# Why symplectic symmetry has N(N+1)/2 generators?

1. Dec 30, 2011

### ndung200790

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.

2. Dec 31, 2011

### fzero

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 lets 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

$$N^2 - \left( \frac{N}{2}\right)^2 - 2 \frac{1}{2} \frac{N}{2}\left(\frac{N}{2} -1\right) = \frac{N(N+1)}{2}.$$