I ##SO(3)## topology

[cianfa72]

TL;DR Summary

About the $SO(3)$ topology as subset/subspace topology from $\mathbb R^9$

$SO(3)$ is a Lie group of dimension 3. It is the set of 3x3 matrices $R$ with the following properties: $$RR^T = R^TR=I, \text{det}(R)=+1$$ There exists a parametrization of $SO(3)$ that maps it on the sphere in $\mathbb R^3$ of radius $\pi$ where the antipodal points are identified. This map is bijective and it basically defines the topology of $SO(3)$ -- i.e. let me say the map is homeomorphism by definition.

Is this the subspace topology from $\mathbb R^9$ on $SO(3)$ as subset of $\mathbb R^9$ ?

 

[pasmith]

I believe these topologies are all ultimately induced by norms on finite dimensional vector spaces, and are therefore topologically equivalent.

 

[cianfa72]

I believe these topologies are all ultimately induced by norms on finite dimensional vector spaces, and are therefore topologically equivalent.

You are basically saying that map is actually a topological embedding of the sphere with antipodal points identified within $\mathbb R^9$ equipped with the standard topology.

How can this be formally proven?

 

[cianfa72]

How can this be formally proven?

I was reasoning as follows: from this thread the condition $RR^T=I$ basically amounts to 9 functions (rational functions of polynomials) smooth on the relevant domain in $\mathbb R^9$. Since they are continuous the preimage of a closet set (either $\{ 0 \}$ or $\{ 1 \}$) is closed in $\mathbb R^9$ as the preimage under the determinant function of $\{ +1 \}$. The intersection of closed sets is closed hence $SO(3)$ is closed in $\mathbb R^9$.

 

[cianfa72]

Consider the following map $\varphi: \mathbb R^3 \to \mathbb R^9$
$$\begin{pmatrix}
\cos \theta + \hat {\theta_1^2}(1 - \cos \theta) & \hat {\theta_3} \sin \theta + \hat {\theta_1} \hat{\theta_2}(1 - \cos \theta) & -\hat {\theta_2}\sin \theta + \hat {\theta_1} \hat{\theta_3} (1 - \cos \theta) \\
- \hat {\theta_3} \sin \theta + \hat {\theta_2} \hat{\theta_1} (1 - \cos \theta) & \cos \theta + \hat {\theta_2^2} (1 - \cos \theta) & \hat {\theta_1} \sin \theta + \hat {\theta_2} \hat{\theta_3} (1 - \cos \theta) \\
\hat {\theta_2} \sin \theta + \hat {\theta_3} \hat{\theta_1} (1 - \cos \theta) & - \hat {\theta_1} \sin \theta + \hat {\theta_3} \hat{\theta_2} (1 - \cos \theta) & \cos \theta + \hat {\theta_3^2} (1 - \cos \theta)
\end{pmatrix}$$
where $\hat {\theta_i} = {\theta_i}/ {\theta}, \theta = \sqrt {\theta_1^2 + \theta_2^2 + \theta_3^2}$. It is defined everywhere on $\mathbb R^3$ and it is smooth hence continuous.

Restricting the domain to the ball $B^3 \subset \mathbb R^3$ of radius $\pi$, the image of $\varphi|_{B^3}$ is $SO(3)$. Both $B^3$ and $SO(3)$ are Hausdorff compact (using subspace topologies from $\mathbb R^3$ and $\mathbb R^9$ respectively) and $\left . \varphi \right |_{B^3}$ is a quotient map (antipodal points on the surface $S^2$ of $B^3$ map on the same $SO(3)$ matrix). Therefore when it comes to the quotient space $B^3/\sim$ one gets an homeomorphism with $SO(3)$.

Does it makes sense ?

 

[fresh_42]

Consider the following map $\varphi: \mathbb R^3 \to \mathbb R^9$
$$\begin{pmatrix}
\cos \theta + \hat {\theta_1^2}(1 - \cos \theta) & \hat {\theta_3} \sin \theta + \hat {\theta_1} \hat{\theta_2}(1 - \cos \theta) & -\hat {\theta_2}\sin \theta + \hat {\theta_1} \hat{\theta_3} (1 - \cos \theta) \\
- \hat {\theta_3} \sin \theta + \hat {\theta_2} \hat{\theta_1} (1 - \cos \theta) & \cos \theta + \hat {\theta_2^2} (1 - \cos \theta) & \hat {\theta_1} \sin \theta + \hat {\theta_2} \hat{\theta_3} (1 - \cos \theta) \\
\hat {\theta_2} \sin \theta + \hat {\theta_3} \hat{\theta_1} (1 - \cos \theta) & - \hat {\theta_1} \sin \theta + \hat {\theta_3} \hat{\theta_2} (1 - \cos \theta) & \cos \theta + \hat {\theta_3^2} (1 - \cos \theta)
\end{pmatrix}$$
where $\hat {\theta_i} = {\theta_i}/ {\theta}, \theta = \sqrt {\theta_1^2 + \theta_2^2 + \theta_3^2}$. It is defined everywhere on $\mathbb R^3$ and it is smooth hence continuous.

