Divided symmetrization and quasisymmetric functions

Abstract

Motivated by a question in Schubert calculus, we study the interplay of quasisymmetric polynomials with the divided symmetrization operator, which was introduced by Postnikov in the context of volume polynomials of permutahedra. Divided symmetrization is a linear form which acts on the space of polynomials in n indeterminates of degree $n-1$. We first show that divided symmetrization applied to a quasisymmetric polynomial in m indeterminates can be easily determined. Several examples with a strong combinatorial flavor are given. Then, we prove that the divided symmetrization of any polynomial can be naturally computed with respect to a direct sum decomposition due to Aval–Bergeron–Bergeron, involving the ideal generated by positive degree quasisymmetric polynomials in n indeterminates.

Appendix A: Proof of Lemma 3.7

We want to prove $\big \langle {{\mathbf {x}}^{\mathbf{c}}}\big \rangle _{n}=(-1)^{|S_\mathbf{c}|}\beta (S_\mathbf{c})$ for any $\mathbf{c}\in {\mathcal {W}}_{n}^{'}$. Our proof proceeds in two steps.

• First, via a sequence of moves, we transform any such $\mathbf{c}$ into a weak composition $\mathbf{c'}=(c_1',\ldots ,c_n')\in {\mathcal {W}}_{n}^{'}$ with the properties that $S_\mathbf{c}=S_{\mathbf{c'}}$ and $\sum _{1\le j\le i}(c'_j-1) \in \{0,-1\}$ for all $1\le i\le n$. Furthermore, our moves ensure that $\big \langle {{\mathbf {x}}^{\mathbf{c}}}\big \rangle _{n}=\big \langle {{\mathbf {x}}^{\mathbf{c'}}}\big \rangle _{n}$.

• Second, we compute $\big \langle {{\mathbf {x}}^{\mathbf{c'}}}\big \rangle _{n}$ explicitly by exploiting a relation satisfied by the numbers $(-1)^{|S|}\beta (S)$.

We now furnish details. Let $\mathbf{c}\in {\mathcal {W}}_{n}^{'}$, and assume that there exists an index $i\in [n]$ such that $\mathrm {ht}_i(\mathbf{c})\notin \{0,-1\}$. Let k be the largest such index. Note that we must have $k\le n-1$ as $\mathbf{c}\in {\mathcal {W}}_{n}^{'}$. Consider the move sending $\mathbf{c}$ to a sequence $\mathbf{d}$ as follows:

$$\begin{aligned} (c_1,\ldots ,c_k,c_{k+1},\ldots ,c_n) \mapsto {\left\{ \begin{array}{ll} (c_1,\ldots ,c_k{-}1,c_{k+1}{+}1,\ldots , c_n) &{} \text {if } \mathrm {ht}_k(\mathbf{c})>0,\\ (c_1,\ldots ,c_k{+}1,c_{k+1}{-}1,\ldots , c_n) &{} \text {if } \mathrm {ht}_k(\mathbf{c})<-1. \end{array}\right. } \end{aligned}$$

(A.1)

In the case $\mathrm {ht}_k(\mathbf{c})>0$, the sequence $\mathbf{d}$ is clearly a weak composition of size $n-1$. If $\mathrm {ht}_k(\mathbf{c})<-1$, then the maximality assumption on k along with the fact that $\mathrm {ht}_n(\mathbf{c})=-1$ implies that $c_{k+1}\ge 1$. Thus, the sequence $\mathbf{d}$ is a weak composition of $n-1$ in this case as well. It is easy to see that $S_\mathbf{c}=S_\mathbf{d}$. We show that $\big \langle {{\mathbf {x}}^{\mathbf{c}}}\big \rangle _{n}=\big \langle {{\mathbf {x}}^{\mathbf{d}}}\big \rangle _{n}$. Assume that $\mathrm {ht}_k(\mathbf{c})>0$ and thus $\mathbf{d}=(c_1,\ldots ,c_k-1,c_{k+1}+1,\ldots , c_n)$. We have

$$\begin{aligned} \big \langle {{\mathbf {x}}^{\mathbf{c}}-{\mathbf {x}}^{\mathbf{d}}}\big \rangle _{n}=\big \langle {x_1^{c_1}\cdots x_{k}^{c_k-1}(x_k-x_{k+1})x_{k+1}^{c_{k+1}}\cdots x_n^{c_n}}\big \rangle _{n}. \end{aligned}$$

(A.2)

By our hypothesis that $\mathrm {ht}_k(\mathbf{c})>0$ , we know that $\deg (x_1^{c_1}\cdots x_{k}^{c_k-1})\ge k$. Corollary 3.5 implies that the right hand side of (A.2) equals 0, which in turn implies that $\big \langle {{\mathbf {x}}^{\mathbf{c}}}\big \rangle _{n}=\big \langle {{\mathbf {x}}^{\mathbf{d}}}\big \rangle _{n}$. The case when $\mathrm {ht}_k(\mathbf{c})<-1$ is handled similarly, and we leave the details to the interested reader.

