Hecke Markov Traces and Polynomial Link Invariants — Examples

The one-dimensional representation. By [A2] the image of ${\sigma }_1^m$ is the $1\times 1$ matrix $(-t{)}^m$, so $\det (I_1-{\bar{\rho }}_2({\sigma }_1^m))=1-(-t{)}^m$ is the scalar displayed.

The formula. Substituting step 1.1 into [A1] with $n=2$ gives ${\mathrm{\Delta }}_{\widehat{{\sigma }_1^m}}(t)\doteq (1-t)(1-(-t{)}^m)\mathrm{/}(1-t^2)$, the displayed formula.

The odd cases. For $m=1$: $(1-t)(1+t)\mathrm{/}(1-t^2)=1$, the unknot value. For $m=3$: $(1-t)(1+t^3)\mathrm{/}(1-t^2)=(1+t^3)\mathrm{/}(1+t)=t^2-t+1$, the trefoil value. For $m=-3$: $(1-t)(1+t^{-3})\mathrm{/}(1-t^2)=(1+t^{-3})\mathrm{/}(1+t)=t^{-3}(t^2-t+1)$, which differs from $t^{-2}-t^{-1}+1=t^{-2}(t^2-t+1)$ by the unit $t^{-1}$, the mirror trefoil value; by [A3] these three closures are knots ($m$ odd).

The even cases. For even $m$ the permutation of ${\sigma }_1^m$ is the identity, so by [A3] the closure has two components; the same display gives $\mathrm{\Delta }\doteq (1-t)(1-t^m)\mathrm{/}(1-t^2)$, and for $m=2$ this is $(1-t)(1-t^2)\mathrm{/}(1-t^2)=1-t$, the one-variable Alexander polynomial of the Hopf link in the convention of [A1]. The formula with the factor $(1-t)$ is the classical formula for knots and links, so no separate knot hypothesis is needed for the value; the distinction is only that for even $m$ the closure is not a knot.

The two-component value. The Hopf link is the closure of ${\sigma }_1^2$, whose two components correspond to the two cycles of the identity permutation of $B_2$ by [A3]; its value $1-t$ is the unit multiple $\pm t^k$ representative of the Alexander polynomial of the Hopf link in the normalization of [A1]. This completes the verification of the displayed values.

The trace of the three-crossing braid. By [A4], ${\pi }_3({\sigma }_1{\sigma }_2{\sigma }_1)=T_1{T}_2{T}_1$; by the recursion of [A1] with $n=2$ and $x=y=T_1$, ${\mathrm{tr}}_3(T_1{T}_2{T}_1)=z{\mathrm{tr}}_2(T_1^2)$. By [A2], $T_1^2=(v-1)T_1+v$, so by linearity and [A1], ${\mathrm{tr}}_2(T_1^2)=(v-1)z+v$. Hence $A={\mathrm{tr}}_3(T_1{T}_2{T}_1)=z^2(v-1)+zv$, and $P(\widehat{\beta })=u^{e(\beta )}{\alpha }^{2}A=u^3{\alpha }^2(z^2(v-1)+zv)$ since $e(\beta )=3$ by [A4].

The closure is the Hopf link. The braid identity ${\sigma }_1^{-1}({\sigma }_1^2{\sigma }_2){\sigma }_1={\sigma }_1{\sigma }_2{\sigma }_1$ holds in $B_3$ by cancellation; since ${\sigma }_1^2\in B_2$ embeds as ${\sigma }_1^2\in B_3$, the element ${\sigma }_1^2{\sigma }_2$ is a positive stabilization of ${\sigma }_1^2$, and [A5] shows that the closures of ${\sigma }_1{\sigma }_2{\sigma }_1$ and ${\sigma }_1^2$ are ambient-isotopic. The endpoint permutation of ${\sigma }_1^2$ is $(12{)}^2=1$, so [A5] gives two components. Closing its two positive crossings yields the usual two-crossing positive Hopf diagram, namely the $(2,2)$-torus link. Hence $\widehat{\beta }$ is that Hopf link. In particular $\widehat{\beta }$ is not a knot, and the knot normalization is not used.

