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Post by Admin on Jan 12, 2017 17:01:28 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Feb 16, 2017 6:52:47 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 16, 2017 7:14:37 GMT
Vasco,
I've had a look at your answer now. I think the main question before us whether it is necessary to rotate the infinitesimal vectors at the vertices. I didn't rotate them, so under my construction of asymptotic h-lines, they remained tangent throughout each of the three translations. I have printed out your answer for a closer look.
Gary
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Post by Admin on Feb 16, 2017 10:23:32 GMT
Vasco, I've had a look at your answer now. I think the main question before us whether it is necessary to rotate the infinitesimal vectors at the vertices. I didn't rotate them, so under my construction of asymptotic h-lines, they remained tangent throughout each of the three translations. I have printed out your answer for a closer look. Gary Gary The way I understand this exercise is that in your solution, since the angles of your h-triangle are zero, there is no h-rotation at $a, b ,c$, just a rotation in the PUHP. Vasco |
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Gary
GaryVasco
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Post by Gary on Feb 16, 2017 16:04:30 GMT
Vasco,
That's correct. I think the problem statement could have a weak interpretation ("show" using an example like the one I chose) or a strong interpretation ("show" in the most general construction). I'm going to work towards a more general solution, which is no doubt what you have, but I haven't studied it yet.
Gary
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Post by Admin on Feb 16, 2017 16:12:39 GMT
Gary
So in your document the small table showing h-rotations of the three translations should show them all as zero, is that right?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 16, 2017 17:25:53 GMT
Gary So in your document the small table showing h-rotations of the three translations should show them all as zero, is that right? Vasco Vasco, The rotations are as I showed them. Whether they are h-rotations or not is something I am thinking about. A key element of my answer came from paragraph 3, sentence 2, p. 312: "Note that since $\mathcal{T}^{\delta}_L$ is conformal, it carries dz to an infinitesimal vector $d\widetilde{z}$ (of equal h-length) making the same angle with L as dz." If you translate a vector along the full h-line semicircle and keep the angle with the h-line constant, the vector undergoes a rotation of $\pi$ in $\mathbb{C}$. We can apply this to all three translations. There seems no need to realign the $\xi$-vector at each vertex for the next translation. I have applied this method to your diagram with pencil and paper and it gives a simple solution. The angle between the original $\xi$ at $\alpha$ and the final $\xi$ at $\alpha$ is $\pi - (\alpha + \beta + \gamma)$. It will take me a few minutes to do a new diagram and put the right angles on the translated $\xi$-vector. It's a little harder to plot than it is to draw. I have two further comments. You make a distinction between rotations in the h-plane and the $Poincar\acute{e}$ plane. I was under the impression that the $Poincar\acute{e}$ plane is a representation of the h-plane. I wonder if it would make more sense to make the distinction between rotations in the h-plane and the $\mathbb{C}$ plane. In paragraph 4, p. 2, you rotate $\xi$ clockwise to put it back in its original orientation. Then in paragraph 5, you add $2\pi$. Wouldn't it be equivalent to just rotate $\xi$ counterclockwise by $\pi$ and avoid having to add $2\pi$? Gary P.S. I have expanded my answer now to include the general case. |
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Post by Admin on Feb 17, 2017 6:49:20 GMT
Gary
I agree that the PUHP is a representation of the h-plane, the Poincarites living in the hyperbolic plane see their 'world' differently. For example, an h-line in the PUHP is a semicircle 'sitting' on the horizon. If we travel along this h-line in the PUHP then we end up turning through an angle $\pm\pi$. A Poincarite travels along it in the hyperbolic plane and doesn't turn at all.
I see the PUHP as the upper half plane of $\mathbb{C}$ together with the metric (31) on page 298.
I did this to compare a 360 degree turn in the PUHP with the turning in the hyperbolic plane experienced by the Poincarites.
I have now studied your solution and I like it very much. I think it is a neater solution than mine. At the end I think you just need to say that $\pi-(\alpha+\beta+\gamma)=-E(\Delta)$, to comply with the exercise.
