Linear norm

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

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

and the linear growth condition

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

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

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

Threads

Key lemmas

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

From (2) we of course have

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

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

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

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

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

An equivalent form of the lemma is that

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

We can generalise Lemma 1:

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

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

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

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

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

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


Corollaries

Corollary 0. The eight commutators [math][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][x,y][/math] or its inverse. [math]\Box[/math]

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

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

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

giving the claim. [math]\Box[/math]

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

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

Thus for instance

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

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

[math] \| [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]\Box[/math]

Corollary 3. One has

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

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

[math] \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]\Box[/math]

Corollary 4. One has

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

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

[math]\| [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]\Box[/math]

Corollary 5. One has

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

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

[math]\| [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]\Box[/math]

Corollary 6. One has

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

Proof. From Lemma 2 we have

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

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

[math] \| 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]\Box[/math]

Corollary 7. For any [math]m,k[/math], one has

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

[math]\| 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]\Box[/math]

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

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

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

Iterations

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

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

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

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

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

Proof. From Corollary 5 and hypothesis one has

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

From Corollary 2 we thus have

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

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

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

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

[math] \|[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]\|[x,y]\| = 0[/math].

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

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

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

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

By the triangle inequality, we conclude that

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

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

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

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

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.