Is every polytope combinatorially equivalent to the intersection of a simplex and a linear subspace?

Question

I wonder whether such a result is known, and if so, whether the proof is trivial.

By polytope I mean the convex hull of finitely many points in $\Bbb R^n$. Assume the simplex to be symmetric and centered at the origin, so that the subspace goes through its center. The subspace can have any dimension $k\in\{0,...,n\}$.

Question: Is every polytope combinatorially equivalent to the intersection of a simplex and a linear subspace through its center?

If this is true, are there estimations for the dimension of the simplex needed?

Answer 1

The answer is yes.

The fact that any polytope is affinely equivalent to a section of a simplex is well-known (see the answer by Tobias Fritz).

Now in any simplex with vertices $(v_i)$ we may consider projective transformations via reweighting barycentric coordinates: given positive numbers $(a_i)$, such a transformation is given by $$ \sum \lambda_i v_i \mapsto \frac{\sum a_i \lambda_i v_i}{\sum a_i \lambda_i}.$$

Projective transformations act transitively on the interior of the simplex, so it follows that any polytope is combinatorially equivalent to a section of some regular simplex through its center.

What I don't know is whether every polytope is affinely equivalent to a section of some regular simplex through its center.

Answer 2

Update: This answers the question without the extra condition on the subspace passing through the centre.

The answer is yes. The following stronger statement is an elementary fact of polyhedral geometry:

Every polytope is affinely isomorphic to the intersection of a simplex with an affine subspace.

One of the most standard useful facts of polyhedral geometry is that every polytope $P$ is the projection of a simplex. This is very simple to see: if $(v_i)_{i=1,\ldots,m}$ with $v_i\in\mathbb{R}^n$ are the vertices of $P$, then $P$ consists of precisely those points $x\in\mathbb{R}^m$ which can be written in the form $$x = \sum_{i=1}^m c_i v_i, \qquad \sum_{i=1}^m c_i = 1, \qquad c_i \geq 0.$$ Here, the simplex is the standard simplex in $\mathbb{R}^m$, as per the second and third equation. The first equation defines the linear projection from $\mathbb{R}^m$ down to $\mathbb{R}^n$. So the vertices of the simplex map bijectively to the vertices of $P$, and this assignment extends uniquely to an affine map.

Now let's see what happens to this statement under duality. The dual of a simplex is again a simplex, and the dual of a surjective affine map is an injective affine map. Hence we prove the claim upon using $P^{\vee\vee} = P$ together with the fact that $P^{\vee}$ is the projection of a simplex.

We can also use similar reasoning as above to write $P$ directly as the intersection of a positive orthant with an affine subspace, which is easily seen to be equivalent to the claim. This time, we should use the $H$-representation of $P$. So there is a matrix $A\in\mathbb{R}^{k\times n}$ and a vector $b\in\mathbb{R}^k$ such that $P$ is the set of solutions of the linear system $$A x - b \geq 0.$$ If $P$ is full-dimensional, then $k$ is the number of facets. This inequality shows that $P$ is the intersection of the positive orthant $\mathbb{R}_+^k$ with the affine subspace parametrized by $Ax - b$.