Understanding the 16-cell

[quickfur]

Some people probably already know this, but this has only occurred to me today, and it really helped me understand the structure of a 16-cell.

First of all, although I have quite a good grasp of the hypercube already, I've always had trouble with the 16-cell. Knowing that it was the dual of the hypercube didn't help, since I just couldn't visualize the resultant shape after swapping edges and faces. Looking at wireframe projections of the 16-cell didn't help very much either, there were too many intersecting lines inside the bounding octahedron, and I couldn't really see where the cells are.

However, today it suddenly occurred to me that you can construct a 16-cell by tapering an octahedron in the +W and -W directions. By the tapering of a 3D object X, I mean to stack increasingly smaller copies of X on top of each other along the W-axis until it has shrunk to a point, and taking the resulting 4D trace. (Is there a better term for this process?) Of course, I'm considering the case where the size of the object decreases linearly.

Suppose we take a 3D octahedron and taper it in the +W direction. As we move along the W axis, the 8 triangular faces of the shrinking octahedron traces out 8 tetrahedral cells, ending at the +W vertex of the 16-cell. This gives us half of the 16-cell. Now we repeat this process in the -W direction, and we obtain another 8 tetrahedral cells directly opposite the first 8 cells. They meet at the -W vertex of the 16-cell. Now we have constructed the entire 16-cell.

I thought this was a very neat way of understanding the 16-cell, because it generalizes all the way from 1D. If you start with the 1D line segment (-1,1) on the x-axis, tapering it in the +Y and -Y directions will give you a 2D diamond (a square rotated 45 degrees). Tapering this diamond in the +Z and -Z directions gives you the octahedron. Finally, tapering the octahedron in the +W and -W directions gives the 16-cell. I suppose repeating this process will yield the 5D cross polytope, but I think I'll leave it at that. :-)

[pat]

That is a good way to visualize the 16-cell. It is sort-of the dual of considering the 3-D cross-sections with our realm as the 16-cell is pushed along the W-axis.

[quickfur]

That is a good way to visualize the 16-cell. It is sort-of the dual of considering the 3-D cross-sections with our realm as the 16-cell is pushed along the W-axis.

Yeah, I guess I've seen the cross-sections of the 16-cell before, but it just never clicked in my mind until now.

On that note, any hints on visualizing the 24-cell? :-) I can sorta visualize it using the construction analogous to the rhombic dodecahedron (cut a tetracube into 8 cubical pyramids and attach them to the faces of another tetracube), but I still can't quite see how those 24 octahedra fit together.

[pat]

I don't have any big clues for the 24-cell. All I have is... here's a rotating version and the orangish shape near the center here is some of the cross-sections.

[Polycell]

Some people probably already know this, but this has only occurred to me today, and it really helped me understand the structure of a 16-cell.

First of all, although I have quite a good grasp of the hypercube already, I've always had trouble with the 16-cell. Knowing that it was the dual of the hypercube didn't help, since I just couldn't visualize the resultant shape after swapping edges and faces. Looking at wireframe projections of the 16-cell didn't help very much either, there were too many intersecting lines inside the bounding octahedron, and I couldn't really see where the cells are.

However, today it suddenly occurred to me that you can construct a 16-cell by tapering an octahedron in the +W and -W directions. By the tapering of a 3D object X, I mean to stack increasingly smaller copies of X on top of each other along the W-axis until it has shrunk to a point, and taking the resulting 4D trace. (Is there a better term for this process?) Of course, I'm considering the case where the size of the object decreases linearly.

Suppose we take a 3D octahedron and taper it in the +W direction. As we move along the W axis, the 8 triangular faces of the shrinking octahedron traces out 8 tetrahedral cells, ending at the +W vertex of the 16-cell. This gives us half of the 16-cell. Now we repeat this process in the -W direction, and we obtain another 8 tetrahedral cells directly opposite the first 8 cells. They meet at the -W vertex of the 16-cell. Now we have constructed the entire 16-cell.

I thought this was a very neat way of understanding the 16-cell, because it generalizes all the way from 1D. If you start with the 1D line segment (-1,1) on the x-axis, tapering it in the +Y and -Y directions will give you a 2D diamond (a square rotated 45 degrees). Tapering this diamond in the +Z and -Z directions gives you the octahedron. Finally, tapering the octahedron in the +W and -W directions gives the 16-cell. I suppose repeating this process will yield the 5D cross polytope, but I think I'll leave it at that. :-)

Yes, the HEXADECACHORON is an octahedral bipyramid. Indeed, an <i>n</i>-dimensional cross polytope is a bipyramid based on an (<i>n</i>-1)-dimensional cross polytope. This construction is described by H.S.M. Coxeter in <i>Regular Polytopes</i>. Likewise, an <i>n</i>-dimensional simplex is a pyramid based on an (<i>n</i>-1)-dimensional simplex, and an <i>n</i>-dimensional hypercube (or orthotope) is a prism based on an (<i>n</i>-1)-dimensional hypercube. These constructions guarantee the existence of these three regular polytopes in spaces of any dimension.

Note that an <i>n</i>-dimensional cross polytope (or orthoplex) has 2^<i>n</i> (<i>n</i>-1)-dimensional simplexes as cells. So in 24 dimensions, for example, the orthoplex comprises 16,777,216 23-dimensional simplexes. Things get pretty hairy as the number of dimensions increases, even for fairly low numbers such as 24.

[Polycell]

There is an "edit" button, you know.

[wendy]

Someone forgot to add in the octagonny, or o3x4x3o?

It is the only non-regular clifford group, that the cells are a clifford-tiling.

It also tiles space densely, the dual tiling is made of its isomorphs.

W

[wendy]

Another way to visualise the 16choron is to start with a cube. Inside this, one draws the two tetrahedra, that make the stella octagula.

For a given tetrahedron, there are four cube-corners that make a kind of squat pyramid. These are tetrahedra, that are inclined 60 degrees to the realm of view. There are four of these on the top, and four on the bottom.

The remaining six faces appear on the cube-faces. These are being looked on in a way that makes them entirely flat, so they appear in a hedrix (plane, lit 2-manifold). The six edges are the four square-edges, and the two diagonals. These faces have the same form as a tetrahedron projected into a square, and the third dimension is made by the perpedicular to the chorix (realm, lit 3-manifold).