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Character of a representation

  1. Mar 23, 2009 #1
    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 [tex]\chi (g)[/tex] 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
    [tex]\chi^j (\theta)=\frac{\sin(j+\frac{1}{2} )\theta}{\sin \frac{\theta}{2} }[/tex]
    where j labelled the irreducible representation.

    So at unit matrix [tex]\chi (0) = 2j + 1[/tex] 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 ?
     
  2. jcsd
  3. Mar 23, 2009 #2
    Think about the trace of the identity matrix. Count how many 1's it has down its diagonal.
     
  4. Mar 23, 2009 #3
    The trace for the identity matrix (2j+1)X(2j+1) is 2j+1. That's easy!

    Thank you so much ThirstyDog.
     
  5. Mar 23, 2009 #4
    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.
    [tex]\chi^j (\theta)=\frac{\sin(j+\frac{1}{2} )\theta}{\sin \frac{\theta}{2} }[/tex]

    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 )

    [tex]\chi^j (\theta)= \sum_m D^j[R_3(\theta)]_m^m
    = \sum_{m=-j}^{m=j} e^{-im\theta}

    =\frac{\sin(j+\frac{1}{2} )\theta}{\sin \frac{\theta}{2} }[/tex]

    I don't understand where does the exponential [tex] e^{-im\theta} [/tex] 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.
     
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