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Let $A$ and $B$ be in $\mathbb{H}^2$. I need to prove that the lorentzian dot product between $A$ and $B$ is less than or equal to $-1$. I have no idea where to start.

Euler88 ...
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John.P
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  • No I've given all the information needed. – John.P Nov 22 '15 at 14:40
  • It's the same. I think what you're missing is the fact that A and B are on the hyperbolic plane, therefore A.A and B.B = -1. – John.P Nov 22 '15 at 14:45
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    We're considering the upper sheet of the hyperboloid of 2 sheets. – John.P Nov 22 '15 at 15:05
  • You could have stated you were working with the hyperboloid model. You could have said that means you're considering the points $(t,x,y)$ on the upper sheet of $t^2-x^2-y^2=1$. You could have said the Lorentzian dot product is $(t_1,x_1,y_1)\cdot(t_2,x_2,y_2)=t_1t_2-x_1x_2-y_1y_2$. Anyway, it seems to me the Lorentzian dot product of things on that upper sheet are $\ge1$, not $\le -1$, no? – anon Nov 25 '15 at 23:54
  • @whacka This was actually an assignment question: http://homepages.warwick.ac.uk/~masgar/Teach/2015_MA243/examples7.pdf (the deadline has already passed though). To answer your question, in lectures we defined the Lorentzian dot product to be given by $-t_{1}t_{2}+x_{1}x_{2}+y_{1}y_{2}$. This convention was taken to be consistent with the way other texts and a certain book that the course is following. In particular, we take the first component, (in your case $t_{1}$ and $t_{2}$ ) to be the vertical direction. But yes, I agree that this question does not have enough information. – Irregular User Nov 26 '15 at 10:48

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