The other two trace values. The same recursion gives ${\mathrm{tr}}_3(T_1^2)={\mathrm{tr}}_2(T_1^2)=C=(v-1)z+v$, and with $T_2^{-1}=v^{-1}{T}_2+(v^{-1}-1)$ from [A1], $B={\mathrm{tr}}_3(T_1{T}_2^{-1}{T}_1)=v^{-1}A+(v^{-1}-1)C=(1-v^{-1})z^2+(3-v-v^{-1})z+(1-v)$.

The Jones specialization. By [A5], with $z_0=-1\mathrm{/}(v+1)$ and $u=s$, ${\mathrm{tr}}_3({\pi }_3(\beta )){\mathrm{\mid }}_{z_0}=-(v^2+1)\mathrm{/}(v+1{)}^2$ and $V(\widehat{\beta })=s^3(-\frac{s^2+1}{s}{)}^2(-\frac{v^2+1}{(v+1{)}^2})=-s(s^4+1)=-s^5-s=-t^{1\mathrm{/}2}(t^2+1)$ with $t=s^2$.

The skein identity. Applying $P$ to the three words $x{\sigma }_{2}y={\sigma }_1{\sigma }_2{\sigma }_1$, $x{\sigma }_2^{-1}y={\sigma }_1{\sigma }_2^{-1}{\sigma }_1$ and $xy={\sigma }_1^2$ with $x=y={\sigma }_1$ and using [A3] and [A4], $P(L_{+})=u^3{\alpha }^{2}A$, $P(L_{-})=u{\alpha }^{2}B$ and $P(L_0)=u^2{\alpha }^{2}C$. The skein relation $l^{-1}{P}_{+}-l{P}_{-}=m{P}_0$ of [A3] reduces, after cancelling the common factor $u^2{\alpha }^2$ and multiplying by $s$ with $s^2=v$, to $A-vB=(v-1)C$. Expanding, $A-vB=(v-1)z^2+vz-v[(1-v^{-1})z^2+(3-v-v^{-1})z+(1-v)]=(v-1{)}^{2}z+v(v-1)=(v-1)C$, so the relation holds identically in $\mathrm{\Lambda }$.

Agreement with the two-strand representative. For ${\sigma }_1^2\in B_2$ one computes ${\mathrm{tr}}_2(T_1^2)=C=(v-1)z+v$, so by [A5] $V(\widehat{{\sigma }_1^2})=s^2(-\frac{s^2+1}{s})\frac{v^2+1}{v+1}=-s(s^4+1)$, the same value as step 2.2, as the invariance of [A3] requires for two representatives of the same link. This completes the computation and the cross-check.

The stabilization factors of the raw trace. By [F3] and [F1] ${\mathrm{tr}}_{n+1}({\pi }_{n+1}(\beta {\sigma }_n))={\mathrm{tr}}_{n+1}({\pi }_n(\beta )T_n)=z{\mathrm{tr}}_n({\pi }_n(\beta ))$, and similarly ${\mathrm{tr}}_{n+1}({\pi }_{n+1}(\beta {\sigma }_n^{-1}))=z_{-}{\mathrm{tr}}_n({\pi }_n(\beta ))$. By [F2] the two factors $z$ and $z_{-}$ are distinct, so whenever ${\mathrm{tr}}_n({\pi }_n(\beta ))\mathrm{\ne }0$ the raw trace family takes different values on the two stabilizations and is therefore not Markov invariant.

The normalized family. For the normalization $F(\beta ):=u^{e(\beta )}{\alpha }^{n-1}{\mathrm{tr}}_n({\pi }_n(\beta ))$, [F3] gives $F(\beta {\sigma }_n)=u\alpha zF(\beta )$ and $F(\beta {\sigma }_n^{-1})=u^{-1}\alpha z_{-}F(\beta )$. Both stabilization factors equal $1$ precisely when $u\alpha z=1$ and $u^{-1}\alpha z_{-}=1$; given the first relation $\alpha =(uz{)}^{-1}$, and substituting this into the second gives $z_{-}\mathrm{/}(u^{2}z)=1$, i.e. $u^2=z_{-}\mathrm{/}z$, since $u$ and $z$ are units. For the trivial braid $1\in B_1$ one has $F(1)=1$, so $F({\sigma }_1)$ and $F({\sigma }_1^{-1})$ are the two factors themselves. If either factor differs from $1$, that stabilization changes the value of this witness. Hence a normalization failing either relation is not invariant under both stabilizations; the two stabilized values need not differ from each other.

