Proof of Lemma 3.1.

We calculate

$$ \begin{align*} \langle \alpha^\vee, 2\rho\rangle = \frac 12\left(\langle \alpha^\vee, 2\rho\rangle + \langle s_\alpha(\alpha^\vee), s_\alpha(2\rho)\rangle\right) = \frac 12\langle \alpha^\vee, 2\rho - s_\alpha(2\rho)\rangle. \end{align*} $$

Let

$$ \begin{align*} I := \{\beta\in \Phi^+\mid s_\alpha(\beta)\in \Phi^-\}. \end{align*} $$

Then $s_\alpha (I) = -I$ and $s_\alpha (\Phi ^+\setminus I) = \Phi ^+\setminus I$ . It follows that

$$ \begin{align*} 2\rho - s_\alpha(2\rho) &= \sum_{\beta \in I} \left(\beta - s_\alpha(\beta)\right) + \sum_{\beta\in \Phi^+\setminus I} \left(\beta-s_\alpha(\beta)\right) \\ &=2\sum_{\beta\in I} \beta. \end{align*} $$

Therefore, we obtain

$$ \begin{align*} \langle \alpha^\vee, 2\rho\rangle = \sum_{\beta\in I}\langle\alpha^\vee, \beta\rangle. \end{align*} $$

Certainly, $\alpha \in I$ . Hence,

$$ \begin{align*} \langle \alpha^\vee, 2\rho\rangle = 2+\sum_{\substack{\alpha \neq \beta\in \Phi^+\\s_\alpha(\beta)\in \Phi^-}}\langle\alpha^\vee, \beta\rangle. \end{align*} $$

Now, if $\alpha ,\beta \in \Phi ^+$ and $s_\alpha (\beta ) = \beta -\langle \alpha ^\vee , \beta \rangle \alpha \in \Phi ^-$ , we get $\langle \alpha ^\vee , \beta \rangle \geq 1$ . We conclude

$$ \begin{align*} \langle \alpha^\vee, 2\rho\rangle = 2+\sum_{\substack{\alpha \neq \beta\in \Phi^+\\s_\alpha(\beta)\in \Phi^-}}\langle\alpha^\vee, \beta\rangle\geq 2 + \#\{\beta\in \Phi^+\setminus\{\alpha\}\mid s_\alpha(\beta)\in \Phi^-\} = 1+\ell(s_\alpha). \end{align*} $$

All claims of the lemma follow immediately from this.

Proof. Note that if $p: w_1\rightarrow w_2 = (w_1s_\alpha )^J$ is an edge in ${\mathrm {QB}}(W^J)$ , then by definition,

$$ \begin{align*} \ell(w_2) = \ell(w_1) + 1 - \langle {\mathrm{wt}}(p),2\rho-2\rho_J\rangle. \end{align*} $$

In general, iterate this observation for all edges of p.

Proof. By part (d) of Proposition 3.4, it suffices to consider the case $J=\emptyset $ , that is, the quantum Bruhat graph.

The if condition is part (c) of Proposition 3.4. It remains to show the only if condition. Note that for each $w\in W$ and $\alpha \in \Delta $ , we get a “silly path” of the form

$$ \begin{align*} w\rightarrow ws_\alpha\rightarrow w \end{align*} $$

in ${\mathrm {QB}}(W)$ . Precisely one of the edges is quantum with weight $\alpha ^\vee $ , and the other one is Bruhat with weight $0$ .

If $\mu \geq {\mathrm {wt}}(w_1\Rightarrow w_2)$ , we may compose a shortest path from $w_1$ to $w_2$ with suitably chosen silly paths as above to obtain a path from $w_1$ to $w_2$ of weight $\mu $ .

Proof. Part (a) follows from [Reference Lenart, Naito, Sagaki, Schilling and Shimozono17, Proposition 4.3].

For part (b), observe that we have an automorphism of $\Phi $ given by $\alpha \mapsto -w_0\alpha $ . The induced automorphism of W is given by $w\mapsto w_0 w w_0$ . Since the function ${\mathrm {wt}}(\cdot \Rightarrow \cdot )$ is compatible with automorphisms of $\Phi $ , we get the claim.