Restricting the domain to the ball $B^3 \subset \mathbb R^3$ of radius $\pi$, the image of $\varphi|_{B^3}$ is $SO(3)$. Both $B^3$ and $SO(3)$ are Hausdorff compact (using subspace topologies from $\mathbb R^3$ and $\mathbb R^9$ respectively) and $\left . \varphi \right |_{B^3}$ is a quotient map (antipodal points on the surface $S^2$ of $B^3$ map on the same $SO(3)$ matrix). Therefore when it comes to the quotient space $B^3/\sim$ one gets an homeomorphism with $SO(3)$.

Does it makes sense ?

How does your relation look like? You have get rid of the entire diameters when transforming $B^3$ to $S^2,$ not only the antipodes.

Here are quite a few possible mappings
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
that describes it better. E.g. $SO(3)/SO(2)\cong S^2.$

 

[cianfa72]

How does your relation look like? You have get rid of the entire diameters when transforming $B^3$ to $S^2,$ not only the antipodes.

No, maybe I was sloppy. There is no transformation from $B^3$ to $S^2$.

My point was that the points on $B^3$ with $\theta = \pi$ are identified in couples (antipodal points).

The map $\varphi$ maps all $\mathbb R^3$ to points (i.e. matrices) in $SO(3)$. Restricting the domain to $B^3$ one gets the relevant quotient map and from there the homeomorphism (when taking the quotient space by antipodal points identication).

 

[fresh_42]

No, maybe I was sloppy. There is no transformation from $B^3$ to $S^2$.

My point was that the points on $B^3$ with $\theta = \pi$ are identified in couples (antipodal points).

The map $\varphi$ maps all $\mathbb R^3$ to points (i.e. matrices) in $SO(3)$. Restricting the domain to $B^3$ one gets the relevant quotient map and from there the homeomorphism (when taking the quotient space by antipodal points identication).

Forget that idea with antipodes. It does not work. $SO(3)$ is three-dimensional, $S^2$ two-dimensional. Identifying antipodes does not reduce the dimension.

 

[fresh_42]

Why are you talking of $S^2$ ? My use of $S^2$ was just to refer to the "boundary" of the $B^3$ ball of radius $\pi$.

Me?

... that maps it on the sphere in $\mathbb R^3$ of radius $\pi$ where the antipodal points are identified ...

You started it! And it is not true. Look at the link. Closest you get by identifying points is the double-cover
$$
\mathbb{Z}_2 \rightarrowtail \operatorname{SU}(2,\mathbb{C})\twoheadrightarrow SO(3)
$$
You started with the embedding into $\mathbb{R}^9$ by coordinate functions. That fixes everything else that comes after it.

 

[cianfa72]

You started it! And it is not true. Look at the link. Closest you get by identifying points is the double-cover
$$
\mathbb{Z}_2 \rightarrowtail \operatorname{SU}(2,\mathbb{C})\twoheadrightarrow SO(3)
$$

Sorry, I used the term sphere in $\mathbb R^3$ to mean actually the ball in $\mathbb R^3$ of radius $\pi$.

 

[fresh_42]

The 2-sphere is the quotient $SO(3)/SO(2).$ Not sure, how the ball fits into that image of quotient spaces. A rotation in $\mathbb{R}^3$ is already determined by what it does to the sphere so why complicate things by introducing a radius? Lengths do not change under $SO(3).$

 

[fresh_42]

Maybe I was unclear. Do you agree that the map $\varphi$ in post#5 is surjective and continuous from $\mathbb R^3$ into $SO(3) \subset \mathbb R^9$ ?

If you solve the problem with $\theta=0$ then yes. But I haven't calculated the determinant or orthogonality so I don't know what it does to the ball.

You can simply use the three standard rotations, their coordinate functions, and $SO(3)$ as the group they generate. Then any zeros won't occur and the coordinates function are smooth.

 

[cianfa72]

If you solve the problem with $\theta=0$ then yes. But I haven't calculated the determinant or orthogonality so I don't know what it does to the ball.

