Multilinear orthogonal and symplectic FFT/SFT

Give the polynomial ring on $V^t$ the grading by degree in each vector slot. The group coaction preserves this grading, so its kernel equation for invariance holds separately in every multidegree. Formal Gram coordinates have degree $e_i+e_j$ (degree $2e_i$ on a symmetric diagonal). The surjection in [F1] is multigraded, hence its restriction in degree $\mathbf{1}=(1,\dots ,1)$ is surjective: take a polynomial preimage and retain only its degree-$\mathbf{1}$ component. A monomial of that degree uses each label once, so it is exactly a perfect-matching product of off-diagonal coordinates. There are no such monomials for odd $t$. The degree-$\mathbf{1}$ polynomials on $V^t$ are exactly multilinear forms, and invariance for the coaction is exactly invariance of these forms. This proves the spanning assertion and the odd-degree vanishing.

Using the exact ideals in [F3], the kernel of this restricted map is the degree-$\mathbf{1}$ component of the kernel in [F1], since the map preserves grading. [F2] computes it: for the symmetric ring it is spanned by determinants on disjoint $(m+1)$-element row and column sets times matching monomials in unused labels; for the alternating ring it is spanned by $(2n+2)$-Pfaffians times matching monomials in unused labels. Applying the polynomial map turns these monomials into exactly the contractions in the Statement, with each pair written in increasing order. Thus these spans give every relation, including the symplectic signs.

If $r\le m$ in the symmetric case, there are fewer than $2(m+1)$ labels, so the determinant relation family of step 2.1 is empty and the contractions are independent. If $r\ge m+1$, choose disjoint sets $I,J$ of size $m+1$, expand $\det Z_{I,J}$, and multiply by a matching on the other $2r-2m-2$ labels. Different determinant permutations produce different matchings because $I$ and $J$ are disjoint. Their coefficients are $\pm 1$, so this is a nonzero formal linear combination of distinct matching monomials, but its image is zero. Thus independence fails precisely beyond $r=m$.

If $r\le n$ in the alternating case, there are fewer than $2n+2$ labels, so the relation family is empty. If $r\ge n+1$, choose $2n+2$ labels, expand their principal Pfaffian, and multiply by any matching on the unused labels. Each term of the Pfaffian is a distinct matching with coefficient $\pm 1$. This is a nonzero formal relation whose contraction vanishes. Hence independence holds exactly for $r\le n$. At $r=0$ there is one empty matching, mapping to the nonzero scalar $1$, in both branches.