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Post by Admin on Jan 17, 2017 14:08:23 GMT
Gary Here is my solution to exercise 21.After reading Gary's solution, I have updated part (ii) of my solution, not to correct errors, but to make the solution easier to follow and understand. Vasco |
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Gary
GaryVasco
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Post by Gary on Jan 19, 2017 22:25:16 GMT
Vasco,
For part (ii) I am getting $\frac{-cz + s}{-sz+c}$. Needham shows no minus term in the numerator. Did you encounter this?
Gary
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Post by Admin on Jan 19, 2017 22:28:34 GMT
Gary
No, it all worked out OK for me.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 19, 2017 22:37:49 GMT
Vasco,
Thanks. I rechecked and found the problem in the calculation of $\mathcal{I}_B$.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 21, 2017 6:28:52 GMT
Vasco, Here is a version of Ex. 21. nh.ch6.ex21.pdf (265.63 KB) I got a curious result in (iii) by trying to use (19), p. 288 to solve for [$\mathcal{R}^\phi_i$]. Gary |
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Post by Admin on Jan 21, 2017 8:31:28 GMT
Gary
Before commenting on the details of your solutions to this exercise, I would like to make some observations about the way I understand what Needham is asking us to do in this exercise. In some ways the wording of the exercise seems designed to lead the reader up the garden path. From my perspective I think it would have been better if Needham had written it as three separate exercises, with a separate introduction for each, making it clear what is given and what is to be proved.
What follows is my rewriting of the three parts as separate exercises, which I hope makes it clearer what Needham wants us to do. I will be interested to know if you agree with my interpretation.
Introduction to the exercise
The introduction to the exercise is asking us to prove, in three different ways, that the Möbius transformation which is an h-rotation about the point $i$ in the PUHP can be written as $\mathcal{R}^{\phi}_i=\frac{cz+s}{-sz+c}$ where $c=\cos(\phi/2)$ and $s=\sin(\phi/2)$.
Part (i)
Given the Möbius transformation $\mathcal{R}^{\phi}_i=\frac{cz+s}{-sz+c}$ where $c=\cos(\phi/2)$ and $s=\sin(\phi/2)$, show that $\mathcal{R}^{\phi}_i=i$ and $\{\mathcal{R}^{\phi}_i\}(i)=e^{i\phi}$ and then explain why this result means that $\mathcal{R}^{\phi}$ must be an h-rotation about the point $i$ when applied to the PUHP.
For me the words in italics explain the meaning of Needham's question at the end of part (i): "Why does this prove the result?". Note that we are not to assume that $\mathcal{R}^{\phi}$ is an h-rotation of the PUHP about $i$.
Part (ii)
Given that $\mathcal{R}^{\phi}_i$ is an h-rotation of the PUHP about the point $i$ (We have proved this in part (i)), show that it can be written as $\mathcal{R}^{\phi}_i=\frac{cz+s}{-sz+c}$ where $c=\cos(\phi/2)$ and $s=\sin(\phi/2)$. We know from subsection 7 of chapter 6 from page 306 to the end of paragraph one on page 308, that $\mathcal{R}^{\phi}_i$ can be written as $\mathcal{R}^{\phi}_i=(\mathfrak{R}_B\circ\mathfrak{R}_A)$.
Part (iii)
By describing and explaining the geometric effect of applying $(D\circ \mathcal{R}^{\phi}_i\circ D^{-1})$ to the PUHP deduce that it is a rotation of the PD i.e. $e^{i\phi}z$.
In this part we only know that $\mathcal{R}^{\phi}_i$ is an h-rotation of the PUHP about $i$. We cannot assume that it has the form $\frac{cz+s}{-sz+c}$. This is what we have to prove by expressing $(D\circ\mathcal{R}^{\phi}_i\circ D^{-1})(z)=e^{i\phi}z$ in matrix form and solving for $[\mathcal{R}^{\phi}_i]$.
Comments on your solutions to follow.
Vasco
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Post by Admin on Jan 21, 2017 10:10:48 GMT
Gary
Comments on your solution to part (i) of exercise 21
I don't think that you have completed the exercise as Needham intended. See my post above in this thread regarding interpretationof the exercise.
Comments on your solution to part (ii) of exercise 21
This seems fine to me. As a result of reading your solution to this part of the exercise, I have decided to update my solution to part (ii) to make my proof a bit clearer and easier to follow.
Comments on your solution to part (iii) of exercise 21
In the first line of your answer on page 4 of your document you say:
" $(D\circ\mathcal{R}^{\phi}_i\circ D^{-1})$ takes the PD to the UHP, rotates it around $i$ by an angle $\phi$ and takes the result back to the PD.".
