Mixing
Mobius Geometry identity
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How do you prove that all figures consisting of three distinct points are congruent in Mobius Geometry?
I understand it relates to the Fundamental Theorem of Mobius Geometry. The concepts of which are very hard for me to grasp.
geometry mobius-transformation
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
$endgroup$
add a comment |
$begingroup$
How do you prove that all figures consisting of three distinct points are congruent in Mobius Geometry?
I understand it relates to the Fundamental Theorem of Mobius Geometry. The concepts of which are very hard for me to grasp.
geometry mobius-transformation
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
$endgroup$
$begingroup$
For a (rather understandible) application to quantum computing, take a look to "symmetric entanglement classes for n qubits " by Martin Aulbach that you will find on ResearchGate
$endgroup$
– Jean Marie
Jan 31 at 21:55
add a comment |
$begingroup$
How do you prove that all figures consisting of three distinct points are congruent in Mobius Geometry?
I understand it relates to the Fundamental Theorem of Mobius Geometry. The concepts of which are very hard for me to grasp.
geometry mobius-transformation
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
$endgroup$
How do you prove that all figures consisting of three distinct points are congruent in Mobius Geometry?
I understand it relates to the Fundamental Theorem of Mobius Geometry. The concepts of which are very hard for me to grasp.
geometry mobius-transformation
geometry mobius-transformation
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
edited Jan 31 at 6:52
YuiTo Cheng
2,2634937
2,2634937
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
asked Jan 31 at 3:10
Melissa JohnsonMelissa Johnson
134
134
$begingroup$
For a (rather understandible) application to quantum computing, take a look to "symmetric entanglement classes for n qubits " by Martin Aulbach that you will find on ResearchGate
$endgroup$
– Jean Marie
Jan 31 at 21:55
add a comment |
$begingroup$
For a (rather understandible) application to quantum computing, take a look to "symmetric entanglement classes for n qubits " by Martin Aulbach that you will find on ResearchGate
$endgroup$
– Jean Marie
Jan 31 at 21:55
$begingroup$
For a (rather understandible) application to quantum computing, take a look to "symmetric entanglement classes for n qubits " by Martin Aulbach that you will find on ResearchGate
$endgroup$
– Jean Marie
Jan 31 at 21:55
$begingroup$
For a (rather understandible) application to quantum computing, take a look to "symmetric entanglement classes for n qubits " by Martin Aulbach that you will find on ResearchGate
$endgroup$
– Jean Marie
Jan 31 at 21:55
add a comment |
1 Answer
1
active
oldest
votes
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I suppose there are many ways of looking at this, but here’s how I think of it: take your three distinct complex numbers, ${a,b,c}$. Then you can get a Möbius transformation mapping these to ${0,1,infty}$, namely
$$
zmapsto f(z)=frac{z-a}{z-c},,
$$
except of course that this sent $b$ to $lambda=frac{b-a}{b-c}$ instead of to $1$. Well: just change that $f$ to $frac1lambda f$. You still have $amapsto0$, $cmapstoinfty$, and now, besides, $bmapsto1$.
answered Jan 31 at 3:34
LubinLubin
45.5k44688
$endgroup$
add a comment |
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$begingroup$
I suppose there are many ways of looking at this, but here’s how I think of it: take your three distinct complex numbers, ${a,b,c}$. Then you can get a Möbius transformation mapping these to ${0,1,infty}$, namely
$$
zmapsto f(z)=frac{z-a}{z-c},,
$$
except of course that this sent $b$ to $lambda=frac{b-a}{b-c}$ instead of to $1$. Well: just change that $f$ to $frac1lambda f$. You still have $amapsto0$, $cmapstoinfty$, and now, besides, $bmapsto1$.
answered Jan 31 at 3:34
LubinLubin
45.5k44688
$endgroup$
add a comment |
$begingroup$
I suppose there are many ways of looking at this, but here’s how I think of it: take your three distinct complex numbers, ${a,b,c}$. Then you can get a Möbius transformation mapping these to ${0,1,infty}$, namely
$$
zmapsto f(z)=frac{z-a}{z-c},,
$$
except of course that this sent $b$ to $lambda=frac{b-a}{b-c}$ instead of to $1$. Well: just change that $f$ to $frac1lambda f$. You still have $amapsto0$, $cmapstoinfty$, and now, besides, $bmapsto1$.
answered Jan 31 at 3:34
LubinLubin
45.5k44688
$endgroup$
add a comment |
$begingroup$
I suppose there are many ways of looking at this, but here’s how I think of it: take your three distinct complex numbers, ${a,b,c}$. Then you can get a Möbius transformation mapping these to ${0,1,infty}$, namely
$$
zmapsto f(z)=frac{z-a}{z-c},,
$$
except of course that this sent $b$ to $lambda=frac{b-a}{b-c}$ instead of to $1$. Well: just change that $f$ to $frac1lambda f$. You still have $amapsto0$, $cmapstoinfty$, and now, besides, $bmapsto1$.
answered Jan 31 at 3:34
LubinLubin
45.5k44688
$endgroup$
I suppose there are many ways of looking at this, but here’s how I think of it: take your three distinct complex numbers, ${a,b,c}$. Then you can get a Möbius transformation mapping these to ${0,1,infty}$, namely
$$
zmapsto f(z)=frac{z-a}{z-c},,
$$
except of course that this sent $b$ to $lambda=frac{b-a}{b-c}$ instead of to $1$. Well: just change that $f$ to $frac1lambda f$. You still have $amapsto0$, $cmapstoinfty$, and now, besides, $bmapsto1$.
answered Jan 31 at 3:34
LubinLubin
45.5k44688
answered Jan 31 at 3:34
LubinLubin
45.5k44688
answered Jan 31 at 3:34
LubinLubin
45.5k44688
answered Jan 31 at 3:34
LubinLubin
45.5k44688
45.5k44688
add a comment |
add a comment |
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$begingroup$
For a (rather understandible) application to quantum computing, take a look to "symmetric entanglement classes for n qubits " by Martin Aulbach that you will find on ResearchGate
$endgroup$
– Jean Marie
Jan 31 at 21:55