The triangle inequality for graded real vector spaces

TL;DR

This paper proves that a natural homogeneous norm on graded Lie algebras satisfies the triangle inequality, confirming a key property for these mathematical structures and answering a previously posed question.

Contribution

It establishes the triangle inequality for a natural candidate norm on graded Lie algebras of any length, advancing understanding of their geometric properties.

Findings

The candidate norm satisfies the triangle inequality.

The result applies to graded Lie algebras of any length.

Answers Moskowitz's question affirmatively.

Abstract

In this paper, we prove that a natural candidate for a homogeneous norm on a graded Lie algebra of any length satisfies the triangle inequality which answers Moskowitz's question.

Equations

$${\alpha }_t(v_1,\dots ,v_r)=(t{v}_1,t^2{v}_2,\dots ,t^r{v}_r).$$

$$\Vert X\Vert =({\Vert v_1\Vert }_1^{2r}+{\Vert v_2\Vert }_2^{2r-2}+\cdots +{\Vert v_r\Vert }_r^2{)}^{1/2r},$$

$$X+Y=(v_1+w_1,v_2+w_2,v_3+w_3,v_4+w_4,v_5+w_5).$$

$${\Vert v_1+w_1\Vert }^{10}+{\Vert v_2+w_2\Vert }^8+{\Vert v_3+w_3\Vert }^6+{\Vert v_4+w_4\Vert }^4+{\Vert v_5+w_5\Vert }^{10}\le (\Vert X\Vert +\Vert Y\Vert {)}^{10}.$$

$$\sum _{r=0}^{10}\left(\genfrac{}{}{0ex}{}{10}{r}\right){\Vert v_1\Vert }^{10-r}{\Vert w_1\Vert }^r+\sum _{r=0}^8\left(\genfrac{}{}{0ex}{}{8}{r}\right)$$

$$+\sum _{r=0}^4\left(\genfrac{}{}{0ex}{}{4}{r}\right){\Vert v_4\Vert }^{4-r}{\Vert w_4\Vert }^r$$

$$\le$$

$$\sum _{r=0}^{10}$$

$$A=({\Vert v_1\Vert }^{10}+{\Vert v_2\Vert }^8+{\Vert v_3\Vert }^6+{\Vert v_4\Vert }^4+{\Vert v_5\Vert }^2{)}^{1/10}$$

$$B=({\Vert w_1\Vert }^{10}+{\Vert w_2\Vert }^8+{\Vert w_3\Vert }^6+{\Vert w_4\Vert }^4+{\Vert w_5\Vert }^2{)}^{1/10}.$$

$$252{\Vert v_1\Vert }^5{\Vert w_1\Vert }^5+70{\Vert v_2\Vert }^4{\Vert w_2\Vert }^4+20{\Vert v_3\Vert }^3{\Vert w_3\Vert }^3+6{\Vert v_4\Vert }^2{\Vert v_4\Vert }^2+2\Vert v_5\Vert \Vert w_5\Vert \le 252A^5{B}^5.$$

$$120{\Vert v_1\Vert }^7{\Vert w_1\Vert }^3+120{\Vert v_1\Vert }^3{\Vert w_1\Vert }^7$$

$$+15{\Vert v_3\Vert }^4{\Vert w_3\Vert }^2$$

$$\le$$

$$120A^7{B}^3$$

$${\Vert v_1\Vert }^7{\Vert w_1\Vert }^3+{\Vert v_2\Vert }^{28/5}{\Vert w_2\Vert }^{12/5}+{\Vert v_3\Vert }^{21/5}{\Vert w_3\Vert }^{9/5}$$

$${\Vert v_1\Vert }^3{\Vert w_1\Vert }^7+{\Vert v_2\Vert }^{12/5}{\Vert w_2\Vert }^{28/5}+{\Vert v_3\Vert }^{9/5}{\Vert w_3\Vert }^{21/5}$$

$$56({\Vert v_2\Vert }^5{\Vert w_2\Vert }^3+{\Vert v_2\Vert }^3{\Vert w_2\Vert }^5)$$

$$15({\Vert v_3\Vert }^4{\Vert w_3\Vert }^2+{\Vert v_3\Vert }^2{\Vert w_3\Vert }^4)$$

$$45{\Vert v_1\Vert }^8{\Vert w_1\Vert }^2+45{\Vert v_1\Vert }^2{\Vert w_1\Vert }^8$$

$$+4{\Vert v_4\Vert }^4\Vert w_4\Vert$$

$$\le$$

$$45A^8{B}^2$$

$$10{\Vert v_1\Vert }^9\Vert w_1\Vert +10\Vert v_1\Vert {\Vert w_1\Vert }^9$$

$$+6{\Vert v_3\Vert }^5\Vert w_3\Vert$$

$$\le$$

$$10A^{9}B$$

Full text

1. Introduction

In [1], M. Moskowitz extended the classical theorem of Minkowski on lattice points and convex bodies in $\mathbb{R}^{n}$ to simply connected nilpotent Lie groups $G$ with a $\mathbb{Q}$-structure whose Lie algebra $\mathfrak{g}$ admits a length 2 grading. For this purpose, the author proved the triangle inequality for a certain natural homogeneous norm of the Lie algebra associated with the grading of length 2. Later, in [2], the same author extended the homogeneous norms for which the triangle inequality holds to gradings of length 3 and 4. At the end of the paper [2], the author demonstrated that the method of his proof cannot be carried on to the case of length 5, so it was left as an open problem. However, the arguments in the proof contains some errors. The inequalities (6) and (12) in the proof of Theorem 1 in [2] are not correct for length $r=2,3,$ or $4$.

