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Post by Beginner on Oct 21, 2021 22:44:41 GMT
On page 153 there is provided the following statement (28):
The first part is clear, but it looks to me that `0` stops being a fixed point as soon as `b` not equals `0`, while I don't see how it follows that $a=1$ in this case. For example,
$M(z) = 2z + 3$
has only $\infty$ as a fixed point, but it is not a translation, if I am not missing something very basic. Could someone please clarify this?
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Post by Admin on Oct 21, 2021 23:30:54 GMT
Hi
$M(z)=2z+3$ also has a fixed point at $z=-3$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 22, 2021 4:14:13 GMT
On page 153 there is provided the following statement (28): The first part is clear, but it looks to me that `0` stops being a fixed point as soon as `b` not equals `0`, while I don't see how it follows that $a=1$ in this case. For example, $M(z) = 2z + 3$ has only $\infty$ as a fixed point, but it is not a translation, if I am not missing something very basic. Could someone please clarify this? The parabolic transformation has only one fixed point. From (28), p. 153: $\infty$ is the sole fixed point if and only if $M(z)$ is a translation, $M(z) = z + b$. On p. 169, we find If $M(z)=\frac{az+b}{cz+d}$ is normalized, then we know from (27), p. 152, that it is parabolic if and only if $(a+d) = \pm 2$, in which case the single fixed point is $\xi = (a-d)/2c$. That is all we need to define this class.
Rewrite the second $M$ as $M(z) = \frac{z+b}{0 \cdot z + 1}$. Then a = 1, c = 0, and d = 1. Notice that b does not figure into (27). Substitute these normalized coefficients into (27), $\hspace{5em}\xi = \frac{(1-1) \pm \sqrt{(\pm 2)^2 - 4}}{2 \cdot 0} = \frac{0}{0} = ?$ Given the homogeneous coordinates, is this result equal to $0, \infty$, or undefined? Something doesn't add up. Perhaps it is my algebra.
For $M(z) = 2z + 3$, [using $\xi = (a-d)/2c$ with normalized coefficients], where $N = \frac{1}{\sqrt{ad-bc}}$ [with the given coefficients], $\hspace{3em}N = \frac{1}{\sqrt{2}}$. $\hspace{3em}M(z) = \frac{2N-N}{0} = \frac{\frac{1}{\sqrt{2}}}{0} = \infty$
Gary
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Post by Admin on Oct 22, 2021 10:29:23 GMT
Gary
The equation for the fixed points is $\frac{az+b}{cz+d}=z$ which is the quadratic equation $cz^2+(d-a)z-b=0$, but if $c=0$, this is no longer a quadratic and so (27), which is obtained from the equation for the roots of a quadratic, is not valid in this case (c=0).
Also on page 158, line 2, we see that Needham states that we must specify that there is no homogeneous coordinate corresponding to [0,0] - it is explicitly excluded.
Again since $M(z)=2z+3$ is not normalised this is another reason for not using (27).
Vasco
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Post by Admin on Oct 22, 2021 13:38:35 GMT
It's probably worth noting that the decomposition (3) on page 124 also breaks down when $c=0$ and the new decomposition of the resulting similarity transformation $Az+B$ is:
(i) $z\mapsto Az$ which is an expansion and a rotation
(ii)$z\mapsto z+B$ which is a translation
Vasco
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Post by Beginner on Oct 22, 2021 14:51:38 GMT
Thanks a lot for a clarification! It is actually quite an embarrassing, I don't know how I could not see that :-(. Maybe it is worth removing this thread for an admin, I don't think it really adds any value to this forum.