The limit for $\theta \to 0$ is the Identity matrix $\mathbb 1$, therefore simply define $\varphi (0,0,0)=\mathbb 1$ and you get a surjective continuous map $\varphi : \mathbb R^3 \to SO(3)$ (the target with the subspace topology from $\mathbb R^9$).

 

[fresh_42]

The limit for $\theta \to 0$ is the Identity matrix $\mathbb 1$, therefore simply define $\varphi (0,0,0)=\mathbb 1$ and you get a surjective continuous map $\varphi : \mathbb R^3 \to SO(3)$ (the target with the subspace topology from $\mathbb R^9$).

Yes, that's why I said "solve the problem" and not "correct the error". The way you defined $\hat\theta$ in post #5 you get a singularity at the origin. If it was chess I would have said "sauber setzen!" (set properly)

 

[cianfa72]

Ok, therefore the map $\varphi$ in #5 is actually a topological embedding of the space $B^3/_\sim \cong SO(3)$ into $\mathbb R^9$ with standard topology.

 

[fresh_42]

Ok, therefore the map $\varphi$ in #5 is actually a topological embedding of the space $B^3/_\sim \cong SO(3)$ into $\mathbb R^9$ with standard topology.

I don't know what $B^3/\sim$ even is, let alone has anything to do with $SO(3).$ You have defined it as a map $\varphi\, : \,\mathbb{R}^3\to \mathbb{R}^9.$ Assumed that $\operatorname{im}\varphi = SO(3)$ you still have to determine $\operatorname{ker}\varphi.$

 

[jbergman]

I think an easier way to approach this problem is as follows. Define a map $\varphi(R) : \mathbb R^9 \rightarrow \mathbb R^{10}$ using with the first 9 coordinates the coordinates of $RR^t$ the last coordinate as $\det(R)$ for the 10 coordinates in $\mathbb R^{10}$. Then if you can show that $\varphi$ is constant rank $k$ then we know then we know that the inverse image of the level set $\varphi^{-1}(c)$ is an embedded submanifold with the subspace topology and we can take $c$ such that $RR^t = I, \det(R)=1$ we can take.

Note, this is probably not a constant rank map at the origin, so you really should start with $GL(\mathbb R, 3)$ which is an open subset of $\mathbb R^9$

 

[cianfa72]

I don't know what $B^3/\sim$ even is, let alone has anything to do with $SO(3).$ You have defined it as a map $\varphi\, : \,\mathbb{R}^3\to \mathbb{R}^9.$ Assumed that $\operatorname{im}\varphi = SO(3)$ you still have to determine $\operatorname{ker}\varphi.$

Consider a straight line through the origin of $\mathbb R^3$ with coordinates ($\theta_1, \theta_2, \theta_3$). It is straightforward to check that antipodal points with $\theta = \pi$ map to the same matrix.

Restrict $\varphi$ to the ball $B^3$ of radius $\pi$ endowed with the subspace topology from $\mathbb R^3$. We get a continuous surjective map between Hausdorff compact spaces. This is a quotient map, hence the induced map between $B^3$ with $\theta= \pi$ antipodal points identified and the target $SO(3)$ is homeomorphism.

 

[cianfa72]

Define a map $\varphi(R) : \mathbb R^9 \rightarrow \mathbb R^{10}$ using with the first 9 coordinates the coordinates of $RR^t$ the last coordinate as $\det(R)$ for the 10 coordinates in $\mathbb R^{10}$. Then if you can show that $\varphi$ is constant rank $k$ then we know then we know that the inverse image of the level set $\varphi^{-1}(c)$ is an embedded submanifold with the subspace topology and we can take $c$ such that $RR^t = I, \det(R)=1$ we can take.

I'm not sure whether it actually suffices: in general $RR^T=I$ doesn't imply $R^TR=I$, you have to consider both

 

[jbergman]

I'm not sure whether it actually suffices: in general $RR^T=I$ doesn't imply $R^TR=I$, you have to consider both

It does suffice. There are several arguments for it. The easiest is to see that $R^t = R^{-1}$.

 

[jbergman]

Consider a straight line through the origin of $\mathbb R^3$ with coordinates ($\theta_1, \theta_2, \theta_3$). It is straightforward to check that antipodal points with $\theta = \pi$ map to the same matrix.

Restrict $\varphi$ to the ball $B^3$ of radius $\pi$ endowed with the subspace topology from $\mathbb R^3$. We get a continuous surjective map between Hausdorff compact spaces. This is a quotient map, hence the induced map between $B^3$ with $\theta= \pi$ antipodal points identified and the target $SO(3)$ is homeomorphism.

