Talk:Cantitruncated 5-cell

Cartesian coordinates
Wow, the coordinate take over the stub-article. Perhaps the fractions can be removed strategic use of a constant? Tom Ruen (talk) 20:24, 16 December 2008 (UTC)

Hmmm... maybe they could be consolidated with others in a page Cartesian coordinates of 5-cell uniform family? SERIOUSLY, I don't mind data, and it's good to know it's been calculated, but I can't easily call it useful to have 2 printed pages of radialized coordinates. In fact to me, it's useless trivia as-is, even if I wanted to punch all those expressions into a calculator or computer code, since I'd need the edge, face, cell data along with it for rendering. Tom Ruen (talk) 20:33, 16 December 2008 (UTC)


 * Perhaps it's more useful to give the non-normalized coordinates in (n+1)-space (which will only need 1 line to represent)? The currently coordinates are essentially the non-normalized coordinates, which are the positive permutations of the corresponding truncated/bitruncated/etc. 5-cross, multiplied by a transformation matrix that maps them to n-space with a "nice" orientation (nice as in, they are origin-centered, and the 5-simplex they derive from has its apex along the first coordinate axis, with its base recursively bisected by a coordinate plane).


 * As far as rendering is concerned, a suitable use of a convex hull algorithm along with linear programming algorithms for finding elements of the face lattice makes coordinates extremely useful.&mdash;Tetracube (talk) 22:05, 16 December 2008 (UTC)

Sorry, I don't have a clear opinion what's best, just that what's here now seems excessive to justify in the article. Tom Ruen (talk) 21:31, 19 December 2008 (UTC)


 * For comparison, the coordinates of the polychoron of this page (cantitruncated 5-cell) in 5-space are all permutations of:


 * $$\left(1,\ 1,\ 1+\sqrt{2},\ 1+2\sqrt{2},\ 1+3\sqrt{2}\right)$$


 * The transformation matrix to map these coordinates to 4-space is:


 * $$\left[\begin{array}{ccccc}

-\sqrt{\frac{4}{5}} & \sqrt{\frac{1}{20}} & \sqrt{\frac{1}{20}} & \sqrt{\frac{1}{20}} & \sqrt{\frac{1}{20}}\\ 0 & -\sqrt{\frac{3}{4}} & \sqrt{\frac{1}{12}} & \sqrt{\frac{1}{12}} & \sqrt{\frac{1}{12}}\\ 0 & 0 & -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}}\\ 0 & 0 & 0 & -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$


 * I have deliberately not simplified the radicals in order to show the pattern: each row begins with a negative term of the form $$-\sqrt{a/b}$$ and continues with repetitions of $$\sqrt{1/ab}$$. The values of a and b are (1,2) for the bottom row, (2,3) for the 2nd last row, (3,4) for the 3rd last row, etc.. Furthermore, the same matrix can be applied to all other 5-cell truncates (bitruncates, omnitruncates, cantellates, etc., etc.) to map them from 5-space to 4-space, the 5-space coordinates being the permutations of the non-negative coordinates of the corresponding 5-cube truncate. (E.g., the omnitruncated 5-cell's coordinates in 5-space are the permutations of $$\left(1,\ 1+\sqrt{2},\ 1+2\sqrt{2},\ 1+3\sqrt{2},\ 1+4\sqrt{2}\right)$$, which are the non-negative coordinates of the omnitruncated 5-cube. Multiplying these points by this matrix maps them to 4-space in a "nice" orientation).


 * The matrix is itself is simply the product of 4 plane rotation matrices that successively rotates the 5th coordinate axis into the 4th, the 4th into the 3rd, the 3rd into the 2nd, and the 2nd to the 1st, transforming $$(1,1,1,1,1)$$ into $$(\sqrt{5},0,0,0,0)$$, with the first row dropped since all points will end up with the same coordinate (i.e., this projects 5-space to 4-space). This is because a 5-cube truncate's non-negative coordinates always yield points lying in a hyperplane orthogonal to (1,1,1,1,1).


 * This pattern can be generalized to n-space, in which case the matrix is the product of (n-1) plane rotations that reduce the n-space coordinates into (n-1)-space. This gives a simple derivation method for the coordinates of any n-simplex truncate, including the n-simplex itself.&mdash;Tetracube (talk) 22:21, 19 December 2008 (UTC)


 * I like the concept (particularly if it can be clearly and concisely annotated): a readily-understood operation (rotation by a comprehensible matrix) on a readily-understood operand (permutations of a vector) has obvious advantages over a bargeload of coordinates with no obvious structure. —Tamfang (talk) 22:44, 22 December 2008 (UTC)


 * I'm a little scared by requiring a higher dimension. I don't have anything positive to say on any option so far, so I'll continue to be quiet mostly. Tom Ruen (talk) 23:00, 22 December 2008 (UTC)

The derivation of the matrix is very simple. Consider how one might rotate the vector $$(1,1)$$ to $$(\sqrt{2},0)$$. This is a 45° clockwise rotation expressed by the matrix:


 * $$\left[\begin{array}{cc}

\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \\ -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$

Observe that any scalar multiple of $$(1,1)$$ gets rotated into a multiple of $$(1,0)$$. We use this to our advantage in the next step: how to rotate $$(1,1,1)$$ to $$(\sqrt{3},0,0)$$? There are at least two obvious ways: one is to rotate directly in the plane spanned by $$(1,0,0)$$ and $$(0,1,1)$$. There is a formula for doing this, but it is quite convoluted and it turns out that when it applied to n-simplices, the resulting coordinates are not only "ugly", but also don't have a "nice" orientation. The second way is to rotate $$(1,1,1)$$ into $$(1,\sqrt{2},0)$$, and then rotate this into $$(\sqrt{3},0,0)$$. When applied to n-simplices, the resulting coordinates have the nice property that one vertex (the "apex") will lie along a coordinate axis, and the "base" will oriented such that it is symmetric by reflection across a coordinate plane, and recursively, the "apex" of the base lies along another coordinate axis, and its base is symmetric across another coordinate plane, etc.. Well, to rotate $$(1,1,1)$$ to $$(1,\sqrt{2},0)$$ is easy: we simply extend our 2D matrix thus:


