Mapping the Riemann sphere with unit interval deleted to the unit disk












2














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







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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
















2














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







share|cite|improve this question






















  • 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














2












2








2


2





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










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, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.






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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, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.






    share|cite|improve this answer




























      3














      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, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.






      share|cite|improve this answer


























        3












        3








        3






        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, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.






        share|cite|improve this answer














        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, $(h circ g circ f)(z) = frac{sqrt{1/z - 1} - i}{sqrt{1/z - 1} + i} = left(sqrt{1 - z} - i sqrt{z} right)^2$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered Jan 3 at 23:48









        0x539

        1,047316




        1,047316






























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