# The circle bundle of $S^2$ and real projective space

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.

• Minor point: If you're computing the de Rham cohomology, the coefficients are in $\mathbb{R}$, so you should be getting an $\mathbb{R}$ in dimension $0$ and $3$ and trivial groups elsewhere. What you've actually computed is cohomology with coefficients in $\mathbb{Z}$. Aug 29, 2011 at 14:45
• Since you kindly volunteer to do so, could you please show (as an appendix to your question) the calculation of cohomology you made: it is always nice to have such examples written down explicitly for reference. Aug 29, 2011 at 15:22
• @George: It would be nice to do a project of having all these detailed computations together in some website, to undo some of the over-streamlining effects of modern mathematics, for all to benefit. Just an idea.
– gary
Aug 29, 2011 at 23:51

$$SO_3$$ 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.

• Thank you for your answer. I am sorry, but I am having some trouble parsing your second sentence. Do you mean to say that $v_1=\pi(x)$ where $\pi$ is the projection $SS^2\rightarrow S^2$? And $v_2$ is the element $x$ is representing in $T_\pi(x) S^2$ (which we can see as a subset of $\mathbb{R^3}$). Then take the cross product to obtain the third basis vector? Aug 29, 2011 at 15:05
• @ThomasRot: sorry that sentence must have been mangled by revisions. Hopefully this is clear now. Jan 9 at 8:05
• Thanks it is clear Jan 9 at 11:03

Here's another approach. I think of it as less elementary but it fits into a fairly broad framework of general-nonsense about bundles.

Fact: The Hopf fibration $S^1 \to S^3 \to S^2$ is the circle bundle over $S^2$ with Euler class $+1$. One way to take this is this bundle is classified by a map $S^2 \to B(SO_2)$, where the induced map $H_2(S^2) \to H_2(B(SO_2))$ sends the generator to the generator. Another way to say this is you can decompose $S^3$ into two solid tori $S^1 \times D^2$, and the gluing map sends the $\{1\} \times S^1$ curve in $\partial (S^1 \times D^2)$ to the diagonal $\{(x,x) : x \in S^1\}$ curve in the boundary of the other $S^1 \times D^2$ i.e. the curve of "slope $1$".

$S^3$ is the group of unit quaternions, so the unit complex numbers $S^1$ is a subgroup. Let $\mathbb Z_n \subset S^1$ be the $n$-th roots of unity. Since $S^3/S^1 \simeq S^2$ by the Hopf fibration, there are also bundles, induced by the Hopf fibration:

$$S^1 / \mathbb Z_n \to S^3 / \mathbb Z_n \to S^2$$

But $S^1 / \mathbb Z_n$ is a circle. So general bundle nonsense says $S^3 / \mathbb Z_n$ is the circle bundle over $S^2$ with Euler class $n$. In the case $n=2$ this is $\mathbb RP^3$, in general this space is called the Lens space $L_{n,1}$. In particular, the unit tangent bundle of $S^2$ is known (by Poincare-Hopf, for example) to be the circle bundle over $S^2$ with Euler class $2$.

• The Hopf fibration is the principle $S^1$-bundle associated to the tautological line bundle $\mathcal{O}(-1)$ of $\mathbb CP^1$, so I guess its Eular characteristic should be -1, instead of 1? Apr 20, 2018 at 18:52
• @YilongZhang: it sounds like you are using a not-generally-shared convention. The Hopf fibration is typically consider to be just a map that happens to be a principle bundle. Apr 20, 2018 at 19:30
• @RyanBudney Sorry for reviving an old post, but how does one prove the fact that you metioned about the Euler class and spherical fibrations? Mar 4, 2021 at 23:19
• @ThePuix: Specifically which theorem do you want to prove, and what is your starting point, i.e. what kinds of tools do you allow? Jan 30 at 20:20
• @RyanBudney It was about the fact that you stated in your answer: "The Hopf fibration $S^1 \to S^3 \to S^2$ is the circle bundle over $S^2$ with Euler class $+1$." Someone else explained to me why using classifying spaces and associated bundles; it turned out to be quite nice. If that is what you had in mind as well, then my confusion is cleared up. If you have a slicker proof, I would be interested in that as well. Thanks anyway! Jan 30 at 20:54

Here's is the outline of the argument using quaternions which Ryan alluded to in his answer.

The quaternions are, as a vector space, isomorphic to $\mathbb{R}^4$. They are the set of all things of the form $q=a + bi + cj + dk$ with $a,b,c$, and $d$ real numbers together with the multiplication rules that $i^2 = j^2 = k^2 = ijk = -1$ and $ij = -ji =k$, $jk = -kj = i$, and $ki = -ik = j$. One can check that this notion of multiplication is associative with unit $1 + 0i + 0j + 0k = 1$ and every nonzero element has a multiplicative inverse. If one defines the norm squared $|q|^2$ of a quaternion as $a^2+b^2+c^2+d^2$, then a computation shows that for any two quaterions $q_1$ and $q_2$, we have $|q_1q_2| = |q_1||q_2|$. It follows that the unit sphere in $\mathbb{R}^4$ has the structure of a (noncommutative) Lie group.

Let $V$ be the subset of all quaternions consisting of those with $a=0$,the imaginary quaternions. Then $V$ is isomorphic to $\mathbb{R}^3$ as a vector space.

For each $q\in S^3$ (i.e, $q$ is a unit length quaternion), define the map $A_q:V\rightarrow V$ which sends $v$ to $qv\overline{q}$ (where $\overline{q}$ negates the $i,j,$ and $k$ terms, but leaves the real part alone). Because $|qv\overline{q}| = |q||v||q| = |v|$, $A_q\in SO(3)$.

That means that $q\rightarrow A_q$ is really a homomorphism from $S^3\rightarrow SO(3)$. Now one shows the following: Every element of $SO(3)$ is in the image of $S^3$ under this map and $1$ and $-1\in S^3$ together make up the whole kernel of the map. It follows that $SO(3)$ is diffeomorphic to $S^3/$~ where $q$~$-1*q$. But this is precisely the identification done on $S^3$ to get $\mathbb{R}P^3$ so we must have that $SO(3)$ and $\mathbb{R}P^3$ are diffeomorphic.

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.