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Post by Admin on Oct 19, 2022 19:15:36 GMT
Mondo
Equation 29 gives us the Mobius transformation which sends the three points $q,r,s$ to the three points $\tilde{q},\tilde{r},\tilde{s}$, which is just the general case rather than a Mobius mapping to the three specific points $0,1,\infty$.
Your quote from page 139 is a completely different issue about considering a line as a circle through infinity.
Vasco
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Post by Admin on Oct 19, 2022 19:48:24 GMT
Mondo
If you understand all the things you say you understand then you should be able to see why all the points on the circle are mapped to the real line.
Vasco
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Post by mondo on Oct 19, 2022 19:57:20 GMT
Rereading this chapter now... hoping it will help me connecting the dots. Will be back with an update. Thanks for all the hints so far.
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Post by Admin on Oct 19, 2022 20:09:08 GMT
Mondo
In reply 14 you say that my first point is clear to you. If that is true then you must be able to see why all the points are mapped to the real line. If you can't see it then you can't possibly understand my first point.
Vasco
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Post by mondo on Oct 20, 2022 4:18:28 GMT
Mondo In reply 14 you say that my first point is clear to you. If that is true then you must be able to see why all the points are mapped to the real line. If you can't see it then you can't possibly understand my first point. Vasco Well I think it is, let's look at it again: OK, so let's take one thing at a time. Since any 3 points lie on a unique circle, and we know that all Mobius transformations map a circle to another circle, then all the points on the one circle must be mapped to points on the other circle. 3 points makes a unique circle - clear Mobius transformations map this circle to another circle - clear All points of that circle must be mapped to the other circle - ok so since this particular mapping maps three points of the circle to three points $0,1,\infty$ then we see that the image circle will actually be a circle with infinite radius - a line. A real line in this case. I think it answers my question as to why that generic circle is mapped to a real line. Next, in order to understand why the interior of circle got mapped to the upper halfplane I need to understand (23) on page 149. Author gives a comment before (23) "If $C$ does contain the origin then $\tilde{C}$ has the opposite orientation and the interior of $C$ is mapped to the exterior of $\tilde{C}$" - how to justify it? What is it based one? The inversion subchapter doesn't mention it. Also, I think I found a typo - on page 132, in the middle we can see a statement "Complex inversion , $z -> 1/z$, is conformal" which I agree with. However on the next page, we read "To understand this, first note that, since inversion is anticonformal.." - so either a typo or I miss something again.
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Post by Admin on Oct 20, 2022 7:55:40 GMT
Mondo In reply 14 you say that my first point is clear to you. If that is true then you must be able to see why all the points are mapped to the real line. If you can't see it then you can't possibly understand my first point. Vasco Well I think it is, let's look at it again: OK, so let's take one thing at a time. Since any 3 points lie on a unique circle, and we know that all Mobius transformations map a circle to another circle, then all the points on the one circle must be mapped to points on the other circle. 3 points makes a unique circle - clear Mobius transformations map this circle to another circle - clear All points of that circle must be mapped to the other circle - ok so since this particular mapping maps three points of the circle to three points $0,1,\infty$ then we see that the image circle will actually be a circle with infinite radius - a line. A real line in this case. I think it answers my question as to why that generic circle is mapped to a real line. Next, in order to understand why the interior of circle got mapped to the upper halfplane I need to understand (23) on page 149. Author gives a comment before (23) "If $C$ does contain the origin then $\tilde{C}$ has the opposite orientation and the interior of $C$ is mapped to the exterior of $\tilde{C}$" - how to justify it? What is it based one? The inversion subchapter doesn't mention it. Yes it does. Right in front of your eyes on page 148. Yes you have missed something again. I don't understand why you think this is a typo. I would suggest you read pages 124-126 again. Vasco
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Post by mondo on Oct 20, 2022 17:38:55 GMT
Well I think it is, let's look at it again: 3 points makes a unique circle - clear Mobius transformations map this circle to another circle - clear All points of that circle must be mapped to the other circle - ok so since this particular mapping maps three points of the circle to three points $0,1,\infty$ then we see that the image circle will actually be a circle with infinite radius - a line. A real line in this case. I think it answers my question as to why that generic circle is mapped to a real line. Next, in order to understand why the interior of circle got mapped to the upper halfplane I need to understand (23) on page 149. Author gives a comment before (23) "If $C$ does contain the origin then $\tilde{C}$ has the opposite orientation and the interior of $C$ is mapped to the exterior of $\tilde{C}$" - how to justify it? What is it based one? The inversion subchapter doesn't mention it. Yes it does. Right in front of your eyes on page 148. I think the key to understand it is to imagine what happens when we map point $0$ though witch $C$ passes - it gets mapped to $\infty$. Hence we have a line and everything that is inside the original circle must be mapped "outside" this line, on the far end, since the point on the circle is the closest to image line. Does my explanation sound logical to you Vasco? Also, I think I found a typo - on page 132, in the middle we can see a statement "Complex inversion , $z -> 1/z$, is conformal" which I agree with. However on the next page, we read "To understand this, first note that, since inversion is anticonformal.." - so either a typo or I miss something again. Yes you have missed something again. I don't understand why you think this is a typo. I would suggest you read pages 124-126 again. Vasco Ok so on page 132 it is about $\frac{1}{z}$ - a complex inversion, while on the next page it is about a "geometrical inversion" $\frac{1}{\bar{z}}$ which must be anticonformal since we have an odd number (1) of reflections, correct?
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Post by Admin on Oct 21, 2022 15:56:12 GMT
Mondo
The circle $C$ does not pass through the point zero in general. The points $q,r,s$ on the circle $C$ are mapped to the points $0,1,\infty$. You need to understand the last paragraph on page 148 and then the part on page 149 leading to (23).
Yes
Vasco
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Post by Admin on Oct 21, 2022 16:16:22 GMT
Mondo
Just edited my reply above.
vasco
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