By applying the aforementioned moves repeatedly, one can transform $\mathbf{c}$ into $\mathbf{c'}$ with the property that $\mathrm {ht}_i(\mathbf{c'})\in \{0,-1\}$ for all $i\in [n]$. We are additionally guaranteed that $S_\mathbf{c}=S_\mathbf{c'}$ and $\big \langle {{\mathbf {x}}^{\mathbf{c}}}\big \rangle _{n}=\big \langle {{\mathbf {x}}^{\mathbf{c'}}}\big \rangle _{n}$. Figure 2 shows the path corresponding to $\mathbf{c'}$ where $\mathbf{c}=(0,3,0,0,0,1,3,0)$ is the weak composition from Fig. 1. Explicitly, the moves in going from $\mathbf{c}$ to $\mathbf{c'}$ are $(0,3,0,0,0,1,3,0)\rightarrow (0,3,0,0,0,2,2,0)\rightarrow (0,3,0,0,1,1,2,0)\rightarrow (0,2,1,0,1,1,2,0)$.

Fig. 2

The path corresponding to $\mathbf{c}=(0,2,1,0,1,1,2,0)$ with $S_\mathbf{c}=\{1,4,5,6\}$

The upshot of the preceding discussion is that to compute $\big \langle {{\mathbf {x}}^\mathbf{c}}\big \rangle _{n}$ it suffices to consider $\mathbf{c}\in {\mathcal {W}}_{n}^{'}$ such that $\mathrm {ht}_{i}(\mathbf{c})\in \{0,-1\}$ for all $i\in [n]$. The map $\mathbf{c}\mapsto S_\mathbf{c}$ restricted to these sequences is a 1–1 correspondence with subsets $S\subseteq [n-1]$. We write $S\mapsto c(S)$ for the inverse map, and set ${\mathbf {x}}(S):={\mathbf {x}}^{c(S)}$.

Our goal now is to prove that

$$\begin{aligned} \big \langle {{\mathbf {x}}(S)}\big \rangle _{n}=(-1)^{|S|}\beta (S) \end{aligned}$$

(A.3)

To this end, we proceed by induction on n. When $n=1$, we have $S=\emptyset$. In this case, we have ${\mathbf {x}}(S)=1$, and both sides of the equality in (A.3) equal 1. Let $n\ge 2$ henceforth. Assume further that $S\ne [n-1]$, and let $i\notin S$. Define $S_i=S\cap [i]$ and $S^i=\{j\in [n-i]\;|\;j+i\in S\}$. Then Corollary 3.5 gives

$$\begin{aligned} \big \langle {{\mathbf {x}}(S)}\big \rangle _{n}-\big \langle {{\mathbf {x}}(S\cup \{i\})}\big \rangle _{n}=\left( {\begin{array}{c}n\\ i\end{array}}\right) \big \langle {{\mathbf {x}}(S_i)}\big \rangle _{i} \big \langle {{\mathbf {x}}(S^i)}\big \rangle _{n-i}. \end{aligned}$$

(A.4)

The numbers $(-1)^{|S|}\beta (S)$ also satisfy this identity, because:

$$\begin{aligned} \beta _n(S)+\beta _n(S\cup \{i\})=\left( {\begin{array}{c}n\\ i\end{array}}\right) \beta _i(S_i) \beta _{n-i}\left( S^i\right) . \end{aligned}$$

(A.5)

This has a simple combinatorial proof: given a permutation corresponding to the left hand side, split its 1-line notation after position i, and standardize both halves so that they become permutations on [1, i] and $[1,n-i]$ respectively.

To conclude, note that (A.4) determines all values $\big \langle {{\mathbf {x}}(S)}\big \rangle _{n}$ by induction in terms of the single value $\big \langle {{\mathbf {x}}([n-1])}\big \rangle _{n}$. Now ${\mathbf {x}}([n-1])=x_2\cdots x_n$ and we have $\big \langle {x_2\cdots x_n}\big \rangle _{n}=(-1)^{n-1}$ by Example 3.6. Since $\beta ([n-1])=1$, we have $\big \langle {{\mathbf {x}}([n-1])}\big \rangle _{n}=(-1)^{|[n-1]|}\beta ([n-1])$, which completes the proof.