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Edited to include clarification in comments

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edited Nov 1, 2018 at 17:36

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Note: The following is the result of an email exchange with one of the coauthors of the paper cited in the OP.

In the definition of a reflection-invariant monomial ideal the requirement that the map $\phi: \mathbf{x}^{\mathbf{c}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}}$ be an involution on $\mathrm{MonSoc}(I)$ is not the usual definition of involution, i.e., that the map is its own inverse. Instead, the map $\phi$ is required to be both an involution on and a permutation of $\mathrm{MonSoc}(I)$. This additional requirement is not mentioned in the paper, though upon closer reading, it is used in a number of the proofs.

As for the example in the OP, the map $\phi$ is clearly not a permutation of $\mathrm{MonSoc}(I)$. Moreover, one can show that $I$ is not reflection-invariant for any choice of $\mathbf{K}$. So the ideal $I$ is not reflection-invariant. Also, with this new restriction on permissible involutions it is easy to prove the note following the definition of reflection invariance in the paper in question.

Edit: Thanks to @benblumsmith's comment the heart of the matter is now clear. The map that sends $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}-\mathbf{b}}$ has domain and codomain equal to $\mathrm{MonSoc}(I)$ (as opposed to the set of all monomials in the quotient ring $K[\mathbf{x}]/\left\langle x_1^{K_1+1}, \dots, x_n^{K_n+1}\right\rangle$). So it's clear that the example in the OP is not a counterexample.

Note: The following is the result of an email exchange with one of the coauthors of the paper cited in the OP.

In the definition of a reflection-invariant monomial ideal the requirement that the map $\phi: \mathbf{x}^{\mathbf{c}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}}$ be an involution on $\mathrm{MonSoc}(I)$ is not the usual definition of involution, i.e., that the map is its own inverse. Instead, the map $\phi$ is required to be both an involution on and a permutation of $\mathrm{MonSoc}(I)$. This additional requirement is not mentioned in the paper, though upon closer reading, it is used in a number of the proofs.

As for the example in the OP, the map $\phi$ is clearly not a permutation of $\mathrm{MonSoc}(I)$. Moreover, one can show that $I$ is not reflection-invariant for any choice of $\mathbf{K}$. So the ideal $I$ is not reflection-invariant. Also, with this new restriction on permissible involutions it is easy to prove the note following the definition of reflection invariance in the paper in question.

Note: The following is the result of an email exchange with one of the coauthors of the paper cited in the OP.

In the definition of a reflection-invariant monomial ideal the requirement that the map $\phi: \mathbf{x}^{\mathbf{c}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}}$ be an involution on $\mathrm{MonSoc}(I)$ is not the usual definition of involution, i.e., that the map is its own inverse. Instead, the map $\phi$ is required to be both an involution on and a permutation of $\mathrm{MonSoc}(I)$. This additional requirement is not mentioned in the paper, though upon closer reading, it is used in a number of the proofs.

As for the example in the OP, the map $\phi$ is clearly not a permutation of $\mathrm{MonSoc}(I)$. Moreover, one can show that $I$ is not reflection-invariant for any choice of $\mathbf{K}$. So the ideal $I$ is not reflection-invariant. Also, with this new restriction on permissible involutions it is easy to prove the note following the definition of reflection invariance in the paper in question.

Edit: Thanks to @benblumsmith's comment the heart of the matter is now clear. The map that sends $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}-\mathbf{b}}$ has domain and codomain equal to $\mathrm{MonSoc}(I)$ (as opposed to the set of all monomials in the quotient ring $K[\mathbf{x}]/\left\langle x_1^{K_1+1}, \dots, x_n^{K_n+1}\right\rangle$). So it's clear that the example in the OP is not a counterexample.

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created Oct 22, 2018 at 15:35

984
7
17

Note: The following is the result of an email exchange with one of the coauthors of the paper cited in the OP.

In the definition of a reflection-invariant monomial ideal the requirement that the map $\phi: \mathbf{x}^{\mathbf{c}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}}$ be an involution on $\mathrm{MonSoc}(I)$ is not the usual definition of involution, i.e., that the map is its own inverse. Instead, the map $\phi$ is required to be both an involution on and a permutation of $\mathrm{MonSoc}(I)$. This additional requirement is not mentioned in the paper, though upon closer reading, it is used in a number of the proofs.

As for the example in the OP, the map $\phi$ is clearly not a permutation of $\mathrm{MonSoc}(I)$. Moreover, one can show that $I$ is not reflection-invariant for any choice of $\mathbf{K}$. So the ideal $I$ is not reflection-invariant. Also, with this new restriction on permissible involutions it is easy to prove the note following the definition of reflection invariance in the paper in question.