# Fundamental group of SO(3) and Dirac's scissors

2016-11-25 ( 2016-09-22)## Topology of $SO(3)$ and fixed-distance pairs of points in $S^2$.

Consider the space $X\subset S^2\times S^2$ of all pairs $(A,B)$ of points such that $|A-B|=\sqrt 3$, where the distance is the euclidean distance of $S^2\subset \RR^3$, and $S^2$ is the unit sphere in $\RR^3$.

- (1)
- $X\approx SO(3) \approx \PP^3(\RR)$.

*Proof:*
The fact that $SO(3) \approx \PP^3(\RR)$ is a not-so-easy elementary exercise
(cf. exercise (11.27) and exercise (7.30) of my Geometry Notes,
protected by a single-character password: the number of *non-zero* digits appearing in the password itself; the problem might be ill-posed).
Now, $SO(3)$ acts freely and transitively on $X$, and since $SO(3)$ is compact and $X$ is Hausdorff, $X\approx SO(3)$.
*/qed/*

The following is an easy exercise as well.

- (2)
- If $\gamma\from I=[0,1] \to X$ is a loop such that $\gamma(0)=\gamma(1)$, then the path $\hat \gamma \from I \to X\times I$ defined
as $\hat \gamma(t) = (\gamma (t) , t)$ is a
**simple**path in $X\times I$ (i.e. $t\neq t'\implies \hat\gamma(t) \neq \hat\gamma(t')$ starting from $(\gamma(0),0)$ ending in $(\gamma(1),1)$, with $\gamma(0) = \gamma(1)$. Moreover, for each $t\in I$, $\gamma(t) = (A(t), B(t))$ where $A(t),B(t)$ are two points such that $|A(t) - B(t)|=\sqrt 3$.

Now, a **homotopy** relative endpoints of $\gamma$ into another path $\gamma'$ is a continuous family of paths $h_ s(t)$ such that
$h_ 0(t) = \gamma(t) $ amd $h_ 1(t) = \gamma'(t)$. An element of $\pi_ 1(X)$ is therefore a homotopy class
(rel. endpoints) of loops $I\to X$, and hence a homotopy class of paths rel. endpoints $I\to X\times I$,
such that $h_ s(0) = ((A(0),B(0)),0)$ and
$h_ s(1) = ((A(1),B(1)),1)$ for each $s$, with $A(0) = A(1)$
and $B(0) = B(1)$.

For each integer $n$, let $\FF_ n(S^2)$ denote the **configuration space** of $n$ points in $S^2$, i.e.
$$
\FF_ n(S^2) = \{ (x_ 1, \ldots, x_ n) \in S^2 \times \ldots \times S^2 : i \neq j \implies x_ i \neq x_ j \}~.
$$
By definition, the inclusion $X\subset \FF_ 2(S^2)$ holds. The fundamental group of $\FF_ n (S^2)$ is called
the **pure $n$-stranded braid group** of $S^2$, and is denoted by $PB_ n(S^2)$.

Consider the projection $$ p\from \FF_ 3(S^2) \to \FF_ 2(S^2), $$ defined as $p(A,B,C) = (A,B)$ for any $(A,B,C) \in \FF_ 3(S^2)$. It is well-defined, and for any $(A,B) \in \FF_ 2(S^2)$ its pre-image in $\FF_ 3(S^2)$ is homeomorphic to $S^2 \smallsetminus \{A,B\}$, which has the same homotopy type of $S^1$ (since $A\neq B$).

Now, let $Y\subset \FF_ 3(S^2)$ be defined as follows: $$ Y = \{ (A,B,C) \in \FF_ 3(S^2) : OABC \text{ are coplanar} \} $$ where $O$ is the origin of $\RR^3$.

- (3)
- $Y$ is a deformation retract of $\FF_ 3(S^2)$.

