Is my graph a tree? Let M be a smooth  connected manifold. G is a group act on M cocompactly
and suppose there is a harmonic function $h$ on M with minimal energy.$h:\rightarrow [0,1]$ such that h is nonconstant and there are points in M such that h takes value 0, 1. 
Let $\mathcal{F}=\{g*h|g\in G\}\cup \{1-g*h|g\in G\}$. Any two elements $f,h$ in $\mathcal{F}$ satisfy exactly one of the following relation. 
$f<h$,$ f>h $, $f=1-g*h$,$f<1-g*h$,$f>1-g*h$. 
As $h$ is a smooth function we can choose a regular value say $t>1/2$ which is very close to $1/2$. Let define the wall that is $W_f = \{x\in M |f(x)=t\}$. Notice that any wall divide the manifold in two parts that is 
$W_f^+=\{x\in M|f(x)>t$ and $W_f^-=\{x\in M|f(x)< t\}$. 
Now lets begin with the construction of the graph 
Let $M^0=M\setminus  \cup_{f\in \mathcal{F}}  W_f$. The maximal indecomposable set will form the vertex and the walls will form edge. It is easy to see that each wall $W_f$ is adjacent to exactly two indecomposable set contain in $W_f^+$ and $W_f^{-}$ respectively. So we can construct a graph. Now the question is that if the graph is a tree ?
Now the way to show that is the following if a vertex V is adjacent to two 
edge $W_f$ and $W_g$ then $f<g$ or $f>g$ can only hold so if there is a cycle say $V_1\rightarrow W_{f_1}\rightarrow V_2\rightarrow W_{f_3}\rightarrow ...V_k\rightarrow W_{f_k}\rightarrow V_1$
then we will get a contradiction that $f_1<f_1$. 
I have miserably stuck to show the above for all cases that $V$ is adjacent to $W_f$ and $W_g$ then $f>g$ or $f<g$ that is to rule out 
other inequalities $f+g<1$ or $f+g>1$ etc... 
If any one know this problem or any other idea to show this will be highly appreciated. 
 A: I will try to extend on the idea I mention to solve the problem. So if 
$V$ is adjacent to edge $W_f$ and $W_g$, then we will give case by case 
relation between $f$ and $g$. As walls does not cross each other so $W_g$ is contained in $W_f^-$ of $W_f^+$. So a V is adjacent to $W_f$ and $W_g$ there is four cases assuming $f+g\neq 1$
Case 1: 
        $V\subset W_f^-\cap W_g^+ $ this will imply $f<g$ 
Case 2: 
        $V\subset W_f^+\cap W_g^-$ this will imply $g>f$
Case 3:  $V\subset W_f^+\cap W_g^+$ this will imply $ f+g>1$
Case 4: $V\subset W_f^-\cap W_g^-$ this will imply $f+g<1$
Suppose there is a cycle 
$V_1\rightarrow W_{f_1}\rightarrow V_2\rightarrow W_{f_2}\rightarrow V_3\rightarrow W_{f_3}\rightarrow...V_k\rightarrow W_{f_k}\rightarrow V_1$
First notice there is a special indecomposable set $\{x\in M|1-t\leq f(x)\leq t\}$. It is easy to see that it is indecomposable form the inequalities. 
So a vertex can be special or not special hence we have two cases to contradict that there is a cycle. 
1): V is not special vertex hence case 1 and case 4 is relevant so it will
and we just go on comparing leaving one vertex in succession, we compare only non special vertex. 
2):V is special then case 2, and case 3 is relevant and we go on comparing with special vertex. 
We obtain a contradiction $f_1<f_1$ 
