Two Applications of Topological Fixed Point Theory

The geometric properties of a complete Riemannian manifold $M$ with non-positive curvature are intimately linked with its fundamental group. Specifically, the structure of the isometry group $Isom(M)$ has been a field of intense investigation. It is not easy to ascertain whether a geometric property is, in fact, topological. As an attempt in this direction, we present a new proof of the following version of Cartan’s fixed point theorem (see [6], Chap. 6, Thm. 23, p. 164). A stronger version of the same for precompact subgroups of $Isom(M)$ is known (see [2], Thm. 1.4.6, p. 21).

The above theorem can be proved by considering the centre of mass of the orbit of $\sigma$. In this article, we present a proof using topological fixed point theory. Our approach also reveals the following fact which is interesting by its own. By a pencil of geodesics, we shall mean a (non-intersecting) family of geodesics any two of which lie within a finite Hausdorff distance of one another.

Next, certain domains in the complex space $\mathbb{C}^{n}$ also become complete manifolds with non-positive curvature when equipped with the Bergmann metric. In this vein, we consider the group of conformal automorphisms of a bounded planar domain (with smooth boundary) having finite connectivity (see [5], Def. 4.2.2, p. 85). It is well-known that such a group is finite whenever the connectivity exceeds two (see [5], Thm. 12.2.3, p. 263). We prove the following result about such domains (where $Aut(.)$ denotes the group of conformal automorphisms of the respective domain).

Theorem 1.2 shows that there are further restrictions on the possible order of an element in $Aut(\Omega_{k})$. Hence, the subgroups of $Aut(\Omega_{k})$ are constrained by the value of $k$. For instance, the cyclic group of order $7$ cannot occur as a subgroup of $Aut(\Omega_{k})$ for any $k<6$ (otherwise, we would have $7|(k^{3}-k)$). Note that the upper bound mentioned above does not exclude this possibility as $\#Aut(\Omega_{5})\leq 10$.

By the Cartan-Hadamard Theorem (see [6], Chap. 6, Thm. 22, p. 162), $M$ is diffeomorphic to $\mathbb{R}^{n}$, where $n=\dim M$. We compactify $M$ by adding its ideal boundary $M(\infty)$ to obtain the Gromov compactification

We refer the reader to [1], Chap, 3, pp. 22-32 for the details. In fact, $\bar{M}$ is homeomorphic to the closed unit ball $\mathbb{B}^{n}\subset\mathbb{R}^{n}$.

Next, let $\sigma\in Isom(M)$ have a finite order. It is known that $\sigma$ induces a homeomorphism $\bar{\sigma}$ of $\bar{M}\approx\mathbb{B}^{n}$.

We shall argue inductively to reduce the dimension $n$. By Brouwer’s fixed point theorem, $\bar{\sigma}$ has a fixed point $x\in\bar{M}$. If $x\in M(\infty)$, we observe that (see [1], Chap. 3, p. 30) $\bar{\sigma}$ maps a horosphere centred at $x$ to itself. This horosphere in $\bar{M}$ is diffeomorphic to $\mathbb{B}^{n-1}$. So, we may repeat the whole argument to keep reducing the dimension. At each stage, the restriction of $\bar{\sigma}$ fixes a point lying either in $M$ or on $M(\infty)$.

If the latter case were to occur until the dimension reaches two, the argument would yield a geodesic $\gamma\subset M$ satisfying

But now, $\gamma$ is isometric with $\mathbb{R}$ (equipped with the standard Euclidean metric). Any isometry of the latter either fixes a point or has infinite order. As $\sigma$ is of finite order, we get a fixed point as asserted. This completes the argument.