Removing a set of triangles from a triangulation may leave a set of triangles that no longer forms a triangulation because it violates the requirement that the union of the remaining triangles be homeomorphic to the unit square. For example, removing the star of an interior vertex may leave a polygonal hole.

This fact requires that we expand our dimension statement for S14 to an equivalent statement on subtriangulations.

The boxed equation gives the appropriate generalization: the degrees of freedom of S14 on a subtriangulation are obtained as follows:

- Three Bézier ordinates (equivalent to specifying the tangent plane0 at each vertex.
- The three Bézier ordinates in the interior of each triangle.
- The Bézier ordinate at the midpoint of each edge.
- From the total number so obtained we subtract 2 for each interior edge since across that edge we have to satisfy the differentiability conditions (i.e., the planarity of the corresponding quadrilaterals).

The whole idea can be summarized as follows: Think of S14 as a subspace of the space of continuous piecewise quartic functions that are differentiable at every vertex. The subspace is defined by the two additional differentiability conditions across each interior edge, and the crux of the matter is to prove that these conditions are linearly independent.

To see that the boxed equation is equivalent to the last equation when the triangles do form a triangulation recall the combinatorics of a triangulation:

ThusN = V_{B}+ 2V_{I}- 2, E_{I}= V_{B}+ 3V_{I}- 3, V = V_{B}+ V_{I}, E = E_{B}+ E_{I}.

3V + 3N + E - 2E_{I}= 3(V_{B}+ V_{I}) + 3(V_{B}+ 2V_{I}- 2) + 2V_{B}+ 3V_{I}- 3 - 2(V_{B}+ 3V_{I}- 3) = 6(V_{B}+V_{I}) - 3.

[28-Apr-1997]