2
$\begingroup$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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 $\Delta^k\times \Delta^{n-k}$, $k=0,1,\dots,\lfloor\frac{n}{2}\rfloor$.

$\endgroup$
1
  • $\begingroup$ Excellent! The answer is both easy to understand and to verify. I'm a bit upset I couldn't think of it myself. $\endgroup$
    – Logan M
    Mar 17, 2011 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.