Weight-system integration proof boundary

A degree-$n$ chord diagram is a pairing of $2n$ cyclically ordered endpoints, so only finitely many such diagrams exist. A 4T relation is determined by a background diagram and finitely many endpoint positions, and a 1T relation by one diagram; hence there are finitely many relation vectors. By [F1], each has integer coefficients, so they form the rows of a finite integer matrix $A$. Write $A_n(F):=F^{C_n}\mathrm{/}{\mathrm{rowspan}}_F(A)$ and $W_n(F):={\mathrm{Hom}}_F(A_n(F),F)$. Since $k$ has characteristic zero, $\mathbb{Q}$ embeds in $k$. The rank of $A$ is unchanged by this field extension: a nonzero maximal minor over $\mathbb{Q}$ remains nonzero in $k$, while every larger minor that vanishes over $\mathbb{Q}$ still vanishes in $k$. Thus the row space over $k$ is the scalar extension of the rational row space, and $A_n(k)\cong A_n(\mathbb{Q}){\otimes }_{\mathbb{Q}}k$.

The quotient is finite-dimensional, so extending a rational basis and its dual basis gives $W_n(k)\cong W_n(\mathbb{Q}){\otimes }_{\mathbb{Q}}k$. If $n=0$, the sole diagram is the empty diagram and any weight system is its value $c\in k$; the constant invariant $V(K)=c$ has top symbol $c$. For $n>0$, choose a finite $\mathbb{Q}$-basis ${\varphi }_1,\dots ,{\varphi }_d$ of $W_n(\mathbb{Q})$; finite-dimensional row reduction chooses this basis without a choice axiom. By [F2], for each $i$ there is a rational-valued finite-type invariant $V_i$ with top symbol ${\varphi }_i$. Write the given weight system as $W={\sum }_{i=1}^d{a}_i{\varphi }_i$ with $a_i\in k$, and define $V={\sum }_{i=1}^d{a}_i{V}_i$, viewing each rational value in $k$. Linearity of the finite-difference operator gives top symbol $W$. If $W_n(\mathbb{Q})=0$, the sum is empty and the zero invariant supplies the zero weight system.