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I will like to know how does one compute the Lie algebra of an abstractly given subgroup of a Lie group? Specifically, let $G = \mathrm{SO} ( n + 1, 1 )$ and consider the flow $$ g_t = \begin{pmatrix} \cosh t &0 &- \sinh t\\ 0 &I_n &0\\ - \sinh t &0 &\cosh t \end{pmatrix} $$ and the associated expanding subgroup $H = \{ h \in G \mid g_{-t} h g_t \to e \textrm{ as } t \to \infty \}$. I want to understand how to get the Lie algebra of $H$ though I do know the final answer.

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Try to derivate your condition, with $h=e^{s\xi}$ – Léo Feb 25 '14 at 21:45
@Léo That did not work out for me. Have you checked if it works? – Singhal Feb 26 '14 at 5:56
up vote 2 down vote accepted

$\mathfrak{h}=\lbrace \xi \in \mathfrak{g}, \frac{d}{ds}\vert_{s=0}(g_{-t}e^{s\xi}g_t) \to 0 $ as $t \to \infty \rbrace $

So you make the computation, gives you a condition, and you check who realizes it in $\mathfrak{g}$...

Does it help?

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The above derivative will be $g_{- t} \xi g_t$, right? – Singhal Feb 26 '14 at 14:21
Seems so... What is $\mathfrak{g}$? – Léo Feb 26 '14 at 14:25
Null trace matrix which anticommute with the n+1,1 "identity matrix" – Léo Feb 26 '14 at 14:33
So can you conclude? – Léo Feb 26 '14 at 14:35
Sorry, I misunderstood $\mathfrak{g}$ to be the Lie algebra of the subgroup $H$ and kept making stupid edits. Yes, the trace should be zero as $\mathrm{SO} ( n + 1, 1 )$ is inside $\mathrm{SL} ( n + 2 )$. In addition, matrices $X \in \mathfrak{g}$ should have $X^T I_{n + 1, 1} + I_{n + 1, 1} X = 0$. Ain't it? – Singhal Feb 26 '14 at 14:37

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