The A\_2 affine Weyl group is the symmetry group of the triangulation of the plane by equilateral triangles. As Sean points out, it may be generated by reflections $r\_1, r\_2, r\_3$ about the edges of a single equilateral triangle. Since A\_n is a Coxeter group, every element $\alpha \in A\_2$ may be assigned a length $l(\alpha)$ with respect to the generators $r\_i$. How might one compute this length with respect to this presentation? 


Any generating set has to contain a reflection about a line in each of the three directions ($0, 2\pi/3, 4\pi/3$) of the lattice. Is that condition sufficient?



computing lengths in the A_2 affine weyl group
Ok, here's a reply in "answer" format.


I asked first whether you knew about the Coxeter group presentation because I didn't want to write a bunch of stuff that you already knew (and therefore would not answer your question). I also didn't (and still don't) understand how strong of a concept you mean by "compute". It would have been helpful if you said something to clear up these two issues in your most recent reply.


Anyway here's some basic stuff I can hammer out quickly...


The presentation of a Coxeter group (by definition) is that has a finite list of order 2 generators (called "simple reflections" in the Weyl group case) and otherwise the only relations are to specify the orders of products of pairs of generators, and some pairs can be omitted (=infinite order). Affine Weyl groups are Coxeter groups so all this applies.


An affine Weyl group $W$ is an **infinite** group and there are arbitrarily long reduced expressions (=shortest words). Even though there are only a finite number of *generating* reflections, there are an infinite number of reflections and the hyperplanes they fix (in your $A\_2$ case, they are lines) divide the space up into connected pieces usually called "alcoves" (in your $A\_2$ case, they are the triangles). A fundamental fact is that $W$ acts simply-transitively on the set of these alcoves, which means if you pick a "basepoint" alcove $\mathcal{A}$, you can associate (bijectively) to each element $ w \in W$ the alcove $ w(\mathcal{A})$.


Again, I'm not sure what you mean by "compute", but here's an attempt at an answer anyway:


A "geometric" way to compute the length $\ell(w)$ of $w \in W$ is to count the number of hyperplanes separating $\mathcal{A}$ from $w(\mathcal{A})$. In a $2$-dimensional (or perhaps even $3$-dimensional) case, e.g. $A\_2$, I guess you could quickly compute shortest words for a particular "geometrically defined" element $w$ as follows:


* print out a drawing of the triangulated space and label $\mathcal{A}$ (this is probably the longest step)
* figure out which alcove $w(\mathcal{A})$ is (you should already know this by definition of "geometrically defined")
* count hyperplanes to determine $\ell(w)$ (this is quick and easy)
* label all the alcoves with their lengths, up to and including $\ell(w)-1$ (quick and easy)
* check all 3 simple reflections until you find one sending $w(\mathcal{A})$ to an alcove whose labeled length is $< \ell(w)$ (quick and easy)
* write down that simple reflection on a piece of paper
* now repeat the previous 2 steps for the new alcove
* when you get to the base alcove, you're finished and the list of simple reflections you wrote down is a reduced word


It's a long list of steps because I like to be explicit but it really doesn't take that long.


