The Burau Representations

Well-definedness and basic properties. By [F1] the assignment $x_i\mapsto 1$ extends to a unique homomorphism $\omega :F_n\to \mathbb{Z}$, so $\omega$ is well defined and independent of every choice made in the definition of the standard meridians. A word $x_{i_1}^{{\epsilon }_1}\cdots x_{i_m}^{{\epsilon }_m}$ has image ${\sum }_{j=1}^m{\epsilon }_j$ by the homomorphism law, the kernel of $\omega$ is therefore exactly the set of words of exponent sum $0$, and $\omega (x_1)=1$ shows that $\omega$ is surjective.

Invariance under the Artin action. The set $H=\{\beta \in B_n:\omega \circ \rho (\beta )=\omega \}$ is a subgroup of $B_n$: $\rho$ is a homomorphism, $\rho (1)=\mathrm{id}$ gives $\omega \circ \rho (1)=\omega$, and if $\omega \circ \rho (\beta )=\omega$ and $\omega \circ \rho ({\beta }^{\mathrm{\prime }})=\omega$ then $\omega \circ \rho (\beta {\beta }^{\mathrm{\prime }})=(\omega \circ \rho (\beta ))\circ \rho ({\beta }^{\mathrm{\prime }})=\omega \circ \rho ({\beta }^{\mathrm{\prime }})=\omega$ and $\omega \circ \rho ({\beta }^{-1})=\omega \circ \rho (\beta {)}^{-1}=\omega$, because $\rho (\beta )$ is an automorphism. By [F2] it suffices to show ${\sigma }_i\in H$ for every $i$. For $\rho ({\sigma }_i)$, the images of the basis elements $x_i\mapsto x_i{x}_{i+1}{x}_i^{-1}$, $x_{i+1}\mapsto x_i$ and $x_j\mapsto x_j$ all have exponent sum $1=\omega (x_j)$, so $\omega \circ \rho ({\sigma }_i)$ and $\omega$ agree on the free basis and hence, by [F1], on all of $F_n$. Therefore $H=B_n$, which is the first assertion of clause (2).

Geometric representative. Assume AC and let $h$ be a homeomorphism representative of the braid mapping class of $\beta$, i.e. a boundary-fixed homeomorphism whose mapping class is the image of $\beta$ (Boundary-fixed mapping class group of a punctured disk). By [F3] the automorphism $h_{\ast }$ induced on ${\pi }_1(X,d)=F_n$ equals $\rho (\beta )$, so $\omega \circ h_{\ast }=\omega \circ \rho (\beta )=\omega$ by step 2.1. AC is used only here, through [F3], and the statement of step 2.1 is choice free.

Nonemptiness and paths to the flower. The point $d=(0,1)$ lies in $X$, because $\mathrm{\mid }{q}_i\mathrm{\mid }<1$ for every $i$, so $X\mathrm{\ne }\mathrm{\varnothing }$. By [F1] the map ${\gamma }_a(t):=H(a,t)$ is a continuous path in $X$ from $a$ to $r(a)\in W$ for every $a\in X$. The flower $W$ is path-connected: a point of $W$ lies in some $t_i\cup C_i$ or equals $d$, and by [F2] it is joined to $d$ by a path inside $t_i\cup C_i\subseteq W$, the case $W=\{d\}$ (that is, $n=0$) being trivial.

Convex open neighbourhoods. Fix $x\in X$. If $\mathrm{\mid }x\mathrm{\mid }<1$, the minimum in the statement is over the nonempty set $\{\mathrm{\mid }x-q_i\mathrm{\mid }\}\cup \{1-\mathrm{\mid }x\mathrm{\mid }\}$ and is positive, since $x\mathrm{\ne }{q}_i$ and $\mathrm{\mid }x\mathrm{\mid }<1$; fix $0<r$ below it and put $U=X\cap B(x,r)$. Then $B(x,r)\subseteq \mathrm{int}{D}^2\subseteq D^2$ because $r<1-\mathrm{\mid }x\mathrm{\mid }$, and $q_i\mathrm{\notin }B(x,r)$ for all $i$ because $r<\mathrm{\mid }x-q_i\mathrm{\mid }$; hence $U=B(x,r)$, an open ball. If $\mathrm{\mid }x\mathrm{\mid }=1$ and $n\ge 1$, the finite minimum ${\min }_i\mathrm{\mid }x-q_i\mathrm{\mid }$ is positive because $x\mathrm{\notin }{Q}_n$; fix $0<r$ below it and put $U=D^2\cap B(x,r)$. Then $q_i\mathrm{\notin }B(x,r)$ for all $i$, so $U=X\cap B(x,r)\subseteq X$. If $n=0$, put $U=X=D^2$. In every case $x\in U\subseteq X$, and $U=X\cap V$ with $V$ open in ${\mathbb{R}}^2$ ($V=B(x,r)$, respectively $V={\mathbb{R}}^2$ for $U=X$), so $U$ is open in $X$; and $U$ is convex, being either an open ball, the intersection $D^2\cap B(x,r)$ of two convex sets, or $D^2$.