I don't agree that the UHP is rotated about $i$ through angle $\phi$. $\mathcal{R}^{\phi}_i$ is an h-rotation, it does not rotate the PUHP about $i$ through an angle $\phi$. However I would agree that the hyperbolic plane does undergo a rotation about $i$.
In the same paragraph you say that you couldn't calculate the composition $(D\circ\mathcal{R}^{\phi}_i\circ D^{-1})$ to get $e^{i\phi}z$. I tried it by hand myself and it worked out OK:
I found $(\mathcal{R}^{\phi}_i\circ D^{-1})$ to be equal to $\frac{ze^{i\phi}+i}{ize^{i\phi}+1}$ and then applying $D$ to this gives $e^{i\phi}z$. However as I say in my first post about this exercise above, I do not think we can assume the form of $\mathcal{R}^{\phi}_i$, we have to calculate it.
On page 5, in the paragraph which starts "Under..." you say: "...the concentric circles in the PD must go to concentric h-circles in the PUHP". In the PUHP these circles are not concentric, but look like those in your figure 3 on page 4.
In the next paragraph you say "Rotating the UHP.... should map the image to itself". This is true only if you are h-rotating, not if you are doing ordinary rotation of the PUHP.
I think you are right about (19) on page 288 being a clockwise rotation and so this explains the discrepancy you found and referred to on page 6 of your document.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 21, 2017 21:36:32 GMT
Gary Comments on your solution to part (i) of exercise 21I don't think that you have completed the exercise as Needham intended. See my post above in this thread regarding interpretationof the exercise. Comments on your solution to part (ii) of exercise 21This seems fine to me. As a result of reading your solution to this part of the exercise, I have decided to update my solution to part (ii) to make my proof a bit clearer and easier to follow. Comments on your solution to part (iii) of exercise 21In the first line of your answer on page 4 of your document you say: " $(D\circ\mathcal{R}^{\phi}_i\circ D^{-1})$ takes the PD to the UHP, rotates it around $i$ by an angle $\phi$ and takes the result back to the PD.". I don't agree that the UHP is rotated about $i$ through angle $\phi$. $\mathcal{R}^{\phi}_i$ is an h-rotation, it does not rotate the PUHP about $i$ through an angle $\phi$. However I would agree that the hyperbolic plane does undergo a rotation about $i$. In the same paragraph you say that you couldn't calculate the composition $(D\circ\mathcal{R}^{\phi}_i\circ D^{-1})$ to get $e^{i\phi}z$. I tried it by hand myself and it worked out OK: I found $(\mathcal{R}^{\phi}_i\circ D^{-1})$ to be equal to $\frac{ze^{i\phi}+i}{ize^{i\phi}+1}$ and then applying $D$ to this gives $e^{i\phi}z$. However as I say in my first post about this exercise above, I do not think we can assume the form of $\mathcal{R}^{\phi}_i$, we have to calculate it. On page 5, in the paragraph which starts "Under..." you say: "...the concentric circles in the PD must go to concentric h-circles in the PUHP". In the PUHP these circles are not concentric, but look like those in your figure 3 on page 4. In the next paragraph you say "Rotating the UHP.... should map the image to itself". This is true only if you are h-rotating, not if you are doing ordinary rotation of the PUHP. I think you are right about (19) on page 288 being a clockwise rotation and so this explains the discrepancy you found and referred to on page 6 of your document. Vasco Vasco, All interesting comments. I find this exercise ties together a few loose $M\ddot{o}bius$ ends for me. Regarding part (i), I expanded my answer, which I found to be an interesting exercise. I hope it will not be found to be fantastical. Regarding part (iii), I will take your word for the time being that there is a solution for the part that I was stuck on. I'll return as time permits. On the issue of concentric circles, I suppose I was anticipating that the circles in the UHP, having the same center, would also be spaced equally by the metric of the UHP. That's something I would like to test, given time. If it turns out to be so, would it not be reasonable to speak of h-concentric circles? I agree with your point about the mapping being true only under h-rotating, but we established in (i) that $\mathcal{R}^{\phi}_i$ is an h-rotation. But it doesn't hurt to be more precise. The gray lines in Figure 3 were plotted by applying the h-rotation to the points of the circles and h-lines in the UHP. I'm glad you agree about (19). Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 21, 2017 23:06:17 GMT
