Proof. It suffices to determine which clans $\gamma $ satisfy $V(\gamma )_\bullet \in \mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m})$ , where $V(\gamma )$ is the flag representative of ${{\mathcal O}}_\gamma $ specified in Definition 2.1 above. We observe first that each $e_i$ is an eigenvector for $\mathsf {x}_{p,q}$ and that if $v_i \in \{e_r+e_s,e_r-e_s\}$ with $r \in [p]$ and $p<s \leq n$ , then ${\mathbb {C}}\{v_i,\mathsf {x}_{p,q} v_i\}={\mathbb {C}}\{e_r,e_s\}$ . Thus, $V(\gamma )_i+\mathsf {x}_{p,q} V(\gamma )_i$ is spanned by those standard basis vectors $e_k$ such that one of

1. (A) there is some $j \in [i]$ with $v_j=e_k$ , or

2. (B) there is some $a \in [i]$ with $v_a=e_r+e_s$ and $k \in \{r,s\}$

holds. On the other hand, the standard basis vector $e_k$ is an element of $V(\gamma )_{m_i}$ if and only if one of

1. (A’) there is some $j \in [m_i]$ with $v_j=e_k$ , or

2. (B’) there is some $b \in [m_i]$ with $v_b=e_r-e_s$ and $k \in \{r,s\}$

holds. Indeed, if there is no $j \in [m_i]$ with $v_j=e_k$ , then $e_k \in V(\gamma )_{m_i}$ if and only if there are $a,b \in [m_i]$ with $v_a=e_r+e_s$ , $v_b=e_r-e_s$ and $k \in \{r,s\}$ . In this case, $a<b$ and $c_a=c_b$ by definition of the flag $V(\gamma )_\bullet $ .

For each i, we have $V(\gamma )_i\leq V(\gamma )_{m_i}$ since $m_i\geq i$ . Thus, $V(\gamma )\in \mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m})$ if and only if $V(\gamma )_i+\mathsf {x}_{p,q}V(\gamma )_i \leq V(\gamma )_{m_i}$ for all i. Suppose $e_k\in V(\gamma )_i +\mathsf {x}_{p,q}V(\gamma _i)$ , so we are in either Case (A) or Case (B) from above. In Case (A), the vector $e_k$ is also an element of $V(\gamma )_{m_i}$ by (A’) since $i\leq m_i$ .

In Case (B), there exists $a\in [i]$ such that $v_a=e_r+e_s$ , and we have ${\mathbb {C}}\{e_r,e_s\}\leq V(\gamma )_i +\mathsf {x}_{p,q}V(\gamma _i)$ with $k\in \{r,s\}$ . By definition of the flag $V(\gamma )$ , there exists some $b>a$ such that $c_a=c_b$ and $v_b= e_r-e_s$ . By (B’), ${\mathbb {C}}\{e_r,e_s\}\leq V(\gamma )_{m_i}$ if and only if $b\in [m_i]$ – that is, if $b\leq m_i$ . The proposition follows.

Proof. As each of ${{\mathcal O}}_\sigma $ and ${{\mathcal O}}_\tau $ lies in $\overline {{{\mathcal O}}_\gamma }$ , we have

• • $\gamma (i;-) \leq \sigma (i;-)=0$ for $i \leq p$ and

• • $\gamma (i;+) \leq \tau (i;+)=0$ for $i \leq q$

by Theorem 2.7. These conditions imply that $\gamma $ cannot contain any signs or pairs of positive integers within the first q entries and cannot contain any minus signs or pairs of positive integers within the first p entries. Since $\gamma $ contains at most q natural number pairs, conditions (a) and (b) now follow. Condition (c) follows from (a) and (b) and the fact that $\gamma $ is a clan. The last statement of the lemma follows immediately, as both ${{\mathcal O}}_\sigma $ and ${{\mathcal O}}_\tau $ lie in $\mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m})$ for every Hessenberg vector ${\mathbf m}$ by Proposition 3.1.

Proof. We observe that both $\sigma \leq \gamma _e$ and $\tau \leq \gamma _e$ . Thus, $\gamma $ satisfies each of the conditions (a), (b) and (c’) by Lemma 3.3, and the first assertion of the lemma is proved. To prove the second, we observe that the equality of the various statistics holds in the case of $w=e$ ; cf. Example 2.8. The general case now follows since $\gamma _0(i;-)\leq \gamma _w(i;-)\leq \gamma _e(i;-)$ , $\gamma _0(i;+)\leq \gamma _w(i;-)\leq \gamma _e(i;+)$ , and $\gamma _e(i,j)\leq \gamma _w(i,j)\leq \gamma _0(i,j)$ by Theorem 2.7 in all cases.