Now, for (c), consider a reduced expression

$$ \begin{align*} w_0 w_1 = s_1\cdots s_q. \end{align*} $$

Then, iterating Lemma 3.7, we get

$$ \begin{align*} {\mathrm{wt}}(w_1\Rightarrow 1) \underset{\text{(a)}}=&\ {\mathrm{wt}}(w_0\Rightarrow w_0 w_1) = {\mathrm{wt}}(w_0\Rightarrow s_1\cdots s_q) \\=&\ {\mathrm{wt}}(s_1w_0\Rightarrow s_2\cdots s_q)=\cdots = {\mathrm{wt}}(s_q\cdots s_1 w_0\Rightarrow 1) \\=&\ {\mathrm{wt}}((w_0 w_1)^{-1}w_0\Rightarrow 1) = {\mathrm{wt}}(w_1^{-1}\Rightarrow 1).\\[-36pt] \end{align*} $$

Proof. Let $C>0$ be a constant sufficiently large for the conclusion of Theorem 3.10 to hold. We see that if $x_1\lessdot x_2$ is any cover in $\Omega _J^{-C}$ , then there are only finitely many possibilities for $x_1^{-1} x_2$ , so the length $\ell (x_1^{-1} x_2)$ is bounded. We fix a bound $C'>0$ for this length.

We can pick $C_2>0$ such that for all $x_1 = w\varepsilon ^\mu \in \Omega _J^{-C_2}$ and $x_2\in \widetilde W^J$ with $\ell (x_1^{-1}x_2)\leq C_1 C'$ , we must at least have $x_2 \in \Omega _J^{-C}$ .

We now consider elements $x = w\varepsilon ^\mu \in \Omega _J^{-C_2}$ and $x' = w'\varepsilon ^{\mu '}\in \widetilde W^J$ with $\ell (x^{-1} x')\leq C_1$ .

First, suppose that $x\leq x'$ . We find elements $x= x_1\lessdot x_2\lessdot \cdots \lessdot x_k = x'$ . Note that $k = \ell (x') - \ell (x) \leq \ell (x^{-1} x')\leq C_1$ . By choice of $C'$ , we conclude that $\ell (x^{-1} x_i)\leq C' i\leq C'C_1$ for $i=1,\dotsc ,k$ . Thus, $x_i \in \Omega _J^{-C}$ .

By Theorem 3.10, we get a path from $(w')^J$ to $w^J$ of weight $\mu -\mu '+\mathbb Z\Phi _J^\vee $ . Thus,

$$ \begin{align*}{\mathrm{wt}}(w_1\Rightarrow w_2)\leq \mu-\mu'\quad \pmod{\Phi_J^\vee},\end{align*} $$

which is the estimate we wanted to prove.

Now, suppose conversely that we are given $\mu - {\mathrm {wt}}(w'\Rightarrow w)\geq \mu '\ \pmod {\Phi _J^\vee }$ . By Lemma 3.6, we find a path $(w')^J = w_1\rightarrow w_2\rightarrow \cdots \rightarrow w_k = w^J$ in ${\mathrm {QB}}(W^J)$ of weight $\mu -\mu '+\mathbb Z\Phi _J^\vee $ . Since $\mu -\mu '$ is bounded in terms of $C_1$ , the length k of this path is bounded in terms of $C_1$ as well. By adding another lower bound for $C_2$ , we can guarantee that each such path $w_1\rightarrow \cdots \rightarrow w_k$ can indeed be lifted to $\Omega _J^{-C}$ , proving that $x\leq x'$ .

Proof. Let $w_0(J)\in W_J$ be the longest element. Let $C_2>0$ such that the conclusion of the previous Lemma is satisfied.

If $x\in \Omega _J^{C_2}$ , then $x w_0(J) w_0 \in \Omega _{-w_0(J)}^{-C_2}$ . Moreover, $w_0(J)w_0$ is a length positive element for x, so $\ell (x w_0(J) w_0) = \ell (x) + \ell (w_0(J) w_0)$ . Choosing $C_2$ appropriately, we similarly may assume $x' \in \Omega ^{C}_{J}$ for some $C>0$ and obtain $\ell (x'w_0(J)w_0) = \ell (x') + \ell (w_0(J)w_0)$ . Then, with the right choice of constants and using the automorphism $\alpha \mapsto -w_0\alpha $ of $\Phi $ , we get

