# How do I figure out the features of a cube in 4 dimensions?

This is a question from Gilbert Strang's Introduction to Linear Algebra and MIT OCW

How many corners does a cube have in $4$ dimensions? How many $3$D faces? How many edges? A typical corner is $(0,0,1,0)$. A typical edge goes to $(0,1,0,0)$.

I know that it will have $16$ corners, but I don't know how to figure out the rest.

So far, I have only learned a little about vectors, their components, and linear combinations. How can I apply this knowledge to figure out the features of this cube? The worked example in the book (for this problem) uses matrices and it is also in the middle of the next chapter.

• recommend Regular Polytopes by Coxeter. – Will Jagy Sep 24 '15 at 0:53

Here's a picture of the three-dimensional cube.

You're right about the vertices. What do you notice about the edges? In particular, can you split them into parallel families? What do you notice about the coordinates of points that are connected by an edge?

What do you notice about the faces? Can you split them into parallel faces? Do their points satisfy any nice equations, here involving $x_1,\,x_2,$ and $x_3$?

There are more combinatorial hints to be had as well; we don't really need vectors to think about cubes (I personally think it's more difficult with vectors and actual coordinates for vertices, but that's just me)

EDIT:

OK, so the big idea, if we'd like to parameterize (and therefore count) faces, is dimension. Roughly, we'll think of it as how many degrees of freedom we have. What does that mean? Here's an example.

Let's pick the top line in the sample $2$-face. Let's call it $(*, 1, 1)$, because its end points have coordinates $x_2 = 1 = x_3$; we get to choose either $0$ or $1$ for $x_1$. We have one degree of freedom, which agrees with lines being one-dimensional.

If we were to count edges of the three-dimensional cube this way, then, we have to:

• Pick the coordinate we'll use an $*$ in; we have ${3 \choose 1} = 3$ choices there.

• We also have to pick what we'll make our remaining $3 - 1$ coordinates; we have $2^{3 - 1} = 2^2 = 4$ choices here, since for the $3 - 1$ coordinates, we're choosing between $0$ or $1$.

Thus, we have $3 \cdot 4 = 12$ edges of the one dimensional cube. The fancy way to write this would be to say that we have

$${3 \choose 1} \cdot 2^{3 - 1}$$

faces of dimension $1$ in a $3$-dimensional cube.

How about writing down a two-dimensional face? Well, now we have to choose the spots we'll put two asterisks, and then choose the remaining coordinate, for a total of ${3 \choose 2} \cdot 2^{3 - 2} = 3 \cdot 2 = 6$ faces.

For a four-dimensional cube, the same applies: To specify a $k$-dimensional face, we have to choose where we're going to put $k$ asterisks, and then choose the remaining $4 - k$ coordinates.

Alternatively, and just for fun, you can build cubes inductively (the original post I had intended, but rejected in favor of thinking in terms of coordinates and equations). This utilizes a different style of thinking, it's much more recursive.

• The coordinates of points that are connected by an edge have $0$'s in them. That's about all I can see. – Cherry_Developer Sep 24 '15 at 1:50
• Well, most points have $0$'s in the coordinates; all but one, in fact :) Let's pick the top edge of the "Sample $2$-face". I would call that line $(*, 1, 1)$, since both endpoints have $x_2 = 1 = x_3$. The left vertical line of that same face I would call $(1, 1, *)$ for similar reasons. – pjs36 Sep 24 '15 at 1:54
• What can I do with this information? I feel clueless. – Cherry_Developer Sep 24 '15 at 2:24
• I understand, I've added some info that should give you direction. I've stared at cubes at least a dozen times thinking about how to understand and give notation to the structure; it takes time to digest everything. – pjs36 Sep 24 '15 at 2:47

A corner is any of $\{0,1\}^4$ so $16$.

An edge is any pair of vertices such that they only differ in $1$ coordinate so $32$.

A face is any set of $4$ vertices that only differ in $2$ coordinates so $16$.

An 'inner cube' is any set of $8$ vertices that only differ in $3$ coordinates so $8$.

• Can you please explain how you deduced that a corner is any of $\{0,1\}^{4}$"? – letsmakemuffinstogether Sep 24 '15 at 1:56
• @letsmakemuffinstogether; like (0,0,0,0), (0,0,0,1),...,(1,1,1,1)? – JMP Sep 24 '15 at 1:58
• Yes, please explain the reasoning. Is it merely combinatoric reasoning because there are 2 choices for each coordinate? – letsmakemuffinstogether Sep 24 '15 at 2:00
• @letsmakemuffinstogether; or inductive logic from a point to line to square to cube to 4-cube – JMP Sep 24 '15 at 2:02
• Would you please be able to write out how you got to those results. I am trying to picture it and understand what you mean, but I'm having trouble with it :( – Cherry_Developer Sep 24 '15 at 2:07

If n is the number of dimensions then each corner has d dimensions. For a unit length cube the coordinate value in each dimension may be either 0 or 1. So the number of corners is the number of possible combinations of 0 or 1 in n dimensions which is simply 2^n. Let the number of corners be c so c = 2^n.

From each corner there is an edge extending in each dimension giving cn. However, each edge is shared by 2 corners, regardless of the dimensions => e = cn/2.