# Linear norm

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

$\|xy\| \leq \|x\| + \|y\| \quad (1)$

and the linear growth condition

$\|x^n \| = |n| \|x\| \quad (2)$

for all $x,y \in G$ and $n \in {\bf Z}$.

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

## Key lemmas

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

From (2) we of course have

$\|x^{-1} \| = \| x\| \quad (3)$

Lemma 1. If $x$ is conjugate to $y$, then $\|x\| = \|y\|$.

Proof. By hypothesis, $x = zyz^{-1}$ for some $z$, thus $x^n = z y^n z^{-1}$, hence by the triangle inequality

$n \|x\| = \|x^n \| \leq \|z\| + n \|y\| + \|z^{-1} \|$

for any $n \geq 1$. Dividing by $n$ and taking limits we conclude that $\|x\| \leq \|y\|$. Similarly $\|y\| \leq \|x\|$, giving the claim. $\Box$

An equivalent form of the lemma is that

$\|xy\| = \|yx\| \quad (4).$

We can generalise Lemma 1:

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

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

$x^{in} x^{jn} = s wy \dots wy s^{-1} t zw^{-1} \dots zw^{-1} t^{-1}$

where the terms $wy, zw$ are each repeated $n$ times. By Lemma 1, conjugation by $w$ does not change the norm. From many applications of this and the triangle inequality, we conclude that

$|i+j| n \|x\| = \| x^{in} x^{jn} \| \leq \|s\| + n \|y\| + \|s^{-1} t\| + n \|z\| + \|t^{-1}\|.$

Dividing by $n$ and sending $n \to \infty$, we obtain the claim. $\Box$

## Applications

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

Proof. Each of these commutators is conjugate to either $[x,y]$ or its inverse. $\Box$

Corollary 1. The function $n \mapsto \|x^n y\|$ is convex in $n$.

Proof. $x^n y$ is conjugate to $x (x^{n-1} y)$ and to $(x^{n+1} y) x^{-1}$, hence by Lemma 2

$\| x^n y \| \leq \frac{1}{2} (\| x^{n-1} y \| + \| x^{n+1} y \|),$

giving the claim. $\Box$

Corollary 2. For any $k \geq 1$, one has

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

Thus for instance

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

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

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

giving the claim by Lemma 1. $\Box$

Corollary 3. One has

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

giving the claim by Lemma 1. $\Box$

Corollary 4. One has

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

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

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

giving the claim by Lemma 1. $\Box$