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Geomery ~ hyperbolic ~ Lorentz matrix

Post Icon Posted: Submitted by Jen on 14 June 2008 - 3:17pm.

Joined: 2007-02-12
Posts: 6

Hi

In week 7 we defined motions in H2, we showed that in the Lorentz matrix A, we have entry A_11>0, then we later go on to say that "Since A_11>0, T(1,0,0)=(A_11,A_21,A_31) belongs to H^2. Why don't we need A_11>1?

Cheers in advance.

Jenny

‹ Hyperbolic trivia Any person who professional in math specially in (sets)plz enter and try solve this ›
Post Icon Posted: 15 June 2008 - 11:19am

Joined: 2006-10-10
Posts: 520

Hey,

Hehe, I was wondering that too :) I think that there is meant to be an explanation of it in the appendix B4/B5 of the printed lecture notes.

The condition that the top left entry is strictly positive is equivalent to "q(v) preserving the upper cone" or something, I don't really understand that :S When I find out I'll post on the forum though
Post Icon Posted: 15 June 2008 - 8:06pm

Joined: 2007-10-01
Posts: 178

Not entirely sure, but is it because we are only considering the upper half of the Hyperbolic sheet we need to make sure the first co-ordinate of any vector $ \textbf{v} \in \mathbb{H}^2 $ is positive. Basically since $ \mathbb{H}^2 := \{ (t,x,y) \mathbb{R}^3 | t > 0 \, \& \, x^2 + y^2 - t^2 = -1 \} $ then for $ A \textbf{v}, v=(1,0,0) $ to be in $ \mathbb{H}^2 $ we must have $ (\alpha_{00},\alpha_{10},\alpha_{20}) $ satisfying $ \alpha_{10}^2 + \alpha_{20}^2 - \alpha_{00}^2 = -1 , \, \alpha_{00} >0 $

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