In this post we use the seminal paper of Jones and Singerman to define a regular map; in particular we will give a number of equivalent definitions of a map and then we will focus our attention on the regular maps at the end. It is important to note that, for the purposes of this post, we restrict our attention to orientable surfaces; in a later post we will extend the definition to non-orientable surfaces.

(Note: after you’ve read this post I heartily recommend you go and read the original Jones-Singerman paper  – it’s a cracking bit of mathematical exposition.)

## The naive idea

In what follows think of a map as being a graph $\mathcal{G}=(\mathcal{V}, \mathcal{E})$ drawn on some surface $S$. For instance a triangulation of the plane will do, or a couple of loops on a torus, with vertices at the intersections. The key point is that edges are not allowed to cross; this implies, for instance, that when the surface $S$ is the plane, a map is the same thing as a planar graph.

## Topological maps

Let’s translate the naive idea into the topological setting. Let $\mathcal{E}$ be a collection of topological spaces each of which is homeomorphic to $[0,1]$ or $S^1$ – these are the edges. Let $\mathcal{V}$ be a subset of $\mathcal{G} = \cup_{e\in\mathcal{E}} e$ – these are the vertices. For $\mathcal{G}$ to be map we need some conditions on $\mathcal{E}$ and $\mathcal{V}$ as follows. For a given $e\in\mathcal{E}$ define $\Delta e = e\cap \mathcal{V}$, and $e^\# = e\setminus\Delta e$. Then we require that

• (AG1) if $e$ is homeomorphic to $S^1$ then $|\Delta e| = 1$ (and $e$ is a loop);
• (AG1′) if $e$ is homeomorphic to $[0,1]$ then $\Delta e$ contains either one or both of the end-points of $e$ (and $e$ is a free edge or a segment respectively);
• (AG2) for all distinct $e_1, e_2\in\mathcal{E}, e_1^{\#} \cap e_2^{\#}=\emptyset$;
• (AG3) for any $v\in \mathcal{V}$, at most finitely many $e\in\mathcal{E}$ satisfy $v\in\Delta e$.

A pair $(\mathcal{G}, \mathcal{V})$ satisfying these axioms is known as an allowed graph. We need some more axioms before we can declare that the pair is a topological map.

Before we give the extra axioms, a little note: when we come to consider regular maps, the notions of free edge and loop become entirely uninteresting (as soon as we have a free edge or a loop, any connected regular map must have at most one vertex) however for the categorical equivalences that we wish to consider for general maps, free edges are a required concept.

Now for the extra axioms. Suppose that $(\mathcal{G}, \mathcal{V})$ is an allowed graph, that $\mathcal{S}$ is a connected, oriented surface without boundary, and there is a homeomorphism of $\mathcal{G}$ with a subspace of $\mathcal{S}$. We identify $\mathcal{G}$ with its image in $\mathcal{S}$ and make a couple of definitions:

• define $\mathcal{F}$ to be the set of connected components of $\mathcal{G}\setminus\mathcal{S}$ (these are the faces of the map)
• we define the valency of a point $p\in \mathcal{V}$ to equal
$2|\{e\in\mathcal{E} | e \textrm{ is a loop and }p\in\Delta e\}|+|\{e\in\mathcal{E} | e \textrm{ is a segment and }p\in\Delta e\}|;$
define the valency of a point $p\in\mathcal{G}\setminus\mathcal{V}$ to equal $1$ whenever $p$ is the end-point of a free edge $e$, and to equal $2$ otherwise.

We can now proceed to define three of our four extra axioms.

• (TM1) whenever $p\in\mathcal{G}$ has valency $k$, there is a neighbourhood $N_p$ of $p$ in $\mathcal{S}$ and a homeomorphism $\phi_p: N_p\to D = \{z\in\mathbb{C} \, \mid \, |Z|<1\}$ such that $\phi_p(p)=0$ and $\phi_p(N_p\cap \mathcal{G})=\{z\in\mathbb{C} \, \mid \, z^k\in[0,1)\subseteq \mathbb{R}\}$;
• (TM2) $\mathcal{G}$ is connected (this isn’t necessary, but it makes life a lot easier and there is effectively no loss of generality);
• (TM3) each face $f$ is homeomorphic to the open disc;

Two more definitions:

• We need the concept of a dart (sometimes called a half-edge). I’m only going to define these for allowed graphs satisfying TM1 to TM3 (as opposed to for allowed graphs), so I can give a more naive definition than that of Jones and Singerman. For me a dart is an edge-vertex incident pair. Specifically the set of darts is:  $\Omega=\{(e,v)\in (\mathcal{E},\mathcal{V}) \, \mid \, v\in\Delta e\}$. Think of a dart $(e,v)$ as being an arrow running along the edge $e$ with its head at $v$.
• For $v\in\mathcal{V}$ let $N_v$ be the neighbourhood mentioned in TM1. For a dart $\alpha=(e,v)\in\Omega$ consider a circular arc in $N_v$ which starts on $e$, follows orientation, and ends on the next edge incident with $v$. This arc lies in a unique face of $\mathcal{G}$; this face is called $face(\alpha)$. Now for $f\in\mathcal{F}$ define the valency of $f$ to be $val(f)=\{\alpha\in\Omega \, \mid \, f=face(\alpha)\}.$

Finally, our last axiom.