 * $$A=\left[\begin{array}{ccc}

1 & 0 & 0 \\ 0 & \sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \\ 0 & -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$

Then, to rotate $$(1,\sqrt{2},0)$$ to $$(\sqrt{3},0,0)$$, the following rotation will do the job:


 * $$B=\left[\begin{array}{ccc}

\sqrt{\frac{1}{3}} & \sqrt{\frac{2}{3}} & 0 \\ -\sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} & 0 \\ 0 & 0 & 1 \end{array}\right]$$

Multiplying B and A (since we perform A first then B), we get the following matrix for rotating $$(1,1,1)$$ to $$(\sqrt{3},0,0)$$:


 * $$BA=\left[\begin{array}{ccc}

\sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} \\ -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\ 0 & -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$

A similar line of argument for the 4D case shows that to rotate $$(1,1,1,1)$$ to $$(2,0,0,0)$$, we multiply the following matrix with BA (well, with BA extended to 4D by inserting a left column and top row):


 * $$C=\left[\begin{array}{cccc}

\sqrt{\frac{1}{4}} & \sqrt{\frac{3}{4}} & 0 & 0 \\ -\sqrt{\frac{3}{4}} & \sqrt{\frac{1}{4}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]$$

I'm deliberately leaving the radicals unsimplified so that the pattern of radicals is obvious: we're progressing from $$\sqrt{2}$$ to $$\sqrt{3}$$ to $$\sqrt{4}$$, etc.. Multiplying out the matrices give us, for the 4D case:


 * $$R_4=\left[\begin{array}{cccc}

\sqrt{\frac{1}{4}} & \sqrt{\frac{1}{4}} & \sqrt{\frac{1}{4}} & \sqrt{\frac{1}{4}} \\ -\sqrt{\frac{3}{4}} & \sqrt{\frac{1}{12}} & \sqrt{\frac{1}{12}} & \sqrt{\frac{1}{12}} \\ 0 & -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\ 0 & 0 & -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$

I'll just denote these matrices by $$R_n$$, for conciseness.

It will not be surprising to you that the next step involves left-multiplying with a 5D rotation matrix with the terms $$\sqrt{\frac{1}{5}}$$ and $$\sqrt{\frac{4}{5}}$$. (Again, not simplifying the radicals so that the pattern is obvious.) Multiplying this with $$R_4$$, we obtain:


 * $$R_5=\left[\begin{array}{ccccc}

\sqrt{\frac{1}{5}} & \sqrt{\frac{1}{5}} & \sqrt{\frac{1}{5}} & \sqrt{\frac{1}{5}} & \sqrt{\frac{1}{5}} \\ \sqrt{\frac{4}{5}} & \sqrt{\frac{1}{20}} & \sqrt{\frac{1}{20}} & \sqrt{\frac{1}{20}} & \sqrt{\frac{1}{20}} \\ 0 & -\sqrt{\frac{3}{4}} & \sqrt{\frac{1}{12}} & \sqrt{\frac{1}{12}} & \sqrt{\frac{1}{12}} \\ 0 & 0 & -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\ 0 & 0 & 0 & -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \end{array}\right]$$

At this point, I'd like to point out that:
 * 1) The first row of $$R_n$$ can be ignored, because we'll be rotating points generated by permutations of some vector, and all permutations of any vector with non-negative coordinates will always have the same dot product with any scalar multiple of $$(1,1,1,\ldots)$$. That is to say, all points generated by permutations of the vector will always lie in a hyperplane orthogonal to $$(1,1,1,\ldots)$$ (see permutohedron for a prime example). As a result, the first coordinate of any vector on this hyperplane after rotation by $$R_n$$ will be a fixed constant, which can be dropped to map the vector back to (n-1)-space.
 * 2) Let $$R'_n$$ denote $$R_n$$ with its first row dropped. Then the remaining (n-2) rows are exactly $$R'_{n-1}$$ with a column of 0's prepended on the left.
 * 3) From the second row onwards, the first non-zero entry of each row is of the form $$\sqrt{\frac{a-1}{a}}$$, and continues with repetitions of $$\sqrt{\frac{1}{a(a-1)}}$$ until the end of the row. The bottom row corresponds with a=2, the second last row with a=3, the third last row with a=4, and so forth.

This gives us a generic formula for n dimensions.

Now the reason I write all of this, is to point out that a single matrix works for all truncations of the n-simplex in a given dimension n. So it seems that it would make sense to have a separate article that describes what this matrix is, and then in each uniform polytope article simply give the coordinates in (n+1)-space, with a link to this article for how to map these coordinates back to n-space.&mdash;Tetracube (talk) 19:20, 23 December 2008 (UTC)


 * A separate article makes more sense, even to the degree of giving coordinates there by section, and referencing in each polytope article. Tom Ruen (talk) 19:32, 23 December 2008 (UTC)


 * OK, I've made a draft start here: User:Tetracube/Coordinates of uniform polytopes. Still unsure how to list the coordinates, perhaps using tables to keep the page length sane?&mdash;Tetracube (talk) 23:04, 23 December 2008 (UTC)