# proof of Scott-Wiegold conjecture

Suppose the conjecture were false. Then we have some $w\in {C}_{p}*{C}_{q}*{C}_{r}$ with $N(w)={C}_{p}*{C}_{q}*{C}_{r}$. Let $a$, $b$, $c$ denote the of $w$ onto ${C}_{p}$, ${C}_{q}$, ${C}_{r}$ respectively. Then $a$, $b$, $c$ are all non-trivial as otherwise $N(w)$ would be contained in the kernel of one of the .

For $$ we say that a spin through
$\theta $ consists of a unit vector, $\overrightarrow{u}\in {\mathbb{R}}^{3}$ together with the
rotation of ${\mathbb{R}}^{3}$ through the angle $\theta $ anticlockwise about $\overrightarrow{u}$.
In we have a single spin through the angle ${0}^{\circ}$ and
a single spin through ${360}^{\circ}$. Thus the set of spins
(usually denoted Spin(3)) naturally has the topology^{} of a
3-sphere.

The spin through $\theta $ about a unit vector $\overrightarrow{u}$ has the same underlying rotation as the spin through ${360}^{\circ}-\theta $ about $-\overrightarrow{u}$. Hence there are precisely two spins corresponding to each rotation of ${\mathbb{R}}^{3}$ about the origin.

is well defined on spins as you can compose the underlying rotations and continuity determines which of the two spins is the correct result. For example a ${350}^{\circ}$ spin about $\overrightarrow{u}$ composed with a ${20}^{\circ}$ spin about $\overrightarrow{u}$ is a ${350}^{\circ}$ spin about $-\overrightarrow{u}$ (not a ${10}^{\circ}$ spin about $\overrightarrow{u}$ which would be at the other end of the 3-sphere).

Let $\overrightarrow{n}$ denote the unit vector $(0,0,1)$. Fix an arc, $I$,
on the unit sphere connecting $\overrightarrow{n}$ and $-\overrightarrow{n}$. Let
$\overrightarrow{t}$ be a vector on this arc. Let $\overrightarrow{u}$ be an arbitrary
unit vector. We may define a homomorphism^{}
${\varphi}_{\overrightarrow{t},\overrightarrow{u}}:{F}_{\{a,b,c\}}\to \mathrm{Spin}(3)$
by:

${\varphi}_{\overrightarrow{t},\overrightarrow{u}}:a\mapsto $ the spin through $(\frac{p-1}{2})\frac{{360}^{\circ}}{p}$ (or ${180}^{\circ}$ if $p=2$) about $\overrightarrow{n}$

${\varphi}_{\overrightarrow{t},\overrightarrow{u}}:b\mapsto $ the spin through $(\frac{q-1}{2})\frac{{360}^{\circ}}{q}$ (or ${180}^{\circ}$ if $q=2$) about $\overrightarrow{t}$

${\varphi}_{\overrightarrow{t},\overrightarrow{u}}:c\mapsto $ the spin through $(\frac{r-1}{2})\frac{{360}^{\circ}}{r}$ (or ${180}^{\circ}$ if $r=2$) about $\overrightarrow{u}$

(Here ${F}_{\{a,b,c\}}$ denotes the free group^{} on $a,b,c$).

So ${\varphi}_{\overrightarrow{t},\overrightarrow{u}}(a)$, ${\varphi}_{\overrightarrow{t},\overrightarrow{u}}(b)$ and ${\varphi}_{\overrightarrow{t},\overrightarrow{u}}(c)$ are spins of between ${120}^{\circ}$ and ${180}^{\circ}$, all having non-trivial underlying rotations.

Let $\stackrel{~}{w}$ be a word in ${F}_{\{a,b,c\}}$ representing $w$, such that $a,b,c$ occur in it $1$ Mod $(2p)$ times, $1$ Mod $2q$ times and $1$ Mod $(2r)$ times, respectively.

We have a homomorphism ${\varphi}^{\prime}:{C}_{p}*{C}_{q}*{C}_{r}\to SO(3)$ induced by $\varphi $. If ${\varphi}_{\overrightarrow{t},\overrightarrow{u}}(\stackrel{~}{w})$ has a trivial underlying rotation for some $\overrightarrow{t}$ and $\overrightarrow{u}$, then $N(w)$ will only contain elements in the kernel of ${\varphi}^{\prime}$. In particular, we would have $a,b,c\notin N(w)$. So we may assume we have a map:

$$h:I\times {S}^{2}\to {S}^{2}$$ |

which maps $(\overrightarrow{t},\overrightarrow{u})$ to the unit vector corresponding to ${\varphi}_{\overrightarrow{t},\overrightarrow{u}}(\stackrel{~}{w})$.

By we have $h(\overrightarrow{n},R\overrightarrow{u})=Rh(\overrightarrow{n},\overrightarrow{u})$ for any rotation $R$ about $\overrightarrow{n}$. Thus $h(\overrightarrow{n},\mathrm{\_}):{S}^{2}\to {S}^{2}$ maps latitudes to latitudes (possibly rotating them and / or moving them up or down).

Also $h(\overrightarrow{n},\overrightarrow{n})=-\overrightarrow{n}$, as
${\varphi}_{\overrightarrow{n},\overrightarrow{n}}(a)$, ${\varphi}_{\overrightarrow{n},\overrightarrow{n}}(b)$ and
${\varphi}_{\overrightarrow{n},\overrightarrow{n}}(c)$ are spins of between ${120}^{\circ}$ and
${180}^{\circ}$ anticlockwise about $\overrightarrow{n}$, so the sum of the
angles will be greater than ${360}^{\circ}$. Similarly one may
that $h(\overrightarrow{n},-\overrightarrow{n})=\overrightarrow{n}$. Thus, as $h(n,\mathrm{\_})$ maps latitudes to latitudes, it must be homotopic^{} to a
reflection of ${S}^{2}$.

Again by we have $h(-\overrightarrow{n},R\overrightarrow{u})=Rh(-\overrightarrow{n},\overrightarrow{u})$ for all rotations $R$ about $\overrightarrow{n}$. Hence $h(-\overrightarrow{n},\mathrm{\_}):{S}^{2}\to {S}^{2}$ also maps latitudes to latitudes.

Further, $h(-\overrightarrow{n},\overrightarrow{n})=\overrightarrow{n}$ and $h(-\overrightarrow{n},-\overrightarrow{n})=-\overrightarrow{n}$. Thus $h(-\overrightarrow{n},\mathrm{\_})$ is homotopic to the .

But $h$ gives a homotopy^{} from $h(\overrightarrow{n},\mathrm{\_})$ to $h(-\overrightarrow{n},\mathrm{\_})$, yielding the desired contradiction^{}.

Title | proof of Scott-Wiegold conjecture |
---|---|

Canonical name | ProofOfScottWiegoldConjecture |

Date of creation | 2013-03-22 18:30:31 |

Last modified on | 2013-03-22 18:30:31 |

Owner | whm22 (2009) |

Last modified by | whm22 (2009) |

Numerical id | 5 |

Author | whm22 (2009) |

Entry type | Proof |

Classification | msc 20E06 |