Mixing
A homeomorphism $T$ from extended complex plane to itself preserving cross ratio is a Mobius map.
$begingroup$
Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
If we assume $T$ fixes infinity, can we prove $T$ is affine?
complex-analysis mobius-transformation
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
asked Jan 31 at 20:08
SaikatSaikat
61
$endgroup$
add a comment |
$begingroup$
Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
If we assume $T$ fixes infinity, can we prove $T$ is affine?
complex-analysis mobius-transformation
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
asked Jan 31 at 20:08
SaikatSaikat
61
$endgroup$
add a comment |
$begingroup$
Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
If we assume $T$ fixes infinity, can we prove $T$ is affine?
complex-analysis mobius-transformation
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
asked Jan 31 at 20:08
SaikatSaikat
61
$endgroup$
Cross ratio preserving means $(Ta,Tb,Tb,Td)=(a,b,c,d)$ where $(a,b,c,d)=dfrac{(a-b)(c-d)}{(a-d)(c-b)}$.
If we assume $T$ fixes infinity, can we prove $T$ is affine?
complex-analysis mobius-transformation
complex-analysis mobius-transformation
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
asked Jan 31 at 20:08
SaikatSaikat
61
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
asked Jan 31 at 20:08
SaikatSaikat
61
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
edited Feb 13 at 18:17
Thomas Shelby
4,7362727
4,7362727
asked Jan 31 at 20:08
SaikatSaikat
61
asked Jan 31 at 20:08
SaikatSaikat
61
asked Jan 31 at 20:08
SaikatSaikat
61
61
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
My answer needs two facts about Möbius transforms:
- They preserve cross ratios (this can be checked by a direct calculation).
- A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).
You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.
However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.
We can express
$$M(z) = frac{alpha z + beta}{gamma z + delta}$$
Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.
answered Jan 31 at 20:35
0x5390x539
1,450518
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
My answer needs two facts about Möbius transforms:
- They preserve cross ratios (this can be checked by a direct calculation).
- A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).
You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.
However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.
We can express
$$M(z) = frac{alpha z + beta}{gamma z + delta}$$
Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.
answered Jan 31 at 20:35
0x5390x539
1,450518
$endgroup$
add a comment |
$begingroup$
My answer needs two facts about Möbius transforms:
- They preserve cross ratios (this can be checked by a direct calculation).
- A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).
You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.
However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.
We can express
$$M(z) = frac{alpha z + beta}{gamma z + delta}$$
Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.
answered Jan 31 at 20:35
0x5390x539
1,450518
$endgroup$
add a comment |
$begingroup$
My answer needs two facts about Möbius transforms:
- They preserve cross ratios (this can be checked by a direct calculation).
- A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).
You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.
However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.
We can express
$$M(z) = frac{alpha z + beta}{gamma z + delta}$$
Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.
answered Jan 31 at 20:35
0x5390x539
1,450518
$endgroup$
My answer needs two facts about Möbius transforms:
- They preserve cross ratios (this can be checked by a direct calculation).
- A Möbius transform can map three arbitrary points to any other three arbitrary points (this is not completely obvious, but at least very intuitive since Möbius transforms have 3 independent complex degrees of freedom).
You also need to know that for any fixed pairwise distinct $u, v, w in mathbb{C} cup {infty}$, the function $f : mathbb{C} cup {infty} to mathbb{C} cup {infty}, f(x) = (u, v; w, x)$ is bijective (again an easy calculation). This means that the images of three distinct points $a, b, c$ under your map $T$ already completely determine it.
However, there also is a Möbius $M$ transform that maps $a, b, c$ to $T(a), T(b), T(c)$ and, by virtue of being a Möbius transform preserves cross-ratios. Therefore $M = T$.
We can express
$$M(z) = frac{alpha z + beta}{gamma z + delta}$$
Suppose $M$ fixes $infty$, i.e. $M(infty) = frac{alpha cdot infty + beta}{gamma cdot infty + delta} = infty$. This is the case iff $gamma = 0$. In this case $M(z) = fracalphadelta z + fracbetadelta$ is affine.
answered Jan 31 at 20:35
0x5390x539
1,450518
answered Jan 31 at 20:35
0x5390x539
1,450518
answered Jan 31 at 20:35
0x5390x539
1,450518
answered Jan 31 at 20:35
0x5390x539
1,450518
1,450518
add a comment |
add a comment |
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