Abstract

This erratum corrects the proof of the main result [1, Thm. 5.2] of [1, Sec. 5]. While this result is correct as stated, its proof is flawed. We adopt the notation of [1, Sec. 5]. The total order |$\prec $| is not a term order, so that [1, Lem. 5.3] loses meaning. In particular, [1, Lem. 5.1] is false because the depth |$d(\sigma )$| is not multiplicative. For |$n = 6$|⁠, the elements |$u, v \in \wedge \{ \Theta _{6}, \Xi _{6}\}$| given by |$u=\xi _{1} \theta _{3}\xi _{3} \theta _{4} \theta _{5} \xi _{5}$| and |$v=\theta _{1} \theta _{2}\xi _{2} \theta _{4} \theta _{6} \xi _{6}$| have the following lattice path representations as in [1, Sec. 5]: We correct the proof of [1, Thm. 5.2] as follows. We shall calculate a Gröbner basis for the ideal |$I_{n} = \langle \delta _{n}\rangle \subset \wedge \{\Theta _{n}, \Xi _{n} \}$| where |$\delta _{n}= \sum _{i=1}^{n} \theta _{i} \xi _{i}$| with respect to the lexicographical term order |$<_{\textrm{lex}}$|⁠. For each Motzkin path |$\sigma $| as in [1, Sec. 5], we define |$j(\sigma )$| to be the |$x$|-coordinate where the depth |$d(\sigma )$| is achieved the 1st time. We have |$d(\sigma )=0$| if and only if |$j(\sigma )=0$|⁠. If |$u, v$| are the Motzkin paths (or monomials) in Example 1, then |$j(u)=5$| and |$j(v)=2$|⁠. Given a Motzkin path |$\sigma =(s_{1}, s_{2},\dots , s_{n})$| and |$i \leq n$|⁠, we define |$k_{i}$| to be the difference between the |$y$|-coordinate of the starting point of |$s_{i}$| and |$d(\sigma )$|⁠. For example, |$k_{1}=-d(\sigma )$| and |$k_{j(\sigma )+1}=0$|⁠. For |$i\leq j(\sigma )$|⁠, we introduce the exterior algebra elements ... ... The initial ideal |$\operatorname{in}_{lex}(\delta _{n}^{k})$| with respect the lexicographical term order contains all monomials |$\sigma $| with depth |$d(\sigma )\leq -k$|⁠. We claim that the leading monomial of |$p(\sigma )\delta _{n}^{-d(\sigma )}$| divides the monomial |$\textrm{wt}(\sigma )$|⁠, and we prove this statement for all |$n$| by induction on |$j(\sigma )$|⁠. The base case |$j(\sigma )=0$| is trivial because all monomials belong to the ideal generated by |$\delta _{n}^{0}=1$|⁠. For the inductive step, we remove the 1st step |$s_{1}$| from |$\sigma $| to get a new path |$\tau =(s_{2}, \dots , s_{n})$| involving only the variables |$\theta _{2}, \dots , \theta _{n}, \xi _{2}, \dots , \xi _{n}$|⁠. Notice that |$p(\sigma )=p_{1}(\sigma )p(\tau )$|⁠. We divide proof in three cases according to the 1st step |$s_{1}$|⁠. Case 1:|$s_{1} = (1,1)$|is an up step. We have |$d(\tau )=d(\sigma )-1$|⁠, |$\textrm{wt}(\sigma )=\textrm{wt}(\tau )$|⁠, and ... Case 2:|$s_{1} = (1,0)$|is a horizontal step. We assume that the horizontal step |$s_{1}$| is labelled with |$\theta $|⁠; the other case is identical. We have |$d(\tau )=d(\sigma )$|⁠, |$\textrm{wt}(\sigma )=\theta _{1} \textrm{wt}(\tau )$|⁠, and ... ... Case 3:|$s_{1} = (1,-1)$|is a down step. We have |$d(\tau )=d(\sigma )+1$|⁠, |$\textrm{wt}(\sigma )=\theta _{1}\xi _{1} \textrm{wt}(\tau )$|⁠, |$p(\sigma )=p(\tau )$|⁠, and ... We conclude that |$\textrm{wt}(\sigma ) \in \textrm{in}_{\textrm{lex}}(\delta _{n}^{-d(\sigma )}) \subseteq \textrm{in}_{\textrm{lex}}(\delta _{n}^{k})$| for all |$k \leq -d(\sigma )$| and the proof is complete. The above theorem substitutes [1, Lem. 5.3]. The 2nd part of the proof of [1, Thm. 5.2] is correct and can be left unchanged. The set |$\{p(\sigma )\delta _{n} \mid \sigma \textnormal{ s.t.} d(\sigma )=-1\}$| is a Gröbner basis for the ideal |$I_{n}= \langle \delta _{n} \rangle $| with respect to the lexicographical term order.

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