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Post by mondo on Oct 1, 2022 6:40:23 GMT
The last sentence on page 151 says that the maximum number of fixed points of Mobius transformation given by $M(z) = z$ is two. If we transform it to a quadratic form we get $cz^2 + (d-a)z - b = 0$. My question is how can I now prove that this has at most two fixed points?
In the middle of page 154 there is a requirement for numerator and denominator to be in proportion $(z-q)$ and $(z-s)$ respectively. How was it derived?
Thank you.
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Post by Admin on Oct 1, 2022 9:21:58 GMT
Mondo The last sentence on page 151 says that the maximum number of fixed points of Mobius transformation given by $M(z) = z$ is two. If we transform it to a quadratic form we get $cz^2 + (d-a)z - b = 0$. My question is how can I now prove that this has at most two fixed points? I think it's best not to use the word "transform" here, as it has a special meaning in the book. Let's say instead we can rewrite the equation $M(z)=z$ as a quadratic equation. Also note that it's $M(z)$ that has the fixed points not the quadratic equation. The quadratic equation has two solutions (or roots) because quadratic equations can be written in the form $(z-A)(z-B)=0$ which means that $z=A$ or $z=B$ . In most cases $A$ and $B$ are different but they can be equal and so we can see that the quadratic has at most two roots and so $M$ has at most two fixed points. If $M(q)=0$ then the numerator of $M$ must be zero when $z=q$ and so $aq+b=0$ which means that $b=-aq$ and so the numerator of $M$ must be of the form $az-aq=a(z-q)$ and so the numerator of $M$ must be proportional to $(z-q)$. Similarly since $M(s)$ goes to infinity the denominator of $M$ equals zero when $z=s$ and so we use the same argument to show that the denominator of $M$ is proportional to $(z-s)$. It follows that $M(z)$ is proportional to $\displaystyle\bigg(\frac{z-q}{z-s}\bigg)$. Vasco |
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Post by mondo on Oct 3, 2022 23:40:11 GMT
I think it's best not to use the word "transform" here, as it has a special meaning in the book. Let's say instead we can rewrite the equation $M(z)=z$ as a quadratic equation. Also note that it's $M(z)$ that has the fixed points not the quadratic equation. The quadratic equation has two solutions (or roots) because quadratic equations can be written in the form $(z-A)(z-B)=0$ which means that $z=A$ or $z=B$ . In most cases $A$ and $B$ are different but they can be equal and so we can see that the quadratic has at most two roots and so $M$ has at most two fixed points. Yes that makes sense. Also author says the concept of fixed points is extremely important - what is so important about it? I also read next two subsections and it doesn't seem to be such a breakthrough. If $M(q)=0$ then the numerator of $M$ must be zero when $z=q$ and so $aq+b=0$ which means that $b=-aq$ and so the numerator of $M$ must be of the form $az-aq=a(z-q)$ and so the numerator of $M$ must be proportional to $(z-q)$. Similarly since $M(s)$ goes to infinity the denominator of $M$ equals zero when $z=s$ and so we use the same argument to show that the denominator of $M$ is proportional to $(z-s)$. It follows that $M(z)$ is proportional to $\displaystyle\bigg(\frac{z-q}{z-s}\bigg)$. Vasco Ok understood. Mobius transformation are all about ratios and proportions of these four coefficients. For this "Cross-Ratio" Subsection I have some more questions: 1.The statement (30) and accompanying Figure [27] say what happens to regions lying to left/right of $C$. They don't give any reasoning why it does transform these regions in such a way so I assume it is a meant to be understood by the previous subsections? In particular based on complex inversion, and discussion around figure [1] ? 2. Statement (31) says that a point lies on circle $C$ only if $Im[p,q,r,s] = 0$ why is that true? Book even says it can be immediately deduced from figure [28]. |
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Post by Admin on Oct 4, 2022 12:52:26 GMT
I think it's best not to use the word "transform" here, as it has a special meaning in the book. Let's say instead we can rewrite the equation $M(z)=z$ as a quadratic equation. Also note that it's $M(z)$ that has the fixed points not the quadratic equation. The quadratic equation has two solutions (or roots) because quadratic equations can be written in the form $(z-A)(z-B)=0$ which means that $z=A$ or $z=B$ . In most cases $A$ and $B$ are different but they can be equal and so we can see that the quadratic has at most two roots and so $M$ has at most two fixed points. Yes that makes sense. Also author says the concept of fixed points is extremely important - what is so important about it? I also read next two subsections and it doesn't seem to be such a breakthrough. Mondo
It helps us prove (24), which is important for chapter 6 and also to obtain the results in section VII on pages 162-171.
