Lascoux and Schützenberger [61]  introduced a recurrence on Schubert polynomials called transition.  These transition equations imply that each Schubert polynomial expands as a positive sum of monomials.  This fact was subsequently reproved by the introduction of combinatorial formulas for these coefficients (e.g., [7, 29]), as well as through  geometric arguments (e.g., [46, 52, 10, 47]). Analogous transition equations were also given for double Schubert polynomials  (e.g., [54]). Also of interest are transition equations for Grothendieck polynomials, which are K-theoretic analogues of Schubert polynomials.  Lascoux  [56] gave transition equations for single Grothendieck polynomials (see also [58]). He also described the double Grothendieck version in [57].  All of these formulas follow from specializations of a recurrence on $\beta$-double Grothendieck polynomials, which represent classes in connective K-theory [43].  An algebraic proof of transition for $\beta$-double Grothendieck polynomials was given in <cit.>.
The goal of this section is to give a geometrically motivated proof of this recurrence.
We do so by analyzing the multigraded Hilbert series of matrix Schubert varieties.   The geometric interpretation of transition in Proposition <ref> was previously known to Knutson and Yong in unpublished work (see their slides [80]), as was an unpublished proof of Proposition <ref> that relied on Lascoux's transition formula.

Fix $w\in S_n$, and take a lower outside corner $(a,b)$ of $D(w)$.  As before, let $v=wt_{a,w^{-1}(b)}$.  Recall
\[
\phi(w,z_{a,b})=\{i\in [a-1]:vt_{i,a}>v \text{ and } \ell(vt_{i,a})=\ell(v)+1\}.
\]
Given $U=\{i_1,\ldots,i_k\}$ with $1\leq i_1<i_2<\cdots <i_k<a$, let
\[w_U= v (a\, i_k \,i_{k-1}\, \ldots\, i_1)\]
(where the second permutation is written in cycle notation).  Note that if $U=\emptyset$, then $w_U=v$.

Given $w\in S_n$, let $(a,b)$ be a lower outside corner of $D(w)$.
If $U\subseteq \phi(w,z_{a,b})$ and $U\neq \emptyset$, then $w_U=\vee\{vt_{i,a}:i\in U\}.$

Take $U\subseteq \phi(w,z_{a,b})$ so that $U\neq \emptyset$.
Write $U=\{i_1,\ldots,i_k\}$ where $1\leq i_1<i_2<\cdots < i_k<a$.  Then $b>v(i_1)>v(i_2)>\cdots >v(i_k)$.  We obtain the permutation matrix for $w_U$ from the permutation matrix for $v$ by doing the following:

* change the $1$'s in positions $(i_\ell,v(i_\ell))$ to $0$'s for all $\ell \in [k]$,

* change the $1$ in position $(a,b)$ to a $0$, and

* place $1$'s in positions $(a,v(i_k)),  (i_k,v(i_{k-1})),\ldots,(i_2,v(i_1)),(i_1,b)$.

\[\rk_v(c,d)-\rk_{w_U}(c,d)=
\begin{cases}
1& \text{if } (c,d)\in \displaystyle \bigcup_{\ell \in [k]} [i_\ell,a-1]\times [v(i_\ell),b-1], \\
0& \text{otherwise.}
\end{cases}\]

In particular,
\[\rk_v(c,d)-\rk_{vt_{{i_\ell},a}}(c,d)=
\begin{cases}
1& \text{if } (c,d)\in [i_\ell,a-1]\times[v(i_\ell),b-1], \\
0& \text{otherwise.}
\end{cases}
\]

\[\rk_v(c,d)-\rk_{w_U}(c,d)=\max\{\rk_v(c,d)-\rk_{v t_{a,i}}(c,d):i\in U\}\]
for all $c,d \in [n]$, which implies $\rk_{w_U}(c,d)=\min\{\rk_{v t_{a,i}}(c,d):i\in U\}$ for all $c,d\in [n]$.  Therefore, $w_U=\vee\{vt_{i,a}:i\in U\}$.

