What is the mapping between the hyperriemann sphere and the complex plane?

[EinsteinKreuz]

As many of you know, using the stereographic projection one can construct a homeomorphism between the the complex plane ℂ¹ and the unit sphere S²∈ℝ³. But the stereographic projection can be extended to
the n-sphere/n-dimensional Euclidean space ∀n≥1. Now what I am talking about is the the mapping I: H→ H where H is the space of all Quaternions and I(q) = (1/q) ∀q ∈ H. So this complex manifold is the one-point compactification of H which I will refer to as ◊.
That is, ◊ : H ∪ {∞}. I: 1/(0+0i+0j+0k) ↔ {∞}. So is there an official name for ◊ and has it already been shown that it is topologically equivalent to S⁴? I assume so but if need be I will give a proof attempt in a followup post.
https://en.wikipedia.org/wiki/N-sphere#Stereographic_projection

 

[lavinia]

While it is true that $S^4$ is the 1 point compactification of $R^4$, it is not a complex manifold. In fact the tangent bundle does not even have an almost complex structure. I am not sure if any sphere other than the Riemann sphere can be a complex manifold. That said, the Riemann sphere is a natural extension of the complex plane for the study of complex analysis. You might want to research whether quaternionic analysis naturally extends to the 4 sphere. What about octonian analysis?

 

[EinsteinKreuz]

While it is true that $S^4$ is the 1 point compactification of $R^4$, it is not a complex manifold. In fact the tangent bundle does not even have an almost complex structure. I am not sure if any sphere other than the Riemann sphere can be a complex manifold. That said, the Riemann sphere is a natural extension of the complex plane for the study of complex analysis. You might want to research whether quaternionic analysis naturally extends to the 4 sphere. What about octonian analysis?

My bad. It is a manifold but not a complex manifold. Now I don't know about Octonion analysis, however, the unit Octonions do not form a group as they are non-associative whereas the quaternions are a non-commutative group under multiplication and the unit pure Quaternions are isomorphic to SO(3).

 

[lavinia]

My bad. It is a manifold but not a complex manifold. Now I don't know about Octonion analysis, however, the unit Octonions do not form a group as they are non-associative whereas the quaternions are a non-commutative group under multiplication and the unit pure Quaternions are isomorphic to SO(3).

Right.

Maybe this is interesting.

http://projecteuclid.org/download/pdf_1/euclid.bbms/1102715140

 

[zinq]

Lavinia wrote: "I am not sure if any sphere other than the Riemann sphere can be a complex manifold."

Neither is anyone else! It is known that the only other sphere Sⁿ besides S² that even has an almost complex structure . . .

(((
i.e., a bundle isomorphism

J: T(Sⁿ) →T(Sⁿ)​

such that

J² = -I,​

where

I: T(Sⁿ) → T(Sⁿ)
​

is the identity. In this sense applying J to T(Sⁿ) is "almost" like multiplication by the imaginary unit i applied to the complex plane ℂ.
)))

. . . is S⁶. (This comes from the fact that S⁶ is the underlying topological space of the "pure imaginary" unit octonions.)

But it remains unknown whether S⁶ admits the structure of a complex analytic manifold.

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EinsteinKreuz wrote ". . . the unit pure Quaternions are isomorphic to SO(3)."

Close, but not quite. The unit-length quaternions form the unit sphere in 4-space, known as S³. This is the unique double covering space of the rotation group SO(3) of 3-space. In fact, the underlying topological space of SO(3) is 3-dimensional (real) projective space, P³, which is obtained by identifying antipodal points of S³.