Maximal projections

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Maximal projections

Postby quickfur » Wed Sep 14, 2011 11:26 pm

In the last few days, I noticed something interesting about the maximal projections of the n-cross.

The maximal projection of the n-cross into (n-1) dimensions, as far as I can tell, is always when projecting to the hyperplane orthogonal to (1,1,1,1,...1). The projection envelope is always the convex hull of an (n-1)-simplex superimposed over its dual.

For example, the maximal projection of the octahedron is the projection orthogonal to (1,1,1), which is the convex hull of two dual triangles (i.e., the compound star 6/3), which is the regular hexagon. The maximal projection of the 16-cell is the projection orthogonal to (1,1,1,1), which is the convex hull of two dual tetrahedra (i.e., the stella octangula), which is the cube.

The maximal projection of the 5-cross is the convex hull of two dual 5-cells... as far as I can tell, this figure should be vertex-transitive as well as cell-transitive, although it doesn't appear to be uniform. If I'm not mistaken, this is the 4D catalan which is the dual of the bitruncated 5-cell.

Now, the question I'm grappling with right now is, is the maximal property transitive? That is, is the maximal (n-2)-dimensional projection of the n-cross equal to the maximal (n-2) projection of the maximal (n-1) projection of the n-cross? Is this true of any maximal (n-k)-dimensional projection of the n-cross for k<n? If so, what is the maximal 3D projection of the maximal 4D projection of the 5-cross? Is there a pattern to which projection is maximal in the series of maximal projections of decreasing dimensions?

Why am I interested in this, you ask? That's because ... the maximal 3D projection of the 6-cross, if I'm not mistaken, is none other than the regular icosahedron. I'm very curious to know, if the transitivity property is true, what is the 4D object whose maximal projection is the icosahedron. What is the series of projection planes that eventually results in the icosahedral projection of the 6-cross?
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Re: Maximal projections

Postby Mrrl » Thu Sep 15, 2011 5:13 am

What do you mean by "maximal projection"? With maximal area/volume? If you take 3-cross with diameter 2, then area of its hexagonal projection is sqrt(3) and area of square projection is 2. The most symmetrical projection is probably the smallest one.
And this property is not transitive: take 4-cross. Its best 3-projection is cube, best 2-projection is regular octagon, but best projection of cube is regular hexagon.
I've tried to find 3D-object between 4-cross and regular octagon, but it's not interesting: it is irregual hexagonal bipyramid. I think that for the icosahedron you'll get something irregular too. If you take icosahedron with vertices (R,0,0) and (1,xi,yi) (i=1..5) and opposite to them, then vertices of one of 4D bodies (between icosahedron and 6-cross) are (-R,R,0,0) and (1,1, xi,yi) - and opposite to them. It is something like bipyramid over pentagonal antiprism.
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Re: Maximal projections

Postby Mrrl » Thu Sep 15, 2011 6:18 am

If you want to find projection sequence from 6-cross to icosahedron, there is a way:

Select some coordinate system in 3D and write coordinates of 6 verteces of icosahendron in columns of matrix 3x6:

Code: Select all
sqrt(5) 1 1 1 1 1
0 0 x1 x2 -x2 -x1
0 2 y1 y2 y2 y1


Here x1=2*sin(2*pi/5), x2=2*sin(pi/5), y1=2*cos(2*pi/5), y2=-2*cos(pi/5).
Rows of this matrix give 3 vectors of orthonormal basis of some subspace in 6D. Add 3 rows to complete the basis of 6D (and keep it orthonomal):
Code: Select all
sqrt(5) 1 1 1 1 1
0 0 x1 x2 -x2 -x1
0 2 y1 y2 y2 y1
-sqrt(5) 1 1 1 1 1
0 0 x2 -x1 x1 -x2
0 2 y2 y1 y1 y2


Its columns give you vertices of 6-cross in 6D. Select any sequence of last 3 coordinates and remove them one by one - you'll get coordinates of 5D and 4D bodies on the way to icosahedron.

Well, bases here are not exactly orthonormal - lengths of all vectors are sqrt(10) instead of 1. But it should not be a problem.
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Re: Maximal projections

Postby wendy » Thu Sep 15, 2011 8:35 am

There are things called 'zonohedra', which are projections of þe measure polytope. On the other hand, one has 'eutactic stars' (coxeter, 'regular polytopes'), which are projections of the tegum-or-cross polytope. The eutactic stars are of some interest, because there are polytopes which are restricted to these.

The eutactic stars of six axies (like the dodecagon, and the icosahedron), are the projection of the 6-tegum. There's a drawing of the icosahedron projected into the 6d cube on the front of the dover 1999 reprint of 'twelve essays' by coxeter.

The c600 is itself a eutactic projection of the 4_21 (/6B).

There is also some interesting thing about a polytope in 5 dimensions ( 2/2 = o3o3x3o3o ) which has 12 faces and 20 vertices. The dual of this, gives the o3o3m3o3o, which does not itself tile, but is the face of the dual of 1_22, which does tile. If make a prism of o3o3x3o3o, and put a x3o3o3o3x at the middle, and a vertex at each pole, you get the 72 vertices of the 1_22.
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Re: Maximal projections

Postby quickfur » Sat Sep 17, 2011 1:03 am

Mrrl wrote:What do you mean by "maximal projection"? With maximal area/volume? If you take 3-cross with diameter 2, then area of its hexagonal projection is sqrt(3) and area of square projection is 2. The most symmetrical projection is probably the smallest one.

By "maximal" I mean having the largest volume ratio with the inscribing (n-k)-sphere. That is, the projection P maximizes the ratio:

r = V(P)/V(S^m)

where S^m is the inscribing sphere of P, and V(x) is the m-dimensional bulk of x.