VascoIf $M(q)=0$ then the numerator of $M$ must be zero when $z=q$ and so $aq+b=0$ which means that $b=-aq$ and so the numerator of $M$ must be of the form $az-aq=a(z-q)$ and so the numerator of $M$ must be proportional to $(z-q)$. Similarly since $M(s)$ goes to infinity the denominator of $M$ equals zero when $z=s$ and so we use the same argument to show that the denominator of $M$ is proportional to $(z-s)$. It follows that $M(z)$ is proportional to $\displaystyle\bigg(\frac{z-q}{z-s}\bigg)$. Vasco Ok understood. Mobius transformation are all about ratios and proportions of these four coefficients. For this "Cross-Ratio" Subsection I have some more questions: 1.The statement (30) and accompanying Figure [27] say what happens to regions lying to left/right of $C$. They don't give any reasoning why it does transform these regions in such a way so I assume it is a meant to be understood by the previous subsections? In particular based on complex inversion, and discussion around figure [1] ? Mondo
No, it's all explained in subsection 1 of section V on pages 148-149.
Vasco2. Statement (31) says that a point lies on circle $C$ only if $Im[p,q,r,s] = 0$ why is that true? Book even says it can be immediately deduced from figure [28].
Yes, right. Look at figure [28]. Ask yourself what happens to any point on the circle $C$ when the transformation is applied.
Vasco |
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Post by mondo on Oct 4, 2022 19:30:34 GMT
Yes, right. Look at figure [28]. Ask yourself what happens to any point on the circle $C$ when the transformation is applied. I understand that points on the circle got mapped to points on a real line and hence the imaginary part of them is $0$. However first of all it won't be true for every transformation right? So we can't generalize "we deduced a neat equation for a circle $C$", can we? Also I am a bit confused by the mapping itself, if we send every three points to $0$, $1$ and $\infty$ respectively then their image should not be a line but just $0$, $1$ and $\infty$? |
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Post by Admin on Oct 5, 2022 8:17:46 GMT
Yes, right. Look at figure [28]. Ask yourself what happens to any point on the circle $C$ when the transformation is applied. I understand that points on the circle got mapped to points on a real line and hence the imaginary part of them is $0$. However first of all it won't be true for every transformation right? So we can't generalize "we deduced a neat equation for a circle $C$", can we? Yes we can, because the unique Mobius transformation that maps the 3 given points $q,r,s$ to $0,1,\infty$ is the cross ratio in the middle of page 154, and this maps the unique circle through $q,r,s$ to the real line. Don't forget that circles are mapped to circles by Mobius transformations, which includes straight lines, and so every point on the circle is mapped to a point on the real line by this Mobius transformation. We can choose any circle by choosing the 3 points $q,r,s$ and it will be mapped to the real line by a different Mobius transformation which has the same form $\displaystyle\frac{(z-q)(r-s)}{(z-s)(r-q)}$, but depends on the 3 points. So (31) on page 155 is the equation of the unique circle through $q,r,s$ where $p$ represents any point on this circle. You have misunderstood the situation. I think my answer to your first question answers this question as well. It's not every 3 points but any 3 points. Every set of 3 points gives us a different circle and a different Mobius transformation. This special Mobius transformation maps the circle to the real line every time. The form of the Mobius transformation is the same but each time we change $q,r,s$ we get a different Mobius transformation. Vasco |
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Post by mondo on Oct 9, 2022 3:53:45 GMT
Thank you Vasco, I have two problems: Every set of 3 points gives us a different circle... Vasco This can not be true as we can find another set of points that form the same circle right? The second problem is with: Don't forget that circles are mapped to circles by Mobius transformations, which includes straight lines, and so every point on the circle is mapped to a point on the real line by this Mobius transformation I think I need to reread at least two preceding subsections as I don't see why does $[z,q,r,s]$ map circles (and probably everything else) to a real line? I will edit if I figure it out. |
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Post by Admin on Oct 9, 2022 6:21:30 GMT
Thank you Vasco, I have two problems: Every set of 3 points gives us a different circle... Vasco This can not be true as we can find another set of points that form the same circle right? Mondo