Proof. We have a pair $s<t$ such that $s \leq i<j<t$ and $c_s^w=c_t^w$ if and only if $t=w^{-1}(s)+p$ by definition of $\gamma _w$ .

Proof. It follows immediately from Lemma 3.3 that $\gamma =\gamma _w$ for some $w \in {\mathbf {S}}_q$ . By Proposition 3.1, for each $i \in [q]$ , we have $m_i \geq p+w^{-1}(i)$ . Thus, for each $i\in [q]$ ,

$$\begin{align*}p+w^{-1}(i)\leq m_i \leq p+q \Leftrightarrow w^{-1}(i) \leq m_i -p \leq q. \end{align*}$$

It follows that $(m_1-p,m_2-p,\ldots ,m_q-p)$ is a sequence of positive integers satisfying $m_i-p \leq q$ ; it is also weakly increasing since ${\mathbf m}$ is. Moreover, for each $i \in [q]$ ,

$$ \begin{align*}m_i-p \geq \max\{w^{-1}(j) \mid j \leq i\} \geq i. \end{align*} $$

This concludes the proof.

Proof. Let w be obtained from u by switching j and k. Direct examination shows that for all $a,b \in [n]$ , all of $\gamma _w(a;+) \leq \gamma _u(a;+)$ , $\gamma _w(a;-) \leq \gamma _u(a;-)$ and $\gamma _w(a,b) \geq \gamma _u(a,b)$ hold. Thus, $\gamma _u \leq \gamma _w$ by Theorem 2.7. Assume for contradiction that there exist $a,b \in [n]$ with $a<b$ and $s \in [q]$ such that $c^w_a=c^w_b=s$ and $b>m_a$ . By the definition of the clan $\gamma _w$ , $a \in [q]$ and $b=p+w^{-1}(a)$ . Since ${{\mathcal O}}_{\gamma _u} \subseteq \mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m})$ , it must be that $a \in \{j,k\}$ and so $b \in \{p+w(j),p+w(k)\}$ . However,

$$ \begin{align*}m_a>m_i \geq p+u^{-1}(i)=p+w^{-1}(i)>p+\max\{w^{-1}(j),w^{-1}(k)\} \geq b, \end{align*} $$

giving the desired contradiction.

Proof. By Lemma 3.5, $\gamma _w(i-1, j)=\left |\left \{k<i \mid w^{-1}(k)>j-p\right \}\right |$ . By definition of the Hessenberg vector ${\mathbf m}$ , there exists $a<i$ such that $m_i=w^{-1}(a)+p$ . Our assumptions imply that

$$ \begin{align*}j<m_i=w^{-1}(a)+p \Rightarrow j-p<w^{-1}(a)\quad \text{and} \quad w^{-1}(i)<w^{-1}(a). \end{align*} $$

Let $k \in \{1,2, \ldots , q\}$ such that $j-p<w^{-1}(k) \leq w^{-1}(a)$ . Note that $j-p<w^{-1}(k)$ implies $w^{-1}(i)<w^{-1}(k)$ and $k \neq i$ .

If $k>i$ , then $w^{-1}(k)<w^{-1}(a) $ and $w^{-1}$ contains $w^{-1}(a) w^{-1}(i) w^{-1}(k)$ as a subsequence, contradicting the fact that $w^{-1}$ is $312$ -free. Thus, we must have $k<i$ . This shows that

$$ \begin{align*}\left\{k \in[q] \mid {j-p}<w^{-1}(k) \leq w^{-1}(a)\right\} \subseteq\left\{k<i \mid w^{-1}(k)>j-p\right\}. \end{align*} $$

The sets are actually equal, since

$$ \begin{align*}w^{-1}(a)=\max \left\{w^{-1}(1), \ldots, w^{-1}(i)\right\}.\end{align*} $$

We conclude $\gamma _w(i-1, j)=w^{-1}(a)-(j-p)=m_i-j$ , as desired.

Proof of Theorem 3.11.

By Proposition 3.2, Corollary 3.7 and Corollary 3.9, every irreducible Hessenberg variety defined using $\mathsf {x}_{p,q}$ is equal to $\overline {{{\mathcal O}}_{\gamma _w}}$ for some $\gamma _w\in \mathbf {Clan}_{p,q}^{231}$ . To complete the proof, we show that $\mathrm {Hess}(\mathsf {x}_{p,q}, {\mathbf m}(w)) = \overline {{{\mathcal O}}_{\gamma _w}}$ . It follows immediately from the definition of the Hessenberg vector ${\mathbf m}(w)$ and Proposition 3.1 that ${{\mathcal O}}_{\gamma _w} \subset \mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m}(w))$ . It therefore suffices to show that if ${{\mathcal O}}_\gamma \subset \mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m}(w))$ for some $\gamma =c_1c_2\cdots c_n\in \mathbf {Clan}_{p,q}$ , then $\gamma \leq \gamma _w$ .