We continue with $w = 4721653$, $(a,b) = (5,5)$, and $y = z_{55}$ as in Example <ref>. Recall that $v = 4721563$ and $\phi(w,y) = \{1,3,4\}$. Consider $U = \{1,3\} \subseteq \phi(w,y)$, which corresponds to $\{u_1=5721463=vt_{1,5},u_2=4751263=vt_{3,5} \} \subseteq \Phi(w,y)$.  Then $w_U = v(531) = 5741263 = \vee \{u_1,u_2\}$.  The Rothe BPDs for $v$ and $w_U$ are presented below with the region $[1,4] \times [4,4] \cup [3,4] \times [2,4]$, in which $\rk_v$ and $\rk_{w_U}$ differ, highlighted on both BPDs in orange.

\[\begin{tikzpicture}[x=1.5em,y=1.5em]
\node at (3.5, 7.5) {$v=4721563$};
\fill[fill=orange](3,3)rectangle(4,7);
\fill[fill=orange](1,3)rectangle(3,5);
\draw[step=1,gray, thin] (0,0) grid (7,7);
\draw[color=black, thick](0,0)rectangle(7,7);
\draw[thick,rounded corners] (.5,0)--(.5,3.5)--(7,3.5);
\draw[thick,rounded corners] (1.5,0)--(1.5,4.5)--(7,4.5);
\draw[thick,rounded corners] (2.5,0)--(2.5,.5)--(7,.5);
\draw[thick,rounded corners] (3.5,0)--(3.5,6.5)--(7,6.5);
\draw[thick,rounded corners] (4.5,0)--(4.5,2.5)--(7,2.5);
\draw[thick,rounded corners] (5.5,0)--(5.5,1.5)--(7,1.5);
\draw[thick,rounded corners] (6.5,0)--(6.5,5.5)--(7,5.5);
\end{tikzpicture}
\hspace{3cm}
\begin{tikzpicture}[x=1.5em,y=1.5em]
\node at (3.5, 7.5) {$w_U = 5741263$};
\fill[fill=orange](3,3)rectangle(4,7);
\fill[fill=orange](1,3)rectangle(3,5);
\draw[step=1,gray, thin] (0,0) grid (7,7);
\draw[color=black, thick](0,0)rectangle(7,7);
\draw[thick,rounded corners] (.5,0)--(.5,3.5)--(7,3.5);
\draw[thick,rounded corners] (1.5,0)--(1.5,2.5)--(7,2.5);
\draw[thick,rounded corners] (2.5,0)--(2.5,.5)--(7,.5);
\draw[thick,rounded corners] (3.5,0)--(3.5,4.5)--(7,4.5);
\draw[thick,rounded corners] (4.5,0)--(4.5,6.5)--(7,6.5);
\draw[thick,rounded corners] (5.5,0)--(5.5,1.5)--(7,1.5);
\draw[thick,rounded corners] (6.5,0)--(6.5,5.5)--(7,5.5);
\end{tikzpicture}
\]

The following lemma is a standard exercise.

Fix a $\mathbb Z^d$ grading on $S=\kappa[z_1,\ldots,z_n]$ and an ideal $I$ of $S$ that is homogeneous with respect to the $\mathbb{Z}^d$ grading.  Pick a variable $z_j$ and write $y=z_j$.  If $I$ has a generating set that does not involve $y$, then

\[\hilb(S/(I+(y));\mathbf t)=(1-\mathbf t^{\deg(y)})\cdot \hilb(S/I;\mathbf t).\]

We will also need the following lemma, which records an elementary relationship among Hilbert functions.  We include a proof for completeness.

Suppose that $S$ is a finitely generated, $\mathbb{Z}^d$-graded $\kappa$-algebra and that $I_1,\ldots,I_r$ are homogeneous ideals of $S$.  If $J=\displaystyle \bigcap_{i \in [r]} I_r$, then

\[\hilb(S/J;\mathbf t)=\sum_{\emptyset \neq U\subseteq [r]} (-1)^{\#U-1}\hilb\left (S/\sum_{i\in U}I_i;\mathbf t\right ). \]

