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Roots and root vectors of sp(4,\mathbb{R})

  1. Sep 14, 2013 #1

    MathematicalPhysicist

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    I found that the cartan subalgebra of ##sp(4,\mathbb{R})## is the algebra with diagonal matrices in ##sp(4,\mathbb{R})##.

    Now to find out the roots I need to compute:

    ##[H,X]=\alpha(H) X##

    For every ##H## in the above Cartan sublagebra, for some ##X \in sp(4,\mathbb{R})##

    Now, I know that ##X## is of the form:

    ##\left[ {\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14}\\
    a_{21} & a_{22} & a_{23} & a_{13} \\
    a_{31} & a_{32} & -a_{22} & -a_{12}\\
    a_{41} & a_{31} & -a_{21} & -a_{11}\\
    \end{array}} \right]##

    So if I take ##H=diag(h_{11},h_{22},h_{33},h_{44})##, I am getting the next equality:

    ##\left[ {\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14}\\
    a_{21} & a_{22} & a_{23} & a_{13} \\
    a_{31} & a_{32} & -a_{22} & -a_{12}\\
    a_{41} & a_{31} & -a_{21} & -a_{11}\\
    \end{array}} \right] = \left[ {\begin{array}{ccccc} 0 & (h_{11}-h_{22})a_{12} & (h_{11}-h_{33})a_{13} & (h_{11}-h_{44})a_{14}\\
    (h_{22}-h_{11})a_{21} & 0 & (h_{22}-h_{33})a_{23} &(h_{22}-h_{44}) a_{13} \\
    (h_{33}-h_{11})a_{31} & (h_{33}-h_{22})a_{32} & 0 & (h_{44}-h_{33})a_{12}\\
    (h_{44}-h_{11})a_{41} & (h_{44}-h_{22})a_{31} & (h_{33}-h_{44})a_{21} & 0\\
    \end{array}} \right]##

    Which means that the roots should be ##h_{11}-h_{44} , h_{22}-h_{33} , h_{33}-h_{22}, h_{44}-h_{11}##, and accodingly the root vectors are:
    ##\left[ {\begin{array}{ccccc}0 & 0 & 0 & a_{14}\\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{array}} \right],\left[ {\begin{array}{ccccc}0 & 0 & 0 & 0\\
    0 & 0 & a_{23} & 0 \\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{array}} \right],\left[ {\begin{array}{ccccc}0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 \\
    0 & a_{32} & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{array}} \right], \left[ {\begin{array}{ccccc}0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0\\
    a_{41} & 0 & 0 & 0\\
    \end{array}} \right]## respectively.

    Is this right, or did I forget something?

    Thanks in advance.
     
  2. jcsd
  3. Sep 14, 2013 #2

    fzero

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    What symplectic form are you using? For the standard one,

    $$ \Omega = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix},$$

    a matrix ##X## in ##sp(2n,\mathbb{R})## must satisfy ##\Omega X + X^T \Omega =0##, and hence is of the form

    $$ X = \begin{pmatrix} A & B \\ C & -A^T \end{pmatrix}, ~~~B^T = B,~~~C^T = C.$$

    Your representative doesn't look at all like this, but I can't discount that there is not some other ##\Omega## for which it is reasonable.

    Also, ##sp(4,\mathbb{R})## has rank 2, so there should should be 2 simple roots. You should not end up with 4 linearly independent root vectors.
     
  4. Sep 14, 2013 #3

    MathematicalPhysicist

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    I am using the next form:
    ##\Omega = \begin{pmatrix} 0 & T_n \\ -T_n & 0 \end{pmatrix}##

    Where ##T_n## is the matrix with 1 in the (i,n-i+1) entry and zero in the rest.
     
  5. Sep 14, 2013 #4

    fzero

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    OK, so it looks like ##T_2 = \sigma_1##, in which case, I agree with your ##X##. The rest looks ok, but you should note that we always have ##h_{33}=-h_{22}, h_{44} = -h_{11}##. Then you will find 2 simple roots. It might also be easiest to pick an explicit basis for the Cartan subalgebra to simplify some computations.
     
  6. Sep 15, 2013 #5

    MathematicalPhysicist

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    Ah, yes you're right. Thanks.
     
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