If a polytope has $m$ faces in 3 dimensions, how many faces does its analogous polytope have in four dimensions? Does a formula exist?
For example, if $m=4$, you have a tetrahedron, and the 4-simplex has 10 faces. If $m=6$, you have a cube, and the tesseract has 24 faces. I know that formulas exist to find the number of m-cubes within the boundaries of an n cube, but what I'm asking is how to find the number of faces of any polytope given only $m$.
There are many different processes that let you go from a 3-dimensional polytope to an "analogous" 4-dimensional polytope. To get from a 3-simplex to a 4-simplex you take a join with a point outside the affine span of the 3-simplex. To get from a 3-cube to a 4-cube you take the product with a unit interval. To get from 3-dimensional octahedron to the 4-dimensional "generalised octahedron" that is the unit ball in $\Bbb{R}^4$ under the 1-norm, you take the join with 2 points outside the affine span of the 3-dimensional octahedron. These processes have a different effect on the number of faces, so there is no single formula to calculate the number of faces in the 4-dimensional polytope from the 3-dimensional polytope.