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Post by Admin on Dec 7, 2016 10:16:22 GMT
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Gary
GaryVasco
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Post by Gary on Dec 12, 2016 3:07:44 GMT
Vasco, It didn't occur to me that Needham would mix the Klein model and Poincaré disc. I kept trying to imagine how orthogonal angles could be produced by vertical projection to the hemisphere. This is what I produced after I saw that you had used the PD. nh.ch6.ex12.pdf (95.12 KB) Gary |
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Post by Admin on Dec 16, 2016 16:14:11 GMT
Gary
Reading your comments about exercise 12 of chapter 6, I started to wonder whether we agree on the interpretation of this exercise. Here is the way I see the h-lines in the Klein model:
The h-lines in the PD are stereographically projected to the hemisphere as semicircles which lie in planes which are orthogonal to the disc. If the h-lines in the PD are orthogonal, their stereographic projections onto the hemisphere also intersect orthogonally on the surface of the hemisphere.
When the hemisphere is then projected vertically downwards back on to the disc the straight line chords (the h-lines of the Klein disc) are not orthogonal, purely as a result of the geometry of the vertical projection. If we imagine the two orthogonal semicircles on the surface of the hemisphere as being made of thin wire and our projection downwards to produce the Klein disc as being produced by shining a parallel beam of light orthogonal to the disc from above the hemisphere, then the shadow of the wires will be chords of the disc, but they will not be orthogonal, and this is solely because of the nature of projection from the hemisphere onto the disc.
Is this the way you see it?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 16, 2016 19:15:16 GMT
Gary Reading your comments about exercise 12 of chapter 6, I started to wonder whether we agree on the interpretation of this exercise. Here is the way I see the h-lines in the Klein model: The h-lines in the PD are stereographically projected to the hemisphere as semicircles which lie in planes which are orthogonal to the disc. If the h-lines in the PD are orthogonal, their stereographic projections onto the hemisphere also intersect orthogonally on the surface of the hemisphere. When the hemisphere is then projected vertically downwards back on to the disc the straight line chords (the h-lines of the Klein disc) are not orthogonal, purely as a result of the geometry of the vertical projection. If we imagine the two orthogonal semicircles on the surface of the hemisphere as being made of thin wire and our projection downwards to produce the Klein disc as being produced by shining a parallel beam of light orthogonal to the disc from above the hemisphere, then the shadow of the wires will be chords of the disc, but they will not be orthogonal, and this is solely because of the nature of projection from the hemisphere onto the disc. Is this the way you see it? Vasco Vasco, I think what I wrote in Ex. 12 agrees with your view, but I did err in the presentation in Exercise 13. I wrote "Hence, when projected vertically to the hemisphere (blue), their images (of the Klein h-lines in $\mathbb{C}$ are also h-lines)." I thought at that point that Klein had dubbed them so as projections of the chords, but I see that was wrong. As you point out, they are merely semicircles. But I would also like to point out that h-lines in the PD only project to the top hemisphere if the source of the projection is the south pole, which is opposite to what seems to be Needham's canonical situation (pp. 142, 285). Also, I don't think the proof can rest on the appearance of the intersecting lines from above. I think it has to depend on the stereographic projection of both circles and rays to the hemisphere. It is the semicircular coincidence of the s-projection of h-lines from the PD and the v-projection of chords from the Klein model that completes it. My corrected version reflects this conclusion. I should completely revise it, but for now I have just corrected it. I agree with your second paragraph and I'm going to abandon the proposal I made in Exercise 13. I also made a small revision of Exercise 12. Gary |
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Post by Admin on Dec 17, 2016 13:52:16 GMT
Gary
I don't think it matters whether we stereographically project from the north pole or the south pole as the only effect is to change the hemisphere from north to south or vice versa. As a consequence of the symmetry of the sphere, stereographic projection can be done from any point $p$ on the sphere to a plane at right angles to the great circle through the point $p$. Needham mentions projection from the south pole in subsection 12 on page 322 just before the suggested exercise.
It also seems to me that the proof becomes a purely geometric proof once the diagram has been drawn, and has very little to do with hyperbolic geometry. We have two arcs which intersect each other at right angles and intersect the unit circle at right angles, and we also have their corresponding chords. If the arcs are not orthogonal to each other then the produced chords do not pass through the centres of the circles of which the arcs form a part.
It does however give us a way of finding out geometrically if two chords in the Klein model represent orthogonal h-lines or not as long as we can construct the arcs and measure angles accurately enough.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 17, 2016 17:20:07 GMT
Gary I don't think it matters whether we stereographically project from the north pole or the south pole as the only effect is to change the hemisphere from north to south or vice versa. As a consequence of the symmetry of the sphere, stereographic projection can be done from any point $p$ on the sphere to a plane at right angles to the great circle through the point $p$. Needham mentions projection from the south pole in subsection 12 on page 322 just before the suggested exercise. It also seems to me that the proof becomes a purely geometric proof once the diagram has been drawn, and has very little to do with hyperbolic geometry. We have two arcs which intersect each other at right angles and intersect the unit circle at right angles, and we also have their corresponding chords. If the arcs are not orthogonal to each other then the produced chords do not pass through the centres of the circles of which the arcs form a part. It does however give us a way of finding out geometrically if two chords in the Klein model represent orthogonal h-lines or not as long as we can construct the arcs and measure angles accurately enough. Vasco Vasco, That all makes sense to me. Would you agree that the point of the problem was to show that intersecting Klein chords are orthogonal if and only if they are the secants of two orthogonal PD h-lines? Then would you agree that the point of exercise 13 is to use our understanding of the PD to show that Klein chords are orthogonal to Klein circles if and only if the chords subsume rays of origin-centred circles? Gary |
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Post by Admin on Dec 18, 2016 7:43:24 GMT
Vasco, That all makes sense to me. Would you agree that the point of the problem was to show that intersecting Klein chords are orthogonal if and only if they are the secants of two orthogonal PD h-lines? Gary Gary Just being pedantic, but to avoid any misunderstanding I would say: Intersecting chords on the unit circle are/represent orthogonal KD h-lines if and only if they are the secants of two orthogonal PD h-lines. No, I see this exercise as using the hemisphere model to derive formulae for h-distances on the Klein disc (which is the vertical downward projection of the hemisphere model), by first showing that concentric circles centred at the origin of the KD and rays through the origin of the KD are orthogonal. It all depends on this one fact. Exercise 14 does a similar thing using spherical geometry to derive formulae for another non-conformal model - the projective model. Vasco |
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