It is a well known and lovely result that the maximum number of regions that $\mathbb R^{k}$ (with $k$ positive) can be divided into by $n$ hyperplanes is given by 
$$1+n+\binom{n}{2}+\cdots+\binom{n}{k}.
$$ and occurs when they are in general position. It is clear that the minimum with distinct hyperplanes is $n+1$ (when they are parallel ) and the next smallest is $2n$ when all but one are parallel and also when each pair has the same (nonempty ) intersection.



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> For which $n,k,m$ is it possible to divide $\mathbb R^{k}$ into exactly $m$ regions using $n$ distinct hyperplanes?
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Fir purposes of induction it might be better to drop the requirements that the hyperplanes be distinct.



Division of space by hyper-planes
Let's denote by $S\_{k,n}$ the set of possible integers $m$, such that $\mathbb R^k$ can be divided into $m$ regions by $n$ hyperplanes. If we denote by $S^{P}\_{k,n}$ the set defined similarly but for the projective space $\mathbb {RP}^k$. We have that $S\_{k,n}=S^P\_{k,n+1}$ since every affine arrangement can be lifted to a projective arrangement with the same number of regions by adding the hyperplane at infinity. Similarly any projective arrangement gives rise to affine arrangements with the same number of regions by deleting one of the hyperplanes. The following result solves the problem for $k=2$:



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> **Theorem:** (N. Manturov, ["Classification of arrangements by the number of their cells"](https://link.springer.com/article/10.1007/BF02189305)) We have $m\in S^P\_{2,n}$ if and only if there exists some integer $0\le r\le n-2$ such that
>  $$(n-r)(r+1)+\binom{r}{2}-\min\bigg\{ n-r, \binom{r}{2}\bigg\}\le m\le (n-r)(r+1)+\binom{r}{2}.$$ 
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The proof in that paper involves quite a bit of casework and it seems like it wouldn't easily generalize to higher dimensions, however I find one particular aspect very interesting. The way it is shown that all numbers satisfying these inequalities do work is by exhibiting a very simple family of arrangements $\mathbb B^c\_{a,b}$ which are obtained as the union of a pencil of $a$ lines (all passing through the same point), a simple arrangement of $b$ lines (no three lines concurrent), such that $c$ of the intersection points of the simple arrangement lie on the first pencil. This leads us to a (possibly too optimistic?) conjecture:



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> **Conjecture:** If an arrangement of $n$-hyperplanes in $\mathbb {RP}^k$ has $m$ regions, then there exists an arrangement of $n$-hyperplanes, obtained as the union of $n$ generalized pencils (one for each type), which also has $m$ regions.
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By a generalized pencil, I mean taking two disjoint projective plane planes of dimension $r\_1,r\_2$ in a $(r\_1+r\_2+1)$-dimensional projective space, picking a generic arrangement of hyperplanes in the first space, and taking the span of each hyperplane with the second projective space. We say that the pencil has type $(r\_1,r\_2)$. If this conjecture were to be true, there would be some hope of a proof strategy along the lines of starting with an arbitrary arrangement and somehow deforming it into such a form while preserving the number of regions.