$$ \begin{align*}x\leq x'\iff&xw_0(J)w_0\leq x'w_0(J)w_0\\{\iff}&w_0 w_0(J)\mu - \mathrm{wt}(w' w_0(J) w_0\Rightarrow ww_0(J) w_0)\geq w_0 w_0(J)\mu'\quad \pmod{\Phi_{-w_0(J)}^\vee}\\{\iff}&w_0(J)\mu + \mathrm{wt}(w_0 w'w_0(J)\Rightarrow w_0 ww_0(J))\leq w_0(J)\mu'\quad \pmod{\Phi_J^\vee}\\\underset{[17,\ \text{Proof 4.3}]}{\iff}&w_0(J)\mu + \mathrm{wt}(w\Rightarrow w')\leq w_0(J)\mu'\quad \pmod{\Phi_J^\vee}. \end{align*} $$

Since $w_0(J)\mu \equiv \mu\ \pmod {\Phi _J^\vee }$ , we get the desired conclusion.

Proof. Note that $s_\gamma w = ws_{-w^{-1}\gamma }$ . We have to show that $-w^{-1}\gamma $ is a quantum root and that

$$ \begin{align*} \ell(ws_{-w^{-1}\gamma}) = \ell(w) - \ell(s_{-w^{-1}\gamma}). \end{align*} $$

Step 1. We show that $-w^{-1}\gamma $ is a quantum root using Lemma 3.1. So pick an element $-w^{-1}\gamma \neq \beta \in \Phi ^+$ with $s_{-w^{-1}\gamma }(\beta )\in \Phi ^-$ . We want to show that $\langle -w^{-1}\gamma ^\vee ,\beta \rangle =1$ .

Note that

$$ \begin{align*} s_{-w^{-1}\gamma}(\beta) = \beta+\langle -w^{-1}\gamma^\vee,\beta\rangle w^{-1}\gamma. \end{align*} $$

In particular, $k := \langle -w^{-1}\gamma ^\vee ,\beta \rangle>0$ . It follows from the theory of root systems that

$$ \begin{align*} \beta_i := \beta +i w^{-1}\gamma\in \Phi,\qquad i=0,\dotsc,k. \end{align*} $$

Since $\beta _0 = \beta \in \Phi ^+$ and $\beta _{k} = s_{-w^{-1}\gamma }(\beta )\in \Phi ^-$ , we find some $i\in \{0,\dotsc ,k-1\}$ with $\beta _i\in \Phi ^+$ and $\beta _{i+1}\in \Phi ^-$ . We show that $k\leq 1$ as follows:

• • Suppose $w\beta _i \in \Phi ^+$ . Then $w\beta _{i+1} = w\beta _i+\gamma> \gamma $ . In particular, $w\beta _{i+1}\in \Phi ^+$ . We see that $w\beta _{i+1}\in {\mathrm {inv}}(w)$ , contradicting maximality of $\gamma $ .

• • Suppose $w\beta _{i+1}\in \Phi ^-$ . Then $-w\beta _i = -w\beta _{i+1}+\gamma> \gamma $ . In particular, $-w\beta _{i}\in \Phi ^+$ . We see that $-w\beta _{i}\in {\mathrm {inv}}(w)$ , contradicting maximality of $\gamma $ .

• • Suppose $i\geq 1$ . Then $\gamma - w\beta _i = -w\beta _{i-1}\in \Phi $ . We already proved $w\beta _i\in \Phi ^-$ , so $-w\beta _i\in {\mathrm {inv}}(w)$ . Since also $\gamma \in {\mathrm {inv}}(w)$ , we conclude $\gamma <-w\beta _{i-1}\in {\mathrm {inv}}(w)$ , contradicting the maximality of $\gamma $ .

• • Suppose $i\leq k-2$ . Then $w\beta _{i+2} = w\beta _{i+1} + \gamma \in \Phi $ . Since both $\gamma $ and $w\beta _{i+1}$ are in ${\mathrm {inv}}(w)$ , we conclude that $\gamma < w\beta _{i+2}\in {\mathrm {inv}}(w)$ , which is a contradiction to the maximality of $\gamma $ .

In summary, we conclude $0=i\geq k-1$ , thus $k\leq 1$ . This shows $\langle -w^{-1}\gamma ^\vee ,\beta \rangle =1$ .