Yes sorry, what I meant to write was "every set of 3 points defines a unique circle" or "There is only one circle through a given set of 3 points."The second problem is with: Don't forget that circles are mapped to circles by Mobius transformations, which includes straight lines, and so every point on the circle is mapped to a point on the real line by this Mobius transformation I think I need to reread at least two preceding subsections as I don't see why does $[z,q,r,s]$ map circles (and probably everything else) to a real line? I will edit if I figure it out.
Only points on the circle are mapped to the real line.
On page 154 we find that the unique Mobius transformation that transforms our 3 points to $0,1,\infty$ on the real line is $[z,q,r,s]$, and so all other points on the same circle must also be mapped to the real line.
Vasco |
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Post by mondo on Oct 10, 2022 6:56:41 GMT
Yes sorry, what I meant to write was "every set of 3 points defines a unique circle" or "There is only one circle through a given set of 3 points." Ok. Only points on the circle are mapped to the real line. On page 154 we find that the unique Mobius transformation that transforms our 3 points to $0,1,\infty$ on the real line is $[z,q,r,s]$, and so all other points on the same circle must also be mapped to the real line. From your answers it sounds obvious that $[z,q,r,s]$ maps every point on the circle to real line while other points are mapped in a different way. I have a problem seeing this. For example if in $R$ we look at the equation of a circle $x^2 + y^2 = r^2$ we can see it will generate a set of points that are equally distant from origin and that forms a circle. Now how can we see $[z,q,r,s]$ will map every point on a circle to a real line? More so, how are we sure every other point outside the circle won't get mapped to a real line? Also, we say this unique Mobius transformation maps three given points to $0, 1, \infty$ but we are in complex plane now, so what is $\infty$? In $R$ we could only have a positive or negative infinity but now there is "infinity many infinities" |
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Post by Admin on Oct 10, 2022 11:03:40 GMT
Mondo Yes sorry, what I meant to write was "every set of 3 points defines a unique circle" or "There is only one circle through a given set of 3 points." Ok. Only points on the circle are mapped to the real line. On page 154 we find that the unique Mobius transformation that transforms our 3 points to $0,1,\infty$ on the real line is $[z,q,r,s]$, and so all other points on the same circle must also be mapped to the real line. From your answers it sounds obvious that $[z,q,r,s]$ maps every point on the circle to real line while other points are mapped in a different way. I have a problem seeing this. For example if in $R$ we look at the equation of a circle $x^2 + y^2 = r^2$ we can see it will generate a set of points that are equally distant from origin and that forms a circle. Now how can we see $[z,q,r,s]$ will map every point on a circle to a real line? More so, how are we sure every other point outside the circle won't get mapped to a real line? All your questions above are answered in subsection 1 of Section V on pages 148-149. To understand this you need to read and understand section IV on pages 139-148 where among other things it explains "The point at infinity". Vasco |
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Post by mondo on Oct 17, 2022 21:01:39 GMT
Vasco, Sorry for late response. To understand this you need to read and understand section IV on pages 139-148 where among other things it explains "The point at infinity". ok, this is clear. From your answers it sounds obvious that $[z,q,r,s]$ maps every point on the circle to real line while other points are mapped in a different way. I have a problem seeing this. For example if in $R$ we look at the equation of a circle $x^2 + y^2 = r^2$ we can see it will generate a set of points that are equally distant from origin and that forms a circle. Now how can we see $[z,q,r,s]$ will map every point on a circle to a real line? More so, how are we sure every other point outside the circle won't get mapped to a real line? All your questions above are answered in subsection 1 of Section V on pages 148-149. If you are referring to a fragment then it is just a restate not an explanation. I think the first encounter of this property that circles are mapped to lines and vice verse is on page 127 and figure [3] as a result of complex inversion. However I found I can't find an explanation as to why only the points on the circle are mapped to the real line. It may just require some experience in performing these transformations to see is as clearly as my example with circle equation above. |
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Post by Admin on Oct 18, 2022 7:55:19 GMT
Mondo
On page 148 when Needham talks about the disc he is talking about points inside the circle. It does explain what happens to points not on the circle. You have to use the decomposition of the Mobius transformation and think about what happens to points not on the circle under these transformations.
