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I know for a group $O(N)$, $N$ is the dimensionality of the matrix corresponding to the transformation, but then for Lorentz transformation we say its $O(3,1)$ while the dimensions of transformation matrix is $4\times 4$. So my question is what these numbers stand for? in case of $O(3,1)$ I know that 3 is for three rotations and 1 is for the boost but in general what do they indicate? for example if we have a group $SU(2,2)$ what do 2 and 2 stand for? And what are the dimensions of the matrix that correspond to this representation?

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    $\begingroup$ In $O(3,1)$, the metric is $(1,1,1,-1)$ (or equivalently $(-1,-1,-1,1)$. In $SU(2,2)$ it is $(1,1,-1,-1)$. In general, for $SU(p,q)$ the metric will be diagonal and contain $+1$ $p$ times and $-1$ $q$ times. Note that neither $O(3,1)$ nor $SU(2,2)$ are compact so the $4\times 4$ realization of the group as matrices will not be unitary. $\endgroup$ – ZeroTheHero Apr 16 '17 at 18:00
  • $\begingroup$ @ZeroTheHero: why not flesh that out into an answer? $\endgroup$ – Kyle Kanos Apr 16 '17 at 18:01
  • $\begingroup$ @KyleKanos Done as requested. $\endgroup$ – ZeroTheHero Apr 16 '17 at 18:15
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In $O(3,1)$, the metric is $(1,1,1,-1)$ (or equivalently $(-1,-1,-1,1)$.) Thus $O(3,1)$ matrices preserve the length of vectors, with length defined by $$ x^2+y^2+z^2-t^2\, , $$ for the metric $(1,1,1,-1)$ (with obvious applications to special relativity).

In $SU(2,2)$ the metric is $(1,1,-1,-1)$ and transformations will preserve the (complex) inner product $$ xx^*+yy^*-zz^*-tt^*\, . $$

In general, for $SU(p,q)$ the metric will be diagonal and contain $+1$ $p$ times and $-1$ $q$ times, and will preserve the length of vectors defined by $$ \sum_{k=1}^p x_kx^*_k - \sum_{s=p+1}^{p+q} x_sx_s^*\, . $$

Note that neither $O(3,1)$ nor $SU(2,2)$ are compact so the $4\times 4$ realization of the group as matrices will not be unitary. The most obvious example of this observation is the boost matrix $$ L(\beta)=\left(\begin{array}{cccc} \cosh(\beta)&0&0&\sinh(\beta)\\ 0&1&0&0\\ 0&0&1&0\\ \sinh(\beta)&0&0&\cosh(\beta)\end{array}\right) $$ acting on the vector $(t,x,y,z)$, with $\beta$ is the rapidity parameter. Clearly here $L^{\dagger}(\beta)=L(\beta)\ne L(-\beta)$, with the latter being the inverse of $L(\beta)$.

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