The scheme of nontransverse zero jets has small dimension

On a frame chart the universal section has three coefficients. At a zero, differentiating a change of frame adds $(dG)\mathrm{\Phi }=0$, so the derivative transforms by invertible matrices. Hence its determinant-zero condition glues to the stated closed subscheme $B$. In jet space it is the condition that the value part is zero and the derivative, a $3\times 3$ matrix, is singular. These equations are pulled back by the actual jet-evaluation morphism of [F1].

Singular $3\times 3$ matrices form a variety of dimension eight over any field. On a chart where a chosen $2\times 2$ minor is invertible, a block matrix has rank at most two precisely when its last entry is the forced Schur complement; the other eight entries are free with that minor inverted. These charts cover rank two. For rank at most one, choosing a nonzero entry as pivot expresses the remaining entries from its row and column, leaving five free parameters; the zero matrix is a further point. Thus the dimension is eight, without a characteristic restriction. The bad jets in each twelve-dimensional jet fiber have zero value and this eight-dimensional derivative locus.

Locally split the surjection of [F1] and trivialize $E$ and ${\mathrm{\Omega }}^1$. The inverse image of that bad-jet locus is locally its product with affine space of dimension $N-12$. Its dimension is therefore at most $\dim W+8+N-12=N-1$ by [F3]. A finite such open cover suffices because $W$ is Noetherian and quasi-compact. This proves the bound globally, and the determinant definition proves the specialized derivative assertion.