# Difference between revisions of "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. $\Box$

## Corollaries

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. $[x,y]^2 x$ 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 Lemma 2

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

giving the claim by Lemma 1. $\Box$

Corollary 5. One has

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

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

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

giving the claim by Lemma 1 and Corollary 0. $\Box$

Corollary 6. One has

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

Proof. From Lemma 2 we have

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

Since $x^{-1} [x,y]^2$ is conjugate to $(yx^{-1} y^{-1}) (xyx^{-2} y^{-1} x)$, we have

$\| x^{-1} [x,y]^2 \| \leq \| yx^{-1} y^{-1} \| + \|xyx^{-2} y^{-1} x\|$

and the claim follows from Lemma 1 and (3). $\Box$

Corollary 7. For any $m,k$, one has

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

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

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

where the final equation is by conjugation invariance (Lemma 1). $\Box$

Corollary 8. One has $\|x\| \leq \| [x,y] x \|$.

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

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

By Lemma 1, the RHS is $\frac{1}{2} \|x\| + \frac{1}{2} \| [x,y] x \|$, and the claim follows. $\Box$

## Iterations

Call a pair of real numbers $(\alpha,\beta)$ admissible if one has the inequality

$\| [x,y] \| \leq \alpha \|x\| + \beta \|y \|$

for all $x,y$. Clearly the set of admissible pairs is closed and convex, and if $(\alpha,\beta)$ is admissible then so is $(\alpha',\beta')$ for any $\alpha' \geq \alpha, \beta' \geq \beta$. From Corollary 0 we also see that the set is symmetric: $(\alpha,\beta)$ is admissible if and only if $(\beta,\alpha)$ is.

Writing $[x,y] = y^x y^{-1}$ we see that $(0,2)$ is admissible, and similarly so is $(0,2)$.

Proposition 1. If $(\alpha,\beta)$ is admissible, then so is $(\frac{\alpha+1}{2}, \frac{\beta}{4})$.

Proof. From Corollary 5 and hypothesis one has

$\| [x,y] x\| \leq \|x\| + \frac{1}{2} ( \alpha \|x^2\| + \beta \|y\| ) = (\alpha+1) \|x\| + \frac{\beta}{2} \|y\|$

and hence also

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

From Corollary 2 we thus have

$\| [x,y]\| \leq \frac{\alpha+1}{2} \|x\| + \frac{\beta}{4} \|y\|.$

The map $(\alpha,\beta) \mapsto (\frac{\alpha+1}{2}, \frac{\beta}{4})$ is a contraction with fixed point $(1,0)$. Thus

$\|[x,y]\| \leq \|x\| \quad (4)$.

From symmetry we also see that if $(\alpha,\beta)$ is admissible, then so is $(\frac{\beta+1}{2}, \frac{\alpha}{4})$. The map $(\alpha,\beta) \mapsto (\frac{\beta+1}{2}, \frac{\alpha}{4})$ is a contraction with fixed point (4/7,1/7), thus

$\|[x,y]\| \leq \frac{4}{7} \|x\| + \frac{1}{7} \|y\|$.

## Solution

Note: this argument only requires Lemma 1, Lemma 2, and Corollary 7 from the preceding sections.

Theorem 1 $\|[x,y]\| = 0$.

Proof Let $n$ be a large natural number. Write $f(m,k) := \| x^m [x,y]^k \|$. Let $X_1,\dots,X_{2n}$ be iid random variables, each taking a value of $(-1,0)$ or $(1,-1)$ with equal probability $1/2$. From Corollary 7 one has

$f(m,k) \leq {\bf E} f( (m,k) + X_j)$

for any $(m,k), j$, and in particular on iterating

$f(0, n) \leq {\bf E} f( (0,n) + X_1 + \dots + X_{2n} ).$

By the triangle inequality, we conclude that

$f(0, n) \leq (\|x\|+\|y\|) {\bf E} | (0,n) + X_1 + \dots + X_{2n} |.$

But the random variable $(0,n) + X_1 + \dots + X_{2n}$ has mean zero and variance $O(n)$, hence by Cauchy-Schwarz

$f(0, n) \ll n^{1/2} (\|x\|+\|y\|).$

But the left-hand side is $n \|[x,y]\|$, so on dividing by $n$ and taking limits we obtain the claim.$\Box$

As a consequence of this theorem and the triangle inequality, any seminorm on a group will factor through to its abelianisation.

## Writeup

Here are the linear norm grant acknowledgments.