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If this is in the scope of MO, I would like to gather here the known tactics of Lie algebra integration, since it appear surprisingly hard to find such a compendium, library or any other kind of collections on this topic.

A Lie algebra integration should be a tactic (a functor or maybe something less formal) that assigns a Lie group to any given Lie algebra and preferable a Lie group morphism, to any Lie algebra morphism. In the most general scenario this would be applicable to infinite Lie algebras and over arbitrary fields, but as I said,less general tactics are ok.

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1.) A well known integration method, is of course the integration by path. This technique is functorial (over at least finite dimensional, real Lie algebras AFAIK) and assigns a simply connected Lie group to any given finite dim. Lie algebra. An introduction can be found in the paper "Integrability of Lie brackets" of Crainic and Fernandes and the references therin.

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2 Answers 2

Sometimes one can use the Frobenius theorem which furnishes leaves to involutive distributions. For example: If $\mathfrak h\subset \mathfrak g$ is a Lie subalgebra and $G$ is a Lie group with Lie algebra $\mathfrak g$, then the left invariant vector subbundle of $TG$ generated by $\mathfrak h$ is an involutive distribution. The leaf through $e$ is then a Lie subgroup of $G$ with algebra $\mathfrak h$. See for example 5.2 in 1. Its use for integrating homomorphisms of Lie algebras is described in 5.4 of 1: The graph of the Lie algebra homomorphism integrates to a leaf in the product of groups, whose projection to the first group is a covering.

The Frobenius theorem can also be used to integrate the action of a Lie algebra on a manifold to a local action of a Lie group on the same manifold. This manifold can even be enlarged to carry a global action, but it might be non-Hausdorff then; see 2 (which elaborates on results of R. Palais).

In infinite dimensions, the notion of regular Lie group allows at least to integrate Lie algebra homomorphisms, see 40.3 in 3. Note that there are infinite dimensional Lie algebras (even Banach-) which do not admit any Lie group -- these are called non-enlargeble; see [4]. 43.6 of 3 contains an example of a sub Lie algebra of a Lie algebra with Lie group, which itself does not admit a Lie group.

  • 1 Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008. (pdf)

  • 2 Franz W. Kamber, Peter W. Michor: Completing Lie algebra actions to Lie group actions. Electron. Res. Announc. Amer. Math. Soc. 10 (2004) 1-10. (pdf)

  • 3 Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997. (pdf)

  • [4] MR2261066 (2007k:22020) Neeb, Karl-Hermann(D-DARM) Towards a Lie theory of locally convex groups. (English summary) Jpn. J. Math. 1 (2006), no. 2, 291–468.

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The method of integration that I like most is via the theory of central extensions. To start with note that if $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}_n$, then the subgroup $\langle\exp(\mathfrak{g})\rangle$ of $GL_n$ that is generated by $\exp(\mathfrak{g})$ integrates $\mathfrak{g}$. Thus we are done with integrating $\mathfrak{g}$ if we find a faithful representation of $\mathfrak{g}$ (which exists by Ado's Theorem).

The problem with the above procedure is that the faithful representation of $\mathfrak{g}$ is not very canonical. However, there is a very canonical representation, the adjoint representation $\operatorname{ad}\colon\mathfrak{g}\to \mathfrak{der(\mathfrak{g}})\leq \mathfrak{gl}(\mathfrak{g})$. This is not faithful, the kernel is the center $\mathfrak{z}(\mathfrak{g})$ of $\mathfrak{g}$ and thus we have a canonical central extension $$\mathfrak{z}(\mathfrak{g})\to\mathfrak{g} \to \mathfrak{g}_{\operatorname{ad}}:=\operatorname{ad}(\mathfrak{g}). $$ We now set out to integrate this central extension. If we set $G_{\operatorname{ad}}:=\langle\exp(\mathfrak{g_{\operatorname{ad}}})\rangle\leq\operatorname{Aut}(\mathfrak{g})$, then this is a connected Lie group integrating $\mathfrak{g}_{\operatorname{ad}}$, and so is its universal covering $\widetilde{G}_{\operatorname{ad}}$. Since $\pi_2(\widetilde{G}_{\operatorname{ad}})=\pi_1(\widetilde{G}_{\operatorname{ad}})=0$ (this is the case for any simply connected finite-dimensional Lie group and was used in different disguise by Cartan in one of his first proofs of Lie's Third Theorem) there now exists a unique central extension $$\mathfrak{z}(\mathfrak{g})\to \widehat{G}\to \widetilde{G}_{\operatorname{ad}}$$ that integrates the above central extension of Lie algebras. The latter follows from the exact sequence$$\operatorname{Hom}(\pi_1(K),Z)\to \operatorname{Ext}(K,Z)\xrightarrow{L}\operatorname{Ext}(\mathfrak{k},\mathfrak{z})\to \operatorname{Hom}(\pi_2(K),Z)\oplus\operatorname{Hom}(\pi_1(K),...) $$ (where $L$ is the Lie differentiation and $...$ is some abelian group not of big importance for this discussion) that can be found in the paper of Neeb that was already cited by Peter Michor (and also in Theorem 7.12 of Neeb's "Central extensions of infinite-dimensional Lie groups" Ann. Inst. Fourier (Grenoble) 52 (2002)(5):1365–1442 (MR1935553)).

The nice thing about the latter approach is that

  • it is functorial at each step
  • it also works in infinite dimensions
  • it can (in infinite-dimensions) be used to show that certain Lie algebras do not integrate
  • it can be used in the non-integrable case to show how non-integrable Lie algebras still integrate to etale Lie 2-groups

I could elaborate on any of the above points, but this is perhaps a bit off topic from the original question.

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