Is there any canonical way to find out the Lie group corresponding to a manifold?

Question

Lie groups are manifolds. But it is well known that not all manifolds are Lie groups. E.g. There are no Lie groups that correspond to the 2-sphere. More generally, the fundamental group has to be a freely generated Abelian group/have connected homeomorphic components, etc. cf. this post.

Given a manifold with the above-mentioned properties, one could look up the Cartan classification of semi-simple Lie groups with already-known manifold structures and find the corresponding Lie group, but if he doesn't find his manifold listed he can't comment that it has no Lie group structure. That is because a particular Lie group may have multiple manifold realizations and one of the realizations may not be given in the classification table. E.g. $U(2)$ is diffeomorphic to $S^3 \times S^1$ but this does not constitute an isomorphism of groups although homotopy equivalence is there. So the "canonical" manifold realization available for the semi-simple Lie groups may not always be the manifold whose corresponding Lie group is to be found.

Is there any canonical/algorithmic way to construct a Lie group given a differentiable manifold that actually admits a Lie group structure?

EDIT: If all of this sounds ill-defined I would like the answerer to explain why/how this is so (because atleast intuitively it makes sense to me if not absolutely rigorously: I am not trained in math) and possibly modify my question to something which is well-defined (and interesting to you).

Answer 1

Here is an answer of sorts. I will interpret the undefined term "canonical" in the question as "as commutative as possible." This will still leave some ambiguity, but let's try. It also leaves out of the "canonical" list some of the most naturally occurring groups, such as $SL(n, {\mathbb R})$, so my definition is far from natural in any sense.

Given a Lie group $G$, I will use the notation $G^{can}$ for a "canonical" Lie group diffeomorphic to $G$ as a smooth manifold.

First of all, I will limit myself to connected Lie groups, otherwise, if $G$ is a Lie group with, say, countably many connected components and the connected component of the identity $G_0$, then I will take as $G^{can}$ the direct product $G_0^{can}\times {\mathbb Z}$. (I am leaving it to you to work out the case of disconnected Lie groups with finitely many components.)

Now, given a conected Lie group $G$, let $K< G$ be a maximal compact subgroup (it is unique up to conjugation). The Cartan's theorem and Levi-Malcev decomposition of $G$ yields a diffeomorphism $G\cong K\times {\mathbb R}^n$ for some $n$. The group $K$ (as a manifold) is uniquely determined up to homotopy equivalence by the topology of $G$. I will ignore this ambiguity for a moment and set $G^{can}=K^{can}\times {\mathbb R}^n$ (as a Lie group), in line with making $G^{can}$ as commutative as possible. Here ${\mathbb R}^n$ is understood as the unique simply-connected abelian Lie group, the additive group of the vector space ${\mathbb R}^n$.

Now, the problem is reduced to the setting of compact Lie groups (up to the issue of homotopy-equivalence). We will be helped by a theorem of Toda:

Toda, Hirosi, A note on compact semi-simple Lie groups, Jap. J. Math., new. Ser. 2, 355-360 (1976). ZBL0355.22007.

Toda proves that if $K_1, K_2$ are simply-connected compact Lie groups which are homotopy-equivalent as topological spaces (actually, one just needs isomorphisms of homotopy groups in matching degrees), then $K_1$ is isomorphic to $K_2$ as a Lie group. Thus, among simply-connected compact groups, each homotopy equivalence class contains a unique group, which will be our canonical choice. This leaves out the case of non-simply connected compact Lie groups $K$. The universal cover $\tilde{K}$ splits (as a Lie group) as the direct product of simple compact factors, denoted $\tilde{K}^{ss}$ (the semisimple part of $\tilde{K}$) and some ${\mathbb R}^m$. If there are no compact factors, then $K^{can}$ is again clear: It is the product of $m$ copies of $S^1$ (the group $K$ is again determined by its homotopy type). In general, $K$ will have a finite covering group of the form $\tilde{K}^{ss}\times (S^1)^m$. This one is canonical. Uniqueness for $K$ will fail, for instance, the semisimple compact groups groups $SO(4)$ and $SO(3)\times Sp(1)$ are not isomorphic but are diffeomorphic. (In this case, $\tilde{K}=\tilde{K}^{ss}= Spin(4)= Sp(1)\times Sp(1)$.) Which of these is more "canonical"? The second one, $SO(3)\times Sp(1)$, is more commutative so maybe you will like it better. In any case, you get finitely many options to choose from and I leave it to you to decide which one is more canonical.