Yes, you guys, that's it. My way of seeing it was to note that there is one unbounded region for each lower-dimensional face of the simplex, plus the simplex itself. These correspond to each subset of the n+1 vertices except the null set, hence 2^(n+1) - 1. I'm not sure if that's how Tom arrived at his sum, or some other way. Tom? ----- Meanwhile, I have a question for Fred: What exactly are the "Möbius transformations" of a sphere of dimension n > 2 ? (One guess: Are these, by analogy with n = 1 and n = 2, the isometries of the Poincaré model of (n+1)-dimensional hyperbolic space, as detected on the S^n boundary of that (n+1)-disk? In any case, a rather clever proof. —Dan ----- On Apr 14, 2016, at 5:01 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote: I did consider this, but omitted it earlier since it turned out not to be required. Notice that there is no need to prescribe the shape of the spherical simplex, since some direct Möbius transformation will transform the "octant" into any given proper simplex, while by continuity the number of regions remains constant. For a Euclidean simplex, modify the spherical case by attaching an extra great S^(n-1) , meeting every region except the chosen (closed) simplex and (via symmetry) its antipode. This acts as hyperplane at infinity in an elliptic model of projective space, in which antipodal points map to coincidence. So simplex and antipode map to the same region, leaving 2^(n+1) - 1 regions ----- On Apr 14, 2016, at 12:51 PM, Tom Karzes <karzes@sonic.net> wrote: Ok, let's try this again. How about sum(i=1,n+1) ((n+1) take i) i.e. sum(i=1,n+1) ((n+1)!/((n-i+1)!*i!)) There are just the rows of Pascal's triangle, minus one of the 1 columns, so I believe this is just: 2^(n+1) - 1 -----
On 4/14/16, Dan Asimov <asimov@msri.org> wrote:
Yes — thanks for the correction, Tom:
"...will divide S^n into 2^(n+1) compartments"
is most definitely what I should have written.
—Dan
On Apr 14, 2016, at 1:16 PM, Tom Karzes <karzes@sonic.net> wrote:
Hi Dan,
In the following:
Clearly, doing the same with the right spherical n-simplex (i.e., having all dihedral angles equal to 90º) in the n-sphere S^n will divide S^n into 2^n compartments.
Didn't you mean "...will divide S^n into 2^(n+1) compartments"?
Tom
Dan Asimov writes:
Given a regular (or any) tetrahedron in R^3 with its face planes extended indefinitely will divide R^3 into a certain number of compartments.
A related situation is that of the 2-sphere S^2, where the right spherical triangle (i.e., having all angles equal to 90º) has its 1-faces extended indefinitely to become great circles. Clearly that will divide S^2 into 2^3 = 8 compartments.
Clearly, doing the same with the right spherical n-simplex (i.e., having all dihedral angles equal to 90º) in the n-sphere S^n will divide S^n into 2^n compartments.
NO! This should be 2^(n+1) compartments, as Tom points out.
Puzzle: ------- Given an n-simplex in R^n with its face-hyperplanes extended indefinitely, how many compartments will they divide R^n into ?
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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