Let $\mathbb{F}_n$ be a free group of rank $n \geq 2$ and $w \in \mathbb{F}_n$, viewed as a word on $n$ letters. Then $w$ is a product of commutators if and only if it belongs to the kernel of $\mathbb{F}_n \twoheadrightarrow \mathbb{F}_n^{\text{ab}} \simeq \mathbb{Z}^n$, that is if the sum of exponents on each letter is zero. From now on, we suppose that $w$ satisfies this property.

Question: How to find the least integer $k$ such that $w$ can be written as a product of $k$ 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 $w$; then we we choose a pairing of edges, encoded in a circle graph $\Gamma$ (topologically, a circle with a collection of interior arcs whose endpoints are pairwise distinct), and we construct a closed surface $\Sigma_{\Gamma}$ 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 $\gamma(w)$ of $w$ as the least genus of a surface obtained from $w$ in the previous way.

For example, $\gamma(a^2b^{-1}a^{-1}ba^{-1})=1$.

We then have the following surprising result:

Theorem 1: $w$ can be written as a product of $\gamma(w)$ commutators, and not fewer.

Moreover, Gorenstein and Turner give in their article a simple way to compute the genus $g(\Sigma_{\Gamma})$ from the circle graph $\Gamma$. First, we number the $n$ interior arcs of $\Gamma$, and we define the $n \times n$-matrix $A(\Gamma)$ by saying that the entry $(i,j)$ is equal to $1$ if the $i$-th and $j$-th interior arcs intersect; otherwise, it is equal to $0$. Notice that $A(\Gamma)$ is symmetric and that any diagonal element is zero, so, working in $\mathbb{Z}_2$, it can be written in some basis as a diagonal block matrix (see for example Kaplansky’s book, Linear algebra and geometry):

$A(\Gamma) = \left( \begin{array}{cccc} J & & & \\ & \ddots & & \\ & & J & \\ & & & 0 \end{array} \right)$ where $J = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$.

In particular, notice that $\mathrm{rank}(A(\Gamma))$ is even. Let us define the genus of $A(\Gamma)$ as

$g(A(\Gamma))= \frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2} (A(\Gamma))$.

Finally, we have

Theorem 2: $g(\Sigma_{\Gamma})= \gamma(A(\Gamma))$.

Proof of theorem 1: We prove by induction on $\gamma(w)$ that $w$ can be written as a product of $\gamma(w)$ commutators. If $\gamma(w)=0$, then $w=1$ and there is nothing to prove.

From now on, suppose that our result holds for $\gamma(w) \leq n-1$ and suppose $\gamma(w)=n \geq 1$. Let $\Gamma$ be a circle graph such that $\gamma(w)= g( \Sigma_{\Gamma})$. First, we replace $w$ by a word $w'$ such that each pair of linked letters in $w$ are different in $w'$. For example,

Because $n \geq 1$, there exist two intersecting interior arcs; therefore, up to a cyclic permutation (that does not change a pairing), we may suppose that $w'$ can be written as

$w'= x_1Rx_2Sx_1^{-1} T x_2^{-1} U$.

The pairing $\Gamma$ clearly induces a pairing $\Gamma'$ whose underlying circle graph is the same; in particular, $g(\Sigma_{\Gamma})= g( \Sigma_{\Gamma'})$. Now, we cut and paste $\Sigma_{\Gamma'}$ in the following way:

From a purely algebraic viewpoint, we set $c=Rx_2S$ and we write $w'= x_1cx_1^{-1}TSc^{-1} RU$. Up to a cyclic permutation, we have

$w'= c^{-1}RUx_1cx_1^{-1}TS$.

Then, we set $d= x_1^{-1}U^{-1}R^{-1}$ and we write

$w' = c^{-1}d^{-1}c d URTS= [c^{-1},d^{-1}] \cdot URTS$.

Now, we want to apply our induction hypothesis to $\tilde{w} = URTS$. First, the pairing $\Gamma'$ induces a pairing $\Gamma''$ for $[c^{-1},d^{-1}] \cdot URTS$ just by adding some disjoint interior arcs (because the word is not reduced), but that does not change the surface $\Sigma_{\Gamma'}$ (in the polygon, we just add consecutive identified edges), hence $g(\Sigma_{\Gamma'})=g( \Sigma_{\Gamma''})$. Then, the pairing $\Gamma''$ induces a pairing $\tilde{\Gamma}$ for $URTS$; but the figure below justifies that we have the connected sum $\Sigma_{\Gamma''} \simeq \mathbb{T}^2 \sharp \Sigma_{\tilde{\Gamma}}$.

Therefore,

$\gamma(\tilde{w})= g( \Sigma_{\tilde{\Gamma}}) = g( \Sigma_{\Gamma''})-1 = g( \Sigma_{\Gamma'})-1 = g( \Sigma_{\Gamma})-1=n-1$.

Using our induction hypothesis, $\tilde{w}$ can be written as a product of $n-1$ commutators, so $w$ can be written as a product of $n$ commutators.

Conversely, we prove that if $w$ is a product of $n$ commutators then there exists a pairing $\Gamma$ satisfying $g(\Sigma_{\Gamma})=n$. We consider the case $n=3$, but the argument is completely general. So

$w = [a_1,a_2] \cdot [a_3,a_4] \cdot [a_5,a_6]$.

We use the pairing $\Gamma$:

The associated matrix is $A( \Gamma) = \left( \begin{array}{ccc} J & & 0 \\ & J & \\ 0 & & J \end{array} \right)$ where $J = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$, hence

$g(\Sigma_{\Gamma}) = \frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2}(A(\Gamma)) = 3$. $\square$

Corollary: Let $w \in \mathbb{F}_n$ be a reduced word of length $m$ in the commutator subgroup. Then $w$ can be written as a product of at most $m/4$ commutators.

Proof. Let $\Gamma$ be a pairing of minimal genus, so that $w$ can be written as a product of $\frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2}(A(\Gamma))$ commutators. The circle graph $\Gamma$ has $m/2$ interior arcs (of course, $m$ is even) and $A(\Gamma)$ is a $\frac{m}{2} \times \frac{m}{2}$-matrix, hence

$\gamma(w) = \frac{1}{2} \mathrm{rank}_{\mathbb{Z}_2}(A(\Gamma)) \leq \frac{1}{2} \cdot \frac{m}{2}= \frac{m}{4}$. $\square$

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 $w = x^2yx^{-1}y^{-2}x^{-1}y$.

Step 1: Draw all the possible pairings.

Step 2: Compute the ranks of the associated matrices.

We have $\gamma(A_1)=\gamma(A_2)=\gamma(A_3)=2$ and $\gamma(A_4)=1$. So we are interesting in the fourth pairing.

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

Up to a cyclic permutation, $w$ is therefore

$cx^{-1}y^{-1}c^{-1}yx= [c, (yx)^{-1}]= [xy,x^{-1}y^{-1}]= xyx^{-1}y^{-2}x^{-1}yx$.

Finally,

$w=x[xy,x^{-1}y^{-1}]x^{-1} = [x^2yx^{-1}, y^{-1}x^{-1}]$.

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

$w= [xy,y^{-1}xy^2]$.

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.