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 22, 2021 16:40:46 GMT
Gary The equation for the fixed points is $\frac{az+b}{cz+d}=z$ which is the quadratic equation $cz^2+(d-a)z-b=0$, but if $c=0$, this is no longer a quadratic and so (27), which is obtained from the equation for the roots of a quadratic, is not valid in this case (c=0). Also on page 158, line 2, we see that Needham states that we must specify that there is no homogeneous coordinate corresponding to [0,0] - it is explicitly excluded. Again since $M(z)=2z+3$ is not normalised this is another reason for not using (27). Vasco I agree that Vasco has clarified the construction. It also helps to clarify other things, at least for me. [Archetypal] elliptic, hyperbolic, and parabolic transformations all have a fixed point at infinity, but only the parabolic lacks a second fixed point at 0. (27) is given as a formula that applies to any [normalized] Möbius transformation, except that in the parabolic case, the equation reduces to $\xi = \frac{a-d}{2c}$, which implies that $c \neq 0$ for at least some normalized parabolic Möbius transformations. I think one might form the impression that this is the case for all [normalized] parabolic Möbius transformations. Subsection 5 takes up Möbius transformations where $c = 0$, which entails at least one fixed point at infinity. The four subtypes are projected onto the Riemann sphere. The parabolic transformation is one of these. At this point, one might form the impression that the parabolic transformation entails $c = 0$. Then a careful reading of (28) and the prior paragraph shows that the parabolic type is the pure translation $M(z) = (z+b)$, where $c = 0$. The subsection ends with the statement that each Möbius transformation [not just the ones where $c = 0$] is equivalent, in a certain sense, to one (and only one) of the four types … [for which $c = 0$]. With that statement, one would like to know more about what is implied by equivalent, in a certain sense. Perhaps is is a reference to the list (44) on p. 170 and the list on p. 180.This is a rather complicated construction. Perhaps it leaps off the page to some, but it took me a few readings and Vasco’s clarification to see it. It will take more work to sort out the connections to list (44). Gary
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Post by Admin on Oct 22, 2021 17:34:00 GMT
Gary
This is not true of the transformations you name but of the $\widetilde{M}$ transformations obtained from them.
I think the rest of your comments stem from this same misunderstanding.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 22, 2021 23:33:26 GMT
Gary This is not true of the transformations you name but of the $\widetilde{M}$ transformations obtained from them. I think the rest of your comments stem from this same misunderstanding. Vasco
Vasco,
I do not disagree, but I think this point is where confusion easily arises. In $\S\ 5, \S\S\ 5$ Fixed Points at Infinity, the idea of $\widetilde{M}$ as a Möbius transformation in the new plane created by $F$ and illustrated in [29] has not yet been introduced. Even so, in $\S\S 5$, we are in the realm of $c = 0$, fixed points 0 and $\infty$, and Möbius transformations labeled with unvarnished $M(z)$. These transformations are all similarity transformations and they provide the “simplest, archetypal” examples of elliptic, hyperbolic, loxodromic, and parabolic Möbius transformations. The framework defining the relation of these archetypal examples to ordinary Möbius transformations does not appear until $\S VII$ Visualization and Classification, $\S\S 1$ The Main Idea, which presents $F(z) = \frac{z-\xi_+}{z-\xi_-}$ I did fail to grasp the implications of the archetypal status of the similarity Möbius transformations presented in $\S\S 5$ and their dependence on $F(z)$, but the rest of my comments still strike me as valid up to the last sentence of paragraph 1. The phrase equivalent, in a certain sense now seems to me to be directed at $F(z)$.
Gary
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Gary
GaryVasco
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Post by Gary on Oct 23, 2021 5:12:43 GMT
Vasco, This is a reply to reply 3, Granted. But we should be able to use $\xi = \frac{a-d}{2c}$ applied to $M(z) = \frac{1 \cdot z + b}{0 \cdot z + 1}$, which is the archetypal parabolic Möbius transformation. Or does the parabolic form of (27) only apply in cases where the parabolic form is non-archetypal? That seems odd, but the equation for the parabolic $\xi$ doesn't seem to work out well when applied to the archetypal parabolic form: It seems clear that $a = 1, d = 1$, and $c = 0$. Normalizing, we find that $ad - bc = 1$, so the normalized coefficients are the same as the originals. $\hspace{3em} \xi = \frac{1-1}{2 \cdot 0} = \frac{0}{0}$, which is undefined So it seems that something is still amiss. Gary
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Post by Admin on Oct 23, 2021 8:46:35 GMT
Gary
Replying to reply 9 Subsection 4 Fixed points of chapter 3 is dealing with the general Mobius transformations where both of the fixed points lie in the finite plane, and this means that $c$ must be non-zero (see the first sentence of subsection 5).
