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I am new to differential geometry. I have the following question:

Let $\mathbf{Q}$ denote the set of unit quaternions. I already have proved using the implicit function theorem that $\mathbf{Q}$ is a $3$-manifold. Now, I need to show:

There does not exist a global diffeomorphism $\phi$ between $\mathbf{Q}$ and $\mathrm{SO}(3)$, i.e., show that there does not exist $\phi: \mathbf{Q} \to \mathrm{SO}(3)$, such that $\phi$ is a $C^\infty$ bijection.

Any ideas of how to go about it?

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    $\begingroup$ Here's an idea: the set of unit quaternions is $S^3$. But you can see that $\mathrm{SO}(3)$ is diffeomorphic to $\mathbb{R}P^3$. Now try to find something that distinguishes $\mathbb{R}P^3$ from $S^3$. They have different fundamental groups, but I don't know if that helps in manifolds. $\endgroup$ – stressed out Feb 17 '19 at 23:36
  • $\begingroup$ As a side note, I just checked on the internet that the De Rham cohomologies of $\mathbb{R}P^3$ and $S^3$ are the same unfortunately. $\endgroup$ – stressed out Feb 17 '19 at 23:53
  • $\begingroup$ As a note, global diffeomorphism is not the same as a $C^\infty$ bijection, you also require its inverse to be $C^\infty.$ $\endgroup$ – positrón0802 Feb 18 '19 at 0:11
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The set of unit quaternions is $S^3$. It can be shown using the algebra of quaternions that $q: S^3 \to \mathrm{SO}(3)$ is a double cover with the kernel $\{\pm 1\}$. This means that $\mathrm{SO}(3)$ is diffeomorphic to $S^3/\{\pm 1\}$ which is $\mathbb{R}P^3$. But the fundamental group of $n$-sphere is trivial for $n>1$ while the fundamental group of $\mathbb{R}P^n$ for $n>1$ is $\mathbb{Z}_2$. This means that they can't be diffeomorphic because homeomorphisms preserve fundamental groups.

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Topologically the set of unit quaternions is nothing else than the ordinary $3$-sphere $S^3 \subset \mathbb{R}^4$. It is well-known that $S^3$ is simply connected wheras $SO(3)$ is not simply connected. Hence there cannot exist a homeomorphism (let alone a diffeomorphism) between $S^3$ and $SO(3)$.

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