ndung200790 said:
Please teach me this:
Why Sp(N) symmetry has N(N+1)/2 generators?(QFT of Peskin and Schroeder).
Thank you very much for your kind helping.
A symplectic matrix is an N \times N matrix M (N must be even) that satisfies
M^T \Omega M = \Omega ,~~\Omega =\begin{pmatrix} 0 & I_{N/2} \\ -I_{N/2} & 0 \end{pmatrix},~~~(*)
where I_{I/2} is the N/2 dimensional unit matrix. In terms of a parameterization into N/2 \times N/2 matrices,
M = \begin{pmatrix} A& B \\ C & D \end{pmatrix},
(*) becomes the three equations
A^T D - C^T B = I,~~ A^T C = C^T A,~~ D^T B = B^T C.
The first of these let's us solve for, say D in terms of the other three, so it is equivalent to (N/2)^2 conditions on M. The other equations demand that A^T C and D^T B are symmetric matrices. They therefore determine half of the off-diagonal components of these objects and are each equivalent to
\frac{1}{2} \frac{N}{2}\left(\frac{N}{2} -1\right)
conditions on M.
The number of independent components of the symplectic matrix M is therefore
N^2 - \left( \frac{N}{2}\right)^2 - 2 \frac{1}{2} \frac{N}{2}\left(\frac{N}{2} -1\right) = \frac{N(N+1)}{2}.
