For one dimension less for both parties, I can see that it is a regular hexagon. But I can't even imagine what kind of 3d space is it that can be orthogonal to (1,1,1,1). Is it a cube? Sphere? Can it be anything? Then, what do we obtain if we intersect them?
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Can't give a full answer, but check out https://en.wikipedia.org/wiki/Hyperplane and https://math.stackexchange.com/questions/584006/intersection-of-hypercube-and-hyperplane-features-of-resulting-polytope and https://en.wikipedia.org/wiki/Truncated_tetrahedron – Chris Culter Jun 01 '19 at 07:44
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The shape one obtains is an octahedron. On way to see this is to project $\Bbb R^4\to \Bbb R^3$ parallel to the final coordinate axis. Since for the shape considered the sum of the coordinates is $0$, we can reconstruct the forgotten coordinate of $(x,y,z)$ as $-x-y-z$. So the shape obtained is $$ \{\,(x,y,z)\in[-1,1]^3\mid 1\geq x+y+z\geq-1\,\} $$ This is a cube truncated by two parallel planes, each passing through $3$ vertices of the cube. It is not hard to see that this cuts along two (equilateral) triangles, and that each of the $6$ faces of the cube is cut in half, leaving one (non equilateral) triangle, and the shape is a (non regular) octahedron. The non regularity is due to the fact that restricted to the plane orthogonal to $(1,1,1,1)$, our projection is not an isometry. This can be repaired by applying to our truncated cube a linear transformation that sends the standard basis of $\Bbb R^3$ to three of the vertices of a regular tetrahedron centred at the origin. I'm pretty sure this transformation maps the octahedron to a regular one.
Marc van Leeuwen
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