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Consider the set of all pairs $(\boldsymbol{n},\boldsymbol{v})$ of vectors in $\mathbb{R}^3$ such that $\boldsymbol{n}$ is a vector on the unit sphere centered at the origin and $\boldsymbol{v}$ is a unit vector tangent to the sphere at the point $\boldsymbol{n}.$

i. Introduce a structure of smooth manifold on this set.

ii. Prove that this manifold is diffeomorphic to the group $SO(3).$

To my understanding, this manifold is $S^2 \times S^1,$ which gives a parametrization of $SO(3),$ but it is far from being a diffeomorphism, i.e. the exercise is false: do you agree?

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    $\begingroup$ This manifold is not $S^2 \times S^1$ (it only looks locally like this). Can you expand on the parametrization you chose? Consider the matrix with columns $\boldsymbol{n}$, $\boldsymbol{v}$ and $\boldsymbol{n \times v}$. $\endgroup$
    – Martin
    Jun 15 '13 at 10:36
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The exercise is fine. The manifold described in the exercise is called the unit tangent bundle of $S^2$ and it is not diffeomorphic to $S^2 \times S^1$ [one way to see this is to observe that you could produce a nowhere vanishing vector field on $S^2$ if they were diffeomorphic. This is impossible by the hairy ball theorem.]

Here's a slightly more detailed outline:

By definition the set $M$ is given as a subset of $\mathbb{R}^3 \times \mathbb{R}^3 \ni (\boldsymbol{n}, \boldsymbol{v})$ subject to the equations $$ \begin{align*} 1 & = \boldsymbol{n} \cdot \boldsymbol{n} && \boldsymbol{n} \text{ is a unit vector}\\ 1 & = \boldsymbol{v} \cdot \boldsymbol{v} && \boldsymbol{v} \text{ is a unit vector}\\ 0 & = \boldsymbol{n} \cdot \boldsymbol{v} && \boldsymbol{n} \text{ is perpendicular to }\boldsymbol{v}. \end{align*} $$ Can you use this information to turn $M$ into a manifold? (implicit functions, regular values, etc)

The map $M \to SO(3)$ given by $(\boldsymbol{n},\boldsymbol{v}) \mapsto [\boldsymbol{n},\boldsymbol{v},\boldsymbol{n} \times \boldsymbol{v}]$ is well-defined smooth and bijective. You can exhibit an explicit smooth inverse, so $M$ and $SO(3)$ are diffeomorphic.

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  • $\begingroup$ I was mistaken in thinking I could give a vector, say $\boldsymbol{n}$ and an angle $\theta \in [0,2\pi)$ for the second one, but as you say this cannot be done globally; now I'm left with providing an explicit form for the inverse map from $SO(3)$ $\endgroup$
    – jj_p
    Jun 15 '13 at 20:58
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    $\begingroup$ Well, $[\boldsymbol{n}, \boldsymbol{v},\boldsymbol{n} \times \boldsymbol{v}] \mapsto (\boldsymbol{n},\boldsymbol{v})$ :-) The map sending a matrix from $SO(3)$ to its first two columns. $\endgroup$
    – Martin
    Jun 15 '13 at 21:17
  • $\begingroup$ True and simple. I was thinking more along these lines: fix $(\boldsymbol{n}_0,\boldsymbol{v}_0)$ and then send $SO(3) \ni R \mapsto (R\boldsymbol{n}_0,R\boldsymbol{v}_0)\in M:$ this is continuous map and we can reach every point of $M$ since, once we reached $(\boldsymbol{n}',\boldsymbol{v}')$ we can still perform a rotation leaving $\boldsymbol{n}'$ fixed and therefore rotating $\boldsymbol{v}'$ which is orthogonal to any other orthogonal $\boldsymbol{v}'':$ despite being more involved than yours, do you think it still can work? $\endgroup$
    – jj_p
    Jun 16 '13 at 9:38
  • $\begingroup$ @Nicolo' yes, this works. It's actually the same argument (up to shifting everything to the coordinate system $\boldsymbol{n}_0, \boldsymbol{v}_0, \boldsymbol{n}_0 \times \boldsymbol{v}_0$): I simply decided to choose $\boldsymbol{n}_0 = (1,0,0)^t$ and $\boldsymbol{v}_0 = (0,1,0)^t$... Note also that this is basically the same argument as the one Ted Shifrin suggests. $\endgroup$
    – Martin
    Jun 16 '13 at 9:50
  • $\begingroup$ Hi @Martin, in this part ¨one way to see this is to observe that you could produce a nowhere vanishing vector field on S2 if they were diffeomorphic¨ how do you produces this vector field? $\endgroup$
    – Framate
    Jul 28 '14 at 4:23
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Yes, you're looking at the manifold of unit tangent vectors of $S^2$. You can go the other way: $SO(3)$ acts smoothly and transitively on $S^2$; moreover, it also acts smoothly and transitively on your manifold $M$. But the stabilizer subgroup of a point is trivial, so the map $SO(3)\to M$ is a diffeomorphism.

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