Lawrence–Krammer–Bigelow Representations and Linearity — Examples

The tine crossings are $z_1=(\frac{1}{5},-\frac{3}{20})$ and $z_2=(\frac{1}{5},\frac{1}{10})$; their parallel crossings are $z_1^{\mathrm{\prime }}=(\frac{1}{5},-\frac{13}{100})$ and $z_2^{\mathrm{\prime }}=(\frac{1}{5},\frac{2}{25})$. Their order along $N$ is $z_1,z_1^{\mathrm{\prime }},z_2^{\mathrm{\prime }},z_2$. The handle–tine–noodle loops ${\xi }_1,{\xi }_2$ go clockwise around respectively $p_2,p_3$ and only $p_2$, so $(a_1,a_2)=(-2,-1)$; the noodle–lower-boundary loop has $A_0=-1$. The explicit labelled-loop formula of The lexicographic order on fork-noodle deck monomials gives $(a_{i,j})=\left(\begin{array}{cc}-5& -4\\ -4& -3\end{array}\right)\mathrm{.}$

Parametrize each segment of each track by equal time within its stage of ${\delta }_{i,j}$, and merge their rational breakpoints. The difference path $D$ is nonzero and piecewise affine. Counting crossings of the ray in direction $1+\mathrm{i}\mathrm{/}10$ is exact rational arithmetic: on a segment from $(x,y)$ to $(X,Y)$ solve $y+u(Y-y)=(x+u(X-x))\mathrm{/}10$ and retain the solution only when $0<u<1$ and the real coordinate in that ray direction is positive. The sign is that of $(Y-X\mathrm{/}10)-(y-x\mathrm{/}10)$. For each returning pair the closed difference path has total count $-1$; for each exchanged pair $D$ followed by $-D$ has count $-1$. Thus $(b_{i,j})=\left(\begin{array}{cc}-2& -2\\ -1& -1\end{array}\right),({ϵ}_{i,j})=\left(\begin{array}{cc}-1& 1\\ -1& 1\end{array}\right),$ the sign matrix following from ${ϵ}_{i,j}=-(-1{)}^{b_{i,i}+b_{j,j}+b_{i,j}}$. The four projected intersections are distinct and transverse, and their signed count is $-1+1-1+1=0$.

The four monomial-weighted terms instead give $⟨N,F⟩=-q^{-5}{t}^{-2}+q^{-4}{t}^{-2}-q^{-4}{t}^{-1}+q^{-3}{t}^{-1}=q^{-5}{t}^{-2}(q-1)(1+qt)\mathrm{\ne }0.$ The four exponent vectors are distinct, so this Laurent polynomial is nonzero without any specialization. For comparison, a fork tine placed to the left of $x=\frac{1}{5}$ misses $N$ and has both ordinary count and pairing zero. The two configurations have the same ordinary count, zero, and different LKB values. Hence the proposed determination by ordinary intersection number alone fails.

All intersections lie on $x=\frac{1}{5}$. In increasing order along $N$ the tine points have heights $(-\frac{3}{20},\frac{1}{25},\frac{7}{100},\frac{1}{10})$ and the parallel points have heights $(-\frac{29}{200},\frac{7}{200},\frac{3}{40},\frac{19}{200})$. Thus the combined order is $z_1,z_1^{\mathrm{\prime }},z_2^{\mathrm{\prime }},z_2,z_3,z_3^{\mathrm{\prime }},z_4^{\mathrm{\prime }},z_4$. The loops ${\xi }_i$ that follow the handle and tine to $z_i$ and return to $d_1$ along $N$ have total puncture windings $(-2,-1,-1,-1)$: the first clockwise loop encloses $p_2,p_3$, while each of the other loops encloses only $p_2$. The hairpin changes none of these windings. The clockwise loop $N$ followed by the lower boundary return has $A_0=-1$. The labelled-loop formula $a_{i,j}=a_i+a_j+A_0$ therefore gives $(a_{i,j})=\left(\begin{array}{cccc}-5& -4& -4& -4\\ -4& -3& -3& -3\\ -4& -3& -3& -3\\ -4& -3& -3& -3\end{array}\right)\mathrm{.}$

