The Burau Representations — Examples

The unreduced matrices. For $n=3$ the block of [A1] sits in rows and columns $1,2$ for $B_1$ and in rows and columns $2,3$ for $B_2$, with all other entries those of the identity: $B_1=\left(\begin{array}{ccc}1-t& t& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right)$ and $B_2=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1-t& t\\ 0& 1& 0\end{array}\right)$, as displayed.

The braid relation for the unreduced matrices. Direct matrix multiplication gives $B_1{B}_2=\left(\begin{array}{ccc}1-t& t-t^2& t^2\\ 1& 0& 0\\ 0& 1& 0\end{array}\right)$ and $B_2{B}_1=\left(\begin{array}{ccc}1-t& t& 0\\ 1-t& 0& t\\ 1& 0& 0\end{array}\right)$; multiplying on the right by $B_1$ respectively $B_2$ gives in both cases the matrix $\left(\begin{array}{ccc}1-t& t-t^2& t^2\\ 1-t& t& 0\\ 1& 0& 0\end{array}\right)$, so $B_1{B}_2{B}_1=B_2{B}_1{B}_2$.

The kernel basis and the restricted action. $\sigma (g_1)=t-t=0$ and $\sigma (g_2)=t\cdot t-t^2=0$, so $g_1,g_2\in \ker \sigma$; they are independent, because $a{g}_1+b{g}_2=0$ reads $(at,-a+bt,-b)=0$ with $t\mathrm{\ne }0$, so $b=0$, then $a=0$ by [A5]. They also span: if $x=x_1{e}_1+x_2{e}_2+x_3{e}_3$ satisfies $x_1+t{x}_2+t^2{x}_3=0$, set $b=-x_3$ and $a=-x_2-t{x}_3$. Then $a{g}_1+b{g}_2$ has coordinates $(at,-a+bt,-b)=(-t{x}_2-t^2{x}_3,x_2,x_3)=(x_1,x_2,x_3)$. Thus independence and this explicit spanning prove that $(g_1,g_2)$ is a ${\mathrm{\Lambda }}_1$-basis of $\ker \sigma$. Since $\ker \sigma$ is $B_1$- and $B_2$-invariant by [A3], the matrices act on this basis: using the actions of [A1], $B_1{g}_1=t((1-t)e_1+e_2)-t{e}_1=-t^2{e}_1+t{e}_2=-t{g}_1$ and $B_1{g}_2=t(t{e}_1)-e_3=t^2{e}_1-e_3=t{g}_1+g_2$; likewise $B_2{g}_1=t{e}_1-((1-t)e_2+e_3)=t{e}_1+(t-1)e_2-e_3=g_1+g_2$ and $B_2{g}_2=t((1-t)e_2+e_3)-t{e}_2=-t^2{e}_2+t{e}_3=-t{g}_2$. Reading the two images as columns gives ${\bar{\rho }}_3({\sigma }_1)=\left(\begin{array}{cc}-t& t\\ 0& 1\end{array}\right)$ and ${\bar{\rho }}_3({\sigma }_2)=\left(\begin{array}{cc}1& 0\\ 1& -t\end{array}\right)$, the displayed reduced matrices.

The braid relation for the reduced matrices. Direct multiplication gives $M_1{M}_2=\left(\begin{array}{cc}0& -t^2\\ 1& -t\end{array}\right)$ and $M_2{M}_1=\left(\begin{array}{cc}-t& t\\ -t& 0\end{array}\right)$ for $M_1=\left(\begin{array}{cc}-t& t\\ 0& 1\end{array}\right)$, $M_2=\left(\begin{array}{cc}1& 0\\ 1& -t\end{array}\right)$; multiplying by $M_1$ respectively $M_2$ gives $M_1{M}_2{M}_1=M_2{M}_1{M}_2=\left(\begin{array}{cc}0& -t^2\\ -t& 0\end{array}\right)$, so the reduced matrices satisfy the braid relation.

Reduction of the unreduced matrices. Step 1.3 exhibited the restricted actions of $B_1,B_2$ on the invariant kernel $\ker \sigma$ in the basis $(g_1,g_2)$ as exactly $M_1,M_2$, and step 2.1 verified the braid relation at both levels; hence the displayed reduced matrices are the reduction of the displayed unreduced matrices to the invariant-covector kernel, and they agree with clause (2) of [A2]. AC is inherited through the cited identification of the reduced matrices with the topological representation; the matrix computations are choice free.

The full twist and its powers. By [A1] and ${\mathrm{\Delta }}^2=({\sigma }_1{\sigma }_2{\sigma }_1{)}^2$, the full twist satisfies ${\bar{\rho }}_3({\mathrm{\Delta }}^2)=t^3{I}_2$; hence for every $k\in \mathbb{Z}$ the homomorphism property gives ${\bar{\rho }}_3({\mathrm{\Delta }}^{2k})=(t^3{I}_2{)}^k=t^{3k}{I}_2$.

