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Prove that there is a two-sheeted covering of the Klein bottle by the torus.

OK, so we take the the polygonal representation of the torus and draw a line in the middle as follows:

enter image description here

Then there are two Klein bottles in there, but how do I write down the actual covering map $q:S^1 \times S^1 \to K$?

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Most topologists would be happy just drawing the diagram you've drawn (though the topologists I know prefer to draw on apples), but if you want to do it explicitly then you can as well.

As you know, the torus $S^1\times S^1$ is homeomorphic to $[0,1]\times [0,1]/\equiv$, where $\equiv$ identifies the edges of the square by $(x,0)\equiv(x,1)$ and $(0,y)\equiv(1,y)$. We also define the Klein bottle to be $K=[0,1]\times [0,1]/\sim$, where $\sim$ identifies the edges of the square by $(x,0)\sim(x,1)$ and $(0,y)\sim(1,1-y)$.

For the torus, we have an explicit continuous surjection $$ \pi:[0,1]\times[0,1]\to S^1\times S^1: (x,y)\mapsto\left(e^{i\pi x},e^{i\pi y}\right) $$ using the standard identification of $S^1$ with the unit circle in the complex plane (more a notational convenience than anything else). Note that we now have: $$ (x_1,y_1)\equiv(x_2,y_2)\Longleftrightarrow \pi(x_1,y_1)=\pi(x_2,y_2) $$ In other words, $\pi$ induces a well-defined homeomorphism $([0,1]\times[0,1]/\equiv)\to S^1\times S^1$.

The next step is to interpret your diagram as a map $[0,1]^2\to[0,1]^2$. This map is then going to induce the two-sheeted covering we want. Explicitly, we have: $$ \phi:[0,1]\times[0,1]\to[0,1]\times[0,1]: (x,y)\mapsto \begin{cases} (2x,y) &\mbox{if } x\le\frac12 \\ (2x-1,1-y) & \mbox{if } x\ge\frac12. \end{cases} $$ Composing this map $\phi$ with the projection $\pi_\sim:[0,1]\times[0,1]\to K$, we get a map $\pi_\sim\circ\phi : [0,1]\times[0,1] \to K$.

We claim that this map $\pi_\sim\circ\phi$ induces a two-to-one covering map $$\psi : S^1 \times S^1 \,\,\, = \,\,\, [0,1]\times[0,1]/\equiv \,\,\,\to\,\,\,[0,1] \times [0,1] / \sim \,\,\,= \,\,\,K $$ Proving that $\psi$ is two-to-one means checking $$ |(\psi^{-1}(\{q\})/\equiv)|=2 $$ for each $q \in K$. And to prove that $\psi$ is a covering map it suffices to check that $\psi$ is a local homeomorphism at $p \in S^1 \times S^1$ (ordinarily this is not enough for checking that something is a covering map, but it suffices when the domain and range are compact manifolds). So one has to check something for the points in $[0,1] \times [0,1]$ that form the equivalence class of the relation $\equiv$ corresponding to $p$: the four corner points; or a pair of opposite side points; or an interior point. Namely one must find neighborhoods of those points which, when fitted together under $\equiv$, form an open neighborhood of $p$ that maps homeomorphically onto an open neighborhood of $q=\psi(p)$. Checking these things is the real content of the proof, and I'll leave them as exercises. It's basically what your diagram is telling you.

Now we have a double-cover by $[0,1]\times[0,1]/\equiv$ of $K$. We already remarked that there is a homeomorphism between $S^1\times S^1$ and $[0,1]\times[0,1]/\equiv$; putting these together gives us a double cover of $K$ by $S^1\times S^1$.



I should stress - there's very little content in any of this, and it really is just a way of making your diagram 'rigorous' in some sense. It's good to work thorugh a few examples like this one explicitly, but you'd be bananas to try and be completely rigorous all the time in topology.

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    $\begingroup$ In spite of the upvotes and the acceptance, I'm afraid this answer is not correct. See my answer below. $\endgroup$ Oct 5 '18 at 12:41
  • $\begingroup$ To augment the comment of @LukasLewark, I have corrected this answer. See my edit summary for details of what was done. $\endgroup$
    – Lee Mosher
    Jun 16 '19 at 18:57
  • $\begingroup$ @LeeMosher $\phi$ is not continuous, nor it is well-defined at $(1/2,y)$ for $y\not=1/2$ $\endgroup$
    – Akerbeltz
    Mar 13 '20 at 1:37
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It is my impression that the Banana-image can be understood to give a correct covering - however, the map $f:T\to K$ from the Torus $T$ to the Klein bottle $X$ described in the accepted answer is not a covering map, but merely a continuous map such that all points in the Klein bottle have exactly two preimages.

