Questions asked by Zhihui Zou, 
currently a Ph.D student in Dr. Michael A. Scott's group (Jan 24 2018)
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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?
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<i> Answer  </i>
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The degrees of freedom (d.o.f) of the representation are the 2x2 coefficients per quad.
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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). 
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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.
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So if (1)=(2) then we have refinability.
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<br> (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.)
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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]
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<i> Answer  </i>
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Here  S (=2x2 split via deCasteljau's algorithm) is knot insertion and 
knot insertion leaves the function unchanged. Since PS is already projected
and S leaves PS unchanged, projecting the projected function a second time 
does not change the function (e.g. the tangent plane at the center of the coarse
repreentation is the same as the tangent plane of the refined representation)
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3. For Isogeometric analysis, we hope the refined basis is nested in the previous basis for each refinement. Are the refinability and the nestedness of basis the same thing? In other words, if the refined geometry is the same as the original geometry, does it mean the refined basis has to be nested in the original basis?
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<i> Answer  </i>
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yes to the restated question:
refinability in the sense of dilation equation = nestedness of spaces.
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no to the assertion: the geometry generated by a non-nested "finer" 
(more d.o.f) representation can be the same as the original one.