By Theorem 2.7 and Lemma 3.4, we must prove $\gamma (i, j) \leq \gamma _w(i, j)$ for all $i \in [q]$ and $j \in \{p+1, \ldots , n\}$ . Seeking a contradiction, suppose $\gamma (i, j)>\gamma _w(i, j)$ . We may assume i is minimal with respect to this property. We write ${\mathbf m}={\mathbf m}(w)$ throughout to simplify notation.

Consider first the case $i=1$ . Note that $\gamma (1, j), \gamma _w(1, j) \in \{0,1\}$ , so we must have $\gamma (1, j)=1$ and $\gamma _w(1, j)=0$ . The latter implies $w^{-1}(1)+p \leq j$ by Lemma 3.5. On the other hand, $\gamma (1, j)=1$ implies by Lemma 2.9 that there exists $t>j$ such that $c_1=c_t$ . Now $m_1=w^{-1}(1)+p<t$ , contradicting the assumption that ${{\mathcal O}}_\gamma \subset \mathrm {Hess}(\mathsf {x}_{p, q}, {\mathbf m})$ .

Now, assume $i>1$ . We have both $\gamma (i-1, j) \leq \gamma _w(i-1, j)$ and $\gamma (i, j)>\gamma _w(i, j)$ . By Lemma 2.9, this can only be the case if

(3) $$ \begin{align} \gamma(i, j)-\gamma(i-1, j)=1, \end{align} $$

and

(4) $$ \begin{align} \gamma_w(i, j)-\gamma_w(i-1, j)=0. \end{align} $$

We may furthermore conclude that

(5) $$ \begin{align} \gamma(i-1, j)=\gamma_w(i-1, j) \end{align} $$

since otherwise, $\gamma (i-1, j)<\gamma _w(i-1, j)$ and

$$ \begin{align*}\gamma(i, j)=\gamma(i-1, j)+1 \leq \gamma_w(i-1, j)=\gamma_w(i, j)_{,} \end{align*} $$

contradicting our assumption that $\gamma (i, j)>\gamma _w(i, j)$ .

By Lemma 2.9, equation (3) implies that there exists $t>j$ such that $c_i=c_t$ . From equation (4), we get that

$$ \begin{align*} &\left|\left\{w^{-1}(1), \ldots, w^{-1}(i)\right\} \cap\{j-p+1, \ldots, q\}\right| = \\ & \quad \quad \quad\quad\quad\quad\quad \left|\left\{w^{-1}(1), \ldots, w^{-1}(i-1)\right\} \cap\{j-p+1, \ldots, q\}\right| \end{align*} $$

so $w^{-1}(i) \leq j-p$ , implying $w^{-1}(i)+p \leq j$ .

If $m_i=w^{-1}(i)+p$ , then we have $m_i<t$ , a contradiction to ${{\mathcal O}}_\gamma \subset \mathrm {Hess}(\mathsf {x}_{p, q}, {\mathbf m})$ . We obtain the same contradiction if $j \geq m_i$ so we may now assume both $m_i>w^{-1}(i)+p$ and $j<m_i$ . By Lemma 3.12 and equation (5), $\gamma (i-1, j)=m_i-j$ . This implies there are precisely $m_i-j$ pairs $(a<b)$ such that $a \leq i-1<j<b$ and $c_a=c_b$ . As ${{\mathcal O}}_{\gamma } \subseteq \mathrm {Hess}(\mathsf {x}_{p, q}, {\mathbf m})$ , we have $b \leq m_a \leq m_i$ in each case. There are only $m_i-j$ values b such that $j<b \leq m_i$ , and each such position in the clan $\gamma $ is occupied by $c_b$ for one of the pairs counted by $\gamma (i-1, j)$ . This forces $t>m_i$ , another contradiction. We conclude $\mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m})=\overline {{{\mathcal O}}_{\gamma _w}}$ , as desired.

Proof. Write $w=w(1)\ldots w(n)$ in one-line notation and find k such that $w(k)=n$ . If $n>1$ , let $w^\prime $ be obtained from w by erasing n from the given one-line representation and let ${\mathbf h}^\prime $ be obtained from $w^\prime $ as ${\mathbf h}$ was obtained from w. For $i \in [n]$ , set

$$ \begin{align*}\mathsf{i}\mathsf{n}\mathsf{v}_i(w):=\mid \{j>i\mid w(j)<w(i)\} \mid, \end{align*} $$

and define $\mathsf {i}\mathsf {n}\mathsf {v}_i(w^\prime )$ similarly.

