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
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tensor
SU
SL
GL