Lectures Heisenberg

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The Geometry of the Heisenberg Group H3

Catherine Bartlett

Department of Mathematics (Pure and Applied)Rhodes University, Grahamstown 6140

Honours Presentation

29 October 2012

Catherine Bartlett (Rhodes) The Geometry of H3 Honours Presentation 1 / 18


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Outline

1 Introduction

2 The Heisenberg group

3 The Heisenberg Lie algebra

4 Properties



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Outline

1 Introduction



4 Properties



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Introduction: Lie groups

Lie groups

A smooth manifold G is aLie groupif it is a group and its groupoperations:

1

multiplication:

: GG G, (x, y

)xy

and2 inversion: : G G, x x1

are smooth (i.e. differentiable).

Matrix Lie group

G is amatrix Lie groupif G is a closed subgroup of GL(n, R)



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Introduction: Lie algebras

Lie algebrasALie algebra g is a vector space equipped with a bilinear operation [, ](the Lie bracket) satisfying

[X, Y] =[Y, X] (skew symmetry)

[X, [Y, Z]] + [Y, [Z, X] + [Z, [X, Y]] = 0. (Jacobi identity)

Lie algebra automorphism

A linear isomorphism :g g that preserves the Lie bracket.

That is, for any X, Yg and , R, we have:1 (X+ Y) =(X) + (Y)

2 ([X, Y]) = [(X), (Y)].



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Outline

1 Introduction



4 Properties



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The Heisenberg group

Matrix representation

H3 =

1 x z0 1 y

0 0 1

|x, y, z R

Inverse

If h=

1 x z0 1 y0 0 1

then h1 =

1 x z+xy0 1 y0 0 1



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Outline

1 Introduction



4 Properties



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The Heisenberg Lie algebra

Matrix representation

h3 =

0 x z0 0 y0 0 0

|x, y, z R

Standard basis

E1 =

0 1 00 0 00 0 0

, E2 =

0 0 00 0 10 0 0

, E3 =

0 0 10 0 00 0 0

.

Commutator relations

[E1, E2] =E3, [E1, E3] =0, [E2, E3] =0.



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Outline

1 Introduction



4 Properties



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Nilpotency

Commutator

[h, k] =hk(kh)1

Nilpotent

Define the sequence of groups n(G) by

0(G) = G, n+1(G) = [n(G), G]

G isnilpotentif

n(G) ={1}

for some n N.



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Nilpotency of H3

Proposition

H3 is nilpotent

Proof

Let

m(x1, x2, x3) =

1 x1 x30 1 x20 0 1

Consider X, Y H3:

X =m(x1, x2, x3), Y =m(y1, y2, y3)

Then the commutator

[X, Y] =XYX1Y1 =m(0, 0, x1y2y1x2)



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Nilpotency of H3

which gives us the commutator subgroup:

1(H3) = [H3, H3] ={m(0, 0, k)|k R}

LetZ =m(0, 0, k).

ThenXZ =m(x1, x2, k+x3) =ZX.

Which means that

[X, Z

] =XZ

(ZX

)

1

=1.

We therefore have that the commutator subgroup

2(H3) = [1(H3), H3] ={1}



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Properties

H3 iscompletely solvable

H3 isunimodular

H3 isdiffeomorphicto R3

H3 issimply connected

Centres

Z(H3) =

1 0 z

0 1 00 0 1

|z R

, Z(h3) =

0 0 z

0 0 00 0 0

|z R

.


3


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h3 and R3

are isomorphic

Lie algebra R3

Let x, y R3 with x= (x1, x2, x3) and y= (y1, y2, y3).Define the operation : R3 R3 R3 by

x y= (0, 0, x1y2x2y1).

Then forms a Lie bracket on R3.

The isomorphism

:h3 R3,

0 x1 x30 0 x2

0 0 0

(x1, x2, x3)


Adj i i


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Adjoint representation

Representation

Arepresentationof a Lie group G is a pair (V, ) where

V is a vector space

is a Lie group homomorphism : G GL(V).

Adjoint representation

Theadjoint representationof a Lie group G is defined to be the map

Ad : GGL(g), g Adg.

Here Adg is the g-automorphism given by

Adg :gg, X gXg1.


Adj i i f H


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Adjoint representation of H3

Matrix representation ofAdh

h=

1 x1 x30 1 x20 0 1

Adh =

1 0 00 1 0

x2 x1 1

AdhEi= hEih1

hE1h1 =

0 1 x20 0 00 0 0

hE2h1 =

0 0 x10 0 10 0 0

hE3h1 =

0 0 10 0 00 0 0

=E1x2E3 =E2+ x1E3 =E3


A hi f h


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Automorphism group ofh3

Consider any vector space isomorphism : R3 R3.

is a linear map and can therefore be represented by a 3 3invertible matrix A: (x) =Ax.

This map is a Lie algebra automorphism if it preserves the Lie bracket

on R3, i.e. ifA(x y) =Ax Ay

for any x, y R3.

Let x, y R3 with x= (x1, x2, x3) and y= (y1, y2, y3) and let A= [aij] bean arbitrary invertible 33 real matrix.



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