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For every finite dimensional Lie algebra $g$, there is a unique simply-connected Lie group $G$ whose Lie algebra is $g$. Is this true in the infinite dimensional case?

Francois Ziegler
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Ramand
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    A side note: in finite-dimensional case for the uniqueness of $G$ you also need to require that $G$ is simply-connected, not just connected. – Michael Jan 07 '14 at 16:32

1 Answers1

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No. Van Est and Korthagen (1964) gave perhaps the first example of what they called a "non-enlargible Lie algebra", having no corresponding Lie group.

Needless to say, this kind of question largely depends on the precise definitions adopted for infinite-dimensional Lie groups and Lie algebras. A canonical reference is Milnor's Remarks on infinite-dimensional Lie groups, which can be found on the internet and contains also some positive results:

The author outlines a theory of Lie groups modelled on arbitrary complete locally convex topological vector spaces (CLCTVSs).This category includes everything one would want to call a Lie group, with possibly a few exceptions such as diffeomorphism groups of noncompact manifolds. (...)

Every Lie group as defined here has an associated (topological) Lie algebra. For finite-dimensional groups the converse is true, but this is not so in general (even for Banach Lie algebras). The passage from Lie algebra to Lie group depends on the notion of regularity. (...) All known Lie groups are regular. A connected, simply connected, regular Lie group is uniquely determined (up to isomorphism) by its Lie algebra.

For a more recent survey I would recommend Neeb's Towards a Lie theory of locally convex groups.

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Francois Ziegler
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  • I appreciate you and your nice solution. thank you – Ramand Jan 07 '14 at 21:44
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    Is there any
    criterion say that when a lie algebra is lie algebra of a lie group?     – Ramand Jan 08 '14 at 12:10
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    Not that I know of, in general. But see §VI of Neeb's paper for integrability results in special situations, such as e.g. when $\mathfrak g$ is a quotient, or a central extension, of a Lie algebra that is itself integrable (i.e. that is the Lie algebra of some Lie group). The results are most complete when $\mathfrak g$ is "locally exponential", but there are also some negative (non-integrability) criteria in general. Do you have access to that paper? – Francois Ziegler Jan 08 '14 at 18:55
  • thank you for your remarkable answer. I have not seen the paper. – Ramand Jan 09 '14 at 10:28