I still think that my solution is also valid, but I am still thinking about it and I may need to amend it to make it easier to understand.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 17, 2017 17:55:32 GMT
Gary I agree that the PUHP is a representation of the h-plane, the Poincarites living in the hyperbolic plane see their 'world' differently. For example, an h-line in the PUHP is a semicircle 'sitting' on the horizon. If we travel along this h-line in the PUHP then we end up turning through an angle $\pm\pi$. A Poincarite travels along it in the hyperbolic plane and doesn't turn at all. I see the PUHP as the upper half plane of $\mathbb{C}$ together with the metric (31) on page 298. I did this to compare a 360 degree turn in the PUHP with the turning in the hyperbolic plane experienced by the Poincarites. I have now studied your solution and I like it very much. I think it is a neater solution than mine. At the end I think you just need to say that $\pi-(\alpha+\beta+\gamma)=-E(\Delta)$, to comply with the exercise. I still think that my solution is also valid, but I am still thinking about it and I may need to amend it to make it easier to understand. Vasco Vasco, I see you have "gone native" as a Poincarétian. As a linguistic anthropologist, I appreciate that ability to take the inside view and I plan to cultivate it more myself. What if we think of $\xi$ as the direction in which Ms. Poincaré is looking. She walks forward a few centimeters on ab turning her view gradually clockwise in alignment with ab. To her, she is always looking and going straight. Then she walks backwards turning gradually counterclockwise on bc. Here the analogy seems to break down. She is going straight, but she is no longer looking along the h-line. From c, she turns gradually clockwise again on ca. When she reaches home, is she surprised that she is now looking in a different direction with a clockwise rotation from her original direction; is that a Poincarétian way of looking at the world? It works out nicely in the plot, but I am somewhat at a loss to explain what is happening in h-terms. Your approach of realigning at each vertex seems better for that. I will add the resolution you suggest. Gary |
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Post by Admin on Feb 20, 2017 15:03:29 GMT
Gary
I have corrected my answer to exercise 26 and republished it. I no longer think there is an error in the statement of the exercise and I obtain the same result as you do, that is "...the composition of these three h-translations is an h-rotation about vertex $a$ through angle $E(\Delta)$.".
Since the PUHP is a conformal map of the hyperbolic plane then an h-rotation in the map is an equal h-rotation in the hyperbolic plane, so I don't think she (the Poincarétian) is surprised to find that she is looking in a different direction when she gets back to $c$.
If you think of this situation on the sphere and she starts at the north pole and goes initially south, turning west at the equator and after a few hundred miles goes north again, back to the north pole and finds herself looking south again but she has turned clockwise about the pole through an angle corresponding to the distance she travelled along the equator.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 20, 2017 20:56:01 GMT
Gary I have corrected my answer to exercise 26 and republished it. I no longer think there is an error in the statement of the exercise and I obtain the same result as you do, that is "...the composition of these three h-translations is an h-rotation about vertex $a$ through angle $E(\Delta)$.". Since the PUHP is a conformal map of the hyperbolic plane then an h-rotation in the map is an equal h-rotation in the hyperbolic plane, so I don't think she (the Poincarétian) is surprised to find that she is looking in a different direction when she gets back to $c$. If you think of this situation on the sphere and she starts at the north pole and goes initially south, turning west at the equator and after a few hundred miles goes north again, back to the north pole and finds herself looking south again but she has turned clockwise about the pole through an angle corresponding to the distance she travelled along the equator. Vasco Vasco, The rewrite reads well. I still have a general question. My approach would make more sense if an h-translation along an h-line could be shown to be an h-rotation. This would be the case if the center of rotation were the intersection of two h-lines. We can always imagine that one of these h-lines is a vertical straight line from the center of a semi-circular h-line. Can we regard the horizon itself as the second h-line? Or do we have to imagine a second h-line asymptotic to the vertical h-line? Gary |
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Post by Admin on Feb 21, 2017 13:47:55 GMT
Gary
In my opinion traversing an h-line in the PUHP cannot in itself produce any rotation in the hyperbolic plane since Ms Poincaré is walking in a straight line. It's only the complete traverse round the triangle (i.e. the composition of the three h-translations) that produces the h-rotation. Compare this with the italicised part of the statement of the exercise on page 337.
Also, the vertex $a$ is the centre of the rotation and is the intersection of two h-lines (blue and black in my diagram). When Ms Poincaré gets back to $a$ she finds that she has been h-rotated.
The horizon is sometimes described as the circle at infinity (an h-circle, not an h-line - see page 300 second paragraph from the bottom - "We now possess...")
As I said in an earlier post, for me page 2 of your solution is fine as it stands.
