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I'm trying to find the Bass-Serre tree corresponding to the Baumslag-Solitar group $BS(1,2)$. I could find only one reference from the Internet, which I don't understand, and although I have the corresponding graph of groups, I cannot reconstruct this tree.

Edit: from the comments below, the corresponding tree would be an infinite binary tree but I'm missing the action of the group on that tree.

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For $BS(m, n)$, every vertex has $n$ edges going in and $m$ coming out. The stable letter of your HNN-extension cycles these edges. – user1729 Sep 10 '12 at 10:05
Could you make that as an answer ? Btw, I have trouble in picturing the action of either generator... if the stable letter cycles through the edges, I end up with a cyclic group, not the infinite cyclic group. – MarcSimon Sep 11 '12 at 6:45
I didn't make it an answer because I'm not 100% sure! Also, what I said was incorrect - I meant the non-stable letter cycles through the edges (the stable letter is the edge stabiliser, while the other letter is the vertex stabiliser.) – user1729 Sep 11 '12 at 8:56
From what I understood, the tree is indeed an infinite binary tree. If the relation in $BS(1,2)$ is $ts=st^2$, I thought at first $t$ would shift the vertices which belong to the same level of the tree. But this doesn't seem to work, and I cannot find the action of $s$ – MarcSimon Sep 11 '12 at 9:33
Well, what I am certain about is that the tree looks like what I described (so, yes, an infinite binary tree for your case). The action...I am less sure about. Perhaps the thing to do is to work out what the action is so that the fundamental domain is what we want it to be (a single vertex and a loop edge). – user1729 Sep 11 '12 at 10:12
up vote 1 down vote accepted

I believe it works like this:

Your tree $T$ is an infinite binary tree, where each vertex has one edge going in, and two edges going out.

Let your group be given by $G=\langle a,b\mid bab^{-1}=a^2\rangle$, and let it act on the right for this tree $T$, and pick a "starting vertex".

Find a copy of the real line in this tree, with the "starting vertex" at coordinate $0$. So at each vertex, $b$ sends some vertex along one of its children, left or right. $a$ acts by swapping these choices.

For example, if $b$ pushes the starting vertex along its left child, then $ab$ pushes it along its right child. $b^{-1}ab$ will send a vertex up to its parent, and down the other branch.

A nice set of pictures for this is in Meier's Groups, Graphs, and Trees. He doesn't discuss this action directly, but he shows how the Cayley graph of $G$ projects to the trivalent tree. It is pages 117--118 in my copy.

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I was thinking about something exactly like you said, but then it seemed to me that the action of $b^{-1}$ is not well defined for every vertex. Instead a figured something in which the real line on the extreme left of the tree gets translated as you said, and whatever lies on the right is folded back on the left part (It is actually quite difficult to explain without a picture...). In this view, I agree with you that conjugation translates vertices which are on the same level... – MarcSimon Sep 17 '12 at 15:47
Yes, I think taking "left all the time" works fine. I think that the action I described is correct though, because $a^2$ should act trivially. – user641 Sep 17 '12 at 20:17

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