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

  1. Dec 30, 2011 #1
    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. jcsd
  3. Dec 31, 2011 #2


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    A symplectic matrix is an [itex]N \times N[/itex] matrix [itex]M[/itex] ([itex]N[/itex] must be even) that satisfies

    [tex] M^T \Omega M = \Omega ,~~\Omega =\begin{pmatrix} 0 & I_{N/2} \\ -I_{N/2} & 0 \end{pmatrix},~~~(*) [/tex]

    where [itex]I_{I/2}[/itex] is the [itex]N/2[/itex] dimensional unit matrix. In terms of a parameterization into [itex]N/2 \times N/2[/itex] matrices,

    [tex] M = \begin{pmatrix} A& B \\ C & D \end{pmatrix}, [/tex]

    (*) becomes the three equations

    [tex] A^T D - C^T B = I,~~ A^T C = C^T A,~~ D^T B = B^T C.[/tex]

    The first of these lets us solve for, say [itex]D[/itex] in terms of the other three, so it is equivalent to [itex](N/2)^2[/itex] conditions on [itex]M[/itex]. The other equations demand that [itex] A^T C [/itex] and [itex]D^T B[/itex] are symmetric matrices. They therefore determine half of the off-diagonal components of these objects and are each equivalent to

    [tex] \frac{1}{2} \frac{N}{2}\left(\frac{N}{2} -1\right)[/tex]

    conditions on [itex]M[/itex].

    The number of independent components of the symplectic matrix [itex]M[/itex] is therefore

    [tex] 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}.[/tex]
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