Let $M$ a closed (smooth) manifold. The group $Diff(M)$ of all difeomorphisms of $M$ is infinite dimensional, therefore it is not a Lie group.

Is there a way to associate a Lie algebra to this? If so, is there some concrete descrpition of such Lie algebra?

EDIT: I would welcome some references. According to the question in the comments: the Lie algebra of the Lie group $G$ is defined as Lie algebra of left invariant vetor fields on $G$. Therefore for an infinite dimensional $G$ we would like to have a notion of a vector field. Vector fields are defined as section of the tangent bundle: the fibers of the tangent bundle are tangent spaces. We define the tangent space at a given point $x$ as the set of classes of smooth curves $\gamma:I \to G, \gamma(0)=x$ with equivalence relation $\gamma_1 \sim \gamma_2$ iff for any chart $\varphi$ we have $\frac{d}{dt}(\varphi \circ \gamma_1)(0)=\frac{d}{dt}(\varphi \circ \gamma_2)(0)$. This set is in one to one correspondence with $\mathbb{R}^n$ for $n$-dimensional manifolds and the linear structure is transported via this correspondence. In infinite dimension there are few delicate moments: instead $\mathbb{R}^n$ we need another ,,model space''. Which model space we choose for infinite dimensional Lie groups, in particular for $Diff(M)$?
Moreover while defining tangent space we need derivative: which notion do we use?

  • 5
    $\begingroup$ The Lie Algebra section of the Wikipedia article says "The Lie algebra of the diffeomorphism group of M consists of all vector fields on M equipped with the Lie bracket of vector fields." en.wikipedia.org/wiki/Diffeomorphism#Lie_algebra $\endgroup$ Aug 12, 2017 at 17:55
  • $\begingroup$ What is it that you don't like about infinite dimensional Lie groups? $\endgroup$ Aug 12, 2017 at 21:09
  • $\begingroup$ I made an edit clarifying my doubts $\endgroup$
    – truebaran
    Aug 15, 2017 at 12:30
  • $\begingroup$ This closely related question might be helpful. $\endgroup$
    – ಠ_ಠ
    Oct 25, 2017 at 23:14

1 Answer 1


This question certainly is too broad to be answered completely. Before talking about infinite dimensional Lie groups you need a concept of infinite dimensional manifolds. As you note correctly, you first need model spaces in infinite dimensions, and the technical difficulties that have to be overcome strongly depend how general spaces you want to allow. If you are fine with Banach spaces as model spaces, then the theory of smooth manifolds mostly works out closely parallel to the finite dimensional case. There are introductory books on differential geometry working in that setting, for example Serge Lang's classics. This allows dealing with diffeomorphism groups of finite differneitiability class or with Sobolev type completions. To deal with infinitely differentiable diffeomorphisms, you need manifolds modelled on Frechet-spaces, and there things become much more complicated. But infinite dimensional differential geometry works in far more general settings, see e.g. the book "The Convenient Setting of Global Analysis" by A. Kriegl and P. Michor.

Having infinite dimensional geometry at hand, you can look at infinite dimensional Lie groups, and indeed the Lie algebra of a diffeomorphism group is the Lie algebra of vector fields with the negative of the usual Lie bracket. Intuitively, you can understand that as follows: A curve in $Diff(M)$ can be viewed as a 1-parameter family $\phi_t$ of diffeomorphisms. To get a tangent vector at the point $\phi=\phi_0$ you should differentiate this curve with respect to $t$ at $t=0$. If you look what happens in a point $x\in M$, then $t\mapsto\phi_t(x)$ is a smooth curve starting at $\phi(x)$, so differentiating this at $t=0$ determines a tangent vector at the point $\phi(x)\in M$. Now you check that this can be done smoothly, so you associate to each $x\in M$ a tangent vector at $\phi(x)$ and this defines a section of the pullback bundle $\phi^*(TM)$. With a bit more work you can indeed construct charts with values in a neighborhood of the zero section in the space of smooth sections of $\phi^*(TM)$ and thus identify that vector space as the modelling vector space. From $\phi=id$, you just get smooth sections of $TM$, i.e. vector fields on $M$. (And the exponential map associates to each vector field its local flow.) You can find a lot of information in that direction in the book by Kriegl and Michor mentioned above.

  • $\begingroup$ Thank you for the great answer $\endgroup$
    – truebaran
    Aug 17, 2017 at 20:48

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