Proof of the following: How many $(n-2)$ dimensional faces from a corner of a hypercube

Question

I asked a question earlier regarding the number of $(n-2)$ dimensional faces exiting a corner of an $n$ dimensional hypercube. (For example the number of points in a corner of a square, or the number of edges in the corner of a cube, the number of planes in a corner of a hypercube etc...)

My gut feeling was that the answer is $n$ choose 2 and this is because there are exactly $n$ $n-1$ dimensional hyperfaces meeting at the corner of an $n$ dimensional hypercube and any combination of two of them (i believe) yields a valid $(n-2)$ dimensional face. For example,

Given a cube there are 3 faces meeting at the corner of the cube and therefore there are 3 choose 2 = 3 edges exiting a corner.

For a square there are 2 edges meeting at the corner and there are therefore 2 choose 2 = 1 point in the corner.

However, I did recieve an answer of $2n - 3$ $(n-2)$ dimensional components. This was without explanation but curiously enough:

$$2(3) - 3 = 6 - 3 = 3$$ $$2(2) - 3 = 4 - 3 = 1$$

So this answer appears to work as well for dimensions 2 and 3. Unfortunately I know not how to check for dimension 4. Can somebody answer this with a proof of which formula is correct and why?

Answer 1

Consider the $n$-dimension hypercube with vertices at $(\pm x,\pm x,\ldots,\pm x)$. Its volume $V_n(x)$ is $(2x)^n$. Imagine we expand the hypercube a little bit from $x$ to $x + \delta$, each $m$-dimension face of the hypercube will contribute an extra hypervolume $V_m(x) \delta^{n-m}$ to the new hypercube. This means:

$$V_{n}(x+\delta) = \sum_{m=0}^n N_{n,m} V_{m}(x)\delta^{n-m}$$ where $N_{n,m}$ is the number of $m$-dimension "faces" of a $n$-dimension hypercube.

One the other hand, binomial theorem tell us:

$$V_n(x+\delta) = 2^n(x+\delta)^n = 2^n\sum_{m=0}^n \binom{n}{m} x^{m} \delta^{n-m} =\sum_{m=0}^n 2^{n-m}\binom{n}{m}V_{m}(x)\delta^{n-m} $$

By comparing the coefficients of powers in $\delta$, we immediately get:

$$N_{n,m} = 2^{n-m} \binom{n}{m}$$

In particular,

$$N_{n,n-2} = 2^2 \binom{n}{n-2} = 2^2 \binom{n}{2} = 2n(n-1)$$

Update

Let $N_{n,m}^c$ be the number of $m$-dimension face in contact to a corner in a $n$-dimension hypercube. If you compare the volume of the hypercube $[0,x]^n$ with that of $[0,x+\delta]^n$ like above, you will obtain:

$$N_{n,m}^c = \binom{n}{m}$$

An alternate combinatorial argument go like this. Since a $n$-dimension hypercube has $N_{n,0} = 2^n$ vertices. There are $2^n N_{n,m}^c$ ways of picking a corner and then a $m$-dimension face in contact to that corner. We can arrive the same count by picking the $m$-dimension face first. As a result, we again get:

$$2^n N_{n,m}^c = 2^m N_{n,m} \quad\implies\quad N_{n,m}^{c} = 2^{m-n} N_{n,m} = \binom{n}{m}$$

BTW, you $n$ choose $2$ argument also works. In any event,

$$N_{n,n-2}^c = \binom{n}{n-2} = \binom{n}{2} = \frac{n(n-1)}{2}$$

For tesseract, this is $6 \ne 2 n - 3$. Let's say your tesseract is $\{\;(x,y,z,t) : 0\le x,y,z,t \le 1\;\}$. The six $2$-dimension faces in contact to the origin are:

$$\begin{cases} x = y = 0\text{ face} &\to&\{0\} &\times& \{0\} &\times& [0,1] &\times& [0,1]\\ x = z = 0\text{ face} &\to&\{0\} &\times& [0,1] &\times& \{0\} &\times& [0,1]\\ x = t = 0\text{ face} &\to&\{0\} &\times& [0,1] &\times& [0,1] &\times& \{0\}\\ y = z = 0\text{ face} &\to&[0,1] &\times& \{0\} &\times& \{0\} &\times& [0,1]\\ y = t = 0\text{ face} &\to&[0,1] &\times& \{0\} &\times& [0,1] &\times& \{0\}\\ z = t = 0\text{ face} &\to&[0,1] &\times& [0,1] &\times& \{0\} &\times& \{0\} \end{cases}$$

Answer 2

Another process is to look at what is called the 'vertex-figure', or arrangement of elements at a vertex. This is pretty much the usual way of doing this.

For the tesseract, and kindred figures, this is a simplex of n-1 dimensions. The 2d elements appear as lines. So it's the same as counting the lines of a tetrahedron, ie

  1  tetrahedron      becomes   body of cube         * 16/16  = 1
  4  triangles        become    four cube-faces      * 16/8   = 8
  6  edges            become    six squares          * 16/4   =24
  4  vertices         become    four edges           * 16/2   =32
  1  (-1)             becomes   the vertex itself    * 16/1   =16

When one multiplies all of these numbers by 16, and divides out by the number of vertices the elements have (ie 16, 8, 4, 2, 1 for the list here), one sees that every element of the tesseract is counted. This is done in the last column.

This is often the usual way one counts the surface consist of a polytope from a vertex.