$X$ is path-connected. Let $a,b\in X$. Concatenate the path ${\gamma }_a$ from $a$ to $r(a)$, a path in $W$ from $r(a)$ to $d$, the reverse of a path in $W$ from $r(b)$ to $d$, and the reverse of ${\gamma }_b$; by [F3] the result is a path in $X$ from $a$ to $b$. This uses only the finitely many explicit paths of step 1.1 and no choice principle.

Local path-connectedness and semilocal simple connectivity. The set $U$ of step 1.2 is nonempty and convex, hence path-connected by [F4] and simply connected by [F5]; Given any open neighbourhood $O$ of $x$ in $X$, the subspace topology supplies $r_0>0$ with $X\cap B(x,r_0)\subseteq O$. Further restrict the radius in step 1.2 to be below $r_0$; when $n=0$ use $D^2\cap B(x,r_0\mathrm{/}2)$ instead of the whole disk. This remains convex and open, and gives a path-connected neighbourhood contained in $O$. Thus these neighbourhoods form the required basis and $X$ is locally path-connected at $x$. Moreover every loop in $U$ is null-homotopic in $U$ by the explicit straight-line homotopy of [F5], so the map ${\pi }_1(U,x)\to {\pi }_1(X,x)$ induced by the inclusion is trivial; by [F6] the space $X$ is semilocally simply connected at $x$. Since $x\in X$ was arbitrary, (b) and (c) hold. No choice principle was used anywhere; all minima are over finite sets or over a finite set enlarged by one real number.

Conclusion. Step 1.1 gives $X\mathrm{\ne }\mathrm{\varnothing }$, step 2.1 gives (a), and step 2.2 gives (b) and (c); this is the assertion.

Normal form. Write a nonzero element of ${\mathrm{\Lambda }}_1$ as $p\mathrm{/}{t}^N$ with $p\in \mathbb{Z}[t]$, $N\ge 0$ by [F1], and factor out the largest power of $t$ dividing $p$: there are $d\ge 0$ and $q\in \mathbb{Z}[t]$ with $p=t^{d}q$ and $t\nmid q$, the latter meaning $q(0)\mathrm{\ne }0$. Then $p\mathrm{/}{t}^N=t^{d-N}q$, so every nonzero element has a representative $t^{m}q$ with $m\in \mathbb{Z}$ and $q(0)\mathrm{\ne }0$. This representative is unique: if $t^{m}q=t^{m^{\mathrm{\prime }}}{q}^{\mathrm{\prime }}$ with $q(0),q^{\mathrm{\prime }}(0)\mathrm{\ne }0$, multiply by $t^{-m}$ to get $q=t^{m^{\mathrm{\prime }}-m}{q}^{\mathrm{\prime }}$; if $m^{\mathrm{\prime }}\ge m$, the constant term of the right-hand side equals $q^{\mathrm{\prime }}(0)\mathrm{\ne }0$ when $m^{\mathrm{\prime }}=m$ and vanishes when $m^{\mathrm{\prime }}>m$, while the constant term of $q$ is $q(0)\mathrm{\ne }0$, so $m^{\mathrm{\prime }}=m$ and $q=q^{\mathrm{\prime }}$; the case $m^{\mathrm{\prime }}<m$ is symmetric.

${\mathrm{\Lambda }}_1$ is an integral domain. Suppose $uv=0$ with $u=t^{m}p$, $v=t^{n}q$ nonzero in normal form, so $p(0),q(0)\mathrm{\ne }0$ and $pq\mathrm{\ne }0$ by [F3]. Then $uv=t^{m+n}pq$. If $m+n\ge 0$, the element $t^{m+n}pq$ is a polynomial, and [F2] yields $k\ge 0$ with $t^k{t}^{m+n}pq=t^{k+m+n}pq=0$ in $\mathbb{Z}[t]$; since $t^{k+m+n}\mathrm{\ne }0$ and $\mathbb{Z}[t]$ is a domain by [F3], $pq=0$, a contradiction. If $m+n<0$, the element is $pq\mathrm{/}{t}^{-(m+n)}$, and [F2] yields $k\ge 0$ with $t^{k}pq=0$ in $\mathbb{Z}[t]$, again forcing $pq=0$ by [F3], a contradiction. Hence $uv\mathrm{\ne }0$, so ${\mathrm{\Lambda }}_1$ is a domain.