Here is an exact ray-crossing calculation of the mutual exponents, rather than an inference from their parities. Let $D_{i,j}$ be the difference of the two labelled tracks along ${\delta }_{i,j}$. Merge the rational segment-time breakpoints of the two tracks; the resulting difference is piecewise affine with rational vertices and never zero. Count signed crossings of the ray in direction $1+\mathrm{i}\mathrm{/}10$: for consecutive difference vertices $(x,y),(X,Y)$ a crossing occurs when $g=y-x\mathrm{/}10$ and $G=Y-X\mathrm{/}10$ have opposite signs and, at $u=-g\mathrm{/}(G-g)$, $x+u(X-x)+(y+u(Y-y))\mathrm{/}10>0$. Its sign is positive when $g<0<G$, negative in the reverse case. No difference vertex lies on this ray. If the labels return, this closed difference path has signed count $-1$, so $b=2(-1)=-2$. If they exchange, concatenate the difference path with its negative; the resulting closed path has signed count $-1$, so $b=-1$. Substitution of the listed vertices gives these counts for every pair. The returning case is exactly $z_i$ preceding $z_j^{\mathrm{\prime }}$ along $N$, yielding $(b_{i,j})=\left(\begin{array}{cccc}-2& -2& -2& -2\\ -1& -1& -2& -2\\ -1& -1& -2& -2\\ -1& -1& -1& -1\end{array}\right)\mathrm{.}$ Together with step 1.1 this specifies every monomial $m_{i,j}=q^{a_{i,j}}{t}^{b_{i,j}}$.

The diagonal exponents are $(-2,-1,-2,-1)$. Applying ${ϵ}_{i,j}=-(-1{)}^{b_{i,i}+b_{j,j}+b_{i,j}}$ to step 2.1 gives $({ϵ}_{i,j})=\left(\begin{array}{cccc}-1& 1& -1& 1\\ -1& 1& 1& -1\\ 1& -1& -1& 1\\ -1& 1& -1& 1\end{array}\right)\mathrm{.}$ The two exponent matrices and this sign matrix list all sixteen labelled contributions; no pair is omitted.

Rows two and three of the signed monomial table cancel entry by entry. In row one, columns two and three cancel; in row four, columns three and four cancel. The four remaining terms give $⟨N,F⟩=-q^{-5}{t}^{-2}+q^{-4}{t}^{-2}-q^{-4}{t}^{-1}+q^{-3}{t}^{-1}=q^{-5}{t}^{-2}(q-1)(1+qt)\mathrm{\ne }0.$ Thus geometrically distinct terms really do cancel, although the collected polynomial is nonzero. The sum of the sixteen signs is zero, so the ordinary algebraic intersection number of the projected surfaces vanishes. The deck-labelled polynomial retains information lost by this unweighted count.

The columns for ${\sigma }_1$. Every case with $i<k$ is vacuous for $k=1$. The case ${\sigma }_1{x}_{1,2}=t{q}^2{x}_{1,2}$ gives the first column $(t{q}^2,0,0)$; the case $k+1<j$ applied to $(k,j)=(1,3)$ gives ${\sigma }_1{x}_{1,3}=tq(q-1)x_{1,2}+q{x}_{2,3}$; and the case for $x_{k+1,j}$ with $(k+1,j)=(2,3)$ gives ${\sigma }_1{x}_{2,3}=x_{1,3}+(1-q)x_{2,3}$. Hence $M_1=\left(\begin{array}{ccc}t{q}^2& tq(q-1)& 0\\ 0& 0& 1\\ 0& q& 1-q\end{array}\right)\mathrm{.}$

The columns for ${\sigma }_2$. For $k=2$ the case $i<k$ with $i=1$ gives ${\sigma }_2{x}_{1,2}=(1-q)x_{1,2}+q{x}_{1,3}$; the case $x_{i,k+1}$ with $i=1$ gives ${\sigma }_2{x}_{1,3}=x_{1,2}+t{q}^2(q-1)x_{2,3}$; and the case $x_{k,k+1}$ gives ${\sigma }_2{x}_{2,3}=t{q}^2{x}_{2,3}$. Hence $M_2=\left(\begin{array}{ccc}1-q& 1& 0\\ q& 0& 0\\ 0& t{q}^2(q-1)& t{q}^2\end{array}\right)\mathrm{.}$