Step 2. We show that

$$ \begin{align*} \ell(ws_{-w^{-1}\gamma}) = \ell(w) - \ell(s_{-w^{-1}\gamma}). \end{align*} $$

Suppose this is not the case. Then we find some $\alpha \in \Phi ^+$ such that $w\alpha \in \Phi ^+$ and $s_{-w^{-1}\gamma }(\alpha ) \in \Phi ^-$ . As we saw before, $\langle -w^{-1}\gamma ^\vee ,\alpha \rangle =1$ , so $s_{-w^{-1}\gamma }(\alpha ) = \alpha + w^{-1}\gamma \in \Phi ^-$ . Now, consider the element $ws_{-w^{-1}\gamma }(\alpha ) = w\alpha + \gamma \in \Phi $ . Since $w\alpha \in \Phi ^+$ by assumption, we have $ws_{-w^{-1}\gamma }(\alpha )>\gamma $ , in particular $ws_{-w^{-1}\gamma }(\alpha )\in \Phi ^+$ . We conclude $ws_{-w^{-1}\gamma }(\alpha )\in {\mathrm {inv}}(w)$ , yielding a final contradiction to the maximality of $\gamma $ .

Proof. We first show $\gamma \in {\mathrm {inv}}(ws_\alpha )$ , that is, $s_\alpha w^{-1}\gamma \in \Phi ^-$ .

Aiming for a contradiction, we thus suppose that

$$ \begin{align*} s_\alpha (-w^{-1}\gamma) = \langle \alpha^\vee,w^{-1}\gamma\rangle \alpha - w^{-1}\gamma \in \Phi^-. \end{align*} $$

Then $-w^{-1}\gamma $ is a positive root whose image under $s_\alpha $ is negative. Since $\alpha $ is quantum, we conclude $\langle \alpha ^\vee ,-w^{-1}\gamma \rangle =1$ . Thus, $-\alpha - w^{-1}\gamma \in \Phi ^-$ . Consider the element

$$ \begin{align*}w(\alpha+w^{-1}\gamma) = \gamma+w\alpha \in \Phi.\end{align*} $$

We distinguish the following cases:

• • If $\gamma +w\alpha \in \Phi ^-$ , we get $\gamma <-w\alpha \in {\mathrm {inv}}(w)$ , contradicting maximality of $\gamma $ .

• • If $\gamma +w\alpha \in \Phi ^+$ , we compute

  $$ \begin{align*} ws_\alpha(-w^{-1}\gamma) = -(ws_\alpha w^{-1})\gamma = -s_{w\alpha}(\gamma) = -(\gamma+w\alpha)\in \Phi^-. \end{align*} $$
  In other words, the positive root $-w^{-1}\gamma \in \Phi ^+$ gets mapped to negative roots both by $s_\alpha $ and by $ws_\alpha \in W$ . This is a contradiction to $\ell (w) = \ell (ws_\alpha ) + \ell (s_\alpha )$ (since $w\rightarrow ws_\alpha $ was supposed to be a quantum edge).

In any case, we get a contradiction. Thus, $\gamma \in {\mathrm {inv}}(ws_\alpha )$ .

The quantum edge condition $w\rightarrow ws_\alpha $ implies $\ell (w) = \ell (ws_\alpha ) + \ell (s_\alpha )$ , so ${\mathrm {inv}}(ws_\alpha )\subset {\mathrm {inv}}(w)$ . Because $\gamma $ is maximal in ${\mathrm {inv}}(w)$ and $\gamma \in {\mathrm {inv}}(ws_\alpha )\subseteq {\mathrm {inv}}(w)$ , it follows that $\gamma $ must be maximal in ${\mathrm {inv}}(ws_\alpha )$ as well.

Finally, we have to show $\langle -w^{-1}\gamma ^\vee ,\alpha \rangle \geq 0$ . If this was not the case, then we would get

$$ \begin{align*} \gamma <s_\gamma(-w\alpha) = -w\alpha + \langle w^{-1}\gamma^\vee,\alpha\rangle \gamma\in {\mathrm{inv}}(w), \end{align*} $$

again contradicting maximality of $\gamma $ .

Proof. Since the estimate

$$ \begin{align*} {\mathrm{wt}}(w\Rightarrow 1)\leq&{\mathrm{wt}}(w\Rightarrow s_\gamma w) + {\mathrm{wt}}(s_\gamma w\Rightarrow 1) \\\leq&-w^{-1}\gamma^\vee + {\mathrm{wt}}(s_\gamma w\Rightarrow 1) \end{align*} $$

follows from [Reference Schremmer28, Lemma 4.3], all we have to show is the inequality “ $\geq $ ”.

