Determine the number of 1, 2 and 3-dimensional faces of the polytope $$S= conv \{(\pm 1, \pm 1, 0, 0), (0,0, \pm 1, \pm 1)\}$$ (the convex hull of 8 points in $\mathbb{R}^4$).
I've been struggling with this exercise... What I was able to show is that $S$ has 8 vertices (0-dimensional faces). I started counting the edges (1-dimensional faces) by enumerating the supporting hyperplanes which determine them, but it's a bit messy... I plan to use the Euler-Poincare formula at the end of the solution, but I don't really know what to do at this point.
You might have thought that the $4$ points $( \pm 1 , \pm 1, 0,0)$ form a square, they don't, they do form a $4$-cycle. In the same way a cuboctahedron contains a hexagon if you slice it in half (but does not have any hexagonal faces) so this object has $4$ cycles but no squares.
Pick any two adjacent vertices from $( \pm 1 , \pm 1, 0,0)$ and any two adjacent vertices from $( 0,0,\pm 1 , \pm 1)$ ... these will form a tetrahedron ... indeed there are $16$ such tetrahedra that are the "faces" (codimension $1$) of this object.
This is a $16$-cell ... check it out ... https://en.wikipedia.org/wiki/16-cell