Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

As I have been studying algebraic topology, something that I found puzzling was the existence of finite homotopy groups. For instance, $\pi_{4}(S^{2})\cong\pi_{5}(S^{4})\cong\mathbb{Z}/2\mathbb{Z}$. I was wondering if there was any kind of intuitive reason for why this might be true, and if there are spaces $X$ such that $\pi_{1}(X)$ is finite. Speaking very roughly, it would seem that a finite, nontrivial fundamental group means that if you repeat a closed path enough times, it can be contracted to a point, something which I find rather hard to visualize. So the question is: Is there any intuitive reason for the existence of finite homotopy groups?

share|improve this question
4  
Think about one of the simplest examples, $\pi_1 \mathbb RP^2 \simeq \mathbb Z_2$. The key picture is to think of $\mathbb RP^2$ as a Moebius band capped off with a disc. The fundamental group computation boils down to the fact that the inclusion map $\mathbb Z \simeq \pi_1 \partial M \to \pi_1 M \simeq \mathbb Z$ is multiplication by two. In homology the existence of torsion produces some rather subtle behavior, such as the torsion linking form mathoverflow.net/questions/36987/… –  Ryan Budney Sep 9 '10 at 19:39
1  
One can get any group as $\pi_1(X)$. –  Robin Chapman Sep 9 '10 at 20:09
1  
One result which gives a geometric criterion for finite fundamental group is the Bonnet-Myers theorem. It provides an upper bound on the diameter of the universal cover of a Riemannian manifold whose Ricci curvature has a certain lower bound; in particular, if the Ricci curvature is bounded below at all then the universal cover is compact and $pi_1$ is finite. As an application, it is possible to express the Ricci curvature of a compact Lie group with a bi-invariant metric in terms of the Lie bracket, and consequently any compact semisimple Lie group has finite fundamental group. –  Paul Siegel Sep 10 '10 at 21:28
1  
Check out this post: mathoverflow.net/questions/60877/… –  Manuel Rivera Aug 23 '12 at 14:38

7 Answers 7

up vote 10 down vote accepted

The simplest (to understand) case of finite $\pi_1$ is the group $SO_3$. This can be illustrated using an arm or a belt! $SO_3$ is the group of rotations in space and a based loop in $SO_3$ can be thought of as a description of the motion of an object in such a way that it ends up back where it started. By attaching a strip of paper to the object, it's possible to see this path in space. For example, taking a belt and holding one end fixed whilst moving the other, or moving your hand (your arm forms the "strip").

So: hold your hand out in front of you, this is easiest if you do it palm-up. Keeping it palm-up, rotate it under your arm back to where you started. Your arm is now twisted (hopefully not too badly). Continue moving your hand in the same direction and with your palm facing up but this time over the top of your arm. When your hand gets back to where it started, your arm is now magically untwisted! So two times round the loop gets you back to an untwisted state, thus $2\gamma = 0$. But $\gamma \ne 0$ as evidenced by your twisted arm at the half-way stage.

If you find this difficult to do, here's an alternative way using a belt. Take a belt and twist it once (that is, hold it out straight and imagine an axis along its length, then twist one end all the way around). Now try to straighten it without twisting either end (though you can move either end in space). Can't be done. But if you twist the belt twice then it can.

(There's some funky you-tube videos showing the arm twists. If you get really good at it, you should do it with a beaker of water.)

share|improve this answer
    
Andrew -- great answer! I just edited the 0 out of $SO_3$. –  algori Sep 9 '10 at 21:11
    
Thanks! I suppose I got too caught up in the formalism of homotopy groups, rather than thinking more visually. –  Daniel Miller Sep 10 '10 at 0:28
    
See also John Stillwell's book Naive Lie Theory, in which he explains this "plate trick" maneuver that Andrew describes on p.184 (with a drawing) to illustrate that the group SO(3) is not simply connected. –  Joseph O'Rourke Sep 10 '10 at 0:59
1  
I did this in a seminar once - with a cup of water :) –  David Roberts Sep 10 '10 at 3:26
3  
I did it once (not in a seminar) and broke the glass! My algebraic-topology lecturer used the belt demonstration. –  Loop Space Sep 10 '10 at 6:56

About $\pi_5(S^4)$, this is already in the stable range, so that it is isomorphic to the cobordism group of immersed curves in the plane. That is clearly ${\mathbb Z}/2$ since you can cancel double points pairwise by attaching handles. There are papers from the late 1970s and early 1980s by Koschorke, Sanderson, and Eccles that describe these correspondences. Basically they are Pontryagin Thom constructions. The cobordism group of immersions with specific normal bundles is isomorphic to the homotopy group of a Thom space. WHen the bundle is trivial, the Thom space is trivial. See also a recent preprint of Snaith about the Kervaire invariant 1 problem.

share|improve this answer
    
Oh, and $\pi_4(S^2)$ can be identified with immersed oriented surfaces in 3-space. It is amusing to find the generator. Hint: it is an immersed projection of the standard torus $(e^\theta, e^\phi) \in {\mathbb C}^2$. –  Scott Carter Sep 9 '10 at 22:07
    