For this, we use induction on $\ell (w)$ . If $1\neq w\in W$ , we find by Lemma 3.13 some quantum edge $w\rightarrow ws_\alpha $ with ${\mathrm {wt}}(w\Rightarrow 1) = {\mathrm {wt}}(ws_\alpha \Rightarrow 1)+\alpha ^\vee $ . If $\alpha =-w^{-1}\gamma $ , we are done.

Otherwise, $\gamma \in \mathrm{max}\ \mathrm{inv}(ws_\alpha )$ and $\langle -w^{-1}\gamma ^\vee ,\alpha \rangle \geq 0$ by the previous lemma. By induction, we have

(3.2) $$ \begin{align} {\mathrm{wt}}(w\Rightarrow 1)&={\mathrm{wt}}(ws_\alpha \Rightarrow 1)+\alpha^\vee \notag\\&= {\mathrm{wt}}(s_\gamma ws_\alpha\Rightarrow 1) + \alpha^\vee - (ws_\alpha)^{-1}\gamma^\vee. \end{align} $$

By Lemma 3.16, we get the following three quantum edges:

This allows for the following computation:

(3.3) $$ \begin{align} \ell(s_\gamma ws_\alpha) =&\,\ell(ws_\alpha) + 1-\langle -(ws_\alpha)^{-1}\gamma^\vee,2\rho\rangle \notag\\=&\,\ell(w) + 2-\langle \alpha^\vee,2\rho\rangle - \langle -w^{-1}\gamma^\vee - \langle -w^{-1}\gamma^\vee,\alpha\rangle \alpha^\vee,2\rho\rangle \notag\\=&\,\ell(s_\gamma w) + 1 +(\langle -w^{-1}\gamma^\vee,\alpha\rangle-1)\langle \alpha^\vee,2\rho\rangle. \end{align} $$

We now distinguish several cases depending on the value of $\langle -w^{-1}\gamma ^\vee ,\alpha \rangle \in \mathbb Z_{\geq 0}$ .

• • Case $\langle -w^{-1}\gamma ^\vee ,\alpha \rangle =0$ . In this case, we get a quantum edge $s_\gamma w\rightarrow s_\gamma w s_\alpha $ by Equation (3.3). Evaluating this in Equation (3.2), we get

  $$ \begin{align*} {\mathrm{wt}}(w\Rightarrow 1)&={\mathrm{wt}}(s_\gamma w s_\alpha\Rightarrow 1)+\alpha^\vee - (ws_\alpha)^{-1}\gamma^\vee \\ &\geq{\mathrm{wt}}(s_\gamma w\Rightarrow 1) - s_\alpha w^{-1}\gamma^\vee \\ &={\mathrm{wt}}(s_\gamma w\Rightarrow 1) - w^{-1}\gamma^\vee. \end{align*} $$

• • Case $\langle -w^{-1}\gamma ^\vee ,\alpha \rangle =1$ . In this case, we get a Bruhat edge $s_\gamma w\rightarrow s_\gamma w s_\alpha $ by Equation (3.3). Evaluating this in Equation (3.2), we get

  $$ \begin{align*} {\mathrm{wt}}(w\Rightarrow 1)& ={\mathrm{wt}}(s_\gamma w s_\alpha\Rightarrow 1)+\alpha^\vee - (ws_\alpha)^{-1}\gamma^\vee \\ &\geq{\mathrm{wt}}(s_\gamma w\Rightarrow 1) + \alpha^\vee - s_\alpha w^{-1}\gamma^\vee \\ &={\mathrm{wt}}(s_\gamma w\Rightarrow 1) - w^{-1}\gamma^\vee. \end{align*} $$

• • Case $\langle -w^{-1}\gamma ^\vee ,\alpha \rangle \geq 2$ . We get

  $$ \begin{align*} \ell(s_\gamma ws_\alpha) &\leq \ell(s_\gamma w) + \ell(s_\alpha) \underset{\text{L}3.1}\leq \ell(s_\gamma w) + \langle \alpha^\vee,2\rho\rangle -1 \\ &<\ell(s_\gamma w) + \ell(s_\alpha) \leq \ell(s_\gamma w) +1+\left(\langle -w^{-1}\gamma^\vee,\alpha\rangle-1\right) \langle \alpha^\vee,2\rho\rangle. \end{align*} $$
  This is a contradiction to Equation (3.3).

