# Are a sphere cross a circle ($S^2 \times S$) and the 3-dimensional projective ($\textbf{RP(3)}$) space homeomorphic?

I have some questions about the space $$\textbf{SO(3)}$$ (special orthogonal transformations in 3-dimensions). I understand that this transformations represent rotations in 3D space fixing the origin and also I understand that $$\textbf{SO(3)}$$ is homeomorphic to $$\textbf{RP(3)}$$ (real-projective 3D-space). A proof for this homeomorphism can be found in: prove $$RP^3\cong SO(3)$$

Now imagine an orthogonal base at the origin ($$e_1$$, $$e_2$$, $$e_3$$). To specify a rotation you have to first choose where in the unit sphere you want to send $$e_1$$, once you choose that direction you have to choose where to send $$e_2$$ around a circle, and once you choose that you are done (because of the orthogonality). So with this analysis it seems that the group of rotations should be homeomorphic to $$S^2 \times S$$ (sphere cross the cirle). But also $$\textbf{SO(3)}$$ is homeomorphic to $$\textbf{RP(3)}$$.

My question is: Are $$S^2 \times S$$ and $$RP^3$$ really homeomorphic? and if not where is the mistake on my reasoning?, if they are, is there a more simple homeomorphism to show that (without the use of $$\textbf{SO(3)}$$ and more in the spirit of cutting and gluing)?

Thanks in advance. :)

• I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration. – Daniel Schepler Dec 1 '18 at 0:11

They are not even homotopy equivalent. We have $$\pi_1(\textbf{RP(3)})= \mathbb{Z}_2$$ (see Sammy Black's comment to An intuitive idea about fundamental group of $$\mathbb{RP}^2$$), but $$\pi_1(S^2 \times S^1 )= \mathbb{Z}$$.