Consider a Coxeter group presentation $< s\_1, \ldots s\_n \mid (s\_i s\_j)^{m\_{ij}}>$ with $m\_{ii}=1$ for all $i$. I can prove (using <http://arxiv.org/abs/1011.4255>) that if for each $i$ there are at most two $j\neq i$ with $m\_{ij}< \infty$, then the corresponding Cayley graph is planar. Is this fact known?


In fact I can do a bit more, namely prove that by defining two 2-cells of the cayley complex to be parallel if they have the same boundary, and deleting all but one 2-cells from each parallel class, we obtain a planar 2-complex. This might be interesting as it gives a nice action of the group on the 2-sphere.



Planar Cayley graphs/complexes for coxeter groups
Let $\Gamma$ be the Coxeter group in question. Order the $s\_i$ so that if $m\_{ij}\neq\infty$ then $|i-j|\leq 1$. Now let $P$ be a polygon with edges $e\_i$ and angles $\pi/m\_{i,i+1}$ between consecutive edges. Note that some vertices are 'ideal', meaning that they have interior angle zero.


The polygon $P$ can be realized geometrically in $X$, where $X$ is one of $S^2$, $\mathbb{R}^2$ or $\mathbb{H}^2$, and $\Gamma$ is the corresponding reflection group. There is a corresponding tiling of $X$ by copies of $P$, and the Cayley graph is just the dual graph to this tiling. So one sees that the Cayley graph is indeed planar, and furthermore the action of $\Gamma$ preserves a natural metric.


This is all quite standard. For further details, I suggest looking at Peter Scott's article 'The geometries of 3-manifolds'. So, yes, it is indeed known that these Cayley graphs are planar.