We will proceed by induction on $r$, noting that the case $r=1$ is simply the assertion $\hilb(S/I_1;\mathbf t) = \hilb(S/I_1;\mathbf t)$.  If $r \geq 2$, set $K = \displaystyle \bigcap_{i \in [r-1]} I_i$, and consider the short exact sequence of $S$-modules
\[
0 \rightarrow S/(K \cap I_r) \rightarrow S/K \oplus S/I_r \rightarrow S/(K+I_r) \rightarrow 0,
\]
where the first nontrivial map is $a \rightarrow (a,a)$ and the second is $(b,c) \mapsto b-c$.  Using this short exact sequence, $\hilb(S/J;\mathbf t) = \hilb(S/K;\mathbf t)+\hilb(S/I_r;\mathbf t)-\hilb(S/(K+I_r);\mathbf t)$.
Set $T = S/I_r$.  Then $\hilb(S/(K+I_r);\mathbf t) = \hilb(T/K;\mathbf t)$.  Now $T$ is a finitely generated, $\mathbb{Z}^d$-graded $\kappa$-algebra, and $KT = I_1T \cap \cdots \cap I_{r-1}T$ is an intersection of $r-1$ homogeneous ideals of $T$. Hence, by induction,
\begin{align*}
\hilb(S/I_r;\mathbf t) &- \hilb(S/(K+I_r);\mathbf t) = \hilb(T;\mathbf t)- \hilb(T/KT;\mathbf t) \\
&= \hilb(T;\mathbf t) - \displaystyle \sum_{\emptyset \neq U\subseteq [r-1]} (-1)^{\#U-1}\hilb\left (T/\sum_{i\in U}I_iT;\mathbf t\right )\\
& = \displaystyle \sum_{\substack{\emptyset \neq U\subseteq [r]\\ r \in U}} (-1)^{\#U-1}\hilb\left (S/\sum_{i\in U}I_i;\mathbf t\right ).
\end{align*}
Also by induction,
\begin{align*}
\hilb(S/K;\mathbf t) &= \displaystyle \sum_{\emptyset \neq U\subseteq [r-1]} (-1)^{\#U-1}\hilb\left (S/\sum_{i\in U}I_i;\mathbf t\right ) \\
&= \displaystyle \sum_{\substack{\emptyset \neq U\subseteq [r]\\ r \notin U}} (-1)^{\#U-1}\hilb\left (S/\sum_{i\in U}I_i;\mathbf t\right ).
\end{align*}
\begin{align*}
\hilb(S/J;\mathbf t) &= \displaystyle \sum_{\substack{\emptyset \neq U\subseteq [r]\\ r \in U}} (-1)^{\#U-1}\hilb\left (S/\sum_{i\in U}I_i;\mathbf t\right )+\displaystyle \sum_{\substack{\emptyset \neq U\subseteq [r]\\ r \notin U}} (-1)^{\#U-1}\hilb\left (S/\sum_{i\in U}I_i;\mathbf t\right )\\
& = \displaystyle \sum_{\emptyset \neq U\subseteq [r]} (-1)^{\#U-1}\hilb\left (S/\sum_{i\in U}I_i;\mathbf t\right ),
\end{align*}
as desired.

We now show that multigraded Hilbert series and K-polynomials of Schubert determinantal ideals satisfy a transition recurrence.  Our proof follows from Proposition <ref>, which gave a recurrence on unions of matrix Schubert varieties.

Take $w\in S_n$ and let $(a,b)$ be a lower outside corner of $D(w)$.  Let $\deg(z_{i,j}) = e_i+e_{n+j}$, in which case $(\mathbf{x,y})^{\deg(z_{i,j})} = x_iy_j$.  Then the following hold.

\[\hilb(R/I_w;\mathbf x,\mathbf y)=\left(1-x_ay_b\right)\hilb(R/I_v;\mathbf x,\mathbf y)+x_ay_b\sum_{\emptyset \neq U\subseteq \phi(w,z_{a,b})}(-1)^{\#U-1}\hilb(R/I_{w_U},\mathbf x,\mathbf y).\]

\[\mathcal K(R/I_w;\mathbf x,\mathbf y)=\left(1-x_ay_b\right)\mathcal K(R/I_v;\mathbf x,\mathbf y)+x_ay_b\sum_{\emptyset \neq U\subseteq \phi(w,z_{a,b})}(-1)^{\#U-1}\mathcal K(R/I_{w_U},\mathbf x,\mathbf y).\]

(i) First suppose $\rk_w(a,b)=0$.  In this case, $z_{a,b}\in I_w$ and $I_w=I_v+(z_{a,b})$,  Furthermore, $\phi(w,z_{a,b})=\emptyset$.  The natural generators of $I_v$ do not involve $z_{a,b}$.  Thus, we apply Lemma <ref> and see

\[\hilb(R/I_w;\mathbf x,\mathbf y)=\left(1-x_ay_b\right) \cdot \hilb(R/I_v;\mathbf x,\mathbf y),\]
as desired.

Now assume $\rk_w(a,b)\geq 1$.  Applying Proposition <ref>, $\init_{z_{a,b}}(I_w)= C \cap (N+(z_{a,b}))$, where $\displaystyle C=\bigcap_{i\in \phi (w,z_{a,b})}I_{v t_{a,i}}$ and $N=I_v$.

