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Post by Deleted on Dec 19, 2015 9:20:38 GMT
SECTION III: Hyberbolic Geometry, SUBSECTION 10: The Poincare Disc, p.316
This is the explanation of how the Poincare Disc is obtained. First, Needham outlines the operation z |-> I(z) reflecting the upper half plane across the circle K centered at -i to the (lower) unit disc, then filling in the upper unit disc mapping that portion to itself. He then mentions we are left with an anticonformal mapping, so we should take z |-> z(conjugate) to make the mapping conformal again. All good...
The problem is in the last paragraph stating that both these transformations combine to, say, D(z) which is a Mobius transformation "...D(z) maps i to 0 and -i to infinity..."
Well, clearly when you first process I(z), it sends all infinite points to -i and then the second part of that operation filling in the upper half unit disc sends +i to the origin. We end up with an entire upper half plane compressed into the unit disc but "upside down" with the origin at +i and infinite points at -i. After the conjugate reflection, it flips the picture "right side up" so that now the origin is at -i and all infinite points are at +i.
Since D(z) is clearly the sum of the 2 transformations, it would be more accurate to say "...D(z) maps infinite points to +i and 0 to -i..."
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Post by Admin on Dec 23, 2015 9:34:45 GMT
SECTION III: Hyberbolic Geometry, SUBSECTION 10: The Poincare Disc, p.316 This is the explanation of how the Poincare Disc is obtained. First, Needham outlines the operation z |-> I(z) reflecting the upper half plane across the circle K centered at -i to the (lower) unit disc, then filling in the upper unit disc mapping that portion to itself. He then mentions we are left with an anticonformal mapping, so we should take z |-> z(conjugate) to make the mapping conformal again. All good... The problem is in the last paragraph stating that both these transformations combine to, say, D(z) which is a Mobius transformation "...D(z) maps i to 0 and -i to infinity..." Well, clearly when you first process I(z), it sends all infinite points to -i and then the second part of that operation filling in the upper half unit disc sends +i to the origin. We end up with an entire upper half plane compressed into the unit disc but "upside down" with the origin at +i and infinite points at -i. After the conjugate reflection, it flips the picture "right side up" so that now the origin is at -i and all infinite points are at +i. Since D(z) is clearly the sum of the 2 transformations, it would be more accurate to say "...D(z) maps infinite points to +i and 0 to -i..." Terrence I agree with you that $D(z)$ must map $\infty$ to $+i$ and $0$ to $-i$, but Needham is also correct to say that $D(z)$ maps $i$ to $0$ and $-i$ to $\infty$. The reason he chooses to use these particular points rather than the ones you would have chosen is so that he can very quickly find the equation for $D(z)$ by saying that $D(z)$ must be proportional to $(z-i)/(z+i)$ and then using the fact that $\pm1$ are fixed points of $D(z)$ to obtain $$D(z)=\frac{iz+1}{z+i}$$ This is the same idea as the one on page 154 subsection 6 The Cross-Ratio. It avoids having to use 'brute force' to find $D(z)$ as Needham says at the top of page 317. I hope you find this acceptable. If you agree then we do not need to record this as an error in the book. Vasco |
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Gary
GaryVasco
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Post by Gary on Nov 7, 2016 6:42:46 GMT
SECTION III: Hyberbolic Geometry, SUBSECTION 10: The Poincare Disc, p.316 This is the explanation of how the Poincare Disc is obtained. First, Needham outlines the operation z |-> I(z) reflecting the upper half plane across the circle K centered at -i to the (lower) unit disc, then filling in the upper unit disc mapping that portion to itself. He then mentions we are left with an anticonformal mapping, so we should take z |-> z(conjugate) to make the mapping conformal again. All good... The problem is in the last paragraph stating that both these transformations combine to, say, D(z) which is a Mobius transformation "...D(z) maps i to 0 and -i to infinity..." Well, clearly when you first process I(z), it sends all infinite points to -i and then the second part of that operation filling in the upper half unit disc sends +i to the origin. We end up with an entire upper half plane compressed into the unit disc but "upside down" with the origin at +i and infinite points at -i. After the conjugate reflection, it flips the picture "right side up" so that now the origin is at -i and all infinite points are at +i. Since D(z) is clearly the sum of the 2 transformations, it would be more accurate to say "...D(z) maps infinite points to +i and 0 to -i..." Terrence I agree with you that $D(z)$ must map $\infty$ to $+i$ and $0$ to $-i$, but Needham is also correct to say that $D(z)$ maps $i$ to $0$ and $-i$ to $\infty$. The reason he chooses to use these particular points rather than the ones you would have chosen is so that he can very quickly find the equation for $D(z)$ by saying that $D(z)$ must be proportional to $(z-i)/(z+i)$ and then using the fact that $\pm1$ are fixed points of $D(z)$ to obtain $$D(z)=\frac{iz+1}{z+i}$$ This is the same idea as the one on page 154 subsection 6 The Cross-Ratio. It avoids having to use 'brute force' to find $D(z)$ as Needham says at the top of page 317. I hope you find this acceptable. If you agree then we do not need to record this as an error in the book. Vasco Vasco and Terrence, Here is my take on the exercise at the bottom of p. 316. I'm finally catching up a bit. The mappings have inverses, so the order of presentation depends on style and emphasis rather than logic. Infinity swaps with +i and 0 swaps with -i. I actually found the "brute force" solution easier to manage than the solution I found, which I think is the obvious one. I would like to see the cross-ratio applied to this. nh.ch6.poincare.disc.ex.p.316.pdf (115.27 KB) Gary |
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Post by Admin on Nov 7, 2016 12:40:19 GMT
Gary
I think that Needham's idea was to notice from [35] that $1+i$ and $1-i$ are orthogonal and so deduce that the constant is $i$ without doing any algebraic manipulation, but I agree that the 'brute force' method is not that bad.
