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A friend of mine was recently struggling with visualising a problem involving a cube that had been sliced into 2 equal parts, diagonally in all 3 dimensions. The result leaves a cut surface which is an equilateral triangle.

That got me thinking. If you cut a cube in two diagonally in all three dimensions, you get a 2D surface that's an equilateral triangle. So, if you cut a tesseract diagonally in all 4 dimensions, what is the cut "surface"?

Instinctively, I'm assuming the cut "surface" is actually a 3d shape, and at a guess it might be a tetrahedron, but I just don't have the maths to know how to work it out. Would love both answers and/or educated guesses!

Thanks

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Yes it is a tetrahedron. I would think about this by thinking of the cube in $n$ dimensions as embedded in $\mathbb{R}^n$, and think about the collection of vertices of the form $(0,...,0,1,0, ..., 0)$ where there is a $1$ in the $i$th entry and all the rest are $0$. This is the collection of vertices that intersects the plane $\sum x_i = 1$. If I connect all these vertices, what shape do I get? Well, we can prove it's a regular polyhedron easily: it is invariant under all permutations of the $x_i$. But in fact, this equation is already the defining equation of the regular $n$-simplex. In dimension $3$, this is a triangle. To think about how to extend it up to the next dimension, note that each point of this triangle also satisfies the same equation with an $x_4 = 0$ included in the coordinates. This is the 'base' of the simplex in the next dimension up. By decreasing each of the other $x_i$ by a little bit, we can extend outwards in the $x_4$ direction until $x_1 = x_2 = x_3 = 0$ and $x_4 = 1$, and this is the new vertex of a tetrahedron.

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  • $\begingroup$ Fantastic. Thanks for the detailed explanation. I think I even understood it (mostly). $\endgroup$
    – xtempore
    Jun 16, 2022 at 5:18
  • $\begingroup$ If you have any questions about my answer, just feel free to ask. Otherwise, you're welcome! $\endgroup$ Jun 16, 2022 at 17:55

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