Gary Before commenting on the details of your solutions to this exercise, I would like to make some observations about the way I understand what Needham is asking us to do in this exercise. In some ways the wording of the exercise seems designed to lead the reader up the garden path. From my perspective I think it would have been better if Needham had written it as three separate exercises, with a separate introduction for each, making it clear what is given and what is to be proved. What follows is my rewriting of the three parts as separate exercises, which I hope makes it clearer what Needham wants us to do. I will be interested to know if you agree with my interpretation. Introduction to the exerciseThe introduction to the exercise is asking us to prove, in three different ways, that the Möbius transformation which is an h-rotation about the point $i$ in the PUHP can be written as $\mathcal{R}^{\phi}_i=\frac{cz+s}{-sz+c}$ where $c=\cos(\phi/2)$ and $s=\sin(\phi/2)$. Part (i)Given the Möbius transformation $\mathcal{R}^{\phi}_i=\frac{cz+s}{-sz+c}$ where $c=\cos(\phi/2)$ and $s=\sin(\phi/2)$, show that $\mathcal{R}^{\phi}_i=i$ and $\{\mathcal{R}^{\phi}_i\}(i)=e^{i\phi}$ and then explain why this result means that $\mathcal{R}^{\phi}$ must be an h-rotation about the point $i$ when applied to the PUHP.For me the words in italics explain the meaning of Needham's question at the end of part (i): "Why does this prove the result?". Note that we are not to assume that $\mathcal{R}^{\phi}$ is an h-rotation of the PUHP about $i$. Part (ii)Given that $\mathcal{R}^{\phi}_i$ is an h-rotation of the PUHP about the point $i$ (We have proved this in part (i)), show that it can be written as $\mathcal{R}^{\phi}_i=\frac{cz+s}{-sz+c}$ where $c=\cos(\phi/2)$ and $s=\sin(\phi/2)$. We know from subsection 7 of chapter 6 from page 306 to the end of paragraph one on page 308, that $\mathcal{R}^{\phi}_i$ can be written as $\mathcal{R}^{\phi}_i=(\mathfrak{R}_B\circ\mathfrak{R}_A)$. Part (iii)By describing and explaining the geometric effect of applying $(D\circ \mathcal{R}^{\phi}_i\circ D^{-1})$ to the PUHP deduce that it is a rotation of the PD i.e. $e^{i\phi}z$. In this part we only know that $\mathcal{R}^{\phi}_i$ is an h-rotation of the PUHP about $i$. We cannot assume that it has the form $\frac{cz+s}{-sz+c}$. This is what we have to prove by expressing $(D\circ\mathcal{R}^{\phi}_i\circ D^{-1})(z)=e^{i\phi}z$ in matrix form and solving for $[\mathcal{R}^{\phi}_i]$. Comments on your solutions to follow. Vasco Vasco, I like the suggested rewrites, but I also like the fact that these three problems are grouped together. If they were separated, one would probably have to write more suggestions referring to the previous exercises. Gary |
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Post by Admin on Jan 22, 2017 7:51:46 GMT
Gary
Reading your replies and your expanded answer to part (i), I find that most of our differences are due to semantics which is to be expected when working in this area of non-Euclidean geometry!
Just one thing: in your expanded answer to part (i) in paragraph three you write: "This tells us that an infinitesimal vector at any point...". Since the derivative of $\mathcal{R}^{\phi}_i$ is evaluated at $z=i$ we cannot say at any point, but just vectors emanating from $i$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 22, 2017 15:21:14 GMT
Vasco,
I am glad you caught that. I think I was thinking of the derivative in general when I wrote that. For me the prize in this exercise was learning that the multiplier of the $M\ddot{o}bius$ transformation is equivalent (or at least related) to the derivative. How would you phrase it?
Gary
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Post by Admin on Jan 23, 2017 15:57:23 GMT
Gary
I'll give it some thought and get back to you.
Vasco
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Post by Admin on Jan 24, 2017 15:49:19 GMT
Gary I'll give it some thought and get back to you. Vasco Gary One way of putting it would be as follows: If a Möbius transformation $M$ has two fixed points $\xi_+$ and $\xi_-$, then the multiplier $\mathfrak{m}_+=M'(\xi_+)$ and the multiplier $\mathfrak{m}_-=M'(\xi_-)$. Since these derivatives are the amplitwists of $M$ at its fixed points, then the geometric effect of $M$ in the vicinity of its two fixed points is an amplitwist equal to the multiplier associated with that fixed point. Whether the amplitwist is a pure rotation, pure expansion or a composition of the two, determines the character of this local geometric effect and determines whether we call $M$ elliptic, hyperbolic or loxodromic. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jan 25, 2017 1:17:14 GMT
Vasco,
It's a clear and useful statement. I'm surprised that Needham didn't give this connection more attention. It's a nice connection between the Möbius transformations and his amplitwist theory.
Gary
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Post by Admin on Jan 25, 2017 6:46:51 GMT
Vasco, It's a clear and useful statement. I'm surprised that Needham didn't give this connection more attention. It's a nice connection between the Möbius transformations and his amplitwist theory. Gary Gary He does make a passing reference to it at the end of subsection 3 of chapter 3 at the end of the subsection on page 168. Vasco |
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