The braid relation by direct multiplication. Multiplying the two matrices gives $M_1{M}_2=\left(\begin{array}{ccc}0& t{q}^2& 0\\ 0& t{q}^2(q-1)& t{q}^2\\ q^2& -q^{2}t(q-1{)}^2& -q^{2}t(q-1)\end{array}\right)\mathrm{.}$ Multiplying this product on the right by $M_1$ and on the left by $M_2$ gives, by expansion of the nine entries of each of the two products, $M_1{M}_2{M}_1=M_2{M}_1{M}_2=\left(\begin{array}{ccc}0& 0& t{q}^2\\ 0& t{q}^3& 0\\ t{q}^4& 0& 0\end{array}\right)=:\mathrm{\Delta }\mathrm{.}$ Each entry of the two triple products is a sum of at most three Laurent monomials; collecting the terms in each of the nine positions gives the displayed common value, so the braid relation $\rho ({\sigma }_1)\rho ({\sigma }_2)\rho ({\sigma }_1)=\rho ({\sigma }_2)\rho ({\sigma }_1)\rho ({\sigma }_2)$ holds.

The full twist. The matrix $\mathrm{\Delta }$ has a single nonzero entry in each row and column, so squaring it multiplies the diagonal entries $t{q}^2\cdot t{q}^4=t^2{q}^6$, $(t{q}^3{)}^2=t^2{q}^6$, $t{q}^4\cdot t{q}^2=t^2{q}^6$ and kills all off-diagonal entries: $(\rho ({\sigma }_1)\rho ({\sigma }_2)\rho ({\sigma }_1){)}^2={\mathrm{\Delta }}^2=q^6{t}^2{I}_3,$ which is the value $q^{2n}{t}^2$ of the full-twist scalar at $n=3$. The columns of $\mathrm{\Delta }$ are exactly the values $\mathrm{\Delta }{x}_{2,3}=t{q}^2{x}_{1,2}$, $\mathrm{\Delta }{x}_{1,3}=t{q}^3{x}_{1,3}$, $\mathrm{\Delta }{x}_{1,2}=t{q}^4{x}_{2,3}$ predicted by the half-twist identity of the source Section 3, $\mathrm{\Delta }{x}_{n+1-j,n+1-i}=t{q}^{i+j-1}{x}_{ij}$.

Invertibility and conventions. Since $\det M_1=\det M_2=-q^{3}t$ is a unit of $\mathrm{\Lambda }$, both matrices lie in ${\mathrm{G}\mathrm{L}}_3(\mathrm{\Lambda })$. The matrices above are those of Krammer's fraction-field model with basis $x_{12},x_{13},x_{23}$, not matrices in Bigelow's integral closed-surface basis; the two models are isomorphic as $B_n$-representations only after fraction-field extension for $n\ge 3$, and the translation between the parameters is $t_{\mathrm{K}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{m}\mathrm{e}\mathrm{r}}=-t_{\mathrm{B}\mathrm{i}\mathrm{g}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{w}}$, so the displayed formulas record Krammer's normalization and not Bigelow's.

The endpoint permutation. Since the transpositions $s_i=(ii+1)\in S_n$ satisfy $s_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}$ and $s_i{s}_j=s_j{s}_i$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$, the assignment $\pi :B_n\to S_n$, $\pi ({\sigma }_i)=s_i$, respects the Artin presentation and is a homomorphism. Composing it with the standard permutation representation $S_n\to {\mathrm{G}\mathrm{L}}_n(F)$ gives a linear representation ${\pi }_F:B_n⟶{\mathrm{G}\mathrm{L}}_n(F)$ of degree $n$ over any field.

The kernel is nontrivial. The exponent-sum map $\epsilon :B_n\to \mathbb{Z}$, $\epsilon ({\sigma }_i)=1$, is well defined because every Artin relator has equal total exponent on both sides; hence $\epsilon ({\sigma }_1^2)=2\mathrm{\ne }0$ and ${\sigma }_1^2\mathrm{\ne }1$ in $B_n$. The element ${\sigma }_1^2$ is pure: its image under $\pi$ is $s_1^2=1$. Therefore ${\sigma }_1^2$ is a nontrivial element of $\ker \pi$, and a fortiori of $\ker {\pi }_F$; for $n\ge 2$ the kernel of the permutation representation is the nontrivial pure braid group.

Linear does not mean faithful. Thus for every $n\ge 2$ the group $B_n$ admits the linear representation ${\pi }_F$ of degree $n$ with $\ker {\pi }_F\mathrm{\ne }1$. Linearity of $B_n$ is therefore not witnessed by an arbitrary representation: the content of Every classical braid group is linear lies in the faithfulness of ${\rho }_{\mathrm{L}\mathrm{K}\mathrm{B}}$, not merely in the existence of a representation, and the claim stated above is refuted.