MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a topological group. One general approach to show that $G$ is connected is the following:

For every subgroup $H\leq G$ (not necessarily closed) we have a projection map: $$ \pi: G\rightarrow G/H $$ We may give to the set of left cosets of $H$ in $G$ the quotient topology. In that case $\pi$ is an open map. Now it is easy to see that if $H$ and $G/H$ are connected then $G$ is connected. Using this idea one may show that many of the classical groups are connected. For example, one may show that $U(n)$ is connected using induction and the usual fibration: $$ U(n-1)\rightarrow U(n)\rightarrow S^{2n-1} $$

Here is my question:

Q: Is there another general approach to show that a topological group $G$ is connected which avoids this kind of "devissage"?

(P.S. Of course one may try to prove directly that the space is path connected but in general this is too difficult)

(P.S.S. In the case of a real Lie group, using Cartan's theorem, one sees that the connectedness of $G$ is equivalent to the connectedness of a maximal compact subgroup of $G$, but I guess that this is a very special situation which does not generalize well to topological groups.)

share|cite|improve this question
First comment: If $G$ is Hausdorff then you can't possibly have $H\times G/H$ homeomorphic to $G$ unless $H$ is closed in $G$, for otherwise $G/H$ will not be Hausdorff. Second comment: A continuous section $G/H\to G$ would produce a homeomorphism $H\times G/H\to G$, and it's a little bit hard to imagine that you are interested in any other way of producing one. – Tom Goodwillie Mar 24 '11 at 0:39
Just a formal comment. At MO, shouldn't the question be related to the title a bit more than this one is? – Gerald Edgar Mar 24 '11 at 1:03
Sorry about that, I formulated another question – Hugo Chapdelaine Mar 24 '11 at 12:59
So now all the previous comments (answering a non-existent question) make no sense... – Gerald Edgar Mar 24 '11 at 13:23
There are certain non-locally compact groups which have natural definitions but where connectedness is hard to verify. For instance, the group of invertible elements in some fixed Banach algebra, equipped with the norm topology; or the group of automorphisms of a fixed Banach algebra $A$, equipped with the restriction of the norm topology of ${\mathcal B}(A,A)$. So perhaps one should start by restricting the question to the locally compact case? – Yemon Choi Mar 24 '11 at 19:31

If $A\in GL_n(\mathbb{C})$ then you can just pick any nonzero complex number $\lambda$ such that $\arg(-\lambda)$ is different from the arguments of all the eigenvalues of $A$. Then the path $t\mapsto t\lambda I+(1-t)A$ joins $A$ to $\lambda I$ in $GL_n(\mathbb{C})$, and it is easy to join $\lambda I$ to $I$, so $GL_n(\mathbb{C})$ is connected. Moreover, $GL_n(\mathbb{C})$ is homeomorphic to $U(n)\times\mathbb{R}^{n^2}$ by Gram-Schmidt, so $U(n)$ is connected. I don't know if there are similar proofs for the other classical groups.

share|cite|improve this answer
Hi Neil, thanks a lot for your amazingly simple proof that $GL_n(\mathbf{C})$ is connected! – Hugo Chapdelaine Mar 25 '11 at 0:20
Just as an afterthought, a direct proof that U(n) is connected; diagonalize, and then join each eigenvalue to 1 by moving it round the unit circle. – Yemon Choi Mar 25 '11 at 0:40

Perhaps this is helpful for the locally compact case.

Corollary 3.1.12 in "Topological groups and related structures" by Arhangel'skii and Tkachenko gives a nice characterization of connectedness in the locally compact case: $G$ is connected $\Leftrightarrow$ $G$ has no proper open subgroups $\Leftrightarrow$ Every neighborhood of the identity algebraically generates $G$.

I don't know how practical this is for the groups you have in mind but the generality is nice.

share|cite|improve this answer
Hi Jeremy, well you don't need your group to be locally compact. As long you have a subgroup with an interior point then by homogeneity it is open. – Hugo Chapdelaine Mar 25 '11 at 0:18
And this result is rarely applicable directly. nevertheless, assuming the connectedness you can use it to show the surjectivity of the exponential map $exp:\mathbf{C}[A]\rightarrow \mathbf{C}[A]^{\times}$ where $A$ is an $n$ by $n$ matrix in $\mathbf{C}$. Note that this implies in particular the surjectivity of the exponential map $exp: M_n(\mathbf{C})\rightarrow GL_n(\mathbf{C})$! – Hugo Chapdelaine Mar 25 '11 at 0:30
Hugo, definitely connectedness implies the other two (which are equivalent for arbitrary $G$) for arbitrary $G$ but this does not seem to be what you are interested in. Local compactness (or maybe something slightly weaker) should be needed for the other direction. – Jeremy Brazas Mar 25 '11 at 0:58
Well connected implies that every open neighboorhood which contains $1$ generates the whole group. For the converse, say that your group is not connected then the connected component of the identity, say $G^{0}$, is closed, ah ok, but may be not open. So do you have such an example at hand? – Hugo Chapdelaine Mar 25 '11 at 1:11
@Hugo: for non-locally compact groups, I think the following might give a counter-example: take the additive group ${\bf Q}$ but equip it with the relative topology inherited as a subset of ${\bf R}$. (This is not locally compact.) Then the connected component of the identity is just the identity, but this is not open in the given topology. My guess is that this group should have no proper open subgroups, intuitively because an open neighbourhood will contain too many rationals; and yet the group is disconnected (intersect it with $(-\infty, \sqrt{2})$ and with $(\sqrt{2},\infty)$). – Yemon Choi Mar 25 '11 at 2:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.