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Director product expansion of Lie groups.

  1. Aug 9, 2011 #1
    For discrete groups, we can easily find the decomposition of the direct product of irreducible representations with the help of the character table. All we need to do is multiply the characters of the irreducible representations to get the characters of the direct product representation and then use the orthogonality relations to find the multiplicity of each of the irreps in the direct product representation.

    Is there an easy way to do something similar for Lie groups? Let two representations of a Lie group be given in terms of their highest weights. I am wondering if there is a straightforward and easy technique to find the decomposition of the direct product of those representations into a direct sum of irreducible representations.
    For example for SU(3) group how can we calculate
    3x8=15+6bar+3
    or in terms of the highest weights
    (1,0)x(1,1)=(2,1)+(0,2)+(1,0) ?
     
  2. jcsd
  3. Aug 10, 2011 #2
    For SU(N) the reps can be thought of as tensors (for a physicist, this means indexed objects) and the decomposition to irreps is equivalent to the decomposition of the tensor into its symmetric and antisymmetric parts.

    The basic reason that all SU(N) representations are just multi-indexed objects is because all representations of SU(N) can be constructed from decomposition of products of the fundamental representation, which is a single indexed tensor.

    The fully antisymmetrized N-indexed tensor consists of just a single state - so is a singlet / trivial rep.
    The fully antisymmetrized (N-1)-indexed tensor has N states and corresponds to the anti-fundamental rep.

    The main calculational tool that is used for products of SU(N) is http://en.wikipedia.org/wiki/Young_tableau" [Broken].
    Each Young tableau corresponds to a SU(N) irrep.
    Boxes in Young tableaux correspond to indices in the tensors.
    Symmetrized indices are rows of boxes and antisymmetrised indices are columns.

    Google "young tableaux su(n) product" to find notes that tell you more.
    (e.g. www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter9.pdf)
    Also, Georgi's book "Lie algebras in particle physics" has a good discussion of both tensor and tableau and other methods of computing product decompositions.

    The simplest example is SU(3): 3 * 3 = 6 + \bar{3}
    Code (Text):

    0 * 0 = 00 + 0
                 0
     
    And: 3 * \bar{3} = 8 + 1
    Code (Text):

    0 * 0 = 00 + 0 = 00 + .
        0   0    0   0
                 0
     
    And: 8 * 3 = 15 + \bar{6} + 3
    Code (Text):

    00 * 0 = 000 + 00 + 00 ~= 000 + 00 + 0
    0      = 0   + 00 + 0     0   + 00
                        0
     
    You can check the dimensionalities: e.g.
    00 = 6 because it has the numberings {11, 12, 13, 22, 23, 33}, (these are the independent terms in the symmetric tensor S_{ij}=S_{ji}, i,j=1,2,3).
    0;0 = \bar{3} because it has the numberings {1;2 , 1;3 , 2;3} (these are the independent terms in the antisymmetric tensor A_{ij}=-A_{ji}, i,j=1,2,3).
    00;0 = 8 because it has the numberings {11;2 , 12;2 , 13;2 , 11;3 , 12;3 , 13;3 , 22;3 , 23;3}

    ---

    In other groups, not all representations can be constructed from products of a single basic rep. In which case, if you want to generalize the above, you can have tensors with more than one-type of index (i.e. that eat vectors that lie in different spaces). Or you can use more powerful methods. http://phyweb.lbl.gov/~rncahn/www/liealgebras/book.html" [Broken] is a good starting point.

    The naive method is you can read off the highest weight vector in the product: it's just the sum of the highest weight vectors in the irreps that you're multiplying. You then calculate all of the weights in the product, remove all of the weights in the new highest weight rep and look for the highest remaining weight vector. Subtract that rep and continue until you've accounted for all weights.

    For the more sophisticated methods, you should read Cahn, Georgi or other books.
     
    Last edited by a moderator: May 5, 2017
  4. Aug 10, 2011 #3
    Thank you very much for the very detailed reply.

    I read Georgi's book and have understood how to use the Young Tableaux to find the direct product expansion. But I just want to clear my understanding of how the Tableaux represent a state. For example if we have a tableaux with a numbering 123;33, how to symmetrise and antisymmetrise it?
    First symmetrise the first row. There are 6 permutations.
    123;33 + 132;33 + 321;33 + 213;33 + 231;33 + 312;33
    Now take each of the terms from above and symmetrise the second row(here it is trivial)
    Then take each of that and antisymmetrise(+ for even and - for odd permutations) the first column(here only one permutation which is odd)
    123;33 - 323;13 + ...
    Then take the second column and antisymmetrise it also
    123;33 - 133;32 -(323;13-333;12) + ...

    We take rows one by one and symmetrise, then take columns one by one and antisymmetrise. Am I right about the stuff wrote above?

    "The naive method is you can read off the highest weight vector in the product: it's just the sum of the highest weight vectors in the irreps that you're multiplying."
    - understood. For example in (1,0)x(1,1), we have (2,1) as the highest weight.

    " You then calculate all of the weights in the product, remove all of the weights in the new highest weight rep and look for the highest remaining weight vector. Subtract that rep and continue until you've accounted for all weights."
    Didn't quite get it. Can you plz explain it a bit more, may be with the example (1,0)x(1,1)=(2,1)+(0,2)+(1,0) ?

    By the way thanks for the link to Cahn's book.
     
  5. Aug 10, 2011 #4
    No problems. It's good for me to revise this stuff every so often.

    The tableau (tableaux is the plural) is automatically (a)symmetrised. This is why you can restrict yourself to strictly increasing sequences in the columns and non-decreasing sequences in the rows.
    But if we take the numbers you gave as representing the different states of a system (or indices on the tensor), then yes, what you said seems basically correct.

