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In my course "Introduction To Algebraic Topology" I had following test problem:

Exemplify a topological space with fundamental group $\mathbb{Z}/3\mathbb{Z}$.

I was supposed to use this theorem:

Let $Y$ be a simply connected topological space. If a group $G$ (finite or countable) acts on $Y$ freely and properly discontinuously, then fundamental group of the quotient space $Y/G$ is naturally isomorphic to $G$.

So problem is I just wasn't able to come up with simply connected space such that $\mathbb{Z}/3\mathbb{Z}$ acts on it freely and properly discontinuously.

Any ideas? Thanks!

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    $\begingroup$ You can construct a space using CW complexes: Take a point, attach a loop, and then attach a disk using a three-to-one map. $\endgroup$
    – user296602
    Commented Dec 19, 2017 at 18:30
  • $\begingroup$ @GlebChili If you are still interested in the problem, you might want to check what are called Lens Spaces. It is basicaly what our Lord Shark has said but for an angle $\frac{2π}{p}$ therefore getting a fundamental group of $Z/p$. $\endgroup$
    – Nick A.
    Commented Apr 28, 2018 at 8:22

2 Answers 2

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How about the unit sphere $S^3$? If $A$ is a $2$-by-$2$ rotation matrix for angle $2\pi/n$, then the block matrix $\pmatrix{A&0\\0&A}$ acts on $S^3$ without fixed points.

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There is an explicit construction:

Consider the CW complex given by three $0$ cells, three $1$ cells and a two cell pasted in (this is a filled in triangle.)

Now, considering the circle with a base point, and pasting in this $CW$-complex by identifying all three vertices with the base point (in a three fold way.)

The fundamental group of the resulting space is exactly $\mathbb Z_3$ by construction, since the two cell gives the relation $a^3=1$ on the fundamental group.

This is called the presentation complex by the way.


If you want to construct the Cayley complex (the universal cover of the presentation complex), you will need to take three disks and paste them to the $1$ skeleton of the first construction. $\mathbb Z_3$ will act by cyclic permutation of the $2$-cells, and taking the quotient will give you exactly the presentation complex. Note that this is a simply connected space and that $\mathbb Z_3$ acts freely on the two cells.

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  • $\begingroup$ My mistake Andres Mejia. I meant to upvote but put my phone into my pocket and the vote switched accidentally. I’m locked out from voting now because I’ve waited too long. If you make a minor edit, I believe I can correct my mistake. $\endgroup$ Commented Dec 19, 2017 at 21:55
  • $\begingroup$ Perhaps that will work. By the way, are you at vassar? $\endgroup$ Commented Dec 19, 2017 at 21:58
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    $\begingroup$ It worked. Yes I am at Vassar. $\endgroup$ Commented Dec 19, 2017 at 22:06

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