1
$\begingroup$

A $\textit{hypermap}$ of type $(g,n)$ is a graph embedded in an oriented surface of genus $g$ such that

1) the complement of the graph is the disjoint union of $n$ topological disks labelled from 1 up to $n$; and

2) the vertices are coloured black and white such that every edge is adjacent to a black vertex and a white vertex.

The degree of a face is the number of edges adjacent to the face, counted with multiplicity 2 if both sides of an edge belong to the same face.

Two hypermaps are considered equivalent if there exists an oriented homeomorphism of the underlying surface carrying one graph to the other, preserving cyclic orientations of edges meeting at a vertex, colours of vertices, and face labels.

I include the defintion of hypermaps just for the post to be self contained. I want to count the hypermaps with following conditions. }

The BMS number $B_{g,n}^{(m)}(\mu_1, \ldots, \mu_n)$ is equal to the weighted count of hypermaps of type $(g,n)$ such that

1) white vertices have degree $m$;

2) edges have labels from the set $\{1, 2, \ldots, m\}$ such that the $m$ edges adjacent to a white vertex are cyclically labelled $1, 2, \ldots, m$ anticlockwise;

3) black vertices are only adjacent to edges of one type; and

4) the face labelled $i$ has degree $2m\mu_i$ for $i = 1, 2, \ldots, n$.

The weight of a hypermap $\Gamma$ is $\frac{1}{|\text{Aut } \Gamma|}$.

Now I want to count BMS numbers in genus 1 and $m=2$. I start with $B_{1,1}^2(2)$. I can draw a hpermap without label having 2 white and one black vertex. That is the usual torus. But I it says $B_{1,1}^2(2)=0$? There exist no such labelling statisfying the condition?

Also how can I draw $B_{1,2}^{2}(3,1)$ and I want to count the automorphism of each graph. Algebraic calculation tells me that $B_{1,2}^{2}(3,1)=1$ and $B_{1,1}^{3}(3)=1/3$.

Some reference or picture would really help.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.