Thus, by Lemma <ref>,

\begin{align*}
\hilb(R/I_w;\mathbf x,\mathbf y)&= \hilb(R/\init_y(I_w);\mathbf x,\mathbf y)\\
&=\hilb(R/C;\mathbf x,\mathbf y)+\hilb(R/(N+(z_{a,b}));\mathbf x,\mathbf y)\\
&\quad -\hilb(R/(C+N+(z_{a,b}));\mathbf x,\mathbf y).
\end{align*}

Since $v\leq u$ for all $u\in \phi(w,z_{a,b})$, we have $N\subseteq C$ and so $C+N+(z_{a,b})=C+(z_{a,b})$.  As $C$ has a generating set that does not involve $z_{a,b}$, we apply Lemma <ref> to conclude
\[\hilb(R/(C+N+(z_{a,b}));\mathbf x,\mathbf y)=\hilb(R/(C+(z_{a,b}));\mathbf x,\mathbf y)=\left(1-x_ay_b\right)\hilb(R/C;\mathbf x,\mathbf y).\]

\[\hilb(R/I_w;\mathbf x,\mathbf y)=\left(1-x_ay_b\right)\hilb(R/N;\mathbf x,\mathbf y)+x_ay_b\hilb(R/C;\mathbf x,\mathbf y).\]

By Lemma <ref>,

\[\hilb(R/C;\mathbf x,\mathbf y)=\sum_{ \emptyset \neq U\subseteq \phi(w,z_{a,b})}(-1)^{\#U-1} \hilb(R/\sum_{i\in U} I_{vt_{a,i}};\mathbf x,\mathbf y).\]

If $U\subseteq \phi(w,z_{a,b})$ and $U\neq \emptyset$, Lemma <ref> yields

\[\hilb(R/\sum_{i\in U} I_{vt_{a,i}};\mathbf x,\mathbf y)=\hilb(R/w_U;\mathbf x,\mathbf y),\]

from which the result follows.

(ii) This is an immediate consequence of <ref>.

We may recover transition equations for (double) Schubert polynomials and (double) Grothendieck polynomials from Proposition <ref>.  We start by working with the $\beta$-double Grothendieck polynomials of [26], from which the other formulas follow immediately.

Write $\mathbb Z[\beta][\mathbf x,\mathbf y]=\mathbb Z[\beta][x_1,\ldots,x_n,y_1,\ldots,y_n]$.  There is an action of the symmetric group $S_n$ on $\mathbb Z[\beta][\mathbf x,\mathbf y]$ defined by $w\cdot f=f(x_{w(1)},\ldots,x_{w(n)},y_1,\ldots, y_n)$.  Given $f\in \mathbb Z[\beta][\mathbf x,\mathbf y]$, define
\[\pi_i(f)=\frac{(1+\beta x_{i+1})f-(1+\beta x_i)s_i\cdot f}{x_i-x_{i+1}}.\]

Let $x_i\oplus y_j=x_i+y_j+\beta x_iy_j$.   We define the $\beta$- double Grothendieck polynomial $\mathfrak G^{(\beta)}_w(\mathbf x,\mathbf y)$ as follows:  If $w_0=n \, n-1 \, \ldots \, 1$, then $\mathfrak G^{(\beta)}_w(\mathbf x,\mathbf y)=\displaystyle \prod_{i+j\leq n} (x_i\oplus y_j)$.  Otherwise, given $w\in S_n$ with $w(i)>w(i+1)$, we define $\mathfrak G^{(\beta)}_{ws_i}(\mathbf x,\mathbf y)=\pi_i(\mathfrak G^{(\beta)}_w(\mathbf x,\mathbf y))$.  Because the operators $\pi_i$ satisfy the same braid and commutation relations as the simple reflections $s_i$, the polynomial $\mathfrak G^{(\beta)}_w(\mathbf x,\mathbf y)$ is well defined.  Setting $\beta=0$ and replacing each $y_i$ with $-y_i$ recovers $\mathfrak S_w(\mathbf x,\mathbf y)$.

Setting $\beta=-1$, we may obtain $\mathfrak G^{(-1)}_w(\mathbf x,\mathbf y)$ by taking $\mathcal K(R/I_w;\mathbf x,\mathbf y)$ and substituting $x_i\mapsto (1-x_i)$ and $y_i\mapsto (1-y_i)$ for all $i\in [n]$ (see <cit.> and <cit.>). From Proposition <ref>, we recover <cit.>.