We will show by induction on n that $\mathsf {i}\mathsf {n}\mathsf {v}_i(w)=h_i-i$ for every i, from which the lemma follows. The case $n=1$ is trivial. Assume $n>1$ . We observe that if $i \geq k$ , then $w_j<w_i$ for all $j>i$ (since w is $312$ -free); hence, $\mathsf {i}\mathsf {n}\mathsf {v}_i(w)=n-i=h_i-i$ . If $i<k$ , then

$$ \begin{align*}\mathsf{i}\mathsf{n}\mathsf{v}_i(w)=\mathsf{i}\mathsf{n}\mathsf{v}_i(w^\prime)=h^\prime_i-i=h_i-i, \end{align*} $$

the second equality following from the inductive hypothesis.

Proof. Recall $c_i^w=c_j^w\in {\mathbb {N}}$ if and only if $j=w^{-1}(i)+p$ . Keeping also in mind that $\ell (w)=\ell (w^{-1})$ , we therefore have

$$\begin{align*}|\{k\in {\mathbb{N}} \mid c_s=c_t=k \text{ for } s<i<t<w^{-1}(i)+p \}| = |\{ s<i \mid w^{-1}(s)<w^{-1}(i) \}|, \end{align*}$$

and thus,

$$ \begin{align*} \sum_{i\in [q]} |\{k\in {\mathbb{N}} \mid c_s=c_t=k \text{ for } s<i<t<j \}| = \ell(w_0) - \ell(w^{-1})=\frac{q(q-1)}{2}-\ell(w). \end{align*} $$

We now obtain

$$ \begin{align*} \ell(\gamma_w) &= \sum_{i\in [q]} \left( w^{-1}(i)-i +p \right) - \sum_{i\in [q]} |\{k\in {\mathbb{N}} \mid c_s=c_t=k \text{ for } s<i<t<j \}|\\ &= pq - \left( \frac{q(q-1)}{2}-\ell(w)\right) = \ell(w)+ pq - \frac{q(q-1)}{2}. \end{align*} $$

As $\dim \mathrm {Hess}(\mathsf {x}_{p,q},{\mathbf m}(w)) = \dim {{\mathcal O}}_{\gamma _w}$ by Theorem 3.11, this proves the first equality in (6).

To prove the second, we observe first that if ${\mathbf m}(w)=(m_1,\ldots ,m_n)$ , then

$$ \begin{align*} \mathrm{area}(\pi_{m(w)}) & = \sum_{i=1}^n (m_i-i) \\ & = pq+\sum_{i=1}^q \max\{w^{-1}(k) \mid k \leq i\}- \sum_{i=1}^q i+pn-\sum_{i=q+1}^n i \\ & = p^2+2pq-{{p+q+1} \choose {2}}+{{q+1} \choose {2}}+\sum_{i=1}^q \max\{w^{-1}(k) \mid k \leq i\}-\sum_{i=1}^q i \\ & = pq+\frac{p(p-1)}{2}+\sum_{i=1}^q \max\{w^{-1}(k) \mid k \leq i\}- \sum_{i=1}^q i. \end{align*} $$

We complete the proof by applying Lemma 3.13 to $w^{-1}$ .

Proof. Let $\gamma _w=c_1c_2\cdots c_n \in \mathbf {Clan}_{p,q}$ for some $w\in {\mathbf {S}}_q$ . We have by definition that

$$\begin{align*}c_1\cdots c_p = 12\cdots q+\cdots +, \end{align*}$$

and furthermore that no $-$ signs occur in $\gamma _w$ . This implies that the cover relations of types IA1, IA2 and II do not occur in the restriction of the weak order on $\mathbf {Clan}_{p,q}$ to $[\gamma _e, \gamma _0]$ . Note also that no cover relation of type IB can occur among the clans in $[\gamma _e, \gamma _0]$ since all arcs are either nested or crossing as there is an arc connecting $i<j$ if and only if $j=w^{-1}(i)+p$ . This finishes the proof of our assertion.

Proof of Theorem 4.2.

By Lemma 4.3, the covering relations of the weak order on $[\gamma _e,\gamma _0]$ are given by either a Type IC1 covering relation or by a Type IC2 covering relation.

