Mixing
conformal automorphism $f$ of $D$ that interchanges
$begingroup$
Let $a$ and $b$ be distinct points in the unit disk $D$. Show that there exists a conformal automorphism $f$ of $D$ that interchanges
$a$ and $b$; that is, $f(a) = b$ and $f(b) = a$.
Idea: we know that $g(z)=frac{alpha-z}{1-bar{alpha}z}$ interchanges $0$ and $alpha$ and by composition we can find out the map $f(a) = b$ for any $a$ and $b$ in the unit disk $D$. But how can I get the other way by the same map? Thanks.
complex-analysis conformal-geometry mobius-transformation automorphism-group
edited Jan 30 at 6:37
Martin R
30.6k33558
asked Jan 29 at 15:09
420420
113
$endgroup$
add a comment |
$begingroup$
Let $a$ and $b$ be distinct points in the unit disk $D$. Show that there exists a conformal automorphism $f$ of $D$ that interchanges
$a$ and $b$; that is, $f(a) = b$ and $f(b) = a$.
Idea: we know that $g(z)=frac{alpha-z}{1-bar{alpha}z}$ interchanges $0$ and $alpha$ and by composition we can find out the map $f(a) = b$ for any $a$ and $b$ in the unit disk $D$. But how can I get the other way by the same map? Thanks.
complex-analysis conformal-geometry mobius-transformation automorphism-group
edited Jan 30 at 6:37
Martin R
30.6k33558
asked Jan 29 at 15:09
420420
113
$endgroup$
add a comment |
$begingroup$
Let $a$ and $b$ be distinct points in the unit disk $D$. Show that there exists a conformal automorphism $f$ of $D$ that interchanges
$a$ and $b$; that is, $f(a) = b$ and $f(b) = a$.
Idea: we know that $g(z)=frac{alpha-z}{1-bar{alpha}z}$ interchanges $0$ and $alpha$ and by composition we can find out the map $f(a) = b$ for any $a$ and $b$ in the unit disk $D$. But how can I get the other way by the same map? Thanks.
complex-analysis conformal-geometry mobius-transformation automorphism-group
edited Jan 30 at 6:37
Martin R
30.6k33558
asked Jan 29 at 15:09
420420
113
$endgroup$
Let $a$ and $b$ be distinct points in the unit disk $D$. Show that there exists a conformal automorphism $f$ of $D$ that interchanges
$a$ and $b$; that is, $f(a) = b$ and $f(b) = a$.
Idea: we know that $g(z)=frac{alpha-z}{1-bar{alpha}z}$ interchanges $0$ and $alpha$ and by composition we can find out the map $f(a) = b$ for any $a$ and $b$ in the unit disk $D$. But how can I get the other way by the same map? Thanks.
complex-analysis conformal-geometry mobius-transformation automorphism-group
complex-analysis conformal-geometry mobius-transformation automorphism-group
edited Jan 30 at 6:37
Martin R
30.6k33558
asked Jan 29 at 15:09
420420
113
edited Jan 30 at 6:37
Martin R
30.6k33558
asked Jan 29 at 15:09
420420
113
edited Jan 30 at 6:37
Martin R
30.6k33558
edited Jan 30 at 6:37
Martin R
30.6k33558
edited Jan 30 at 6:37
Martin R
30.6k33558
30.6k33558
asked Jan 29 at 15:09
420420
113
asked Jan 29 at 15:09
420420
113
asked Jan 29 at 15:09
420420
113
113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $a' = 1/overline a, ,b' = 1/overline b$. Take the Mobius transformation $f$ that maps the points $a, b, a'$ to $b, a, b'$. Since $f$ preserves the cross-ratio, we get
$$(a, b; a', b') =
(b, a; b', f(b')) =
(a, b; f(b'), b'),$$
therefore $f(b') = a'$. Since $b$ and $b'$ are symmetric (conjugate) wrt the unit circle $mathcal C$, their images $a$ and $a'$ are symmetric wrt $f(mathcal C)$. In the same way, $b$ and $b'$ are symmetric wrt $f(mathcal C)$.
