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Given an $n$-dimensional polytope $P\subset \mathbb{R}^n$.

For a subset of indices $i_1,\ldots,i_k$ of $\{1,\ldots,n\}$ and reals $a_1,\ldots,a_k$, we denote by $P(x_{i_1}=a_{1},\ldots,x_{i_k}=a_{k})$ the set of point of the polytope $P$ where coordinates $x_{i_j}$ are fixed to some constant $a_j$ (it is the intersection of $P$ with the hyperplanes of equations $x_{i_j}=a_{j}$). Let $f(a_1,\ldots,a_k)$ the $n-k$-volume of $P(x_{i_1}=a_{1},\ldots,x_{i_k}=a_{k})$. I think that this function is piecewise polynomial in the variables $a_j$ and that this function is continuous in its support (i.e. the set where it is not null). These two results seems quite elementary but I didn't find a reference nor a proof.

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Indeed, this it is trivially true for a simplex, and it is also true for a polytope since it is a disjoint union of finitely many simplexes --and a sum of piecewise polynomial functions is piecewise polynomial. – Pietro Majer Dec 19 '12 at 12:38
@Pietro: Your comment takes care of "piecewise polynomial" but what about "continuous on its support"? It seems to me that the volume functions of the simplices might have smaller supports than that of the whole polytope. I suppose one wants to say that discontinuities arise only when a face is parallel to the affine space used for cutting, and that the simplices can be arranged so that those discontinuities for the simplices occur only at the ends of the support (for the whole polytope), but it's not immediately obvious (to me) how to arrange that. – Andreas Blass Dec 19 '12 at 12:47
@Andreas The continuity follows from the following fact about convex polyhedra: the curvature measures of convex polyedra are continuous with respect to the Hausdorff measure; see e.g. "Introduction to geometric Probability" by Klain and Rota. The slices vary continuously and so will their volumes (which are special cases of curvature measure). – Liviu Nicolaescu Dec 19 '12 at 19:49
BTW, $($this function$)^{1/n-k}$ is concave in its support. – Anton Petrunin Dec 20 '12 at 2:41
I think Andreas' hint works with a generic triangulation of the polytope. I've added it as an answer, since it is a bit too long. No convexity is needed; though in the case of a convex polytope we can say that the possible discontinuities of $f$ on the $a$-space are locasted on the boundary of the support of $f$. – Pietro Majer Dec 21 '12 at 14:46

Thank you for your comments. Putting all the things together we have an answer. The proof for the piecewise polynomial part is given by Pietro (but not the continuity). The concavity of $f^{1/n-k}$ gives the continuity (at least in the interior of the polytope which is sufficient for me, today). This concavity is due to the so called Brunn-Minkowski inequality (and the function itself is unimodal) see the proof (for only one hyperplane) in a good reference for my problem : "On the sectional area of convex polytope" by David Avis et al. For your information, in dimension 3 they show that the function is piecewise quadratic using drums...

If you have a more direct answer, please don't hesitate to share it.

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Following Andreas Blass's hints, one possibility to show the continuity is the following. Let $d:=n-k$ be the dimension of the cutting affine spaces.

First, let's say that a face of a polytope is a bad face if it spans an affine space containing a $d$ dimensional subspace parallel to the cutting space; otherwise, call it a nice face. If all faces of a polytope are nice, let's say it's a nice polytope, after all.

For a nice simplex, one checks that the function $f$ is continuous on the whole $a$-space.

By genericity, any polytope can be subdivided into nice simplexes, plus a number of simplexes whose bad faces are included into some bad faces of the polytope.

In other words, a generic triangulation produces a simplicial complex where no bad faces are introduced. This shows that the points of discontinuity of the function $f$ of the polytope are exactly the projection in the $a$-space of its bad faces.

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