Proposition: Two pieces, a clean intersection, and no single relation

The local diagonal piece ${\Lambda }_{\mathrm{d}}$ and the closure of the flow-out piece (eq:Lflow) each have dimension sixteen. They meet at

in dimension fifteen.

Their tangent intersection is exactly the tangent space of that intersection. Thus the two pieces intersect cleanly.

Their tangent sum has dimension seventeen. Consequently no single smooth sixteen-dimensional Lagrangian submanifold can contain both full local pieces at their intersection.

Proof. The parameters for ${\Lambda }_{\mathrm{d}}$ are: $a$ with four components, the nonzero null separation $b-a$ with three components, $y$ with four components, $\theta$ with one component, and $\kappa$ with four components. Their total is sixteen.

For the flow-out closure, replace the unrestricted four-dimensional $\kappa$ by a three-dimensional nonzero null covector and add the parameter $\tau$. The dimension is again sixteen.

At $\tau =0$ the base incidence is $a=z$. The further equation $p(\kappa )=0$ gives a fifteen-dimensional intersection in the diagonal piece.

A tangent vector to the diagonal piece has $\delta r=0$, with no constraint on $\delta \kappa$. A tangent vector to the flow-out closure at $\tau =0$ satisfies

To lie in both tangent spaces it must therefore have $\delta \tau =0$ and satisfy the linearized null condition on $\delta \kappa$. These are precisely the tangent conditions of the fifteen-dimensional intersection.

The tangent sum has dimension $16+16-15=17$. A smooth sixteen-dimensional submanifold containing both pieces would have to contain this seventeen-dimensional tangent sum, which is impossible.