3

This should be a very basic algebraic topology question. The other day I was thinking about the fact that $P^2(R)$ has $\pi_1 = Z/2Z$.

On the other hand I thought to myself how something like this can never happen for, say, an open subset of the real plane $R^2$. It's a very intuitive fact but I can't prove it.

So I guess my general question is: can open subsets of $R^2$ have torsion elements in $\pi_1$?

Related to this: is there a classification of the homotopy types of open subsets of the plane?

user125763
  • 327
  • I think you want to ask about weak homotopy types, not homotopy types: I shudder to think about what the homotopy type of say, $\mathbb{R}^2 \setminus \mathbb{Q}^2$ is... – Qiaochu Yuan Feb 16 '14 at 05:51
  • What do you mean by weak homotopy type? Just that there is a weak homotopy equivalence between the two? Also, $Q^2$ is not closed in $R^2$ so $R^2 - Q^2$ shouldn't really enter the picture (but I guess you could take $R \times C$, where $C$ is the Cantor set). – user125763 Feb 16 '14 at 05:59
  • Sorry, yes, I meant something like $\mathbb{R}^2$ minus a $2$-dimensional Cantor set. And yes, that's what I mean by weak homotopy type. – Qiaochu Yuan Feb 16 '14 at 05:59
  • QY: Do you have a good example to understand the difference between weak and strong homotopy type for a pathological space? I'm not too familiar with this - whenever I think about homotopy theory I always assume everything to be a CW complex. (also - I cannot comment on MO so I might as well ask here: do you have a concise reference to learn about this stuff about Stone-Cech and $C^*$-algebras?) – user125763 Feb 16 '14 at 06:07
  • Weak homotopy type is the concept that lets you assume that everything is a CW complex (every space is weakly homotopy equivalent to a CW complex). For example, any space which is totally disconnected has the weak homotopy type of a discrete space (e.g. the Cantor set). I don't have a reference. – Qiaochu Yuan Feb 16 '14 at 06:10

3 Answers3

5

The new question is very different from the original one. The answer to it is that every noncompact connected surface is homotopy-equivalent to a bouquet of circles, see proofs and references here. In particular, fundamental group is free and, hence, torsion-free.

Thus, the answer to the last question is: Yes, there is a description, namely they all are disjoint unions of bouquets of circles.

A (quite a bit) more difficult theorem is that the fundamental group of every open connected subset of $R^3$ is still torsion-free. In dimensions $\ge 4$ this is, of course, false.

Community
  • 1
Moishe Kohan
  • 97,719
  • thanks - I wanted to strike through the old question, rather than delete it, but I couldn't find how. – user125763 Feb 16 '14 at 06:20
  • I don't believe that this is true in the generality you're claiming. An open subset of $\mathbb{R}^2$ need not be homotopy equivalent to the disjoint union of its connected components; for example, as mentioned in the comments, it could be $\mathbb{R}^2$ minus $\mathbb{R}$ times a Cantor set. – Qiaochu Yuan Feb 16 '14 at 08:21
  • @Qiaochu: What I said is correct, proven by Whitehead in much greater generality than I stated here. He also proved that any open n-manifold is homotopy equivalent to its subcomplex of dimension $n-1$. This is from his paper on immersion a of 3-manifolds (around 1960, I will find the exact date) but the result I mentioned is general. – Moishe Kohan Feb 16 '14 at 10:31
  • @Quiaochu: Whitehead's paper is from Proceeding of LMS, 1961, lemma 2.1. – Moishe Kohan Feb 16 '14 at 10:42
  • Incidentally, complement to R times cantor set has homotopy type of a countable discrete set. This is completely trivial. In my bouquet of circles I allow no circles, i.e. just a single point. Maybe this was unclear. Pathological examples you are worrying about do not occur in the context of manifolds. – Moishe Kohan Feb 16 '14 at 15:16
  • @studiosus: you're right, my apologies. I was confused about what the complement of a Cantor set looks like. – Qiaochu Yuan Feb 17 '14 at 17:25
2

I think it is still possible to have $2$-torsion. The special orthogonal group $SO(n)$ is orientable, yet has a fundamental group of $\mathbb{Z}/2\mathbb{Z}$ for all $n\geq 3$. To see that $SO(n)$ is orientable, note that $SO(n)$ has a Lie group structure, so its tangent bundle is trivial. Hence, the top exterior power of the cotangent bundle is trivial, meaning that $SO(n)$ has a non-vanishing form of top-degree.

Peter Crooks
  • 2,051
  • ah crap. But is what I think makes sense in $R^2$ true? Can opens inside it have any $2$-torsion? I was hoping this wasn't something which boiled down to a big theorem. (like a classification of the homotopy types of open subsets of the real plane) – user125763 Feb 16 '14 at 05:30
  • Note that $\text{SO}(3) \cong \mathbb{RP}^3$ and all odd-dimensional real projective spaces are orientable as well (as well as all having fundamental group $\mathbb{Z}_2$ in dimension at least $2$). – Qiaochu Yuan Feb 16 '14 at 05:34
  • Thanks guys - I guess that's why I couldn't remember the connection between the two things: there isn't one! – user125763 Feb 16 '14 at 05:41
0
  1. If $M$ is a connected smooth orientable $2$ manifold ( with countable basis) then it has a Riemannian structure. Locally a Riemannian structure can be written as $ds^2 = f(x,y) (d x^2 +dy^2)$. That means $M$ is a smooth orientable $2$ manifold with a conformal structure locally isomorphic to the plane. It follows that $M$ has a complex structure ( it is a Riemann surface).
  1. Every connected Riemann surface $M$ has a simply connected cover. If $M$ is not compact then its universal cover $\tilde M$ can be taken to be upper half plane $\mathcal{H}$. Now $M$ will be isomorphic to $\tilde M$ modulo a totally discontinued group $\Gamma$ of automorphism of $\tilde M$. The important facts are : $\Gamma$ acts on $\mathcal{H}$ without fixed points and $\pi_1(M) \simeq \Gamma$.
  1. Now we have to show that if $\gamma \in PSL(2, \mathbb{R})$ ( the group of analytic automorphisms of $\mathcal{H}$), and $\gamma$ has no fixed points, then $\gamma$ is not of torsion. But now recall that every matrix in $SL(2, \mathbb{R})$ is either conjugated to $\left(\begin{matrix}1&n\\ 0&1\end{matrix}\right)$, or $\left(\begin{matrix}\cos \theta &- \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right)$, or $\left(\begin{matrix}t &0 \\ 0& \frac{1}{t}\end{matrix}\right)$. Only the second type can be torsion in $PSL(2, \mathbb{R})$ ( if $t=-1$ the image in $PSL(2,\mathbb{R})$ is $1$), but they also have $2$ (conjugate) fixed points, so one will lie in $\mathcal{H}$).

It would be a bit more delicate to show that $\Gamma$ is free, not clear to me now.

orangeskid
  • 53,909