Mapping the Riemann sphere with unit interval deleted to the unit disk
Let [0, 1] ⊂ C stand for the closed interval between 0 and 1, and Ω =
C [0, 1] ∪ {∞} a subset of the Riemann sphere. Find a biholomorphic map of Ω on the unit disc.
The Riemann sphere with the real unit interval deleted is not simply connected, so I don't see how there can be such a mapping, but if someone could give me a hint it would be much appreciated.
complex-analysis
asked Jan 3 at 23:30
David Warren
586313
add a comment |
Let [0, 1] ⊂ C stand for the closed interval between 0 and 1, and Ω =
C [0, 1] ∪ {∞} a subset of the Riemann sphere. Find a biholomorphic map of Ω on the unit disc.
The Riemann sphere with the real unit interval deleted is not simply connected, so I don't see how there can be such a mapping, but if someone could give me a hint it would be much appreciated.
complex-analysis
asked Jan 3 at 23:30
David Warren
586313
The Riemann sphere without the unit interval deleted is simply connected (in the same way as the punctured sphere is simply connected).
– 0x539
Jan 3 at 23:38
Do you need an explicit map or is it sufficient to prove existence?
– 0x539
Jan 3 at 23:38
either would help, but I'm looking for a specific map
– David Warren
Jan 3 at 23:43
add a comment |
Let [0, 1] ⊂ C stand for the closed interval between 0 and 1, and Ω =
C [0, 1] ∪ {∞} a subset of the Riemann sphere. Find a biholomorphic map of Ω on the unit disc.
The Riemann sphere with the real unit interval deleted is not simply connected, so I don't see how there can be such a mapping, but if someone could give me a hint it would be much appreciated.
complex-analysis
asked Jan 3 at 23:30
David Warren
586313
Let [0, 1] ⊂ C stand for the closed interval between 0 and 1, and Ω =
C [0, 1] ∪ {∞} a subset of the Riemann sphere. Find a biholomorphic map of Ω on the unit disc.
The Riemann sphere with the real unit interval deleted is not simply connected, so I don't see how there can be such a mapping, but if someone could give me a hint it would be much appreciated.
complex-analysis
complex-analysis
asked Jan 3 at 23:30
David Warren
586313
asked Jan 3 at 23:30
David Warren
586313
asked Jan 3 at 23:30
David Warren
586313
asked Jan 3 at 23:30
David Warren
586313
asked Jan 3 at 23:30
David Warren
586313
586313
The Riemann sphere without the unit interval deleted is simply connected (in the same way as the punctured sphere is simply connected).
– 0x539
Jan 3 at 23:38
Do you need an explicit map or is it sufficient to prove existence?
– 0x539
Jan 3 at 23:38
either would help, but I'm looking for a specific map
– David Warren
Jan 3 at 23:43
add a comment |
The Riemann sphere without the unit interval deleted is simply connected (in the same way as the punctured sphere is simply connected).
– 0x539
Jan 3 at 23:38
Do you need an explicit map or is it sufficient to prove existence?
– 0x539
Jan 3 at 23:38
either would help, but I'm looking for a specific map
– David Warren
Jan 3 at 23:43
The Riemann sphere without the unit interval deleted is simply connected (in the same way as the punctured sphere is simply connected).
– 0x539
Jan 3 at 23:38
The Riemann sphere without the unit interval deleted is simply connected (in the same way as the punctured sphere is simply connected).
– 0x539
Jan 3 at 23:38
Do you need an explicit map or is it sufficient to prove existence?
– 0x539
Jan 3 at 23:38
Do you need an explicit map or is it sufficient to prove existence?
– 0x539
Jan 3 at 23:38
either would help, but I'm looking for a specific map
– David Warren
Jan 3 at 23:43
either would help, but I'm looking for a specific map
– David Warren
Jan 3 at 23:43
add a comment |
1 Answer
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The existence of such a mapping follows immediately from the Riemann mapping theorem, since it basically says that any simply connected subset of the Riemann sphere which does not contain at least two points of the sphere is biholomorphically equivalent to the unit disk.
I will construct such a mapping in three steps:
Firstly, $f(z) = 1/z$ maps $mathbb{C} cup {infty} setminus [0, 1]$ to $mathbb{C} cup {infty} setminus [1, infty] = mathbb{C} setminus [1, infty)$.
Secondly, $g(z) = sqrt{z - 1}$ maps $mathbb{C} setminus [1, infty)$ to the upper half plane ${z in mathbb{C} | operatorname{Im} z > 0}$.
Finally, $h(z) = frac{z-i}{z + i}$, the Cayley transform, maps the upper half plane to the unit disk.
