# Fixed points of maps defined on Teichmüller space

Let $$\mathcal{T}_A$$ be a Teichmüller space of the sphere $$S^2$$ with a finite set $$A$$ of marked points, and suppose that $$f \colon \mathcal{T}_A \to \mathcal{T}_A$$ is a holomorphic map that has a periodic cycle (that is not a fixed point).

In the case, when $$A = 4$$, $$\mathcal{T}_A$$ is biholomorphic to $$\mathbb{D}$$ and, therefore, such a map $$f$$ should be conjugated to a rotation map on $$\mathbb{D}$$ (for instance, by Denjoy–Wolff theorem). Thus, $$f$$ has a fixed point.

What about the case when $$|A| > 4$$? Can we say that $$f$$ necessarily has a fixed point if it has a periodic cycle?

As any holomorphic map on $$\mathcal{T}_A$$, the map $$f$$ should be $$1$$-Lipschitz. Therefore, since it has a cycle, then there exists a $$f$$-invariant compact subset of $$\mathcal{T}_A$$. Moreover, since $$\mathcal{T}_A$$ is geodesic, then $$f$$ should preserve geodesics between points of the hyperbolic cycle. But it is not clear for me whether these arguments lead to a conclusion….

• When you say "map", do you mean biholomorphic? Every biholomorphic map of Teichmuller space of dimension $>1$ will come from the mapping class group and, hence, will have a fixed point if it has a periodic orbit (NRP). Commented Jun 19 at 14:35
• I am talking about holomorphic maps but I do not assume that they are injective. If f is a holomorphic map of the unit disk that is not an automorphism then every orbit either converges to a unique fixed point inside the disk or to a unique point on the boundary. In the last case periodic cycles are not possible, so I believe my claim is correct...
– A B
Commented Jun 19 at 14:43

Let $$T$$ denote the Teichmuller space of a hyperbolic Riemann surface of finite conformal type. Suppose that $$f: T\to T$$ is a holomorphic map which has a periodic orbit $$Z$$, i.e. a finite invariant subset on which $$f$$ acts as an order $$n$$ cyclic permutation. Set $$g=f^n$$, then $$g$$ fixes $$Z$$ pointwise. Since $$f$$ and $$g$$ commute, $$f(F_g)\subset F_g$$, where $$F_g$$ is the fixed-point set of $$g$$ in $$T$$. It is nonempty since $$Z\subset F_g$$. I claim that $$F_g$$ is a convex subset of $$T$$. Indeed, take two points $$p, q\in F_g$$. Then the Teichmuller geodesic segment $$pq$$ between $$p, q$$ is mapped via $$g$$ to a path $$c$$ in $$T$$ of length $$\le d(p,q)$$ (since $$g: T\to T$$ is a holomorphic map, it weakly decreases the Kobayashi-Teichmuller distance). But $$c$$ connects $$p$$ and $$q$$, hence, its length $$=d(p,q)$$, i.e. $$c$$ is a geodesic segment equal to $$pq$$; thus, $$pq$$ has to be fixed by $$g$$ pointwise. Thus, $$pq\subset F_g$$. (Most likely, $$F_g$$ is a submanifold but we do not need this.) Convexity of $$F_g$$ implies that it is contractible. The restriction of $$f$$ to $$F_g$$ is a periodic homeomorphism of order $$n$$. It is now a classical argument going back to Nielsen that $$f$$ has a fixed point in $$F_g$$. It goes as follows. Set $$F:=F_g$$ and $$h:=f|_F$$. Suppose first that $$n$$ is prime. A finite cyclic group of prime order cannot act freely on a contractible and locally contractible finite-dimensional locally compact metrizable space (since finite nontrivial groups have infinite cohomological dimension). Hence, some nontrivial element of $$\langle h\rangle$$ fixes a point in $$F$$. But, since $$n$$ is prime, this element generates the cyclic group $$\langle h\rangle$$. The general case is proven by induction on the number of prime factors of $$n$$. Suppose that $$n=pq$$ where $$p$$ is prime. Then $$h^q$$ has order $$p$$ and, hence, has nonempty fixed-point set $$F'=F_{h^q}$$ in $$F$$. But $$F'$$ is again convex, hence, a contractible and locally contractible subset, invariant under the action of $$h$$. Thus, we reduced the problem to finding a fixed point of $$h^p$$ in $$F'$$. The order of $$h^p$$ is $$q and we continue inductively.
• Thank you! But don't you require that $d_{\tau}g$ is not the identity for every $\tau \in F_g$ in the first part of your argument? And can you provide a reference for the Nielsen's argument or, perhaps, elaborate a bit more on the fact that a finite prime order group cannot act freely on a contractible finite-dimensional CW-complex?
• @AB: What is $d_\tau g$? As for the nonexistence of free actions of cyclic group actions on contractible finite-dimensional cell complexes, see my answer on MSE here, namely, the second argument. Commented Jun 25 at 8:31
• By $d_{\tau}g$, I mean the differential of $g$ at the point $\tau$. Because in order to state that $F_g$ is a submanifold, you need to apply the inverse function theorem?