OK -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rcs@xmission.com Sent: Wednesday, September 02, 2015 8:55 AM To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: [EXTERNAL] [math-fun] test Just a test message. --Rich _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Trial, shell, or river in Hampshire UK ? WFL On 9/2/15, Cordwell, William R <wrcordw@sandia.gov> wrote:
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-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of rcs@xmission.com Sent: Wednesday, September 02, 2015 8:55 AM To: math-fun@mailman.xmission.com Cc: rcs@xmission.com Subject: [EXTERNAL] [math-fun] test
Just a test message. --Rich
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I. Consider the vertices V of an equilateral triangle as a metric space with all distances equal to 1. In the cartesian power V^n, we take the metric is taken to be D(x=(x_1,...,x_n), y=(y_1,...,y_n)) := sqrt((x_1-y_1)^2+...+(x_n-y_n)^2), where all x_j and y_k belong to V. So D(x,y) is the sqrt of the number of dimensions in which the coordinates differ. Question: Clearly, V^n embeds isometrically in R^2n, since V embeds isometrically in the plane. But is there a smaller dimension for which V embeds isometrically? Let E(n) = E_V(n) be the least dimension in which V^m embeds.* What is E(n) ??? ----- II. By permuting the coordinates 1,...,n of V^n, and independently permuting the 3 elements of each of the n factors V, it follows that the isometry group of V^n contains the subgroup S_n x S_3 of size 3(n!). Is S_n x S_3 equal to the full isometry group Isom(V^n) ? ----- III. Consider an imperfect model of the hyperbolic plane constructed as follows: (*) Let p, q be positive integers with (p-2)(q-2) > 4. Start with countably many disjoint regular (planar) p-gons of side = 1. Identify their edges to make a polyhedral surface that is topologically a plane, such that each vertex is surrounded by q of the p-gons, and every edge of any p-gon is identified with an edge of another p-gon. Call the resulting polyhedral surface H = H(p,q). Define its metric via D(x,y) := the infimum of all lengths of paths on H between x and y. (There is actually always a unique shortest path.) A. What is the least dimension E(p,q) = E_H(p,q) in which H(p,q) embeds* isometrically into the euclidean space of dimension E(p,q)??? B. Set V_H(p,q) denote the set of vertices of H(p,q), with its metric D'(x,y) defined as D'(x,y) = the length of the shortest edge-path in H(p,q) between x and y. (So all distances will be integers.) What is the least dimension E_V_H(p,q) for which V_H(p,q) embeds* isometrically into the euclidean space of dimension E_V_H(p,q) ??? ----- —Dan __________________________________________________________________________ * Note: When — as in questions I and IIB — we embed a *discrete* metric space X isometrically in some R^m, via some function f: X -> R^m , we require the *euclidean* distance of any two points f(x), f(y) of the subset f(X) in R^m to equal the original distance between x and y in X. But when — as in question IIA — we embed a *connected* metric space Y isometrically in some R^m, via some function f: Y -> R^m , we require the *intrinsic* distance of any two points f(x), f(y) of the subset f(Y) in R^m to equal the original distance between x and y in X. This "intrinsic distance" is the infimum of the lengths in R^m of all paths between f(x) and f(y) *that lie completely within* f(Y).
On Sep 2, 2015, at 1:33 PM, Dan Asimov <asimov@msri.org> wrote:
I. Consider the vertices V of an equilateral triangle as a metric space with all distances equal to 1.
In the cartesian power V^n, we take the metric is taken to be
D(x=(x_1,...,x_n), y=(y_1,...,y_n)) := sqrt((x_1-y_1)^2+...+(x_n-y_n)^2),
where all x_j and y_k belong to V.
So D(x,y) is the sqrt of the number of dimensions in which the coordinates differ.
Question: Clearly, V^n embeds isometrically in R^2n, since V embeds isometrically in the plane. But is there a smaller dimension for which V embeds isometrically?
Let E(n) = E_V(n) be the least dimension in which V^m embeds.* What is E(n) ??? -----
II. By permuting the coordinates 1,...,n of V^n, and independently permuting the 3 elements of each of the n factors V, it follows that the isometry group of V^n contains the subgroup S_n x S_3 of size 3(n!). Is S_n x S_3 equal to the full isometry group Isom(V^n) ? ----- . . . . . .
Correction: In question II I rashly ignored the fact that n independent S_3's form the group (S_3)^n. So the group that is clearly a subgroup of Isom(V^n) is in fact S_n x (S_3)^n , of size n! x 6^n. That's the group I suspect is the full isometry group of V^n, but am not sure. —Dan
participants (5)
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Cordwell, William R -
Dan Asimov -
Fred Lunnon -
rcs@xmission.com -
rwg