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Post by Admin on Nov 21, 2016 14:26:59 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Dec 8, 2016 23:34:08 GMT
Vasco, My solutions to Exercise 11, Ch. 6 nh.ch6.ex11.pdf (151.23 KB) What, exactly, does "intrinsic" mean in this context? I have to say the profusion of maps and planes provide wonderful opportunities for confusion: h-space, Beltrami UHP, PD, Beltrami hemisphere, Euclidean circle in BUHP, sphere, projection from sphere to $\mathbb{C}$. I suppose it just takes practice. Gary |
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Post by Admin on Dec 9, 2016 9:19:57 GMT
Vasco, My solutions to Exercise 11, Ch. 6 View AttachmentWhat, exactly, does "intrinsic" mean in this context? I have to say the profusion of maps and planes provide wonderful opportunities for confusion: h-space, Beltrami UHP, PD, Beltrami hemisphere, Euclidean circle in BUHP, sphere, projection from sphere to $\mathbb{C}$. I suppose it just takes practice. Gary Gary As a result of your communication pointing out an algebraic error in my solution to exercise 11, I have corrected it and published the corrected version. Here, for me, "intrinsic" means "to them", "them" being the inhabitants of the sphere. So the sentence becomes: "Let the inhabitants of the sphere of radius $R$ draw a circle of (to them) radius $\rho$". This is what we (inhabitants of the earth) do when we draw a circle on the ground. If we think of the plane in which the circle lies as cutting through the sphere (earth), then we can imagine removing the the cap we have cut off. The distance from the top of the cap in any direction along the geodesic to the edge of this cap is the intrinsic radius of the circle with intrinsic centre at the top of the cap. The circle can also be thought of as a Euclidean circle whose Euclidean centre lies at the centre of the bottom of the cap, and whose Euclidean radius is equal to the radius of the bottom of the cap. I agree with you about the "wonderful opportunities for confusion". All I can suggest is to think of these in the same way you think of maps in an atlas, even though they are not all on a Euclidean plane (eg the hemisphere model). So we have the spherical geometry and hyperbolic geometry and maps of these two gemetries. The spherical geometry is perhaps easier to think about because we live as inhabitants of a spherically geometric world, which we think of as Euclidean for most everyday (small distance) purposes. When the distances get larger we have to use spherical geometry or resort to maps. By the way, although all the credit goes to Beltrami, these planes etc are named after Poincaré and Klein and I think that to avoid adding to the confusion mentioned above we should stick to using these names and make appropriate references to Beltrami in the credits. Do you agree? I will now go and look at your solution to exercise 11. Vasco |
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Post by Admin on Dec 9, 2016 17:44:19 GMT
Gary
I have had a quick look at your document, and although I also have some comments on Solution 1, I would like to make some comments on solution 2 since it takes a similar approach to mine:
on the first line you write "$d\widehat{s}$" is an infinitesimal arc length in the PD".
In my opinion, $d\widehat{s}$ is in the hyperbolic plane.
Then you write "The infinitesimal $ds$ is in the PUHP".
In my opinion $ds$ is in the PD.
If I am right about the above then you cannot substitute $\rho$ for $|z|$.
This looks like the start of an interesting discussion.
Vasco
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Post by Admin on Dec 10, 2016 7:57:10 GMT
Gary
In your solution 1 to part (i) of the exercise you seem to be using the Möbius transformation $D(z)$ to transform distances, but it transforms points not distances.
In part (ii) of the exercise I don't think there is any requirement to introduce a coordinate system. Also, in figure 1 I think $\rho$ should be the distance along half of the arc not the chord, and the angle at the centre would then be correctly labelled as $\rho/R$, otherwise it doesn't follow.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 11, 2016 1:24:47 GMT
Vasco,
I think this is a topic where I should not hesitate to let you take the lead, so I'm not sure how much I can contribute to a substantive discussion. I take it from your nice metaphor that in problem 11 that "(intrinsic) radius" should be a radius "intrinsic to the surface of the sphere" as opposed to a radius of a circle in which both circle and radius are drawn on a plane intersecting the sphere.
Regarding the names of the planes and maps, when I wrote it, I was referring to p. 298, title of Subsection 4 "Beltrami's Hyperbolic Plane", but then I see on p. 301 that the hyperbolic map plotted on the BHP is called the PUHP ($\mathbb{P}$ 2, ln 3), so that is fine by me. I agree. I wonder whether the convention is the same in Italian texts.
Your comments on my second solution to 11 (i):
It does appear that $d\widehat{s}$ is in the hyperbolic plane. Exercise 11 (i) refers to an h-circle "in the hyperbolic plane" $\textit{represented}$ as an Euclidean circle in the PD. So it appears that the statement on p. 309 does not apply to this situation and I can delete my solution 1. The implied transformation here seems to be
$\hspace{5em}$E-circle in PD $\rightarrow$ h-circle in h-plane (offstage?)
Regarding part (ii)
I agree with your statement that $\rho$ can not be substituted for $|z|$ and I have fixed it. At this point, I hope I have a better understanding of the subject. I feel as though I do.
In my solution to part 1, I was using the Möbius transformation only to transfer the origin point of the PD to the origin of the PUHP. I think that part was correct, but there is likely no application to part (i) of the problem.
I am replacing my answer with the repaired version.
Gary
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Post by Admin on Dec 20, 2016 17:06:49 GMT
Gary
I have just noticed a typo in my solution to this exercise and I have corrected it and replaced the published solution with the corrected version.
The upper limit of the integral is $2\pi$ not $\pi$.
Vasco
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