Let be a free group of rank and , viewed as a word on letters. Then is a product of commutators if and only if it belongs to the kernel of , that is if the sum of exponents on each letter is zero. From now on, we suppose that satisfies this property.

**Question:** How to find the least integer such that can be written as a product of commutators?

In their article *Applications of topological graph theory to group theory*, Gorenstein and Turner find a simple solution closely related to the theory of closed surfaces. More precisely, we first construct a polygon whose edges are labelled and oriented by the letters of ; then we we choose a pairing of edges, encoded in a *circle graph* (topologically, a circle with a collection of interior arcs whose endpoints are pairwise distinct), and we construct a closed surface by pairwise indentifying the edges of our polygon following the circle graph (see the figure below). Of course, the surface depends on the circle graph we chose; in particular, we define the *genus* of as the least genus of a surface obtained from in the previous way.

For example, .

We then have the following surprising result:

**Theorem 1:** can be written as a product of commutators, and not fewer.

Moreover, Gorenstein and Turner give in their article a simple way to compute the genus from the circle graph . First, we number the interior arcs of , and we define the -matrix by saying that the entry is equal to if the -th and -th interior arcs intersect; otherwise, it is equal to . Notice that is symmetric and that any diagonal element is zero, so, working in , it can be written in some basis as a diagonal block matrix (see for example Kaplansky’s book, *Linear algebra and geometry*):

where .

In particular, notice that is even. Let us define the *genus* of as

.

Finally, we have

**Theorem 2:** .

**Proof of theorem 1:** We prove by induction on that can be written as a product of commutators. If , then and there is nothing to prove.

From now on, suppose that our result holds for and suppose . Let be a circle graph such that . First, we replace by a word such that each pair of linked letters in are different in . For example,

Because , there exist two intersecting interior arcs; therefore, up to a cyclic permutation (that does not change a pairing), we may suppose that can be written as

.

The pairing clearly induces a pairing whose underlying circle graph is the same; in particular, . Now, we cut and paste in the following way:

From a purely algebraic viewpoint, we set and we write . Up to a cyclic permutation, we have

.

Then, we set and we write

.

Now, we want to apply our induction hypothesis to . First, the pairing induces a pairing for just by adding some disjoint interior arcs (because the word is not reduced), but that does not change the surface (in the polygon, we just add consecutive identified edges), hence . Then, the pairing induces a pairing for ; but the figure below justifies that we have the connected sum .

Therefore,

.

Using our induction hypothesis, can be written as a product of commutators, so can be written as a product of commutators.

Conversely, we prove that if is a product of commutators then there exists a pairing satisfying . We consider the case , but the argument is completely general. So

.

We use the pairing :

The associated matrix is where , hence

.

**Corollary:** Let be a reduced word of length in the commutator subgroup. Then can be written as a product of at most commutators.

**Proof.** Let be a pairing of minimal genus, so that can be written as a product of commutators. The circle graph has interior arcs (of course, is even) and is a -matrix, hence

.

Notice that the proof of Theorem 1 is effective, that is it gives an algorithm to express a word as product with a minimal number of commutators. Let us apply this algorithm to the word .

**Step 1:** Draw all the possible pairings.

**Step 2:** Compute the ranks of the associated matrices.

We have and . So we are interesting in the fourth pairing.

**Step 3:** Cut and paste the surface of minimal genus.

Up to a cyclic permutation, is therefore

.

Finally,

.

Notice that this decomposition is not unique, since we also have

.

Even more surprisingly, Stallings and Gersten noticed that such ideas may be applied in order to prove the fundamental theorem of algebra, following an intuition of Gauss! See their article* On Gauss’s first proof of the fundamental theorem of algabra*.