Take the embedded superdegree per direction on a tensor product cell

[pbrubeck]

The problem

TensorProductElement.embedded_superdegree adds the degree of each factor.

UFL defines the embedded superdegree as the degree of the smallest Lagrange
space that holds this element. The docstring of that property says that the
Lagrange space of degree 1 on a quadrilateral holds the polynomial xy. A
Lagrange space on a cell that is not a simplex is therefore a space of a given
degree in each direction, not a space of a given total degree.

A Q7 space on a quadrilateral * interval cell has degree 7 in each direction.
The smallest Lagrange space that holds it is the Q7 space itself, so the
embedded superdegree is 7. The property gives 14, because it adds 7 and 7.

The same space on a hexahedron gives 7, because a hexahedron element is not a
tensor product element. Two cells that hold the same space therefore give
different answers.

UFL estimates a quadrature degree from this property. For
inner(grad(u), grad(v))*dx on the extruded cell, UFL gives 26 in place of 14.
FIAT then applies 26 to each direction of the cell.

The change

embedded_superdegree now takes the largest embedded superdegree of the
factors. The definition of the property does not change.

Effect

An extruded mesh of quadrilaterals, with the fix in FEniCS/ufl#506.

form	points before	points after	flops before	flops after
Q7 inner(grad(u), grad(v))*dx	3375	729	33,516,981	15,941,187
NCE7 inner(curl(u), curl(v))*dx	3375	729	253,603,812	115,918,062

Each matrix agrees with a rule of degree 32 to machine precision.

AI declaration: written with Claude Code (Claude Opus 5).

[mscroggs]

Looks correct to me for tensor product elements