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

The Cartan matrix in order to find the roots.

  1. Dec 10, 2012 #1
    Hi All,

    I've been following a group theory course which I am struggling with at the moment. I'm from 3 different books (Georgi, Cahn and also Jones). I'm trying to understand section 8.7 and 8.9 in the book by Georgi.

    I (think I) understand that any pair of root vectors of a simple Lie algebra has an SU(2) subalgebra (as Georgi explains on page 93).

    And due to equation 8.47, 8.48, and 8.49 (sorry I don't know how to make equations in Latex), we can say (I think) that the off-diagonal elements of the 2x2 cartan matrix are equal to 2*m, where m denotes the projection of the total angular momentum (see equation 3.23).

    Therefore, if we look at the illustration on page 118 which illustrates how to find all the roots for G_2, we see that the simple roots are represented by the boxes alpha_1 == (2 -1) (thus m_1 = -1/2) and alpha_2 == (-3 2) (thus m_2 = -3/2).

    I think that until here I understand it (please correct me if I'm wrong).

    And then Georgi writes: "We know that the simple roots alpa_1 and alpha_2 are the highest weights of spin 1 representations for their respectives SU(2)s.

    Why is this the case? As far as I know they are simple roots and thus they are the lowest weights? And I don't understand what he means with spin 1 representation? I've look at equation 3.31 but that doesn't clarify it for me.

    Now he says m_1 is the bottom of a doublet and m_2 is the bottom of the quartet. I'm not sure why you can just assume that's the case?
  2. jcsd
  3. Dec 11, 2012 #2
    Spin 1 representation: In case of su(2), the spin algebra, there is exactly one irrep per dimension. The 1-dim irrep corresponds to heighest weight j=0, i.e. it describes a spin 0 system and is therefore called the spin 0 representation. The 2-dim irrep corresponds to spin 1/2, the 3-dim irrep to spin 1, and so on. In general dimension of irrep = 1+2j.
  4. Dec 11, 2012 #3
    Hi CuriosusNN,

    Thanks for your reply. I understand what you're saying but I can't see that how the simple roots are a 3 dimensional representation?

    In fact, I thought that alpha_1 would have highest weight j = 1/2 (because it is a doublet) and alpha_2 would have highest weight j = 3/2 (because it is a quartet).

    Any help would be much appreciated because I'm worried that I'm fundamentally misunderstanding something.

  5. Dec 11, 2012 #4
    I haven't yet had time to dig deeper myself, but I think something is completely wrong with your assumption that the Cartan Matrix should be dependent on the Jz quantum number m. This cannot be true since the Cartan matrix is independent of any choice of representation. However, the allowed range of m is restricted by the chosen representation of su(2) labelled by j, i.e. -j ≤ m ≤ j, and even in a given representation we have states of different m which can each be reached by operating with the step operators E_+-\alpha up or down in the range from -j to j.
  6. Dec 11, 2012 #5
    Now I'm thoroughly confused...

    We know that for SU(2):

    J_3 acting on the "state" m gives "eigenvalue" m (equation 3.7):

    And that is exactly how he construct the Cartan matrix (see equation 8.47).

    And I don't see how else he constructs that root diagram on page 118?

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

Similar Discussions: The Cartan matrix in order to find the roots.
  1. Order of a Matrix (Replies: 1)