I want to outline some arguments concerning Sylow groups that are based on results in this paper of Conder, Potočnik and Širán. My arguments will look slightly different from those of CPS because I will restrict myself to orientable surfaces, and will define the group $G$ which is associated with my regular map according to the previous posts on this blog (the group in CPS is constructed differently – it is twice as large as ours).

For the purposes of this post all maps will be finite, hence (as we proved earlier) the corresponding surface is compact and we can define the Euler characteristic $\chi$. Let $(G, \mathcal{V}, \mathcal{S})$ be a regular topological map of type $(m,n)$ (where these are true orders). Recall that, when the map is regular we can think of $G$ as acting on the map via homeomorphisms of the surface (we can’t do this for non-regular maps), and this is how we will think of $G$ in what follows..

Write $V$ (resp. $E$, $F$) for the number of vertices (resp. edges, faces) in our map. I assert that the following follows immediately from the work of Jones & Singerman that we have already discussed:

$\chi=V-E+F=|G|(\frac1{m}+\frac1{n}+\frac12) = -|G|\frac{mn-2m-2n}{2mn}.$

The first equality is the formula for the Euler characteristic, the second follows by regularity, the third by rearranging. Note that the second equality is based on the fact (which will be useful in its own right) that a vertex (resp. edge, resp. face) stabilizer is a cyclic subgroup of order $m$ (resp. $2, n$).

Now write $[m,n]$ (resp. $(m,n)$) for the lcm (resp. gcd) of $m$ and $n$. Then we can rearrange the above equation to obtain that

$\frac{|G|}{[m,n]} = \frac{-2(m,n)\chi}{mn-2m-2n} = \frac{-2\chi}{[m,n]-2\frac{m+n}{(m,n)}}.$

The two equations displayed so far immediately yield some nice consequences. In what follows, if $k$ is an integer, $p$ a prime, write $|k|_p=p^s$ if $p^s$ is the highest power of $p$ that divides $k$.

Proposition 1: Let $p$ be a prime dividing the order of $G$. Either $p$ divides $\chi$ or a Sylow $p$-subgroup of $G$ is cyclic.

Proof: All of the statements that we make depend only on the equations given above. Let $P$ be a Sylow $p$-subgroup of $G$. Suppose that $p\nmid \chi$. Then either $p\mid[m,n]$ or $|G|_p=2$. But in the latter case $P$ is cyclic and we’re done. So we assume from here on that $p\mid [m,n]$.

Suppose that $|G|_p=|[m,n]|_p$. In this case $|G|_p=|m|_p$ or $|G|_p=|n|_p$; suppose, without lost of generality that $|G|_p=|m|_p$. Since we know that $G$ contains a cyclic group of order $m$ (a vertex stabilizer) we conclude that $P$ is cyclic as required.

Suppose that $|G|_p\neq |[m,n]|_p$; then $p\mid \frac{|G|}{[m,n]}$. Since $p\nmid\chi$ we conclude that $p=2$ and $\chi$ is odd. But this implies that $m$ and $n$ are odd and so $|G|_2=2$ and $P$ is cyclic as required.

QED

Now it is well-known in group theory that any group with a cyclic Sylow $2$-subgroup is solvable. Thus Prop. 1 implies that any map on an orientable surface with odd Euler characteristic corresponds to a solvable group. Great!! Except that no orientable surfaces with odd Euler characteristic exist: in the orientable case $\chi=2-2g$ where $g$ is the genus of the surface. (Incidentally I’d like a good explanation of that. I really don’t understand how $\chi$ and $g$ work in the world of topology.)

CPS have great success in using Prop. 1 to classify regular maps, with most of their attention focussed on non-orientable surfaces. But that story is for another day. For now let me do some bookwork and write down an extension of Prop. 1 that might come in handy to someone someday.

Proposition 2: Let $p$ be a prime dividing the order of $G$. Either $p^2$ divides $\chi$ or a Sylow $p$-subgroup of $G$ has a cyclic subgroup of index $p$.