The image of the center. Since $Z(B_3)=⟨{\mathrm{\Delta }}^2⟩$ by [A2], the image of the center is $⟨{\bar{\rho }}_3({\mathrm{\Delta }}^2)⟩=⟨t^3{I}_2⟩$. The matrix $t^3{I}_2$ has infinite order by [A3], so $⟨t^3{I}_2⟩$ is infinite cyclic. Moreover ${\bar{\rho }}_3({\mathrm{\Delta }}^{2k})=t^{3k}{I}_2=I_2$ holds if and only if $t^{3k}=1$, which by [A3] happens if and only if $3k=0$, that is, if and only if $k=0$; hence ${\bar{\rho }}_3$ is injective on $Z(B_3)$.

The specialization at $t=-1$. Apply the homomorphism $\epsilon$ of [A4] entrywise to ${\bar{\rho }}_3({\mathrm{\Delta }}^2)=t^3{I}_2$: the result is ${\bar{\rho }}_3(-1)({\mathrm{\Delta }}^2)=(-1{)}^3{I}_2=-I_2\mathrm{\ne }{I}_2$; so ${\mathrm{\Delta }}^2\mathrm{\notin }\ker {\bar{\rho }}_3(-1)$, even though its square ${\mathrm{\Delta }}^4$ has image $(-I_2{)}^2=I_2$. Over ${\mathrm{\Lambda }}_1$ the same conclusion is step 2.1: ${\bar{\rho }}_3({\mathrm{\Delta }}^{2k})\mathrm{\ne }{I}_2$ for every $k\mathrm{\ne }0$, so neither ${\mathrm{\Delta }}^2$ nor any nonzero power ${\mathrm{\Delta }}^{2k}$ lies in the kernel of ${\bar{\rho }}_3$. AC is inherited through the cited agreement theorem; the scalar and matrix computations are choice free.

Specialization of the generators. Applying the augmentation $\epsilon$ entrywise to ${\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}$ gives the homomorphism $\mathrm{\Theta }:={\epsilon }_{\ast }\circ {\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}:B_n\to {\mathrm{GL}}_n(\mathbb{Z})$, since $\epsilon$ is a unital ring homomorphism [A2]. On the generator, $\epsilon (1-t)=0$ and $\epsilon (t)=1$, so the block of $B_i$ becomes $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and the identity entries stay $1$; hence $\mathrm{\Theta }({\sigma }_i)=E_i$, the permutation matrix of the transposition $(ii+1)$, namely the matrix swapping the $i$-th and $(i+1)$-st coordinates.

Factorization through ${\pi }_n$. The matrices $E_i$ satisfy $E_i^2=I$, $E_i{E}_{i+1}{E}_i=E_{i+1}{E}_i{E}_{i+1}$ and $E_i{E}_j=E_j{E}_i$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$, because they are the matrices of the corresponding permutations of the coordinate basis. By the Coxeter presentation and von Dyck [A3] there is a homomorphism $\mathrm{\Phi }:S_n\to {\mathrm{GL}}_n(\mathbb{Z})$ with $\mathrm{\Phi }(s_i)=E_i$, the natural permutation representation on ${\mathbb{Z}}^n$; then $\mathrm{\Phi }\circ {\pi }_n$ and $\mathrm{\Theta }$ are homomorphisms $B_n\to {\mathrm{GL}}_n(\mathbb{Z})$ agreeing on the generators, hence equal. So the specialization factors through the surjection ${\pi }_n$ and is exactly the permutation representation.

Invariant line and the specialized sequence. Every permutation matrix fixes $v$, so $\mathbb{Z}v$ is a trivial submodule. Put $N=\ker \sigma$. Since $\sigma (e_1)=1$, every $x\in {\mathrm{\Lambda }}_1^n$ has the unique decomposition $x=(x-\sigma (x)e_1)+\sigma (x)e_1$, giving ${\mathrm{\Lambda }}_1^n=N\oplus {\mathrm{\Lambda }}_1{e}_1$. Consequently $N\cap (t-1){\mathrm{\Lambda }}_1^n=(t-1)N$, and $N\mathrm{/}(t-1)N$ injects into ${\mathbb{Z}}^n$. Its image is the sum-zero lattice: one inclusion follows by evaluating $\sigma (x)=0$ at $t=1$; conversely, if $\bar{x}\in {\mathbb{Z}}^n$ has sum zero, take its constant-coordinate lift $x$ and replace it by $x-\sigma (x)e_1$, which is in $N$ and still reduces to $\bar{x}$. Multiplication $a\mapsto (t-1)a$ is an isomorphism ${\mathrm{\Lambda }}_1\to (t-1){\mathrm{\Lambda }}_1$, because ${\mathrm{\Lambda }}_1$ is a domain and $t-1\mathrm{\ne }0$ by Units, powers and the domain property of the Laurent polynomial ring. Thus the specialized target is $(t-1){\mathrm{\Lambda }}_1\mathrm{/}(t-1{)}^2{\mathrm{\Lambda }}_1\cong \mathbb{Z}$, where the class of $(t-1)a$ maps to $\epsilon (a)$; the map $(t-1)\sigma$ becomes the sum functional. This proves the asserted specialized exact sequence, without assuming that an arbitrary specialization preserves injectivity.