These two pairs of symmetric points uniquely determine the circle, therefore $f(mathcal C) = mathcal C$. (If $a, b$ and the origin $O$ are not collinear, then $f(mathcal C)$ has to be a circle with the center at the intersection of $aa'$ and $bb'$, which is $O$, and with the radius $sqrt{ |a| cdot |a'|} = 1$. If $a, b, O$ are collinear, the center is found from a linear equation.)
answered Jan 30 at 1:03
MaximMaxim
6,1181221
$endgroup$
add a comment |
$begingroup$
As you already noticed, for $alpha in Bbb D$ the Möbius transformation
$$
T_alpha(z) = frac{alpha - z}{1- bar alpha z}
$$
is an automorphism of $ Bbb D$ which interchanges the points $0$ and $alpha$. This can be used to construct an automorphism interchanging two given points $a, b in Bbb D$: With $c = T_a(b) = frac{a- b}{1- bar a b}$ the Möbius transformation
$$
f = T_a^{-1} circ T_c circ T_a
$$
has the desired properties:
$$
begin{matrix}
& T_a & & T_c & & T_a^{-1}\
a & to & 0 & to & c & to & b\
b & to & c & to & 0 & to & a
end{matrix}
$$
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a' = 1/overline a, ,b' = 1/overline b$. Take the Mobius transformation $f$ that maps the points $a, b, a'$ to $b, a, b'$. Since $f$ preserves the cross-ratio, we get
$$(a, b; a', b') =
(b, a; b', f(b')) =
(a, b; f(b'), b'),$$
therefore $f(b') = a'$. Since $b$ and $b'$ are symmetric (conjugate) wrt the unit circle $mathcal C$, their images $a$ and $a'$ are symmetric wrt $f(mathcal C)$. In the same way, $b$ and $b'$ are symmetric wrt $f(mathcal C)$.
These two pairs of symmetric points uniquely determine the circle, therefore $f(mathcal C) = mathcal C$. (If $a, b$ and the origin $O$ are not collinear, then $f(mathcal C)$ has to be a circle with the center at the intersection of $aa'$ and $bb'$, which is $O$, and with the radius $sqrt{ |a| cdot |a'|} = 1$. If $a, b, O$ are collinear, the center is found from a linear equation.)
answered Jan 30 at 1:03
MaximMaxim
6,1181221
$endgroup$
add a comment |
$begingroup$
Let $a' = 1/overline a, ,b' = 1/overline b$. Take the Mobius transformation $f$ that maps the points $a, b, a'$ to $b, a, b'$. Since $f$ preserves the cross-ratio, we get
$$(a, b; a', b') =
(b, a; b', f(b')) =
(a, b; f(b'), b'),$$
therefore $f(b') = a'$. Since $b$ and $b'$ are symmetric (conjugate) wrt the unit circle $mathcal C$, their images $a$ and $a'$ are symmetric wrt $f(mathcal C)$. In the same way, $b$ and $b'$ are symmetric wrt $f(mathcal C)$.
These two pairs of symmetric points uniquely determine the circle, therefore $f(mathcal C) = mathcal C$. (If $a, b$ and the origin $O$ are not collinear, then $f(mathcal C)$ has to be a circle with the center at the intersection of $aa'$ and $bb'$, which is $O$, and with the radius $sqrt{ |a| cdot |a'|} = 1$. If $a, b, O$ are collinear, the center is found from a linear equation.)
answered Jan 30 at 1:03
MaximMaxim
6,1181221
$endgroup$
add a comment |
$begingroup$
Let $a' = 1/overline a, ,b' = 1/overline b$. Take the Mobius transformation $f$ that maps the points $a, b, a'$ to $b, a, b'$. Since $f$ preserves the cross-ratio, we get
$$(a, b; a', b') =
(b, a; b', f(b')) =
(a, b; f(b'), b'),$$
therefore $f(b') = a'$. Since $b$ and $b'$ are symmetric (conjugate) wrt the unit circle $mathcal C$, their images $a$ and $a'$ are symmetric wrt $f(mathcal C)$. In the same way, $b$ and $b'$ are symmetric wrt $f(mathcal C)$.