In this paper, we claim that for the grading of lengths 3, 4 and 5, the homogeneous norms defined as in [2] satisfy the the triangle inequality. For the sake of conciseness, the complete proof will be provided only for the most complicated case, the case of length 5. The gradings of length 3 and 4 can be done with the same manner. Our proof is quite elementary. The main tools in our proof are only Hölder’s inequality and Muirhead’s inequality. We refer the reader to [3] for the latter inequality.

A Lie algebra $\mathfrak{g}$ is said to be graded if there is a finite family of subspaces $V_{1},\dots,V_{r}$ with $\mathfrak{g}\ =\ V_{1}\oplus\dots\oplus V_{r}$ satisfying $[V_{i},V_{j}]\subseteq V_{i+j}$ for all $i,j$ (here, $V_{k}=0$ if $k>r$). The integer $r$ is its length. For a graded Lie algebra $\mathfrak{g}$, we define, for $t\in\mathbb{R}^{\times}$,

[TABLE]

Then $\alpha_{t}$ is Lie algebra automorphism of $\mathfrak{g}$. Moreover, if $\mathfrak{g}$ is a graded Lie algebra, then $\mathfrak{g}$ must be nilpotent (see [1] and [2]).

As in [1] and [2], we define a homogeneous norm on a graded Lie algebra $\mathfrak{g}$ as a function $\left\lVert\cdot\right\rVert:\mathfrak{g}\rightarrow\mathbb{R}$ satisfying the following conditions.

1. (i)

$\left\lVert\cdot\right\rVert\geq 0$  and is [math] only at [math]. 2. (ii)

$\left\lVert X\right\rVert=\left\lVert-X\right\rVert$  for all $X\in\mathfrak{g}$. 3. (iii)

$\left\lVert\alpha_{t}(X)\right\rVert=\ |t|\left\lVert X\right\rVert$  for all $t\in\mathbb{R}$ and $X\in\mathfrak{g}$. 4. (iv)

$\left\lVert X+Y\right\rVert\leq\left\lVert X\right\rVert+\left\lVert Y\right\rVert$  for all $X,Y\in\mathfrak{g}$.

Any graded Lie algebra $\mathfrak{g}$ possesses a natural candidate for a homogeneous norm as follows: for $X=(v_{1},\dots,v_{r})$, let

[TABLE]

where $\left\lVert\cdot\right\rVert_{i}$ is the Euclidean norm on each $V_{i}$ (we may suppress the subscript for norms). This norm satisfies properties (i) and (ii). However, the the homogeneity property (iii) holds if and only if $r\leq 2$. In [1], it was shown that the subadditivity property (iv) (or the triangle inequality) holds when $r=2$. Our main result is that the triangle inequality holds for $r=3,4$ and $5$.

Theorem 1**.**

$\left\lVert X+Y\right\rVert\leq\left\lVert X\right\rVert+\left\lVert Y\right\rVert$*  for  $r\leq 5$.*

2. Proof of the main theorem

We prove the main theorem for length $r=5$ only and one will see that the similar arguments can be used to prove for length $r=3$ and $4$ with slightly modification but easier.

Proof of the main theorem: Let $X=(v_{1},v_{2},v_{3},v_{4},v_{5})$ and $Y=(w_{1},w_{2},w_{3},w_{4},w_{5})$. Then

[TABLE]

By taking $10$th powers, we need to prove that

[TABLE]

Expanding both sides by using the binomial theorem, applying the Schwartz inequality to all norms, and canceling yield

[TABLE]

where

[TABLE]

and

[TABLE]

Consider the middle terms of each summation, we have

[TABLE]

which, in fact, holds by Holder’s inequality (with $p=q=2$) since the largest coefficient dominates the others.

Next we show that

[TABLE]

We note that taking $\left\lVert v_{4}\right\rVert=\left\lVert v_{5}\right\rVert=\left\lVert w_{4}\right\rVert=\left\lVert w_{5}\right\rVert=0$ does not effect this inequality. By Hölder’s inequality, we have the two following inequalities:

[TABLE]

Thus, we only need to show that

[TABLE]

These two inequalities follow from Muirhead’s inequality because $(28/5,12/5)$ majorizes $(5,3)$ and $(21/5,9/5)$ majorizes $(4,2)$. Hence, the equality (2.1) holds.

By the similar arguments (with some changes on exponents), we can prove another two inequalities:

[TABLE]

and

[TABLE]

It remains to show that $210\left\lVert v_{1}\right\rVert^{6}\left\lVert w_{1}\right\rVert^{4}+210\left\lVert v_{1}\right\rVert^{4}\left\lVert v_{1}\right\rVert^{6}\leq 210A^{7}B^{3}+210A^{3}B^{7}$. We can see that this is obvious by taking $\left\lVert v_{i}\right\rVert=\left\lVert w_{i}\right\rVert=0$ for all $i\neq 1$. ∎

Remark**.**

The above proof was done by using elementary tools including Hölder’s inequality and Muirhead’s inequality. We strongly believe that the proof can be extended to any length $r\geq 6$ with the same manner. The crucial steps must be to match up all suitable terms from the left and the right of inequalities, so that Hölder’s inequality and Muirhead’s inequality can be used.