# $SO(3)$ Lie group

I'm a little stuck at the moment where to go next with this. I know that there is a fact that there is a curve in $SO(3)$, beginning and ending at the identity which cannot be deformed to the constant curve, but when traversed twice can be deformed.

I want to be able to explain this clearly. Can someone explain this to me or help me out a little please?

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Well, this is just a consequence of the fact that the fundamental group of SO(3) is the cyclic group of order 2. Equivalently, the universal cover of SO(3) is the group SU(2), and the kernel is $\{\pm 1\}$. – M Turgeon May 7 '12 at 14:25

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– M Turgeon May 7 '12 at 14:25

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This has nothing to do with Lie algebra... – Olivier Bégassat May 7 '12 at 14:32

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– t.b. May 7 '12 at 15:17

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@MTurgeon Touche. – Neal May 7 '12 at 16:25
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## 2 Answers

I find it easiest to see that $SO(3)$ has a nontrivial order-two element of its homotopy group by thinking concretely $SO(3)$. A rigid motion of $\mathbb{R}^3$ that fixes the origin may be described by an axis of rotation and a (positively oriented) angle of rotation between $-\pi$ and $\pi$. Note that if we rotate about an axis by an angle greater than $\pi$, this is the same as rotating about the same axis by a negative angle greater than $-\pi$.

We may thus parametrize $SO(3)$ by an axis through the origin and a distance along that axis in $[-\pi,\pi]$. This is a three-ball --- but wait, there's more! A rotation of $\pi$ about an axis is the same as a rotation by $-\pi$ about the axis, so to complete the picture of $SO(3)$, we need to quotient the ball by antipodal identification.

Modulo concerns about topological structure (take it for granted that two rotations about similar axis, of similar angles, are "nearby" in $SO(3)$), we have constructed a homeomorphism between $SO(3)$ and the three-ball of radius $\pi$ modulo antipodal boundary identification, $\mathbb{R}P^3$.

Now "it's clear"* why we have a nontrivial order-two element of $\pi_1SO(3)$. Take the circle defined by traversing $\mathbb{R}P^3$ from the south pole to the north pole through the origin. It is already "stretched taut" --- no deformation can pull it down to a point. But if the path traversed twice can be contracted to a point; it's a quick exercise to draw a path homotopic to the loop traveled twice and contract it down to the north/south pole.

This may also give some motivation for why Dirac's belt trick works. The belt represents a loop in $SO(3)$ and the "twist" at any point along the length of the belt indicates the angle of rotation of the element of $SO(3)$ through which the loop passes. We start with a flat belt (a path mapping onto the origin) and then begin twisting one end while keeping the other fixed, until the second end has passed through one full rotation. Now the second end is as flat as the first, indicating that the path is closed, but the the belt is twisted in the middle.

Try as you might, you cannot get the loops out of the belt without rotating either end. But if you do the same thing again, rotate the second end through a second full rotation, now you can untwist the belt without rotating either end.

• Perhaps intuitively clear, but described precisely in BenjaLim's answer.
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Neal has already explained to you why $SO(3) \cong \Bbb{R}P^3$. Therefore to understand why there is a loop $a$ in the fundamental group of $SO(3)$, it suffices to prove that $\pi_1(\Bbb{R}P^3)$ has an element of order 2. To do this view $\Bbb{R}P^3$ with the usual $CW$ - structure. Now by the Van - Kampen theorem we know that attaching a 3-cell and higher to a path connected space $X$ does not change the fundamental group, so that $\pi_1(\Bbb{R}P^3) \cong \pi_1(\Bbb{R}P^2)$. Applying the Van Kampen theorem again shows that $\pi_1(\Bbb{R}P^2)$ is the free group on one generator $a$ subject to the relation $a^2 =1$, which is nothing more than $\Bbb{Z}/2\Bbb{Z}$. Choosing your basepoint to be the identity shows that indeed there is a non-contractible loop at the identity in $SO(3)$.

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