These two pairs of symmetric points uniquely determine the circle, therefore $f(mathcal C) = mathcal C$. (If $a, b$ and the origin $O$ are not collinear, then $f(mathcal C)$ has to be a circle with the center at the intersection of $aa'$ and $bb'$, which is $O$, and with the radius $sqrt{ |a| cdot |a'|} = 1$. If $a, b, O$ are collinear, the center is found from a linear equation.)
answered Jan 30 at 1:03
MaximMaxim
6,1181221
$endgroup$
Let $a' = 1/overline a, ,b' = 1/overline b$. Take the Mobius transformation $f$ that maps the points $a, b, a'$ to $b, a, b'$. Since $f$ preserves the cross-ratio, we get
$$(a, b; a', b') =
(b, a; b', f(b')) =
(a, b; f(b'), b'),$$
therefore $f(b') = a'$. Since $b$ and $b'$ are symmetric (conjugate) wrt the unit circle $mathcal C$, their images $a$ and $a'$ are symmetric wrt $f(mathcal C)$. In the same way, $b$ and $b'$ are symmetric wrt $f(mathcal C)$.
These two pairs of symmetric points uniquely determine the circle, therefore $f(mathcal C) = mathcal C$. (If $a, b$ and the origin $O$ are not collinear, then $f(mathcal C)$ has to be a circle with the center at the intersection of $aa'$ and $bb'$, which is $O$, and with the radius $sqrt{ |a| cdot |a'|} = 1$. If $a, b, O$ are collinear, the center is found from a linear equation.)
answered Jan 30 at 1:03
MaximMaxim
6,1181221
answered Jan 30 at 1:03
MaximMaxim
6,1181221
answered Jan 30 at 1:03
MaximMaxim
6,1181221
answered Jan 30 at 1:03
MaximMaxim
6,1181221
6,1181221
add a comment |
add a comment |
$begingroup$
As you already noticed, for $alpha in Bbb D$ the Möbius transformation
$$
T_alpha(z) = frac{alpha - z}{1- bar alpha z}
$$
is an automorphism of $ Bbb D$ which interchanges the points $0$ and $alpha$. This can be used to construct an automorphism interchanging two given points $a, b in Bbb D$: With $c = T_a(b) = frac{a- b}{1- bar a b}$ the Möbius transformation
$$
f = T_a^{-1} circ T_c circ T_a
$$
has the desired properties:
$$
begin{matrix}
& T_a & & T_c & & T_a^{-1}\
a & to & 0 & to & c & to & b\
b & to & c & to & 0 & to & a
end{matrix}
$$
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
$endgroup$
add a comment |
$begingroup$
As you already noticed, for $alpha in Bbb D$ the Möbius transformation
$$
T_alpha(z) = frac{alpha - z}{1- bar alpha z}
$$
is an automorphism of $ Bbb D$ which interchanges the points $0$ and $alpha$. This can be used to construct an automorphism interchanging two given points $a, b in Bbb D$: With $c = T_a(b) = frac{a- b}{1- bar a b}$ the Möbius transformation
$$
f = T_a^{-1} circ T_c circ T_a
$$
has the desired properties:
$$
begin{matrix}
& T_a & & T_c & & T_a^{-1}\
a & to & 0 & to & c & to & b\
b & to & c & to & 0 & to & a
end{matrix}
$$
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
$endgroup$
add a comment |
$begingroup$
As you already noticed, for $alpha in Bbb D$ the Möbius transformation
$$
T_alpha(z) = frac{alpha - z}{1- bar alpha z}
$$
is an automorphism of $ Bbb D$ which interchanges the points $0$ and $alpha$. This can be used to construct an automorphism interchanging two given points $a, b in Bbb D$: With $c = T_a(b) = frac{a- b}{1- bar a b}$ the Möbius transformation
$$
f = T_a^{-1} circ T_c circ T_a
$$
has the desired properties:
$$
begin{matrix}
& T_a & & T_c & & T_a^{-1}\
a & to & 0 & to & c & to & b\
b & to & c & to & 0 & to & a
end{matrix}
$$
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
$endgroup$
As you already noticed, for $alpha in Bbb D$ the Möbius transformation
$$
T_alpha(z) = frac{alpha - z}{1- bar alpha z}
$$
is an automorphism of $ Bbb D$ which interchanges the points $0$ and $alpha$. This can be used to construct an automorphism interchanging two given points $a, b in Bbb D$: With $c = T_a(b) = frac{a- b}{1- bar a b}$ the Möbius transformation
$$
f = T_a^{-1} circ T_c circ T_a
$$
has the desired properties:
$$
begin{matrix}
& T_a & & T_c & & T_a^{-1}\
a & to & 0 & to & c & to & b\
b & to & c & to & 0 & to & a
end{matrix}
$$
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
answered Jan 30 at 6:26
Martin RMartin R
30.6k33558
30.6k33558
add a comment |
add a comment |
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