    Each proper numbering of the tableau corresponds to a particular state. That state is the particular (anti)symmetric combination of a tensor product of fundamental states that you get with the algorithm you outlined. Of course, you only need to do the (anti)symmetrization once with a symbolic state an use that result for all states. SU(2) is easiest, because any pair of antisymmetrized indices/states is equivalent to the singlet/trivial state so you only need to worry about symmetrization. I've https://www.physicsforums.com/attachment.php?attachmentid=37904&stc=1&d=1313025840" on this.

    Denoting the symmetries of an arbitrary tensor is not easy and in fact, Young Tableaux are often used for this. See the computer tensor algebra packages xAct and Cadabra, which both use the index canonicalization algorithm given in http://arxiv.org/abs/0803.0862.

    The action of a Lie group element, G, on the direct product of elements, L1 and L2, in the two reps, R1 and R2, is
    G(L1 x L2) = G(L1) x G(L2).
    So the action of a Lie algebra element is a derivative
    A(L1 x L2) = A(L1) x L2 + L1 x A(L2).

    Now assume that L1 and L2 are weight vectors of weights l1 and l2 respectively. Then L1 x L2 is a weight vector since
    H(L1 x L2) = H(L1) x L2 + L1 x H(L2) = l1 L1 x L2 + l2 L1 x L2 = (l1 + l2) L1 x L2
    Now you can generate all weights of R1 x R2 by calculating all of the product weights - although I think that the multiplicities have to be calculated separately - see Cahn.

    Let's take it as a fact that not all weights can be reached from any other weight in the product representation can be obtained using the raising and lowering operators of the Lie algebra. I think that this is basically a consequence of Schur's lemma (but don't quote me on this) and is the fundamental reason that semisimple Lie algebra reps decompose into direct sums. Then all we need to do is find the irreps in the product rep and we don't need to prove that they form a direct sum. Using the fact that irreps are 1:1 with highest weight vectors, we just:
    (1) find the highest weight vector in the list of weights we computed.
    (2) Compute the weights in the irrep corresponding to that highest weight and remove them from the list.
    (3) If the list is empty then stop, else go to (1).

    I can't be bothered doing the detailed calc for the SU(3) (1,0)x(1,1)=(2,1)+(0,2)+(1,0) problem. It's long, but not too hard.
    This naive method is helped if you have a computer algebra system that can do the grunt work. See for example the post branching rules: https://www.physicsforums.com/showthread.php?p=2609024
    Also the links and discussions on
    https://www.physicsforums.com/showthread.php?t=457338
    and
    https://www.physicsforums.com/showthread.php?t=517180
    but probably using the more advanced/sophisticated methods given in textbooks it better in the long run.
     

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  6. Aug 10, 2011 #5
    I forgot to mention, that if you only have a few simple calculations to do, then the http://www-math.univ-poitiers.fr/~maavl/LiE/form.html" [Broken] is really handy.

    For the (1,0)x(1,1) decomposition of su(3)~A2 example:
    1) select "Tensor product decomposition" and "A, 2", click proceed.
    2) Set the highest weights to [1,0] and [1,1], click start.
    Output: 1X[2,1] +1X[0,2] +1X[1,0]
    ie (2,1)+(0,2)+(1,0).

    http://davidsd.org/2010/03/lie-group-computations-with-python/" [Broken].
     
    Last edited by a moderator: May 5, 2017
  7. Aug 18, 2011 #6
    Now I think the method of symmetrization/antisymmetrization I wrote earlier was not correct. From Georgi's book I get the following impression. For example in the Young Tableau 1232;23 first two columns have two elements each. We need to symmetrize them, taking the column as a single unit. The last two columns have one element only. We symmetrize them also. (I earlier thought that I need to symmetrize among all the elements with in a row which was wrong)

    Below I use the Young Tableau notation to denote tensors

    symmetrizing, taking the first two columns as a single unit

    1232;23 + 2132;32

    Now symmetrizing the last two columns

    1232;23 + 1223;23 + 2132;32 + 2123;32

    Now we antisymmetrize with in each column separately.

    1232;23 - 2232;13 - 1332;22 + 2332;12

    + ....

    So I guess for SU(N) we just generalize the above. For SU(4), take columns with the three elements and symmetrize them, taking a column as a single unit. Take columns with the two elements and symmetrize them, again taking a column as a single unit. Take columns with a single element and symmetrize them. Finally antisymmetrize with in each column.
     
  8. Aug 18, 2011 #7
    "The action of a Lie group element, G, on the direct product of elements, L1 and L2, in the two reps, R1 and R2, is
    G(L1 x L2) = G(L1) x G(L2).
    So the action of a Lie algebra element is a derivative
    A(L1 x L2) = A(L1) x L2 + L1 x A(L2).

    Now assume that L1 and L2 are weight vectors of weights l1 and l2 respectively. Then L1 x L2 is a weight vector since
    H(L1 x L2) = H(L1) x L2 + L1 x H(L2) = l1 L1 x L2 + l2 L1 x L2 = (l1 + l2) L1 x L2
    Now you can generate all weights of R1 x R2 by calculating all of the product weights - although I think that the multiplicities have to be calculated separately - see Cahn."

    I never realized these facts, even though I used all these implicitly when constructing Gellmann's meson octets etc from fundamental reps! Thank you very much.
     
  9. Aug 18, 2011 #8
    I also understand how you take the highest weight, then lower it and find all the weights reachable using the lowering/raising operators, then look for the highest weight left in the remaining space and continue the process...

    By the way, the web interface to LiE is cool. Thanks again :-)
     
  10. Aug 20, 2011 #9
    No probs, glad I could help!
     
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