I'm confused. I've seen some materials saying that the torus and Klein bottle do not have the same fundamental group. However, although I understand the standard presentations of both groups (the torus has a presentation of $\langle a, b \mid aba^{-1}b^{-1}=1\rangle$, while the Klein bottle has a presentation of $\langle a, b \mid abab^{-1}=1\rangle$), I do not understand why the two groups are not isomorphic.

By composing the standard covering map of $\mathbb{Z}\times\mathbb{Z}$ onto the torus with the standard double-covering map of the torus onto the Klein bottle, one can produce a covering map of $\mathbb{Z}\times\mathbb{Z}$ onto the Klein bottle, with each copy of the Klein bottle corresponding to a region of $\mathbb{Z}\times\mathbb{Z}$ with dimension $0.5\times1$. It is easy to see that the map of $\mathbb{Z}\times\mathbb{Z}$ can be horizontally stretched to make a map mapping each unit square to the Klein bottle, akin to well-known covering map to the torus. Then, using lifting correspondences, it can be seen, by similar logic to the torus covering map, that the fundamental group of the Klein bottle must be isomorphic to $\mathbb{Z}\times\mathbb{Z}$ (so that, via the double covering map from the torus to the Klein bottle, the induced homomorphism between fundamental groups from that of the torus to that of the Klein bottle would map the fundamental group of the torus isomorphically into the group $\mathbb{Z}\times2\mathbb{Z}$).

Where am I going wrong?

  • $\begingroup$ I don't understand your argument so I can't say where it's going wrong, but if you take the given presentation for the Klein bottle, and abelianize it, you get that $\pi_1(K)^{ab}$ has some $2$-torsion, whereas $\pi_1(T)^{ab}$ doesn't; so they can't be isomorphic. $\endgroup$ Jan 11, 2020 at 16:56
  • $\begingroup$ You are using $\mathbb Z \times \mathbb Z$ in a nonsensical manner when you write about "the standard covering map of $\mathbb Z \times \mathbb Z$ onto the torus". Did you intend to write "the standard covering map of $\mathbb R \times \mathbb R$ onto the torus"? Perhaps this is at the root of your confusion. $\endgroup$
    – Lee Mosher
    Jan 11, 2020 at 18:12
  • $\begingroup$ one is abelian and the other is not $\endgroup$
    – William
    Jan 11, 2020 at 21:03

1 Answer 1


I'm assuming you mean a covering map from $\mathbb{R}\times \mathbb{R}$ to the torus or Klein bottle, rather than $\mathbb{Z} \times \mathbb{Z}$.

It's true that there are universal coverings $p_T\colon\mathbb{R}^2 \to T$ and $p_K\colon\mathbb{R}^2 \to K$, but their groups of deck transformations are different.

In the case of $T$, $p_T$ is the quotient map for the equivalence relation $(x, y)\sim (x + m, y+n)$, so the deck transformations $Aut(p_T)$ are determined by $(m,n)\in\mathbb{Z}\times\mathbb{Z}$. In particular if $t_x, t_y\colon \mathbb{R}^2 \to \mathbb{R}^2$ are defined by $t_x(x, y) = (x+1, y)$ and $t_y(x, y) = (x, y+1)$ then one shows that

$$\begin{array}\\ Aut(p_T) &= \{ f\colon \mathbb{R}^2 \to \mathbb{R}^2\ |\ f(x,y) = (x+n, y+m)\text{ for some }n, m\in\mathbb{Z} \} \\ &= \langle t_x,t_y\rangle \cong \mathbb{Z}\times \mathbb{Z}\end{array}$$

Hence by covering space theory, $\pi_1(T) \cong Aut(p_T) \cong \mathbb{Z} \times \mathbb{Z}$.

However deck transformations of $p_K$ are not quite the same. We will use the rectangle $I \times [0, \frac{1}{2}]$ as the fundamental domain, and we want to identify the left and right sides with the same orientation, and the top and bottom sides with opposite orientation. We will still have $(x, y) \sim (x+m,y)$, but from the perspective of our rectangle we have $(x, 0) \sim(1-x, \frac{1}{2})$, which is then equivalent to $(-x,\frac{1}{2})$; in general our equivalence relation is $(x, y) \sim ((-1)^nx + m, y+\frac{n}{2})$ for $m, n\in \mathbb{Z}\times\mathbb{Z}$. Then $t_x$ is still a deck transformation, but we can't just translate up by $\frac{1}{2}$ because it's not compatible with our equivalence relation, so we introduce $t_{y/2}(x, y) =(-x, y+\frac{1}{2})$. Then it's still possible to show that $Aut(p_K)$ is generated by $t_x$ and $t_{y/2}$, but these elements don't commute: $$ t_xt_{y/2}(x, y) = t_x(-x, y+\frac{1}{2}) = (-x + 1, y + \frac{1}{2}) = t_{y/2}(x-1,y) = t_{y/2}t_x^{-1}(x, y) $$

The result is that $\pi_1(K) \cong Aut(p_K) \cong \langle a, b\ |\ abab^{-1} = 1 \rangle$. This group is not isomorphic to $\mathbb{Z}\times \mathbb{Z}$ because it is non-abelian.

In terms of these generators the homomorphism $\pi_1(T) \to \pi_1(K)$ sends $t_x$ to $t_x$, and $t_y$ to $t_{y/2}^2$. Note that indeed $t_x$ and $t_{y/2}^2$ commute: $t_xt_{y/2}^2 = t_{y/2}t_x^{-1}t_{y/2} = t_{y/2}^2t_x$.


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