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Post by Admin on Oct 21, 2016 6:26:36 GMT
Gary
Here is my attempt at this suggested exercise:
Figure 35a on page 316 shows a mapping of the entire upper half-plane into the unit disc by means of the inversion
$z\mapsto\widetilde{z}=\mathcal{I}_K(z)$,
where $K$ is the illustrated circle centred at $-i$ and passing through $\pm1$.
In order for this disc to represent the hyperbolic plane, it must inherit its metric from the upper half-plane.
In other words we must define the h-separation $\mathcal{H}\{\widetilde{a},\widetilde{b}\}$ of any two points in the disc to be the h-separation $\mathcal{H}\{a,b\}$ of their preimages in the upper half-plane.
Any h-circle in the plane can be defined by its two end-points $p,q$ on the horizon, and the h-separation of these two points is $\mathcal{H}\{p,q\}$.
Inversion in $K$ maps this h-line through $p$ and $q$ to the arc of a circle inside the unit circle and orthogonal to it at the points $\widetilde{p}$ and $\widetilde{q}$, the images of $p$ and $q$.
By definition $\mathcal{H}\{\widetilde{p},\widetilde{q}\}=\mathcal{H}\{p,q\}$, and so the h-lines of the disc are precisely the images of the h-lines in the upper half-plane.
Vasco
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Nov 16, 2016 19:37:20 GMT
Gary Here is my attempt at this suggested exercise: Figure 35a on page 316 shows a mapping of the entire upper half-plane into the unit disc by means of the inversion $z\mapsto\widetilde{z}=\mathcal{I}_K(z)$, where $K$ is the illustrated circle centred at $-i$ and passing through $\pm1$. In order for this disc to represent the hyperbolic plane, it must inherit its metric from the upper half-plane. In other words we must define the h-separation $\mathcal{H}\{\widetilde{a},\widetilde{b}\}$ of any two points in the disc to be the h-separation $\mathcal{H}\{a,b\}$ of their preimages in the upper half-plane. Any h-circle in the plane can be defined by its two end-points $p,q$ on the horizon, and the h-separation of these two points is $\mathcal{H}\{p,q\}$. Inversion in $K$ maps this h-line through $p$ and $q$ to the arc of a circle inside the unit circle and orthogonal to it at the points $\widetilde{p}$ and $\widetilde{q}$, the images of $p$ and $q$. By definition $\mathcal{H}\{\widetilde{p},\widetilde{q}\}=\mathcal{H}\{p,q\}$, and so the h-lines of the disc are precisely the images of the h-lines in the upper half-plane. Vasco Vasco, It appears to be correct as far as it goes, but I think you might add "and conjugation" or "and $z \rightarrow \overline{z}$ right after "Inversion in $K$" in S -2. Figure [35a] illustrates only the inversion, but I think mention of the second step is necessary or at least helpful in proving the assertion. Gary |
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Post by Admin on Nov 17, 2016 7:34:12 GMT
Gary Here is my attempt at this suggested exercise: Figure 35a on page 316 shows a mapping of the entire upper half-plane into the unit disc by means of the inversion $z\mapsto\widetilde{z}=\mathcal{I}_K(z)$, where $K$ is the illustrated circle centred at $-i$ and passing through $\pm1$. In order for this disc to represent the hyperbolic plane, it must inherit its metric from the upper half-plane. In other words we must define the h-separation $\mathcal{H}\{\widetilde{a},\widetilde{b}\}$ of any two points in the disc to be the h-separation $\mathcal{H}\{a,b\}$ of their preimages in the upper half-plane. Any h-circle in the plane can be defined by its two end-points $p,q$ on the horizon, and the h-separation of these two points is $\mathcal{H}\{p,q\}$. Inversion in $K$ maps this h-line through $p$ and $q$ to the arc of a circle inside the unit circle and orthogonal to it at the points $\widetilde{p}$ and $\widetilde{q}$, the images of $p$ and $q$. By definition $\mathcal{H}\{\widetilde{p},\widetilde{q}\}=\mathcal{H}\{p,q\}$, and so the h-lines of the disc are precisely the images of the h-lines in the upper half-plane. Vasco Vasco, It appears to be correct as far as it goes, but I think you might add "and conjugation" or "and $z \rightarrow \overline{z}$ right after "Inversion in $K$" in S -2. Figure [35a] illustrates only the inversion, but I think mention of the second step is necessary or at least helpful in proving the assertion. Gary Gary At the point on page 316 where the exercise is posed Needham talks of the unit disc, and I understand this as referring to the mapping of the PUHP to the unit disc by means of the inversion in $K$. Needham has not defined the mapping to be the Poincaré disc. In the statement of the exercise Needham refers to the h-lines of the upper half-plane being mapped to h-lines in the unit disk (not the PD). I can see that when thinking about this exercise the reader might think of the unit disc as being the PD, but I don't think it matters. The exercise does not rely on the unit disc being the PD. One is just a reflection in the real axis of the other. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Nov 17, 2016 16:17:15 GMT
Vasco
On rereading, I agree.
Gary
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Post by Admin on Jan 29, 2017 10:07:26 GMT
Gary
As a result of some of our recent discussions which caused me to go back over certain parts of the text, I decided that my attempt at the suggested exercise near the top of page 316, was incorrect. Here is my new attempt:
Given any two points $a$ and $b$ on the PUHP and their images $\widetilde{a}$ and $\widetilde{b}$ on the PD, we can draw the unique h-line through $a$ and $b$. Since the transformation described in the first paragraph of page 316, from the PUHP to the PD, maps circles to circles and is anticonformal (preserves angles), then the h-line in the PUHP is mapped to the h-line in the PD through the points $\widetilde{a}$ and $\widetilde{b}$, as illustrated in figure 35a on page 316.
If we define the h-separation $\mathcal{H}\{\widetilde{a},\widetilde{b}\}$ of any two points on the PD to be equal to the h-separation $\mathcal{H}\{a,b\}$ of their preimages in the PUHP, then it follows that the h-lines of the disc are precisely the images of h-lines in the upper half-plane.
Vasco
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