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Post by Admin on Oct 13, 2016 9:17:19 GMT
Gary
Since $p$ is a point on $M$ then $\mathfrak{R}_M(p)=p$ and $\mathfrak{R}_M$ also swaps the two ends of $L$ on the horizon. Since these two points are also the ends of the asymptotic lines to $L$ and $p$ is a point on both asymptotic lines, then it follows that $\mathfrak{R}_M$ also swaps the two asymptotic lines.
So since $\mathfrak{R}_M$ is anticonformal the angle between any one of the asymptotic lines and $M$ at $p$ is carried to the same angle between the other asymptotic line and $M$. It follows that $M$ bisects the angle at $p$ contained by the asymptotic lines.
Vasco
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Oct 13, 2016 18:17:10 GMT
Gary Since $p$ is a point on $M$ then $\mathfrak{R}_M(p)=p$ and $\mathfrak{R}_M$ also swaps the two ends of $L$ on the horizon. Since these two points are also the ends of the asymptotic lines to $L$ and $p$ is a point on both asymptotic lines, then it follows that $\mathfrak{R}_M$ also swaps the two asymptotic lines. So since $\mathfrak{R}_M$ is anticonformal the angle between any one of the asymptotic lines and $M$ at $p$ is carried to the same angle between the other asymptotic line and $M$. It follows that $M$ bisects the angle at $p$ contained by the asymptotic lines. Vasco Vasco, It works for me on an intuitive level, but isn't there a need for some explanatory principle prior to "then it follows that $\mathfrak{R}_M$ also swaps the two asymptotic lines"? How do we know it is sufficient that there are inversions of the asymptotic lines at p and the ends of L? I thought of (1) sliding p along M towards q, or (2) rotating L around the center of M in order to get p to approximate q or the intersection of M with L. With (1), the asymptotics would have to shrink. With (2), L would have to shrink. Gary |
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Post by Admin on Oct 14, 2016 9:27:51 GMT
Gary Since $p$ is a point on $M$ then $\mathfrak{R}_M(p)=p$ and $\mathfrak{R}_M$ also swaps the two ends of $L$ on the horizon. Since these two points are also the ends of the asymptotic lines to $L$ and $p$ is a point on both asymptotic lines, then it follows that $\mathfrak{R}_M$ also swaps the two asymptotic lines. So since $\mathfrak{R}_M$ is anticonformal the angle between any one of the asymptotic lines and $M$ at $p$ is carried to the same angle between the other asymptotic line and $M$. It follows that $M$ bisects the angle at $p$ contained by the asymptotic lines. Vasco Vasco, It works for me on an intuitive level, but isn't there a need for some explanatory principle prior to "then it follows that $\mathfrak{R}_M$ also swaps the two asymptotic lines"? How do we know it is sufficient that there are inversions of the asymptotic lines at p and the ends of L? I thought of (1) sliding p along M towards q, or (2) rotating L around the center of M in order to get p to approximate q or the intersection of M with L. With (1), the asymptotics would have to shrink. With (2), L would have to shrink. Gary Gary Don't forget that here we are essentially reflecting circles in the circle $M$ and so angle magnitudes are preserved. Let's call the two asymptotic lines $A_1$ and $A_2$. Then $\mathfrak{R}_M(A_1)$ is the h-line which passes through $p$ and is orthogonal to the horizon at the other end of $L$. But this is the definition of $A_2$, and so $A_2=\mathfrak{R}_M(A_1)$. Done. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 14, 2016 23:35:43 GMT
Vasco, It works for me on an intuitive level, but isn't there a need for some explanatory principle prior to "then it follows that $\mathfrak{R}_M$ also swaps the two asymptotic lines"? How do we know it is sufficient that there are inversions of the asymptotic lines at p and the ends of L? I thought of (1) sliding p along M towards q, or (2) rotating L around the center of M in order to get p to approximate q or the intersection of M with L. With (1), the asymptotics would have to shrink. With (2), L would have to shrink. Gary Gary Don't forget that here we are essentially reflecting circles in the circle $M$ and so angle magnitudes are preserved. Let's call the two asymptotic lines $A_1$ and $A_2$. Then $\mathfrak{R}_M(A_1)$ is the h-line which passes through $p$ and is orthogonal to the horizon at the other end of $L$. But this is the definition of $A_2$, and so $A_2=\mathfrak{R}_M(A_1)$. Done. Vasco Vasco, This is nice. I recommend including it if you write up a document on the recommended exercises in Ch 6. Gary |
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