Take $w\in S_n$ and let $(a,b)$ be a lower outside corner of $D(w)$.  Then
\[\mathfrak G^{(\beta)}_w(\mathbf x,\mathbf y)=(x_a\oplus y_b) \mathfrak G_v^{(\beta)}(\mathbf x,\mathbf y)+(1+\beta(x_a\oplus y_b))\sum_{ \emptyset \neq U\subseteq \phi(w,z_{a,b})}\beta^{\#U-1}\mathfrak G_{w_U}^{(\beta)}(\mathbf x,\mathbf y).\]

Making the substitutions $x_i\mapsto (1-x_i)$ and $y_i\mapsto (1-y_i)$ into  Proposition <ref><ref> yields

\begin{align*}
\mathfrak G^{(-1)}_w(\mathbf x,\mathbf y)&=(x_a+ y_b-x_ay_b) \mathfrak G_v^{(-1)}(\mathbf x,\mathbf y)\\&+(1-(x_a+ y_b-x_ay_b))\sum_{ \emptyset \neq U\subseteq \phi(w,z_{a,b})}(-1)^{\#U-1}\mathfrak G_{w_U}^{(-1)}(\mathbf x,\mathbf y).
\end{align*}

Thus, the $\beta=-1$ case follows immediately from Proposition <ref><ref>.

We now derive the general equation from the $\beta=-1$ case. As a shorthand, write
\[\mathfrak G^{(-1)}_u(-\beta\mathbf x,-\beta \mathbf y)=\mathfrak G^{(-1)}_u(-\beta x_1,\ldots, -\beta x_n,-\beta y_1,\ldots,-\beta y_n).\]
Substituting $x_i\mapsto -\beta x_i$ and $y_i\mapsto -\beta y_i$ for all $i$  yields

\begin{align*}
\mathfrak G^{(-1)}_w(-\beta\mathbf x,-\beta\mathbf y)&=-\beta(x_a\oplus y_b)\mathfrak G_v^{-1}(-\beta\mathbf x,-\beta\mathbf y)\\
&\quad +(1+\beta(x_a\oplus y_b))\sum_{ \emptyset \neq U\subseteq \phi(w,z_{a,b})}(-1)^{\#U-1}\mathfrak G_{w_U}^{(-1)}(-\beta\mathbf x,-\beta\mathbf y).
\end{align*}

We have $\mathfrak G^{(-1)}_u(-\beta\mathbf x,-\beta \mathbf y)=(-\beta)^{\ell(u)}\mathfrak G^{(\beta)}_u(\mathbf x,\mathbf y)$ for all $u\in S_n$. Thus,

\begin{align*}
(-\beta)^{\ell(w)}\mathfrak G^{(\beta)}_w(\mathbf x,\mathbf y)&=-\beta(x_a\oplus y_b) (-\beta)^{\ell(v)}\mathfrak G_v^{\beta}(\mathbf x,\mathbf y)\\
&\quad +(1+\beta(x_a\oplus y_b))\sum_{ \emptyset \neq U\subseteq \phi(w,z_{a,b})}(-1)^{\#U-1}(-\beta)^{\ell(w_U)}\mathfrak G_{w_U}^{(\beta)}(\mathbf x,\mathbf y).
\end{align*}

If $U\subseteq \phi(w,z_{a,b})$, then $\ell(w_U)=\ell(w)+\#U-1$. Furthermore, $\ell(v)+1=\ell(w)$.  Thus,

\begin{align*}
(-\beta)^{\ell(w)}\mathfrak G^{(\beta)}_w(\mathbf x,\mathbf y)&=(-\beta)^{\ell(w)}(x_a\oplus y_b) \mathfrak G_v^{(\beta)}(\mathbf x,\mathbf y)\\
&\quad +(-\beta)^{\ell(w)}(1+\beta(x_a\oplus y_b))\sum_{ \emptyset \neq U\subseteq \phi(w,z_{a,b})}\beta^{\#U-1}\mathfrak G_{w_U}^{(\beta)}(\mathbf x,\mathbf y),
\end{align*}

from which the claim follows.

By making appropriate specializations of the equation in Corollary <ref>, we may recover transition equations for (double) Schubert and (double) Grothendieck polynomials.
In particular, we recover Theorem <ref>.

