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It is well known that $\mathbb{C}P(n+1)$ is diffeomorphic to $U(n+1)/U(n) \times U(1)$.

Since $U(n) \times U(1)$ is a close Lie-subgroup of $U(n+1)$, the canonical projection $p:U(n+1) \to U(n+1)/U(n) \times U(1)$ is a fiber bundle with fiber $U(n) \times U(1)$.

So, for $n=1$, $\mathbb{C}P(2)$ is diffeomorphic to $U(2)/U(1) \times U(1)$.

From the long exact sequence for the fiber bundle $U(1) \times U(1) \to U(2) \to U(n+1)/U(n) \times U(1)$, we have that $\pi_n (\mathbb{C}P(2))$ is isomorphic to $\pi_n(U(2)) \cong \pi_n(S^3)$ for $n \geq 3$.

But it is well known that there is a fiber bundle $S^1 \to S^5 \to \mathbb{C}P(2)$ is a fiber bundle.

From the long exact sequence for the fiber bundle $S^1 \to S^5 \to \mathbb{C}P(2)$ we have that $\pi_n (\mathbb{C}P(2))$ is isomorphic to $\pi_n(S^5)$ for $n \geq 3$.

Now we have a contracdition.

What is false?

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I believe you have confused the two different indexing conventions for complex projective space. – Alexander Woo Jan 20 2012 at 3:16
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It's $CP^n$ not $CP^{n+1}$ that's diffeo to $U(n+1)/(U(n)\times U(1))$. In particular $U(2)/(U(1)\times U(1))$ is $CP^1$, not $CP^2$. – Vitali Kapovitch Jan 20 2012 at 3:18
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You've got to work on this a little. I am guessing you mean "contradiction" rather than "contraction." You need a more informative title. Meanwhile, Simon and Garfunkle: youtube.com/watch?v=uRv4S0BPMik – Will Jagy Jan 20 2012 at 3:18
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Actually Garfunkel. The lyrics I had in mind: "I don't know what is real, I can't touch what I feel, And I hide behind the shield of my illusion." – Will Jagy Jan 20 2012 at 3:26
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In general, sentences of the form "It is well known that..." are good places to look for an error in an argument that gives absurd conclusions. – Terry Tao Jan 20 2012 at 4:21
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closed as too localized by Yemon Choi, Andres Caicedo, Qiaochu Yuan, Ryan Budney, Alain Valette Jan 20 2012 at 7:02

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