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For a real rectangular matrix $S$, maximize $p^{T}Sq$ over separate sign choices for $p$ and $q$. With $x=(p,q)$, the symmetric block matrix $A=\frac{1}{2}\left(\begin{array}{cc}0& S\\ S^T& 0\end{array}\right)$ satisfies $x^{T}Ax=p^{T}Sq$. Its zero diagonal blocks allow the Krivine rounding scheme to linearize the cross-block expected objective.

For bipartite binary quadratic optimization, if $v^{\ast }$ is the sign optimum and $p_{\mathrm{SDP}}^{\ast }$ its semidefinite relaxation of binary quadratic optimization value, then $c_K{p}_{\mathrm{SDP}}^{\ast }\le v^{\ast }\le p_{\mathrm{SDP}}^{\ast }$. The upper bound follows from rank-one lifting. For the lower bound, Krivine rounding scheme preprocessing followed by Gaussian hyperplane rounding has expected objective exactly $c_K{p}_{\mathrm{SDP}}^{\ast }$, and every rounded vector is feasible. Some outcome reaches at least this expectation. No independence of rounded coordinates or positive semidefiniteness of the original bipartite objective is needed.

The Krivine rounding scheme yields $c_K=0.56109985\dots$ as its guaranteed objective factor. It is obtained by normalizing $\sinh t=1$, so $t=\mathrm{arsinh}(1)$ and $c_K=2t/\pi$. This proof gives a valid universal factor for the bipartite sign problem, not a proof that it is optimal.

Put $t=\log (1+\sqrt{2})=\mathrm{arsinh}(1)$, so $c_K=2t/\pi$. The power series coefficients of the hyperbolic sine preprocessing areThus $f_j=|g_j|$ for every $j\ge 0$, and the series converge on the full interval. Applying coefficient-dominated entrywise positivity gives $Y⪰0$.

Every diagonal entry belongs to one of the diagonal blocks and equals $f(X_{ii})=f(1)=\sinh t$. Because $e^t=1+\sqrt{2}$ and $e^{-t}=\sqrt{2}-1$, this is $(e^t-e^{-t})/2=1$. ThereforeThe matrix lies in the elliptope and admits a Gram matrix representation by unit vectors. This preprocessing is the Krivine rounding scheme; the equality of absolute coefficients is what preserves positive semidefiniteness even though the cross-block sine coefficients alternate in sign.