As a consequence of Corollary <ref>, by setting $\beta=0$ we obtain

\[\mathfrak G^{(0)}_w(\mathbf x,\mathbf y)=(x_a+y_b) \mathfrak G_v^{(0)}(\mathbf x,\mathbf y)+\sum_{  i\in \phi(w,z_{a,b})}\mathfrak G_{w_{\{i\}}}^{(0)}(\mathbf x,\mathbf y).\]

We have $w_{\{i\}}=vt_{i,a}$ for all $i\in \phi(w,z_{a,b}).$  Furthermore,
$\mathfrak S_u(\mathbf x,\mathbf y)=\mathfrak G^{(0)}_u(\mathbf x,-\mathbf y)$.  Thus, the result follows immediately by substituting $y_i\mapsto -y_i$ for all $i\in [n]$.

§ FURTHER INQUIRY

Computations in this section were assisted by Macaulay2 [35] and Sage [75].  All computations were performed over the field $\kappa = \QQ$.  As usual, let $R=\kappa[z_{1,1}, \ldots, z_{n,n}]$.

Closely related to the study of Gröbner degenerations is, of course, the study of Gröbner bases themselves.  The following problem remains open:

Describe the generating sets of the Schubert determinantal ideal $I_w$ that can occur as a Gröbner basis under some diagonal term order $\sigma$.

Even when $\init_\sigma(I_w)$ is radical, explicit Gröbner bases are only known in a special case governed by pattern avoidance [45]. One of the two patterns in $S_5$ that does not fall under that result is $21543$.  When $w =21543$ and $\sigma$ is lexicographic order on left-to-right reading order of the variables, already there are Gröbner generators of higher degree than any Fulton generator of $I_w$.  The ideal $\init_\sigma(I_{w})$ has a reduced Gröbner basis containing nine elements, including one of degree five, though $I_{21543}$ has only eight Fulton generators, one of degree one and seven of degree three.

The situation looks especially complicated when there is some diagram $D \subseteq [n] \times [n]$ so that $\#\{\mathcal B\in \bpd(w):D(\mathcal B)=D\}>1$, in which case the authors expect there will always be different Gröbner bases arising from different diagonal term orders.  For example, if $w = 214365$, take $\sigma$ to be lexicographic order on left-to-right reading order and $\sigma'$ to be lexicographic order on the variables ordered starting from $z_{11}$, progressing down the column $1$, then down column $2$, and so on.  Then
\[
\init_\sigma (I_w) = (z_{11} ,z_{12}z_{21}z_{33}, z_{12}z_{21}z_{34}z_{43}z_{55}, z_{12}z_{23}z_{31}z_{34}z_{43}z_{55}, z_{13}z_{21}^2z_{32}z_{34}z_{43}z_{55})
\]
\[
\init_{\sigma'} (I_w) = (z_{11} ,z_{12}z_{21}z_{33}, z_{12}z_{21}z_{34}z_{43}z_{55}, z_{21}z_{32}z_{13}z_{34}z_{43}z_{55}, z_{31}z_{12}^2z_{23}z_{34}z_{43}z_{55}).
\]
Both  $(z_{11}, z_{12}, z_{21}^2)$ and $(z_{11}, z_{12}^2, z_{21})$ are primary ideals giving multiplicity two at the prime ideal $(z_{11}, z_{12}, z_{21})$ (as predicted by Theorem <ref>), but, as we see above, they contribute to distinct initial ideals.  We see from the initial ideals that the reduced Gröbner bases under $\sigma$ and under $\sigma'$ both have five elements, one each of degrees one, three, five, six, and seven; however, the Gröbner bases themselves are distinct.  The complexity of this example may be surprising in light of the fact that $I_{214365}$ has only three Fulton generators, one each of degrees one, three, and five.

Moreover, not only can diagonal initial ideals fail to be radical, they can even have embedded associated primes. For example, with $\sigma$ and $\sigma'$ as above, $\init_\sigma(I_{2143675})$ has $53$ associated primes, $43$ of which are height four and $10$ of which are of height five, and $\init_{\sigma'}(I_{2143675})$ has $49$ associated primes,  $43$ of which are of height four and six of which are of height five.  Because  $\spec(R/\init_\sigma(I_{2143675}))$ and $\spec(R/\init_{\sigma'}(I_{2143675}))$ are equidimensional, all height five associated primes are embedded. The authors do not know if the condition $\#\{\mathcal B\in \bpd(w):D(\mathcal B)=D\}\leq 1$ for all $D \subseteq [n] \times [n]$ precludes embedded primes, nor even if it is possible for some diagonal initial ideal of $I_w$ to have embedded primes while another does not.

