When is a polyhedron uniquely determined by its projections? Let $P$ denote a polyhedron in $\mathbb{R}^D$ defined by the intersection of $k$ halfspaces, so $P = \{x\in\mathbb{R}^D : Ax\le b\}$ for $A\in\mathbb{R}^{k\times D}$, $b\in\mathbb{R}^k$. For a subset of coordinates $S\subseteq\{1,\ldots, D\}$, let $P_S$ denote the projection of $P$ onto $\mathrm{span}(\{e_i : i\in S\})$. I want to know when $P$ is uniquely determined by projections of the form $P_S$. If enforcing that $P$ is bounded makes things simpler, that is fine as well.
For concreteness, I am interested in statements of the form "If $P$ is the intersection of $k$ halfspaces in $D$ dimensions, then it is uniquely determined by projections of the form $P_S$ for $|S|\le m$", where $m$ is something that depends on $k$ and $D$.
Klee's theorem seems relevant (see, e.g., https://mathoverflow.net/questions/15612/do-plane-projections-determine-a-convex-polytope) but it does not say anything about uniqueness.
 A: If $P$ is the intersection of $k > 0$ halfspaces in $D > 0$ dimensions, then it is not necessarily uniquely determined by all projections of the form $P_S$ for $|S| < D$. This still holds true even if $P$ is restricted to be bounded.
Counterexample for the unbounded case: Choose $k$ distinct unit vectors $v_1, \ldots, v_k$ all of whose coordinates are positive. Let $$P := \bigcap_{i=1}^k\{x \in \mathbb R^d: \langle x, v_i\rangle \leq 1\}$$ be a convex polyhedron. Then $P_{|S|} = \mathbb R^{|S|}$ for all $S \subsetneq \{1, \ldots, D\}$. A different choice of $v_i$ changes $P$, but yields the same projections.
Counterexample for the bounded case: First, $k > D$ is required for $P$ to exist.
Let $P_0$ be the convex hull of $\{0, e_1, \ldots, e_D\}$, let $v$ be any vector in the interior of $P_0$. Then $P_v$, defined as the convex hull of $\{v, e_1, \ldots, e_D\}$, has the same projections as $P_0$. Both $P_v$ and $P_0$ are simplices definable as the intersection of $D+1 \leq k$ hyperplanes.