In any case, we get a contradiction or the required conclusion, finishing the proof.

Proof.

1. (a) Consider the Cartan matrix

   $$ \begin{align*} C_{\alpha,\beta} := \langle \alpha^\vee,\beta\rangle,\qquad \alpha,\beta\in {\mathrm{cl}}(J). \end{align*} $$
   This must be the Cartan matrix associated to a certain Dynkin diagram, namely the subdiagram of the affine Dynkin diagram of $\Phi _{{\mathrm {af}}}$ with set of nodes given by J. We know that this must coincide with the Dynkin diagram of a finite root system by the fact that J is spherical. Hence, $C_{\bullet ,\bullet }$ is the Cartan matrix of a finite root system. Both claims follow immediately from this observation.

2. (b) Let $\varphi $ denote the map

   $$ \begin{align*} \varphi : \Phi_J^+\rightarrow \Phi_{\mathrm{af}}^+,\qquad \alpha\mapsto (\alpha,\Phi^+(-\alpha)). \end{align*} $$
   By (a), the map is injective. For each root $\alpha \in {\mathrm {cl}}(J)$ , we certainly have $\varphi (\alpha )\in \Phi _J^+$ .

   Now, for an inductive argument, suppose that $\alpha \in \Phi _J^+, \beta \in {\mathrm {cl}}(J)$ and $\alpha +\beta \in \Phi $ satisfy $\varphi (\alpha )\in \Phi _J^+$ . We want to show that $\varphi (\alpha +\beta )\in \Phi _J^+$ .

   We have $(\alpha ,\Phi ^+(-\alpha )),(\beta ,\Phi ^+(-\beta ))\in \Phi _J^+$ , hence

   $$ \begin{align*} (\alpha+\beta,\Phi^+(-\alpha) + \Phi^+(-\beta))\in \Phi_J^+. \end{align*} $$
   Hence, it suffices to show that $\Phi ^+(-\alpha ) + \Phi ^+(-\beta ) = \Phi ^+(-\alpha -\beta )$ .

   If $\beta \in \Delta $ , this is clear. Hence, we may assume that $\beta =-\theta $ , where $\theta $ is the longest root of the irreducible component of $\Phi $ containing $\alpha ,\beta $ . Then $\alpha -\theta \in \Phi $ implies $\alpha \in \Phi ^+$ and $\alpha -\theta \in \Phi ^-$ . We see that $\Phi ^+(-\alpha ) + \Phi ^+(\theta ) = \Phi ^+(-\alpha +\theta )$ holds true.

Proof. Write

$$ \begin{align*} I := \{\beta\neq \gamma\in \Phi^+_J\mid s_\beta(\gamma)\notin \Phi^+_J\}. \end{align*} $$

Then

$$ \begin{align*} {}^J{}\ell(s_\beta w) =&\, \#\{\gamma\in \Phi^+_J\mid w^{-1}s_\beta(\gamma)\in \Phi^-\} \\=&\,\#\{\gamma\in \Phi^+_J\setminus (I\cup\{\beta\})\mid w^{-1}s_\beta(\gamma)\in \Phi^-\} + \#\{\gamma\in I\mid w^{-1}s_\beta(\gamma)\in \Phi^-\}. \end{align*} $$

Since $s_\beta $ permutes the set $\Phi ^+_J\setminus (I\cup \{\beta \})$ , we get

$$ \begin{align*} \ldots =& \,\#\{\gamma\in \Phi^+_J\setminus (I\cup\{\beta\})\mid w^{-1}\gamma\in \Phi^-\} + \#\{\gamma\in I\mid w^{-1}s_\beta(\gamma)\in \Phi^-\}. \end{align*} $$

Note that if $\gamma \in I$ , then $\langle \beta ^\vee ,\gamma \rangle>0$ and thus

$$ \begin{align*} w^{-1}s_\beta(\gamma) = w^{-1}\gamma - \langle \beta^\vee,\gamma\rangle w^{-1}\beta> w^{-1}\gamma. \end{align*} $$

We obtain

$$ \begin{align*} &\#\{\gamma\in \Phi^+_J\setminus (I\cup\{\beta\})\mid w^{-1}\gamma\in \Phi^-\} + \#\{\gamma\in I\mid w^{-1}s_\beta(\gamma)\in \Phi^-\} \\\leq&\,\#\{\gamma\in \Phi^+_J\setminus (I\cup\{\beta\})\mid w^{-1}\gamma\in \Phi^-\} + \#\{\gamma\in I\mid w^{-1}\gamma\in \Phi^-\} \\=&\,{}^J{}\ell(w)-1.\\[-38pt] \end{align*} $$

Proof. We fix an element $\lambda \in \mathbb Z\Phi ^\vee $ that is dominant and sufficiently regular (the required regularity constant follows from the remaining proof).

