A topological space $X$ who can be $drawn$, whose fundamental group is $S_3$

Question

There exists an example of a topological space $X$ who can be $drawn$, whose fundamental group is $S_3$? I know that every group is the fundamental group of a topological space, but I need a concrete example.

Answer 1

An interesting question. In general you get $\langle a,b,c|a^m=b^n=c^l=abc\rangle$ as fundamental group of the $3$-manifold $\mathcal{L}/(m,n,l)$ where $\mathcal{L}$ is the connected Lie-group of orientation preserving isometries of the plane $P$ and $(m,n,k)=\langle a,b,c|a^m=b^n=c^l=abc\rangle$ is the orientation preserving transformations of a triangle with angles $\frac{\pi}{m}, \frac{\pi}{n}, \frac{\pi}{l}$ in the Euklidean, spherical or hyperbolic plane $P$, depending opon $\frac{1}{m}+\frac{1}{n}+\frac{1}{l}$ $ =1$, $>1$ or $<1$, respectively. (see Gems, Computers and Attractors for 3-Manifolds by Sóstenes Lins).

You should also read ON THE 3-DIMENSIONAL BRIESKORN MANIFOLDS M(p,q,r)by John Milnor.

But there is an easier way.

Maybe you should think about $\langle a,b,c|a^2=b^2=c^3=abc=e\rangle$ or $\langle s_1, s_2 \mid s_1^2 = s_2^2 = (s_1s_2)^3 = e\rangle$ (see groupprops.subwiki.org) in terms of van Kampen to construct a space with this fundamental group.

First take $S^1\vee S^1$ (one cycle for each $s_i$) and three copies of $D^2$ to form the relation. The boundary of the first $D^2$ is glued to the cycle of $s_1$ by going twice around. Also the boundary of the second $D^2$ is glued to the cycle of $s_2$ going twice around. Then the boundary of the third $D^2$ goes three times around the whole $S^1\vee S^1$.

This should give your fundamental group. It will be hard to visualize, but I think it is possible by drawing things in a transparent way.