group and Γ is a discrete central subgroup. Conversely, every such quotient has the same Lie algebra as G~.

Facts & Assumptions

Given: Countable choice and a connected real Lie group G.

[L1]

There is a covering homomorphism p:G~G with G~ connected and simply connected (Universal covering Lie group).

[L2]

A covering homomorphism is a surjective homomorphism and a covering map (Covering homomorphisms of Lie groups).

[L3]

Under [A1], discrete subgroups are closed embedded zero-dimensional Lie subgroups (Discrete subgroups are closed embedded zero-dimensional Lie subgroups).

Proof

technique · identify the kernel and factor the covering
1.1

Let Γ=kerp. A fiber of a covering is discrete, so Γ is discrete; it is normal because it is a kernel. For fixed γΓ, the map xxγx1 is continuous from connected G~ into the discrete space Γ, hence constant. At the identity its value is γ, so γ is central.

L1L2algebra
2.1

The fibers of p are exactly the cosets of Γ. Hence p factors through a bijective homomorphism p:G~/ΓG. Covering charts for p give the quotient its unique smooth structure for which the quotient projection is a local diffeomorphism, and in those charts p and its inverse are smooth. Thus p is a Lie-group isomorphism.

L2step 1.1
3.1

Conversely, let Γ be a discrete central subgroup of a simply connected Lie group G~. It is closed and embedded by [L3]. Choose an identity neighborhood meeting Γ only in the identity and shrink it so that distinct translates are disjoint. Its translates furnish smooth quotient charts, making G~G~/Γ a covering homomorphism. Its identity differential is an isomorphism, so the two groups have the same Lie algebra. The trivial subgroup and one-point group are included. Countable choice is used exactly through [L3]; steps 1.1–2.1 need no additional choice.

A1L3algebra

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