It seems intuitive that the higher the genus of a graph the more vertices and edges it would have to have. Is it known if the minimum number of vertices must be larger for graphs of higher genus?
Asked
Active
Viewed 287 times
2 Answers
2
Yes, this is inevitable simply because every graph on $n$ vertices has genus at most that of $K_n$ (take a drawing of $K_n$ on the appropriate surface and delete edges to get the graph you're interested in). So for every genus $g$, the minimum number of vertices needed for a graph of genus $g$ is the smallest $n$ such that $K_n$ has genus $g$, and this is obviously a strictly increasing function of $g$.
In fact $g(K_n)=\lceil\frac1{12}(n-3)(n-4)\rceil$, so by finding the smallest $n$ for which this formula gives $g$, you can work out the minimum number of vertices required.
Especially Lime
- 46,692
1
There is at least an upper bound on the genus in terms of the number of vertices and edges in the graph. Let $\gamma(G)$ be the genus of a graph $G=(V,E)$ and $\beta(G)=|E|-|V|+1$ be its Betti number. Then the following holds: $$\gamma(G)\leq\left\lceil\frac{\beta(G)}{2}\right\rceil$$
For a reference on how to prove this see for example this paper