Now, a covering relation of Type IC1 on clans in $[\gamma _e, \gamma _0]$ is of the form

(7) $$ \begin{align} \gamma_w = c_1 \cdots c_{w^{-1}(i)+p} \cdots c_{w^{-1}(i+1)+p} \cdots c_n \xrightarrow{\; s_i \;} \gamma' = c_1 \cdots c_{w^{-1}(i+1)+p} \cdots c_{w^{-1}(i)+p} \cdots c_n, \end{align} $$

for some $i\in [q-1]$ such that $w^{-1}(i) < w^{-1}(i+1)$ . Since the resulting clan is obtained from $\gamma _w$ by interchanging i and $i+1$ in the one-line notation for w, we see that $\gamma ' = \gamma _{s_i w}$ . Similarly, a covering relation of type IC2 on clans in $[\gamma _e, \gamma _0]$ is of the form

(8) $$ \begin{align} \gamma_w = c_1 \cdots c_ic_{i+1} \cdots c_n \xrightarrow{\;s_i\;} \gamma'=c_1 \cdots c_{i+1} c_i \cdots c_n, \end{align} $$

for some $i\in \{p+1,\dots , n-1\}$ such that $w(i-p)<w(i-p+1)$ . In this case, we have $\gamma ' = \gamma _{ ws_{i-p}}$ since the resulting clan is obtained by interchanging the entries in the positions $i-p$ and $i-p+1$ in the one-line notation for w. In conclusion, we see that, for $w,v\in {\mathbf {S}}_q$ , if the clan $\gamma _w$ is covered by the clan $\gamma _v$ in the weak order, then w is covered by v in either the right weak order or the left weak order on ${\mathbf {S}}_q$ .

We proceed to prove the converse statement. Let $w,v\in {\mathbf {S}}_q$ be two permutations. Let $\gamma _w=c_1^wc_2^w \cdots c_n^w$ denote the (unique) clan corresponding to w, which is defined by

• • $c_i^w= +$ for all $q<i \leq p$ , and

• • $c_i^w=c_{p+w^{-1}(i)}^w=i$ for all $i\in [q]$ .

Let $\gamma _v$ denote the unique clan corresponding to v, defined in a similar manner. Now, we assume that w is covered by v in the left weak order on ${\mathbf {S}}_q$ . Hence, $s_iw= v$ holds for some $i\in [q-1]$ . After writing w and $s_iw$ in their one-line notations, we see that the covering relation $w\leq _L v$ corresponds to the covering relation in (7). Likewise, if w is covered by v in the right weak order in such a way that $ws_i= v$ for some $i\in [q-1]$ , then the covering relation in (8) holds. Hence, we proved that $\gamma _w \preceq \gamma _v$ if and only if $w\leq _L v$ or $w\leq _R v$ , as desired.

Proof. Since $\varphi (s_i) = w_0 s_i w_0 = s_{n-i}$ for all $i\in [n-1]$ , the first assertion of the lemma is obvious. Next, if $s_{i_1}\cdots s_{i_r}$ is a reduced expression for $v\in \mathbf {S}_{q}$ , then $s_{n-i_1}\cdots s_{n-i_r}$ is a reduced expression of $w_0 v w_0$ . In particular,

$$\begin{align*}v = s_{i_1}\cdots s_{i_r} \Rightarrow \varphi(v) = s_{n-i_{r}}s_{n-i_{r-1}}\cdots s_{n-i_1}, \end{align*}$$

so $\ell (\varphi (v))=\ell (v)$ . Finally, $\mathrm {Supp}(u)\subseteq \{s_1, \ldots , s_{{q-1}}\}$ and $\mathrm {Supp}(\varphi (v)) \subseteq \{s_{p+1}, \ldots , s_{n-1}\}$ . Since $q\leq p$ , we obtain the final assertion.

Proof. Recall from Lemma 4.9 that $\varphi $ maps ${\mathbf {S}}_q = \left < s_1, \ldots , s_{q-1} \right>$ to $\left < s_{p+1}, \ldots , s_{n-1} \right>$ and note that the intersection of these subgroups is the trivial group since $q\leq p$ . Thus, if $u_1,u_2,v_1,v_2\in {\mathbf {S}}_q$ such that $u_1\varphi (v_1) = u_2\varphi (v_2)$ , then

$$ \begin{align*} u_2^{-1}u_1 &= \varphi(v_2)\varphi(v_1)^{-1} \in {\mathbf{S}}_q \cap \left< s_{p+1}, \ldots, s_{n-1} \right> = \left<e\right>\\ & \quad\quad\Rightarrow u_1=u_2 \;\text{ and }\; \varphi(v_1)=\varphi(v_2), \end{align*} $$

and injectivity of the map follows. Since $\mathrm {Supp}(u)\cap \mathrm {Supp}(\varphi (v))=\emptyset $ , any reduced word for $u\varphi (v)$ is the product of a reduced word of u in ${\mathbf {S}}_q$ and a reduced word for $\varphi (v)$ in $\left < s_{p+1}, \ldots , s_{n-1} \right>$ . Thus, $\ell (u\varphi (v)) = \ell (u)+\ell (\varphi (v)) = \ell (u)+\ell (v)$ , as desired.