This connection sounds very intriguing. Could you go into a little more detail, or point me to a particular place to start reading? –  j.c. Sep 9 '10 at 22:19
1  
It looks like the 1980 paper "Multiple points of codimension one immersions" by Peter John Eccles is pretty accessible. I'll start there www.springerlink.com/index/2504061088521677.pdf –  j.c. Sep 9 '10 at 22:29

Some neat examples of finite fundamental groups G of 3-manifolds are given by the quotients of the sphere S3 of unit quaternions by a finite subgroup G, such as the quaternion group {1, -1, i, -i, j, -j, k, -k} or order 8, or the double covers of orders 24, 48, 120 of the rotations of the platonic solids. The case of the group of order 120 gives the Poincare 3-sphere, which has the same homology of the 3-sphere but different fundamental group.

share|improve this answer
    
And at a more banal level, one can get any finite cyclic group in the same manner :-) –  Robin Chapman Sep 9 '10 at 20:32

To distill some points made in other comments, you must understand $\pi_1$ and covering space theory simultaneously, as they are two sides of the same coin and each informs the other.

Ryan's description of $RP^2$ as a Mobius band capped off by a disk hints at the process to produce a space $X$ satisfying $\pi_n(X)=G$ for any group (abelian if $n>1$) $G$. (If finite $\pi_1$ is hard to visualize, think about a space with $\pi_1$ the real numbers, or the circle!) The basic idea is to take a wedge of $n$ spheres and attach some $n+1$ disks. The idea is that the $n$-spheres generate and the $n+1$ disks introduce relations.

Note that $SO(3)$ is obtained from $RP^2$ by attaching a $3$-disk which doesn't affect $\pi_1$ but does $\pi_2$. So $\pi_2(RP^2)=Z$ and $\pi_2(SO(3))=0$.

But your question hints at more subtle kinds of examples, namely $\pi_n(S^k)$ when $n>k$. Those are finite (except for some cases when $n=2k-1$) and non-trivial for many $n$ and $k$. Note that there are no $n$ or $n+1$ disks around, so these groups are non-zero for interesting reasons. Sorting out what the $\pi_n(S^k)$ are and why is a beautiful and impossibly difficult branch of mathematics. Scott hints at some geometric approaches using immersions and embeddings. There are many other ways to study these groups.

share|improve this answer

A lot of standard examples have $\pi_1(X)=C_2$ or have a $C_2$ as a factor. I think it's harder to visualize spaces with larger cyclic $\pi_1$, so here's a simple example.

Pick a positive integer $n$. We'll construct a space with $\pi_1(X)=C_n$. Take the closed unit disc $D$ in the plane: $D=\{(x,y):x^2+y^2\le1\}$. Identify points on the circumference an angular distance of $2\pi/n$ apart. We get a quotient space $X$, where $(\cos t,\sin t)$ and $(\cos(t+2\pi/n),\sin(t+2\pi/n))$ are identified. In $X$, the map from $[0,2\pi/n]$ to $X$ taking $t$ to $(\cos t,\sin t)$ is a loop $\lambda$. Tracing out $\lambda$ $n$ times gives the whole circumference. This big loop can be contracted by passing it over the centre of the disc $D$, but no smaller positive integer multiple of $\lambda$ can be contracted.

Now, intuiting higher homotopy groups is a much harder matter ....

share|improve this answer

One approach is to understand this as part of the geometry of universal covering spaces. Such spaces are simply connected (even sometimes contractible), but may have finite groups acting on them sufficiently "nicely" (and there's the catch) so that the quotient inherits the finite group as its fundamental group. In fact if the long exact sequence of a fibration (of homotopy groups) applies, you can force a fundamental group to coincide with a discrete group as fibre.

share|improve this answer

Perhaps it isn't necessary to add another answer, but here is one more nevertheless.

  1. Given any discrete group $G$, form the classifying space $BG$. A map of $S^1$ to $BG$ is equivalent to giving a principal $G$-bundle over $S^1$. Such a bundle can formed by gluing the ends of the trivial bundle $[0,1]\times G$ with a twist given by an element of $G$. So in this way, we get $\pi_1(BG)=G$.

  2. Of course, invoking $BG$ is perhaps violating the spirit of the question. When $G$ is finite, we can approximate this by something more concrete. Let $V$ be a faithful unitary representation of $G$ with no trivial summands (e.g. the complement of $\mathbb{C}$ in the regular representation). After replacing $V$ by $V\oplus V$ if necessary, we can assume that $\dim V>1$. Then the unit sphere $S\subset V$ will be simply connected, and $G$ will act on it without fixed points. So the quotient $S/G$ will have fundamental group $G$. This is the same construction used (implicitly) in many of the other answers.

share|improve this answer
    
I think you need a more restrictive hypothesis on G. This is not the case for every compact Lie group, $BS^1 = \mathbb{C}P^\infty$ which is simply connected, hence $\pi_1(BS^1)=0$. It is true for finite groups and $\mathbb{Z}$. –  Sean Tilson Sep 13 '10 at 20:14
    
Good point. $\mbox{}$ –  Donu Arapura Sep 19 '10 at 13:28

Your Answer

 
discard

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.