Characterize the permutations $w \in S_n$ for which there is some diagonal term order $\sigma$ so that $\init_\sigma(I_w)$ has embedded primes. Relatedly, characterize the permutations $w \in S_n$ so that $\init_\sigma(I_w)$ has embedded primes for all diagonal term orders $\sigma$.

We state below two problems related to Subsection <ref>.

Characterize the Cohen–Macaulay (or sequentially Cohen–Macaulay) ASM varieties.

Characterize the sets of permutation $w_1, \ldots, w_r \in S_n$ with $J = \displaystyle \bigcap_{i \in [r]} I_{w_i}$ for which $\spec(R/J)$ is Cohen–Macaulay (or sequentially Cohen–Macaulay).

In light of Lemma <ref>, a complete solution to Problem <ref> would imply a complete solution to Problem <ref>.  Equidimensional ASMs that are not Cohen–Macaulay (hence also not sequentially Cohen–Macaulay) appear as soon as $\asm(5)$ and even as a union of just two matrix Schubert varieties: If $w_1 = 34512$ and $w_2 = w_1^{-1} = 45123$, then
\[
I_{w_1} \cap I_{w_2} = (z_{11},z_{12},z_{21}, z_{22})+((z_{13},z_{23}) \cap (z_{31},z_{32})),
\]
which defines (up to affine factors) a standard first example of a scheme that is equidimensional but not Cohen–Macaulay.  One easily checks the equality $I_{w_1} \cap I_{w_2} = I_A$ for
\[
A = \begin{pmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
1 & 0 & -1 & 0 & 1\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}.
\]

An example of an ASM that is not even equidimensional is $A = \begin{pmatrix} 0 & 0 & 1 & 0\\
1 & 0 & -1 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \end{pmatrix}$, which satisfies $I_A = I_{4123} \cap I_{3412} = (z_{11}, z_{12}) + ((z_{13}) \cap (z_{21}, z_{22})).$  Because $\spec(R/I_A)$ is not equidimensional, it is not Cohen–Macaulay.  However, the Stanley–Reisner complex $\Delta(I_A)$ is shellable in the nonpure sense of <cit.>; therefore, $\spec(R/I_A)$ is sequentially Cohen–Macaulay.

The questions of which ASMs or, more generally, which unions of matrix Schubert varieties are Cohen–Macaulay (or sequentially Cohen–Macaulay) are wide open and, in the opinion of the authors, quite interesting.

Finally, we consider a problem concerning integer partitions.

Fix $N \in \NN$, and consider the standard grading on $R$ and any diagonal term order $\sigma$.  Characterize the partitions of $N$ that arise as a vector of multiplicities of the minimal primes of $\init_\sigma(J)$ for some $J = \displaystyle \bigcap_{i \in [r]} I_{w_i}$ for permutations $w_1, \ldots, w_r$ all of the same Coxeter length and satisfying $e(R/J) = N$.

Using Theorem <ref>, one can fix a monomial Schubert determinantal ideal $I_w$ and integer $m \in \ZZ_+$ and construct an intersection $J$ of Schubert determinantal ideals so that the multiplicity of $\spec(R/I_w)$ along $\spec(R/\init_\sigma(J))$ is exactly $m$ (provided $n$ is sufficiently large).  To give an example of this construction, it will be convenient to work in $S_{\infty} = \displaystyle \bigcup_{n \in \ZZ_+} S_n$ with the usual embeddings $S_n \hookrightarrow S_{n+1}$.

If $I_w = (z_{11}, z_{12}, z_{21})$, then we may begin with $w=321 \in S_3$ and choose the lower outside corner $(a,b) = (2,1)$. We multiply by the $3$-cycle $(234) = (a(b^{-1})(n+1))$ to obtain $w' = 3142 \in S_4$.  We have constructed $w'$ to have a blank tile at $(3,2)$, where pipes $2$ and $w^{-1}(1) = 3$ cross in the Rothe BPD of $w$, so that, with $y = z_{32}$, we have $C_{y, I_{w'}} = I_w$.  Similarly, we may multiply $w'$ by the $3$-cycle $(345)$ to obtain $w'' = 31254 \in S_5$, which satisfies $C_{y', I_{w''}} = I_{w'}$.  It follows from Theorem <ref> that, if $J = I_{32145} \cap I_{31425} \cap I_{31245}$, then $\mult_{I_w}(R/\init_\sigma(J)) = 3$.  (At each step, if the newly created lower outside corner is $(a',b')$, then multiplying by the $3$ cycle $(a'(b'^{-1})(n+1))$ creates a new permutation whose unique droop of pipe $b'$ recovers the previous permutation via Lemma <ref>.)  The cost of growing the multiplicity at $I_w$ to $3$ in this way is multiplicity $2$ at $(z_{11},z_{12}, z_{32})$ and multiplicity $1$ at $(z_{11},z_{12}, z_{43})$ as well as multiplicity $1$ at $(z_{11},z_{12},z_{13})$ coming from the unique droop of pipe $3$ in the Rothe BPD of $31254$.

