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9 votes
1 answer
657 views

Must an inverse limit of simply connected groups be simply connected?

While the fundamental group $\pi_1$ preserves products, it is not true in general that an inverse limit of simply connected topological spaces is simply connected. I would like to know if similar ...
Jeremy Brazas's user avatar
12 votes
2 answers
883 views

Does almost every pair of elements in a compact Lie group generates the connected component?

It is known that almost every pair of elements in a connected compact Lie group (topologically) generates the group. Obviously this isn't true for non-connected groups but Given a compact Lie ...
Jack the Ripper's user avatar
4 votes
2 answers
2k views

A question about unitary and anti-unitary matrices

The question is the following: Let $U:\mathbf{C}^n\to \mathbf{C}^n$ be a unitary operator; let $\tilde{U}:\mathbf{C}^n\to\mathbf{C}^n$ be an antiunitary operator. Can one deform $U$ to $\tilde{U}$ ...
jacaboul's user avatar
  • 327
6 votes
1 answer
249 views

Growth function of locally compact groups

Every locally compact second countable group $G$ has a regular left-invariant measure $h$, the Haar measure. On the other hand the Birkhoff–Kakutani Theorem asserts that such groups also admit a ...
Alessandro Carderi's user avatar
0 votes
1 answer
434 views

Reference request: Any connected Lie group has a countable base for its topology

I am looking for a reference for the assertion in the title. This assertion is proved in a comment of user nfdc23 to this question. Has any proof of this assertion been published?
Mikhail Borovoi's user avatar
4 votes
0 answers
640 views

Closed subgroups of a connected Lie group

Is it true that for any closed subgroup $H$ of a connected Lie group $G$, the group of connected components $\pi_0(H)$ is finite or countable? (inspired by the comment of nfdc23 to this question ).
Mikhail Borovoi's user avatar
2 votes
0 answers
321 views

Surjective homomorphisms of non-connected Lie groups

Let $\psi\colon B\to C$ be a homomorphism of real Lie groups, where the group $C$ is connected. Let $B^0$ denote the identity component of $B$, and we set $\pi_0(B)=B/B^0$, then $\pi_0(B)$ is a ...
Mikhail Borovoi's user avatar
3 votes
1 answer
141 views

Measure on orbits of $N$ under conjugation by $H$

Let $G$ be a locally compact topological group with closed subgroups $H, N$ and $H$ normalizing $N$. Then $H$ acts continuously on $N$ by conjugation. If it will help, assume that $N$ is nilpotent, ...
D_S's user avatar
  • 6,180
2 votes
0 answers
82 views

Uniquely divisible neighborhoods of identity in topological groups

Let $G$ be a (finite dimensional real) Lie group, and let $A\subset G$ be an open neighborhood of identity. If $A=\operatorname{Exp}(\mathcal{A})$ is the injective range of the exponential map from a ...
Bedovlat's user avatar
  • 1,959
4 votes
3 answers
2k views

Why is the unipotent radical of a parabolic subgroup unimodular?

Let $G$ be a connected, reductive group defined over a local field $F$. Let $P$ be a parabolic subgroup of $G$ which is defined over $F$, and let $N = \mathscr R_u(P)$ be the unipotent radical of $P$....
D_S's user avatar
  • 6,180
2 votes
0 answers
62 views

Codistal subgroups of locally compact groups

Let $G$ be a topological group and let $H$ be a closed subgroup of $G$. Say $H$ is codistal in $G$ if the translation action of $G$ on the coset space $G/H$ is distal (meaning that no non-diagonal ...
Colin Reid's user avatar
  • 4,728
4 votes
0 answers
264 views

When can a locally compact group be approximated by discrete subgroups?

This question is about partitioning a (locally) compact group into cells by using discrete subgroups. Let $G$ be a locally compact group. (I am really most interested in the case where $G$ is a ...
Jason Rute's user avatar
  • 6,287
7 votes
4 answers
2k views

Topological structure of SO(n) as a product

I’m interested in the question for which $n$ the special orthogonal group is homeomorphic to the product $$ \mathrm{SO}(n) \approx S^{n-1} \times \mathrm{SO}(n-1). $$ Allen Hatcher [1, p. 293 f.] ...
Florian Oppermann's user avatar
3 votes
1 answer
142 views

What's the topology on the mapping space $Map_H(G, Y)$ when $G$ is not finite

When $G$ is a finite group and $H$ a closed subgroup of it, the sets of right cosets $H\backslash G$ has the discrete topology on it. Let $Y$ be a $H-$space. We have the $G-$homeomorphism \begin{...
Megan's user avatar
  • 1,040
3 votes
1 answer
380 views