Proof: Again all of the statements that we make depend only on the equations given above. Let $P$ be a Sylow $p$-subgroup of $G$. Suppose that $p^2\nmid \chi$; there are three possibilities: (a) $|G|_p=p$; or (b) $p=2$ and $|G|_2\leq 4$; or (c) $p\mid [m,n]$. The first two cases yield the result immediately, hence we assume that (c) holds.

Suppose that $|G|_p\leq p \cdot |[m,n]|_p$. In this case, WLOG, $|G|_p\leq p\cdot |m|_p$ or $|G|_p=|n|_p$. Since we know that $G$ contains a cyclic group of order $m$ (a vertex stabilizer) the result follows.

Finally suppose that $|G|_p > p \cdot |[m,n]|_p$. Then $p=2$ and $\chi=2a$ for some odd integer $a$. But now if either $m$ or $n$ are even, then $[m,n]-2\frac{m+n}{(m,n)}$ is even and we conclude that $|G|_2=2$ and the result follows. On the other hand if $m$ and $n$ are both odd, then $|G|_2=2$ and the result follows.

QED

I am most interested in applying Prop. 2 in the case $p=2$. In this situation we know explicitly what the group $G$ looks like (see for instance this paper, and the papers it cites). I wanted to write a load more stuff on this but I’m out of time for now. More anon.

In the earlier post What is a regular map? we stated that the three categories AM, TM and RM were equivalent, but we did not complete the proof. The main task that we left open was to demonstrate how, if given an element of AM, we can construct (in a natural way) an element of RM. This construction is the topic of today’s post.

Before we get going, a caveat: to properly prove the stated categorical equivalences I really need to ensure that morphisms behave properly. As it is I’m going to content myself with an inspection of the objects in each category (go the original paper of Jones and Singerman for a full treatment).

Let $(G, \Omega, x, y)$ be an element of AM(m,n) and assume that $m$ and $n$ are the true orders of the elements $y$ and $z=y^{-1}x$. Recall that $G$ is a quotient of the group

$\Gamma = \Gamma(m,n)=\langle x, y,z \, \mid \, x^2=y^m=z^n=xyz=1\rangle.$

## Some Riemannian geometry

Let us consider the group

$\Gamma = \Gamma(l,m,n)=\langle x, y,z \, \mid \, x^l=y^m=z^n=xyz=1\rangle.$

where we assume, without loss of generality, that $l\leq m \leq n$. The group $\Gamma$ is well known in Riemannian geometry: it is the $(l,m,n)$-triangle group. It has a natural action on a simply connected Riemann surface, preserving a tessellation of that surface by isometric geodesic triangles with angles of size $\frac{\pi}l, \frac{\pi}m$ and $\frac{\pi}n$. (I will not define terms like tesselation or geodesic triangle here; your intuition is likely to be good enough to understand what is going on.) J&S allow for the possibility that $l, m, n$ are not all finite, but I won’t bother with that generality here.

The classical Uniformization theorem of Poincaré states that there are precisely three simply connected Riemann surfaces and, what is more, that they can only be tessellated by triangles of a certain form. The breakdown is as follows:

1. The unit sphere $S^2$ in $\mathbb{R}^3$ (a surface of positive constant curvature with the Euclidean metric): this can be tessellated by isometric triangles with angles $\frac{\pi}l, \frac{\pi}m$ and $\frac{\pi}n$ if and only if $\frac1{l} + \frac1{m} + \frac1{n} >1$. This condition allows us to explicitly list all possible tesselations; in all cases the associated group $\Gamma$ is finite:

• $(l,m,n)=(2,2,n)$. The tesselation looks like a ring of $n$ vertices around the equator of the sphere; with vertices at each pole. The group $\Gamma\cong D_n$ the dihedral group of order $2n$.
• $(l,m,n)=(2,3,3)$. The tesselation is identical to the tetrahedron, and $\Gamma\cong A_4$.
• $(l,m,n)=(2,3,4)$. The tesselation is identical to the octahedron, and $\Gamma\cong S_4$.
• $(l,m,n)=(2,3,5)$. The tesselation is identical to the icosahedron, and $\Gamma\cong A_5$.

