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Given the $n$-cube in $\mathbb{E}^n$ with vertices $(\pm1,\dotsc,\pm1)$, I'd like to find the equations for the hyperplanes in $\mathbb{E}^n$ that are the mirrors of reflection symmetries of the cube.

Essentially, given an $n$-dimensional cube, there are $n$ hyperplanes such that reflection through these planes generates the entire automorphism group of the $n$-cube. They correspond to a reflection through a vertex, an edge, a face, and so on throughout the dimensions. For example, the $1$-cube is just the line segment $[-1,1]$. Then the hyperplane is just the point $x=0$. When $n=2$, we just have a square and the two hyperplanes are the lines $y=x$ and $x=0$.

For the $3$-cube, I calculated the automorphisms that generate the automorphism group explicitly and then used those to determine the hyperplane equations. I found that the three planes are $x=0, y=x,$ and $z=y$.

For any more dimensions, I wasn't able to work it out explicitly and so all I have is a conjecture based on the previous work. The same equations are used as we move up in dimension, just adding a single new equation each time in the newest variable. My conjecture is that given variables $(x_1,\dotsc,x_n)$, then the equations of the hyperplanes of reflection symmetries will be $$ \begin{align*} x_1 &= 0 \\ x_2&= x_1\\ &\vdots \\x_n &= x_{n-1} \end{align*} $$ But I'm not sure how to go about proving that this is the case. Does anyone have any suggestions?

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Surely $x=z$ must be a valid hyperplane in the $3$-cube case? In general I'd expect to see all $x_i=x_j$ (for $i\ne j$), seeing as the cube is invariant under re-ordering the coordinates. –  Matt Pressland Dec 10 '12 at 14:43
    
I'm looking for the three that generate every other hyperplane. –  chris Dec 10 '12 at 14:49
    
Oh, sorry, you did say that. (Although if I'm being pedantic, you shouldn't really say "the", there are other choices of generators). –  Matt Pressland Dec 10 '12 at 15:01

1 Answer 1

up vote 2 down vote accepted

These are the mirrors of the hyperoctahedral group. There are $n^2$ mirrors in total divided into two orbits: $n(n-1)$ mirrors of the form $x_j \pm x_k = 0$ for $1 \leq j < k \leq n$ and an additional $n$ mirrors of the form $x_k = 0$.

Indeed, the full group is generated by $n$ reflections: $n-1$ reflections in the mirrors $x_k - x_{k+1} = 0$ for $k < n$ together with the reflection in $x_n = 0$.

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Perfect. I didn't realize there was a name for cubes and cross-polytopes together, but that makes sense because they're dual. Thanks. –  chris Dec 10 '12 at 15:13

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