\[
\begin{tikzpicture}[x=1.5em,y=1.5em]
\node at (1.5, 3.5) {$w = 321$};
\draw[step=1,gray, thin] (0,0) grid (3,3);
\draw[color=black, thick](0,0)rectangle(3,3);
\draw[thick,rounded corners,color=blue] (.5,0)--(.5,0.5)--(3,0.5);
\draw[thick,rounded corners,color=blue] (1.5,0)--(1.5,1.5)--(3,1.5);
\draw[thick,rounded corners,color=blue] (2.5,0)--(2.5,2.5)--(3,2.5);
\end{tikzpicture}
\hspace{2cm}  \begin{tikzpicture}[x=1.5em,y=1.5em]
\node at (2, 4.5) {$w' = 3142$};
\draw[step=1,gray, thin] (0,0) grid (4,4);
\draw[color=black, thick](0,0)rectangle(4,4);
\draw[thick,rounded corners,color=blue] (.5,0)--(.5,2.5)--(4,2.5);
\draw[thick,rounded corners,color=blue] (1.5,0)--(1.5,0.5)--(4,0.5);
\draw[thick,rounded corners,color=blue] (2.5,0)--(2.5,3.5)--(4,3.5);
\draw[thick,rounded corners,color=blue] (3.5,0)--(3.5,1.5)--(4,1.5);
\end{tikzpicture}
\hspace{2cm} \begin{tikzpicture}[x=1.5em,y=1.5em]
\node at (2.5, 5.5) {$w'' = 31254$};
\draw[step=1,gray, thin] (0,0) grid (5,5);
\draw[color=black, thick](0,0)rectangle(5,5);
\draw[thick,rounded corners,color=blue] (.5,0)--(.5,3.5)--(5,3.5);
\draw[thick,rounded corners,color=blue] (1.5,0)--(1.5,2.5)--(5,2.5);
\draw[thick,rounded corners,color=blue] (2.5,0)--(2.5,4.5)--(5,4.5);
\draw[thick,rounded corners,color=blue] (3.5,0)--(3.5,0.5)--(5,0.5);
\draw[thick,rounded corners,color=blue] (4.5,0)--(4.5,1.5)--(5,1.5);
\end{tikzpicture}
\]

In this case, we have constructed the partition $7 = 3+2+1+1$ coming from $7 = e(R/J)$ and $3$, $2$, $1$, and $1$ the multiplicities at the distinct minimal primes of $R/\init_\sigma(J)$.  The intersection $J' = I_{3214} \cap I_{3142} \cap I_{2413} \cap I_{4123} \cap I_{4123} \cap I_{2341}$ gives $7 = e(R/J') = 3+1+1+1+1$. The complete list of possible partitions arising in this way, however, is not immediately obvious.

If $\lambda$ is a partition whose conjugate has distinct parts, it is possible to construct an ideal $J=\displaystyle \bigcap_{i\in[r]}I_{w_i}$ for which the vector of multiplicities is $\lambda$.  Fix a partition $\mu=(\mu_1,\ldots,\mu_r)$ with distinct parts and let $w_i$  be the simple transposition $s_{\mu_i}$.  Then $\bpd(w_i)$ has exactly $\mu_i$ elements and $\{D(\mathcal B):\mathcal B\in \bpd(w_i)\}=\{\{(i,i)\}:i\in[\mu_i]\}$.  Thus, if $\lambda=(\lambda_1,\ldots, \lambda_{\mu_1})$ is the conjugate of $\mu$ and $\sigma$ is a diagonal term order on $R$,  the multiplicity of $\spec(R/(z_{i,i}))$ along $\spec(R/\init_\sigma(J))$ is exactly $\lambda_i$.

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