The Kontsevich integral and universality — Examples and Counterexamples

The diagram $D_1$ has one chord and no other chords, so that chord is isolated. In degree one there is no 4T generator, since 4T has two chords. Thus [F2] gives a valid framed weight system with $w(D_1)=2$.

Every functional on the unframed quotient must vanish on $D_1$, because [F1] imposes $D_1=0$. If $w$ factored through that quotient, it would satisfy $w(D_1)=0$, contrary to $w(D_1)=2\mathrm{\ne }0$ in $\mathbb{Z}$. Hence the framed weight system does not descend through 1T.

The unique degree-one circle diagram has an isolated chord, so [F3] sets it to zero. In degree two, the two chord pairings are exactly $D_{\times }$ and $D_{\parallel }$: their four endpoints either alternate or do not. By [F2], $w_2$ evaluates them as $1$ and $0$, so it is a nonzero functional on the unframed quotient.

The unknot normalization gives $a_2(U)=0$. For the positive Hopf link $H_2=\widehat{{\sigma }_1^2}$, switching one crossing cancels an adjacent positive and negative half twist and gives the two-component unlink; smoothing gives $\widehat{{\sigma }_1}=U$. A local type-I curl on one component of the two-component unlink has both crossing signs isotopic to that unlink and smoothing equal to the three-component unlink. The skein relation therefore gives $z{\mathrm{\nabla }}_{U_3}=0$. Multiplication by the indeterminate $z$ is injective in $\mathbb{Z}[z]$, because a polynomial $f=\sum a_j{z}^j$ with $zf=0$ has every coefficient $a_j=0$; hence ${\mathrm{\nabla }}_{U_3}=0$. Applying the same curl argument to the unknot gives ${\mathrm{\nabla }}_{U_2}=0$. Thus ${\mathrm{\nabla }}_{H_2}=z$. For $T=\widehat{{\sigma }_1^3}$, switching one crossing cancels an adjacent inverse pair and gives $U$, while smoothing gives $H_2$. Hence ${\mathrm{\nabla }}_T-1=z{\mathrm{\nabla }}_{H_2}=z^2$, so $a_2(T)=1$.

Let $W$ be the functional supplied by [F4] for $a_2$. Its degree-two restriction is ${\tilde{w}}_2$ by [F2] and complex linearity; the only lower-degree summand is degree zero because [F3] makes degree one zero. Since $a_2(U)=0$ and $I_0(U)=1$ by [F5], the value of $W$ on the degree-zero unit is $-{\tilde{w}}_2(I_2(U))$. Therefore $W(I_{\le 2}(K))={\tilde{w}}_2(I_2(K))-{\tilde{w}}_2(I_2(U))$, which equals $a_2(K)$ by [F4]. This verifies the displayed universal-series recovery for every knot type.