I see the non brute force method as an application of the cross-ratio formula in the middle of page 154, where $q,r,s$ are $i,\pm1,-i$ respectively.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 7, 2016 21:07:37 GMT
Gary I think that Needham's idea was to notice from [35] that $1+i$ and $1-i$ are orthogonal and so deduce that the constant is $i$ without doing any algebraic manipulation, but I agree that the 'brute force' method is not that bad. I see the non brute force method as an application of the cross-ratio formula in the middle of page 154, where $q,r,s$ are $i,\pm1,-i$ respectively. Vasco Vasco, Yes, I see it now. I actually tried the cross-ratio formula first and got the right answer, but I didn't recognize it. But I don't understand how you deduce that the constant is i from the fact that $1+i$ and $1-i$ are orthogonal. It it just that $(1-i)/(1+i)$ has arg = Pi/2 and multiplying by $i$ gives arg = 0 to D(1)? Gary |
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Post by Admin on Nov 7, 2016 22:06:55 GMT
Gary I think that Needham's idea was to notice from [35] that $1+i$ and $1-i$ are orthogonal and so deduce that the constant is $i$ without doing any algebraic manipulation, but I agree that the 'brute force' method is not that bad. I see the non brute force method as an application of the cross-ratio formula in the middle of page 154, where $q,r,s$ are $i,\pm1,-i$ respectively. Vasco Vasco, Yes, I see it now. I actually tried the cross-ratio formula first and got the right answer, but I didn't recognize it. But I don't understand how you deduce that the constant is i from the fact that $1+i$ and $1-i$ are orthogonal. It it just that $(1-i)/(1+i)$ has arg = Pi/2 and multiplying by $i$ gives arg = 0 to D(1)? Gary Gary If $k\frac{1-i}{1+i}=1$ then $k(1-i)=(1+i)$ and so since $k$ must rotate $(1-i)$ through $\pi/2$ to get $(1+i)$, then $k$ must equal $i$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 7, 2016 23:51:31 GMT
Vasco, Yes, I see it now. I actually tried the cross-ratio formula first and got the right answer, but I didn't recognize it. But I don't understand how you deduce that the constant is i from the fact that $1+i$ and $1-i$ are orthogonal. It it just that $(1-i)/(1+i)$ has arg = Pi/2 and multiplying by $i$ gives arg = 0 to D(1)? Gary Gary If $k\frac{1-i}{1+i}=1$ then $k(1-i)=(1+i)$ and so since $k$ must rotate $(1-i)$ through $\pi/2$ to get $(1+i)$, then $k$ must equal $i$. Vasco Vasco, Thanks. I thought you were somehow doing it by inspection of a plot. Even so, I hadn't thought of it like that. I just did the algebra. Gary |
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Post by Admin on Nov 8, 2016 6:50:57 GMT
Gary If $k\frac{1-i}{1+i}=1$ then $k(1-i)=(1+i)$ and so since $k$ must rotate $(1-i)$ through $\pi/2$ to get $(1+i)$, then $k$ must equal $i$. Vasco Vasco, Thanks. I thought you were somehow doing it by inspection of a plot. Even so, I hadn't thought of it like that. I just did the algebra. Gary Gary I was doing it by inspection of a plot - figure 35 on page 316 - I just did a bit of algebra above to explain it. Vasco |
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