Timeline for Find the fixed geodesic of an orientation-preserving isometry of the $3D$ hyperboloid model
Current License: CC BY-SA 3.0
9 events
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S Dec 8, 2017 at 22:57 | history | suggested | j0equ1nn | CC BY-SA 3.0 |
the real answer was in the comments do I summarized them in the answer
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Dec 8, 2017 at 19:10 | review | Suggested edits | |||
S Dec 8, 2017 at 22:57 | |||||
Dec 7, 2017 at 23:08 | vote | accept | j0equ1nn | ||
Dec 7, 2017 at 23:08 | comment | added | j0equ1nn | Ah okay, I see what you mean. Thanks. | |
Dec 7, 2017 at 22:27 | comment | added | Igor Rivin | It's in the link in my answer. | |
Dec 7, 2017 at 21:59 | comment | added | j0equ1nn | That makes sense, but let's say I give you a specific matrix. Can we say exactly what the pair of light like vectors are? I have not found any literature on a conformal map from $\overline{\mathcal{H}^3}$ to $\mathcal{I}^3\cup\big(\big\{p\in\mathbb{R}^{3,1}\mid\mathrm{n}(p)=0\big\}/\sim\big)$. | |
Dec 7, 2017 at 21:34 | comment | added | Igor Rivin | @j0equ1nn The two endpoints you mention in the OP give two (light-like) vectors. Their span is your geodesic. | |
Dec 7, 2017 at 20:39 | comment | added | j0equ1nn | That works fine for points but is awkward to do on an entire geodesic in the upper half-plane. I'm looking for a way of characterizing the plane in $\mathbb{R}^4$ that intersects with $\mathcal{I}^3$ to give $g$. Since this depends only on the matrix entries, there should be some function $\mathrm{PSL}(2,\mathbb{C})\rightarrow\mathbb{R}^4\times\mathbb{R}^4$ that outputs this pair of vectors. | |
Dec 7, 2017 at 18:59 | history | answered | Igor Rivin | CC BY-SA 3.0 |