# Homeomorphic to the disk implies existence of fixed point common to all isometries?

A fellow grad student was working on this seemingly simple problem which appears to have us both stuck. (The problem naturally came up in his work so isn't from the literature as far as we know).

Let $M$ be a metric space homeomorphic to the closed unit disk $D^n\subset \mathbb{R}^n$. We call such a metric space an $n$-cell. Let $\mbox{Isom}(M)$ be the group of bijective isometries $M\rightarrow M$.

If $M$ is an $n$-cell, does there exist a point $p\in M$ such that $\varphi(p)=p$ for all $\varphi\in \mbox{Isom}(M)$?

Clearly such a $p$ need not be unique.

So far, the best attempt has been to consider a set which is invariant under isometries, as follows.

Let $\partial M$ be the boundary of $M$ and define a function $f\colon M\rightarrow \mathbb{R}$ by $f(x)=\sup_{y\in\partial M}\{d(x,y)\}$ which is continuous, and as $M$ is compact must attain its minimum say $m$. Then, let $$A=\{p\in M\mid f(p)=m\}.$$ That is, $A$ is the (non-empty) set of points in $M$ which minimise the maximum distance to the boundary of $M$.

It should be clear that if $\varphi$ is an isometry on $M$, then $\varphi(A)=A$, and one would hope that $A$ is in fact a single point (or at least fixed pointwise instead of just setwise). However, proving this is not clear. There is obviously something else missing here as the topology on the $n$-cell is crucial. For instance an annulus has no such fixed point and the set $A$ would be the inner boundary circle.

It's possible the above set up isn't the right way to tackle the problem. It's also possible that there exists a counterexample and there is some $M$ with no fixed point. Any help is appreciated.

• You may already know this, but the similar statement "If $M$ is an $n$-cell, then for all $\phi \in Isom(M)$ there exists a point $p \in M$ such that $\phi(p) = p$" is implied by Brouwer's Fixpoint Theorem.
– GMB
Sep 20, 2013 at 19:05
• @GregBodwin Yes it's certainly similar. In some sense the conditions have been both strengthened and weakened. On the one hand we ask that a point is fixed by a set of functions instead of just one, and on the other hand we only consider a special class (in some cases finite) of maps rather than all continuous maps. Given that the result relies on the topology of a cell, I expect that a proof might need to use Brouwer's F.P Theorem, or share similarities to a proof of Brouwer. Sep 20, 2013 at 19:12
• The set $A$ can be all of $M$. Consider a ball with the Euclidean metric capped at $1$. So I'm not optimistic that investigating $A$ will lead to a proof. Sep 20, 2013 at 22:09
• Is there any possibility that such a isometry comes from a isometry on the ball? Something like a conjugation Sep 24, 2013 at 1:13
• Would you at all be interested in the overly special smooth case where the metric comes from a Riemannian metric? I've thought about that case briefly and think I can prove the existence of a fixed point, but I figured I'd ask if you cared about this before pursuing it further. Oct 1, 2013 at 19:23

Contrary to our intuition the answer is negative: There are smooth actions of compact groups $G\times D^n\to D^n$ (even finite groups) on closed $n$-disk $D^n$ (for $n$ sufficiently large) so that the action is isometric with respect to some Riemannian metric on $D^n$ and has no point in $D^n$ fixed by the group $G$. By considering the distance function defined by the Riemannian metric, one obtains actions of compact groups of isometries of the resulting metric space, which do not have a fixed point in $D^n$.

Here is where the examples are coming from:

1. Given a smooth manifold $M$ (possibly with boundary) and a smooth action of a compact group $G$ on $M$, denoted $G\times M\to M$, there exists a $G$-invaraint Riemannian metric $ds^2$ on $M$. This standard fact is proven by taking an arbitrary Riemannian metric $ds_o^2$ on $M$ and averaging it under the action of the group $G$. If $G$ is finite, the averaging procedures is just $$ds^2= \frac{1}{|G|} \sum_{g\in G} g^*(ds_o^2).$$ If $G$ is compact one replaces the sum with the integral over the Haar measure of $G$.

