Simplicial complex with fundamental group $\mathbb{Z}/3\mathbb{Z}$

Question

Is there any example of a simplicial complex with fundamental group $\mathbb{Z}/3\mathbb{Z}$?

(Preferably a “simple” example with as few vertices as possible.)

So far I learnt about $RP^2$, with fundamental group $\mathbb{Z}/2\mathbb{Z}$.

Thanks.

Answer 1

Let us see why Ted Shifrin's suggestion works. So let $X$ be the topological space obtained from the following identifications:

$\hskip2in$

Then, for instance, taking the barycentric subdivision of the triangle, it is clear that $X$ is a simplicial complex.

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Edit: As pointed out by Cheerful Parsnip the barycentric subdivision of the triangle doesn't turn $X$ into a simplicial complex but the second barycentric subdivision does: prove that each triangle in the second barycentric subdivision embeds in $X$ and that if any two distinct such triangles intersect, their intersection is either an edge or a point; this last property fails if we use the first barycentric subdivision.

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Let us see why $\pi_1(X)\cong\mathbb Z/3\mathbb Z$.

Let $Y$ be the space obtained by taking three copies of the above triangle $T,T',T''$;(say $T'$ and $T''$ have edges $a',b',c'$ and $a'',b'',c''$, respectively, each having the same orientation as in the drawing above), then gluing $a,a',a''$ respecting orientations, and finally gluing $b,b',b''$ and $c,c',c''$ the same way. Then $Y$ is a $3$-fold covering of $X$(can you see why?).

It is easy to see(a drawing will help) that $Y$ is homeomorphic to the space obtained by taking a sphere $S$ and a closed disk $D$ and gluing $\partial D$ along the equator of $S$.

Let $V$ be the union of the upper hemisphere of $S$ and $D$. Then $V$ is homemorphic to a sphere, $V\cup S= Y$ and $V\cap S$ is a disk, thus $Y$ is simply connected because of Seifert-Van Kampen theorem.

Thus we get that as $Y$ is a $3$-fold covering of $X$, $\pi_1(X)$ must have size $3$, and thus it is isomorphic to $\mathbb Z/3\mathbb Z$.

If you want to do this for any $n>0$ ($n$ not necessarily prime), do the same construction(using an $n$-gon instead), and show that there is a deck transformation of order $n$.