Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why are the casimirs independent of the representation

  1. May 5, 2009 #1
    Question in the title, ie why is [tex] Tr(T_{a_1}T_{a_2}...T_{a_n}) [/tex] independent of which representation we choose, where the Ts are a matrix representation of the group generators.
     
  2. jcsd
  3. May 5, 2009 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's not true.

    For example, if you pick a positive integer m, there is a representation where every group element has trace m: this representation sends every group element to the identity matrix acting on an m-dimensional space.

    And given any matrix representation [itex]\rho[/itex] of a group, one can construct a new representation [itex]\rho'[/itex] by

    [tex]
    \rho'(g) = \left[ \begin{matrix}{\rho(g) & 0 \\ 0 & \rho(g)} \end{matrix} \right]
    [/tex]

    and under this representation, [itex]Tr\, \rho'(g) = 2 Tr \, \rho(g)[/itex].



    You're making some extra, relevant assumptions -- what are they?
     
  4. May 5, 2009 #3
    Ah I see. But couldnt we also have

    [tex]
    \rhosingle-quote(g) = \left[ \begin{matrix}{\rho(g) & 0 \\ 0 & 0} \end{matrix} \right]
    [/tex]

    as a higher dimensional rep with the same trace?

    I think what I meant is that given a 2 different matrix reps of the same dimension, the casimirs are the same. But they can change when you change the dimension of the rep, thus they can be used to label reps. Thanks
     
  5. May 6, 2009 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Different representations with the same dimension can still have that trace be different. For example, consider

    [tex]\rho'(g) = \rho(g)^2[/tex]

    and

    [tex]\rho'(g) = (\det g)^{-1}\rho(g)[/tex]




    Anyways, I went to look up Casimir on wikipedia; if that's what you're talking about, then I think you are misunderstanding things. The Casimir invariant is an element of the universal enveloping algebra of the Lie algebra, and its representation under [itex]\rho[/itex] is a matrix.

    Wikipedia states that by Schur's lemma, for any irreducible representation, [itex]\rho(\Omega)[/itex] is proportional to the identity matrix. That constant of proportionality can be computed with a formula involving traces; maybe that's what you're thinking of?
     
  6. May 6, 2009 #5
    Yeah I think Im confusing things a little.

    Casimirs can be constructed from the generators by:

    [tex]
    d_{a_1 a_2...a_n} = Tr(T_{a_1}T_{a_2}...T_{a_n})
    [/tex]

    But is is their eigenvalues that label the reps. So I think what I really meant to ask is why can the eigenvalues of the casimirs can be used to label reps of different dimension.

    Sorry for my confusion
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why are the casimirs independent of the representation
  1. Casimir operators (Replies: 5)

Loading...