Hurwitz automorphism theorem

I recently discovered that the number of conformal automorphisms of a compact genus $g\geq 2$ Riemann surface is finite and it has an upper bound of $84(g-1)$. This is known as the Hurwitz’s Automorphism Theorem. There are a number of ways to prove this theorem, but we shall do so using the language of orbifolds. The standard approach is to see that the automorphism group $\text{Aut}(X)$ is finite by showing that it is discrete. Then, we can apply the Riemann-Hurwitz formula to count and obtain the upper bound.

Hurwitz’s Automorphism Theorem: If $X$ is a compact Riemann surface of genus $g\geq 2$, then the group $\text{Aut}(X)$ of conformal automorphisms of $X$ has order $\vert \text{Aut}(X) \vert \leq 84(g-1)$.

Finiteness

By the uniformisation theorem (see this post), $X$ is conformally isomorphic to the quotient $\mathbb{H} / \Gamma$, where $\mathbb{H}$ is the upper half plane and $\Gamma$ is a free and discrete subgroup of $\text{Aut}(\mathbb{H}) = PSL(2,\mathbb{R})$. Basic covering space theory tells us that the automorphism group of $X$ is isomorphic to the quotient group $N(\Gamma)/\Gamma$, where $N(\Gamma)$ is the normalizer of $\Gamma$ in $\text{Aut}(\mathbb{H})$.

Suppose $\text{Aut}(X)$ is not discrete, then the normalizer $N(\Gamma)$ is not a Fuchsian group. (Fuchsian is another way of saying discrete in the context of subgroups of $PSL(2,\mathbb{R})$.) What could $\Gamma$ be so that this condition is satisfied? Recall that any non-trivial element $\gamma$ of $PSL(2,\mathbb{R})$ is either

• elliptic, i.e. conjugate to a rotation $z \mapsto e^{i\theta} z$ on the unit disk $\mathbb{D}$,
• parabolic, i.e. conjugate to a translation $z \mapsto z+a$ on $\mathbb{H}$ for some $a \in \mathbb{R}$, or
• hyperbolic, i.e. conjugate to a dilation $z \mapsto bz$ on $\mathbb{H}$ for some $b \in (0,\infty)\backslash \{1 \}$.

Since $\Gamma$ acts freely, every non-trivial element $\gamma$ can’t have fixed points and therefore either parabolic or hyperbolic. When $\Gamma$ is generated by a single element $\gamma$, then…

• if $\gamma$ is a parabolic, e.g. $\gamma(z)=z+a$, then $N(\Gamma)$ contains $\{ z \mapsto z+c \vert c \in \mathbb{R} \}$ and is non-discrete;
• if $\gamma$ is a hyperbolic, e.g. $\gamma(z) =bz$, then $N(\Gamma)$ contains $\{ z\mapsto cz \vert : c \in (0,\infty) \}$ and is also non-discrete.

It takes some computational work, but we can show that when $N(\Gamma)$ is non-discrete, $\Gamma$ must be “elementary” and therefore cyclic. (A more comprehensive study of Fuchsian groups can be found in the book of Beardon¹.) However, in the two cases above, the corresponding Riemann surface $X = \mathbb{H} / \Gamma$ is not compact, which leads to a contradiction.

A slightly different approach is the following. The normaliser is contained in the semigroup $\text{End}(\Gamma) = \{ g \in \Gamma \vert g\Gamma g^{-1} \subset \Gamma \}$. Suppose $\text{End}(\Gamma)$ is not discrete. We then have a sequence of distinct elements $g_n$ in $\text{End}(\Gamma)$ converging to some $g_{\infty} \in \text{End}(\Gamma)$. Note that $(g_n^{-1} \circ g_\infty)^{-1} \Gamma (g_n^{-1} \circ g_\infty)$ is a subset of $g_\infty^{-1} \Gamma g_\infty$. Since $g_\infty^{-1} \Gamma g_\infty$ is discrete and $g_n^{-1} \circ g_\infty \to Id$, then for sufficiently large $n$, $g_n^{-1} \circ g_\infty$ would commute with every element in $\Gamma$, which implies that every non-trivial element in $\Gamma$ must have the same fixed point in $\bar{\mathbb{D}}$ and must be of the same type. This is only possible when $\Gamma$ is abelian, so $\text{End}(\Gamma)$ is discrete.