Note that this is necessary, but not a sufficient condition for $f$ to be a covering. For $f$ to be a double covering map, every point $x\in K$ must have an open neighborhood $U$ such that there exist a homeomorphism $h: U\times \{0,1\}\to f^{-1}(U)$ with (*) $f\circ h|_{U\times\{i\}}$ a homeomorphism from $U\times\{i\}$ to $U$ for $i\in\{1,2\}$.

For the given map, everything is satisfied except condition (*). Thus it is a quite cool example of something that is nearly a covering map!

If you fold along the middle line, then a point $x\in K$ that's on the image of the fold line does have a small open neighborhood $U$ such that $f^{-1}(U)$ is homeomorphic to two copies of $U$. However, restricted to one of these copies, $f$ is not a homeomorphism, because it is two-to-one for points in $U$ not lying on the fold.

It's quite subtle! Another way to see that this is not a covering map goes as follows. If it were, because it is a double cover, there would be exactly one non-identical deck transformation $g: T\to T$, which maps every $y\in T$ to the unique $g(y) \neq y$ with $f(g(y)) = f(y)$. For $y$ in the interior of the square, and not on the fold, $g(y)$ is the reflection of $y$ across the fold. So when you pick a sequence of $y_i$'s approaching a point $y$ on the fold, then $y_i$ and $g(y_i)$ approach one another. By continuity of $g$, we must have $g(y) = y$! But for $y$ on the fold, $g(y)$ is on the boundary of the square.

If you have trouble seeing the mistake, here's a simpler map that fail to be a covering in the same way: let $k: S^1\to S^1$ be defined as $k(z) = z^2$ for $\text{Im} z\geq 0$, and $k(z) = z^{-2}$ for $\text{Im} z \leq 0$. Every point has two preimages, but it's not a covering...

Finally, here's how to construct a map $m: T\to K$ that is a double covering map. Instead of folding along the line (identifying $(x,y)$ with $(1-x, y)$), take the glide reflection (identifying $(x,y)$ with $(1-x, y + \frac{1}{2} \pmod{1})$. I'll let you work out the details...

Note that this fits in well with Amitai Yuval's answer.

I realize my answer is some years late, but I didn't want to leave this uncommented.

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One way to define the torus is as the quotient $T=\mathbb{R}^2/\mathbb{Z}^2$, where $\mathbb{Z}^2$ acts on $\mathbb{R}^2$ by translations. Thus, for any space $X$, specifying a map $f:T\to X$ is equivalent to specifying a map $\overline{f}:\mathbb{R}^2\to X$, which satisfies $\overline{f}\circ g=\overline{f}$ for any $g\in\mathbb{Z}^2$.

One way to define the Klein bottle is as the quotient $K=\mathbb{R}^2/G$, where $G$ is a group of symmetries which contains $\mathbb{Z}^2$. Thus, the natural projection $\pi:\mathbb{R}^2\to K$ descends to the desired double cover $p:T\to K$.

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    $\begingroup$ what would be that $G$ exactly? $\endgroup$
    – rmdmc89
    Sep 9 '19 at 1:33
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    $\begingroup$ @rmdmc89 It's the group of homeomorphisms from $\mathbb{C}$ to $\mathbb{C}$ generated by $a: z\mapsto z+i$ and $b: z\mapsto \overline{z}+\frac{1}{2}+i$. See Exercise 17.9 parts (e), (f), (g) and (i) in Czes Kosniowski's $\textit{A First Course in Algebraic Topology}$. $\endgroup$
    – WLOG
    May 7 '20 at 1:52
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I would like to give my formula. Let us think of $T^2$ as $S^1\times S^1$ and $S^1$ as the unit circle in $\mathbb{C}$. Let $f$ act on $T^2$ by $f(z,w)=(-z, \bar{w})$, then $f^2$ is the identity. $f(z,w)\sim(z,w)$ gives an equivalence relation on $T^2$ and $T^2/\sim$ provides the desired two-fold covering map from $T^2$ to the Klein bottle, as illustraded in the figure by OP.

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