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Group representations/branching rules question

  1. Mar 4, 2010 #1
    Hi,

    there exist tables with group theory breaking patters and branching rules such as
    [tex]E_7 \rightarrow E_6 \times U(1)[/tex]
    [tex]133 \rightarrow 78 + 1 + 27 + \bar{27}[/tex]
    [tex]56 \rightarrow 1 + 1 + 27 + \bar{27}[/tex]

    What is missing is the U(1) charge. Does anyone know how to obtain it?
    Perhaps it could be done by using a character formula but then I would need the characters of these representations.

    Any help would be appreciated.
     
  2. jcsd
  3. Mar 30, 2010 #2
    Hi Xino,

    I'm only just learning a lot of this myself, but since no one else answered....
    My naive way of getting maximal regular subgroups with U(1) factors follows from how it happens in field theory.

    If we want a single U(1) factor, then we want a element, h, in the Cartan subalgebra that commutes with all simple roots except for the one that we want to break - call that root ax (ie the one corresponding to the node on the Dynkin diagram that we want to delete) and its corresponding root vector (Lie algebra element) ex.
    We can normalise h so that ex has charge +q: [h,ex] = h(ex) = q ex and h(ei)=0 for i!=x.
    If we have the weight M = m1 a1 +...+ mx ax +...+ mn an then it's weight vector obeys
    h|M> = (m1 h(a1) +...+ mn h(an))|M> = q |M>
    So, to get the charge of any weight you just need to look at the coefficient of ex in its simple root expansion.

    I've got an incomplete Mathematica program for calculating weights (it's fairly simple to do in almost any language, there's old papers that give algorithms in ALGOL and Fortran...). It took me a while to get around to ironing out a couple of bugs - that's why my reply to your post is so late.

    My conventions for En are that the branch point is 3 from the end so we have the Dynkin diagram:

    Code (Text):

    (1)--(2)-...-(n-3)--(n-2)--(n-1)
                    |
                   (n)
     
    For the 56 of E7 I find the weights (written as Dynkin coefficients and arranged by layer)
    {1,0,0,0,0,0,0}
    {-1,1,0,0,0,0,0}
    {0,-1,1,0,0,0,0}
    {0,0,-1,1,0,0,0}
    {0,0,0,-1,1,0,1}
    {0,0,0,0,-1,1,1} , {0,0,0,0,1,0,-1}
    {0,0,0,0,0,-1,1} , {0,0,0,1,-1,1,-1}
    {0,0,0,1,0,-1,-1} , {0,0,1,-1,0,1,0}
    {0,0,1,-1,1,-1,0} , {0,1,-1,0,0,1,0}
    {0,0,1,0,-1,0,0} , {0,1,-1,0,1,-1,0} , {1,-1,0,0,0,1,0}
    {-1,0,0,0,0,1,0} , {0,1,-1,1,-1,0,0} , {1,-1,0,0,1,-1,0}
    {-1,0,0,0,1,-1,0} , {0,1,0,-1,0,0,1} , {1,-1,0,1,-1,0,0}
    {-1,0,0,1,-1,0,0} , {0,1,0,0,0,0,-1} , {1,-1,1,-1,0,0,1}
    {-1,0,1,-1,0,0,1} , {1,-1,1,0,0,0,-1} , {1,0,-1,0,0,0,1}
    {-1,0,1,0,0,0,-1} , {-1,1,-1,0,0,0,1} , {1,0,-1,1,0,0,-1}
    {-1,1,-1,1,0,0,-1} , {0,-1,0,0,0,0,1} , {1,0,0,-1,1,0,0}
    {-1,1,0,-1,1,0,0} , {0,-1,0,1,0,0,-1} , {1,0,0,0,-1,1,0}
    {-1,1,0,0,-1,1,0} , {0,-1,1,-1,1,0,0} , {1,0,0,0,0,-1,0}
    {-1,1,0,0,0,-1,0} , {0,-1,1,0,-1,1,0} , {0,0,-1,0,1,0,0}
    {0,-1,1,0,0,-1,0} , {0,0,-1,1,-1,1,0}
    {0,0,-1,1,0,-1,0} , {0,0,0,-1,0,1,1}
    {0,0,0,-1,1,-1,1} , {0,0,0,0,0,1,-1}
    {0,0,0,0,-1,0,1} , {0,0,0,0,1,-1,-1}
    {0,0,0,1,-1,0,-1}
    {0,0,1,-1,0,0,0}
    {0,1,-1,0,0,0,0}
    {1,-1,0,0,0,0,0}
    {-1,0,0,0,0,0,0}
    Rewrite the above as twice the coefficients in the simple root expansion (right multiply by two times the inverse of the Cartan matrix):
    {3,4,5,6,4,2,3}
    {1,4,5,6,4,2,3}
    {1,2,5,6,4,2,3}
    {1,2,3,6,4,2,3}
    {1,2,3,4,4,2,3}
    {1,2,3,4,2,2,3} , {1,2,3,4,4,2,1}
    {1,2,3,4,2,0,3} , {1,2,3,4,2,2,1}
    {1,2,3,4,2,0,1} , {1,2,3,2,2,2,1}
    {1,2,3,2,2,0,1} , {1,2,1,2,2,2,1}
    {1,2,3,2,0,0,1} , {1,2,1,2,2,0,1} , {1,0,1,2,2,2,1}
    {-1,0,1,2,2,2,1} , {1,2,1,2,0,0,1} , {1,0,1,2,2,0,1}
    {-1,0,1,2,2,0,1} , {1,2,1,0,0,0,1} , {1,0,1,2,0,0,1}
    {-1,0,1,2,0,0,1} , {1,2,1,0,0,0,-1} , {1,0,1,0,0,0,1}
    {-1,0,1,0,0,0,1} , {1,0,1,0,0,0,-1} , {1,0,-1,0,0,0,1}
    {-1,0,1,0,0,0,-1} , {-1,0,-1,0,0,0,1} , {1,0,-1,0,0,0,-1}
    {-1,0,-1,0,0,0,-1} , {-1,-2,-1,0,0,0,1} , {1,0,-1,-2,0,0,-1}
    {-1,0,-1,-2,0,0,-1} , {-1,-2,-1,0,0,0,-1} , {1,0,-1,-2,-2,0,-1}
    {-1,0,-1,-2,-2,0,-1} , {-1,-2,-1,-2,0,0,-1} , {1,0,-1,-2,-2,-2,-1}
    {-1,0,-1,-2,-2,-2,-1} , {-1,-2,-1,-2,-2,0,-1} , {-1,-2,-3,-2,0,0,-1}
    {-1,-2,-1,-2,-2,-2,-1} , {-1,-2,-3,-2,-2,0,-1}
    {-1,-2,-3,-2,-2,-2,-1} , {-1,-2,-3,-4,-2,0,-1}
    {-1,-2,-3,-4,-2,-2,-1} , {-1,-2,-3,-4,-2,0,-3}
    {-1,-2,-3,-4,-4,-2,-1} , {-1,-2,-3,-4,-2,-2,-3}
    {-1,-2,-3,-4,-4,-2,-3}
    {-1,-2,-3,-6,-4,-2,-3}
    {-1,-2,-5,-6,-4,-2,-3}
    {-1,-4,-5,-6,-4,-2,-3}
    {-3,-4,-5,-6,-4,-2,-3}