• (TM4) For all $f\in\mathcal{F}, 1\leq val(f)<\infty$.

A triple $(\mathcal{G}, \mathcal{V}, \mathcal{S})$ satisfying AG1 to AG3 and TM1 to TM4 is called a topological map. Define the type of the map to be $(m,n)$ where $m$ is the l.c.m of $val(v) ( v\in\mathcal{V})$ and $n$ is the l.c.m of $val(f) ( v\in\mathcal{F})$; we allow $m$ and/or $n$ to be infinite when the l.c.m. does not exist. The map is said to have finite type if $m$ and $n$ are finite; the map is said to be finite if $\Omega$ is finite (J&S state that a map is finite exactly when $\mathcal{S}$ is compact; I’d like a proof of that).

Some examples: regular tesselations of the plane $\mathcal{S}=\mathbb{R}^2$ by triangles, squares, and hexagons give infinite maps of type $(6,3), (4,4)$ and $(3,6)$ respectively. Platonic solids give finite maps on the sphere. And so on and so on.

We define morphisms of maps $(\mathcal{G}_i, \mathcal{V}_i, \mathcal{S}_i)$  just as one would expect. Note, in particular, that all branch-points of the associated covering $\mathcal{S}_1\to\mathcal{S}_2$  have finite order. We then have a category TM of topological maps; define TM(m,n) to be the subcategory of all maps of type $(r,s)$ where $r\mid m$ and $s\mid n$.

## Algebraic maps

We can define algebraic maps much more easily; they are a quadruple $(G,\Omega, x,y)$ where $\Omega$ is a set and $x,y\in G such that

• (AG1) $x^2=1$;
• (AG2) $G=\langle x, y\rangle$ is transitive on $\Omega$;

We define the type of an algebraic map $\mathcal{A}$ to be$(m,n)$ where $m$ is the order of $y$ and $n$ is the order of $z=y^{-1}x$. Similarly to before the map is said to have finite type if $m$ and $n$ are finite; the map is said to be finite if $\Omega$ is finite.

We define morphisms in the obvious way – they are permutation group morphisms mapping the distinguished elements $x$ and $y$ to the corresponding distinguished elements. Thus we have a category AM of all algebraic maps and, just as before, a subcategory AM(m,n) to be the subcategory of all maps of type $(r,s)$ where $r\mid m$ and $s\mid n$.

Now, a trivial observation: the group $G$ has a presentation of the following form: $\langle x, y, z \, \mid \, x^2=y^m=z^m=xyz=\cdots = 1 \rangle$. Let us define the following “universal group”: $\Gamma=\Gamma(m,n)= \langle x, y, z \, \mid \, x^2=y^m=z^m=xyz= 1 \rangle$; then $G=\Gamma/ N$ for some normal subgroup $N\unlhd \Gamma$.

In fact, we can define the concept of a universal algebraic map of type $(m,n)$: it is the algebraic map $(\Gamma, \Gamma, x, y)$ – here the set $\Omega=\Gamma$ where we simply ignore the group operation. Note that we allow one or both of $m$ and $n$ to be equal to $\infty$. Now one can see immediately that any algebraic map in AM(m,n) has the following form: $(\Gamma/ M^*, \Gamma/ M, xM^*, yM^*)$. We must clarify our notation: here $M$ is any subgroup of $\Gamma$, and $M^*$ is the core of $M$; that is, it is the intersection of all conjugates of $M$.

A consequence of the preceding paragraph is that every algebraic map in AM(m,n) is prescribed by (the conjugacy class of) the subgroup $M$; we call $M$ the map subgroup; observe that $M/M^* < G=\Gamma/M^*$ and $M$ acts as a stabilizer in the action on $\Omega$.

## Connecting topological maps and algebraic maps

It turns out that the category TM(m,n) and the category AM(m,n) are equivalent. For now we will show how, given an element of TM(m,n), one can construct an element of AM(m,n).

Suppose, then, that we have a topological map $\mathcal{M}=(\mathcal{G}, \mathcal{V}, \mathcal{S})$. We will define our algebraic map in terms of a permutation group acting on $\Omega$, the set of darts.

Define the element $x$ as the element that maps a dart $(e,v)$ to $(e,v')$ where, if $e$ is a segment or a loop, then $v'\neq v$ while, if $e$ is a free edge, then $v'=v$. (We think of $x$ as the element which maps each dart to its “opposite”.) Clearly $x^2=1$.

Now the  the element $y$ is the element that maps a dart $(e,v)$ to $(e',v)$ where $e'$ is the “next” edge incident with $v$ as one follows a circular path around the vertex $v$ following orientation. Clearly $y$ has order $m$ where $m$ is the l.c.m. of $val(v) ( v\in\mathcal{V})$ (as required).

