How can I prove that $O(3,1)$ and $O(1,3)$ are the same group?

  • $\begingroup$ Just for clarification: do you mean $SO(3,1)$ and $SO(1,3)$ instead of $O(3,1)$ and $O(1,3)$? $\endgroup$
    – Hunter
    Feb 8, 2014 at 12:59
  • 1
    $\begingroup$ Not necessarily. $\endgroup$
    – rainman
    Feb 8, 2014 at 13:03

2 Answers 2

  1. Define the Lie group
    $$\tag{1} O(p,q)~:=~ \{\Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) ~|~\Lambda^T\eta\Lambda= \eta \} $$ of pseudo-orthogonal real matrices $\Lambda$ for the metric $$\tag{2} \eta_{\mu\nu}~=~{\rm diag} (\underbrace{+1,\ldots,+1}_{p~\text{times}},\underbrace{-1,\ldots, -1}_{q~\text{times}}), \qquad n~=~p+q.$$

  2. The groups $O(p,q)=O(q,p)$ are equal since the overall sign of the metric $\eta_{\mu\nu}\to -\eta_{\mu\nu}$ does not matter in the definition $$\tag{3}\Lambda^T\eta\Lambda~=~ \eta. $$


The matrices $M$ in $O(3,1)$ and $O(1,3)$ are defined by the condition $$ M G M^T = G $$ for $$ G=G_{1,3} ={\rm diag} (1,1,1,-1)\text{ and } G=G_{3,1} = {\rm diag} (1,-1,-1,-1)$$ respectively. I use the convention where the first argument counts the number of $+1$'s in the metric tensor and the second one counts the negative numbers $-1$ that follow.

But these two groups only differ by a permutation of the entries.

First, note that it doesn't matter whether we have a "mostly plus" or "mostly minus" metric. If you change the overall sign of the metric via $G\to -G$, $MGM^T = G$ will remain valid.

Second, the two groups only differ by having the "different signature coordinate" at the beginning or at the end. But it may be permuted around. If $M$ obeys the first condition of $O(1,3)$, $MG_{1,3}M^T =G_{1,3}$, you may define $$ M' = P M P^{-1} $$ where $P$ is the cyclic permutation matrix $$ P = \pmatrix{0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0}$$ and it is easy to see that $M'$ will obey $$ M' G_{3,1} M^{\prime T} = G_{3,1} $$ simply because $$ M' G_{3,1} M^{\prime T} = PMP^{-1} G_{3,1} P^{-1T} M^T P^T $$ but $P^{-1}=P^T$ and, crucially, $$ P^{-1} G_{3,1} P = -G_{1,3} $$ So all the $P$'s will combine or cancel and one gets it.

One should try to get through every step here but the reason why the groups are isomorphic is really trivial: they are the groups of isometries of the "same" spacetime, one that has three spacelike and one timelike dimension, and they only differ by the convention how we order the coordinates and whether we use a mostly-plus or mostly-minus metric. But such changes of the notation don't change the underlying "physics" so the groups of symmetries of the "same" object parameterized differently must be isomorphic.

  • 2
    $\begingroup$ Alhamdulillah! Great! $\endgroup$
    – rainman
    Feb 8, 2014 at 13:19

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