## Hot answers tagged topological-groups

98

The 2-adic rationals $\mathbb{Q}_2$ and the 3-adic rationals $\mathbb{Q}_3$ are homeomorphic, because each one is a countable disjoint union of Cantor sets. They are also isomorphic as groups if you assume the axiom of choice, because they are both fields of characteristic 0 and therefore vector spaces over $\mathbb{Q}$ (of the same cardinal dimension). ...

49

(In response to suggestion of Johannes Hahn): As indicated in my comment, and developed further in j.p.'s comment, a fairly intuitive "explanation" is provided by the fact that Lagrange's Theorem tells us that an element $x$ of a finite group $G$ is already central in $G$ once it commutes with more than half the elements of $G$ (since $C_{G}(x)$ is a ...

48

The answer is no in general, but this is a rather deep fact.
Theorem: (Nikolov, Segal) If $G$ is any compact Hausdorff topological group, then every finitely generated (abstract) quotient of $G$ is finite.
N. Nikolov and D. Segal, Generators and commutators in finite groups; abstract quotients of compact groups, arXiv, http://arxiv.org/abs/1102.3037

47

The OP says:
" ...Recently, Palais wrote about it in the Notices but he only treats the old story from the 1950s and seems not to be aware of Olver’s facts."
Actually, I am aware of Olver's work and also Sören Illman's contribution.
Sören is an old friend and wrote to me somewhat miffed that I did not mention his work on the problem. What he proved was a ...

44

Sorry for the necromancy, but the following was too cute to resist:
Let
$A$ be $Z_4$ with the discrete topology
$B$ be $Z_4$ with the indiscrete topology
$C$ be $Z_2 \times Z_2$ with the discrete topology
$D$ be $Z_2 \times Z_2$ with the indiscrete topology.
Then $A \times D$ is not isomorphic to $B \times C$ as a topological group, but the the ...

41

Andreas shot first, but I still encourage everybody to have a look at the lemma on p.263 of R. Alperin,Locally compact groups acting on trees and property $T$. Monatsh. Math. 93 (1982), no. 4, 261–265: any homomorphism from a locally compact group to $\mathbb{Z}$, is continuous. This answers Florent's question.

36

Here is an example: a product of infinitely many $\mathbb{RP}^\infty$'s.
The crucial thing thing to see is that $\mathbb{RP}^\infty$ (or, easier to see, its universal cover $S^\infty$) has a group structure whose underlying group is a vector space of dimension $2^{\aleph_0}$. This is not hard: the total space $S^\infty$ of the universal ...

32

Here is a direct proof for free groups.
Let $x_1,\dots,x_m$ be the generators of our group. Consider a word $x_{i_n}^{e_n}\dots x_{i_2}^{e_2}x_{i_1}^{e_1}$ where $e_i\in\{\pm 1\}$ and there are no cancellations (that is, $e_k=e_{k+1}$ if $i_k=i_{k+1}$).
I'm going to map this word to a nontrivial element of $S_{n+1}$, the group of permutations of ...

31

Andrew's answer is right, but I'll just throw in a few comments since "topological" homotopy invariants are of great interest to me. Here Paul Fabel has shown that $\pi_{1}^{top}$ on the Hawaiian earring is not a topological group. It turns out that multiplication can fail to be continuous even for some reasonably nice spaces (like locally simply connected ...

30

Edit (Oct 12, 2015):
My original answer contained an error, as noticed by user47958.
I now record only those parts of what I wrote which
might be relevant to a future complete solution.
(What I am writing here is NOT a complete solution.)
I order topologies so that $\mathcal T\subseteq \mathcal T'$
means that $\mathcal T'$ is finer than $\mathcal T$.
I let ...

24

Subsets of a group that are closed with respect to any Hausdorff group topology
are called unconditionally closed.
Clearly, all algebraic sets are unconditionally closed, where a subset of a group $G$ is called algebraic if it is an intersection of finite unions of the sets of solutions to some equations with coefficient from $G$.
A.A.Markov proved that ...

23

Given two spaces $X$ and $Y$, how to define the mapping space betweeen them, i.e. what topology should we put on the set of maps between them?
If $X$ is compact and $Y$ a metric space, this is quite easy as one can put a metric on $Map(X,Y)$: For $f,g\in Map(X,Y)$ define their distance just to be the maximum of the distances between $f(x)$ and $g(x)$ as ...

22

...if $H$ is a Lie group, then $G \to G/H$ is locally trivial. Is the claim true, and if so, what is the reference?
Yes, it is true. See the Corollary in section 4.1 of: "On the Existence of Slices for Actions of Non-compact Lie Groups", which you can download here: http://vmm.math.uci.edu/ExistenceOfSlices.pdf
This is a paper originally published in ...

21

No. Take $G=SO(5)$ and $H=SO(3)$, both of which have fundamental group $\mathbb{Z}/2$. I claim that there is no continuous map $f: G\to H$ which induces the identity homomorphism.
If there were, then $f$ would induce a nontrivial homomorphism $f_\ast: H_1(G;\mathbb{Z}/2)\to H_1(H;\mathbb{Z}/2)$, and a graded ring homomorphism $f^\ast: ...

20

Here is a counter-example with $G$ homeomorphic to $\mathbb R^2$. Let $f:\mathbb R\to\mathbb R$ be a discontinous additive homomorphism (constructed using a Hamel basis of $\mathbb R$ over $\mathbb Q$). Define a group operation $*$ on $\mathbb R^2$ by
$$
(x,y)*(x',y') = (x+x'e^{f(y)},y+y') .
$$
This groups is a semidirect product of $\mathbb R$ and $\mathbb ...

