Questions asked by Zhihui Zou, currently a Ph.D student in Dr. Michael A. Scott's group (Jan 24 2018)

• 1. In section 4, you mentioned the refinability as,"Refinability means that applying B-spline subdivision to the control net c and then applying the Algorithm yields the same surface as applying the Algorithm followed by subdividing the resulting Bézier patches by deCasteljau’salgorithm". My question is why refinability is equivalent to the commutativity? I thought refinability means the geometry doesn't change for every refinement. Am I right? If so, how is it related to commutativity?

The degrees of freedom (d.o.f) of the representation are the 2x2 coefficients per quad.
The question (challenge) is whether one can (1) find a (linear) operator to increase (refine) the dof but retain the geometry when applying the algorithm ( = the "dilation equation(s)" of refinability).
We know that the geometry is retained when (2) applying deCasteljau's algorithm (a 2x2 split called "S" in that paragraph) to each patch since it is knot insertion. Knot insertion leaves the function unchanged.
So if (1)=(2) then we have refinability.

(Refinement of the BB-form of the surface is not useful, because it does not preserve the built-in smoothness when treating all BB-coefficients as d.o.f.)

• 2. In your paper, there is a statement [ as underlined by the red line in the following screenshot.] Could you please explain a little bit why "PSPS = SPS since S keeps a projected function unchanged"? I don't quite understand it. Screenshot: Commutativity follows since, by definition of a projection, PPS=PS and [red underlined] PSPS = SPS since S keeps the projected function unchanged. [end red]