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]

Corollaries

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] \| ). }[/math]

Proof. [math]\displaystyle{ [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 Lemma 2

[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]

Corollary 5. One has

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

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

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

giving the claim by Lemma 1 and Corollary 0. [math]\displaystyle{ \Box }[/math]

Corollary 6. One has

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

Proof. From Lemma 2 we have

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

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

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

and the claim follows from Lemma 1 and (3). [math]\displaystyle{ \Box }[/math]

Corollary 7. For any [math]\displaystyle{ m,k }[/math], one has

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

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

[math]\displaystyle{ \| 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} \|), }[/math]

where the final equation is by conjugation invariance (Lemma 1). [math]\displaystyle{ \Box }[/math]

Corollary 8. One has [math]\displaystyle{ \|x\| \leq \| [x,y] x \| }[/math].

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

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

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

Iterations

Call a pair of real numbers [math]\displaystyle{ (\alpha,\beta) }[/math] admissible if one has the inequality

[math]\displaystyle{ \| [x,y] \| \leq \alpha \|x\| + \beta \|y \| }[/math]

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

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

Proposition 1. If [math]\displaystyle{ (\alpha,\beta) }[/math] is admissible, then so is [math]\displaystyle{ (\frac{\alpha+1}{2}, \frac{\beta}{4}) }[/math].

Proof. From Corollary 5 and hypothesis one has

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

and hence also

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

From Corollary 2 we thus have

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

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

[math]\displaystyle{ \|[x,y]\| \leq \|x\| \quad (4) }[/math].

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

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

Solution

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

Theorem 1 [math]\displaystyle{ \|[x,y]\| = 0 }[/math].

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

[math]\displaystyle{ f(m,k) \leq {\bf E} f( (m,k) + X_j) }[/math]

for any [math]\displaystyle{ (m,k), j }[/math], and in particular on iterating

[math]\displaystyle{ f(0, n) \leq {\bf E} f( (0,n) + X_1 + \dots + X_{2n} ). }[/math]

By the triangle inequality, we conclude that

[math]\displaystyle{ f(0, n) \leq (\|x\|+\|y\|) {\bf E} | (0,n) + X_1 + \dots + X_{2n} |. }[/math]

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

[math]\displaystyle{ f(0, n) \ll n^{1/2} (\|x\|+\|y\|). }[/math]

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

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