Metabolic Seifert form

An integral Seifert form on a free abelian group $L$ of rank $2g$ is metabolic if there is a rank-$g$ direct summand $M\subset L$ on which the entire bilinear form vanishes:Over a field, replace the direct summand by a $g$-dimensional vector subspace. Such a half-rank vector subspace is a metabolizer. In a basis starting with a basis of the metabolizer, the Seifert matrix has a zero upper-left $g$-by-$g$ block.

The knot-theoretic nonsingularity condition concerns the matrix $V-V^T$, which defines a unimodular intersection pairing; $V$ itself need not be an invertible matrix. For example,has a metabolizer spanned by the first basis vector and has $\det (V-V^T)=1$, despite $\det V=0$. Requiring $V$ itself to be nonsingular would exclude this legitimate metabolic example. Algebraic concordance of knots identifies two forms when their difference is metabolic.

A metabolizer for an integral Seifert form $\theta$ on a free abelian group $L$ of rank $2g$ is a rank-$g$ direct summand $M\subseteq L$ on which the form vanishes:This is the data witnessing a metabolic Seifert form. Over a field, the corresponding definition uses a half-dimensional vector subspace. The same term also denotes a half-dimensional isotropic subspace for a metabolic bilinear form.

For $L={\mathbb{Z}}^2$ with Seifert matrix $V=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, the subgroup spanned by the first basis vector is a metabolizer. Its self-pairing is zero and it has the complementary summand spanned by the second basis vector. The subgroup generated by twice the first basis vector is also isotropic and has rank one, but is not a direct summand: its quotient in $L$ has torsion.

Thus half rank alone does not suffice over the integers. The direct-summand condition ensures that a basis of the metabolizer extends to a basis of the whole lattice, exposing a zero $g$-by-$g$ block in the Seifert matrix.