Linear norm

This is the wiki page for understanding seminorms of linear growth on a group [math]\displaystyle{ G }[/math] (such as the free group on two generators). These are functions [math]\displaystyle{ \| \|: G \to [0,+\infty) }[/math] that obey the triangle inequality

[math]\displaystyle{ \|xy\| \leq \|x\| + \|y\| \quad (1) }[/math]

and the linear growth condition

[math]\displaystyle{ \|x^n \| = |n| \|x\| \quad (2) }[/math]

for all [math]\displaystyle{ x,y \in G }[/math] and [math]\displaystyle{ n \in {\bf Z} }[/math].

We use the usual group theory notations [math]\displaystyle{ x^y := yxy^{-1} }[/math] and [math]\displaystyle{ [x,y] := xyx^{-1}y^{-1} }[/math].

Threads

• https://terrytao.wordpress.com/2017/12/16/bi-invariant-metrics-of-linear-growth-on-the-free-group/, Dec 16 2017.
• Bi-invariant metrics of linear growth on the free group, II, Dec 19 2017.

Key lemmas

Henceforth we assume we have a seminorm [math]\displaystyle{ \| \| }[/math] of linear growth. The letters [math]\displaystyle{ s,t,x,y,z,w }[/math] are always understood to be in [math]\displaystyle{ G }[/math], and [math]\displaystyle{ i,j,n,m }[/math] are always understood to be integers.

From (2) we of course have

[math]\displaystyle{ \|x^{-1} \| = \| x\| \quad (3) }[/math]

Lemma 1. If [math]\displaystyle{ x }[/math] is conjugate to [math]\displaystyle{ y }[/math], then [math]\displaystyle{ \|x\| = \|y\| }[/math].

Proof. By hypothesis, [math]\displaystyle{ x = zyz^{-1} }[/math] for some [math]\displaystyle{ z }[/math], thus [math]\displaystyle{ x^n = z y^n z^{-1} }[/math], hence by the triangle inequality

[math]\displaystyle{ n \|x\| = \|x^n \| \leq \|z\| + n \|y\| + \|z^{-1} \| }[/math]

for any [math]\displaystyle{ n \geq 1 }[/math]. Dividing by [math]\displaystyle{ n }[/math] and taking limits we conclude that [math]\displaystyle{ \|x\| \leq \|y\| }[/math]. Similarly [math]\displaystyle{ \|y\| \leq \|x\| }[/math], giving the claim. [math]\displaystyle{ \Box }[/math]

An equivalent form of the lemma is that

[math]\displaystyle{ \|xy\| = \|yx\| \quad (4). }[/math]

We can generalise Lemma 1:

Lemma 2. If [math]\displaystyle{ x^i }[/math] is conjugate to [math]\displaystyle{ wy }[/math] and [math]\displaystyle{ x^j }[/math] is conjugate to [math]\displaystyle{ zw^{-1} }[/math], then [math]\displaystyle{ \|x\| \leq \frac{1}{|i+j|} ( \|w\| + \|z\| ) }[/math].

Proof. By hypothesis, [math]\displaystyle{ x^i = s wy s^{-1} }[/math] and [math]\displaystyle{ x^j = t zw^{-1} t^{-1} }[/math] for some [math]\displaystyle{ s,t }[/math]. For any natural number [math]\displaystyle{ n }[/math], we then have

[math]\displaystyle{ x^{in} x^{jn} = s wy \dots wy s^{-1} t zw^{-1} \dots zw^{-1} t^{-1} }[/math]

where the terms [math]\displaystyle{ wy, zw }[/math] are each repeated [math]\displaystyle{ n }[/math] times. By Lemma 1, conjugation by [math]\displaystyle{ w }[/math] does not change the norm. From many applications of this and the triangle inequality, we conclude that

[math]\displaystyle{ |i+j| n \|x\| = \| x^{in} x^{jn} \| \leq \|s\| + n \|y\| + \|s^{-1} t\| + n \|z\| + \|t^{-1}\|. }[/math]

Dividing by [math]\displaystyle{ n }[/math] and sending [math]\displaystyle{ n \to \infty }[/math], we obtain the claim. [math]\displaystyle{ \Box }[/math]

Applications

Corollary 0. The eight commutators [math]\displaystyle{ [x^{\pm 1}, y^{\pm 1}], [y^{\pm 1}, x^{\pm 1}] }[/math] all have the same norm.