For $w\in W$ , we consider the element $w\varepsilon ^\lambda \in \widetilde W$ . Then there exist uniquely determined elements $w'\varepsilon ^{\lambda '}\in {}^J{}{W_{\mathrm {af}}}$ and $y \in \widetilde W_J$ such that

$$ \begin{align*} w\varepsilon^\lambda = y\cdot w' \varepsilon^{\lambda'}. \end{align*} $$

We define ${}^J{}\pi (w) := (w', \lambda -\lambda ')$ and check that it has the required properties.

1. (0) $w'\in {}^J{}W$ : Since $\widetilde W_J$ is a finite group, we may assume that $\lambda '$ is superregular and dominant as well. For $(\alpha ,k)\in J$ , we have

   $$ \begin{align*} (w'\varepsilon^{\lambda'})^{-1}(\alpha,k) = ((w')^{-1}\alpha, k+\langle \lambda', (w')^{-1} \alpha\rangle)\in \Phi^+_{{\mathrm{af}}} \end{align*} $$
   because $w'\varepsilon ^{\lambda '}\in {}^J{}{W_{\mathrm {af}}}$ . Since $\lambda '$ is superregular and dominant, we have
   $$ \begin{align*} ((w')^{-1}\alpha, k+\langle \lambda', (w')^{-1} \alpha\rangle)\in \Phi^+_{{\mathrm{af}}}\iff (w')^{-1}\alpha\in \Phi^+. \end{align*} $$
   This proves $w'\in {}^J{}W$ .

2. (1) If $w\in {}^J{}W$ , then ${}^J{}\pi (w) = (w,0)$ : The proof of (0) shows that $w\varepsilon ^\lambda \in {}^J{}{W_{\mathrm {af}}}$ so that $w\varepsilon ^\lambda = w'\varepsilon ^{\lambda '}$ .

3. (2) Let $w\in W$ and $\beta \in \Phi _J^+$ . We have to show

   $$ \begin{align*} {}^J{}\pi(s_\beta w) = (w', \lambda - \lambda' + \Phi^+(-\beta)w^{-1}\beta^\vee). \end{align*} $$
   Put
   $$ \begin{align*} b := (\beta, \Phi^+(-\beta))\in \Phi_{{\mathrm{af}}}^+. \end{align*} $$
   By Lemma 3.21, we have $ b \in (\Phi _{\mathrm {af}})_J^+$ . The projection of
   $$ \begin{align*} r_{ b}w\varepsilon^\lambda = s_\beta w \varepsilon^{\lambda + \Phi^+(-\beta)w^{-1}\beta^\vee} \in \widetilde W_J\cdot w\varepsilon^\lambda \end{align*} $$
   onto ${}^J{}{W_{\mathrm {af}}}$ must again be $w'\varepsilon ^{\lambda '}$ . We obtain
   $$ \begin{align*} {}^J{}\pi(s_\beta w) = (w', \lambda + \Phi^+(-\beta)w^{-1}\beta^\vee-\lambda') \end{align*} $$
   as desired.

   For the second claim, it suffices to observe that

   $$ \begin{align*} \varepsilon^{w(\lambda-\lambda')} = w\varepsilon^\lambda \varepsilon^{-\lambda'} w^{-1} = y w' \varepsilon^{\lambda'} \varepsilon^{-\lambda'} w^{-1} = y \underbrace{w' w^{-1}}_{\in W_J}\in \widetilde W_J. \end{align*} $$

The fact that ${}^J{}\pi $ is uniquely determined (in particular, independent of the choice of $\lambda $ ) can be seen as follows: If $w\in {}^J{}W$ , then ${}^J{}\pi (w)$ is determined by (1). Otherwise, we find $\beta \in \Phi _J^+$ with $w^{-1}\beta \in \Phi ^-$ . We multiply w on the left with $s_\beta $ , and iterate this process, until we obtain an element in ${}^J{}W$ . This process will terminate after at most ${}^J{}\ell (w)$ steps with an element in ${}^J{}W$ . Now, for each of these steps, we can use property (2) to determine the value of ${}^J{}\pi (w)$ .