2. The Euclidean plane $\mathbb{R}^2$ (a surface of zero constant curvature with the Euclidean metric): this can be tessellated by isometric triangles with angles $\frac{\pi}l, \frac{\pi}m$ and $\frac{\pi}n$ if and only if $\frac1{l} + \frac1{m} + \frac1{n}= 1$. Once again we can explicitly list all possible tesselations.

• $(l,m,n)=(3,3,3)$. The tesselation looks like the standard honeycomb of equilateral triangles; the group $\Gamma\cong (\mathbb{Z}\times \mathbb{Z})\rtimes C_3$.
• $(l,m,n)=(2,4,4)$. This looks like the standard tesselation by squares, with each square split into two; here $\Gamma\cong (\mathbb{Z}\times \mathbb{Z})\rtimes C_4$
• $(l,m,n)=(2,3,6)$. this looks like the honeycomb of equilater triangles with each triangle split into two; here $\Gamma\cong S_6$.

3. The Hyperbolic plane $\mathbb{H}$ (a surface of negative constant curvature with a hyperbolic metric): this can be tessellated by isometric triangles with angles $\frac{\pi}l, \frac{\pi}m$ and $\frac{\pi}n$ if and only if $\frac1{l} + \frac1{m} + \frac1{n}= 1$. In this case there are an infinite number of possibilities – this is where things get interesting! The smallest triple $(l,m,n)$ where we lie in the hyperbolic plane is $(2,3,7)$; the associated group $\Gamma$ and its quotients are known as Hurwitz groups; they have a huge literature.

In each case the group $\Gamma$ acts by isometries on the simply connected Riemann surface (i.e. it preserves the associated metric); what is more $\Gamma$ acts conformally (angles are also preserved) and totally discontinuously (every group element moves every point of the surface “a good distance”).

## Making a Universal Riemann map

Let us first construct a Riemann map for the universal algebraic map in AM(m,n). Take the group $\Gamma = \Gamma(2,m,n)=\Gamma(m,n)$.  (Beware: in the algebraic maps context we are not allowed to assume that $m\leq n$ as the distinguished element $y$ is of order $n$. Of course $\Gamma(m,n)\cong \Gamma(n,m)$ so the discussion of the previous section still applies.)

We have just seen that $\Gamma(m,n)$ preserves a tesselation of triangles with angles $\frac{\pi}2, \frac{\pi}m$ and $\frac{\pi}n$ on $\mathcal{U}$, the corresponding simply connected Riemann surface. Here is a little piece of the tesselation (taken from J&S), with the actions of $x,y$ and $z$ inscribed:

One can imagine the rest of the surface being covered by repeated copies of  this diamond such that every tesselation-vertex is surrounded either by angles of size $\frac{\pi}{n}$ or by angles of size $\frac{\pi}{m}$.

We construct a Riemann map on the simply connected surface $\mathcal{U}$ as follows: The vertices are those tesselation-vertices which are surrounded by angles $\frac{\pi}m$ (in the figure $b$ and $bx$ are two such); the edges are simply unions of those lines in the tesselation that join them (in the figure the vertical line segment is one such).

The edges of the map we have constructed are all geodesics; they all meet at vertices at angles of $\frac{\pi}{m}$, faces are regular $n$-gons; in other words we have a a Riemann map of type $(m,n)$. This map is known as the universal Riemann map of type $(m,n)$ (clearly it corresponds to the universal algebraic map).

The dual map – of type $(n,m)$ – can be constructed by taking vertices to be the tesselation-vertices which are surrounded by angles $\frac{\pi}{n}$ and so on.

## The general Riemann map

Recall that an algebraic map of type $(m,n)$ is completely determined by an associated mapping subgroup $M$ – any subgroup of $\Gamma(m,n)$. The group $M$ preserves the tesselation described above and it acts totally discontinuously on the simply connected Riemann surface $\mathcal{U}$; we can, therefore construct the quotient Riemann surface $\mathcal{S}=\mathcal{U}/M$ on which will be preserved the quotient tesselation.

