Why is Lie's Third Theorem difficult?

Question

Recall the following classical theorem of Cartan (!):

Theorem (Lie III): Any finite-dimensional Lie algebra over $\mathbb R$ is the Lie algebra of some analytic Lie group.

Similarly, one can propose "Lie III" statements for Lie algebras over other fields, for super Lie algebras, for Lie algebroids, etc.

The proof I know of the classical Lie III is very difficult: it requires most of the structure theory of Lie algebras.

But why should it be difficult? For example, for finite-dimensional Lie algebra $\mathfrak g$ over $\mathbb R$, the Baker-Campbell-Hausdorff formula (the power series given by $B(x,y) = \log(\exp x \exp y)$ in noncommuting variables $x,y$; it can be written with only the Lie bracket, no multiplication) converges in an open neighborhood of the origin, and so defines a unital associative partial group operation on (an open neighborhood in) $\mathfrak g$. What happens if one were to try to simply glue together copies of this open neighborhood?

Alternately, are there natural variations of Lie III that are so badly false that any easy proof of Lie III is bound to fail?

Answer 1

That gluing together of group chunks, constructed from the BCH formula is precisely more or less what Serre does to prove the theorem (in the first proof he gives in) his book on Lie groups and Lie algebras. [Serre, Jean-Pierre. Lie algebras and Lie groups. 1964 lectures given at Harvard University. Second edition. Lecture Notes in Mathematics, 1500. Springer-Verlag, Berlin, 1992. viii+168 pp.]

Answer 2

This question is related to the following rather deep result: if $G$ is a finite dimensional Lie group, then $\pi_2(G)=0$.

The analogous statement fails for Lie algebroids, as does Lie III. The paper of Tseng and Zhu shows that this is no accident.

Lie III may be interpreted as saying the following: the foliation of the space of $\frak{g}$-connections on the 1-simplex associated to infinitesimal gauge transformation has a Hausdorff leaf space.

A connection is a one-form $A\in\mathcal{A}=\Omega^1([0,1],\frak{g})$ with values in the Lie algebra $\frak{g}$. The infinitesimal gauge action is given by the formula

$\delta_XA=dX+[A,X]$,

where $X\in\Omega^0([0,1],\frak{g})$: these vectors span an integrable distribution in the tangent space of $\mathcal{A}$.

(This leaf space is then the simply connected Lie group associated to $G$.) I was taught this point of view by Raoul Bott.

Answer 3

There are Lie algebras which are not the Lie algebra of an algebraic group. (See here.) That rules out some proofs, but it doesn't seem like it would be a problem for the sort of analytic proof you are proposing.

Answer 4

I think Serre says somewhere (probably in the same reference people have mentioned) that the third theorem is deeper (which doesn't necessarily mean more difficult) because it is false for general Banach-modeled Lie groups and Lie algebras.

Answer 5

You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using Ado's theorem, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

I posted a related question in math.stackexchange.com

https://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theorem

Answer 6

Now there is a rather simple proof of III Lie theorem using the momentum map. It is in

Gijs M. Tuynman. An elementary proof of Lie’s Third Theorem. Publications de l’U.E.R. Mathematiques Pures et Appliquees, I.R.M.A. Univ. Lille 1 34 (1994), X1-X4,

see a copy in Google Cache (an original PDF seems unavailable online).

The author says in the end that almost the same proof is given in Gorbatsevich The construction of a simply connected Lie group with a given Lie algebra (here in Russian; an English version also exists).

Answer 7

@Theo, doubtless in the year since you answered this question, someone who can produce such impressive notes would have gotten all the details of a thorough answer sorted out, but my own proof of Lie III from my own forthcoming exposition on Lie theory, for what it's worth, it is basically a turning on its head of the proof that two simply connected connected Lie groups are isomorphic and runs roughly as follows. Once we have the local Lie group, form the set of all formal products of the form $\gamma = \Pi_j \exp\left(X_j(\tau)\right)$ where $X_j(\tau)$ are a $C^1$ paths in the local Lie group, of course small enough that CBH applies to all pairs of its products. Then you can show that all continuous deformations of this path through the set of formal products can be written, thanks to the Hadamard formula, as $\Pi_j \exp\left(X_j(\tau)\right) \Pi_j \exp\left(\delta X_j(\tau)\right)$, i.e. you can shuffle all the variation to one end of the product. Now restrict the variations $\delta X_j$ to be small enough so that the CBH formula applies to each stage of the n-fold product $\Delta = \Pi_j \exp\left(\delta X_j(\tau)\right)$. Thus, even though we have not rigourously defined the "value" of the formal product, now the concept of small continuous, but finite variations of the path that leave the "value" of the formal product unchanged is meaningful - it is any such variation such that, as calculated by CBH, $e = \Delta = \Pi_j \exp\left(\delta X_j(\tau)\right)$ where $e$ is the group identity. So now we can define two of our formal products to be equivalent if one can find a finite sequence of small, continous variations that leave the products "value" unchanged, in the sense defined above, that step-by-step deforms one formal product to the other. We thus condense the big set of formal products into its equivalence classes modulo the equivalence just defined. This approach does two things: $\mathbf{(i)}$ it irons out any inconsitency that might arise from an element's having horribly many potential representations as different formal products and $\mathbf{(ii)}$ the set of equivalence classes, as a set of homotopy classes is simply connected by construction. So now we just check that this beast is a Lie group and we are done: in my exposition this is easy, because I use essentially a converse of Satz 1 of Freudenthal's 1941 "Die Topologie der Lieschen Gruppen Als Algebraisches Phanomen. I" to define a Lie group - I show one can take essentially the properties listed in Freudenthal's theorem, use them as axioms and show that a group that is a simply connected manifold builds itself from them. What we have done, of course, is to build the unique, simply connected connected Lie group that has the Lie algebra "input" to our proof".

I believe my approach is somewhat like the J.P. Serre's 1964 lectures approach cited in the first answer.