Proof. We argue first that $\psi _w(u,v) \in W(\gamma _w)$ for all $(u,v)\in \mathcal {S}(w y_0)$ . Given $(u,v)\in \mathcal {S}(w y_0)$ , let $u=s_{a_1}s_{a_2}\cdots s_{a_r}$ and $v=s_{b_1}s_{b_2}\cdots s_{b_t}$ be reduced words for u and v, respectively. By assumption,

$$\begin{align*}wy_0 =s_{a_1}s_{a_2}\cdots s_{a_r} s_{b_1}s_{b_2}\cdots s_{b_t} \end{align*}$$

is a reduced word for $wy_0$ . Manipulating this expression and using the fact that $y_0 s_i y_0 = s_{q-i}$ for all i implies

$$\begin{align*}s_{a_r} \cdots s_{a_2}s_{a_1} w s_{q-b_t}\cdots s_{q-b_2}s_{q-b_1} = y_0 \end{align*}$$

with $\ell (y_0) = \ell (w) + r+t$ . In particular, this expression yields a chain of length $r+t$ in the two-sided weak order on ${\mathbf {S}}_q$ :

$$ \begin{align*} w \xrightarrow{ s_{a_1} } s_{a_1} w &\xrightarrow{\; s_{a_2} \; } s_{a_2}s_{a_1} w \rightarrow \cdots \xrightarrow{\; s_{a_r}\;} s_{a_r} \cdots s_{a_2}s_{a_1} w = u^{-1}w \\ & \xrightarrow{\; s_{q-b_t} \;} u^{-1}w s_{q-b_t} \rightarrow \cdots \xrightarrow{\; s_{q-b_2} \;} u^{-1}w s_{q-b_t}\cdots s_{q-b_2} \xrightarrow{\; s_{q-b_1} \;} y_0. \end{align*} $$

In this chain, left multiplication by $s_{a_k}$ is a cover in the left weak order on ${\mathbf {S}}_q$ and corresponds to a cover of type IC1 on clans. This cover of type IC1 on clans is labeled by the simple reflection $s_{a_k}\in {\mathbf {S}}_n$ . Right multiplication by $s_{b_k}$ is a cover in the right weak order on ${\mathbf {S}}_q$ and corresponds to a cover of type IC2 on clans. This cover of type IC2 on clans is labeled by the simple reflection $s_{n-b_k}= \varphi (s_{b_k}) \in {\mathbf {S}}_n$ . By Theorem 4.2 and definition of the W-set, it follows that

$$\begin{align*}u\varphi(v) = s_{a_1} s_{a_2 } \cdots s_{a_r} s_{n-b_t} \cdots s_{n-b_2}s_{n-b_1} \in W(\gamma_w), \end{align*}$$

as desired.

To complete the proof, it suffices by Lemma 4.10 to show that $\psi _w$ is surjective. We proceed by induction on the nonnegative integer $\ell (wy_0)=\ell (y_0)- \ell (w)$ . If $\ell (wy_0)=0$ , then $w=y_0$ , $W(\gamma _w) = W(\gamma _0) = \{e\}$ , and $\mathcal {S}(wy_0) = \mathcal {S}(e) = \{(e,e)\}$ . Thus, our claim holds trivially in this case.

Suppose now that $w\in {\mathbf {S}}_q$ such that $\ell = \ell (wy_0)>0$ and $\psi _{w'}$ is surjective for all $w'\in {\mathbf {S}}_q$ such that $\ell (w'y_0) = \ell -1$ . Since the W-set of $\gamma _w$ is obtained by multiplying the labels of the weak order cover relations along a saturated path from $\gamma _w$ to $\gamma _0$ , if $x\in W(\gamma _w)$ , there exists $w' \in {\mathbf {S}}_q$ and $x'\in W(\gamma _{w'})$ such that $\gamma _w \xrightarrow {\;s_i\;} \gamma _{w'}$ and $x=s_i x'$ . By Theorem 4.2, $w'$ is a cover of w in the two-sided weak order on ${\mathbf {S}}_q$ , so $\ell (w') = \ell (w)+1$ . This in turn implies $\ell (w'y_0) = \ell (wy_0)-1 = \ell -1$ , and the induction hypothesis implies that there exists $(u,v)\in \mathcal {S}(w'y_0)$ such that $x'=u\varphi (v)$ .

