I'm looking for readable references on calculating the growth rates of surface groups. 
There's some approach done briefly in page 159 of de la Harpe's "Topics in Geometric Group Theory", who cites:



> 
> Cannon 1980, The growth of the closed surface groups and the compact hyperbolic Coxeter groups, Cannon
> 
> 
> 


However, this seems to be unpublished work and unavailable in the web. And de la Harpe's account seems not detailed enough for me to fill the gaps or understand.


Is this calculation done anywhere else? With hopefully more details or an alternative approach



Growth rates of surface groups
Just to make the method as concrete as possible, I'll compute the growth rate for the fundamental group $G$ of a surface of genus two. The Cayley graph of $G$ is the 1-skeleton of a tiling of the hyperbolic plane by regular octagons, with eight octagons meeting at each vertex.


Given an element $g\in G$, let $|g|$ denote its word length, i.e. the number of edges in the shortest path from $1$ to $g$ in the Cayley graph. It is not hard to prove that every octagon $O$ of the tiling has exactly one ***base vertex*** of minimum word length, whose length will be denoted $|O|$. Then $O$ has exactly one vertex of length $|O|$, two vertices each of lengths $|O|+1$, $|O|+2$, and $|O|+3$, and one vertex of length $|O|+4$. Given a vertex $g\in G$ and an octagon $O$ that contains it, let
$$
t(g,O) = |g|-|O|
$$
so $t(g,O)\in \{0,1,2,3,4\}$, with $t(g,O)=0$ if and only if $g$ is the base vertex for $O$.


Now, each vertex $g\in G$ is contained in eight octagons, giving us eight corresponding numbers $t(g,O)$. This lets us classify the non-identity elements of $G$ into four types:


* *Type 1:* These are vertices which are the base vertex for six octagons and satisfy $t(g,O)=1$ for the remaining two octagons. For example, all the vertices adjacent to the identity vertex have this type.
* *Type 2:* These are vertices which are the base vertex for six octagons and satisfy $t(g,O\_1)=1$ and $t(g,O\_2)=2$ for the remaining two octagons.
* *Type 3:* These are vertices which are the base vertex for six octagons and satisfy $t(g,O\_1)=1$ and $t(g,O\_2)=3$ for the remaining two octagons.
* *Type 4:* These are vertices which satsify $t(g,O)=4$ for some octagon $O$. Then $g$ is a base vertex for only five octagons and satisfies $t(g,O\_1)=t(g,O\_2)=1$ for the remaining two octagons.


If $g\in G$, every vertex $h$ adjacent to $g$ is either a ***predecessor*** or ***succesor*** of $g$, depending on whether $|h|<|g|$ or $|h|>|g|$. (Note that $|h|\ne |g|$ since the Cayley graph is bipartite.) Note that vertices of types 1, 2, and 3 have one predecessor and seven successors, and vertices of type 4 have two predecessors and six successors. Specifically:


* Vertices of type 1 have two successors of type 2 and five successors of type 1.
* Vertices of type 2 have one successor of type 3, one successor of type 2, and five successors of type 1.
* Vertices of type 3 have one successor of type 4, one successor of type 2, and five successors of type 1.
* Vertices of type 4 have two successors of type 2, and four successors of type 1.


For each $n\geq 1$, let $\textbf{v}\_n$ be the vector $4$-vector counting the number of elements of each type of word length $n$. For example, $\textbf{v}\_1=(8,0,0,0)$ and $\textbf{v}\_2=(40,16,0,0)$. Then $\textbf{v}\_n$ satisfies the recurrence relation
$$
\textbf{v}\_{n+1} = \begin{bmatrix}5 & 5 & 5 & 4 \\ 2 & 1 & 1 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1/2 & 0\end{bmatrix} \textbf{v}\_n
$$
where the $1/2$ comes from the fact that each vertex of type 4 has two predecessors, so we must divide by two to avoid overcounting. For example,
$$
\textbf{v}\_3=(280,96,16,0),\quad \textbf{v}\_4=(1960,672,96,8),\quad \textbf{v}\_5=(13672,4704,672,48)
$$
Note that the sum of the entries of $\textbf{v}\_n$ is the number of elements $\sigma\_n$ of length $n$, so we are getting
$$
\sigma\_1=8,\quad \sigma\_2=56,\quad \sigma\_3=392,\quad \sigma\_4=2736,\quad \sigma\_5=19096,\quad\ldots
$$
which agrees with the [OEIS entry](http://oeis.org/A063812) for this sequence.


The largest eigenvalue of the $4\times 4$ matrix above is the largest root of $x^4-6x^3-6x^2-x+1$, which is approximately 6.97984. This is the growth rate of $\sigma\_n$ (the spherical growth function), and is hence also the growth rate of $\beta\_n$ (the growth function for balls).


All of this analysis continues to work as the genus increases. For genus $n$, the Cayley graph is a tiling of the hyperbolic plane by $4n$-gons, with $4n$ polygons meeting at every vertex. There are $2n$ types of vertices and one gets a $2n\times 2n$ matrix whose largest eigenvalue is the growth rate. For example, the growth rate for a surface group of genus three is the largest eigenvalue of the matrix
$$
\begin{bmatrix}9 & 9 & 9 & 9 & 9 & 8 \\ 2 & 1 & 1 & 1 & 1 & 2 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/2 & 0\end{bmatrix}
$$


**Edit:** Perhaps unsurprisingly, the growth rate for a surface group turns out to be *very close* to the growth rate of a free group with the same number of generators. The following table shows the differences.
$$
\begin{array}{r|ccccc}
\text{genus} & 2 & 3 & 4 & 5 \\
\hline
\text{growth rate} & 7-\epsilon & 11-\epsilon & 15-\epsilon & 19-\epsilon \\
\text{value of }\epsilon & 2.0\times 10^{-2} & 6.8\times 10^{-5} & 8.7\times 10^{-8} & 5.9\times 10^{-11}
\end{array}
$$