The explicit witness. Take $u=1$ and $\alpha =z^{-1}$ in a ring containing $\mathrm{\Lambda }$ with $z$ inverted. Then the positive factor is $u\alpha z=1$, while the negative factor is $u^{-1}\alpha z_{-}=z_{-}\mathrm{/}z=\frac{z+1-v}{vz}$, which is not $1$: the equality $\frac{z+1-v}{vz}=1$ would give $z+1-v=vz$, i.e. $(1-v)(z+1)=0$, and both factors $1-v$ and $z+1$ are nonzero in the domain $\mathrm{\Lambda }$. Concretely, for $\beta =1\in B_1$ one has ${\mathrm{tr}}_1(1)=1$ and hence $F({\sigma }_1)=1$ while $F({\sigma }_1^{-1})=z_{-}\mathrm{/}z\mathrm{\ne }1$ in $B_2$; the two closures are the same unknot, so $F$ is not an invariant of the closure, which refutes the displayed statement. The corrected normalization is the one of [F4], whose two factors are both $1$.

The trace of ${\sigma }_1^3$. By [A1] and $T_1^2=(v-1)T_1+v$, $T_1^3=T_1{T}_1^2=(v-1)T_1^2+v{T}_1=(v-1)((v-1)T_1+v)+v{T}_1=((v-1{)}^2+v)T_1+v(v-1)=(v^2-v+1)T_1+v(v-1)$. By linearity and [A2], ${\mathrm{tr}}_2(T_1^3)=(v^2-v+1)z+v(v-1)$, the displayed trace value.

The Jones value. At $z_0=-1\mathrm{/}(v+1)$, ${\mathrm{tr}}_2(T_1^3){\mathrm{\mid }}_{z_0}=\frac{-(v^2-v+1)+v(v-1)(v+1)}{v+1}=\frac{v^3-v^2-1}{v+1}$. By [A3] with $n=2$ and $e({\sigma }_1^3)=3$, $V(\widehat{{\sigma }_1^3})=s^3(-\frac{s^2+1}{s})\frac{v^3-v^2-1}{v+1}=-s^2(s^2+1)\frac{s^6-s^4-1}{s^2+1}=-s^2(s^6-s^4-1)=-s^8+s^6+s^2$, i.e. $-t^4+t^3+t$ for $t=s^2$.

Normalization and skein. $V(\text{unknot})=1$ and the skein relation $s^{-2}{V}_{+}-s^2{V}_{-}=(s-s^{-1})V_0$ hold by [A3]; in the variable $t=s^2$ this is $t^{-1}{V}_{+}-t{V}_{-}=(t^{1\mathrm{/}2}-t^{-1\mathrm{/}2})V_0$. The closure of ${\sigma }_1^3$ is the right-handed trefoil by [A4], and applying $t\mapsto t^{-1}$ to the value $-t^4+t^3+t$ of step 2.1 gives $-t^{-4}+t^{-3}+t^{-1}$, the mirror polynomial from [A4]. The value of step 2.1 itself agrees with the trefoil table value in the cited survey; no inverse-variable table comparison is needed.

The cross-check. For the $2$-braid representative ${\sigma }_1^2$ one computes ${\mathrm{tr}}_2(T_1^2)=(v-1)z+v$ from [A1] and [A2], so at $z_0$ its value is $(v^2+1)\mathrm{/}(v+1)$ and by [A3] $V(\widehat{{\sigma }_1^2})=s^2(-\frac{s^2+1}{s})\frac{v^2+1}{v+1}=-s(s^4+1)=-s^5-s$, which is different from $-s^8+s^6+s^2$; by [A5] the closure of ${\sigma }_1{\sigma }_2{\sigma }_1$ is the closure of that positive stabilization of ${\sigma }_1^2$, so ${\sigma }_1{\sigma }_2{\sigma }_1$ is not a second braid representative of the trefoil and the two computations are consistent with the invariance of [A3].