We now know that $\text{Aut}(X)$ is discrete. Once we know that $\text{Aut}(X)$ is discrete, it is easy to upgrade it to finiteness. Any discrete subset of a compact space is finite. Since $X$ is compact, the orbit of $\text{Aut}(X)$ acting on any point $x \in X$ is finite.

There are other ways to prove finiteness. Another way that works in higher dimensional compact hyperbolic Riemannian manifolds is what is often called Bochner-Yano theorem² in differential geometry.

Upper Bound

We will proceed in a straightforward manner using the language of orbifolds. (See this, or this, or the talk I gave a while ago.) By finiteness, the quotient $Y = X / \text{Aut}(X)$ is a compact 2-D hyperbolic Riemannian orbifold. The quotient map $q: X \mapsto Y$ is an orbifold covering map of degree $\vert \text{Aut}(X) \vert$ satisfying the Riemann-Hurwitz formula

The upper bound $84(g-1)$ can be obtained once we show that the Euler characteristic of $Y$ is bounded above by $-\frac{1}{42}$.

The orbifold $Y$ does not admit any corner reflectors because $Y$ is compact. Suppose $Y$ has genus $h$ and $N$ cone points of order $m_1, m_2, \ldots m_N \geq 2$. The Euler characteristic of $Y$ is equal to

Our goal now is to find the values of $h$, $N$ and $m_j$’s such that $\chi(Y)$ is the maximum number possible. If $h\geq 2$, then $\chi(Y) \leq -2$. If $h=1$, there must be at least one cone point of order $m_1 \geq 2$, which then yields $\chi(Y) \leq 2-2 - (1-\frac{1}{2}) \leq - \frac{1}{2}$. Now, suppose $Y$ has genus $h=0$.

• If $N \geq 5$, then $\chi(Y) = 2- \sum_{j=1}^5 (1-\frac{1}{m_j}) \leq 2 - \sum_{j=1}^5 \frac{1}{2} \leq - \frac{1}{2}$.
• If $N =4$, then since the signature $(2,2,2,2;)$ yields a parabolic orbifold instead, the highest possible $\chi(Y)$ for a hyperbolic $Y$ would be when $Y$ has signature $(2,2,2,3;)$ with Euler characteristic $\chi(Y) = 2- \frac{1}{2} - \frac{1}{2} - \frac{1}{2} - \frac{2}{3} = -\frac{1}{6}$.
• If $N=3$, then we can again perform a simple case-by-case analysis on the orders $m_1, m_2, m_3$ and we should easily find that the highest possible $\chi(Y)$ would be when $Y$ has signature $(2,3,7;)$ with Euler characteristic $\chi(Y) = 2- \frac{1}{2} - \frac{2}{3}-\frac{6}{7} = -\frac{1}{42}$.
• If $N \leq 2$, then by the classification of 2-D orbifolds, $Y$ is either bad or elliptic.

Therefore, $\chi(Y) \leq -\frac{1}{42}$. When this upper bound is attained, the orbifold $Y$ is a topological sphere and has signature $(2,3,7)$. In fact, we can endow $Y$ with the unique complex structure $\mathbb{P}^1$ such that the projection $q: X\to \mathbb{P}^1$ is a holomorphic branched covering map with three critical values of multiplicity $2$, $3$, and $7$ respectively.

We say that $X$ is a Hurwitz surface if $\text{Aut}(X)$ is a group of order $84(g-1)$. I have to emphasise that we have not actually shown that the upper bound is sharp, so a priori we do not know if Hurwitz surfaces actually exist. In fact, it does not exist for genus $2$, and it has been found that a Hurwitz surface always has genus $3$ or higher.

References

1: A. Beardon. The Geometry of Discrete Groups. Springer-Verlag, New York, 1995.
2: S. Bochner, K. Yano. Curvature and Betti numbers. Annals of Mathematics Studies, 32, Princeton University Press, 1953.
3: A. Hurwitz. Über algebraische Gebilde mit Eindeutigen Transformationen in sich. Mathematische Annalen, 41 (3): 403–442, 1893.