Yes, a rotation can be described as a point in the closed ball of radius π with antipodal points identified on the boundary. But, I am not sure you have shown that your map has the properties you desire, subjectivity and a map from the quotient is injective.

 

[cianfa72]

But, I am not sure you have shown that your map has the properties you desire, subjectivity and a map from the quotient is injective.

Indeed. For $\theta = \pi$ if you switch $(\theta_1,\theta_2,\theta_3)$ to $(-\theta_1,-\theta_2, -\theta_3)$ you get from $\varphi$ the same $SO(3)$ matrix. $\varphi$ is surjective and continuous from the closed ball $B^3$ to $SO(3)$ both with the subspace topologies. In the quotient space you get identifying antipodal points on the $B^3$ boundary, the induced $\bar \varphi$ is bijective and continuous (in the quotient topology). Morever the quotient is compact since $B^3$ is and $SO(3)$ is Hausdorff compact in the subspace topology from $\mathbb R^9$, hence $\bar \varphi$ is homeomorphism.

 

[cianfa72]

For $\theta = \pi$ if you switch $(\theta_1,\theta_2,\theta_3)$ to $(-\theta_1,-\theta_2, -\theta_3)$ you get from $\varphi$ the same $SO(3)$ matrix. $\varphi$ is surjective and continuous from the closed ball $B^3$ to $SO(3)$ both with the subspace topologies. In the quotient space you get identifying antipodal points on the $B^3$ boundary, the induced $\bar \varphi$ is bijective and continuous (in the quotient topology). Morever the quotient is compact since $B^3$ is and $SO(3)$ is Hausdorff compact in the subspace topology from $\mathbb R^9$, hence $\bar \varphi$ is homeomorphism.

We can use the (group) surjective homomorphism $R: S^3 \to SO(3)$ from prove $RP^2 \cong SO(3)$.

$S^3$ is isomorphic to the unit quaternions $q=a\mathbf1 + b \mathbf i + c \mathbf j + d \mathbf k, a^2 + b^2 + c^2 + d^2=1$. Morever any unit quaternion can be written as: $$\cos \left (\frac {\theta} {2} \right ) + \sin \left (\frac {\theta} {2} \right ) \left ( \hat {\theta_1} \mathbf i + \hat {\theta_2} \mathbf j + \hat {\theta_3} \mathbf k \right ), \hat {\theta_1^2} + \hat {\theta_2^2} + \hat {\theta_3^2} = 1 $$
Substituting it into the $R$ map above, one gets the explicit map $\varphi$ in my post #5. It turns out to be surjective from the closed ball $B^3$ to $SO(3)$.

 

[jbergman]

We can use the (group) surjective homomorphism $R: S^3 \to SO(3)$ from prove $RP^2 \cong SO(3)$.

$S^3$ is isomorphic to the unit quaternions $q=a\mathbf1 + b \mathbf i + c \mathbf j + d \mathbf k, a^2 + b^2 + c^2 + d^2=1$. Morever any unit quaternion can be written as: $$\cos \left (\frac {\theta} {2} \right ) + \sin \left (\frac {\theta} {2} \right ) \left ( \hat {\theta_1} \mathbf i + \hat {\theta_2} \mathbf j + \hat {\theta_3} \mathbf k \right ), \hat {\theta_1^2} + \hat {\theta_2^2} + \hat {\theta_3^2} = 1 $$
Substituting it into the $R$ map above, one gets the explicit map $\varphi$ in my post #5. It turns out to be surjective from the closed ball $B^3$ to $SO(3)$.

Overall this looks correct to me. There are a lot of details that need to be filled in, though, to make this more rigorous.

 

[WWGD]

Ok, therefore the map $\varphi$ in #5 is actually a topological embedding of the space $B^3/_\sim \cong SO(3)$ into $\mathbb R^9$ with standard topology.

I guess you want to use here that a continuous bijection between compact and Hausdorff is a homeomorphism?

 

[cianfa72]

I guess you want to use here that a continuous bijection between compact and Hausdorff is a homeomorphism?

Yes, exactly (by using quotient topology on $B^3/_\sim$ from closed ball $B^3$ topology and subspace topology from $\mathbb R^9$ on $SO(3)$.

