# The intersection of the facets on the polyhedron

Given a point set $$E=\{\alpha_j\}_{j=1}^m\subset \mathbb{N}^{n}$$ ($$1\leq m< \infty$$).

Define the polyhedron $$\mathcal{N}(E)$$ to be the convex hull of the set $$\begin{equation*} \bigcup_{j=1}^m \left(\alpha_j + \mathbb{R}^n_{+}\right), \end{equation*}$$ where $$\mathbb{R}^n_+=\{x\in \mathbb{R}^n:x_j\geq 0, \, 1 \leq j\leq n\}$$. If there are $$n$$ facet (the face has dimension $$n-1$$) $$F_1,\,F_2,\,\cdots,\,F_n$$ on $$\mathcal{N}(E)$$ satisfying $$\bigcap_{j=1}^n F_j\neq \emptyset.$$

How to prove that for $$n\geq 3$$, if the intersection of $$n$$ facets is nonempty, then it must be a single point?

For a general convex polytopes, this claim is not feasible. For a 4-dimensional example, see Show that each edge of the cyclic polytope $C_4(6)$ is contained in either three or four facets, and either three or four 2-faces.

I think the concepts of redundant face maybe helpful. And I can only prove the case $$n=3$$, in the following answer (I post it to avoid a length question) I show if three facets intersect, then they must coincide with a single point.

• Is it true that $\mathcal{N}(E) = \mathbb{R}^n_{+} + \Gamma(E)$ where $\Gamma(E)$ is convex hull of $E?$
– dsh
Oct 8, 2023 at 11:06
• yeah and we can assume that $E$ is finite.
– cbi
Oct 8, 2023 at 11:07
• Then, I guess, you can find all facets of $\mathcal{N}(E)$ from facets of $\Gamma(E)$ and $\mathbb{R}^n_{+}$ and maybe apply counterexample you provided.
– dsh
Oct 8, 2023 at 11:09
• It is different. That counterexample is a bounded closed polytope.
– cbi
Oct 8, 2023 at 11:12
• Thanks, I thought facets $\mathcal{N}(E)$ is union of some subset of facets of $\Gamma(E)$ (which is also bounded polytope) and of some subset of facets of $\mathbb{R}^n_{+}.$
– dsh
Oct 8, 2023 at 11:16

This answer is only for the case $$n=3$$.
Since the affine hull of a facet is a 2-dimensional plane, there exist $$w_j\in \mathbb{R}^n$$ ($$n=3$$) such that the equation of these (support) plane is $$w_{j}\cdot x=1$$, $$j=1,2,3$$, with $$\mathcal{N}(E)\subset\bigcap_{j=1}^3\{x:w_j\cdot x\geq 1\}.$$ In our framework, one can easily show that $$w_j\in \mathbb{R}^{n}_+$$ (in higher dimension this fact is also true). Then our claim is equivalent to the linear independence of $$\{w_1,w_2,w_3\}$$. If they are not linear independent, we show there is a contradiction. Firstly, we can assume that there exist $$\lambda_j$$, $$j=1,2$$ such that $$w_3=\lambda_1 w_1+\lambda_2 w_2.$$ Let $$\beta\in \cap_{j=1}^3 F_j$$ is a point on $$\mathcal{N}(E)$$. Then $$\beta\cdot w_j=1$$. Thus $$\lambda_1+\lambda_2=1$$. Without loss of generality, we can always assume that $$\lambda_1,\lambda_2$$ are non-negative. (Indeed, if there is a negative, say $$\lambda_1$$, then we rewrite the relation of $$w_j$$ as $$w_2=\frac{1}{\lambda_2}(w_3-\lambda_1 w_1)=:\mu_1 w_3+\mu_2 w_1$$
with $$\mu_j$$ non-negative.)
Now we assume $$\lambda_j$$ are non-negative, however, since $$w_3$$ is the normal vector of the (proper) facet $$F_3$$, we can choose $$\alpha\in F_3$$ and $$\alpha$$ not belong to $$F_1,F_2$$, then $$w_j\cdot\alpha>1$$ for $$j=1,2$$. And $$1=w_3\cdot\alpha=\lambda_1 w_1\cdot \alpha+\lambda_2 w_2\cdot \alpha>\lambda_1+\lambda_2=1.$$ A contradiction.