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

  • #1
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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.
 

Answers and Replies

  • #2
fzero
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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]##
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.
 
  • #3
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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.
 
  • #4
fzero
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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.
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.
 
  • #5
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Ah, yes you're right. Thanks.
 

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