A Comparison between $\pi_{1}$ of cohomology and cohomology of $\pi_{1}$

Let we have a complex of abelian topological or lie groups $$\ldots \to G_{n}\to G_{n+1}\to \ldots$$ such that the image of $G_{n}$ is a closed subgroup of $G_{n+1}$. Then we have a complex of ...
Ali Taghavi's user avatar
6 votes
2 answers
392 views

Union of conjugates of a closed subgroup of a compact group

Let $G$ be a compact Hausdorff group, $H \leq G$ a closed subgroup of infinite index in $G$. Is it possible that the conjugates of $H$ cover some open neighbourhood of $1$ in $G$ (or the whole of $G$...
Pablo's user avatar
  • 11.3k
7 votes
2 answers
2k views

Conditions for a topological group to be a Lie group

In flipping through the Springer lecture notes on Serre's 1964 'Lie Algebras and Lie Groups' lectures at Harvard, I found this pair of suprising results (page 157): Let $G$ be a locally compact group....
Joshua Seaton's user avatar
1 vote
2 answers
481 views

Cross section for closed Lie subgroup in a Lie group

Let $G$ be a Lie group and $H$ a closed Lie subgroup. Is there an explicit way to construct a local cross section of $H$ in $G$ so that $\pi: G\to G/H$ is a fiber bundle?
ruhi's user avatar
  • 23
20 votes
2 answers
2k views

Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)?

The isometry group of a metric space is a topological group (with the compact open topology). The isometry group of a Riemann Manifold is a Liegroup. (Thm. of Steenrod-Myers) So, is every topological ...
archipelago's user avatar
  • 2,974
6 votes
3 answers
1k views

$\pi_1$ Sequence of Topological Groups

Consider a connected topological group $G$ (not necessarily Lie). You have some maps $G\times G\to G$, such as projection to either summand, or multiplication $(g,h)\mapsto gh$. Now let's look at a ...
Chris Gerig's user avatar
  • 17.5k
9 votes
2 answers
717 views

Topology on extensions of topological groups

Let $G$ and $H$ be two topological groups and let $\mathcal{E}:0 \to G \to E \to H \to 0$ be an extension of abstract groups. Is there a way to introduce a topology on $E$ such that $\mathcal{E}$ ...
jap's user avatar
  • 125
2 votes
2 answers
423 views

Lie (and topological) group extensions of $\mathbb{R}^2$ by $\mathbb{R}$

What are all the non-split Lie (and topological) group extensions $0 \to \mathbb{R} \to G \to \mathbb{R}^2 \to 0$? Here, $\mathbb{R}$ and $\mathbb{R}^2$ are regarded as Lie (and topological) groups ...
jap's user avatar
  • 125
2 votes
1 answer
571 views

On homeomorphic compact connected topological groups

I wish to thank Professor Claudio Gorodski for his very helpful answers to my question on the webcite: If compact connected Lie groups are homeomorphic as topological space, are they isomorphic as Lie ...
sife's user avatar
  • 491
17 votes
0 answers
1k views

What groups are Lie groups?

We know how to tell if a topological group is a Lie group: this was famously asked by Hilbert and answered gloriously by Gleason, Montgomery and Zippin in the 50s (a locally compact topological group ...
Mariano Suárez-Álvarez's user avatar
5 votes
3 answers
1k views

On closed totally disconnected subgroups of connected real Lie groups

So the following statement seems to be obvious but I don't see how to prove it: Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete? Note that ...
Hugo Chapdelaine's user avatar
8 votes
1 answer
730 views

Status of Hilbert-Smith conjecture and H-S conjecture for Hölder actions

The Hilbert-Smith conjecture states that If $G$ is a locally compact group which acts effectively on a connected manifold as a topological transformation group then is $G$ a Lie group. It was ...
Zarathustra's user avatar
  • 1,414
-3 votes
4 answers
5k views

What is the situation with Hilbert's Fifth Problem?

The common knowledge in this regard seems to be that Hilbert's Fifth Problem was completely solved by Gleason, Montgomery, and Zippin. However, such wisdom was contested by Peter Olver: Olver, Peter ...
Elemer E Rosinger's user avatar
12 votes
4 answers
2k views

Which compact groups have finitely many irreducible representations of each dimension?

If my understanding is correct, this is true of sufficiently nice nonabelian Lie groups (see Ben Webster's answer below), and any finite group. On the other hand, this is false for any infinite ...
Qiaochu Yuan's user avatar
6 votes
0 answers
2k views

Fourier transforms via Kurzweil-Henstock integral on locally compact commutative groups

Is it possible to define Fourier transforms on locally compact commutative groups using the Kurzweil-Henstock integral instead of the Lebesgue integral?
teil's user avatar
  • 4,351

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