And this property is not transitive: take 4-cross. Its best 3-projection is cube, best 2-projection is regular octagon, but best projection of cube is regular hexagon.[...]

You're right, it can't be transitive.

In that case, though, I'm curious, what are the maximal k-dimensional projections of (n+k)-cross polytopes, for some fixed k. For k=2, the maximal projections seem to be just 2(n+k)-gons. What about k=3? It seems we should have some interesting variations here: the maximal projection of the 3-cross is the 3-cross itself, but what about the 4-cross and 5-cross? The 6-cross gives the icosahedron using my definition of maximal (I think); what do the other crosses give?
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Re: Maximal projections

Postby quickfur » Sat Sep 17, 2011 1:11 am

On further thought, I wonder if this could just be a variation of "n mutually-repulsive electric charges inside a sphere", except that this time we have n sticks whose tips repel each other, and each stick is hinged at the origin. Going up to the next dimension's cross polytope is equivalent to adding another stick to the system.

The 3D case may be quite easy to model using a physics simulator, i think. Would be interesting to find out what kind of configurations come out. :)
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Re: Maximal projections

Postby Mrrl » Sat Sep 17, 2011 9:17 am

Best 3-projection for 4-cross should be a cube. And I guess that for 5-cross it will be 5-antiprism, but probably it will be not the uniform one.
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Re: Maximal projections

Postby quickfur » Sat Sep 17, 2011 5:12 pm

Mrrl wrote:Best 3-projection for 4-cross should be a cube. And I guess that for 5-cross it will be 5-antiprism, but probably it will be not the uniform one.

I agree that best 3-projection for 4-cross should be a cube... although i suspect the repelling-sticks simulation will probably end up with a square antiprism instead.
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Re: Maximal projections

Postby Mrrl » Sat Sep 17, 2011 5:55 pm

Square-antiprism has no opposite vertices, so you can't get it by sticks simulation.
Best 5-antiprism for the projection of 5-cross has height h=R*2/sqrt(3) (so it's inscribed in the same cylinder (with h=sqrt(2)*r) as a cube).
Last edited by Mrrl on Sat Sep 17, 2011 6:53 pm, edited 1 time in total.
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Re: Maximal projections

Postby quickfur » Sat Sep 17, 2011 6:52 pm

Mrrl wrote:square-antiprism has no opposite vertices, so you can't get it by sticks simulation

Hmm you are right!! so maybe you will get a cube from the sticks simulation after all. Very interesting! I wonder if you can get a dodecahedron from it too. Maybe I should write a program to run the simulation!
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Re: Maximal projections

Postby Mrrl » Mon Sep 19, 2011 9:32 am

There is one interesting thing about maximal projections.
Two 3D candidates for 5-cross are 5-antiprism (coordinates (a*cos(2*pi*k/5),a*sin(2*pi*k/5),b) and opposite, where a^2+b^2=1) and inscribed elongated 4-bipyramid (coordinates (0,0,1), (0,c,d), (0,-c,d), (c,0,d), (-c,0,d) and opposite, where c^2+d^2=1). Parameters of bodies for the maximal volume are: b=sqrt(1/3) for the first and d=(sqrt(6)-1)/5 for the second.
But there are some conditions for coordinates of the cross projections: if vertices have coordinates (xi,yi,zi), then vectors (x1,x2,x3,x4,x5), (y1,y2,y3,y4,y5) and (z1,z2,z3,z4,z5) must be orthogonal and have equal length. For our two bodies we have conditions 5*b^2=5/2*a^2 and 2*c^2=4*d^2+1. If we solve these equations, we'll get b=sqrt(1/3) and d=(sqrt(6)-1)/5 - the same values as for the maximal bodies :)
BTW, 5-antiprism is 2.5% better than elongated 4-bipyramid: it has volume 1.97433*R^3 vs. 1.92937*R^3.
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Re: Maximal projections

Postby quickfur » Mon Sep 19, 2011 2:52 pm

Mrrl wrote:There is one interesting thing about maximal projections.
Two 3D candidates for 5-cross are 5-antiprism (coordinates (a*cos(2*pi*k/5),a*sin(2*pi*k/5),b) and opposite, where a^2+b^2=1) and inscribed elongated 4-bipyramid (coordinates (0,0,1), (0,c,d), (0,-c,d), (c,0,d), (-c,0,d) and opposite, where c^2+d^2=1). Parameters of bodies for the maximal volume are: b=sqrt(1/3) for the first and d=(sqrt(6)-1)/5 for the second.
But there are some conditions for coordinates of the cross projections: if vertices have coordinates (xi,yi,zi), then vectors (x1,x2,x3,x4,x5), (y1,y2,y3,y4,y5) and (z1,z2,z3,z4,z5) must be orthogonal and have equal length. For our two bodies we have conditions 5*b^2=5/2*a^2 and 2*c^2=4*d^2+1. If we solve these equations, we'll get b=sqrt(1/3) and d=(sqrt(6)-1)/5 - the same values as for the maximal bodies :)
BTW, 5-antiprism is 2.5% better than elongated 4-bipyramid: it has volume 1.97433*R^3 vs. 1.92937*R^3.

Cool! I'll have to look into that sometime... I'm going to be busy for the next few weeks so I won't be able to play with this until later.
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Re: Maximal projections

Postby Mrrl » Mon Sep 19, 2011 3:17 pm

Sorry, I was wrong. For the elongated bipyramid volume is maximal when d=(sqrt(13)-1)/6, there V=2.021535*R^3, and it is a projection of 5-cross when d=sqrt(6)/6. But volume for this value is V=2.01833*R^3, that is still more than the volume of 5-antiprism...
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