There are two possible cases to consider: the cover $\gamma _w \prec \gamma _{w'}$ is either of type IC1 or IC2. If $\gamma _w \xrightarrow {\;s_i \;} \gamma _{w'}$ is a cover in the weak order on clans of type IC1, then the proof of Theorem 4.2 implies $i\in [q-1]$ and $w'=s_iw$ . Our assumptions also yield

(10) $$ \begin{align} \ell(x) = \ell(x')+1 &\Rightarrow \ell(s_iu\varphi(v) ) = \ell(u\varphi(v))+1\notag \\ &\Rightarrow \ell(s_i u) + \ell(v) = \ell(u) + \ell(v)+1\notag \\ &\Rightarrow \ell(s_i u) = \ell(u)+1, \end{align} $$

where the second implication follows from Lemma 4.10. This shows $(s_iu,v)\in \mathcal {S}(wy_0)$ since $s_i u v= s_i w'y_0 = wy_0$ and

$$\begin{align*}\ell(w'y_0) = \ell(u)+\ell(v) \Rightarrow \ell (y_0) - \ell(w) -1 = \ell(u)+\ell(v) \Rightarrow \ell(wy_0) = \ell(s_iu) + \ell(v) \end{align*}$$

by (10) above. Now $x= s_i u \varphi (v) = \psi _w (s_i u,v)$ , so $\psi _w$ is surjective in this case.

If $\gamma _w \xrightarrow {\;s_i \;} \gamma _{w'}$ is a cover in the weak order on clans of type IC2, then the proof of Theorem 4.2 implies $i\in \{p+1, \ldots , n-1\}$ and $w' = w s_{i-p}$ with $\ell (w')$ . Note that $s_i$ commutes with $u\in {\mathbf {S}}_q$ and recall that $s_{i} = \varphi (s_{n-i})$ . Thus,

(11) $$ \begin{align} x = s_i x'=s_i u \varphi(v) = u s_i \varphi(v) = u \varphi(s_{n-i}) \varphi(v) = u \varphi(vs_{n-i}) \end{align} $$

by Lemma 4.8. Our assumptions also imply

(12) $$ \begin{align} \nonumber\ell(x) = \ell(x')+1 &\Rightarrow \ell(u \varphi(vs_{n-i})) = \ell(u\varphi(v))+1 \\ \nonumber&\Rightarrow \ell(u) + \ell(vs_{n-i}) = \ell(u) + \ell(v)+1 \\ &\Rightarrow \ell(vs_{n-i}) = \ell(v)+1, \end{align} $$

where the first implication follows from (11) and the second from Lemma 4.10. This shows $(u,vs_{n-i})\in \mathcal {S}(wy_0)$ since $uvs_{n-i} = w' y_0 s_{n-i} = w' s_{i-p} y_0 = wy_0$ and

$$\begin{align*}\ell(w'y_0)= \ell(u)+\ell(v) \Rightarrow \ell(y_0)-\ell(w)-1 = \ell(u)+\ell(v) \Rightarrow \ell(wy_0) = \ell(u) + \ell(vs_{n-i}) \end{align*}$$

by (12) above. Using (11), we conclude $x=\psi _w(u, vs_{n-1})$ so $\psi _w$ is indeed surjective.

Proof. By Theorem 2.11, the polynomial representative of the cohomology class of $\overline {{{\mathcal O}}_{\gamma _w}}$ is given by the formula

$$ \begin{align*} \mathfrak{S}(\gamma_w) := \sum_{x\in W(\gamma_w)} \mathfrak{S}_x, \end{align*} $$

where $\mathfrak {S}_x$ is the Schubert polynomial indexed by the permutation $x\in {\mathbf {S}}_{n}$ . By Theorem 4.12, each $x\in W(\gamma _w)$ can be written $x = u \varphi (v)$ for a unique $(u,v)\in \mathcal {S}(wy_0)$ . The result now follows immediately, as u and $\varphi (v)$ have disjoint supports by Lemma 4.9, so $\mathfrak {S}_{u\varphi (v)} = \mathfrak {S}_u \mathfrak {S}_{\varphi (v)}$ (see, for example, [Reference ManivelMan01, Corollary 2.4.6].)

Proof. It follows from Proposition 4.14 that

$$ \begin{align*} \mathfrak{S}_{s_m} \mathfrak{S}({\gamma_w})= \sum_{(u,v) \in \mathcal{S}(wy_0)} \mathfrak{S}_{s_m} \mathfrak{S}_{u\varphi(v)}. \end{align*} $$

We apply Monk’s formula to each product $\mathfrak {S}_{s_m}\mathfrak {S}_{u \varphi (v)}$ and obtain

$$ \begin{align*} \mathfrak{S}_{s_m}\mathfrak{S}_{u\varphi(v)} =\displaystyle \sum_{\substack{j \leq m<k \\ \ell\left(u\varphi(v) t_{j k}\right)=\ell(u)+\ell(v)+1}} \mathfrak{S}_{u\varphi(v)t_{j k}}. \end{align*} $$

We aim to show each summand appearing on the RHS above will not appear in the corresponding sum for another pair from $\mathcal {S}(wy_0)$ .

