Theorem: Projection rank and the excess correction

The map $J$ has constant rank fourteen on the retained matching manifold. Its local fibers have dimension one.

Its local image is an isotropic fourteen-dimensional immersed submanifold of $T^{\ast }W$, contained in $N^{\ast }N$ with the zero section allowed. It is not a Lagrangian submanifold of $T^{\ast }W$.

Accordingly, neither the rank-two tangent defect nor the fibers of $J$ justify the specification's clean-excess-two claim.

Proof. For fixed $w\in N$, the differential of $J$ in the seven frequency variables has image

of dimension six.

The base projection of $J$ is the identity on $N$. Its differential maps the eight base tangent directions injectively onto $T_{w}N$ in the base component of $T(T^{\ast }W)$. These directions are independent of the six purely vertical image directions. Thus

The rank is constant.

A vector in $\ker \mathrm{d}J$ must have zero base component. The attachment frequency variation must then be zero, because its $z$ component cannot be canceled by cone covectors. The remaining kernel is the one-dimensional cone-frequency relation

The constant-rank theorem gives one-dimensional local fibers.

Every covector in (eq:sum-map) annihilates $T_{w}N$. Hence the image lies in the conormal bundle of $N$. Let ${\lambda }_{\mathrm{can}}$ be the tautological one-form on $T^{\ast }W$. For every tangent vector to ${\mathcal{F}}_{\mathrm{match}}$, its base component belongs to $T_{w}N$, so

Differentiating gives $J^{\ast }\mathrm{d}{\lambda }_{\mathrm{can}}=0$. Since $J$ has constant rank, its local image is isotropic.

A Lagrangian submanifold of $T^{\ast }W$ has dimension $\dim W=16$. The image has dimension fourteen, so it is not Lagrangian.