Denote by $\text{SO}_0(1,4)$ the identity component of the special linear isometry group $\mathrm{SO}(1,4)$ of the LorentzMinkowski space $\mathbb{R}_1^5$, that is, of $$\text{SO}(1,4)=\left\{X\in\text{SL}(5,\mathbb{R})\mid X^tI_{1,4}X=I_{1,4}\right\},\quad\text{where}\;I_{1,4}:=\text{diag}(1,1,1,1,1).$$ Is there any explicit matrix expression for a general element of this Lie group? Maybe using quaternions or spin representation?

$\begingroup$ Does en.wikipedia.org/wiki/Iwasawa_decomposition help...? $\endgroup$– Paul LevyMay 25, 2017 at 15:03

$\begingroup$ I'm not so sure what the question is. Iwasawa decomposition provides one particular parameterization, is it what's meant by "matrix expression"? $\endgroup$– YCorMay 25, 2017 at 16:12

1$\begingroup$ Wrong equality $\mathrm{SO}_0(1,4)=\mathrm O(1,4)\cap\mathrm{SL}(5,\mathbf R)$ now corrected. (RHS still has two components.) $\endgroup$– Francois ZieglerMay 25, 2017 at 20:37

3$\begingroup$ @paulgarrett $SL_2(H)$ is isogeneous to $SO(5,1)$, not $SO(4,1)$ (check the dimension!... actually I found the same error in Witte's book too). $\endgroup$– YCorMay 25, 2017 at 23:41

2$\begingroup$ The question is vague and the answer, of course, depends on the intended use of this parametrization. In addition to the Cartan and Iwasawa decompositions mentioned by others, there are also the exponential map and the Cayley transform (both parametrize suitably defined generic elements forming a dense subset of the group in terms of the Lie algebra). $\endgroup$– Victor ProtsakMay 26, 2017 at 0:52
2 Answers
The Cartan decomposition: $$ \mathrm{SO}_0(1,4)=\left\{\left( \begin{array}{cc} 1&0\\\hline 0&A \end{array}\right) \exp\left(\begin{array}{cc} 0&{}^tb\\\hline b&0 \end{array}\right): A\in\mathrm{SO}(4), b\in\mathbf R^{4\times1} \right\}, $$ (where the exponential is easily computed explicitly in terms of hyperbolic sines and cosines) boils this down to your favorite parametrization of $\mathrm{SO}(4)$.

$\begingroup$ I think this was what I was looking for. I'll work on this decomposition to see if it works for me. $\endgroup$– EduMay 26, 2017 at 10:00

$\begingroup$ (The computed exponential is in e.g. arxiv.org/abs/1103.0156, (30) and (35).) $\endgroup$ May 26, 2017 at 13:37
Yes, there should be an explicit expression. Let me sketch how to get it. Start with the Iwasawa decomposition to write your matrix $M$ as a product of three matrices $M = KAN$ where $N$ is nilpotent, $A$ is abelian and $K$ is a matrix from the maximal compact subgroup of $\mathrm{SO}_0(1,4)$ which is $\mathrm{S}(\mathrm{O(1)} \times \mathrm{O}(4))_0 \simeq \mathrm{SO}(4).$ Now the elements of $\mathrm{SO}(4)$ can be parametrized by twotuples of quaternions. The explicit form of these matrices depends on your symmetric form as was kindly noted by YCor in the comments.

1$\begingroup$ There's already one problem: the maximal split torus of $SO(4,1)$, for the given choice of representation (i.e., the choice of $I_{1,4}$), cannot be chosen in diagonal form (and $N$ doesn't take an upper triangular form either). $\endgroup$– YCorMay 25, 2017 at 15:33

$\begingroup$ Ah, yes. You are right. I'm so used to antidiagonal forms for orthogonal groups that I always assume them. I'll edit. $\endgroup$ May 25, 2017 at 15:36

2$\begingroup$ The maximal compact subgroup of a connected Lie group is connected. In your case it will just be $\mathrm{SO}(4)$ (the maximal compact in $\mathrm{SO}(1,4)$ is $\mathrm{S}(\mathrm{O}(1)\times\mathrm{O}(4))$). $\endgroup$– YCorMay 25, 2017 at 16:01

$\begingroup$ @YCor Thank you. I've overlooked the zero index. $\endgroup$ May 25, 2017 at 16:09

1$\begingroup$ It's not a symmetric space, at least in the usual sense. Maybe it's a pseudoRiemannian symmetric space in some reasonable sense. Don't know what you mean by "the maximal subgroup". Notation $K$ evokes a compact group but $SO(1,3)$ is certainly not compact. $\endgroup$– YCorMay 26, 2017 at 12:09