The elementary geometric half twist, its support disc, and its opposite

Let $n\in \mathbb{N}$ and let $Q=(q_1,\dots ,q_n)$ be the base configuration of Geometric braids in the disc with setwise endpoints, with $h=\frac{1}{4(n+1)}$ and $q_j=((2j-n-1)h,0)$. Fix an index $i$ with $1\le i\le n-1$. The two adjacent base points are

$q_i=((2i-n-1)h,0),q_{i+1}=((2i+1-n)h,0),$

their midpoint is $m_i:=(q_i+q_{i+1})\mathrm{/}2=((2i-n)h,0)=q_i+(h,0)$, and their support disc is the open disc

$U_i:=\{w\in {\mathbb{R}}^2:\parallel w-m_i{\parallel }_2<\frac{3}{2}h\}\mathrm{.}$

The support disc contains the two base points $q_i,q_{i+1}$, whose distance from $m_i$ is $h$, and it contains no other base point: for $k<i$ one has $\parallel q_k-m_i{\parallel }_2=(2(i-k)+1)h\ge 3h$ and for $k>i+1$ one has $\parallel q_k-m_i{\parallel }_2=(2(k-i)-1)h\ge 3h$, both exceeding $\frac{3}{2}h$. Moreover $U_i\subseteq D^{\circ }$, because every point of $U_i$ has norm at most $\parallel m_i{\parallel }_2+\frac{3}{2}h\le (n-2)h+\frac{3}{2}h<(n+1)h<1$. Finally $U_i\cap U_j=\mathrm{\varnothing }$ whenever $\mathrm{\mid }i-j\mathrm{\mid }>1$, since then the midpoints are at distance $2\mathrm{\mid }i-j\mathrm{\mid }h\ge 4h>3h$, twice the radius.

The positive half twist. Define $\rho \text{\hspace{0.17em}⁣}:I\to {\mathbb{R}}^2$, the anticlockwise diamond path, by

$\rho (t):=\{\begin{array}{ll}(2th-h,-2th),& 0\le t\le \frac{1}{2},\\ (2th-h,2th-2h),& \frac{1}{2}\le t\le 1.\end{array}$

Thus $\rho (0)=(-h,0)$, $\rho (\frac{1}{2})=(0,-h)$, $\rho (1)=(h,0)$, and $\rho$ is continuous, piecewise linear, and satisfies $\parallel \rho (t){\parallel }_2\le h$ for all $t$ with $\rho (t)\mathrm{\ne }0$ throughout. The elementary half twist at $i$, written ${\sigma }_i$, is the tuple of motions

$({\sigma }_i{)}_i(t):=m_i+\rho (t),({\sigma }_i{)}_{i+1}(t):=m_i-\rho (t),({\sigma }_i{)}_k(t):=q_k(k\mathrm{\notin }\{i,i+1\})\mathrm{.}$

It is a braid based at $Q$ (Geometric braids in the disc with setwise endpoints, Continuity of a map of topological spaces at a point and globally): the two moving points stay in $U_i$ and are antipodal about $m_i$, so they are distinct; every other point is fixed and lies outside $U_i$; and the endpoint permutation is the transposition of $i$ and $i+1$, since $({\sigma }_i{)}_i(1)=m_i+(h,0)=q_{i+1}$ and $({\sigma }_i{)}_{i+1}(1)=m_i-(h,0)=q_i$.

The words positive and anticlockwise refer to the fixed orientation of the plane of the disc and to the fixed planar projection whose horizontal axis contains $Q$: the moving pair turns through the half turn anticlockwise, the label $i$ passing below its midpoint and the label $i+1$ above. Together with the first-under-second stacking convention of Stacking of geometric braids is a well-defined associative operation on isotopy classes this fixes the sign convention for every signed crossing on this page and its companion.

The opposite half twist. The opposite (clockwise) half twist at $i$ is defined by the reflected path

${\rho }^{-}(t):=({\rho }_1(t),-{\rho }_2(t)),$

through the motions $({\sigma }_i^{-}{)}_i(t):=m_i+{\rho }^{-}(t)$, $({\sigma }_i^{-}{)}_{i+1}(t):=m_i-{\rho }^{-}(t)$ and $({\sigma }_i^{-}{)}_k(t):=q_k$ for $k\mathrm{\notin }\{i,i+1\}$. It is again a braid supported in $U_i$ with endpoint permutation the transposition of $i$ and $i+1$; its pair turns through the same half turn clockwise, the label $i$ passing above the midpoint. The two motions are not equal, and they are not related by a reparametrisation of the height: they are the two signed crossings of the pair.

The opposite motion is the group inverse. For a braid $\beta$ the reversed braid ${\bar{\beta }}_j(t)=z_{\pi (\beta {)}^{-1}(j)}(1-t)$ represents $[\beta {]}^{-1}$ in the group $G_n$ (The isotopy classes of geometric braids based at $Q$ form a group, and the endpoint permutation is a homomorphism). For $\beta ={\sigma }_i$ one computes $\overline{{\sigma }_i}={\sigma }_i^{-}$: reversing the height and relabelling by the transposition of $i$ and $i+1$ sends $\rho (t)=m_i+({\rho }_1,{\rho }_2)$ to the motion with relative path $-\rho (1-t)$, which is the reflected path ${\rho }^{-}$ traversed from $t=0$ to $t=1$. Hence

$[{\sigma }_i^{-}]=[{\sigma }_i{]}^{-1}\text{in }{G}_n\mathrm{.}$

Elementary cases. For $n\le 1$ there is no index $i$ with $1\le i\le n-1$, and there are no elementary half twists; the assertions above are vacuous. For $n\ge 2$ and $1\le i\le n-1$ both $[{\sigma }_i]$ and $[{\sigma }_i^{-}]$ are nonidentity elements of $G_n$, since their endpoint permutations are transpositions, which are not the identity permutation.

Two letters do not commute in general. For adjacent indices the supports $U_i$ and $U_{i+1}$ overlap and no commutation is asserted; Far commutativity of elementary geometric half twists proves commutativity only for disjoint supports.