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Cayley Graphs on Billiard Surfaces

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 Added by Jason Schmurr
 Publication date 2019
  fields
and research's language is English




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In this article we discuss a connection between two famous constructions in mathematics: a Cayley graph of a group and a (rational) billiard surface. For each rational billiard surface, there is a natural way to draw a Cayley graph of a dihedral group on that surface. Both of these objects have the concept of genus attached to them. For the Cayley graph, the genus is defined to be the lowest genus amongst all surfaces that the graph can be drawn on without edge crossings. We prove that the genus of the Cayley graph associated to a billiard surface arising from a triangular billiard table is always zero or one. One reason this is interesting is that there exist triangular billiard surfaces of arbitrarily high genus , so the genus of the associated graph is usually much lower than the genus of the billiard surface.



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265 - Jason Schmurr 2015
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Let $G$ be a group and $Ssubseteq G$ its subset such that $S=S^{-1}$, where $S^{-1}={s^{-1}mid sin S}$. Then {it the Cayley graph ${rm Cay}(G,S)$} is an undirected graph $Gamma$ with the vertex set $V(Gamma)=G$ and the edge set $E(Gamma)={(g,gs)mid gin G, sin S}$. A graph $Gamma$ is said to be {it integral} if every eigenvalue of the adjacency matrix of $Gamma$ is integer. In the paper, we prove the following theorem: {it if a subset $S=S^{-1}$ of $G$ is normal and $sin SRightarrow s^kin S$ for every $kin mathbb{Z}$ such that $(k,|s|)=1$, then ${rm Cay}(G,S)$ is integral.} In particular, {it if $Ssubseteq G$ is a normal set of involutions, then ${rm Cay}(G,S)$ is integral.} We also use the theorem to prove that {it if $G=A_n$ and $S={(12i)^{pm1}mid i=3,dots,n}$, then ${rm Cay}(G,S)$ is integral.} Thus, we give positive solutions for both problems 19.50(a) and 19.50(b) in Kourovka Notebook.
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