Rational splitting and integral failure. Over $\mathbb{Q}$ every $x\in {\mathbb{Q}}^n$ is $({\sum }_i{x}_i\mathrm{/}n)v+(x-({\sum }_i{x}_i\mathrm{/}n)v)$ with the second summand of sum zero, and $\mathbb{Q}v\cap \{x:{\sum }_i{x}_i=0\}=0$ because $nav=0$ forces $a=0$; both summands are preserved by the permutation action, and $\mathbb{Q}v$ is trivial, so the second summand is the reduced permutation representation. Over $\mathbb{Z}$, an element of $\mathbb{Z}v+\{x:{\sum }_i{x}_i=0\}$ has coordinate sum $na$ for some $a\in \mathbb{Z}$, so the sum is contained in $\{x:{\sum }_i{x}_i\equiv 0<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>n\}$, and conversely $x$ with $n\mid {\sum }_i{x}_i$ is $av+y$ with $a=({\sum }_i{x}_i)\mathrm{/}n$ and $y$ of sum zero; the containment is proper because $(1,0,\dots ,0)$ has sum $1$ and $n\ge 2$. Hence the rational splitting is not an integral direct sum.

The case $n=2$. For $n=2$ the sum-zero lattice is $\mathbb{Z}(1,-1)$, on which the transposition $(12)$ acts by $x\mapsto -x$, the sign representation; this is the reduced summand of step 4.1. No choice principle is used.

The invariant line. By [A1], ${\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )v=v$ for every $\beta \in B_n$; hence ${\mathrm{\Lambda }}_{1}v=\{av:a\in {\mathrm{\Lambda }}_1\}$ is a $B_n$-invariant line in the finite free module $W$.

No invariant complement. Suppose, for contradiction, that $C\subseteq W$ is a ${\mathrm{\Lambda }}_1$-submodule with $W={\mathrm{\Lambda }}_{1}v\oplus C$ and ${\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )(C)\subseteq C$ for every $\beta \in B_n$. Let $\pi :W\to {\mathrm{\Lambda }}_{1}v$ be the projection along $C$, so $\pi$ is ${\mathrm{\Lambda }}_1$-linear, $\pi (v)=v$, and $\pi$ is $B_n$-equivariant: writing $x=av+c$ with $a\in {\mathrm{\Lambda }}_1$, $c\in C$, invariance of $C$ and $v$ give ${\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )x=av+{\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )c$ with ${\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )c\in C$, so $\pi ({\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )x)=av=\pi (x)$. Writing $\pi (x)=\lambda (x)v$ defines a ${\mathrm{\Lambda }}_1$-linear functional $\lambda :W\to {\mathrm{\Lambda }}_1$ with $\lambda (v)=1$ and $\lambda ({\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )x)=\lambda (x)$ for all $\beta \in B_n$ and $x\in W$.

The contradiction. Since $\lambda ({\rho }_n^{\mathrm{m}\mathrm{a}\mathrm{t}}(\beta )x)=\lambda (x)$ for every $\beta$, evaluating on $\beta ={\sigma }_i$ gives $\lambda B_i=\lambda$ for every $i$; by the classification in [A1] there is $c\in {\mathrm{\Lambda }}_1$ with $\lambda =c\sigma$ as row vectors. Then $1=\lambda (v)=c\sigma (v)=c(1+t+\cdots +t^{n-1})$, so $1+t+\cdots +t^{n-1}$ has the multiplicative inverse $c$ in ${\mathrm{\Lambda }}_1$. This contradicts [A2] for $n\ge 2$. Hence the invariant line ${\mathrm{\Lambda }}_{1}v$ has no $B_n$-invariant complement, and the refuted statement fails already for $n=2$. This is the integral obstruction behind the caveat of The reduced and unreduced Burau representations have the same kernel that its splitting $K^n=\ker \sigma \oplus Kv$ is only a field statement. AC is inherited from the cited same-kernel proposition; the module and matrix computations are choice free.