 

[lavinia]

TL;DR Summary: About the $SO(3)$ topology as subset/subspace topology from $\mathbb R^9$

$SO(3)$ is a Lie group of dimension 3. It is the set of 3x3 matrices $R$ with the following properties: $$RR^T = R^TR=I, \text{det}(R)=+1$$ There exists a parametrization of $SO(3)$ that maps it on the sphere in $\mathbb R^3$ of radius $\pi$ where the antipodal points are identified. This map is bijective and it basically defines the topology of $SO(3)$ -- i.e. let me say the map is homeomorphism by definition.

Is this the subspace topology from $\mathbb R^9$ on $SO(3)$ as subset of $\mathbb R^9$ ?

Without checking the parameterization, assuming it induces a bijection from the closed ball modulo antipodal points on its boundary onto the SO(3) matrices, if it is is continuous(proof?) then you get a continuous bijection from a compact space (proof?) onto SO(3) as a subset of R^9. AS @WWGD points out, since SO(3) is Hausdorff in R^9 (proof?) this map is a homeomorphism.

 

[WWGD]

Without checking the parameterization, assuming it induces a bijection from the closed ball modulo antipodal points on its boundary onto the SO(3) matrices, if it is is continuous then you get a continuous (proof?) bijection from a compact space (proof?) onto SO(3) as a subset of R^9. AS @WWGD points out, since SO(3) is Hausdorff in R^9 (proof?) this map is a homeomorphism.

All rotations can be represented by the (closed) 3-ball $B^3$ of radius, say 1 , which is Compact, being closed and bounded. Since the quotient map (Identifying an angle with its negative) is, by definition, continuous, the quotient is the continuous image of the compact ball B^3, thus compact. SO(3) is actually a manifold, thus metrizable, thus Hausdorff. I think that does it.For rigor, just need to show the map is a bijection, which you argued in your first post. That should do it.

 

[cianfa72]

Without checking the parameterization, assuming it induces a bijection from the closed ball modulo antipodal points on its boundary onto the SO(3) matrices, if it is is continuous(proof?)

Yes, it is. Consider the overall space $(\theta_1, \theta_2, \theta_3)$, that's $\mathbb R^3$. The map $\varphi$ in post #5 from $\mathbb R^3$ to $\mathbb R^9$ is trivially continuous using standard euclidean topologies on both domain and target (note that its value at $(0,0,0)$ is defined as the identity matrix $I$). Then the restriction $\varphi|_{B^3}$ to subsets of domain (the closed ball $B^3$) and target (SO(3)) in the subspace topologies respectively is continuous as well. When it comes to the quotient $B^3/_{\sim}$, $\varphi|_{B^3}$ descends/passes to the quotient as bijective continuous map $\tilde {\varphi}$ using the quotient topology on it.

 

[lavinia]

@cianfa72

This may be obvious but I thought I'd mention it anyway.

If one identifies SU(2) with S^3 then the covering map SU(2)→SO(3) maps the upper hemisphere of S^3 onto SO(3) with antipodal points remaining only on the equatorial 2 sphere. The upper hemisphere is a closed 3 ball in Euclidean space. This gives you the topology of SO(3) as you described in the parameterization of SO(3) as a subset of R^9.

Aside:
I thought that @fresh_42 's exact sequence SO(2) →SO(3) →S^2 is instructive because it represents the 2 sphere as the set of cosets of SO(2) in SO(3) with the quotient topology. It follows since SO(3) is a Lie group and SO(2) is a closed subgroup that SO(3) is a principal SO(2) bundle over the 2 sphere.

By looking at the action of SO(3) on the 2 sphere by rotations, it becomes clear that this principal SO(2) bundle is the tangent circle bundle of S^2, the set of tangent vectors of length 1 in each tangent plane. Proof involves the same type of argument that you used except in reverse. One starts with the SO(3) matrices and maps them onto the tangent circle bundle of S^2 which can be viewed as a subset of R^6.

Second Aside:
if one splits the 2 sphere into upper and lower hemispheres then its tangent circle bundle also splits into two pieces and each piece is the Cartesian product of a closed disk with a circle, D^2 × S^1. Their common boundary is S^1× S^1 which is just the set of tangent circles along the equator.

If one thinks of D^2 × S^1 as a circle of closed disks then one sees that it is topologically a solid torus and the boundary being a circle of circles is just a torus. So the tangent circle bundle is two solid tori pasted together along their boundaries and therefore, SO(3) is also homeomorphic to two solid tori pasted together along their boundaries.

This point of view gets complicated because the same is true for any oriented circle bundle over the 2 sphere e.g. the Hopf fibration so in order to distinguish SO(3) from the others, one needs to know how the two solid tori are pasted together. Different pastings give different spaces. Finding this pasting is an instructive exercise.