When, at the end of subsection 4, Needham talks about the exceptional case when the two fixed points coalesce into one fixed point, he is still talking about a quadratic, with two repeated roots that lie in the finite plane, and so where $c$ is non-zero.
The last sentence of subsection 4 is just saying that in the cases where there is only one fixed point the transformation is called parabolic, regardless of the value of $c$.
We can't use the formula he gives there for the repeated root when $c=0$ because it has been derived using a quadratic equation, assuming that $c$ is non-zero.
Subsection 5 deals specifically with the cases where $c=0$ i.e. when $M(z)=az/d+b/d=Az+B$, and shows that when $M$ is a translation ($A=1$) then it only has one fixed point, and so by the definition at the end of subsection 4 referred to above, it is parabolic.
Vasco
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Post by Admin on Oct 23, 2021 10:22:59 GMT
Replying to reply 8.
Gary
Yes I misinterpreted what you were saying in reply 8. Apologies for that. However I don't think the phrase you mention is directed at $F$.
Do you agree that the words "each Mobius transformation" in the last sentence on page 153 refer to general Mobius transformations and not to the special ones when $c=0$?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 23, 2021 17:54:35 GMT
Gary Replying to reply 9 Subsection 4 Fixed points of chapter 3 is dealing with the general Mobius transformations where both of the fixed points lie in the finite plane, and this means that $c$ must be non-zero (see the first sentence of subsection 5). When, at the end of subsection 4, Needham talks about the exceptional case when the two fixed points coalesce into one fixed point, he is still talking about a quadratic, with two repeated roots that lie in the finite plane, and so where $c$ is non-zero. The last sentence of subsection 4 is just saying that in the cases where there is only one fixed point the transformation is called parabolic, regardless of the value of $c$. We can't use the formula he gives there for the repeated root when $c=0$ because it has been derived using a quadratic equation, assuming that $c$ is non-zero. Subsection 5 deals specifically with the cases where $c=0$ i.e. when $M(z)=az/d+b/d=Az+B$, and shows that when $M$ is a translation ($A=1$) then it only has one fixed point, and so by the definition at the end of subsection 4 referred to above, it is parabolic. Vasco Vasco,
This seems to sort it out nicely. The initial phrase "In this case" of the last sentence of Section V, Subsection 4 seems ambiguous between any case with a single fixed point and the case with a single fixed point where the quadratic applies. The grammar suggests the more restrictive interpretation, but perhaps that was not intended.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 23, 2021 17:56:14 GMT
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Post by Admin on Oct 23, 2021 18:25:27 GMT
Gary Replying to reply 9 Subsection 4 Fixed points of chapter 3 is dealing with the general Mobius transformations where both of the fixed points lie in the finite plane, and this means that $c$ must be non-zero (see the first sentence of subsection 5). When, at the end of subsection 4, Needham talks about the exceptional case when the two fixed points coalesce into one fixed point, he is still talking about a quadratic, with two repeated roots that lie in the finite plane, and so where $c$ is non-zero. The last sentence of subsection 4 is just saying that in the cases where there is only one fixed point the transformation is called parabolic, regardless of the value of $c$. We can't use the formula he gives there for the repeated root when $c=0$ because it has been derived using a quadratic equation, assuming that $c$ is non-zero. Subsection 5 deals specifically with the cases where $c=0$ i.e. when $M(z)=az/d+b/d=Az+B$, and shows that when $M$ is a translation ($A=1$) then it only has one fixed point, and so by the definition at the end of subsection 4 referred to above, it is parabolic. Vasco Vasco, This seems to sort it out nicely. The initial phrase of Section V, Subsection 4 seems ambiguous between any case with a single fixed point and the case with a single fixed point where the quadratic applies. The grammar suggests the more restrictive interpretation, but perhaps that was not intended. Gary
Gary Which initial phrase are you referring to, or do you mean the last phrase? I agree that the last sentence in subsection 4 on page 152 is a bit ambiguous. Vasco
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