\sum_{\scriptscriptstyle \overset{{{\bf{k}}}}
{i,i',j,j' \not\in {{\bf{k}}}}} p_{11\cdots 1}(ii'{{\bf{k}}}) \\
&=& \sum_{{{\bf{n}}}} C_{{{\bf{n}}}}^2 \, \nu_{i,{{\bf{n}}}} \nu_{i',{{\bf{n}}}} \, 
\left( 1 - \nu_{j,{{\bf{n}}}} \right) \left( 1 - \nu_{j',{{\bf{n}}}} \right),\end{aligned}$$ and the reverse is true for $p_{0011}(ii'jj')$.

The Schwarz inequality $0 \le \xi(ii'jj') \le 1$ imposes bounds on $\xi(ii'jj')$, the cosine of the hyper-angle between the vectors. Substituting [[**\[scalar\]**]{}]{} into [[**\[pi-od-signs1\]**]{}]{} yields $$\begin{aligned}
\pi^{od}(x'_1x'_2;x_1x_2) &=& 
\sum_{\scriptscriptstyle
i<i' \ne j<j'}
s(ii') s(jj') \nonumber\\
&&\left[p_{1100}(ii'jj') \,p_{0011}(ii'jj')\right]^{1/2}
\xi(ii'jj') \nonumber\\
&&\left(
\psi_i(x'_1)\psi_{i'}(x'_2)-\psi_{i'}(x'_1)\psi_i(x'_2)\right)\nonumber\\
&&\left(
\psi_j(x_1)\psi_{j'}(x_2)-\psi_{j'}(x_1)\psi_j(x_2)\right),
\label{final_pi-od}\end{aligned}$$ in which only $\xi(ii'jj')$ retains exponential complexity: $$\xi(ii'jj') = \frac{
\sum' p^{1/2}_{11\cdots 1}(ii'{{\bf{k}}}) p^{1/2}_{11\cdots 1}(jj'{{\bf{k}}})}
{\left(
\sum' p_{11\cdots 1}(ii'{{\bf{k}}}) \, \,
\sum' p_{11\cdots 1}(jj'{{\bf{k}}}) 
\right)^{1/2}},$$ where the primed sums are over all ${{\bf{k}}}$ with $i,i',j,j' \not\in {{\bf{k}}}$.

Inserting 4-OPs like $$p_{1111}(ii'kl) = \sum_{{{\bf{n}}}} C_{{{\bf{n}}}}^2 \, \nu_{ii',{{\bf{n}}}} \,
\nu_{kl,{{\bf{n}}}}; \qquad k < l \neq i,i',j,j'$$ in place of the $N$-OPs in $\xi$ reduces the complexity of $\pi^{od}$ to algebraic. The resulting approximation, $$\xi(ii'jj') \approx \frac{
\sum''_{k<l} p^{1/2}_{1111}(ii'kl) p^{1/2}_{1111}(jj'kl)}
{\left(
\sum''_{k<l} p_{1111}(ii'kl) \, \,
\sum''_{k<l} p_{1111}(jj'kl) 
\right)^{1/2}},
\label{xi-4}$$ is not variational, but obeys the 0,1 bounds of the Schwarz inequality. It is exact for $N=4$ within the PDC, and scales as $M^4$. In [[**\[xi-4\]**]{}]{} the doubly primed sums are over the indices $k<l$, which must differ 