Today I felt like computing the integral cohomology of the unit circle bundle of the tangent bundle of $S^2$. For completeness, it is defined by $SS^2=\{x\in TS\colon ||x||=1\}$, where we use the standard Riemannian metric on $S^2$. The cohomology of circle bundles, or more generally sphere bundles, can be computed by the Gysin sequence http://en.wikipedia.org/wiki/Gysin_sequence. By using that the euler class of $S^2$ is 2 times the generator of $H^2(S^2)$, I think I succeeded in this computation (if someone wants to check it, I'd be more than happy the give the details), and I find that
$H^0(SS^2)=\mathbb{Z}$
$H^1(SS^2)=0$
$H^2(SS^2)=\mathbb{Z}/2\mathbb{Z}$
$H^3(SS^2)=\mathbb{Z}$.
This equals the cohomology of real projective space $\mathbb{R}P^3$ (see http://topospaces.subwiki.org/wiki/Cohomology_of_real_projective_space). I was wondering if these spaces are actually homeomorphic, and if there is a nice explicit way of describing the homeomorphism.
Edit: Jason DeVito pointed out a mistake in my formulation.
Edit 2: As asked, here is the calculation of the cohomology groups. The Gysin sequence is for the sphere:
$\rightarrow H^n(S^2)\rightarrow H^n(SS^2)\rightarrow H^{n-1}(S^2)\rightarrow H^{n+1}(S^2)\rightarrow$
The middle map is taking the cup product with the euler class, which is just mapping a generator of $H^{n-1}$ to the twice the generator of $H^{n+1}$. Of course this is only nonzero if $n=1$.
The exact sequence breaks down for $n=0$ to
$0\rightarrow H^0(S^2)\rightarrow H^0(SS^2)\rightarrow 0$
Which gives the isomorphism $H^0(SS^2)=\mathbb{Z}$ of course. For n=1 we get a sequence
$0\rightarrow H^1(SS^2)\rightarrow\mathbb{Z}\rightarrow^2\mathbb{Z}$.
Because the kernel of the map "cupping with the Euler class" has as kernel $0$, and the map before that is injective, we find $H^1(SS^2)=0$. We also have, just after this point in the sequence
$0\rightarrow \mathbb{Z}\rightarrow^2\mathbb{Z}\rightarrow H^2(SS^2)\rightarrow 0$.
Thus we find that $H^2(SS^2)=\mathbb{Z}/2\mathbb{Z}$. At $n=3$ we find
$0\rightarrow H^3(SS^2)\rightarrow H^2(S^2)\rightarrow 0$
Which gives the remaining non zero cohomology group. All the other groups vanish because the $H^q(SS^2)$ are sandwiched between higher homology groups of the two sphere, which are all zero.
Best Answer
$SO_3$ is the space of triples $(v_1,v_2,v_3)$ of elements of $\mathbb R^3$ which are an oriented orthonormal basis.
Given an element $x \in SS^2$, you construct on orthonormal pair in $\mathbb R^3$. $v_1$ is the point of $S^2$ your vector $x$ is tangent to, i.e. $v_1 = \pi(x)$. $v_2$ would be the vector in $\mathbb R^3$ that is the image of $x \in T_{v_1}S^2$ under the inclusion of vector-spaces $T_{v_1}S^2 \subset \mathbb R^3$, i.e. it comes from the inclusion $SS^2 \to S^2 \times \mathbb R^3$ then projecting onto the fiber.
But given $v_1$ and $v_2$ orthonormal, $v_3 = v_1 \times v_2$, the cross product.
So that's essentially why $SO_3$ and $SS^2$ are diffeomorphic / homeomorphic.
There's a lot of fun ways to see $\mathbb RP^3$ and $SO_3$ are diffeomorphic. There are arguments using the quaternions. I prefer the exponential map $T_ISO_3 \to SO_3$ -- consider it restricted to balls of various radius and stop at the first radius where the function is onto.