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Consider the torus $T = \Bbb S^1 \times \Bbb S^1$. Show that $T$ does not have the fixed point property. A space $X$ is said to have the fixed point property if for any continuous map $f : X \to X$ there exists $x \in X$ such that $f(x) = x$.

I think I've figured out why the torus doesn't have this proprety, but I cannot construct an explicit map for it. If we consider the identification

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then we would need to construct a map such that it wont intersect the diagonal

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and I think that a map like this blue line would work since it should not intersect the diagonal ever? Can this be constructed explicitly?

enter image description here

Walker
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5 Answers5

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You need a map $\mathbb T \rightarrow \mathbb T$, not just a line segment in the torus. I am not sure what criterion you intend to invoke with the diagonal.

Let’s consider an easier example. Let us show that the circle $\mathbb S^1$ does not have the fixed point property. Geometrically it is clear why: we can just rotate the circle a bit and this will not have a fixed point. Let $r:\mathbb S^1 \rightarrow \mathbb S^1$ be such a rotation.

Now one can identify the torus $\mathbb T$ up to homeomorphism with the topological space $\mathbb S^1 \times \mathbb S^1$. Since $r$ doesn’t have a fixed point, the continuous map $$r\times\operatorname{id}:\mathbb S^1\times \mathbb S^1 \rightarrow \mathbb S^1 \times \mathbb S^1$$ doesn’t have a fixed point.

If you insist on defining the torus via gluing the sides of a square and don’t want to use the identification above, we can still write down essentially the same map as I just constructed. The idea is to just translate everything in the square a tiny bit to the right (points which map out of the right bound will be mapped into the left part of the square again). But this way we have to explicitly show that the map is welldefined and continuous, which is not hard but tedious.

Jonas Linssen
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If a space X does not have the fixed point property then any product XxY also does not have it.

This reduces your problem to showing that the circle does not have the fixed point property. You can surely do that.

Mariano Suárez-Álvarez
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$T=S^1\times S^1$, so define $f:T\to T$ by $f(z,w)=(iz,iw)$. It clearly has no fixed points.

Mark
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  • What is this map doing to the torus? – Walker Oct 16 '22 at 15:24
  • That's a rotation in each coordinate. Similarly, you can rotate just $1$ of the coordinates, i.e $(z,w)\to (iz,w)$, it will work too. Note that $S^1$ itself doesn't have the fixed point property, because of rotation maps. – Mark Oct 16 '22 at 15:25
  • Is there some intuition for this? I'm taking each point on the torus and rotating it in some direction? – Walker Oct 16 '22 at 15:28
  • Well, you can imagine the torus as $S^1\times S^1$, right? (this definition is actually intuitive, although it is formally a subset of $\mathbb{R^4}$) There are two coordinates, each one is identified with the circle $S^1$. You take a point $(z,w)$, and rotate both $z$ and $w$ on the corresponding circle by angle $\frac{\pi}{2}$. – Mark Oct 16 '22 at 15:31
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What about $f(e^{2i \pi x}, e^{2i \pi y}) = (e^{2i \pi \left(x+\frac{\pi}{2}\right)}, e^{2i \pi y})$?

This is deduced from the fact that already $\mathbb S^1$ doesn't have the fixed point property as can be seen by rotating it around its center.

mathcounterexamples.net
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If we consider the torus $T=S^1\times S^1$ as the identification space $T=I\times I/\sim$ of the unit square as shown above, then $f([t,s])=[t+\frac{1}{2}-\lfloor{t+\frac{1}{2}}\rfloor,s]$ is a homeomorphism without fixed points and it is not the identity map. (All points of the identity map is fixed point). Here, $\lfloor x\rfloor$ denotes the floor of the real number $x$.

Bob Dobbs
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