## Subspace of $\mathbb{R}^n$ spanned by the image of convex $(n-1)$-polyhedra under the face-counting map

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Fix $n \in \mathbb{N}$. A convex polyhedron $C$ in $\mathbb{R}^n$ is the convex hull of finitely many points with nonempty interior. For $H$ a supporting hyperplane, ie $C$ is contained in one of the two closed half-spaces bounded by $H$, we call $H \cap C$ a $j$-face of $C$, where $j$ is the affine dimension of $H \cap C$. By convention, $\varnothing$ is called a $-1$-face of $C$ and $C$ an $n$-face of itself.

Define a function $F$ from the set of convex polyhedra to $\mathbb{R}^{n+2}$ by coordinates, so that $F(C) = (a^C_{-1}, ..., a^C_n)$, where $a^C_j$ is the number of $j$-faces of $C$ for $j=-1,...,n$. Let $W$ be the affine subspace of $\mathbb{R}^{n+2}$ generated by $\operatorname{im} F$.

It's clear that $a^C_{-1}=1$ and $a^C_n=1$. Euler's formula $\displaystyle \sum_{j=-1}^n (-1)^j a^C_j = 0$ (which may be more familiar as the Euler characteristic $V+E-F=2$ in the case of $n=3$) is a third affine relation between the $a^C_j$'s. Hence, $\operatorname{dim}W \le n-1$.

Is it always true for any n that $\operatorname{dim}W = n-1$? Put differently, for any $n$, are the three equations above the only affine relationships that must be satisfied by $a^C_j$'s for all convex polyhedra $C \subset \mathbb{R}^n$, or is there some $n$ in which there is another relation?

I seem to recall an affirmative answer to this, but I can't remember how it was solved or where I found it.

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## 1 Answer

Yes this is always true. One just needs to exhibit enough polyhedra so that the span of their f-vectors is $n-1$ dimensional. Such a family is given by the polyhedra $\Delta^k\times I^{n-k}$, where $\Delta^k$ is the $k$-simplex and $I^{k}$ is the $k$-cube.

The same argument can be used that the corresponding dimension for simplicial polyhedra is $\lfloor \frac{n}{2}\rfloor +1$, and so the only affine relations are the Dehn-Sommerville relations. One looks at the family $\Delta^k\times \Delta^{n-k}$, $k=0,1,\dots,\lfloor\frac{n}{2}\rfloor$.

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Excellent! The answer is both easy to understand and to verify. I'm a bit upset I couldn't think of it myself. – The Cheese Stands Alone Mar 17 2011 at 12:06
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