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I would like to know the Lie algebra of the Lorentz group $SO(1, n)$. Can you tell me, what the answer is? Thank you in advance!

Niklas
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  • The matrix form of the answer depends on the choice of the matrix $J$ defining the quadratic form. The latter can be chosen in diagonal form $x_1^2-\sum_{j\ge 2}x_j^2$. But it is often convenient to use instead the form $x_1x_2+\sum_{j\ge 3}x_j^3$ (or alternatively $x_1x_n+\sum_{j=2}^{n-1}x_j^2$), because the 1-dimensional maximal split torus then can be chosen diagonal in this basis. In both cases the explicit matrix form of the Lie algebra can be computed using a block decomposition and the formula given in Tsemo Aristide's answer. – YCor Dec 11 '16 at 01:16

2 Answers2

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Consider the $n\times n$-matrix $J$ such that $a_{ij}=0$ if $i\neq j$, $a_{ii}=1, i<n, a_{nn}=-1$. $A\in SO(n,1)$ if and only if $A^tJA=I$. Thus the Lie algebra of $SO(n,1)=\{M\in M(n,R), M^tJ+JM=0\}$.

Tsemo Aristide
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Lie algebra of general $SO(N)$ group consists of skew-symmetric matrices with zero trace with ordinary matrix commutator as Lie bracket. You can easily deduce it by differentiating group conditions $A^TA = E$ and $\mbox{det}A = 1$ at zero: $d(A^TA) = dA^T(0) + dA(0) = 0$.

If it is necessary, look at more details about Lorentz group's Lie algebra on Wikipedia.

Hasek
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