It's a fact: the fundamental group of any topological group is abelian.
Let X be a topological group, with identity element e. Let f and g be paths at e(continuous maps from the unit interval I into X, whose value at both 0 and 1 is e). Denote the ordinary multiplication of paths with a stop: f.g is just the path obtained by following f, and then following g, and reparametrising appropriately to obtain another path. Now, there is another natural multiplication on paths, induced by the group structure of X: denote this by *, where (f*g)(t)=f(t)g(t), where juxtaposition represents the group multiplication. Since multiplication is a continuous operation, this is indeed a path at e. Now, I put it to you that this operation is well-defined not only on paths, but on homotopy classes of paths: suppose f is homotopic to f~ via a path homotopy F, and g is homotopic to g~ via a path homotopy G. Then let H(s,t)=F(s,t)G(s,t); this is again a path homotopy, between f*g and f~*g~. Which is all well and good.
Now, let ce denote the constant path at the identity. It should not surprise you to learn that for any paths f and g at e (not path classes, mind you), f.g=(f.ce)*(ce.g). Now let's pass to path homotopy classes: [f].[g]=[f.ce]*[ce.g]=[f]*[g]. Eureka! The new operation * is absolutely identical to the old operation . on the level of path classes. Which is good. We might as well forget about . altogether, and treat our fundamental group now as having its operation defined by *. Which makes everything very easy indeed. Inversion is a continuous group operation, so for any path f at e, the inverse path f-1 defined by f-1(t)=(f(t))-1 is again a path at e. To show that the fundamental group of X is abelian, one must find a homotopy between f*g and g*f for any paths f and g at e.
But this is the very essence of simplicity! We want to get our homotopy H(s,t) by multiplying f(t)g(t) on the left by something that is the identity when s=0 and is f-1(t) when s=1, to kill off the left f; and we want to multiply on the right by something that is the identity when s=0, and is f(t) when s=1. Furthermore, we need to fix H(s,0)=H(s,1)=e for all s. There's an obvious way to accomplish this. Just let H(s,t)=f-1(st)f(t)g(t)f(st). When s=0, this is just f(t)g(t). When s=1, this is g(t)f(t).
And we're done!
It's a snap.
Posted by aloysius at November 15, 2003 05:21 PM | TrackBack |Oddly enough, this was a problem I just did out of Bredon. This proof is MUCH nicer than the one Bredon suggests. He has you do this:
Show that f*g(t) is homotopic to f(t)g(t) where f*g(t) is doing one loop then the other and f(t)g(t) is the group product.
Use the same idea to show f*g(t) is homotopic to g(t)h(t).
Combine to show Abelian.
The Homotopy you have to use looks like this:
H(t,s)={f(2t/(1+s)) for t in 0 to (1-s)/2
H(t,s)={f(2t/(1-s))g(2t/(1+s)-(1-s)/(1+s)) for t in (1-s)/2 to (1+s)/2
H(t,s)={g(2t/(1+s)-(1-s)/(1+s)) for t in (1+s)/2 to 1
That's the first one. The second one just switches the f and g positions in the middle.
Nevertheless, you method rocks way harder than this stupid mess.
Now, back to the email!
Posted by: Eric E on November 17, 2003 06:22 PMI feel pretty darn good about this...I showed it to my roommate earlier today, and he'd never seen it done this way before. We tried drawing pictures, and what my homotopy does is pretty funky...It's like it's sucking up the first f loop bit by bit, and shitting it out the back of g. This may actually have been an original thought...I'll keep checking, and see.
Posted by: Luke on November 17, 2003 11:49 PMOf course they commute. I see them on the train every morning.
(Hi, Eric!)
Posted by: Graham on November 20, 2003 10:43 AMHi, i am sushmita, currently studying in Indian Institute of Technology, India, Delhi. I am doing a project in compact topological groups, was wondering if u could give me some examples of compact topological groups with proof
Posted by: Sush on April 21, 2004 01:29 PMWell, the nicest examples of compact topological groups are compact Lie groups: SO(n,R), SU(n), Sp(n), and tori, direct products of U(1) (which is also the circle). Or finite groups with the discrete topology. Topologically these are great because they're compact smooth (even real-analytic) manifolds, and things like their fundamental groups are easy to extract from (for SO and SU) the fibrations we get by considering their actions on spheres. And they have nice representation theories.
One compact topological group that isn't a Lie group is the p-adic integers, for any prime p: they actually form a compact topological ring. There are a bunch of ways to define them...It's probably easiest to use the prime p=2. As a set, the 2-adics are the set of all infinite sequences of 0's and 1's. We put a metric on this: define the distance between any two sequences a_n and b_n to be 2^(-k) if k is the smallest integer with a_k not equal to b_k. For example, the distance between (01010...) and (00010...) would be 2^(-1) (the first spot in the sequence is the zeroth). Topologically, this makes the 2-adics homeomorphic to the Cantor set, which is just about the most un-manifoldy thing you can find without being topologically just silly. There's a lot on p-adics out there...Wikipedia's got an article describing p-adic arithmetic here:
http://en.wikipedia.org/wiki/P-adic_numbers
..along with several constructions of them. Actually proving that the p-adics are homeomorphic to Cantor sets is a bit tedious; one way to do it is to think of the Cantor set as being the set of all reals in the unit interval whose ternary expansions contain only the digits 0 and 2...Then it's easy to think up a bijection between the 2-adics and the Cantor set: given a 2-adic sequence, map it to the real number you get by replacing the digit 1 with the digit 2 everywhere it occurs, and vice-versa. Then it's just a matter of showing the map is a homeomorphism; it's probably easiest to think of the inverse map, going from the Cantor set to the 2-adics; if you can show that that map is continuous, then any continuous bijection from a compact space to a Hausdorff space (and the 2-adics, being metric, are certainly Hausdorff) is automatically a homeomorphism without having to check that the map in the other direction is also continuous.
Being a Cantor set, the 2-adics are totally disconnected: so from an algebraic topology viewpoint, they're horrible. The only maps into them from the interval, or spheres, or simplices, are constants, since the image of any connected set (like an interval, sphere or simplex) is connected, and the only connected subsets of the 2-adics are singletons. So they have uncountably many components, trivial homotopy groups, zeroth homology group free abelian on uncountably many generators, and trivial higher homology and cohomology groups.
Another example of compact topological groups would be compact linear algebraic groups: consider the orthogonal group of nxn matrices over an arbitrary field k, even of prime characteristic, and give it the Zariski topology, making it an affine variety. I've no idea what the algebraic topology looks like there...It might be interesting to find out. It's probably completely horrible. The connected components of a variety are irreducible, I think...
Anyhow, Lie groups are by far the nicest, and the ones I know most about.
Posted by: aloysius on April 21, 2004 06:32 PM