Fundamental group of SO(3) and Dirac's scissors · DL Ferrario's Test Web Page

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

## 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 $SO(3)$.