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, $\overrightarrow{u}\in {ℝ}^3$ together with the rotation of ${ℝ}^3$ 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 ${ℝ}^3$ 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 $\tilde{w}$ 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 ${\varphi }_{\overrightarrow{t},\overrightarrow{u}}(\tilde{w})$ 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 ${\varphi }_{\overrightarrow{t},\overrightarrow{u}}(\tilde{w})$.

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