##
The Generic Dimension of S^{1}_{4}

This page illustrates an actual proof using the
Bernstein-Bézier form. It concerns the space of
*C*^{1} quartics for which a minimal determining
set is constructed in reference 8 of the bibliography. There it is shown
that

(1)

where *V* is the
number of vertices and *sigma* is the number of *singular vertices* in the
triangulation.

Here we show that *generically*

(2)

The proof is by induction in the number of vertices.
Clearly (2) holds for the case of just one triangle.
Suppose (2) holds for *V* vertices and assume our
triangulation *T* has *V+1* vertices. Then consider a
boundary vertex *v* which is such that
removing its star *S*_{v} (i.e., the set of all triangles having
*v as a vertex*) leaves a triangulation.
(As an exercise you may want to show that it is always possible
to find such a boundary vertex.) The triangulation remaining
after removing *S*_{v} has *V*
vertices and a minimal determining set of *6V-3* points
by the induction hypotheses. We have to show that we can
expand this set to a minimal determining set on all of
*T* by adding 6 points in *S*_{v}.
Since we are considering the generic
case only we may assume that no pair of edges on any vertex is
parallel.

Figure 1. nearby illustrates the situation. The vertex
*v* is indicated by the red MDS symbol (a plus sign in a
circle). Bézier ordinates at the points marked with
green triangles are implied by the continuity and
differentiability conditions across the bottom of
*S*_{v} where it joins the remaining triangles of
*T*. Internal smoothness conditions determine
Bézier ordinates at the points marked with red
triangles. The points marked with yellow MDS symbols must be
in the minimal determining set since they enter no smoothness
conditions at all. Finally, setting the Bézier
ordinates at the two points marked with blue MDS signs implies
all remaining B&ezier; ordinates. There is a total of six new
points in the minimal determining set. It's clear that adding
these points form a determining set. To see that it is minimal
it suffices to check that for each of the new points $P$ one can
construct a cardinal spline where the Bézier ordinate
corresponding to *P* is 1 and all others are zero.

### A nontrivial exercise

Ask yourself how the above proof fails if some pairs of
edges sharing a vertex are parallel. Then ask yourself how to
fix the problem. If you are impatient read reference 8 in the
bibliography.

[15-Mar-1999]