19 votes
Accepted

Does $\Bbb Z[X]$ determine $X$?

The answer is no. If $X$ is a simplicial set then $\mathbb Z[|X|]\cong|\mathbb Z[X]|$. That is, your topological abelian group freely generated by the realization of $X$ is homeomorphic to the ...
Tom Goodwillie's user avatar
14 votes
Accepted

Are there extremally disconnected groups?

Yes: every locally compact ED group is discrete. Indeed, for topological groups, ED passes to quotients and open subgroups. Let by contradiction $G$ be a non-discrete ED locally compact group. Passing ...
YCor's user avatar
  • 60.1k
11 votes
Accepted

Homotopic but not equivariantly homotopic maps

For any $G$-space the $G$-equivariant maps $[EG,X]_G$, also known as the homotopy fixed points $X^{hG}$, are a Borel homotopy invariant of $X$ meaning that it is an invariant of $G$-equivariant maps ...
Connor Malin's user avatar
  • 5,191
9 votes

Groups with no (proper) closed subgroups?

The only discrete groups without any non-trivial proper closed subgroups are the cyclic groups of prime order. For the non-discrete case, we have a characterization of such subgroups. Theorem: Let $G$ ...
Joseph Van Name's user avatar
9 votes

Homotopic but not equivariantly homotopic maps

$\newcommand{\RP}{\mathbb{RP}}$Connor Malin's answer is excellent. Derived from that, here is a small example: Let $G = C_2$, the group with two elements, let $X = S^1$ with antipodal action, and let $...
Steve Costenoble's user avatar
7 votes
Accepted

Extreme amenability of topological groups and invariant means

The action $G\curvearrowright\beta G$ is continuous iff $G$ is discrete, so for nondiscrete groups it is not true that $G$ is extremely amenable iff this action has a fixed point. What one should look ...
Alessandro Codenotti's user avatar
7 votes
Accepted

Non-continuous group homomorphism from p-adic field to C*

Since $\mathbf C^\times$ is a divisible group, Zorn’s lemma tells us for every abelian group $G$ and subgroup $H$ that each group homomorphism $H\to \mathbf C^\times$ extends (somehow) to a group ...
KConrad's user avatar
  • 49.5k
6 votes
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Is every contractible homogeneous space of a connected Lie group homeomorphic to a Euclidean space?

The main references would be Theorem 1.2 of Sergey Antonyan's paper and Theorem 32.5 in Markus Stroppel's book. Topological group $G$ is called almost connected if the factor group $G/G_0$ of $G$ ...
Shijie Gu's user avatar
  • 1,936
6 votes
Accepted

Topologies that turn the real numbers into a compact Hausdorff topological group

A compact abelian group $A$ is the same as the Pontryagin dual of a discrete abelian group $B$. The group $A$ is divisible ($nA=A$ for all $n\ge 1$) if and if $B$ is torsion-free, and $A$ is torsion-...
YCor's user avatar
  • 60.1k
6 votes
Accepted

Does every (Abelian) Polish group have a nontrivial locally compact subgroup?

No. Here's a counterexample. Every 3-adic integer $x\in\mathbb Z_3$ has a unique representation as $x=\sum_{n\geq 0} a_n 3^n$ with balanced ternary digits $a_n\in\{-1,0,1\}.$ Define $f(x)=\sum_{n\geq ...
Colin McQuillan's user avatar
5 votes
Accepted

$\operatorname{Spaces}/BG$ $\sim$ $\operatorname{Spaces}^G$ $\sim$ $??(\Omega G)$

If $A$ is a braided ∞-group, the delooping $\def\B{{\sf B}}\B A$ is an ∞-group. Consider the ∞-category of spaces equipped with an action of the ∞-group $\B A$. Since $\B Ω G≃G$, this ∞-category is ...
Dmitri Pavlov's user avatar
4 votes
Accepted

Examples of non-discrete, cocompact subgroups

You can find many such examples among groups acting on trees. Let $T$ be a $k$-regular tree and let $G$ be the subgroup of $\operatorname{Aut}(T)$ of automorphisms stabilizing each of the 2 parts of ...
Tom De Medts's user avatar
  • 6,494
4 votes

Is this submonoid of the isometry group on $\Bbb Q_2$ closed to inverses?

Unfortunately I lack the reputation to comment, and the question is extremely hard to parse so I can’t be all that confident (silver lining: more space to ramble). To me it appears that if you fix up ...
Ophelia's user avatar
  • 91
4 votes

Does $\Bbb Z[X]$ determine $X$?

As folks have correctly pointed out, my previous positive answer was not right. (And I have deleted it.) And Simon Henry observes that one can't even recover a finite set $S$ from $\mathbb Z[S]$ in a ...
Nicholas Kuhn's user avatar
4 votes
Accepted

Is there a purely topological definition of $\text{Spin}(p,q)$?

