# centralizers in hyperbolic manifolds are cyclic

I am having trouble seeing why this statement is true: "If S admits a hyperbolic metric, then the centralizer of any non-trivial element of $\pi(S)$ is cyclic. In particular, $\pi(S)$ has trivial center".

This is on page 23 of Mapping Class Groups by Farb, Margalit.

I have seen some arguments online that do this explicitly by listing the possible elements of $\pi_1(S)$ as deck transformations on $H^2$, the covering space of $S$. However, the book I am using seems to have a different proof. The argument it gives is that if $a$ is centralized by $b$, then $a, b$ have the same fixed points, which I understand mostly. Then the authors claim that since the action of $\pi_1(S)$ on $S$ is discrete, then the centralizer of $a$ in $\pi_1(S)$ is infinite cyclic. I understand that the action is discrete but do not understand why this implies that the centralizer should be infinite cyclic...

Also, I understand that if $S$ had non-trivial center, then $\pi_1(S) = Center(\alpha) = \mathbb{Z}$. The book then claims that this implies that $S$ has infinite volume, which is a contradiction. But I do not understand why $\pi_1(S)= \mathbb{Z}$ implies that $S$ has infinite volume...

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a and b are hyperbolic isometries, and since they have the same fixed points, they have the same axis. Thus we can map the group they generate to R, using the translation length function. This is an injective homomorphism, so this group is a discrete subgroup of R. – user641 Oct 8 '12 at 7:21

Since $a$ and $b$ have the same fixed points, either they both correspond to hyperbolic isometries (if there are two fixed points) or they correspond to parabolic elements (if there is one fixed point).
Suppose $a$ is parabolic, so that the centralizer consists of parabolic elements which fix the same point $p$ on the boundary of $H^2$. In an upper half space model with $p$ at infinity, these parabolic elements look like Euclidean translations. Since the centralizer is discrete, there is a shortest translation. This translation will generate the centralizer.
If $a$ is hyperbolic, then the centralizer is a group of hyperbolic translations that preserve a geodesic in $H^2$ and act as a translation on this geodesic. Again, since the centralizer is discrete, there is an element which translates the geodesic a shortest distance. This element will generate the centralizer.
Now if $\pi_1(S)$ is cyclic then it is either generated by a parabolic element or generated by a hyperbolic element. In either case a fundamental domain for the action of $\pi_1(S)$ on $H^2$ will have infinite area. Think about what a fundamental domain will look like in an upper half space model. In the parabolic case, it will be a region between two "vertical" geodesics. In the hyperbolic case it will be the region between two "concentric" geodesics. In either case, it will contain half of a Euclidean disk centered at a point in the boundary of $H^2$ and this has infinite hyperbolic area.