Existence and unicity of a complete bounded cell in a generic hyperplane arrangement.

Question

Let $n>d$ be integers and $H_1,\ldots,H_n$ be hyperplanes in $\mathbb{R}^d$ in generic position. By generic position I mean that if we change slightly their position, then the configuration does not change, or equivalently $\dim H_{i_1}\cap\cdots\cap H_{i_k} =d-k$ when $k\leq d$ and the intersection is empty if $k>d$. We call cells the connected components of $\mathbb{R}^d\setminus \bigcup_i H_i$. We say that a cell is complete if it has $n$ facets. I have two questions:

1. Does it exist always a complete bounded cell?
2. If there exist a bounded complete cell, is it unique?

Edit: The answer to both question is no. I give the counter examples in the answer section.

Edit 2: I let here the ideas I was following when I tried to prove uniqueness. It might be a useful point of view for related questions.

For any subset $I=\{i_1,\ldots,i_{d+1}\}\subset[[1,n]]$, there is a unique simplexe $\Delta_I$ whose facets are contained in the hyperplanes $H_{i_1},\ldots,H_{i_{d+1}}$. The collection of simplices $\Delta_I$ together with the inclusion form a poset $(\mathcal{P},\subset)$. We will note $\max\mathcal{P}$ the set of maximal elements of $\mathcal{P}$.

I think one can prove that if $H_1,\ldots,H_n$ contains the $n$ facets of a polytope $P$ (a complete bounded cell), then $P=\bigcap_{\Delta\in\max\mathcal{P}} \Delta$. If this is true then we have proved the unicity.

I think I have a proof for $\bigcap_{\Delta\in\max\mathcal{P}}\Delta\subset P$. I am now searching for the reverse inclusion.

Answer 1

The answer to both questions is no.

For the first, consider the following arrangement of 5 lines in $\mathbb{R}^2$. None of the bounded cells has $5$ facets.

For the second, consider the following situation with $7$ planes in $\mathbb{R}^3$. We cut a cube by an hyperplane passing by the middle point of $6$ edges as in the picture. It separates the cube in two polytopes with $7$ facets each. They are both complete bounded cells of the arrangements of the hyperplanes supporting the face of the cube and the hyperplane intersecting the cube. This configuration is not generic because there are parrallele hyperplanes. But if one moves a bit the hyperplanes then this becomes a generic position and I still have $2$ bounded complete cells.

[Edit] There exist a exemple even more simple then the previous one. To have a counter example we need to find a polytope with $(n-1)$ facet and a plane meeting all the facets. In the previous example $n$ was $7$, but we can low that to $n=5$: