A fixed point domain is a subset $F \subset \mathbb{R}^n$ in which Brouwer’s Fixed Point Theorem is true. Can we find fixed point domain in $\mathbb{R}^n$ other than the space homeomorphic to $D^n$? Is there any non-trivial example?

More specifically, can we classify all fixed point domain in $\mathbb{R}$, or even $\mathbb{R}^n$?

  • $\begingroup$ In effect, are you asking whether Brouwer's Fixed Point Theorem is sharp? That is, whether we can prove the existence of a fixed point for continuous maps on another class of subsets of $\mathbb{R}^n$ that aren't necessarily homeomorphic to $D^n$? $\endgroup$ – erfink Oct 24 '17 at 3:49
  • $\begingroup$ Yes that's exactly what I'm asking. I have trouble finding non-trivial examples. $\endgroup$ – William Sun Oct 24 '17 at 3:50
  • $\begingroup$ Will have to give this a little more thought, but a few initial impressions: 1) $F$ must be bounded 2) $F$ must be closed 3) 1&2 give $F$ compact, 4) $F$ must be connected $\endgroup$ – erfink Oct 24 '17 at 3:59
  • $\begingroup$ A generalized version of this fixed point theorem provides the existence of a fix point from any convex compact subset $K$ of a Euclidean space to $K$ itself. If your impressions are right, does it mean we have classified all $F$s in $\mathbb{R}$? $\endgroup$ – William Sun Oct 24 '17 at 4:06
  • $\begingroup$ en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem a slightly more general topological theorem. $\endgroup$ – erfink Oct 24 '17 at 4:13

I assume this is what you are referring to:

Definition. Let $X$ be a topological space. We will say that $X$ has the Fixed Point Property (FPP) if for any continuous mapping $f:X\to X$ there is $x\in X$ such that $f(x)=x$.

With that definition the Brouwer's Fixed Point Theorem can be reduced to:

Brouwer's Fixed Point Theorem. Closed disk $D^n\subseteq\mathbb{R}^n$ has FPP.

But there are more nonhomeomorphic subspaces of $\mathbb{R}^n$:

Lemma. Let $X$ be a FPP space. If $A\subseteq X$ is a retract of $X$ then $A$ has FPP.

Proof. Take the retraction $r:X\to A$, the inclusion $i:A\to X$ and consider any continuous function $f:A\to A$. Then we have an induced $F=i\circ f\circ r$ and since $F:X\to X$ then $F$ has a fixed point, say $x_0\in X$. We have $x_0=F(x_0)=i(f(r(x_0))$ and since the image of $i$ is $A$, then $x_0\in A$. It can be easily checked that $x_0$ is a fixed point of $f$. $\Box$

Non-trivial Example. Take $D^2$ and perform "squeezing" along its equator towards the center of the ball. The resulting space $X$ is the wedge sum (or bouquet) of two disks $X\simeq D^2\vee D^2$. It is a retract of $D^2$ and thus it has FPP by the lemma. But it is not homeomorphic to any ball because the base point becomes a cut point while disks have no cut points.

The problem of classifying FPP spaces is still open and not much has been done in the field (seems to be very hard). For more information read this: https://en.wikipedia.org/wiki/Fixed-point_property

EDIT. If you are looking for a non-contractible FPP subspace of $\mathbb{R}^n$ then have a look at this:

Pathological FPP space

This space, call it $\mathcal{K}$ can be formally defined as follows:

$$K := \{(x,y)\in\mathbb{R}^2\ |\ \lVert(x,y)\rVert=1\mbox{ and }y\leq 0\}$$ $$\forall_{v,w\in\mathbb{R}^2}\ I(v,w):=\{tv+(1-t)w\ |\ t\in[0,1]\}$$ $$\forall_{n\in\mathbb{N}}\ K_n:= I\big((1, 0), (-1, n^{-1})\big)$$ $$\mathcal{K}=K\cup\bigcup_{n=1}^{\infty}K_n$$

The proof that $\mathcal{K}$ has FPP and it is not contractible can be found in D.R. Smart, "Fixed Point Theorems", Theorem 3.2.3.

  • $\begingroup$ It is still contractible. So I don't consider this as non-trivial. $\endgroup$ – Anubhav Mukherjee Oct 24 '17 at 11:45
  • $\begingroup$ @AnubhavMukherjee Well, I do. The OP explicitely asks about non-homeomorphic space, not non-homotopic. $\endgroup$ – freakish Oct 24 '17 at 11:49
  • $\begingroup$ But do you know anything non-homotopic? Look at the example I gave in the comment section. Atleast can we tell anything for manifold? $\endgroup$ – Anubhav Mukherjee Oct 24 '17 at 11:50
  • $\begingroup$ @AnubhavMukherjee I've updated the answer. Unfortunately I don't know much about manifolds in the context of FPP. $\endgroup$ – freakish Oct 24 '17 at 14:49
  • 1
    $\begingroup$ have a look at this post mathoverflow.net/questions/284292/… $\endgroup$ – Anubhav Mukherjee Oct 24 '17 at 14:54

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