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3 updated and made more precise to address the criticism

(updated; apologies for way too much room left for interpretation in the original post)

More precisely, $\sigma(c)$ records the order of (Euclidean) distances $d(x,e_k)$ at which the standard basis vectors $e_k$ are from every $x\in c$. E.g. for $\sigma(c)=()=12 \dots n$, the identity permutation, one has $d(x,e_1)$<$d(x,e_2)$<...<$d(x,e_n)$, and the arrangement hyperplanes on the boundary of this $c$ are $H_{k,k+1}$ for $k=1,\dots ,n -1$.

Let $\ell$ be a general position line in $H$. Then $\ell$ intersects $\binom{n}{2}+1$ cells. Two of these intersections are unbounded; let us denote the corresponding cells $c_0$ and $c_t$, respectively. As we follow $\ell$ from $c_0$ to $c_t$, from one cell to the next, the corresponding permutation $\sigma(c)$ \sigma(c) = i_1 i_2...i_n$changes from$\sigma(c_0)$to$\sigma(c_t)$; namely, it gets multiplied by$(i_k,j_k)$(i_k,i_{k+1})$ whenever we cross $H_{i_k,j_k}$. H_{i_k,i_{k+1}}$. One can also view this as flipping the adjacent entries, i.e. applying the transposition$(k,k+1)$to the sequence$\sigma(c)$. Assume that$\sigma(c_0)$is the identity permutation. Then$\ell$gives us an expression for specifies a sequence of flips$\sigma(c_t)$:(k,k+1)$ that need to be applied to obtain $$\sigma(c_t)=(i_0,j_0)\cdot \sigma(c_t)=n, n-1, n-2, \ldots 1. E.g. for n=3 there are two such sequences: • (i_1,j_1)\cdot\ldots\cdot 12), (i_t,j_t).$$23), (12)
• (23), (12), (23)
• Question. Are all the expressions sequences of $\sigma(c_t)$ into $\binom{n}{2}$ transpositions flips leading from $1,2, \dots, n$ to$n, n-1, n-2, \ldots 1$ realized by general position lines in $H$? EDIT: Possibly with several labelings of $H_{ij}$, not just one fixed. Note that $\sigma(c_t)$ is not arbitrary, as $\ell$ has to intersect each $H_{ij}$ on its way from $c_0$ to $c_t$.

I guess that the answer is yes they do, but to show this, one might need to apply some machinery of root systems I am not familiar with.

2 rectifucation

Let $\mathcal{A} =A_{n-1}$ be the $A_{n-1}$ arrangement in $\mathbb{R}^{n}$, i.e. the set of hyperplanes $H_{ij}$ specified by equations $x_i=x_j$, for $1\leq i\leq j\leq n$. It is well-known that the connected components $c$ of the complement of the intersection of $\mathcal{A}$ with the affine hyperplane $H$={$x\mid \sum_i x_i=1$} (let me call them cells) are in 1-to-1 correspondence with permutations $\sigma(c)\in S_{n}$, so that crossing the hyperplane $H_{ij}$ corresponds to multiplication by the transposition $(i,j)$.

Let $\ell$ be a general position line in $H$. Then $\ell$ intersects $\binom{n}{2}+1$ cells. Two of these intersections are unbounded; let us denote the corresponding cells $c_0$ and $c_t$, respectively. As we follow $\ell$ from $c_0$ to $c_t$, from one cell to the next, the corresponding permutation $\sigma(c)$ changes from $\sigma(c_0)$ to $\sigma(c_t)$; namely, it gets multiplied by $(i_k,j_k)$ whenever we cross $H_{i_k,j_k}$. Assume that $\sigma(c_0)$ is the identity permutation. Then $\ell$ gives us an expression for $\sigma(c_t)$:
$$\sigma(c_t)=(i_0,j_0)\cdot (i_1,j_1)\cdot\ldots\cdot (i_t,j_t).$$

Question. Are all the expressions of $\sigma(c_t)$ into $\binom{n}{2}$ transpositions realized by general position lines in $H$? EDIT: Possibly with several labelings of $H_{ij}$, not just one fixed. Note that $\sigma(c_t)$ is not arbitrary, as $\ell$ has to intersect each $H_{ij}$ on its way from $c_0$ to $c_t$.

I guess that the answer is yes they do, but to show this, one might need to apply some machinery of root systems I am not familiar with.

1

# lines through A_n reflection arrangement and permutations

Let $\mathcal{A} =A_{n-1}$ be the $A_{n-1}$ arrangement in $\mathbb{R}^{n}$, i.e. the set of hyperplanes $H_{ij}$ specified by equations $x_i=x_j$, for $1\leq i\leq j\leq n$. It is well-known that the connected components $c$ of the complement of the intersection of $\mathcal{A}$ with the affine hyperplane $H$={$x\mid \sum_i x_i=1$} (let me call them cells) are in 1-to-1 correspondence with permutations $\sigma(c)\in S_{n}$, so that crossing the hyperplane $H_{ij}$ corresponds to multiplication by the transposition $(i,j)$.

Let $\ell$ be a general position line in $H$. Then $\ell$ intersects $\binom{n}{2}+1$ cells. Two of these intersections are unbounded; let us denote the corresponding cells $c_0$ and $c_t$, respectively. As we follow $\ell$ from $c_0$ to $c_t$, from one cell to the next, the corresponding permutation $\sigma(c)$ changes from $\sigma(c_0)$ to $\sigma(c_t)$; namely, it gets multiplied by $(i_k,j_k)$ whenever we cross $H_{i_k,j_k}$. Assume that $\sigma(c_0)$ is the identity permutation. Then $\ell$ gives us an expression for $\sigma(c_t)$:
$$\sigma(c_t)=(i_0,j_0)\cdot (i_1,j_1)\cdot\ldots\cdot (i_t,j_t).$$

Question. Are all the expressions of $\sigma(c_t)$ into $\binom{n}{2}$ transpositions realized by general position lines in $H$?

I guess that the answer is yes they do, but to show this, one might need to apply some machinery of root systems I am not familiar with.