The three bullet points are crucial to understanding this.
Vasco
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Post by mondo on Oct 19, 2022 0:37:25 GMT
Mondo On page 148 when Needham talks about the disc he is talking about points inside the circle. It does explain what happens to points not on the circle. You have to use the decomposition of the Mobius transformation and think about what happens to points not on the circle under these transformations. The three bullet points are crucial to understanding this. Vasco It is really hard to get. For now I only try to understand why points on the circle get mapped to points on the real line. When you say I should use decomposition do you mean I should decompose eq.29 $[z,q,r,s] = \frac{(z-q)(r-s)}{(z-s)(r-q)}$? I think in some other thread we agreed that we get this equation from elementary operations (3) and not the other way around right? Also from (29) I can't tell i.e what kind of inversion is it (in regards to what?). On page 27 where inversion is discussed, we always do it in regards to i.e a circle $K$ or some line. Page 148 which you suggested has some comments on what happens to $C$ depending on its relation to origin. However I would like to get that algebraically from equations like $[z,q,r,s]$ |
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Post by Admin on Oct 19, 2022 9:46:21 GMT
Mondo
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.
Using this fact it follows that:
Since in our particular case the circle through the points $q,r,s$ is transformed to a straight line (the real axis, which we take to be a circle through the point at infinity), then the Mobius transformation that maps the points $q,r,s$ on the circle, to the points $0,1,\infty$, on another "circle" (the real axis), must map all the points on the circle to points on the straight line (the real axis).
Vasco
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Post by mondo on Oct 19, 2022 17:35:31 GMT
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. Clear, also here you implicitly assume lines are special kind of circles. Since in our particular case the circle through the points $q,r,s$ is transformed to a straight line (the real axis, which we take to be a circle through the point at infinity), then the Mobius transformation that maps the points $q,r,s$ on the circle, to the points $0,1,\infty$, on another "circle" (the real axis), must map all the points on the circle to points on the straight line (the real axis). Yes I know that because the book says it on the page 155, right below figure [27] "This in turn gives us a more vivid picture...that maps the oriented circle $C$.. to the real axis". But what I am looking for is a proof why equation [29] does it? I mean, why it maps every point from the circle to a real line. I can see that for those three points $q,r,s$ since for example $M(r) = \frac{(r-q)(r-s)}{(r-s)(r-q)} = 1$. Can we get the confidence just by the way this mapping was created? After all we enforced mapping of three points to a points on a real axis so this is quite suggestive what will happen to the rest of them. Or maybe let me ask differently, if there were no figure [28] and accompanying description I quoted above, would you or a seasoned geometer be able to tell what eq. (29) does? Additionally, second to last paragraph on page 139 says "Conversely, if the image curve passes through $0$ then the original curve passed through point $\infty$" hmm but wait, (29) is supposed to be a generic formula that maps every circle to a real axis and not only those which pass through $\infty$ right? |
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