I have made a couple of small changes to page 2 of my solution document to make my explanation a bit clearer.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 21, 2017 16:46:30 GMT
Gary In my opinion traversing an h-line in the PUHP cannot in itself produce any rotation in the hyperbolic plane since Ms Poincaré is walking in a straight line. It's only the complete traverse round the triangle (i.e. the composition of the three h-translations) that produces the h-rotation. Compare this with the italicised part of the statement of the exercise on page 337. Also, the vertex $a$ is the centre of the rotation and is the intersection of two h-lines (blue and black in my diagram). When Ms Poincaré gets back to $a$ she finds that she has been h-rotated. The horizon is sometimes described as the circle at infinity (an h-circle, not an h-line - see page 300 second paragraph from the bottom - "We now possess...") As I said in an earlier post, for me page 2 of your solution is fine as it stands. I have made a couple of small changes to page 2 of my solution document to make my explanation a bit clearer. Vasco Vasco, On rereading your answer, I have understood it for the first time. It makes good sense to me. But I still find the hyperbolic plane mysterious. You say "an h-line in the PUHP cannot in itself produce any rotation in the hyperbolic plane since Ms Poincaré is walking in a straight line", but Needham says "in hyperbolic geometry the composition of these three h-translations is an h-rotation about vertex a through angle $\mathcal{E}(\Delta)$", and one can see the parts of the eventual composite rotation in the changing direction of $\xi$ as it traverses each side of the triangle. I don't find the question of whether translations on h-lines are rotations discussed in this chapter, but there is at least one hint: On p. 314, Figure [33b] we can see the correspondence between the rotations $\alpha$ and $\beta$ about the origin of the UHP. The same angles appear in the shaded asymptotic triangle. The angles of the triangle are unquestionably h-rotations about their vertices. Do these same angles at the origin (or the horizon in general) not sweep out rotations equivalent to h-translations in the h-line? Is there some reason we do not call them h-rotations? I understand that the horizon is infinitely distant from any point in the UHP, so perhaps that is why there is no readily available place to put a center of rotation. Yet in [33b], Needham drew the angles from the origin, which is on the horizon. Is that vertex at the origin actually in $\mathbb{C}$ and not in the h-plane. If so, it still has consequences for the h-plane. Has something been left unresolved in the model of the UHP? Gary |
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Post by Admin on Feb 22, 2017 8:07:03 GMT
Gary
I take that back, I was wrong and you are right. Looking at the hyperbolic plane from outside (extrinsically), we can see the turning as we move round the triangle, but from the (intrinsic) view of Ms P. she thinks she is not turning and is surprised when she arrives back home to find she has been h-rotated about her starting point.
I think that Figure 33b can be looked at as the PUHP or as the Euclidean complex plane. I see the specified angles $\alpha$ and $\beta$ at the origin in 33b as resulting from Euclidean geometry in the complex plane. I agree that a suitable h-rotation about the vertex at $\alpha$ would rotate the illustrated semicircular h-line through this same vertex into the vertical h-line, and similarly for h-rotation about the vertex at $\beta$. Also if we do an h-translation along the h-line from the $\beta$ vertex to the $\alpha$ vertex and then back to the $\beta$ vertex via the third vertex at infinity, we find that this has produced an h-rotation about the $\beta$ vertex equal to $\pi-(\alpha+\beta)$, as expected since the third angle at infinity is zero.
Yes, I don't think we can put a centre of rotation on the horizon for the reasons you state, and, as I write in my answer above, the origin is in the complex plane. This construction of the triangles and angles at the origin is only so that we can calculate the limits of integration for the integral near the top of page 315, and for no other reason in my opinion. I don't think therefore that anything has been left unresolved in the PUHP model of the hyperbolic plane.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 22, 2017 23:23:05 GMT
Vasco,
That helps.
I suppose one can compare the experience of Ms. P traveling from point $b$ to point $c$ to our Euclidean experience of sitting in a back-facing seat in a train and looking out the window at an angle of Pi/4 (or any acute angle) from the line traveled. At the end of traversing a segment of a straight-line track, we will still be looking at the same angle to the direction of travel. The same can be said for Ms. P traveling on (traversing) a segment of an h-line. The poor thing doesn't realize that she has been rotated (from our external perspective) and if she were to shift her gaze to a direction orthogonal to her line of travel, she would be looking at the straight line of an intersecting orthogonal h-line (assuming someone had painted the line or a second track crossed hers orthogonally), just as we would be looking at a perpendicular Euclidean line.
In asking myself regarding [33], p. 314, "Why does it work to calculate the area of a pseudosphere using angles whose vertex in the real line of $\mathbb{C}$ appears to be transposed onto the horizon but is not actually on the horizon and whose distances use the Euclidean metric?" I suppose it is because the horizon actually is the real line. Can we think of it as the real line showing through from behind the h-plane? The h-metrics don't work on the horizon because y = 0.
Gary
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