SO(n) = orthogonal matrix with determinant 1.
SO==special orthogonal

Dim(SO(n))=n(n-1)/2

For X a defining rep of SO(3):

(Xab)ij=delta(i,a)delta(j,b)-delta(j,a)delta(i,b)

=>Commutation relations:

[Xab,Xgd]=delta(b,g)Xad-delta(a,d)Xbg-delta(a,g)Xbg+delta(b,d)Xad

Xab=eabgLg for eabg=antisymmetric tensor.

SO(4) is special. It is the only SO(n) which decomposes. SO(4) ~= SO(3) x SO(3)

Decomposition:

Xij=-Xji=SO(3) sub algebra

Define Lk as: Xij=i/2*fijkLk

Xi4=-X4i

Define Ki as:-iXij=Ki

Define J(1)=1/2(L+K) (SU(2) group)

Define J(2)=1/2(L-K) (Another SU(2) group)
Note that [J(1)i,J(2)i]=0

=> D(j1,j2) labeling ok.

For arbitrary D(j1,j2) j1+j2=integer => rep exists.

Example Of SO(4) symmetry in nature:

SO(4) is equivalent to the symmetries of a planet orbiting a star or
electron around a hydrogen atom (classically)
Define: M=1/2m(PxL - LxP)-kX/r where

- X= vector of current direction
- P=current momentum
- L=current angular momentum
- m = reduced mass of system Note that L*M=0 and [M,H]=0

S0(2) = rotational symmetry of a circle. G={g(theta)|0<=theta<2Pi}

g(th1)g(th2)=g(th1+th2) if th1+th2<2Pi, g(th1+th2-2Pi) if th1+th2=>2Pi

Defining representation:

g(th)=

|cos(th) -sin(th)|

|sin(th) cos(th)|

=D(th)

Transpose(D)D=Id.=> D real and unitary.

This is an abelian group so its irreps must be 1-d.

M^DM=D~=

|e^ith 0 |

|0 e^-ith|

with M=1/sqrt(2)*

|1 -i|

|-i 1|

Irreps of SO(2): Dm(th)=e^imth for m any integer.

D=D1+D-1

D0 is the trivial rep.

for M!=0 g(th)-> Dm(th) is |m| to 1 mapping.
only D1 and D-1 are faithful.

Integral(G,dg)=1. Here dg=dth/2Pi.

Generator:

D(dth)=

|1 -dth|

|dth 1 |

+O(dth^2)

=Id-idthJ+O(dth^2)

where J=

| 0 -i |

| i 0 |

Note that adjoint(J)=J.

characters are orthogonal.

Integral(G,Xm*(g)Xn(g)dg)=delta(m,n)

<=> Integral(0,2Pi,e^imth * e^-inth dth/2Pi)=delta(m,n)

Completeness:

f(g)=Sum(-infinity,infinity,ame^-imth)

f(th)=Sum(m,cme^-imth) whre cm=Integral(0,2Pi,e^imth * f(th)dth/2Pi)

This is basically the Fourier series.

SO(1,1)=

|cosh t sinh t|

|sinh t cosh t|

-infinity<t<infinity

For SO(2) x'^2+y'^2=x^2+y^2

for S0(1,1) x'^2-y'^2=x^2-y^2

SO(1,1) -> Fourier transform

The conjugate rep. to a rep in SO(m) is found by filling in the missing spaces in a young diagram.

For example the conjugate to:

### #

is:

## ###

source

jl@crush.caltech.edu index

Linear_Transformation_Group

group_generation

euclidean_group

poincare_group

covering_group

unitary_group

lie_algebra

lie_group

adjoint

tensor

SU

SL

GL