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Consider $\mathbb{CP}^1 =\frac{\mathbb{C}^2\setminus \{ 0\}}{\mathbb{C}^*}$, one dimensional projective Hilbert space. I was wondering if someone could help me about proving $\mathbb{CP}^1 \cong S^2$

Here $\mathbb{C}^*$ acts on $\mathbb{C}^2\setminus \{ 0\}$, so we can construct $\frac{\mathbb{C}^2\setminus \{ 0\}}{\mathbb{C}^*}$.

Mahtab
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    What you described is ${\mathbb C}P^0$, not ${\mathbb C}P^1$. – Moishe Kohan Mar 05 '24 at 13:29
  • @MoisheKohan Sorry, I meant $\frac{\mathbb{C}^2\setminus { 0}}{\mathbb{C}^*}$. – Mahtab Mar 05 '24 at 13:41
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    Ok, what does the symbol $\cong$ stand for? Is it a homeomorphism, a diffeomorphism, a biholomorphism? And tell us what do you know about this problem (which is a standard exercise in several areas of math). – Moishe Kohan Mar 05 '24 at 13:52
  • @MoisheKohan I think the symbol $\cong$ denotes a diffeomorhism. I am a student in physics and I'm stuying symmetris in quantum mechanics. The underlying space in quantum mechanics is the projective Hilbert space $\mathbb{P}\mathcal{H}=\frac{\mathcal{H\setminus { 0}}}{\mathbb{C}^*}$ and I'm reading a book in which the author mentioned this problem. I don't know a reference about it and I really appreciate if you could introduce me a good reference about projective spaces and $\mathbb{CP}^n$. Thank you. – Mahtab Mar 05 '24 at 14:06
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    @Mahtab Look at this problem. – Sachchidanand Prasad Mar 05 '24 at 14:31
  • @SachchidanandPrasad Thank you very much for the link. – Mahtab Mar 07 '24 at 05:20

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