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The group $\Bbb{Z}^3$ can be represented as $\Bbb{Z}^3=\langle{a,b,c | [a,b],[b,c],[a,c]}\rangle$, where $[a,b]$ means $ab=ba$.

All these commutators reminded me of the torus fundamental group, $\pi_1(T^2)=\langle a,b | [a,b]\rangle$ and I tried to come up with some quotient space of tori that will represent $\Bbb{Z}^3$.

First approach: "intuitively" gluing tori

Let $X$ be the quotient space of two tori modulo the horizontal circle, that is, two tori glued on the horizontal circle. We get $\pi_1(X)=\langle a,b,c | [a,b],[a,c]\rangle$ where $a,b$ generate the first torus and $a,c$ generate the second torus, and $a$ is the shared horizontal generating loop.

To add the relation $[b,c]$, we need to somehow glue a third torus with generators $b,c$ (the vertical generating loops in the two other tori) to the space $X$. This confused me, so I turned to a safer way of dealing with the problem.

Second approach: fundamental polygon

Drawing the fundamental polygon of $\Bbb{Z}^3=\langle{a,b,c | [a,b],[b,c],[a,c]}\rangle$, without making any assumptions in how it "should" look, I realized $\Bbb{Z}^3$ is obtained by gluing three tori, like in the first approach, where the three squares representing each torus are glued together as three adjacent sides of a cube. This gives a more concrete way to think about this space. Two tori are glued at the horizontal circle, as before, and the third torus... It's horizontal generating loop is identified with the vertical generating loop in one torus, and its vertical loop is identified with the vertical loop in the other torus, just as "intuitively" gluing tori implied.

Alternatively, we can think of it differently, taking the identified sides to take different roles in each torus. First, two tori are identified on a loop which is horizontal in one and vertical in the other. This gives one torus inside the other torus,

enter image description here

and now we need to add the third torus whose horizontal loop is $a$ and its vertical loop is $c$. Well, that is more imaginable than the other way. It will involve some kind of "folding" of the big torus, since now both $a,c$ are facing the same direction in space, and their faces should be orthogonal for them to generate another torus, but it seems possible.

After some headache, I think I found what I was looking for: a representation of $\Bbb{Z}^3$ as the fundamental group of glued tori. Is there an easier approach to obtaining and imagining that space? Does it even have an embedding into $\Bbb{R}^3$?

One thing that bothered me along the way is that the fundamental polygon of the three glued tori, after some manipulations, brought me to $1=abca^{-1}c^{-1}b^{-1}=ada^{-1}d^{-1}$ which is exactly the fundamental polygon of a single torus. What does that mean? Saying the spaces are homotopic seems false, but I didn't find an error...

I have little background in topology, so please try to keep it simple.

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  • $\begingroup$ $\mathbb{Z}^3 \cong \pi_1(S^1 \times S^1 \times S^1)$, i.e., the fundamental group of the $3$-torus. $\endgroup$ – Dan Rust Oct 7 '17 at 17:24
  • $\begingroup$ Also, note that $\mathbb{Z}^3$ is not the fundamental group of any surface. Check out the classification of closed surfaces to read further. $\endgroup$ – Dan Rust Oct 7 '17 at 17:27
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    $\begingroup$ I suppose you mean $Z^3$ as a group rather than a topological space. Yes, $Z^3$ is the fundamental group of a 2-dimensional complex obtained by gluing three 2-dimensional tori (it is the 2-dimensional skeleton of $T^3$ equipped with some natural cell-complex structure). It can be also described as the presentation complex of the standard presentation of $Z^3$ as in line 1 of your question. No, it does not embed in $R^3$. $\endgroup$ – Moishe Kohan Oct 7 '17 at 17:28
  • $\begingroup$ @DanRust isn't the 3-torus a surface? $\endgroup$ – Whyka Oct 7 '17 at 17:43
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    $\begingroup$ @Whyka A surface means a two-dimensional manifold, while the 3-torus is three-dimensional. $\endgroup$ – lisyarus Oct 7 '17 at 17:46

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