Word length in the surface groups

Question

I want to know if there are some results about the title of this question.

Let $G$ be an orientable closed surface group with genus $n$ greater than 1. We know it has a canonical presentation. $$G=\left< a_1,b_1\dots,a_n,b_n\mid [a_1,b_1]\cdots[a_n,b_n]\right>.$$ Then, we can get a definition about the word length of $\forall ~x\in G$ as $$|x|_S=\min\{m\mid x=x_1\cdots x_m,x_i\in S\cup S^{-1}\},$$ where $S=\{a_1,b_1,\dots,a_n,b_n\}$. For these surface groups, Dehn gave a famous theorem about the identity element (refer to Stillwell's answer in Dehn's algorithm for word problem for surface groups for more details). On this topic, I have two questions.

1. Is there a sufficient and necessary description about the shortest words representing $\forall ~x\in G$?
   From Dehn's theorem, we know the word must be freely reduced and can't contain a segment with length greater than $\frac{1}{2}\cdot 4n=2n$, which is a subword of a cyclic permutation of $[a_1,b_1]\cdots[a_n,b_n]$. However, this two conditions are necessary but not sufficient. For example, we have the following counterexample.
   Let $n=2$. Words $(b_1a_1^{-1}b_1^{-1}a_2a_1b_1a_1^{-1}b_1^{-1})^{-1}$ and $a_2b_2a_2^{-1}a_2^{-1}b_2^{-1}a_1$ satisfying the two conditions have different lengths but represent the same element. $$(b_1a_1^{-1}b_1^{-1}a_2a_1b_1a_1^{-1}b_1^{-1})^{-1}=(a_1b_1a_1^{-1}b_1^{-1})^{-1}(b_1a_1^{-1}b_1^{-1}a_2)^{-1}=(a_2b_2a_2^{-1}b_2^{-1})(b_2a_2^{-1}b_2^{-1}a_1)=a_2b_2a_2^{-1}a_2^{-1}b_2^{-1}a_1.$$
2. Are there any results about the word lengths of $|x|_S$ and $|x^2|_S$ for $\forall 1\ne x\in G$? Or more generally, about the word length of $|x|_S$ and $|x^k|_S$ for $k>1$?

Answer 1

Considering the usual tiling of the $\mathbb{H}^2$ associated to a given surface group, classical arguments of small cancellation theory can be used to prove that:

Lemma 1. Let $\alpha,\beta$ be two combinatorial geodesics with the same endpoints. There exists a sequence of combinatorial geodesics $\gamma_0=\alpha, \ldots, \gamma_n=\beta$ such that each $\gamma_{i+1}$ is obtained from $\gamma_i$ by circling a polygon (i.e. replacing a subset $\zeta$ with a path $\xi$ where $\zeta \xi^{-1}$ is the boundary of a polygon).

Lemma 2. Let $\alpha$ be an arbitrary combinatorial path. There exists a sequence of combinatorial path $\gamma_0=\alpha, \ldots, \gamma_n$ such that $\gamma_n$ is a geodesic and each $\gamma_{i+1}$ is obtained from $\gamma_i$ by circling a polygon or by removing a backtrack.

At the level of words, you get an elementary operation: replacing half of a relation with the other half; and a word will have minimal length iff you cannot apply a free reduction after a sequence of elementary operations.

Regarding your second question, a naive guess would be the following. Say a word is cyclically geodesic if all its cyclic permutations are geodesic. Then it is plausible that every element $g$ is conjugate to a cyclically geodesic word $w$ and that $w^k$ is geodesic for every $k \geq 1$. Then, $|g|= k \cdot |w| + \mathrm{cst}$ for every $k \geq 1$. But this has to be checked. Geometrically, this essentially means that a non-trivial element admits an axis. As shown by Kapovich for small cancellation groups, if I remember correctly, this is true up to taking a sufficiently large (but uniform) power. Here, for surface groups, taking a power may not be necessary.