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Since the space $\Gamma(TM)$ of all vector fields on a smooth manifold $M$ is a real Lie algebra with respect to the usual commutator bracket, I was curious if in fact it is the Lie algebra of some infinite dimensional Lie group? Would it be the Lie algebra of the automorphism group $\text{Aut}(M)$, if the latter group can be realized as some appropriate infinite dimensional manifold?

More generally, can any infinite dimensional Lie algebra be realized as the Lie algebra of an infinite dimensional Lie group? If not, can this be "corrected" by considering a larger category of generalized smooth spaces, such as diffeological spaces, or Frolicher spaces?

I personally know almost nothing about infinite dimensional Lie groups and manifolds; I have only studied the finite-dimensional theory. This is just something I was curious about.

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    Great question. Given a vector field $X$ what would $\exp(tX)$ be ? There are some problems if the $1$-parameted group of diffeomorphism of $X$ determined by $X$ is not defined for all $t \in \mathbb{R}$. – orangeskid May 17 '15 at 08:27
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    As an additional comment which is less known: Lie algebra of the group $Aut(M,L)$ where L is a Line bundle with a hermitian metric with curvature satisfies In $F_L=-i\omega$ is isomorphic with the space of solutions $f\in C_0^\infty (M, \mathbb C)$ of the équation $\mathcal Df=0$ where $\mathcal D$ is the Lichnerowicz operator –  Feb 19 '18 at 11:38

2 Answers2

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Morally speaking, the Lie algebra of vector fields is the Lie algebra of $\text{Diff}(M)$, the diffeomorphism group of $M$. The relationship between these is less tight than in the finite-dimensional case: for example,

  1. The exponential map can fail to be defined at any nonzero time (as mentioned by orangeskid in the comments), and
  2. Even when defined (say on a compact manifold), the exponential map can fail to be a local diffeomorphism at the identity.

It is also not true that infinite-dimensional Lie algebras are Lie algebras of infinite-dimensional Lie groups in general (see, for example, this MO question), and I have the impression that this is a genuine phenomenon and not just an artifact of working with too restrictive of a notion of infinite-dimensional Lie group, although I don't know much about this.

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Qiaochu Yuan
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To add to Qiaochu Yuan's answer, I found out that in synthetic differential geometry (which admits a well-adapted model containing the category of smooth manifolds as a fully subcategory) it is easy to prove that the Lie algebra of vector fields is associated to the diffeomorphism group. Working in such a smooth topos, let $M$ be a space, $R$ the line object, and $D=\{d \in R : d^2 = 0 \}$ the "walking tangent vector".

The tangent bundle is $M^D \to M$ where the projection is the evaluation at zero map. By currying (product-exponential adjunction), a vector field on $M$ is an infinitesimal transformation of the space; more precisely a map $X: D \to M^M$ such that $X(0)=1_M$. In other words, the vector fields $\mathfrak{X}(M)$ on $M$ coincide with the tangent space at the identity to the monoid $M^M$ of endomorphisms of $M$. Moreover, if $M$ is infinitesimally linear, then any vector field $X$ on $M$ satisfies $X(-d) \circ X(d) = 1_M$ and is thus invertible. So we have particular $T_1 (M^M) = T_1 \mathrm{Aut}(M)$.

So $\mathfrak{X}(M)$ is the Lie algebra of both the endomorphism monoid of $M$, and its diffeomorphism group (provided $M$ is infinitesimally linear).

If you're curious about SDG, Kock's excellent introductory text is available online for free.

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