TITLE: The circle bundle of $S^2$ and real projective space
QUESTION [31 upvotes]: 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.

REPLY [4 votes]: The way I understands the unit tangent bundle $T_1S^2$ is through the complex line bundle on $\mathbb CP^1\cong S^2$.
As you know $T_1S^2$ has Eular class +2, I claim that it is the unit circle bundle of the line bundle $\mathcal{O}(2)$ on $\mathbb CP^1$. This is a consequence of the fact that isomorphism class of oriented $S^1$-bundles on $S^2$ is in bijection to isomorphism class of complex line bundle on $\mathbb CP^1$, which is also admits bijection to $\mathbb Z$ ("isomorphism" here should be understood as in the corresponding category). That is, there is a commutative diagram.
$\require{AMScd}$
\begin{CD}
\{complex \ line \ bundle \ on\  \mathbb CP^1\}@>{c_1}>> \mathbb Z\\
@VVV @VV=V\\
\{oriented \ S^1bundle \ on \ S^2\}  @>{e}>> \mathbb Z
\end{CD}
Where $c_1$ is by associating the 1st Chern class, and $e$ is by associating the Eular class, and the first vertical map is by associating the unit circle bundle.
Now, it turns down to study the line bundle $\mathcal{O}(2)$. A remark here is that the Eular class of the line bundle $\mathcal{O}(2)$ is also 2, as one may see this as taking an irreducible quadric polynomial as global section, it vanishes on $\mathbb CP^1$ at exactly 2 points (for example $z_0z_1$ vanishes at $[1,0],[0,1]$). Similarly for each $n\in \mathbb Z$, the line bundle $\mathcal{O}(n)$ has Eular class equal to $n$, by considering dual bundle when $n$ negative.
We start by understanding the topology of tautological line bundle $\mathcal{O}(-1)$ on $\mathbb CP^1$, which arises in nature. The total space of the bundle is given by $$E= \{(z_1,z_2;w_1,w_2)\in\mathbb CP^1\times \mathbb C^2|(w_1,w_2)=t(z_1,z_2),\ \exists t\in \mathbb C \}$$
with each fiber a complex line over each point $[z_1,z_2]\in \mathbb CP^1$ which is spanned by itself. and $E=\mathbb C P^1\cup \mathbb C^2-\{0\}$ is the blowup at the origin of $\mathbb C^2$.
Now if we restrict the bundle to the unit circle on each fiber, we get $$\{|w_1|^2+|w_2|^2=1\}=S^3\to \mathbb CP^1$$
$$(w_1,w_2)\mapsto [w_1,w_2]$$
which is exactly the Hopf fibration, in particular, we have showed that the Eular class of Hopf fibration is $-1$.
Now, what happen its tensor product $O(-1)\otimes O(-1)= O(-2)$? 
On each fiber, which is the complex line $L_{[w_1,w_2]}\subset \mathbb C^2$, the tensor product undergoes $z\mapsto z^2$, by properly choosing the scale, and identifying $L_{[w_1,w_2]}\cong \mathbb C$. We see that this globalize a $\mathbb Z_2$ action on $\mathbb C^2-\{0\}$, the total space with zero section removed. If we restricted to $S^3$ is antipodal map, we will get a $S^1$-bundle $S^3/\mathbb Z_2\to \mathbb CP^1$, which is the unit circle bundle of the complex line bundle $\mathcal{O}(-2)$, with Eular class equal to $-2$.
Finally, to obtain total space of $\mathcal{O}(2)$, we need to take the dual of $\mathcal{O}(-2)$. Geometrically, one can remove the zero section and send $z\mapsto z^{-1}$ by identifying each fiber with $\mathbb C-\{0\}$. Especially, this map preserves the unit sphere and therefore $T_1S^2$ is diffeomorphic to $S^3/\mathbb Z_2$, as desired.