Andy wrote: << On Thu, Dec 22, 2011 at 2:26 PM, Dan Asimov <dasimov@earthlink.net> wrote: << The "shrine problem" is, given a family S of shapes (let's say certain subsets of R^n) for which it makes sense to talk about a continuous curve [0,1] -> P,

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This requires not just a "family" S, but a "topological space" S. In the cases below, you specify the set S, but not the intended topology on S. It's not at all clear to me what the intended topology is on, say, the set of embedded arcs in the plane. Are you considering them just as subsets of the plane, or as parameterized arcs, so that two parameterizations of the same arc are considered different points of S? Assuming the latter for the moment, at least two topologies occur to me that seem natural; the topology of pointwise convergence and the topology of uniform convergence. Which is intended in the problem?

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The set of shapes S doesn't need to have a topological structure in order to define a continuous curve in S. A continuous curve (sheesh, I keep screwing up the notation! Bleeeh) in say the set S of embedded arcs in R^2 could just be defined by the natural condition that: Given any continuous map F: I x [0,1] -> R^2 (where I := [0,1]also) such that the restriction of F to any I x {t} is an embedding, we will say that s: [0,1] -> S via s(t) := F(I x {t}) is a continuous curve in S. << << c_n) S := all smooth Riemannian metrics on the n-disk D^n. (In this case, maps into R^n above must be replaced with maps into D^n.) These problems have a curious status:

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I'm confused; a smooth Riemannian metric on D^n is not a subset of D^n. So I don't know how to make sense of the requirement that P(s) is a member of s. P(s) is a point in D^n; S is a smooth Riemannian metric on D^n.

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That's the reason for the parenthetical sentence "In this case . . .." Here we need only that P: S -> D^n (subject to the 1) continuity condition and the 2) equivariance with respect to isometries f: D^n -> D^n between two (possibly identical) Riemannian structures on D^n. --Dan ________________________________________________________________________________________ It goes without saying that .