Distinct powers of $t$. The elements $t^m=t^m\cdot 1$ and $1=t^0\cdot 1$ are in normal form, since the constant polynomial $1$ has $1\mathrm{\ne }0$. By uniqueness in step 1.1, $t^m=1$ forces $m=0$ and $q=1$. More generally $t^m=t^{m^{\mathrm{\prime }}}$ forces $t^{m-m^{\mathrm{\prime }}}=1$, hence $m=m^{\mathrm{\prime }}$.

Units. If $u=\pm t^m$, then $u\cdot (\pm t^{-m})=1$, so $u$ is a unit. Conversely let $u=t^{m}p$ in normal form be a unit, with inverse $v=t^{n}q$ in normal form; then $t^{m+n}pq=1$. Applying the normal form uniqueness of step 1.1 to $t^{m+n}pq$ and to $1=t^0\cdot 1$ gives $m+n=0$ and $pq=1$ in $\mathbb{Z}[t]$. By [F4] the unit $p$ of $\mathbb{Z}[t]$ is one of the two constants $\pm 1$; hence $u=\pm t^m$.

The sum $1+t+\cdots +t^{n-1}$. For $n\ge 2$ put $f=1+t+\cdots +t^{n-1}$, a polynomial with constant term $1$ and at least two nonzero coefficients. In normal form $f=t^{0}f$. If $f$ were a unit, step 2.3 would give $f=\pm t^m$ for some $m$; two elements equal in ${\mathrm{\Lambda }}_1$ have the same normal-form exponent and polynomial by step 1.1, so $m=0$ and $f=\pm 1$, contradicting that $f$ has at least two nonzero coefficients. Hence $f$ is not a unit, which is (d); claims (a), (b), (c) are steps 2.1, 2.2 and 2.3. No choice principle is used.

The kernel and its quotient. The kernel $K=\ker \omega$ is a normal subgroup of ${\pi }_1(X,d)$, and by [F5] the map $\omega$ induces an isomorphism ${\pi }_1(X,d)\mathrm{/}K\to \mathrm{im}\omega =\mathbb{Z}$. Hence ${\pi }_1(X,d)\mathrm{/}K$ is infinite and the index of $K$ is infinite: a finite index would make the quotient finite.

Existence and uniqueness of the cover. The four hypotheses of [F1] hold by The punctured disk is path-connected, locally path-connected and semilocally simply connected, so the subgroup $K$ is realised by a based connected covering $p:(\tilde{X},\tilde{d})\to (X,d)$ with $p_{\ast }{\pi }_1(\tilde{X},\tilde{d})=K$; by the bijection [F2] any two such based coverings are uniquely based-isomorphic. This defines the Burau infinite cyclic cover.

Regularity, deck group and the generator $t$. Since $K$ is normal, [F3] makes $p$ regular, and [F4] together with step 1.1 gives $\mathrm{Deck}(\tilde{X}\mathrm{/}X)\cong {\pi }_1(X,d)\mathrm{/}K\cong \mathbb{Z}$; by [F7] the fibre is in bijection with the set of right cosets of $K$, so the fibre is infinite. Fixing the isomorphism as the one induced by $\omega$, the positive generator of $\mathbb{Z}$ corresponds to a deck transformation $t$; by [F6] the monodromy of $t$ raises the total winding of loops by $1$. The group generated by $t$ is all of $\mathrm{Deck}(\tilde{X}\mathrm{/}X)$, so $\mathrm{Deck}(\tilde{X}\mathrm{/}X)=\{t^k:k\in \mathbb{Z}\}$ with $t^{-1}$ the negative generator, and the left-action convention on the total space is the one recorded in Deck transformations and the deck-transformation group of a covering.

Each deck transformation induces an automorphism. By [F3] each $T_{t^k}$ is a homeomorphism, so by [F2] it induces $(T_{t^k}{)}_{\ast }:=H_1((T_{t^k}{)}_{\mathrm{\#}})$, an endomorphism of $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$. Since $T_{t^k}\circ T_{t^{-k}}=\mathrm{id}=T_{t^{-k}}\circ T_{t^k}$ by [F3], functoriality in [F2] gives $(T_{t^k}{)}_{\ast }\circ (T_{t^{-k}}{)}_{\ast }=\mathrm{id}$ and $(T_{t^{-k}}{)}_{\ast }\circ (T_{t^k}{)}_{\ast }=\mathrm{id}$; hence each $(T_{t^k}{)}_{\ast }$ is an automorphism of $M_{\mathrm{r}\mathrm{e}\mathrm{d}}$, and $t_{\ast }:=(T_t{)}_{\ast }$ is a unit of the ring ${\mathrm{End}}_{\mathbb{Z}}(M_{\mathrm{r}\mathrm{e}\mathrm{d}})$ with inverse $(T_{t^{-1}}{)}_{\ast }$.