Recently I became interested in trying to visualize the fact that $\pi_1(\text{SO}(3)) = \mathbb{Z}/2\mathbb{Z}$. For whatever reason, the plate trick doesn't do it for me, so I've been looking for something else. I know that $\text{SO}(3)$ is homeomorphic to $P^3(\mathbb{R})$, and one concrete visualization of this space is as the solid unit ball with opposite points on the boundary identified. This model allows a straightforward visualization of the fundamental group, but what it doesn't seem to easily allow for is a corresponding visualization of the action of $\text{SO}(3)$. (In other words, it doesn't seem to be easy to visually verify that we are actually working in $\text{SO}(3)$.)

So I tried the following instead: $\text{SO}(3)$ acts simply transitively on the unit tangent bundle of $S^2$, so $UT(S^2)$ ought to be homeomorphic to $\text{SO}(3)$. This satisfies both my requirements: it is easy to visualize paths in this space, and it is also easy to see the action of $\text{SO}(3)$.

Now I would just appreciate someone confirming that I've got the details right: in this model, a representative of the nontrivial homotopy class of paths in $\pi_1(\text{SO}(3))$ is the path which parallel transports a unit vector once around a great circle, correct? How should I visualize the homotopy between the square of this path and the trivial path? (This might boil down to the plate trick or the belt trick, but at least it'll be stated in a language that I can understand.)

  • 8
    $\begingroup$ $P^3(\mathbb R)$ you can think of as a compact unit ball with antipodal points on the boundary identified. Every point in the ball is a vector (direction + magnitude). The direction vector will be the eigenvector of an element of $SO_3$, and its magnitude (multiplied by $\pi$) will be the amount you rotate about that axis. That's how the two are related. More formally what I'm describing is the exponential map for the Lie group $SO_3$. $\endgroup$ – Ryan Budney Jan 17 '11 at 16:14
  • $\begingroup$ Just scratching at the surface a bit: your construction reminds me of Penrose flags; I suspect that you can find descriptions/pictures in his book Spinors and Space-time vol 1. $\endgroup$ – Willie Wong Jan 17 '11 at 16:22
  • $\begingroup$ This might be helpful: math.mit.edu/~pokman/18.904/handout1116.pdf $\endgroup$ – PrimeNumber Jan 17 '11 at 16:26
  • $\begingroup$ @Ryan: many thanks. I figured something like this was true but wasn't sure of the details. $\endgroup$ – Qiaochu Yuan Jan 17 '11 at 16:29
  • $\begingroup$ A slightly displeasing aspect of the identification of $\mathrm{SO}(3)$ with $UT(S^2)$ is that it is not canonical but depends on the choice of one unit tangent vector. This is not the case for the identification with $P^3(\mathbb{R})$. $\endgroup$ – Rasmus Jan 17 '11 at 18:52

Here's a bit of an elaboration on my earlier comment.

You can see $SO_3$ as the quotient of a unit $3$-ball by the antipodal map on the boundary. This is realized explicitly by the exponential map for $SO_3$. The tangent space to the identity of $SO_3$ is all $3\times 3$ matrices $H$ such that $H^T=-H$. Up to conjugation by an element of $SO_3$, such a matrix looks like

$$\left(\begin{matrix} 0 & \theta & 0 \\ -\theta & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\right)$$

which under exponentiation is converted into rotation by $\theta$ about the $z$-axis. Every element of $SO_3$ is rotation by some angle about a fixed axis since the characteristic polynomial has $1$ as a root -- so this is a rather explicit description of how the exponential map is onto. If you choose the length of a matrix $[a_{i j}]$ to be $\sqrt{ \sum_{i j} a_{i j}^2}$, then technically $SO_3$ is the quotient of the ball of radius $\pi/\sqrt{2}$, with antipodal points on the boundary identified. But that's small potatoes.

The 3-ball with antipodal points identified, as you've seen has fundamental group $\Bbb Z_2$. In particular you can think of the 3-ball with antipodal points identified as being built up from the 1-ball with antipodal points identified, union a 2-cell (this gives a CW-decomposition of the 2-ball with antipodal points identified) then union a 3-cell. So the fundamental group all "happens" in the 2-ball with antipodal points identified. The 1-skeleton generates the fundamental group and in our model (under exponentiation) corresponds to one full rotation by $2\pi$ about any chosen axis. The relation corresponds to the 2-cell attachment and you can literally interpret it as the plate trick in a "sufficiently nice" model. Let me explain.

The trouble with the plate trick is it uses a human arm. Human arms you can think of as carrying a core underlying piece of information, a map $f : [0,1] \to SO_3$. $f(t)$ represents the orientation in space of the part of the arm distance $t$ from the shoulder, so $0$ represents the shoulder distance, and $1$ the palm of the hand. There are technical issues -- if you choose the framing corresponding to the positions of the appropriate bone in the arm, you don't get a continuous map, just a piecewise-constant map -- but if you consider the orientations of the ligaments between the bones it becomes a continuous map. So the trouble with the plate trick is the map $f$ isn't the only constraint on the arm. The arm is embedded in space. Bones are rigid, there's only so much ligaments are capable of doing. So you could be rightly concerned that the plate trick is a demonstration of a peculiarity of anatomy, rather than something about $SO_3$.

Alright, but formally what's going on with the plate trick? As mentioned, a configuration of the arm we're representing by $f$, which is an element of the path space $PSO_3$ of continuous paths in $SO_3$ with $f(0) = I \in SO_3$ fixed.

There is the path-loop fibration:

$$\Omega SO_3 \to PSO_3 \to SO_3$$

and $PSO_3$ is contractible. The plate trick "is" the induced map $\pi_1 SO_3 \to \pi_0 \Omega SO_3$ from the homotopy long exact sequence of the above fibration, which is guaranteed to be an isomorphism by the path-loop fibration. Specifically, elements of $\pi_1 SO_3$ you are thinking of as motions of your hand. The path-loop fibration is telling you that you can realize the motion as a motion of your arm (or at least, a path in $PSO_3$ but the fact that people can do the plate trick says you can actually realize it as a motion of your arm).

There is of course an element that's a bit misleading in all this. The arm isn't complicated enough to allow for you to lift the product of two generators of $\pi_1 SO_3$ and witness the null homotopy -- since the arm allows a fixed complexity, the null-homotopy happens as you lift the concatenation.

But you can see from the rather explicit exponential map above how the null homotopy actually works, as it's sitting in the image of the 2-ball with antipodal points identified.

How is that?

  • 3
    $\begingroup$ Thank you very much for the details. I think I've also more or less figured out how everything works in UT(S^2) - it's nice to have multiple pictures. (The reason I prefer UT(S^2) to the plate trick is that it is, for me, easier to visualize, as well as to describe in words: I think of a path in it as a little tank moving around S^2 while waving its turret around.) $\endgroup$ – Qiaochu Yuan Jan 17 '11 at 20:41
  • $\begingroup$ How do you prove that all matrices in Lie algebra of $SO_3$ have the form stated up to conjugation by an element from $SO_3$? $\endgroup$ – user0810 Aug 3 '14 at 1:24
  • $\begingroup$ @user810: One way is to compute the Lie algebra for $SO_2$ and then argue that something in the Lie algebra for $SO_3$ has a zero eigenvalue, so it splits off an orthogonal factor in the $SO_2$ Lie algebra. $\endgroup$ – Ryan Budney Aug 3 '14 at 5:56

Does the following help?


There is a silent video from which these stills are taken: it is 60MB and is available at


I have not had the facilities to improve this.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.