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

[MathematicalPhysicist]

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.

 

[fzero]

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.

 

[MathematicalPhysicist]

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.

 

[fzero]

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.

 

[MathematicalPhysicist]

Ah, yes you're right. Thanks.