Let $P$ be a simple convex polytope in $\mathbb{R}^d$ (that is, any vertex belongs to exactly $d+1$ facets). Given the collection of outer normals to facets of $P$, combinatorics of $P$ may be different, of course. But is this information enough to reconstruct number of faces of $P$ of all dimensions? If yes, what is specific procedure to do this? I am looking at what happened when we move facets parallel and pass through `singularity', i.e. not-simple polytope, and on first glance it looks that $f$-vector preserves, but I am always not sure with such things. And this is not good proof anyway, I would prefer more direct argument. 


If the statement is however false, I wonder for which specific collections it is still true. 



$f$-vector of simple convex polytope via directions of facets
Consider a simplex $S$ in five dimensions, and one more face normal $\pi$ in general position with respect to $S$ (no hyperplane perpendicular to $\pi$ passes through more than one vertex of $S$). Intersect $S$ with a halfspace $H$ perpendicular to $\pi$, and translate $H$ continuously towards $S$, starting from a position where it contains all of $S$.


Then when the intersection with $H$ first cuts off one vertex of $S$, the one cut-off vertex is replaced by five new vertices (on the edges of $S$ connecting the cut-off vertex to the others) so the total number of vertices becomes $6-1+5=10$. When the intersection with $H$ cuts off two vertices of $S$, they are replaced by eight new vertices (again, one for each edge of $S$ that connects one of the two cut-off vertices to the four remaining ones), so the total number of vertices becomes $6-2+8=12$. And when the intersection with $H$ cuts off three vertices of $S$, they are replaced by nine new vertices, so the total number of vertices becomes again $6-3+9=12$. All of these are simple $5$-polytopes.


So no, the $f$-vector is not preserved. Not even the number of vertices is preserved.


