4
$\begingroup$

It is hard for me to understand that $SO(3)$ has $\mathbb{Z}_2$ as the fundamental group. The problem is that, it is hard to image a mechanism such that there is a non-contractible loop which when repeated becomes contractible.

So, is there any simpler example?

$\endgroup$
1

3 Answers 3

4
$\begingroup$

The projective plane $\mathbb{R} P^2$. There is, almost by definition, a covering map $S^2 \mapsto \mathbb{R} P^2$. Any path in $S^2$ connecting the north pole to the south pole descends to a non-contractible loop in $\mathbb{R} P^2$, which follows from covering space theory.

$\endgroup$
3
$\begingroup$

The smallest example is the real projective plane, $\Bbb{RP}^2$, which has fundamental group $\Bbb Z/2$. This is the only closed surface with this fundamental group, and $\Bbb{RP}^3 = SO(3)$ is the only 3-manifold with that as its fundamental group. But there are a lot of manifolds with fundamental group $\Bbb Z/2$; infinitely $n$-manifolds, $n$ fixed and at least 4, have fundamental group $\Bbb Z/2$.

Torsion in the fundamental group is just a thing that happens. It can even happen in open subsets of Euclidean space, provided you're in $\Bbb R^4$ - the fact that you have to go up to 4 might explain why it's hard for us to picture it. Actually, every finitely presented group is the fundamental group of a closed 4-manifold; in particular every finite group.

If you want to understand the picture of how torsion can happen, I really strongly suggest thinking about $\Bbb{RP}^2$ through one of its many forms (for this purpose maybe the best depictions are as a sphere with its antipodal points identified, or a Mobius band with a disc glued on along the boundary.)

$\endgroup$
3
$\begingroup$

For any $n > 1$, $\mathbb{RP}^n$ has fundamental group $\mathbb{Z}_2$ as it has $S^n$ as its two-fold universal cover.

One can visualise the non-trivial loop as follows. Let $x \in S^n$ be on the equator $S^{n-1} \subset S^n$, and consider a path $\gamma$ from $x$ to $-x$ along a great semicircle. The image of $\gamma$ in $\mathbb{RP}^n$, say $\gamma'$, is a closed loop but by construction, it does not lift to a loop in the universal covering space. Therefore, $\gamma'$ determines a non-zero element of $\pi_1(\mathbb{RP}^n)$. Once you know that $\pi_1(\mathbb{RP}^n) = \mathbb{Z}_2$, this is the only such element.

$\endgroup$
1
  • 1
    $\begingroup$ This happens to overlap with the OP's examples: the group manifold of $SO(3)$ is $\mathbb{RP}^3$ $\endgroup$ Sep 21, 2015 at 16:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .