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Sam Hopkins
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The answer to my questions is that I was wrong. I'll switch to the more standard notation of having $D_n$ act on $\{ -n, -n+1, \ldots, -2, -1, 1, 2, \ldots, n-1, n \}$ in order to match the reference of Proctor.

Here is the simplest example to demonstrate my error. Take $n \geq 4$ and look at $D_n \cdot [2]$. Then my belief above was that we should have $(2,1) > (1,-2)$, since $(2,1) > (1,-2)$ and $(2,-1) > (-1,-2)$ are both valid inequalities in type $B$. But this is wrong; any type $D$ transposition which lowers $(2,1)$ takes it below $(1,-2)$. In terms of weights, we are taking about $e_2+e_1$ and $-e_2+e_1$. They differ by $2 e_2 = (e_2+e_1) + (e_2-e_1)$, which is a sum of positive roots. However, subtracting either $e_2+e_1$ or $e_2-e_1$ from $e_2+e_1$ gives $0$ and $2 e_1$ respectively, neither of which are in the $D_n$ orbit of $e_2+e_1$.

The paper of Proctor which Sam Hopkins pointed me to gives the correct criterion in Theorem 5D: Let $I=\{ i_1 < i_2 < \cdots < i_k \}$ and $J=\{ j_1 < j_2 < \cdots < j_k \}$ be two $k$-element subsets of $\{ \pm 1, \pm 2, \ldots, \pm n \}$, both of which have the property of not containing both $a$ and $-a$. Then $I \leq J$ if and only if two conditions hold:

  • $i_r \leq j_r$ and

  • If there is a set of indices $p$, $p+1$, \ldots, $q$$p, p+1, \ldots, q$ such that $\{ |i_p|, |i_{p+1}|, \ldots, |i_q| \}$ and $\{ |j_p|, |j_{p+1}|, \ldots, |j_q| \}$ are each permutations of $\{ 1,2,\ldots, q-p+1 \}$, then the number of negative elements among $(i_p, i_{p+1}, \ldots, i_q)$ and $(j_p, j_{p+1}, \ldots, j_q)$ must have the same parity.

At the moment, I don't see how to make this second condition sound like a rank matrix condition.

The answer to my questions is that I was wrong. I'll switch to the more standard notation of having $D_n$ act on $\{ -n, -n+1, \ldots, -2, -1, 1, 2, \ldots, n-1, n \}$ in order to match the reference of Proctor.

Here is the simplest example to demonstrate my error. Take $n \geq 4$ and look at $D_n \cdot [2]$. Then my belief above was that we should have $(2,1) > (1,-2)$, since $(2,1) > (1,-2)$ and $(2,-1) > (-1,-2)$ are both valid inequalities in type $B$. But this is wrong; any type $D$ transposition which lowers $(2,1)$ takes it below $(1,-2)$. In terms of weights, we are taking about $e_2+e_1$ and $-e_2+e_1$. They differ by $2 e_2 = (e_2+e_1) + (e_2-e_1)$, which is a sum of positive roots. However, subtracting either $e_2+e_1$ or $e_2-e_1$ from $e_2+e_1$ gives $0$ and $2 e_1$ respectively, neither of which are in the $D_n$ orbit of $e_2+e_1$.

The paper of Proctor which Sam Hopkins pointed me to gives the correct criterion in Theorem 5D: Let $I=\{ i_1 < i_2 < \cdots < i_k \}$ and $J=\{ j_1 < j_2 < \cdots < j_k \}$ be two $k$-element subsets of $\{ \pm 1, \pm 2, \ldots, \pm n \}$, both of which have the property of not containing both $a$ and $-a$. Then $I \leq J$ if and only if two conditions hold:

  • $i_r \leq j_r$ and

  • If there is a set of indices $p$, $p+1$, \ldots, $q$ such that $\{ |i_p|, |i_{p+1}|, \ldots, |i_q| \}$ and $\{ |j_p|, |j_{p+1}|, \ldots, |j_q| \}$ are each permutations of $\{ 1,2,\ldots, q-p+1 \}$, then the number of negative elements among $(i_p, i_{p+1}, \ldots, i_q)$ and $(j_p, j_{p+1}, \ldots, j_q)$ must have the same parity.

At the moment, I don't see how to make this second condition sound like a rank matrix condition.

The answer to my questions is that I was wrong. I'll switch to the more standard notation of having $D_n$ act on $\{ -n, -n+1, \ldots, -2, -1, 1, 2, \ldots, n-1, n \}$ in order to match the reference of Proctor.

Here is the simplest example to demonstrate my error. Take $n \geq 4$ and look at $D_n \cdot [2]$. Then my belief above was that we should have $(2,1) > (1,-2)$, since $(2,1) > (1,-2)$ and $(2,-1) > (-1,-2)$ are both valid inequalities in type $B$. But this is wrong; any type $D$ transposition which lowers $(2,1)$ takes it below $(1,-2)$. In terms of weights, we are taking about $e_2+e_1$ and $-e_2+e_1$. They differ by $2 e_2 = (e_2+e_1) + (e_2-e_1)$, which is a sum of positive roots. However, subtracting either $e_2+e_1$ or $e_2-e_1$ from $e_2+e_1$ gives $0$ and $2 e_1$ respectively, neither of which are in the $D_n$ orbit of $e_2+e_1$.

The paper of Proctor which Sam Hopkins pointed me to gives the correct criterion in Theorem 5D: Let $I=\{ i_1 < i_2 < \cdots < i_k \}$ and $J=\{ j_1 < j_2 < \cdots < j_k \}$ be two $k$-element subsets of $\{ \pm 1, \pm 2, \ldots, \pm n \}$, both of which have the property of not containing both $a$ and $-a$. Then $I \leq J$ if and only if two conditions hold:

  • $i_r \leq j_r$ and

  • If there is a set of indices $p, p+1, \ldots, q$ such that $\{ |i_p|, |i_{p+1}|, \ldots, |i_q| \}$ and $\{ |j_p|, |j_{p+1}|, \ldots, |j_q| \}$ are each permutations of $\{ 1,2,\ldots, q-p+1 \}$, then the number of negative elements among $(i_p, i_{p+1}, \ldots, i_q)$ and $(j_p, j_{p+1}, \ldots, j_q)$ must have the same parity.

At the moment, I don't see how to make this second condition sound like a rank matrix condition.

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David E Speyer
  • 156.2k
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The answer to my questions is that I was wrong. I'll switch to the more standard notation of having $D_n$ act on $\{ -n, -n+1, \ldots, -2, -1, 1, 2, \ldots, n-1, n \}$ in order to match the reference of Proctor.

Here is the simplest example to demonstrate my error. Take $n \geq 4$ and look at $D_n \cdot [2]$. Then my belief above was that we should have $(2,1) > (1,-2)$, since $(2,1) > (1,-2)$ and $(2,-1) > (-1,-2)$ are both valid inequalities in type $B$. But this is wrong; any type $D$ transposition which lowers $(2,1)$ takes it below $(1,-2)$. In terms of weights, we are taking about $e_2+e_1$ and $-e_2+e_1$. They differ by $2 e_2 = (e_2+e_1) + (e_2-e_1)$, which is a sum of positive roots. However, subtracting either $e_2+e_1$ or $e_2-e_1$ from $e_2+e_1$ gives $0$ and $2 e_1$ respectively, neither of which are in the $D_n$ orbit of $e_2+e_1$.

The paper of Proctor which Sam Hopkins pointed me to gives the correct criterion in Theorem 5D: Let $I=\{ i_1 < i_2 < \cdots < i_k \}$ and $J=\{ j_1 < j_2 < \cdots < j_k \}$ be two $k$-element subsets of $\{ \pm 1, \pm 2, \ldots, \pm n \}$, both of which have the property of not containing both $a$ and $-a$. Then $I \leq J$ if and only if two conditions hold:

  • $i_r \leq j_r$ and

  • If there is a set of indices $p$, $p+1$, \ldots, $q$ such that $\{ |i_p|, |i_{p+1}|, \ldots, |i_q| \}$ and $\{ |j_p|, |j_{p+1}|, \ldots, |j_q| \}$ are each permutations of $\{ 1,2,\ldots, q-p+1 \}$, then the number of negative elements among $(i_p, i_{p+1}, \ldots, i_q)$ and $(j_p, j_{p+1}, \ldots, j_q)$ must have the same parity.

At the moment, I don't see how to make this second condition sound like a rank matrix condition.

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