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I recently noticed a few things in some recent questions on MO:

1) the fundamental group of $S^2$ minus, say, 4 points, is $\langle a,b,c,d\ |\ abcd=1\rangle$.

2) The fundamental group of a torus minus a point is $\langle a,b,c\ |\ [a,b]c=1\rangle$.

I was just wondering if you have a manifold $M$, and you know a presentation of its fundamental group, can you quickly get a presentation of the fundamental group of $M$ minus $n$ points?

Of course, this could merely be a coincidence. These are both surfaces, and both their fundamental groups only have one relation. But perhaps there is more to it than that?

Any and all helpful comments appreciated,


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Are you sure that $\pi_1(T^2- \{pt\})$ has that fundamental group? Because $T^2 - \{pt\}$ is homotpic to $S^1 \vee S^1$ and then its fundamental group need to be $\mathbb{Z}^{*2}$. – Gianluca Faraco Jun 27 at 19:50

1 Answer 1

up vote 16 down vote accepted

The basic tool here is Van Kampen's theorem. Let $M$ be the manifold, and let $M' = M\setminus {x}.$ Let $D$ be a small ball around $x$. Then $M = M' \cup D,$ and $M \cap D = D\setminus {x}$ is homotopic to $S^{n-1}$ (here $n$ is the dimension of $M$).

So Van Kampen's theorem says that $\pi_1(M) = \pi_1(M')*_{\pi_1(S^{n-1})} \pi_1(D)$. If $n > 2,$ then $S^{n-1}$ is simply connected, and of course $\pi_1(D)$ is simply connected, and so this reduces to $\pi_1(M) = \pi_1(M')$; in other words, deleting points doesn't change $\pi_1$ in dimensions $> 2$. (You can probably convince yourself of this directly: imagine you have some loop contracting in a 3-dimensional space; then if you remove a point, you can always just perturb the contraction slightly so that it misses the point. Of course, this is not the case in two dimensions.)

If $n = 2$, then $\pi_1(S^1) = \mathbb Z$ is infinite cyclic, while $\pi_1(D)$ is trivial again. So we see that $\pi_1(M)$ is obtained from $\pi_1(M')$ by killing a loop. One can be more precise, of course (and I will restrict myself to the orientable case, so as to make life easier): if you have a genus $g$ surface, with $r>0$ punctures, and also $g + r > 1$, then $\pi_1$ is a free group on $2 g + r - 1$ generators. (Every puncture after the first adds another independent loop, namely the loop around that puncture.) If $g \geq 1,$ and $r = 0$, then $\pi_1$ is obtained via Van Kampen as above: you begin with a free group on $2g$ generators for $M'$, and then you kill off the loop around the puncture when you fill it in (so you get the standard presentation for the $\pi_1$ of a compact orientable surface of genus $g \geq 1$). (Conversely, going from the compact surface $M$ to the once-punctured surface $M'$ does not add any generators to $\pi_1$, but gets rid of a relation.) If $g = 0$ and $r \leq 1,$ then you have either a disk ($r = 1$) or a sphere ($r = 0$) and so $\pi_1$ is trivial.

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Thank you Matt. I was hoping there was something really cool going on behind the scenes here, but as you point out, this only really matters in dimension 2, and then it's just Van-Kampen. – user641 Aug 9 '10 at 4:44

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