The graded Euler characteristics have the Jones normalization

Fix a quantum degree and tensor the finite free complex with $\mathbb{Q}$. For an integer matrix $d$, the kernel after tensoring is $(\ker d)\otimes \mathbb{Q}$: if $d(a\mathrm{/}r)=0$ with $a$ integral and $r\mathrm{\ne }0$, then $d(a)=0$ because the target is torsion-free. The image after tensoring is $\mathrm{im}(d)\otimes \mathbb{Q}$ by clearing denominators, so rationalized homology is $H^i(C)\otimes \mathbb{Q}$ and has dimension ${\mathrm{rank}}_{\mathbb{Z}}{H}^i(C)$. Rank-nullity in this bounded rational complex gives $\dim C^i=\dim H^i+\dim \mathrm{im}{d}^i+\dim \mathrm{im}{d}^{i-1}$; alternating over $i$ cancels the image dimensions. The same argument applies to the reduced complex, whose terms are finite-rank free.

On a state $v$, summing $q^{\sum \mathrm{deg}(a)}$ over its $k(v)$ independent labels gives $(q+q^{-1}{)}^{k(v)}$, so the unreduced chain Euler sum is ${\sum }_v(-1{)}^{\mathrm{\mid }v\mathrm{\mid }-n_{-}}{q}^{\mathrm{\mid }v\mathrm{\mid }+n_{+}-2n_{-}}(q+q^{-1}{)}^{k(v)}$. With $q=-A^{-2}$, one has $q+q^{-1}=\delta$ and, term by term, $(-1{)}^{\mathrm{\mid }v\mathrm{\mid }-n_{-}}{q}^{\mathrm{\mid }v\mathrm{\mid }+n_{+}-2n_{-}}{A}^{-(n-2\mathrm{\mid }v\mathrm{\mid })}=(-A^3{)}^{-w}$. The state contribution is consequently $(q+q^{-1})(-A^3{)}^{-w}{A}^{n-2\mathrm{\mid }v\mathrm{\mid }}{\delta }^{k(v)-1}$. For the reduced cube, the based label contributes degree zero by [F4], leaving $(q+q^{-1}{)}^{k(v)-1}$ in place of $(q+q^{-1}{)}^{k(v)}$, so its state contribution is $(-A^3{)}^{-w}{A}^{n-2\mathrm{\mid }v\mathrm{\mid }}{\delta }^{k(v)-1}$.

Summing the state contributions in step 2.1 gives the two displayed formulas for the chain Euler sums. Step 1.1 identifies those sums with the Euler characteristics of integral cohomology. When $D$ is the crossingless unknot, $n_{+}=n_{-}=0$ and $k=1$, so the unreduced value is $q+q^{-1}$ and the reduced value is $1$.