Homework 1 Problem 6


Consider a rigid body in $\mathbb{R}^3$ which is free to rotate in any direction about its center of mass, the latter fixed in space. The configuration manifold for this system $-$ i.e., the manifold comprising all possible orientations of this body relative to a fixed reference orientation $-$ is called $\text{SO}(3)$. Clearly, any particular reorientation of the body with respect to the reference configuration can be specified by a single axis about which to rotate and an angle through which to rotate about this axis. Since axes (i.e. directions) in $\mathbb{R}^3$ can be identified with points on the sphere $\mathbb{S}^2$, and angles can be identified with points on the circle $\mathbb{S}^1$, is $\text{SO}(3)$ globally diffeomorphic to $\mathbb{S}^2 \times \mathbb{S}^1$? If not, are the two manifolds locally diffeomorphic? Justify your response in words.


$\text{SO}(3)$ is not globally diffeomorphic to $\mathbb{S}^2 \times \mathbb{S}^1$, however the two manifolds are locally diffeomorphic.

To prove the two manifolds are not globally diffeomorphic, we assume there exists a global diffeomorphism $F$ between the two and come up with a contradiction. A point on $\mathbb{S}^1$ can be identified with a tangent vector at a point in $\mathbb{S}^2$. A point in $\text{SO}(3)$ corresponds to a point in $\mathbb{S}^2 \times \mathbb{S}^1$ via the diffeomorphism $F$. We are identifying this point in $\mathbb{S}^2 \times \mathbb{S}^1$ with a point in $\mathbb{S}^2$ and a tangent vector at that point. Since $F$ is diffeomorphic, we can make this assignment to all points of $\mathbb{S}^2$ such that the $\mathbb{S}^1$ part (the tangent vector) varies smoothly over $\mathbb{S}^2$. We have thus created a smooth non-zero vector field on $\mathbb{S}^2$. However, it is a well-known fact from lectures (and topology) that a smooth non-zero vector field does not exist on $\mathbb{S}^2$. Hence there can be no global diffeomorphism between the two manifolds.

To show that $\mathbb{S}^2 \times \mathbb{S}^1$ is locally diffeomorphic to $\text{SO}(3)$ , we come up with local diffeomorphism(s) between the two. I give two such local diffeomorphisms.

Since both manifolds are $3$ dimensional, there exists diffeomorphisms $f$, $g$ and $h$, such that:

\begin{align} f: & ~ \text{SO}(3) \rightarrow \text{D} \subseteq \mathbb{R}^3 \\ g: & ~ \mathbb{S}^2 \times \mathbb{S}^1 \rightarrow \text{E} \subseteq \mathbb{R}^3 \\ h: & ~ \text{D} \rightarrow \text{E} \end{align}

Then the required local diffeomorphism is $f \circ g \circ h^{-1}$ .

Another local diffeomorphism, $m$ can be constructed as suggested by the question. A point in $\text{SO}(3)$ is identified with a point on $\mathbb{S}^2$ (the axis) and a point on $\mathbb{S}^1$ (the ‘anti-clockwise’ rotation. The direction of rotation has to be specified for $m$ to be a well-defined map). This map is not globally 1-1, as the rotation by angle $\theta$ about axis $\hat{n}$ is the same as rotation by $\left(2\pi - \theta\right)$ about $-\hat{n}$. This map is locally 1-1 and onto, and as we ‘move’ smoothly over the points of one manifold, we also move smoothly over the second.

Scott's content with this explanation, and with the idea that any two manifolds of the same dimension (by our classroom definition) are locally diffeomorphic, but the discussion below is interesting…


I'm not convinced that they are locally diffeomorphic. Your first example of such a diffeomorphism implies that any pair of n-dimensional manifolds are locally diffeomorphic; is that true? Well, it's already known that for two manifolds to be locally diffeomorphic they must have the same dimension. So your claim would boil down to manifolds are locally diffeomorphic if and only if they have the same dimension. Granted, I can't come up with any obvious counter examples - but you can understand my reluctance.

Your second example may also be flawed. It seems to me that the suggested mapping has trouble near $\theta = 0$. Specifically, let $f$ be the suggested mapping from $\mathbb{S}^2 \times \mathbb{S}^1 \to SO(3)$. Consider an open ball $U$ around the point $(\hat{n}, 0)$. Note that no matter how small you make $U$, it will contain an infinitude of Identity matrices. In other words, under this map there is no local surjection from $\mathbb{S}^2 \times \mathbb{S}^1$ to $SO(3)$ at $(\hat{n}, 0)$.


I think any two manifolds of dimensions 1,2 or 3 are locally diffeomorphic, whereas any two manifolds of dimension greater than 3 are locally homeomorphic. We know any two homeomorphic manifolds of dimension less than 4 are diffeomorphic. By definition of manifold, $f$ and $g$ are local homeomorphisms, and hence diffeomorphisms. By the same logic, h is a homeomorphism from an open subset of $\mathbb{R}^3$ to another open subset of $\mathbb{R}^3$, and hence a diffeomorphism.

The second map is more informal, and it might be worth discussing it further. I do not exactly understand what you mean there though. Especially, can you define what you mean by failing to be a local surjection?


I think you are right about the first one. In the second example, I mean the suggested map is not a surjection between an open ball around $(\hat{n},0)$ and an open ball around $I \in SO(3)$ (identity matrix). The reason is an open ball around $(\hat{n},0)$ contains infinite mappings to $I$: each direction near $\hat{n}$ with a 0 rad rotation is an identity matrix. Since the suggested map is not surjective, it cannot be a homeomorphism.


Locally, an open ball from one manifold is being mapped to an open set in the other. If I think of the range of that open ball, it is surjective (or onto). I do not understand why any number of 0 radians should cause a problem? Are you thinking of saying the map is not locally injective? As long as the range of the open ball is open, we are fine. And the range is open, as different axes's will map to different points on the other manifold.


Indeed, I was thinking of injective. The open ball around $(\hat{n},0)$ will map to the identity matrix infinity times.


Lets see what happens when we map an open ball around $(\hat{n},0)$ from $SO(3)$ to the other manifold. The open ball will contain many values of angles and axes, and from what I infer, you seem ok when angle is not zero. From what I understand, you think the problem is that when we have two different axes with rotation of 0 degrees, which will be the case in any open ball around $(\hat{n},0)$ ? Lets take two such points in $SO(3)$, say $(\hat{n_1},0)$ and $(\hat{n_2},0)$. On $\mathbb{S}^2 \times \mathbb{S}^1$, the 0 angle will map to the same point on $\mathbb{S}^1$ for both those points. However, since the axes are different, they will map to different points on $\mathbb{S}^2$. Hence the map in injective. We have already agreed it is surjective. Hence its bijective, and therefore the inverse map is also bijective.


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