1. ## Fundamental group

Hello everybody!

I have a problem!! What is the fundamental group of $\displaystyle \mathbb{R}^4\setminus S^1$?

(where $\displaystyle S^1$ is the circle, i.e. $\displaystyle S^1=\{(x,y)\in\mathbb{R}^2|x^2+y^2=1\}$)

2. Originally Posted by miccoli
Hello everybody!

I have a problem!! What is the fundamental group of $\displaystyle \mathbb{R}^4\setminus S^1$?

(where $\displaystyle S^1$ is the circle, i.e. $\displaystyle S^1=\{(x,y)\in\mathbb{R}^2|x^2+y^2=1\}$)

Hmmm....isn't $\displaystyle \mathbb{R}^4\setminus S^1$ still contractible? I base this idea on the one that in 4-dimensional euclidean space one can untie one's shoe laces without tearing the laces or pulling them "out of the loop" (since there's no actual loop in R^4 in this case!).
Thus the fund. group would be the trivial one...but I'm not really sure.

Tonio

3. Hello!
The fundamental group of $\displaystyle R^4\setminus S^1$ is the trivial group, but I'm trying to proof this.
Can you explain more clearly your idea?

Thanks!

4. Originally Posted by miccoli
Hello!
The fundamental group of $\displaystyle R^4\setminus S^1$ is the trivial group, but I'm trying to proof this.
Can you explain more clearly your idea?

Thanks!

This is just a vague idead, please do try to find more formal stuff: what's the problem with $\displaystyle \mathbb{R}^3\setminus S^1$ for its fund. group to be trivial? That any closed path looping around $\displaystyle S^1$ isn't null-homotopic, right?
Well, such a thing cannot happen in $\displaystyle \mathbb{R}^4\setminus S^1$ since any closed path "looping" around $\displaystyle S^1$ isn't actually looping around, since the extra dimension allows to "pull" out such a closed path without tearing $\displaystyle S^1$!
This is kindda anti-intuitive and, of course, pretty hard to visualize, but my dearly remembered topology teacher prof. Farajoun once made it pretty clear to me by means of the Klein bottle. He said something like: do you see where the bottle's arm "gets into" the bottle after twisting around from its lower part? Well, there is NOT such an "openning or introducing" gash there in the bottle as you'd expect it to be with a 3-dimensional bottle, since that arm is "getting" into the bottle exactly at the 4th dimension, so there is no gash, gap or crack there in the bottle!
Well, trying to come up to terms with this ideas is that in knot theory we were told that just as two concentric circles cannot be "untied" in 2 dimensions (think of a table) but they're easily untied in 3 dimensions (just pull up any of the circles aways from the table), two knotted or tied circles in 3 dimensions cannot be untied in 3 dim's but they can in 4 dom's.
This is the idea I have and the one I offer you to chew over. Good luck!

Tonio