Proof. Each of these commutators is conjugate to either [math]\displaystyle{ [x,y] }[/math] or its inverse. [math]\displaystyle{ \Box }[/math]

Corollary 1. The function [math]\displaystyle{ n \mapsto \|x^n y\| }[/math] is convex in [math]\displaystyle{ n }[/math].

Proof. [math]\displaystyle{ x^n y }[/math] is conjugate to [math]\displaystyle{ x (x^{n-1} y) }[/math] and to [math]\displaystyle{ (x^{n+1} y) x^{-1} }[/math], hence by Lemma 2

[math]\displaystyle{ \| x^n y \| \leq \frac{1}{2} (\| x^{n-1} y \| + \| x^{n+1} y \|), }[/math]

giving the claim. [math]\displaystyle{ \Box }[/math]

Corollary 2. For any [math]\displaystyle{ k \geq 1 }[/math], one has

[math]\displaystyle{ \| [x,y] \| \leq \frac{1}{2k+2} (\| [x^{-1},y^{-1}]^k x^{-1} \| + \| [x,y]^k x \|). }[/math]

Thus for instance

[math]\displaystyle{ \| [x,y] \| \leq \frac{1}{4} (\| [x^{-1},y^{-1}] x^{-1} \| + \| [x,y] x \|). }[/math]

Proof. [math]\displaystyle{ [x,y]^{k+1} }[/math] is conjugate both to [math]\displaystyle{ x(y[x^{-1},y^{-1}]^k x^{-1}y^{-1}) }[/math] and to [math]\displaystyle{ (y^{-1} [x,y]^k xy)x^{-1} }[/math], hence by Lemma 2

[math]\displaystyle{ \| [x,y] \| \leq \frac{1}{2k+2} ( \| y[x^{-1},y^{-1}]^k x^{-1} \| + \| (y^{-1} [x,y]^k xy)x^{-1}\|) }[/math]

giving the claim by Lemma 1. [math]\displaystyle{ \Box }[/math]

Corollary 3. One has

[math]\displaystyle{ \|[x,y]^2 x\| \leq \frac{1}{2} ( \| x y^{-1} [x,y] \| + \| xy [x,y] \| ).$ '''Proof'''. \lt math\gt [x,y]^2 x }[/math] is conjugate both to [math]\displaystyle{ y (x^{-1} y^{-1} [x,y] x^2) }[/math] and to [math]\displaystyle{ (x[x,y]xyx^{-1}) y^{-1} }[/math], hence by Lemma 2

[math]\displaystyle{ \displaystyle \|[x,y]^2 x\| \leq \frac{1}{2} ( \|x^{-1} y^{-1} [x,y] x^2\| + \|x[x,y]xyx^{-1}\|) }[/math]

giving the claim by Lemma 1. [math]\displaystyle{ \Box }[/math]

Corollary 4. One has

[math]\displaystyle{ \| [x,y] x\| \leq \frac{1}{4} ( \| x^2 y [x,y] \| + \| xy^{-1} x [x,y] \| ). }[/math]

Proof [math]\displaystyle{ ([x,y] x)^2 }[/math] is conjugate both to [math]\displaystyle{ y^{-1} (x [x,y] x^2 y x^{-1}) }[/math] and to [math]\displaystyle{ (x^{-1} y^{-1} x [x,y] x^2) y }[/math], hence

[math]\displaystyle{ \| [x,y] x\| \leq \frac{1}{4} ( \| x [x,y] x^2 y x^{-1} \| + \| x^{-1} y^{-1} x [x,y] x^2 \| ), }[/math]

giving the claim by Lemma 1. [math]\displaystyle{ \Box }[/math]