20

If this was the case, then $\pi_{n-2}(SO(n))\cong \pi_{n-2}(SO(n-1))$. The first group is in the stable range and hence one of $\mathbb{Z},\mathbb{Z}/2$ or 0; in particular every surjective endomorphism of it is an isomorphism. Now the orthonormal frame bundle of $S^{n-1}$ is a fiber sequence $SO(n-1)\to SO(n) \to S^{n-1}$ which gives a long exact sequence
...

19

Update: A bit of a digital paper chase led me, via David Robert's thesis (note that in the latest version, it is Chapter 5, section 2 that is most relevant), to this paper on the arXiv. The last sentence of the abstract is:
These hoop earring spaces provide a simple class of counterexamples to the claim that $\pi_{1}^{top}$ is a functor to the category ...

18

You can also use the fact that free groups are linear. For instance if $$a=\begin{pmatrix}
1 & 2 \\\\
0 & 1 \\ \end{pmatrix}$$ and $b$ is the transpose of $a$, a simple ping-pong argument shows that $a$ and $b$ generate a free subgroup of $SL(2,Z)$. Now any element of $SL(2,Z)$ will be non-trivial in reduction mod $p>>1$.

18

I believe the answer is yes. Let's begin by recalling that if one wants to show that a locally compact group $G$ is of type I, it suffices to show that $G$ contains a "large" compact subgroup $K$, in the sense that for every $\pi \in \hat{G}$ and $\sigma \in \hat{K}$, the multiplicity of $\sigma$ in $\pi|_K$ is finite. This is how Harish-Chandra showed that ...

17

Your questions are related to Bohr compactification, a left adjoint to the inclusion of compact (= compact Hausdorff) groups into all topological groups. A discrete group G can be embedded into a compact group iff the natural map from G to its Bohr compactification is an injection. Such groups are called "maximally almost periodic". Take a look at this ...

16

There is a beautiful and very short proof due to Hempel of the residual finiteness of both free groups and fundamental groups of closed surfaces. See his paper
J. Hempel, Residual finiteness of surface groups, Proc. Amer. Math. Soc. 32 (1972), 323.

16

There is a large literature about this, see "non-topologizable groups". These are, by definition, groups for which the only Hausdorff group topology is discrete. There are various examples, the first of which were obtained by Olshanskii and Shelah (see here for references) around 1980.
An observation is that for a group $G$, we have the equivalence between:
...

16

Sorry for resurrecting such an old question, but I think we can give a much simpler proof here. We'll reduce the problem from $G$ to the Bohr compactification $B\mathbf{Z}$ of $\mathbf{Z}$, then from $B\mathbf{Z}$ to the profinite completion $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ of $\mathbf{Z}$, and then we'll argue directly.
Let $\phi:G\to\mathbf{Z}$ be a ...

16

Another line of reasoning: If $SO(n)$ were homeomorphic to the product $S^{n-1}\times SO(n-1)$ then $S^{n-1}$ would be a retract (not deformation retract!) of $SO(n)$, hence $S^{n-1}$ would be an H-space since a retract of an H-space is an H-space, assuming the retract contains the identity element, which we can arrange here by a suitable choice of ...

15

For each $\alpha \in S^1$, the map $\varphi_{\alpha} \colon \mathbb Z \to S^1$, given by $n \mapsto \alpha^n$, induces a topology $\tau_{\alpha}$ on $\mathbb Z$.
A basis of neighborhoods of $0$ for $\tau_{\alpha}$ is given by the sets
$$U_{n,\alpha} := \left\{k \in \mathbb Z \mid |\alpha^k-1| < \frac1n \right\},
\quad n \in \mathbb N.$$
I denote by ...

15

The book I always look at for such things is Nachbin, The Haar Integral, which is short, and has a whole chapter on Integration on Locally Compact Homogeneous Spaces.
And a plus: he gives you a choice of reading the proof of the existence and uniqueness of the Haar integral according to Weil or according to Henri Cartan.

15

The two are naturally dual lattices. The fundamental group of a torus $T$ can be canonically identified with the group (known as the cocharacter lattice) of $\it homomorphisms$ from the circle group to $T$, or equivalently the kernel of the (universal cover=exponential map) homomorphism from the Lie algebra $t$ to $T$. The characters of the torus on the ...

14

The general phenomenon you're looking for is called automatic continuity. This is the line of research that stemmed from results regarding the continuity of homomorphisms from (R,+) to (R,+). The two principal general results illustrating the phenomenon of automatic continuity are the following.
Theorem (S. Banach). Any Baire measurable homomorphism between ...

14

The following article gives you a lot of information on the question you are asking:
On Homeomorphism Groups and the Compact-Open Topology,
Jan J. Dijkstra
http://www.cs.vu.nl/~dijkstra/research/papers/2005compactopen.pdf
http://www.jstor.org/pss/30037630
The answer is in general "no".

14

You can use Magnus series to prove this. I believe this argument is, at most, a slight variant of one in a paper of his (but it's been a number of years).
If your free group has n generators, take the noncommutative power series ring $R = \mathbb{F}_2\langle\langle x_1,\ldots,x_n\rangle \rangle$, with 2-sided ideal of definition $I = (x_1,\ldots,x_n)$. ...

Only top voted, non community-wiki answers of a minimum length are eligible