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So i was checking out Cauchy's rigidity theorem, which basically says that any convex polyhedron is rigid. This statement is however not true for a non-convex polyhedron. There may be some continuous deformation that leaves sides and faces intact, but changes angles between faces. A well known example of this by Klaus Steffen can be found in the image.
Now Idjad Sabitov proved further that, although such a continuous deformation changes angles between faces, it does conserve the volume enclosed by those faces.
Here's where i get confused, because this seems to suggest that if we have two polyhedra that are congruent in faces, either both convex or non-convex, and we have some continuous deformation from one to the other that leaves sides and faces intact, then their volumes must remain the same.
However, i seem to have found a counterexample here:
http://one-sick-psi-guy.blogspot.nl/2008/07/mythbuster-collapsible-cube.html
When the top of the cube is twisted, it folds in on itself nicely, seemingly keeping sides and faces intact. So any two states of the cube seem to be both non-convex, and congruent in faces, so should have the same volume. This obviously is not the case.
What am i missing here? Is this deformation non-continuous? How is a "continuous deformation" even defined?
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>>8454741
folding cube in question
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Read up on everything you mentioned and came to the same conclusion you did. I agree that this volume is clearly changing while the faces remain rigid. Do you have the original Connelly article / Bellows conjecture? Maybe we don't understand something about the class of non-convex polyhedra they're using.
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>>8454782
I found the original proof:
https://www.math.ucdavis.edu/~deloera/MISC/BIBLIOTECA/trunk/Connelly/Connelly3.pdf
In it they speak of a "flex" of a surface (triangulated).
For the definition of a "flex" i find the following article which defines it for graphs representing polyhedra
http://erikdemaine.org/papers/LinkageTR/paper.pdf
with Connelly as co-writer, leading me to think this must be the proper definition.
However, this doesn't help at all. As you can see in the definition in the image, we have that a "flex" is defined as a set of continuous functions (in 3-space) such that
[eqn]p(t) = (p_1(t),\dots,p_n(t))[/eqn]
[eqn]p(0) = p[/eqn]
[eqn]||p_i(t) - p_j(t)|| \text{ is constant over $t$ if $p_i$ and $p_j$ are connected}[/eqn]
All three seem to apply in our case.
However, in the beginning of the first article, it does say "Consider a triangulated polyhedral surface S in three-space". The top and bottom of our cube are not triangles, but surely we can replace the top and bottom with some pyramid shape, and nothing would change in the "bellows" part of the polyhedron.
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>>8454836
Forgot the image
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>>8454836
Their definitions are what I thought it was. I would assume our intuition is wrong, but I'm pretty sure if we compute the volumes of those translations they would be different.
I'll work through the rest of paper when I have time next week. I'm procrastinating my real work atm...
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Imagine a square box with corners at :
(1,0,0), (0,1,0), (-1,0,0), (0,-1,0)
(1,0,sqrt(2)), (0,1,sqrt(2)), (-1,0,sqrt(2)), (0,-1,sqrt(2))
so the box edges are all of length sqrt(2).
Now lower the top of the box and turn it counterclockwise
while maintaing the edge lengths. Let "a" denote the angle
of rotation of the top as the box height
z changes from sqrt(2) down to zero. This gives
|(cos(a),sin(a),z) - (1,0,0)|^2 = 2
(cos(a)-1)^2 + sin(a)^2 + z^2 = 2.
Now if we split one of the sides into two triangles,
the diagonal will be of length 2, and that also needs to be
maintained as we lower and swist the top:
|(cos(a),sin(a,z) - (0,-1,0)|^2 = 4
cos(a)^2 + (sin(a)+1)^2 + z^2 = 4
Subtracting one equation from the other leads to
sin(a)+cos(a)=1
which is only true when a=0 or a=pi/2. Thus something
needs to give between z=sqrt(2) and z=0.
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>>8455813
solving
(cos(a)-1)^2 + sin(a)^2 + z^2 = 2.
gives a = arccos(z^2/2).
Plugging that expression for "a" into
sqrt(cos(a)^2 + (sin(a)+1)^2 + z^2)
shows that this function is does not have a constant value 2. It needs to stretch about 10%.
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>>8454742
paper is not reliable giving you bends in what should be rigid sides, making up for the area like that chocolate bar illusion
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