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Let the unit tangent bundle be defined as follows:

$$T^1S^2=\{(p,v)\in \mathbb R^3 \times \mathbb R^3 | |p|=|v|=1 \text{ and } p \bot v \}$$

Let $SO(3)$ be the group of rotations of $\mathbb R^3$. Apparently, $SO(3)$ is in bijection with $T^1S^2$.

My question is:

If $N$ is a point on $S^2$, say the north pole, does the rotation in $SO(3)$ moving $N$ to $p$ along $v$ correspond to $(p,v)$ in $T^1 S^2$?

Put the other way around:

Does the matrix $(p,v, p \times v)$ corresponding to $(p,v)$ represent the rotation around the axis $p$? And if so, is the angle somehow represented by $v$?

Later added

The reason why I think there should be geometric meaning to this bijection or at least some insight to be gained is that finding the bijection was an exercise in a book I am reading.

If there was no insight to be gained the exercise would be more or less purely computational and not very insightful.

self-learner
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4 Answers4

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"Bijection" is a very weak statement. Any manifold of positive dimension is in bijection with $\mathbb{R}$. In fact $\text{SO}(3)$ is diffeomorphic to the unit tangent bundle, but this diffeomorphism isn't canonical; you need to fix a point $(p, v)$ in the unit tangent bundle, and then the diffeomorphism is given by the natural action of $\text{SO}(3)$ on the unit tangent bundle acting on this point. ($\text{SO}(3)$ acts on $S^2$ by rotations and this action is smooth so it extends to an action on the tangent bundle. The induced maps on tangent vectors are isometries, so it restricts to an action on the unit tangent bundle.)

In other words, the unit tangent bundle is a principal homogeneous space for $\text{SO}(3)$. A simpler example of this phenomenon is that the circle $S^1$ is a principal homogeneous space for $\text{SO}(2)$, so in particular they are diffeomorphic, but to pick a diffeomorphism you need to pick a point of $S^1$ to serve as the identity.

Qiaochu Yuan
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  • Thanks, +1. (This is probably offtopic, I just wonder, can $T^1 S^n$ be diffeomorphic to a Lie group for $n>2$?) – Peter Franek Jun 28 '14 at 19:04
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    @Qiaochu: There is, in fact, a canonical diffeomorphism (for some definition of canonical): Map $(p,v)\in T^1 S^2$ to the matrix with columns $p,v,$ and $p\times v$. (And +1 from me) – Jason DeVito - on hiatus Jun 28 '14 at 19:23
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    @Peter: It never is diffeo to a Lie group for $n>2$. The point is that $T^1 S^n$ is simply connected, so if a Lie group, it must be semisimple. But all semisimple Lie groups have $H^3$ nontrivial, even rationally. This immediately implies, since $T^1 S^n$ is $n-1$ connected, that $n\leq 4$. When $n=3$, $T^1 S^3 = S^3 \times S^2$ has nontrivial $\pi_2$, so can't be Lie group. When $n=4$, $T^1 S^4$ is rationally $6$-connected. – Jason DeVito - on hiatus Jun 28 '14 at 19:27
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    @Jason: I would call that an identification of the unit tangent bundle with the space of orthonormal frames in $\mathbb{R}^3$, which is also a principal homogeneous space for $\text{SO}(3)$. But you still need to choose such an orthonormal frame in order to turn a matrix as you describe it into an element of $\text{SO}(3)$. – Qiaochu Yuan Jun 28 '14 at 21:09
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    @Jason: I think you want "compact and simply connected"; compactness is of course crucial here. – Qiaochu Yuan Jun 29 '14 at 00:04
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    @Qiaochu: I figured you'd object to my map as "canonical" because of the choice of basis ;-). I, personally, am of the mindset that $\mathbb{R}^n$ comes with a God-given orthonormal basis (and to me, the notation $SO(3)$ means the orthogonal group relative to the usual inner product on $\mathbb{R}^3$), so I still view my map as "canonical". Now, if we were, say, talking about $V$ being the degree at most 2 polynomials over $\mathbb{R}$ with $L^2$ inner product, I'd agree that my map is non-canonical. Also, you're right compactness is an essential ingredient - thanks! – Jason DeVito - on hiatus Jun 29 '14 at 21:09
  • Don't we need to pick both an orientation on the circle and a point on the circle to ensure that the diffeomorphism with $SO(2)$ canonical? – Mozibur Ullah Jan 16 '24 at 05:22
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You can use the poincare representation of $SO(3)$, which gives a bijective and differentiable mapping between $T_1 S^2$ and $SO(3)$. The representation is like this: first an element of $SO(3)$ can be identified as an orthonormal frame of $R^3$, and then each point $(p,v)\in T_1S^2$ gives an orthonormal frame: $\{p, v, p\times v\}$.

Simon Zhu
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If you want to think about rotations then the diffeomorphism that you should consider is the one that to a point $(p,v)$ in the unit tanget space corresponds the rotation in $R^3$ with axis $p$ and angle of rotation $v$.

inquisitor
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I'm not trying to get the bounty or anything I'm just trying to understand so I'll go ahead and post some more thoughts on this:

The following three rotations form a generating set for $SO(3)$:

$$R_x = \left ( \begin{array}{ ccc } 1 & 0 & 0 \\ 0 & \cos \Theta & - \sin \Theta \\ 0 & \sin \Theta & \cos \Theta \end{array}\right ) $$

$$R_y = \left ( \begin{array}{ ccc } \cos \Theta & 0 & - \sin \Theta \\ 0 & 1 & 0 \\ \sin \Theta & 0 & \cos \Theta \end{array}\right ) $$

$$R_z = \left ( \begin{array}{ ccc } \cos \Theta & - \sin \Theta & 0 \\ \sin \Theta & \cos \Theta & 0\\ 0 & 0 & 1 \\ \end{array}\right ) $$

We recall that the bijection $b: T^1 S^2 \to SO(3)$ maps $(\vec{p}, \vec{v})$ to the matrix $(\vec{p}, \vec{v}, \vec{p} \times \vec{v})$. If we try to write $R_z$ in tis form we see that

$$ \vec{p} = \left ( \begin{array}{ c } \cos \Theta \\ \sin \Theta \\ 0 \end{array}\right )$$ and $$ \vec{v} = \left ( \begin{array}{ c } - \sin \Theta\\ \cos \Theta \\ 0 \end{array}\right )$$

We see that $\vec{p}$ is the vector in the $x-y$-plane forming an angle of $\Theta $ with the $x$-axis. The vector $\vec{v}$ is the vector $\vec{p}$ multiplied by $i$, that is, forming an angle of $\Theta + 90$ degrees with the $x$-axis.

If we could show that $b$ is not just a bijection but a homomorhism then I'm hoping to establish that the information about the rotation is encoded into $(\vec{p}, \vec{v})$ and because it's obvious what $R_z, R_y$ and $R_x$ map to so it becomes clear for all elements in $SO(3)$.

self-learner
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