Now we construct a Riemann map on the surface $\mathcal{S}$ in precisely the same way as the previous section. Job done.

## Equivalence of categories

The fact that the categories AMTM and RM are equivalent is demonstrated by Prop. 5.3 of J&S. Start with a topological map $\mathcal{M}$; we described in an earlier post how to construct an algebraic map $\mathcal{A}$ from $\mathcal{M}$; we now know how to construct a Riemann map $\mathcal{M'}$ from $\mathcal{A}$. The key point is that the two maps $\mathcal{M}$ and $\mathcal{M'}$ are isomorphic (where we view both of them as objects in the category TM).

In the earlier post What is a regular map? we stated a side result without proving it. Let’s do that now.

Prop. A topological map $(\mathcal{G}, \mathcal{V}, \mathcal{S})$ is finite if and only if the surface $\mathcal{S}$ is compact.

Proof. Suppose that the surface $\mathcal{S}$ is compact and the map is infinite; in other words $\Omega$ is infinite. Since the valency of all the vertices is finite (AG3) we conclude that $|\mathcal{V}|$ is infinite. This implies that there is an accumulation point on $\mathcal{S}$ for the set $\mathcal{V}$. Let $v_1, v_2, \dots$ be a converging sequence of vertices; (TM1) implies that for $n$ large enough,  these vertices must all be joined to each other and, what is more, they must all have valency 2 (think of the edges between them forming a single line on which they all lie). Then, for $n$ large enough every $v_n$ lies on two darts $\alpha_n, \beta_n$ and (provided we label darts appropriately) it is clear that $face(\alpha_n) = face(\alpha_m)$ for $n$ and $m$ large enough. This is a contradiction of (TM4).

Now for the converse; we suppose that the map is finite. Let $f$ be a face of the map; (TM4) implies that there are a finite number of darts $\alpha$ for which $f=face(\alpha)$. We list these darts:$\alpha_1=(v_1, e_1), \dots, \alpha_n=(v_n, e_n)$. Now (TM3) implies that $f$ is homeomorphic to an open disc; it is clearly sufficient to show that $f\cup e_1 \cup \cdots \cup e_n$ is homeomorphic to a closed disc (since then the surface $\mathcal{S}$ is homeomorphic to a finite union of compact surfaces and so is itself compact).

Now why is $f\cup e_1 \cup \cdots \cup e_n$ is homeomorphic to a closed disc? First of all, there is the possibility that $n=1$ and the edge $e_1$ is a loop. In this case the result is clear. Suppose this is not the case – then all edges are segments or free-edges. With a little thought it should be clear that the edges $e_i$ which are segments form a closed loop; the free-edges can be thought of as spikes coming off this closed loop, “poking into” the face $f$. This union of segments clearly forms the boundary of $f\cup e_1 \cup \cdots \cup e_n$; the required homeomorphism can then be obtained by “pinching” the face round each of the free-edges, and extending it smoothly to the union of segments. Hopefully the principle is clear.

QED

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.

This blog is intended to document the progress we make in understanding the concept of a regular map. If we’re lucky, it might lead to a paper. In the short term, most of our posts will discuss results already in the literature; as time goes by, we may discuss possible avenues for new research.

The great thing about regular maps (and it’s one of the reasons why we decided to investigate them together, despite our disparate mathematical backgrounds) is that one can approach them from a number of different avenues: via (for instance) algebraic topology, combinatorial and algebraic graph theory, hyperbolic geometry, surface groups and Riemann surfaces, or geometric group theory. None of us are experts in all of these areas so we may well intersperse posts on regular maps with posts discussing useful background concepts.

Anyone should feel free to comment on what we write; we’d be very glad if regular-map-experts were to visit this blog occasionally and tell us when we’re talking nonsense. If you would like to write a post on any subject relating to regular maps, then that would be splendid. You can email us as follows: