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Exercise: Show that $\mathcal{P}_n$ is free over $NH_n$ of rank $n!$.

Exercise: Show that $NH_n$ is free over $\mathcal{P}_n$of rank $n!$.

Let $\mathcal{P}_n$ be the polynomial ring $k\left[x_1, x_2, \ldots, x_n\right]$.

If $w=s_{i_1}\cdots s_{i_k}$ is a reduced expression for $w\in S_n$, the braid relations imply that the element $$\partial_w=\partial_{i_1}\cdots\partial_{i_k}$$ is well defined. Moreover, by (1) we have $$\partial_u\partial_v=\begin{cases}\partial_{uv}&\mbox{if }\ell(u)+\ell(v)=\ell(uv),\\ 0&\mbox{otherwise}.\end{cases}$$

Example: Consider $S_3=\{e,s_1,s_2,s_1s_2,s_2s_1,s_1s_2s_1$, so the Schubert polynomials belong to $\mathcal{P}_3=F[x_1,x_2,x_3]$. \begin{align*} S_{s_1s_2s_1}&=x_2x_3^2\\ S_{s_2s_1}&=\partial_1(x_2x_3^2)=\frac{x_2-x_1}{x_2-x_1}x_3^2=x_3^2\\ S_{s_1s_2}&=\partial_2(x_2x_3^2)=\frac{x_2x_3^2-x_3x_2^2}{x_3-x_2}=x_2x_3\\ S_{s_2}&=\partial_1\partial_2(x_2x_3^2)=\partial_1(x_2x_3)=x_3\\ S_{s_1}&=\partial_2\partial_1(x_2x_3^2)=\partial_2(x_3^2)=x_2+x_3\\ S_e&=\partial_2\partial_1\partial_2(x_2x_3^2)=\partial_2(x_3)=1. \end{align*}

Let $\mathcal{P}_n$ be the polynomial ring $k\left[x_1, x_2, \ldots, x_n\right]$. The symmetric group $S_n$ acts on $\mathcal{P}_n$ from the left by the formula $${}^\pi f = f\left(x_{\pi\left(1\right)}, x_{\pi\left(2\right)}, \ldots, x_{\pi\left(n\right)}\right)$$ for all $\pi \in S_n$ and $f \in \mathcal{P}_n$. We let $s_1, s_2, \ldots, s_{n-1}$ denote the $n-1$ adjacent transpositions generating $S_n$ (so $s_i = \left(i,i+1\right)$ for each $i$).

If $w=s_{i_1}\cdots s_{i_k}$ is a reduced expression for $w\in S_n$, the braid relations (and Matsumoto's theorem for the standard presentation of $S_n$ as a Coxeter group) imply that the element $$\partial_w=\partial_{i_1}\cdots\partial_{i_k}$$ is well defined. Moreover, by (1) we have $$\partial_u\partial_v=\begin{cases}\partial_{uv}&\mbox{if }\ell(u)+\ell(v)=\ell(uv),\\ 0&\mbox{otherwise}.\end{cases}$$

Example: Consider $S_3=\{e,s_1,s_2,s_1s_2,s_2s_1,s_1s_2s_1\}$, so the Schubert polynomials belong to $\mathcal{P}_3=F[x_1,x_2,x_3]$. \begin{align*} S_{s_1s_2s_1}&=x_2x_3^2\\ S_{s_2s_1}&=\partial_1(x_2x_3^2)=\frac{x_2-x_1}{x_2-x_1}x_3^2=x_3^2\\ S_{s_1s_2}&=\partial_2(x_2x_3^2)=\frac{x_2x_3^2-x_3x_2^2}{x_3-x_2}=x_2x_3\\ S_{s_2}&=\partial_1\partial_2(x_2x_3^2)=\partial_1(x_2x_3)=x_3\\ S_{s_1}&=\partial_2\partial_1(x_2x_3^2)=\partial_2(x_3^2)=x_2+x_3\\ S_e&=\partial_2\partial_1\partial_2(x_2x_3^2)=\partial_2(x_3)=1. \end{align*}

Exercise: Verify that the following relations hold in $NC_n$: $$\partial_i\partial_j=\partial_j\partial_i\mbox{ if }|i-j|>1\;\;\;\mbox{and}\;\;\; \partial_i\partial_{i+1}\partial_i=\partial_{i+1}\partial_i\partial_{i+1}$$ for $i<n-1$.

\textbf{Proposition:} The Schubert polynomial $S_w$ is homogeneous of degree $\ell(w)$.

\textbf{Proof:} Note that $\deg(x^\delta)=1+2+\cdots+(n-1)={n\choose 2}=\ell(w_0)$. Now, let $\mathcal{P}_n^d$ be the degree $d$ component of $\mathcal{P}_n$. Then, $$\partial_i:\mathcal{P}_n^d\longrightarrow\mathcal{P}_n^{d-1}$$ so $$\deg (S_w)=\deg (\partial_{w^{-1}w_0}(x^\delta))={n\choose 2}-\ell(w^{-1}w_0)={n\choose 2}-\ell(w_0)+\ell(w)=\ell(w).$$ $\square$

Exercise: Verify that the following relations hold in $NC_n$: $$\partial_i\partial_j=\partial_j\partial_i\mbox{ if }|i-j|>1,$$ and $$\partial_i\partial_{i+1}\partial_i=\partial_{i+1}\partial_i\partial_{i+1} \mbox{ if } i<n-1.$$

Proposition: The Schubert polynomial $S_w$ is homogeneous of degree $\ell(w)$.

Proof: Note that $\deg(x^\delta)=1+2+\cdots+(n-1)={n\choose 2}=\ell(w_0)$. Now, let $\mathcal{P}_n^d$ be the degree $d$ component of $\mathcal{P}_n$. Then, $$\partial_i:\mathcal{P}_n^d\longrightarrow\mathcal{P}_n^{d-1}$$ so $$\deg (S_w)=\deg (\partial_{w^{-1}w_0}(x^\delta))={n\choose 2}-\ell(w^{-1}w_0)={n\choose 2}-\ell(w_0)+\ell(w)=\ell(w).$$ $\square$