This post is a continuation of the previous post, which has attracted a large number of comments. I’m recording here some calculations that arose from those comments (particularly those of Pace Nielsen, Lior Silberman, Tobias Fritz, and Apoorva Khare). Please feel free to either continue these calculations or to discuss other approaches to the problem, such as those mentioned in the remaining comments to the previous post.

Let {F_2} be the free group on two generators {a,b}, and let {\| \|: F_2 \rightarrow {\bf R}^+} be a quantity obeying the triangle inequality

\displaystyle \| xy\| \leq \|x \| + \|y\|

and the linear growth property

\displaystyle \| x^n \| = |n| \| x\|

for all {x,y \in F_2} and integers {n \in {\bf Z}}; this implies the conjugation invariance

\displaystyle \| y^{-1} x y \| = \|x\|

or equivalently

\displaystyle \| xy \| = \| yx\|

We consider inequalities of the form

\displaystyle \| xyx^{-1}y^{-1} \| \leq \alpha \|x\| + \beta \| y\| \ \ \ \ \ (1)

or

\displaystyle \| xyx^{-2}y^{-1} \| \leq \gamma \|x\| + \delta \| y\| \ \ \ \ \ (2)

for various real numbers {\alpha,\beta,\gamma,\delta}. For instance, since

\displaystyle \| xyx^{-1}y^{-1} \| \leq \| xyx^{-1}\| + \|y^{-1} \| = \|y\| + \|y\|

we have (1) for {(\alpha,\beta) = (2,0)}. We also have the following further relations:

Proposition 1

  • (i) If (1) holds for {(\alpha,\beta)}, then it holds for {(\beta,\alpha)}.
  • (ii) If (1) holds for {(\alpha,\beta)}, then (2) holds for {(\alpha+1, \frac{\beta}{2})}.
  • (iii) If (2) holds for {(\gamma,\delta)}, then (1) holds for {(\frac{2\gamma}{3}, \frac{2\delta}{3})}.
  • (iv) If (1) holds for {(\alpha,\beta)} and (2) holds for {(\gamma,\delta)}, then (1) holds for {(\frac{2\alpha+1+\gamma}{4}, \frac{\delta+\beta}{4})}.

Proof: For (i) we simply observe that

\displaystyle \| xyx^{-1} y^{-1} \| = \| (xyx^{-1} y^{-1})^{-1} \| = \| y^{-1} x^{-1} y x \| = \| y x y^{-1} x^{-1} \|.

For (ii), we calculate

\displaystyle \| xyx^{-2}y^{-1} \| = \frac{1}{2}\| (xyx^{-2}y^{-1})^2 \|

\displaystyle = \frac{1}{2} \| (xyx^{-2}y^{-1} x) (yx^{-2} y^{-1}) \|

\displaystyle \leq \frac{1}{2} (\| xyx^{-2}y^{-1} x\| + \|yx^{-2} y^{-1}\|)

\displaystyle \leq \frac{1}{2} ( \| x^2 y x^{-2} y^{-1} \| + 2 \|x\| )

\displaystyle \leq \frac{1}{2} ( 2 \alpha \|x\| + \beta \|y\| + 2 \|x\|)

giving the claim.

For (iii), we calculate

\displaystyle \| xyx^{-1}y^{-1}\| = \frac{1}{3} \| (xyx^{-1}y^{-1})^3 \|

\displaystyle = \frac{1}{3} \| (xyx) (x^{-2} y^{-1} xy) (xyx)^{-1} (x^2 y x^{-1} y^{-1}) \|

\displaystyle \leq \frac{1}{3} ( \| x^{-2} y^{-1} xy\| + \| x^2 y x^{-1} y^{-1}\| )

\displaystyle = \frac{1}{3} ( \| xy x^{-2} y^{-1} \| + \|x^{-1} y^{-1} x^2 y \| )

\displaystyle \leq \frac{1}{3} ( \gamma \|x\| + \delta \|y\| + \gamma \|x\| + \delta \|y\|)

giving the claim.

For (iv), we calculate

\displaystyle \| xyx^{-1}y^{-1}\| = \frac{1}{4} \| (xyx^{-1}y^{-1})^4 \|

\displaystyle = \frac{1}{4} \| (xy) (x^{-1} y^{-1} x) (y x^{-1} y^{-1}) (xyx^{-1}) (xy)^{-1} (x^2yx^{-1}y^{-1}) \|

\displaystyle \leq \frac{1}{4} ( \| (x^{-1} y^{-1} x) (y x^{-1} y^{-1}) (xyx^{-1}) \| + \|x^2yx^{-1}y^{-1}\| )

\displaystyle \leq \frac{1}{4} ( \|(y x^{-1} y^{-1}) (xy^{-1}x^{-1})(x^{-1} y x) \| + \gamma \|x\| + \delta \|y\|)

\displaystyle \leq \frac{1}{4} ( \|x\| + \|(xy^{-1}x^{-1})(x^{-1} y x) \| + \gamma \|x\| + \delta \|y\|)

\displaystyle = \frac{1}{4} ( \|x\| + \|x^{-2} y x^2 y^{-1} \|+ \gamma \|x\| + \delta \|y\|)

\displaystyle \leq \frac{1}{4} ( \|x\| + 2\alpha \|x\| + \beta \|y\| + \gamma \|x\| + \delta \|y\|)

giving the claim. \Box

Here is a typical application of the above estimates. If (1) holds for {(\alpha,\beta)}, then by part (i) it holds for {(\beta,\alpha)}, then by (ii) (2) holds for {(\beta+1,\frac{\alpha}{2})}, then by (iv) (1) holds for {(\frac{3\beta+2}{4}, \frac{3\alpha}{8})}. The map {(\alpha,\beta) \mapsto (\frac{3\beta+2}{4}, \frac{3\alpha}{8})} has fixed point {(\alpha,\beta) = (\frac{16}{23}, \frac{6}{23})}, thus

\displaystyle \| xyx^{-1}y^{-1} \| \leq \frac{16}{23} \|x\| + \frac{6}{23} \|y\|.

For instance, if {\|a\|, \|b\| \leq 1}, then {\|aba^{-1}b^{-1} \| \leq 22/23 = 0.95652\dots}.