The sigma sequence is antitone (componentwise decreasing).

The first and second components separately form decreasing sequences and both are eventually zero. These are the formal versions of conditions (15.2) and (15.3).

There exists \(K\) such that \(\sigma _X(k) = 0\) for all \(k \ge K\).

The components of the sigma sequence satisfy interlacing inequalities. If \(k\) is even, then \(b_X(k+1) \le a_X(k)\) and \(a_X(k+1) \le b_X(k)\). If \(k\) is odd, the inequalities are reversed.

If \(X \in \mathrm{Pi}\), the successive drops \(a_X(k)-a_X(k+1)\) and \(b_X(k)-b_X(k+1)\) satisfy the alternating inequalities (15.6) and (15.7).

For a polarized chromosome, the drop in the first sigma component from \(k\) to \(k+1\) counts genes in \(X^{(k)}\) that are positive in the alternating basis; the drop in the second component counts the corresponding negative genes.

The alternating inequalities imply useful global comparisons: later drops are bounded by the initial drop, and telescoping gives \(a_1-a_i \le b_0-b_{i-1}\) for \(i\ge 1\).

The inequalities between \(b_0\) and \(a_1\) detect the sign of a rank-one polarized gene. Summing this test over a chromosome produces a negative gene whenever \(b_0{\gt}a_1\).

If \(X {\lt} Y\) in \(\mathrm{Pi}\), then \(\sigma _X(k) \le \sigma _Y(k)\) componentwise for all \(k\) with strict inequality for some \(k\).