How to draw automorphisms of a Cayley graph

113 Views Asked by At

In the figure below, I have drawn the undirected Cayley graph of the Cayley graph of $\mathbb{Z}_5 \times \mathbb{Z}_5$ with respect to the generating set $S=\{(0,1),(1,0)\}$.

Can someone please guide me to draw automorphisms of this graph? At least some automorphisms ?

Thanks a lot in advance.

Figure 1

1

There are 1 best solutions below

4
On

An automorphism $f$ of a graph $G$ is a permutation of the set $V$ of its vertices. So if we have a drawing $d$ of a graph $G$ in a space $X$, a natural way to draw $f$ is to draw an image of its graph, that is a set $\Gamma=\{(d(v), d(f(v)):v\in V\}$. A problem is that if the space $X$ is a plane then the set $\Gamma$ should be drawn (or, at least, visualized) in a four-dimensional space. Charles H. Hinton wrote that it can done via special skills, which can be developed by freeing our imagination of objects of “elements of the self”, related to our vision and location, in order to imagine the things as they are, for instance, to see not only their surfaces but also inner points. This aim can be achieved by special imagination exercises.

But even four dimensional drawing of an automorphism of the given graph would be non-perfect, because, using Euler’s formula we can easily show that the considered graph is non-planar, so we need three dimensions for its crossing-free drawings in a Euclidean space, thus the corresponding drawings of its automorphisms should be done in a six-dimensional space.

Also we can very imperfectly draw an automorphism $f$ of the given graph in a plane, using the numbers of the vertices of the graph as their coordinates, that is a drawing of $f$ is a set $\{(x,f(x)): x=1,\dots, 25\}$.

But before drawing $f$, we have to find it. I remark that the group of automorphisms of the given graphs it rather big and contains the following generators: a swap, which maps $(i,j)$ to $(j,i)$ for all $(i,j)$, a shift, which maps $(i,j)$ to $(i+1\pmod 5,j)$ for all $(i,j)$, and a reflection, which maps $(i,j)$ to $(5-i,j)$ for all $(i,j)$. I have a looking plausible sketch of a proof that this set of generators generates the whole group of automorphisms of the given graph.