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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.

For a field $K$ of characteristic different from two, the Witt group $W(K)$ classifies nondegenerate symmetric bilinear forms after hyperbolic parts are discarded. Two forms represent the same class if adding suitable hyperbolic planes of a quadratic form makes them isometric. The group operation is orthogonal direct sum.

A metabolic bilinear form has a totally isotropic subspace of half its dimension. In this characteristic it is a sum of hyperbolic planes of a quadratic form and represents zero. The inverse element of the equivalence class of $B$ is represented by $-B$: in $B\oplus (-B)$, the vectors $(v,v)$ form such a half-dimensional totally isotropic subspace.

Over the real numbers, Sylvester's law of inertia reduces a nonsingular form to $p$ positive squares and $q$ negative squares. A positive-negative pair is a hyperbolic plane of a quadratic form, so cancelling pairs leaves exactly the signature $p-q$. Conversely, equal signatures leave the same remainder. Thusunder the signature. This records how much definite form remains after all positive-negative cancellation.^[ref]

Over the complex numbers, every nonzero diagonal coefficient can be rescaled to one, since it has a square root. Two unit squares already form a hyperbolic plane of a quadratic form, using the independent isotropic vectors $(1,i)$ and $(1,-i)$. Pairs therefore cancel, and only the parity of the dimension remains:Odd dimension cannot disappear, because adjoining a hyperbolic plane of a quadratic form changes dimension by two. These are symmetric bilinear forms over the complex numbers, rather than Hermitian forms; confusing the two would give a different classification.^[ref]