2. Thus, it remains to find compact groups acting smoothly on $D^n$ without a fixed point. The first examples of this type were constructed by Conner and Richardson, in the case when $G$ is the icosahedral group, i.e., the group $I$ of orientation preserving symmetries of the regular 3-dimensional icosahedron. Later on, Oliver constructed an example of a smooth action of $SO(3)$ on $D^8$ which does not have a fixed point (in particular, the subgroup $I<SO(3)$ does not fix a point in $D^8$). You can find a detailed description of Oliver's example, references and further information in the survey:

M.Davis, A survey of results in higher dimensions, In "The Smith Conjecture", (editors: J. W. Morgan and H. Bass), Academic Press, New York, 1984, https://people.math.osu.edu/davis.12/old_papers/survey.pdf

On the positive side, if you equip $D^n$ with a Riemannian metric of nonpositive curvature $ds^2$, so that the boundary is convex, then the isometry group of $(D^n, ds^2)$ does have a fixed point in $D^n$: This is a corollary of Cartan's Fixed Point theorem. You can find a proof of the latter for instance in Petersen's "Riemannian Geometry" book. I think, Kirill in his answer was trying to reproduce the proof of Cartan's theorem, without the nonpositive curvature assumption (which, of course, cannot be done).

• Nice answer, it is very interesting. Can I ask: what is Cartan's fixed point theorem? The only thing I could find was this, and it mentions it only briefly. Oct 8, 2013 at 12:24
• @Kirill: Suppose that $X$ is a complete simply-connected Riemannian manifold of nonpositive curvature with convex boundary. Then the isometry group of $X$ has a fixed point. It is usually stated for manifolds without boundary, but the same proof goes through if you have convex boundary. Check out Petersen's book or this blog: amathew.wordpress.com/tag/cartan-fixed-point-theorem Oct 8, 2013 at 12:27
• Thanks for the great answer. The example seems pretty accessible which is surprising - I was expecting any counterexamples to be much more pathological (ie not smooth) metric spaces. Oct 8, 2013 at 12:54

This is only probably correct wrong. I'm going to assume the manifold is a Riemannian manifold, diffeomorphic to an $n$-disk, I'm not sure if this makes much difference to you. I think there is a problem with this if $M$ has more than one minimal geodesic between some pairs of points.

Consider the function "mean squared distance from a given point": $$f(x) = \int_M d(x,y)^2\,dy.$$

Given a vector $v$ tangent to $M$ at $x$ and another point $y$ on $M$, we know that $\partial_v d(x,y) = 0$ when $v$ is orthogonal to $t$, the unit tangent vector to the minimal geodesic going from $x$ to $y$, and that $\partial_t d(x,y) = -1$.

Thus for an arbitrary vector $v$ tangent at $x$, $\partial_v d(x,y) = -(v,t)$, and so: $$\partial_v f(x) = -2\int_M dy\,d(x,y)(v,t),$$ where $t\in TM_x$ is a function of the variable of integration $y$, and the Hessian of $f$ is $$H(w,v) = \partial_w \partial_v f(x) = 2\int_M dy\,(w,t)(v,t),$$ $$H = 2\int_M t^*\otimes t^*.$$

This Hessian is positive definite everywhere on $M$, and so $f$ has no degenerate critical points on $M$, and is thus a Morse function. Morse theory implies that if $m_k$ is the number of critical points of $f$ of index $k$ (index of a critical point is the number of decreasing directions), we have $$\sum_{0\leq k\leq n} (-1)^k m_k = \chi(M),$$ where $\chi(M)$ is the Euler characteristic of $M$ (the alternating sum of its Betti numbers).

Finally, $H$ is positive definite, so $m_k = 0$ for $k>0$, and $\chi(M)=1$ because $M$ is diffeomorphic to the unit disk. Therefore $$m_0 = 1,$$ and so $f$ achieves its minimum at precisely one point, which must be invariant under all isometries because it is a level set of $f$, which is a function invariant under isometries.

• Since $d(x,y)^2\geq 0\Rightarrow f\geq 0$ and $d(x,y)^2 = 0$ at $x=y$, your argument should imply that the unique minimum of $f$ is obtained at $y$? But since $y$ is arbitrary, this would imply all of $M$ is fixed under isometries.
– Neal
Oct 3, 2013 at 13:13
• @Neal I want $y$ to be a variable of integration, not a fixed arbitrary point on $M$. $f$ is a function of $x$ only. Oct 3, 2013 at 13:14
• Oh, duh. Maybe rewrite the paragraph after the first integral? "$y$ fixed" threw me off.
– Neal
Oct 3, 2013 at 13:16
• Does this not prove that such a fixed point is unique? There exist spaces which have more than a single fixed point (Take a disk with a 'bump' somewhere off center - it has a fixed line going through the center and the bump) Oct 3, 2013 at 13:32
• @Kirill Don't you need to assume convexity so that geodesics don't get kinked when they slam into the boundary? Actually I think the interior being a geodesic space implies convexity, so that's probably a moot point.
– Neal
Oct 3, 2013 at 13:55