markandmath

Near the beginning of the book A Primer on Mapping Class Groups it’s shown that the center of the fundamental group of a hyperbolic surface of finite volume must be trivial. I had a hard time understanding the proof until I got help from someone. There’s some important and basic stuff involved in it.

Let $M$ be a hyperbolic surface, not necessarily of finite volume (for now). So $M$ is the quotient of $\mathbb{H}^2$ by a subgroup $\Gamma$ of $\mathrm{PSL}(2,\mathbb{R})$ acting freely and properly discontinuously, where $\Gamma \simeq \Pi_{1}(M)$ . What we’ll show is:

If $g \in \Gamma,$ not equal to the identity, and $H$ is a subgroup of $\Gamma$ which contains $g$ and satisfies that every element of $H$ commutes with $g,$ then $H$ must be cyclic.

In other words such an $H$ exists only if it really must. I’ll say later how this implies the trivial center thing if $M$ has finite volume.

So, fix our $g \in \Gamma.$ Suppose $h \in \Gamma$ commutes with $g,$ and isn’t the identity.

Fact numero uno: $g$ and $h$ have the same fixed points in $\partial \mathbb{H}^2.$

Proof: Say $x \in \partial \mathbb{H}^2$ is a fixed point of $g.$ Then $h(x) = h(g(x)) = g(h(x),$ meaning that if $x$ is a fixed point of $g$ then so is $h(x).$ Remember from the classification of isometries that every isometry has zero (elliptic), one (parabolic), or two (hyperbolic) fixed points on the boundary, and nothing in $\Gamma$ is elliptic. If we’re in the case that both $g$ and $h$ are hyperbolic, then say their fixed points are the reds and the blues respectively.

circle2

From the above, we know that $h$ sends fixed points of $g$ to fixed points of $g,$ so it must send the red points to the red points. We want it to be that the red and blue points are actually the same. If not, then either $h$ fixes three points, so is the identity, or $h^2$ fixes three points and is the identity. But even in the latter case we must have that $h$ is the identity. This follows from thinking about how any hyperbolic isometry is conjugate to the typical one: $z \mapsto az.$ This kind of reasoning shows that the fixed points of $h$ and $g$ are the same, with there not being much different when one (and hence both) are parabolic. $\Box$

So, they have the same fixed points at infinity. The next observation is that this means $g$ and $h$ leave invariant the same “copy of $\mathbb{R}$ “. In the case that they’re hyperbolic, it’s the geodesic joining the two fixed points. In the case that they’re parabolic, it’s any horocycle tangent to the fixed point.

geodesic_horocycle — The red line is a geodesic which might be sent by a hyperbolic isometry to itself, the blue circle is a horocycle which might be sent by a parabolic isometry to itself.

Each of those is isometric to $\mathbb{R}.$ In the case of the vertical line geodesic in the upper half plane, the isometry with $\mathbb{R}$ is given by natural log, in the case of a horocycle, just think of one of the horocycles tangent to the point at infinity in the upper half plane, so a horizontal line. At height $1,$ that horocycle really is exactly $\mathbb{R}.$

Fact numero dos: The $H \leq \Gamma$ as above gives rise to a group of isometries of $\mathbb{R},$ with its usual metric, which is cyclic.

Proof: As we saw above, each element of $H$ gives an isometry of $\mathbb{R}.$ And this has to actually be a translation of $\mathbb{R},$ as opposed to a reflection or something. As usual you see this by thinking about the “typical” hyperbolic and parabolic isometries, $z \mapsto az$ and $z \mapsto z + b,$ and knowing that really any hyperbolic and parabolic isometry behaves the same way since it’ll be conjugate to one of those.

So, $H$ gives a group of translations $\{x \mapsto x + a_i\}.$ It turns out we can pick an $i_0$ so that $0 < |a_{i_0}| \leq |a_i|$ for all non-zero $a_i.$ If we weren’t able to do that, it would contradict the discreteness of $\Gamma,$ i.e. that it’s supposed to act freely and properly discontinuously. Now we want to prove that $x \mapsto x + a_{i_0}$ generates the group. Fix any $i,$ assuming for now that $a_i > 0.$ Choose the smallest $n$ so that $a_i \leq na_{i_0} < 2a_i.$ If $na_{i_0}$ is strictly greater than $a_i,$ then their difference satisfies $na_{i_0} - a_i < a_{i_0},$ but $x \mapsto x + na_{i_0} - a_i$ is in the group, which contradicts $a_{i_0}$ being the smallest.

cyclic_translations — The red point is at $a_i.$

So $na_{i_0} = a_i,$ and we conclude that the group is cyclic. $\Box$

And that’s it! Since $H$ restricted to the geodesic or the horocycle is cyclic, $H$ must actually be cyclic. Though you might have to think about that for a little bit.

So why does that mean the center of $\Gamma$ is trivial if $M$ has finite volume? Well, if the center is nontrivial then it has some non-identity element $g,$ and everything in $\Gamma$ commutes with $g,$ so $\Gamma$ must be cyclic. But that means $\Gamma$ is “too small” for $M$ to be finite volume. For example, the fundamental regions might look like the following.

infinite_vol_fund_regions — The left is a fundamental region of the group generated by $z \mapsto az.$ The right is a fundamental region of the group generated by $z \mapsto z + b.$

The left is a fundamental region of the group generated by $z \mapsto az.$ The right is a fundamental region of the group generated by $z \mapsto z + b.$

Each of those has infinite volume. And for the general case of any hyperbolic/parabolic isometry, you again just say it’s the same picture “up to conjugation”.

This kind of restriction on commuting elements can be thought of as support for the saying that the fundamental group of a hyperbolic surface has to be complicated. And this gives a way of seeing that the torus can’t be given a hyperbolic structure, without using the Gauss-Bonnet theorem, since the torus has fundamental group $\mathbb{Z} \times \mathbb{Z}.$ Since this is abelian (and not cyclic, regardless of what we think the volume should be) by what we’ve seen here it’s too simple for hyperbolic aspirations.