6

I've come up with some examples to apply the Hurewicz theorem to compute $H_1(X)$.

This is only interesting if $\pi_1(X)$ is not abelian. The only examples of $X$ such that $\pi_1(X)$ not abelian I can come up with are $\vee_i S^1$ and $\Sigma_g$ the surface of genus $g$ for $g > 1$.

Does anyone know any other examples, preferably easy ones? Many thanks for your help!

Community
  • 1
Rudy the Reindeer
  • 46,359
  • Very easy examples would be products of the ones you know already :) – t.b. Jul 31 '11 at 16:44
  • your first comment: I was looking for something more interesting, yet not too difficult. : ) – Rudy the Reindeer Jul 31 '11 at 16:48
  • @gary: the Hurewicz theorem doesn't say that... – Qiaochu Yuan Jul 31 '11 at 16:54
  • 3
    More seriously (I temporarily confused myself): Take $S^3$ identified with the unit quaternions. The quaternions $\pm 1, \pm i, \pm j, \pm k$ form the quaternion group $Q$. The quotient $S^3/Q$ has fundamental group $Q$. Interesting examples also arise from knot complements (try to compute the fundamental group of the complement of the trefoil knot, for example). – t.b. Jul 31 '11 at 16:56
  • 7
    @Matt: are you aware that every group is the fundamental group of some space? It's not particularly hard to find a finite $2$-complex whose fundamental group is any finitely presented group, either (start with a wedge of circles, then glue in $2$-cells corresponding to the relations), and their homology isn't difficult to compute. – Qiaochu Yuan Jul 31 '11 at 16:58
  • @Qiaochu: No, I wasn't aware of that, thanks. – Rudy the Reindeer Jul 31 '11 at 17:00
  • Would it be appropriate to make this a community wiki question? – Rudy the Reindeer Jul 31 '11 at 17:00
  • @Matt: ah. Then see http://math.stackexchange.com/questions/36775/is-the-group-of-rational-numbers-the-fundamental-group-of-some-space . – Qiaochu Yuan Jul 31 '11 at 17:06
  • 2
    @Qiaochu's answer here explains his comment in somewhat more detail and lhf gives the precise reference to Hatcher. – t.b. Jul 31 '11 at 17:06

4 Answers4

6

One more example: complement to any non-trivial knot in $S^3$ has non-abelian $\pi_1$.

Grigory M
  • 17,478
4

I am late to the party here, but take any finitely presented group, say $G=\langle x_1,\ldots , x_n: r_1,\ldots r_m \rangle$. Take a bouquet of circles embedded in 4-space and a tubular neighborhood thereof. This is a space with fund. group free of rank $n$. Now carve out a tubular neighborhood of the words represented by the relations, and sew in $D^2 \times S^2$s in their place. There is plenty of room to do this in 4-space. So you can construct a $4$-manifold with $\pi_1$ being any finitely presented group that you like.

t.b.
  • 78,116
Scott Carter
  • 2,439
1

Try the cube with a twist

Take the quotient space of the cube $I^3$ obtained by identifying each square face with opposite square via the right handed screw motion consisting of a translation by 1 unit perpendicular to the face, combined with a one-quarter twist of its face about it's center point.

I think it is a problem in Hatcher's book somewhere

Community
  • 1
Juan S
  • 10,268
0

Consider the figure 8. Using Van Kampen's theorem, you know that its fundamental group is the free product of the additive group integers with itself. Said group is nastily non-abelian.

ncmathsadist
  • 49,383