Difference between revisions of "Linear norm"

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=== Applications ===
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== Applications ==
  
 
'''Corollary 1'''.  The function <math>n \mapsto \|x^n y\|</math> is convex in <math>n</math>.
 
'''Corollary 1'''.  The function <math>n \mapsto \|x^n y\|</math> is convex in <math>n</math>.
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==== Iterations ====
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== Iterations ==

Revision as of 13:35, 21 December 2017

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]


Applications

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] \| ).$ '''Proof'''. \ltmath\gt[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

[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]


Iterations