Now let’s start thinking about 4d Platonic solids. We’ve seen the 4-cube… what else is there? Well, in 3d we can take a cube and build an octahedron as shown here. The same trick works in any dimension. In *n* dimensions, we get something called the *n*-dimensional **cross-polytope**, which has one corner at the center of each (*n*-1)-dimensional face of the *n*-cube.

**Puzzle 1.** What’s a 2d cross-polytope?

It’s worth noting the relationship between cubes and cross-polytopes is symmetrical. In other words, we can also build an *n*-cube by putting one corner at the center of each (*n*-1)-dimensional face of the *n*-dimensional cross-polytope! For example:

It gets tiring to say ‘(*n*-1)-dimensional face’, so people use the word **facet** for an (*n*-1)-dimensional face of an *n*-dimensional thing.

Now let’s think a bit more carefully about what happens in 4 dimensions. Since the 4-cube has 8 facets (each being a cube), the 4d cross-polytope must have 8 corners. And since the 4-cube has 16 corners, the 4d cross-polytope must have 16 facets. This is why it’s also called the **16-cell**.

It also has other names. Amusingly, the Simple English Wikipedia says:

In four dimensional geometry, a

16-cellis a regular convex polychoron, or polytope existing in four dimensions. It is also known as thehexadecachoron. It is one of the six regular convex polychora first described by the Swiss mathematician Ludwig Schläfli in the mid-19th century. Conway calls it anorthoplexfor ‘orthant complex’, as well as the entire class of cross-polytopes.

Simple English, eh? That would really demoralize me if I were a non-native speaker.

### The 4d cross-polytope

But let’s sidestep the fancy words and think about what the 4d cross-polytope looks like. To draw a cross-polytope in *n* dimensions, we can draw the *n* coordinate axes and draw a dot one inch from the origin along each axis in each direction. Then connect each dot to every other one *except* the opposite one on the same axis. Then erase the coordinate axes.

In 3 dimensions you get this:

It may not look like much, but it’s a perspective picture of the vertices and edges of an octahedron, or 3d cross-polytope.

**Puzzle 2.** How many line segments going between red dots are in this picture? These are the edges of the 3d cross-polytope.

**Puzzle 3.** How many triangles with red corners can you see in this picture? These are the triangular faces of the 3d cross-polytope.

Now let’s do the same sort of thing in 4 dimensions! For this we can start with 4 axes in the plane, each at a 45° angle from the next. We can then draw a dot one inch from the origin along each axis in each direction… and connect each dot to each other *except* the opposite one on the same axis. We get this:

If we then erase the axes, we get this:

This a perspective picture of a 4d cross-polytope!

**Puzzle 4.** How many line segments going between red dots are in this picture? These are the edges of the 4d cross-polytope.

**Puzzle 5.** How many triangles with red corners can you see in this picture? These are the triangular 2-dimensional faces of the 4d cross-polytope.

Let’s say that 4d polytope has:

• 0-dimensional **vertices**,

• 1-dimensional **edges**,

• 2-dimensional **faces**, and

• 3-dimensional **facets**.

In general, the **facets** of an *n*-dimensional thing are its (*n*-1)-dimensional parts, while the parts of every dimension below *n* are often called **faces**. But in 4d we have enough words to be completely unambiguous, so let’s use the words as above. And in 3d, let’s use face in its traditional sense, to mean a 2d face.

So, as long as I talk only about 3d and 4d geometry, you can be sure that when I say **face** I mean a 2-dimensional face. When I say **facet**, I’ll mean a 3-dimensional face.

**Puzzle 6.** What shape are the facets of the 4d cross-polytope?

### 4-cube versus 4d cross-polytope

On top you see the 4-cube. At right, the 4d cross-polytope. Both are projected down to the plane in the same way.

So, the 4d cross-polytope has

vertices: one centered at each cubical facet of the 4-cube. To see how this works, mentally move the cross-polytope up and put it on top of the 4-cube.

On the other hand, the 4d cross-polytope has

^{4}= 16

facets: one for each corner of the 4-cube.

And this is a general pattern. As I already observed, the *n*-dimensional cross-polytope has one vertex in the middle of each facet of the *n*-cube, and vice versa. For this reason we say they are **Poincaré dual** to each other, or simply **dual**. The *n*-cube has

*n*

vertices and

^{n}

facets, but for the *n*-dimensional cross-polytope it’s the other way around.

### Figure credits and more

The picture of the octahedron in cube and cube in octahedron are from Frederick J. Goodman, who has written a book about this stuff called *Algebra: Abstract and Concrete*.

The other images are on Wikimedia Commons, and all have been released into the public domain except this one:

which was made by Markus Krötzsch.

