k-frames done right

Higher-dimensional geometry (previously "Polyshapes").

k-frames done right

Postby PWrong » Wed Dec 09, 2009 7:32 am

I'd like to fromalise the idea of k-frames somewhat. Given any shape with a lot non-smooth points, we'd like a way to determine all the possible frames of the shape. Obviously we can't start from the minimal frame, because there are infinity ways to "fill in" a shape. So we have to start with the maximal frame and fid a sequence of operators that will each reduce the frame by one. The first operator is equivalent to the usual boundary operator. Ideally we'd like all the operators to be defined the same way.

I'm thinking we could use the characteristic function somehow (the function that returns 1 if x is in the shape and 0 otherwise). Differentiability would be involved and also manifold charts might come in handy.
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Re: k-frames done right

Postby PWrong » Wed Dec 09, 2009 10:46 am

The shapes we're looking at (cubes, cylinders etc) are not smooth manifolds, in fact I don't think they're manifolds at all. We usually assume they're manifolds because it's possible to get an arbitrarily good approximation to a cube or cylinder with a smooth manifold. A manifold must be locally homeomorphic to Euclidean space at every point. The cube is locally homeomorphic to R^n everywhere except at the points on the frame.

So for a non-manifold X, we define f(X), the frame of X, as the subset of points where X is not locally homeomorphic to Euclidean space.
Equivalently, f(X) is the smallest subset of X such that X-f(X) is a smooth manifold.

This covers max-frames, for example the interior of a disc is locally homeomorphic to R2, but the boundary isn't.
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Re: k-frames done right

Postby Keiji » Wed Dec 09, 2009 12:45 pm

Wouldn't this definition only generate the 3-frame, 2-frame and 0-frame cubes, skipping out the 1-frame?
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Re: k-frames done right

Postby PWrong » Thu Dec 10, 2009 12:33 am

The edges of a 2-frame cube aren't homeomorphic to R^2 because there's a right angle there. So the frame of a 2-frame is the set of edges, which is the 1-frame.
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Re: k-frames done right

Postby Keiji » Thu Dec 10, 2009 3:22 am

Just because there's a right angle on the surface doesn't mean it doesn't map. That's just like folding a piece of paper. Only at the vertices is there an issue.
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Re: k-frames done right

Postby PWrong » Thu Dec 10, 2009 4:13 am

My bad, somehow I got the idea that homeomorphisms have to be smooth. In fact they just have to be continuous and the inverse has to be continuous. This means cubes and cylinders are indeed manifolds, but not smooth manifolds. To be smooth it has to be diffeomorphic to Euclidean space. So my definition of frame sends everything to the empty set. Even at the vertices, because you can unfold them carefully with a bit of careful stretching.

For the new definition we just replace homeomorphic with diffeomorphic. Unfolding an edge is continuous but not a smooth map, so they count.
f(X) is the subset of points where X is not locally diffeomorphic to Euclidean space.

This can easily be turned into a more rigorous definition involving charts, but I won't go into that here.
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Re: k-frames done right

Postby PWrong » Sun Dec 13, 2009 6:35 am

What we need now is a kind of Leibnitz rule for the Cartesian product. That is,
f(AxB) = fAxB U AxfB
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