Reflection-invariant monomial ideals and Alexander duality First we give some definitions from Section 3 of the paper Monomials, Binomials, and Riemann-Roch by Manjunath and Sturmfels and then we restate a claim from that paper offered without proof. Finally we provide an example that seems to contradict that claim. The question is
Question: Is the example given below a counterexample to the claim? And if not, why not?
Definitions
Fix an Artinian monomial ideal $I$ of a polynomial ring $K[\mathbf{x}] = K[x_1, \dots, x_n]$. A monomial $\mathbf{x}^{\mathbf{b}}$ is a socle monomial of $I$ if $\mathbf{x}^{\mathbf{b}} \notin I$ and $x_i\mathbf{x}^{\mathbf{b}} \in I$ for all $i$. Let $\mathrm{MonSoc}(I)$ be the set of all socle monomials of $I$.
Def: $I$ is reflection invariant if there is a canonical monomial $\mathbf{x}^{\mathbf{K}}$ such that the map that sends a monomial $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}}$ is an involution on $\mathrm{MonSoc}(I)$.
Following these definitions the authors note the following.
The Claim
Claim: $I$ is reflection invariant with canonical monomial $\mathbf{x}^{\mathbf{K}}$ if and only if the monomial ideal generated by $\mathrm{MonSoc}(I)$ equals the Alexander dual $I^{[\mathbf{K} + \mathbf{e}]}$ where $\mathbf{e} = (1,1,\dots, 1)$.
The (Counter?) Example
Let $I = \langle a^4,~ab^2,~b^3,~a^3c,~abc,~c^3 \rangle \subset K[a,b,c]$ and let $\mathbf{K} = (3,2,2)$. Then 
$$\mathrm{MonSoc}(I) = \left\{a^{3}b,~a^{2}c^{2},~b^{2}c^{2}\right\}.$$ 
By (the constructive proof of) Proposition 5.2 in this paper
 the ideal $J = \langle a^4, a^2b, b^3, ac, b^2c, c^3 \rangle$ is the unique Artinian ideal with
$$\mathrm{MonSoc}(J) = \left\{\mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}} \mid \mathbf{x}^{\mathbf{b}} \in \mathrm{MonSoc}(I)\right\}.$$
Moreover, the same algorithm can be used to show that $I$ is the unique Artinian ideal with 
$$\mathrm{MonSoc}(I) = \left\{\mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}} \mid \mathbf{x}^{\mathbf{c}} \in \mathrm{MonSoc}(J)\right\}.$$
In particular, the map that sends a monomial $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}}$ is an involution on $\mathrm{MonSoc}(I)$, so $I$ is reflection-invariant with canonical monomial $\mathbf{x}^{\mathbf{K}}$. We now get a contradiction to the claim above by computing the Alexander dual (in Macaulay2, for example) and noting that the minimal generators of $I^{[(4,3,3)]}$ are $\{a^4bc,~a^2b^3c,~ab^2c^3\} \neq \mathrm{MonSoc}(I)$.
Again, the question is
Question: Is the example just given a counterexample to the above claim? And if not, why not?
 A: 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. 
