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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





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.

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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 $\mathbb{Z}$. –  Jason DeVito Aug 29 '11 at 14:45
@Jason, thank you. –  Thomas Rot Aug 29 '11 at 14:55
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. –  Georges Elencwajg Aug 29 '11 at 15:22
@Georges : of course –  Thomas Rot Aug 29 '11 at 15:40
@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 '11 at 23:51

3 Answers 3

up vote 6 down vote accepted

$SO_3$ 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 a pair of orthonormal basis for it. $v_1$ is the point of $S^2$ your vector $x$ is tangent to, $x \in T_{v_1}S^2$ and $v_2$ 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 $T_{v_1}S^2 \subset \mathbb R^3$.

But given $v_1$ and $v_2$ orthonormal, $v_3 = v_1 \times v_2$.

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.

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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? –  Thomas Rot Aug 29 '11 at 15:05
Dear Ryan, this is a correct nice answer but I find your notations confusing. You cannot write $v \in T_xS^2$ because $x$ is not a point on the sphere, but on the circle bundle.I suppose $x,v,v_2$ all denote the same vector in the tangent plane at $v_1$? –  Georges Elencwajg Aug 29 '11 at 15:07
Er, yes, my morning TeX typesetting can be a little sloppy. I believe I reversed a $v_1$ and $x$ in the earlier write-up. This is hopefully more comprehensible. –  Ryan Budney Aug 29 '11 at 15:42
Dear Ryan, Thank you for your answer. It is clear to me now. –  Thomas Rot Aug 31 '11 at 16:24

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 $2$.

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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.

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Thank you for your answer. It is very lucidly written. –  Thomas Rot Aug 31 '11 at 16:25

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