    The breaking you describe, E7 -> E6 * U(1), is what you get from deleting the first node in the Dynkin diagram. So we care about how many a1's are in each weight and read off the charges
    3
    1
    1
    1
    1
    1 , 1
    1 , 1
    1 , 1
    1 , 1
    1 , 1 , 1
    -1 , 1 , 1
    -1 , 1 , 1
    -1 , 1 , 1
    -1 , 1 , 1
    -1 , -1 , 1
    -1 , -1 , 1
    -1 , -1 , 1
    -1 , -1 , 1
    -1 , -1 , -1
    -1 , -1
    -1 , -1
    -1 , -1
    -1 , -1
    -1
    -1
    -1
    -1
    -3
    The weights with charge ±3 are {±1,0,0,0,0,0,0} and thus both correspond to the trivial E6 rep {0,0,0,0,0,0}. There are 27 weights with charge +1 and 27 weights with charge -1. You can check that they exactly match the E6 reps {1,0,0,0,0,0} and {0,0,0,0,1,0} resp.

    You can do the same thing with the 133 dimensional adjoint rep of E7.
    The 1 + 78 of E6 have U(1) charge 0.
    The 27 and \bar{27} have charges -1 and +1 respectively (but that can be swapped by a different choice of normalisation of h)

    Hope that helps,
    Simon

    PS. Sage can also calculate branching rules:
    http://www.sagemath.org/doc/reference/sage/combinat/root_system/weyl_characters.html
    http://sporadic.stanford.edu/bump/trac_8442/
     
    Last edited: Mar 30, 2010
  4. Mar 30, 2010 #3

    garrett

    User Avatar
    Gold Member

    Nicely explained, Simon.
     
  5. Mar 30, 2010 #4
    Thanks Garrett,

    It's good to know that someone who's had experience with these calculations didn't see anything too bad in what I wrote. Any hints on how to simplify my approach are welcome, as I'll be doing a lot of these calculations in the near future.
     
  6. Mar 31, 2010 #5
    Hi,

    first of all, many thanks for taking your time and writing such a helpful and extensive answer. In the meantime I had come up with a different (and less elegant way) to get the same results except for a relative charge normalization.

    My way of doing this was as follows. I set up a projection matrix which projects weights of E7 to weights of E6. The projection of the 56 contains among others the 27 of E6. It is clear that all weights of the 27 must have the same charge which can be obtained (similar to the way you do it) as a scalar product of the weight and a to be determined charge vector. By demanding that the scalar product of each of the 27 weights with some charge vector gives 1 (that is how I normalized the charge), I can solve for the charge vector. Given this charge vector, I can compute the charge of any weight. This is essentially what you do, just backwards.

    In case anyone is interested, my Projection matrix is:
    P={{0, 0, 0, 0, 1, 1, 0},
    {0, 0, 0, 1, 0, 0, 0},
    {0, 1, 1, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 1},
    {0, 0, 1, 1, 1, 0, 0},
    {1, 0, 0, 0, 0, 0, 0}}

    (I set it up according to the conventions in the Slansky paper "Group theory for unified model building". That means it is defined by requiring that highest weights are projected on highest weights.)

    The charge vector is q={0, 0, 2, 1, 0, 1, 1}.
    So for example the weight w={0, 0, 0, 0, 0, 1, 0} is projected to P.w={1,0,0,0,0,0} which is a 27 weight of E6.
    q.w=1 so it has charge 1.

    These results differ from yours in that I use a different convention for the Cartan Matrix. (My Dynkin diagram is reflected along the vertical. When trying to reproduce your results I therefore at first removed a wrong node and got the breaking to SO(12) instead...)

    One note: Charge normalization does matter; not absolute but relative. The representations you get from breaking to the 56 should have the same normalization as the ones branching off from the 133.

    Best wishes!
     
  7. Apr 3, 2010 #6
    PS: I should acknowledge that I arrived at the posted summary after private correspondence with Simon.
     
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