Suppose there exist pairs $(u_1,v_1)$ and $(u_2,v_2)$ in $\mathcal {S}(w)$ such that

(14) $$ \begin{align} u_1\varphi(v_1)t_{jk} = u_2\varphi(v_2)t_{j'k'} \end{align} $$

for some $j,k,j',k'$ such that $j\leq m<k$ and $j'\leq m<k'$ . To complete the proof, we show that (14) implies $u_1=u_2$ and $v_1=v_2$ .

Write $x_1=u_1\varphi (v_1)$ and $x_2=u_2\varphi (v_2)$ for the remainder of the proof. We begin with a few observations. By construction, each permutation $x_i=u_i\varphi (v_i)$ with $ i\in \{1,2\}$ satisfies

(15) $$ \begin{align} [ q] =\{x_i(1), \ldots, x_i(q)\} \end{align} $$

and

(16) $$ \begin{align} x_i(a)=a \text{ for all } q+1\leq a \leq p, \end{align} $$

and

(17) $$ \begin{align} [n]\setminus [p] = \{x_i(p+1), \ldots, x_i(n)\}. \end{align} $$

We obtain the one-line notation of $x_1t_{jk}$ from that of $x_1$ by exchanging the entries in positions j and k and similarly for $x_2t_{j'k'}$ . These observations imply that pairs $j<k$ and $j'<k'$ satisfying (14) must fall into one of the following cases:

1. 1. $j,j'\in [q]$ , $k,k'\in [n]\setminus [q]$ ,

2. 2. $j,j',k,k'\in [q]$ ,

3. 3. $j,j',k,k'\in \{q+1, \ldots , p\}$ ,

4. 4. $j,j'\in \{q+1, \ldots , p\}$ and $k,k'\in [n]\setminus [p]$ , and

5. 5. $j,j',k,k'\in [n]\setminus [p] $ .

Note that cases (3) and (4) do not arise when $p=q$ .

We begin with Case (1). In this case, the equality (14) and equation (15) imply

$$ \begin{align*} \{k\} &= ([n]\setminus [q]) \cap \{(x_1t_{jk})^{-1}(1), \ldots, (x_1t_{jk})^{-1}(q)\}\\ &=([n]\setminus [q]) \cap \{(x_2t_{j'k'})^{-1}(1), \ldots, (x_2t_{j'k'})^{-1}(q)\} = \{k'\} \end{align*} $$

so $k= k'$ . Similarly, using (14), (16) and (17), we have

$$ \begin{align*} \{j\} &= [q] \cap \{(x_1t_{jk})^{-1}(q+1), \ldots, (x_1t_{jk})^{-1}(n)\}\\ &=[q] \cap \{(x_2t_{j'k'})^{-1}(q+1), \ldots, (x_2t_{j'k'})^{-1}(n)\} = \{j'\} \end{align*} $$

so $j=j'$ . Now (14) implies $x_1=x_2$ , and thus, $u_1=u_2$ and $v_1=v_2$ by Lemma 4.10. Case (4) follows by similar reasoning, so we omit it to avoid repetition.

Now suppose we are in the setting of Case (2). Then (14) becomes $u_1t_{jk}\varphi (v_1) = u_2t_{j'k'}\varphi (v_2)$ since $t_{jk}$ and $t_{j'k'}$ commute with $\varphi (v_1)$ and $\varphi (v_2)$ . By Lemma 4.9, we also know that both of the following sets

$$\begin{align*}\operatorname{Supp}(u_1t_{jk}) \cap \operatorname{Supp}(\varphi(v_1))\quad\text{and}\quad \operatorname{Supp}(u_2t_{j'k'})\cap \operatorname{Supp}(\varphi(v_2)) \end{align*}$$

are empty. Consequently, $\operatorname {Supp}(\varphi (v_1)) = \operatorname {Supp}(\varphi (v_2)) $ implying that $\varphi (v_1)= \varphi (v_2)$ , and hence, $v_1 = v_2$ by Lemma 4.8. It now follows from the definition of the set $\mathcal {S}(wy_0)$ that $u_1=u_2$ . Indeed, we have

$$ \begin{align*}u_1v_1 = wy_0 = u_2v_2 \text{ and } v_1=v_2 \Rightarrow u_1v_1 = u_2v_1 \Rightarrow u_1=u_2.\end{align*} $$

The proof of Case (5) is almost identical to that of (2), so we omit it to avoid repetition.

Finally, we consider Case (3). The equality (14) and equation (16) immediately imply that $j=j'$ and $k=k'$ . Thus, $x_1=x_2$ , and we conclude $u_1=u_2$ and $v_1=v_2$ as before. This finishes the proof of our theorem.