$f, g, h$ are biholomorphic on their respective domains, so $h circ g circ f$ does the job. Explicitly, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.
answered Jan 3 at 23:48
0x539
1,047316
add a comment |
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1 Answer
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The existence of such a mapping follows immediately from the Riemann mapping theorem, since it basically says that any simply connected subset of the Riemann sphere which does not contain at least two points of the sphere is biholomorphically equivalent to the unit disk.
I will construct such a mapping in three steps:
Firstly, $f(z) = 1/z$ maps $mathbb{C} cup {infty} setminus [0, 1]$ to $mathbb{C} cup {infty} setminus [1, infty] = mathbb{C} setminus [1, infty)$.
Secondly, $g(z) = sqrt{z - 1}$ maps $mathbb{C} setminus [1, infty)$ to the upper half plane ${z in mathbb{C} | operatorname{Im} z > 0}$.
Finally, $h(z) = frac{z-i}{z + i}$, the Cayley transform, maps the upper half plane to the unit disk.
$f, g, h$ are biholomorphic on their respective domains, so $h circ g circ f$ does the job. Explicitly, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.
answered Jan 3 at 23:48
0x539
1,047316
add a comment |
The existence of such a mapping follows immediately from the Riemann mapping theorem, since it basically says that any simply connected subset of the Riemann sphere which does not contain at least two points of the sphere is biholomorphically equivalent to the unit disk.
I will construct such a mapping in three steps:
Firstly, $f(z) = 1/z$ maps $mathbb{C} cup {infty} setminus [0, 1]$ to $mathbb{C} cup {infty} setminus [1, infty] = mathbb{C} setminus [1, infty)$.
Secondly, $g(z) = sqrt{z - 1}$ maps $mathbb{C} setminus [1, infty)$ to the upper half plane ${z in mathbb{C} | operatorname{Im} z > 0}$.
Finally, $h(z) = frac{z-i}{z + i}$, the Cayley transform, maps the upper half plane to the unit disk.
$f, g, h$ are biholomorphic on their respective domains, so $h circ g circ f$ does the job. Explicitly, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.
answered Jan 3 at 23:48
0x539
1,047316
add a comment |
The existence of such a mapping follows immediately from the Riemann mapping theorem, since it basically says that any simply connected subset of the Riemann sphere which does not contain at least two points of the sphere is biholomorphically equivalent to the unit disk.
I will construct such a mapping in three steps:
Firstly, $f(z) = 1/z$ maps $mathbb{C} cup {infty} setminus [0, 1]$ to $mathbb{C} cup {infty} setminus [1, infty] = mathbb{C} setminus [1, infty)$.
Secondly, $g(z) = sqrt{z - 1}$ maps $mathbb{C} setminus [1, infty)$ to the upper half plane ${z in mathbb{C} | operatorname{Im} z > 0}$.
Finally, $h(z) = frac{z-i}{z + i}$, the Cayley transform, maps the upper half plane to the unit disk.
$f, g, h$ are biholomorphic on their respective domains, so $h circ g circ f$ does the job. Explicitly, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.
answered Jan 3 at 23:48
0x539
1,047316
The existence of such a mapping follows immediately from the Riemann mapping theorem, since it basically says that any simply connected subset of the Riemann sphere which does not contain at least two points of the sphere is biholomorphically equivalent to the unit disk.
I will construct such a mapping in three steps:
Firstly, $f(z) = 1/z$ maps $mathbb{C} cup {infty} setminus [0, 1]$ to $mathbb{C} cup {infty} setminus [1, infty] = mathbb{C} setminus [1, infty)$.
Secondly, $g(z) = sqrt{z - 1}$ maps $mathbb{C} setminus [1, infty)$ to the upper half plane ${z in mathbb{C} | operatorname{Im} z > 0}$.
Finally, $h(z) = frac{z-i}{z + i}$, the Cayley transform, maps the upper half plane to the unit disk.
$f, g, h$ are biholomorphic on their respective domains, so $h circ g circ f$ does the job. Explicitly, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.
answered Jan 3 at 23:48
0x539
1,047316
edited 2 days ago
answered Jan 3 at 23:48
0x539
1,047316
answered Jan 3 at 23:48
0x539
1,047316
answered Jan 3 at 23:48
0x539
1,047316
1,047316
add a comment |
add a comment |
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The Riemann sphere without the unit interval deleted is simply connected (in the same way as the punctured sphere is simply connected).
– 0x539
Jan 3 at 23:38
Do you need an explicit map or is it sufficient to prove existence?
– 0x539
Jan 3 at 23:38
either would help, but I'm looking for a specific map
– David Warren
Jan 3 at 23:43