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8/10/2019 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
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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
2 The Heisenberg group
3 The Heisenberg Lie algebra
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
2 The Heisenberg group
3 The Heisenberg Lie algebra
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
2 The Heisenberg group
3 The Heisenberg Lie algebra
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
2 The Heisenberg group
3 The Heisenberg Lie algebra
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
.
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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)
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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.
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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
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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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