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.—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.—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.—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?—Tetracube (talk) 23:04, 23 December 2008 (UTC)