Now define $z=y^{-1}x$; with a little thought (and some diagram drawing) it is not too hard to work out that the action of $z$ is to fix the faces, while mapping a dart onto the “next” dart as one proceeds round the face following orientation. One sees immediately that the order of $z$ is  $n$, the l.c.m of $val(f) ( v\in\mathcal{F})$ (as required).

We have constructed our algebraic map, as promised; we will refer to it from here on as $Alg(\mathcal{M})$. One thing should be made clear: the group that we have described is most certainly not an automorphism group for the underlying topological map since the given dart permutations cannot, in general, be chosen to act as homeomorphisms of the underlying surface.

## Riemann maps

Consider the subcategory RM(m,n) of TM(m,n) consisting of all elements for which the underlying surface is a Riemann surface, for which the edges of the graph are all geodesics in the corresponding metric, and where the angle between edges at a given vertex is constant. It is a surprising and hugely important result that, in fact, RM(m,n) and TM(m,n) are equivalent. In other words, in considering a topological map we are allowed to assume this extra structure.

The proof can be found in J&S; I’ll give a brief outline here, but won’t go into details (maybe in a future post…) Note, first, that in the previous section we showed that TM(m,n) embeds into AM(m,n). We’ve noted (it is a triviality) that RM(m,n) embeds into TM(m,n). To prove that RM(m,n) and TM(m,n) are equivalent, then, it is sufficent to prove that AM(m,n) embeds into RM(m,n).

To do this we start with an element of AM(m,n); observe first that the corresponding group is isomorphic to $\Gamma(m,n)/N$ for some normal group $N$. Now observe that $\Gamma(m,n)$ preserves a tesselation of the hyperbolic plane (provide $m$ and $n$ are not too small, in which case we have a tesselation of the plane or the sphere); the quotient $\Gamma(m,n)/N$ therefore, preserves a tesselation on a quotient space, i.e. a tesselation on a particular Riemann surface. This tesselation is precisely equivalent to a graph inscribed upon the Riemann surface; thus, given an element of AM(m,n) we can construct an element of RM(m,n) as required.

## Automorphism groups

We were a little vague about morphisms earlier, so let us firm things up: an automorphism of a topological map (i.e. of an element of TM) is a homeomorphism of the surface $\mathcal{S}$ which restricts to a graph automorphism of $\mathcal{G}$. If we are assuming that our element in fact lies in RM (as we can), then we can assume that the action on the surface is by isometry with respect to the Riemannian metric. An automorphism of an algebraic map (i.e. of an element of AM) is just a permutation group automorphism which fixes the distinguished elements $x$ and $y$.

For $\mathcal{M}$ in TM, write $TAut(\mathcal{M})$ for the automorphism group of $\mathcal{M}$ and write $Aut(\mathcal{M})$ for the automorphism group of $Alg(\mathcal{M})$, the corresponding algebraic map. It is easy to see that $Aut(\mathcal{M}) \cong TAut(\mathcal{M})/K$ where $K$ is the set of all elements of $TAut(\mathcal{M})$ which fix every dart. Now the key result concerning automorphisms is the following (the proof of which takes a paragraph, so we leave it as an exercise):

Prop. 1 Let $\mathcal{M}$ be a topological map, with $Alg(\mathcal{M})=(G,\Omega, x,y)$, or let $\mathcal{M}=(G,\Omega, x,y)$ be an algebraic map. Then

1. $Aut(\mathcal{M})$ acts faithfully on $\Omega$ as the centralizer of $G$ in $Sym(\Omega)$;
2. $Aut(\mathcal{M})$ acts semi-regularly on $\Omega$, i.e. the stabilizer of every point of $\Omega$ is trivial.

In the case of algebraic maps, one can come at these things from a different angle; it turns out that for $\mathcal{M}$ in AM(m,n), we have $Aut(\mathcal{M}) \cong N_\Gamma(M)/M$ where $M$ is the map subgroup of $\Gamma$, the universal group.

## Regularity

Call an algebraic map $\mathcal{M}$ regular if $Aut(\mathcal{M})$ acts transitively on $\Omega$. Now the following are equivalent:

1. $\mathcal{M}$ is regular;
2. the mapping subgroup $M$ is normal in $\Gamma$;
3. $(G,\Omega)$ is a regular permutation group (i.e. $G$ is transitive on $\Omega$, and every stabilizer is trivial).

The first two equivalences are clear. The third follows from Prop. 1 and the group theory fact  that the centralizer of a regular subgroup of $Sym(\Omega)$ is semi-regular. (Indeed to prove Prop. 1 one uses a more general fact: namely that the centralizer of a transitive subgroup of $Sym(\Omega)$ is semi-regular.)

Now suppose that $\mathcal{M}$ is a topological map. If $Alg(\mathcal{M})$ is regular, then the discussion of the previous section implies that $Aut(\mathcal{M})$ acts regularly on the darts of $\mathcal{M}$ and, in this case, we say that $\mathcal{M}$ is regular.

All that remains is to classify the regular maps (he says with a smile)… Once one has got a handle on the category RM it becomes clear that this is a question in hyperbolic geometry. But that will have to wait for another day.