What is the derivation of the character formula for SU(2) representation?

[matematikawan]

I'm trying to understand this paper on the representation of SU(2).

I know these definitions:
A representation of a group G is a homomorphism from G to a group of operator on a vector space V. The dimension of the representation is the dimension of the vector V.
If D(g) is a matrix realization of a representation, the character \chi (g) is the trace of D(g).


The paper I'm reading state that the dimension of the representation is the character evaluated on the unit matrix. (***)

I try to confirm this with the character formula for SU(2) which is given as
\chi^j (\theta)=\frac{\sin(j+\frac{1}{2} )\theta}{\sin \frac{\theta}{2} }
where j labelled the irreducible representation.

So at unit matrix \chi (0) = 2j + 1 which is the correct dimension for the irreducible representation.

My question is how do we go about proving (***). I can't find the literature that proved this statement. Any clues ?

 

[ThirstyDog]

Think about the trace of the identity matrix. Count how many 1's it has down its diagonal.

 

[matematikawan]

The trace for the identity matrix (2j+1)X(2j+1) is 2j+1. That's easy!

Thank you so much ThirstyDog.

 

[matematikawan]

Really sorry I have to ask again. I'm already clear of my initial problem. My problem now is to understand the derivation for the character formula of SU(2) which is given by.
\chi^j (\theta)=\frac{\sin(j+\frac{1}{2} )\theta}{\sin \frac{\theta}{2} }

One book I'm reading now derived the above formula in the context of SO(3) as follows ( I think it should be ok because SU(2) and SO(3) share the same Lie algebra )

\chi^j (\theta)= \sum_m D^j[R_3(\theta)]_m^m <br /> = \sum_{m=-j}^{m=j} e^{-im\theta} <br /> <br /> =\frac{\sin(j+\frac{1}{2} )\theta}{\sin \frac{\theta}{2} }

I don't understand where does the exponential e^{-im\theta} comes from?

Again any clues for this?


I'm in a different time zone. It is about 2am now. I have to :zzz: and hope someone could help.