For more on cross-polytopes, see this:

• Cross-polytope, Wikipedia.

“In general the faces of an n-dimensional thing are its (n-1)-dimensional parts, while the parts of every dimension below n are called facets.”

I am under the impression that the opposite convention is more common. It seems that “facet” is usually reserved for the (n-1)-dimensional faces?

Oh my gosh—you’re right!

I’ll change my article. Since I’ll mainly be talking about 3d and 4d polytopes, I’ll use

faceto mean a 2d facet in both cases (since in 3d geometry that’s traditional), andfacetto mean a 3d facet of a 4d polytope.(Don’t forget to also change the uses of “face” below that point! There are several that need fixing.)

Thanks! I’ve tried to fix the whole article.

Someday I should continue and finish this series. I never got too far into the promised land of 4d polytopes.

P1: A square. The Wikipedia link at the end kind of gives it away. ;-)

P2: 12. 6 vertices, each with 4 lines, with each shared by 2 vertices = 6 * 4 / 2 = 12.

P3: 8. 6 vertices, each with 4 triangles, with each shared by 3 triangles = 6 * 4 / 3 = 8. Though again, “octa”hedron kind of gives it away…

P4: 24. As in P2, 8 vertices, each with 6 lines, each shared by 2 vertices = 8 * 6 / 2 = 24.

P5: 32. This one’s a bit tricky, in that counting the triangles is not completely obvious. I broke it down by taking triangles from each vertex formed by edges 1 through 5 lines apart. But there are no triangles formed by lines 3 apart (since that would connect diagonally opposite vertices). So you end up with 5 (1 apart) + 4 (2 apart) + 0 (3 apart) + 2 (4 apart) + 1 (5 apart) = 12. Then, as in P3, 8 vertices, each with 12 triangles, each shared by 3 vertices = 8 * 12 / 3 = 32.

P6: Tetrahedron. The 3d extension of the 2d triangle.

Somewhat off topic: In reading further on this fascinating subject, I came across a notion I’d never heard of:

http://en.wikipedia.org/wiki/Abstract_polytope

This (in addition to being an interesting generalization) implicitly cleans up something that’s been bothering me for years:

https://en.wikipedia.org/wiki/Euler_characteristic

Namely, for convex polytopes, it’s 0 for even dimensions, but 2 for odd dimensions. But if you include the “greatest” rank (n, the whole object, of which there is always 1) and the “least” rank (-1, the “null face”, of which there is also always 1), then the “abstract Euler characteristic” (to coin a term) is 0 for convex polytopes in all dimensions.

Since a convex polytope is contractible, its Euler characteristic as a topological space is the same as that of a point—namely, 1. (The Euler characteristic of a finite set generalizes the cardinality of a finite set in this way.) To get this answer, you have to count the

n-dimensional ‘body’ of the polytope, and it gives If we leave it out, we get the Euler characteristic of the boundary of the polytope, namelyThere’s also a modified version of Euler characteristic called ‘reduced Euler characteristic’, where we subtract 1. If we calculate this for a point we get 0, and similarly that’s what we get for a convex polytope.

In future parts of “Symmetry and the Fourth Dimension” will you be discussing the quaternion generators for the 4-D regular polytopes, comparing the SO(4) generators to the SU(2) x SU(2) generators, for example? For the 3-D regular polyhedrons, Coxeter gives three generators, i, j, and a quaternion in the imaginary prime in terms of cosines and sines of p, q, and h. Is there a use for these generators for the 4-D case?

Yes, I’ll be talking about quaternionic descriptions of 4d polytopes. For now, you can read my article Platonic solids in all dimensions. Make sure you make it down to the section ‘Platonic solids and the quaternions’.

[…] Puteți afla mai multe despre acest lucru în seria mea “solide platonice și a patra dimensiune”: partea 1, partea 2, partea 3, partea 4, partea 5, Partea 6, partea 7, parte 8, parte 9, parte 10, parte 11 , parte 12, și parte 13. […]

It’s been a long time since this installment of the series. I fear it might have died before cutting deeper into the meaty parts of 4D geometry. Is there hope to taste those in this series?

I feel a bit guilty about this. I’d planned on a lot more, but then I lost energy. I can’t promise to revive this series. You can however, look at the series of shorter articles that it was supposed to expand on!

• John Baez, 4d.

They’re in reverse chronological order, so go to the bottom and climb back up!

Thank you for the pointers. Its a bummer, it would have been of great help understanding the finite subgroups of SU(2).

I wrote a little about 4d Platonic solids and finite subgroups of SU(2) here:

• Platonic solids in all dimensions: Platonic solids and the quaternions.

But there’s a lot more to say!