*Proof:*
For $(A,B,C) \in \FF_ 3(S^2)$, since $A,B,C$ are disjoint, there is a unique plane for $A,B,C$.
Let $\eta$ be the plane parallel to such a plane, passing through the origin $O$,
and $l$ the line by $O$ orthogonal to $\eta$. The line $l$ passes through the circumcenter of the
triangle $ABC$ (why?). Now, let $P_ \eta$ denote the projection of $A,B,C$ in the plane $\eta$,
orthogonal to $\eta$. Since $A,B,C$ are not on $l$, the projections $A'=P_ \eta A$, $B'=P_ \eta B$
and $C' = P_ \eta C$ are different from $O$ and distinct. Let the retraction
$r \from \FF_ 3(S^2) \to Y$ be defined as
$$
r(A,B,C) = ( \dfrac{A'}{|A'|}, \dfrac{B'}{|B'|}, \dfrac{C'}{|C'|} ) \in Y.
$$
By construction, if $(A,B,C) \in Y$, then $A'=A$, $B'=B$ and $C'=C$, and
therefore $r(A,B,C) = (A,B,C)$.
Hence, $r$ is a retraction.
To show that it is a deformation retraction,
consider for $t\in [0,1]$ the paths
$$
(A(t),B(t),C(t)) = ( (1-t)A+ t A', (1-t)B+tB', (1-t)C + tC' ),
$$
and divide each component by its norm. It is path in $\FF_ 3(S^3)$ joining
$(A,B,C)$ (for $t=0$) to $r(A,B,C)$ (for $t=1$), which depends continuously on $A,B,C$,
and hence it gives the homotopy between $ir$ and the identity of $\FF_ 3(S^2)$.
*/qed/*

As the next step, let $\hat X \subset Y$ be the subspace defined as follows: $$ \hat X = \{ (A,B,C) \in Y : \text{ $ABC$ is equilateral } \} \subset Y \subset \FF_ 3(S^2). $$

- (4)
- $\hat X$ is a deformation retract of $Y$.

*Proof:*
Given $(A,B,C) \in Y$, let $\eta$ denote the plane containing $OABC$.
There are in $S^2\cap\eta$ exactly three points $A'$, $B'$ and $C'$
such that $A'=A$, $|B'-A'| = |C'-A'| = | C' - B'|$ and
the order of $ABC$ in $S^2\cap \eta$ is the same as the order
of $A'B'C'$. If we define

$r(A)=A'$, $r(B)=B'$, $r(C)=C'$ in $S^2\cap \eta$,
it is a continuous function $r\from Y \to \hat X$,
which can be shown to be a deformation retraction.

*/qed/*

Finally, consider the projection (see Fadell-Neuwirth 1962) of $\FF_ 3(S^2) \to \FF_ 2(S^2)$, defined by $$ (A,B,C) \mapsto (A,B)~. $$ It sends $\hat X \subset \FF_ 3 (S^2)$ to the subset $X\subset \FF_ 2(S^2)$, since two vertices of an equilateral triangle inscribed in a unit circle have distance $\sqrt 3$. It is one-to-one, continuous, from a compact space to a Hausdorff space, and hence $$ \hat X \approx X ~. $$

As a consequence, we have proved that

- (5)
- $SO(3)$ is a deformation retract of the configuration space of $3$ points on a $2$-sphere $\FF_ 3(S^2)$. Hence $\pi_ 1(\FF_ 3(S^2)) = \pi_ 1(SO(3)) = \ZZ_ 2$.

## Pure braids in $\FF_3 (S^2)$ and the fundamental group of $SO(3)$.

As we have seen, $X \approx SO(3)$ is a deformation retract of $\FF_ 3 (S^2)$. Its fundamental group can be therefore computed as the fundamental group of $\FF_ 3(S^3)$, i.e. the group of pure $3$-stranded braids in $S^2$. One can represent it as strings joining two concentric (which cannot rotate) spheres, or any such settings. A semi-popular example is as in the pictures below.

The problem is to unknot the yellow braid, **without** any rotation of the scissors (and without any cut). I've tried to take pictures of the process, with
no success.

Further reading: Fred Cohen and Jonathan Pakianathan notes on *Configuration spaces and braid groups*, where they give a homotopy equivalence between $\FF_ 3(S^2)$ and $SO(3)$.