Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have the following question regarding group actions on trees to which I suspect the answer to be "yes", but it could very well be that extra conditions are required (it is certainly true for free actions on trees):

Let $G$ be a finitely generated group acting by simplicial automorphisms on a minimal simplicial tree $T$ (here, minimal means that $T$ does not have a $G$-invariant proper subgraph, and it implies that $T$ is cocompact). We call $g\in G$ elliptic if it fixes a point in $T$ and hyperbolic if not.

Moreover, let $R\subset T$ be an infinite ray based at a vertex $x_0\in V(T)$. Since $T$ is cocompact and therefore has only finitely many orbits of vertices, when walking along $R$ we eventually find a vertex $y\in V(R)$ that is a $G$-translate of some 'earlier' vertex, i.e. $y=gy_0$ for some $y_0\in [x_0,y)\subset R$ and $g\in G$. However, we do not know whether $g$ is elliptic or hyperbolic.

Question: Does there exist $y\in V(R)$ such that $y=gy_0$ for some $y_0\in [x_0,y)\subset R$ and some hyperbolic $g\in G$?

I ended up considering two cases and I have sorted out one of the two. The other one, however, seems to break up into many other cases again that I cannot yet handle. I decided not to post my partial result in order not to lead people in the (possibly) wrong direction from the very beginning.

share|improve this question
    
Sorry my mobile cut off my comment. I meant to say if R is a half ray of the axis and $y=g^2x_0$, then $y=gy_0$ with $y_0=gx_0$, so you must exclude that case. –  Benjamin Steinberg Aug 7 '13 at 14:38
    
Can you use the fact that the action is cofinite on the oriented edges of T? –  staylor Aug 7 '13 at 14:45
add comment

1 Answer 1

up vote 5 down vote accepted

The answer is yes. For the proof, the ray $R$ contains three distinct points in order, $y_0,y_1,y_2$, such that $y_1 = g_1 y_0$ and $y_2 = g_2 y_1$ for some elements $g_1,g_2 \in G$. If either of $g_1,g_2$ is hyperbolic, you are done. Suppose that both of $g_1,g_2$ are elliptic. The midpoint $p_1$ of $[y_0,y_1]$ must be fixed by $g_1$, and so $g_1$ takes the segment $[y_0,p_1]$ to the segment $[y_2,p_1]$. Similarly, $g_2$ fixes the midpoint $p_2$ of $[y_1,y_2]$ and takes $[y_1,p_2]$ to $[y_2,p_2]$. Consider $g = g_2 g_1$ which takes $y_0$ to $y_2$, and note by construction that $g[y_0,y_2] \cap [y_0,y_2] = \lbrace y_2 \rbrace$, so $g$ is hyperbolic and $[y_0,y_2]$ is a fundamental domain for the axis of $g$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.