Characterize words in the free group on n letters corresponding to homotopy classes of simple closed curves in the sphere with n+1 punctures.

Question

I consider X = a sphere with n+1 punctures, i.e. a plane with n points $p_i$ removed and choose a base point x (different from the $p_i$).

A classical way to setup an isomorphism between the homotopy group of X wrt x, a.k.a. $\pi_1(X,x)$, and the free group $F_n$ on n letters, a.k.a. $\mathbb{Z}\ast\cdots\ast\mathbb{Z}$, is to choose n disjoint simple curves $\gamma_i$ from x to $p_i$ and to each curve $\gamma_i$ associate a closed loop that follows $\gamma_i$ starting from $x$ up to a point close to $p_i$, makes one circular turn around $p_i$ in the positive orientation, and then follows $\gamma_i$ to the basepoint.

For a technical reason it is a good idea to require that the circular ordering of the set of curves $\gamma_i$ near $x$ is the same as the ordering of the index $i$ (see the example below the question).

Question : what is the set of words in $F_n$ that are, under the isomorphism, the homotopy classes of the simple* closeds loop based on $x$? (I suppose this is known but I do not know where to look.) (*simple means injective)

For example for n=3, I can get the word "abc" but it seems that I cannot get "cba".

Note: this set will be invariant by inversion and by the action of the mapping class group of (X,x), in particular it is invariant by inner conjugacies.

Answer 1

My formulation, characterizing a subset S of $F_n$, was a bit vague and as Moishe Kohan pointed out may mean several things. I will stress two kind of answers I prefer :

• A. Give an algorithm taking a word and telling whether or not it belongs to S.
• B. Give an algorithm generating all the words in S one by one (possibly with repetition)

and maybe a third kind,

• C. Seeing S as a subset of the set of vertices of the Cayley tree of $F_n$ w.r.t. the chosen generating set, give a geometric description of S.

(though the word geometric is ambigous...) In all cases, the simpler, the better.

As indicated by Charlie Frohman in the comments, an answer of type A can be found in the article of Birman and Series "An algorithm for simple curves on surfaces" (the authors explain their algorithm is not the first to be published). Let me describe it below in the particular case I am intereseted in. Note that in [BS] the order of multiplication in the group is backwards to reflect composition of functions but I prefer to multiply forward to reflect concatenation of paths.

Let $a, \ldots, z$ denote the generators as I described them in the question: their order matters. (There is an obvious abuse of notations we do not necessarily consider the case of 26 generators.) Let $\overline a, \ldots, \overline z$ denote their inverses. Order them as follows: $(a, \overline a, \ldots, a, \overline z)$. Consider a word $w$ that is cyclically reduced (reduced means no two consecutive letters are inverses of each other, and cyclically means that the first letter is considered as consecutive to the last). Then $w\in S$ iff the following product reduces to the empty word. Let $k$ be the length of $w$. The product is the product from left to right of the following $2k$ words, taken in a precise order we specify below. The words are obtained by iterating cycling on $w$ and on its inverse. By cycling I mean removing the first letter and appending it at the end. The order is given by lexicographically ordering the words.

I think an algorithm of type B is given by taking the $n!$ words whose letters are strictly increasing for the order $a,\ldots,z$ and letting the mapping class group act on them. This action has well described finite sets of generators.

In the particular case n=2 the description is simpler as pointed out by Moishe Kohan. Paraphrasing: "all conjugates of the words $1,a^{±1},b^{±1},(ab)^{±1}$". It is not obvious that the above two algorithms are equivalent to this description.