Is the space $\mathbb{PR}^3$ homeomorphic to $\mathbb{PR}^2\times S^1$?

Question

In this Wiki article it is described how the $SO(3)$ is homeomorphic to the projective space $\mathbb{RP}^3$. I would suggest another way which I hope it works.

On $S^2$ one may take any direction (not vector) as an axis of rotation, and for each such axis we have the rotation about it as an $S^1$. Since the set of all directions is topologised as the projective plane $\mathbb{PR}^2$, so at least it must be true that $SO(3)$ is some fiber bundle $(E,\pi,\mathbb{PR}^2)$ of $S^1$ over $\mathbb{PR}^2$. But is it possible to assume that $SO(3) \overset{?}{\cong} \mathbb{PR}^2 \times S^1$ (a trivial bundle) ? and hence, is the space $\mathbb{PR}^3$ homeomorphic to $\mathbb{PR}^2\times S^1$ ?

Answer 1

The space $\mathbb{RP}^3$ has mod-$2$ cohomology ring $\mathbb Z_2[x]/x^4$ while $\mathbb{RP}^2\times S^1$ has mod-$2$ cohomology ring $\mathbb Z_2[x]/x^2\otimes\mathbb Z_2[y]/y^2$. These two rings are non-isomorphic, so these spaces are not even homotopy equivalent.

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Alternatively:

• just comparing $H^1(-,\mathbb Z_2)$ is enough.
• calculate fundamental groups. $\mathbb{RP}^3$ has fundamental group $\mathbb Z/2$, while $\mathbb{RP}^2\times S^1$ has fundamental group $\mathbb Z\times\mathbb Z/2$.

Answer 2

The universal covering space of $\mathbb{P}\mathbb{R}^3$ is $S^3$ and of $\mathbb{P}\mathbb{R}^2\times S^1$ is $S^2\times\mathbb{R}$. $S^3 \ncong S^2\times\mathbb{R}$, since $S^3$ is compact and $S^2\times\mathbb{R}$ not.

Answer 3

$\mathbb P\mathbb R^2$ thus $\mathbb P\mathbb R^2\times S^1$ is not orientable, but $SO(3)$ certainly is as a Lie group.

The way you try to parametrize $SO(3)$ has a few problems.

• The identity element $I_3$ corresponds to infinitely many elements in $\mathbb P\mathbb R^2\times S^1$.
• An axis and an angle doesn't immediately produce an element in $SO(3)$: you need to know which direction to go. If instead of $\mathbb P\mathbb R^2$, we use $S^2$ to parametrize the axes, then we may go clockwise along the axis by convention but this won't give a one-to-one parametrization of $SO(3)$ but a covering space. We may try to associate a direction to go for each line instead of a direction vector, but this won't be consistent according to the non-orientability of $\mathbb P\mathbb R^2$.

So you can't consider $SO(3)$ as a fibre bundle over $\mathbb P\mathbb R^2$ as described. But there is a way to do it: embed the Lie group $S^1$ into $SO(3)$ through a fixed axis, and $SO(3)$ would be a $S^1$-principle bundle over the coset space. This should be related to the Hopf fibration of $S^3$, the universal cover of $SO(3)$.