# Slicing a tesseract

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

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.