4

What are some examples of groups that can not be given a smooth structure such that they become a Lie Group?

Edit: To be a bit more specific, I was hoping that somebody could give an example of a finite dimensional topological group that is a topological manifold but does not admit a smooth structure making it into a Lie Group.

Memeozuki
  • 1,043

2 Answers2

6

Any group is a Lie group if you give it the discrete topology. The better question is whether a topological group has a smooth structure that makes it a Lie group.

Local compactness is obviously necessary (because you want finite dimensions), so any non-locally compact group will be an example.

Generally, most locally compact groups are Lie groups. This question is essentially Hilbert's 5th Problem, which has been solved: https://en.wikipedia.org/wiki/Hilbert's_fifth_problem

TYS
  • 463
  • 1
    Why should most locally compact groups be locally Euclidean? I would have trouble thinking of more than, say, the p-adics, but this seems more like a personal failure than a statement that most groups really are manifolds. –  Jun 13 '15 at 17:20
  • 1
    As @MikeMiller said, you really need to add the requirement that the group has the homeomorphism type of a manifold to exclude profinite groups and so forth. – Jim Belk Jun 13 '15 at 19:22
  • Are there topological groups that are topological manifolds that cannot be made into smooth manifolds? – Memeozuki Jun 14 '15 at 00:52
  • 2
    @Abraham: no. This is part of the solution to Hilbert's fifth problem. See, for example, https://terrytao.wordpress.com/2011/10/08/254a-notes-5-the-structure-of-locally-compact-groups-and-hilberts-fifth-problem/. – Qiaochu Yuan Jun 14 '15 at 03:25
1

You can take any uncountable Cartesian product of rationals $Q$ with discrete topology. This have discrete topology again (pre-images of $\emptyset \times \emptyset \times \cdots \times \{x\} \times \emptyset \times \emptyset \cdots, x\in Q$) are one-point elements in the product). Thus it has dimension 0 as a topological manifold. It is a group. It is a topological group (addition is bi-continuous and inversion is continuous, because of the discrete topology). Moreover, it is locally compact. But it is not second countable.

Svata
  • 19
  • This answer is incorrect. $\emptyset\times A = \emptyset$ for any set $A$. Infinite products of a discrete space will not be discrete. This fails already for ${0,1}^{\Bbb N}$. – s.harp Jul 23 '23 at 09:14
  • Sorry, I wanted to write this trivial example. Take uncountable Cartesian product of the discrete group of rational numbers and equip the product with the discrete topology. Since it is not second countable, it can not be a differentiable manifold... – Svata Jul 25 '23 at 12:41
  • Then I understand the point of the answer better. But still I'm skeptical about allowing non-second countable topological manifolds while excluding this for the differentiable case. I have not seen anybody do that. – s.harp Jul 25 '23 at 13:10