Here is a copy of my answer at MathStackExchange. Almost nobody does this. There is a discussion in the Wikipedia article on spinor groups but it has too many mistakes and too few references. The only ...
Moishe Kohan's user avatar
  • 9,664
4 votes
Accepted

Classifying space of centralizer

$\newcommand{\S}{\mathcal S}$ The answer is yes, always. Let's compute $\Omega(map(BG,BH),Bf)$. The idea is to realize the map $* \xrightarrow{Bf} map(BG,BH)$ as a map of the form $\Gamma(BG,-)$ where ...
Maxime Ramzi's user avatar
  • 13.3k
4 votes
Accepted

In a topological group, is $G/A\to G/B$ a covering map if $A$ is open in $B$?

Let $C$ be the intersection of the conjugates $bAb^{-1}$. Assume the additional hypothesis that $C$ is open in $B$. This holds, for example, when $B$ is locally connected. The set $B\backslash C$, ...
Tom Goodwillie's user avatar
3 votes

Compact subgroups of a linear group over non-Archimedean local field

The group $G={\rm GL}(n,F)$ acts transitively on $\mathcal O$-lattices of $F^n$. The stabilizer of the standard lattice $L_o = {\mathcal O}^n$ is $K={\rm GL}({\mathcal O})$. It is an open compact ...
Paul Broussous's user avatar
3 votes

Is every contractible homogeneous space of a connected Lie group homeomorphic to a Euclidean space?

I don't have an answer. Note that $H$ need not be compact. For example, $G$ could be the universal covering of $SL_2\mathbb R$, and $H$ could be a $1$-dimensional closed subgroup. Here are some more ...
Linus's user avatar
  • 553
3 votes
Accepted

Does every locally compact group G contain a maximal open subgroup P which is a pro-Lie group?

After thinking about the problem for some time, I came up with a counter-example, so as YCor wrote in his comment, it is indeed true that there is not always a maximal open pro-Lie subgroup. Let $A$ ...
Cosine's user avatar
  • 559
3 votes
Accepted

Is every compact quasisimple group a Lie group?

The group $G/Z(G)$ being simple, is a Lie group by Peter-Weyl. Hence it is either finite or connected. If it is finite, $G$ has center of finite index, hence has a finite derived subgroup. Since $G$ ...
YCor's user avatar
  • 60.1k
2 votes

Willis theory for discrete groups?

One situation where you can apply tdlc group theory in a nontrivial way to discrete groups is if the group $G$ that you are interested in has a commensurated subgroup $H$ (meaning, all conjugates of $...
Colin Reid's user avatar
  • 4,678
2 votes

orbits in locally compact group

These are exactly (a) all locally compact groups in which every element is torsion, and (b) the circle group. (a) All groups in which every element is torsion work. This involves many discrete groups (...
YCor's user avatar
  • 60.1k
2 votes
Accepted

Open conjugacy classes in a second countable profinite group

Yes, there exist profinite groups $G$ with a conjugacy class of empty interior and consisting of elements of finite order, generating $G$ as an abstract group. Let $H$ be a nonabelian finite simple ...
YCor's user avatar
  • 60.1k
2 votes
Accepted

Uniqueness of left-invariant Borel probability measure on compact groups

Uniqueness does hold, as a direct consequence of Halmos's theorem that Haar measure is completion regular. Here are pointers to relevant parts of Fremlin's book. "Completion regular" is ...
Colin McQuillan's user avatar
2 votes

Is there a purely topological definition of $\text{Spin}(p,q)$?

What about the complex point of view? The complex analogue of $SO(p,q)$ (complex matrices preserving a nondegenerate bilinear form and having determinant $1$) depends only on $p+q$, since all complex ...
Tom Goodwillie's user avatar
1 vote

Uniqueness of left-invariant Borel probability measure on compact groups

[This was too long to be a comment, but is basically just a recontextualization of the other answer.] If you use Baire measures rather than Borel measures (which is probably the more natural setting ...
Cameron Zwarich's user avatar
1 vote

Parametrization of topological algebraic objects

Please regard this as a tentative answer since I am not sure that it corresponds to the spirit of your query. Many of the situations that you use to illustrate your query can be coded by using the ...
memorial's user avatar
  • 396
1 vote
Accepted

Continuity of central character

Let $K\subset G$ be a compact open subgroup such that $V^K\ne0$. Now $Z$ acts on $V^K$ by the central character $\omega_\pi$, and the action is trivial on $Z\cap K$. Thus $\omega_\pi$ is trivial on ...
Kenta Suzuki's user avatar
  • 1,547

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