Proof. We first show the equation

$$ \begin{align*} {\mathrm{wt}}(w_1'\Rightarrow w_2')+\mu_2 = {\mathrm{wt}}(w_1'\Rightarrow w_2). \end{align*} $$

Induction by ${}^J{}\ell (w_2)$ . The statement is trivial if $w_2\in {}^J{}W$ . Otherwise, we find some $\alpha \in {\mathrm {cl}}(J)$ with $w_2^{-1}\alpha \in \Phi ^-$ . Because $(w_1')^{-1}\alpha \in \Phi ^+$ , we obtain from Lemma 3.7 that

$$ \begin{align*} {\mathrm{wt}}(w_1'\Rightarrow w_2) =& {\mathrm{wt}}(w_1'\Rightarrow s_\alpha w_2) - \Phi^+(-\alpha)w_2^{-1}\alpha^\vee. \end{align*} $$

By Lemma 3.24, we have

$$ \begin{align*} {}^J{}\pi(s_\alpha w_2) = (w^{\prime}_2, \mu_2 + \Phi^+(-\alpha)w_2^{-1}\alpha^\vee). \end{align*} $$

Using the inductive hypothesis, we get

$$ \begin{align*} {\mathrm{wt}}(w_1'\Rightarrow w_2) &= {\mathrm{wt}}(w_1'\Rightarrow s_\alpha w_2) - \Phi^+(-\alpha)w_2^{-1}\alpha^\vee \\ &={\mathrm{wt}}(w_1'\Rightarrow w_2') + \mu_2 + \Phi^+(-\alpha)w_2^{-1}\alpha^\vee - \Phi^+(-\alpha)w_2^{-1}\alpha^\vee \\ &={\mathrm{wt}}(w_1'\Rightarrow w_2') + \mu_2. \end{align*} $$

This finishes the induction.

It remains to prove the inequality

$$ \begin{align*} {\mathrm{wt}}(w_1'\Rightarrow w_2)-\mu_1\leq {\mathrm{wt}}(w_1\Rightarrow w_2). \end{align*} $$

The argument is entirely analogous, using [Reference Schremmer28, Lemma 4.3] in place of Lemma 3.7.

Proof. Part (a) follows readily from the definition. Let us prove part (b). We focus on the first identity, as the proof of the second identity is analogous.

Up to replacing $\alpha $ by $-\alpha $ , which does not change the reflection $s_\alpha $ nor the value of

$$ \begin{align*} \chi_J(\alpha)w_1^{-1}\alpha^\vee, \end{align*} $$

we may assume $\alpha \in \Phi _J^+$ . Now, write

$$ \begin{align*} {}^J{}\pi(w_1) = (w_1', \mu_1),\qquad {}^J{}\pi(w_2) = (w_2', \mu_2). \end{align*} $$

Then ${}^J{}\pi (s_\alpha w_1) = (w_1', \mu _1 + \Phi ^+(-\alpha )w_1^{-1}\alpha ^\vee )$ . Thus,

$$ \begin{align*} {}^J{}{\mathrm{wt}}(s_\alpha w_1\Rightarrow w_2) &= {\mathrm{wt}}(w_1'\Rightarrow w_2') - \mu_1 - \Phi^+(-\alpha)w_1^{-1}\alpha^\vee + \mu_2 \\ &={}^J{}{\mathrm{wt}}(w_1\Rightarrow w_2) - \Phi^+(-\alpha)w_1^{-1}\alpha^\vee \\ &= {}^J{}{\mathrm{wt}}(w_1\Rightarrow w_2) +\chi_J(\alpha)w_1^{-1}\alpha^\vee \end{align*} $$

as $\alpha \in \Phi _J^+$ .

Now, we prove part (c). Again, we only show the first inequality. If $w_1\beta \in \Phi _J$ , the inequality follows from part (b). Otherwise, we use (a) and [Reference Schremmer28, Lemma 4.3] to compute

$$ \begin{align*} {}^J{}{\mathrm{wt}}(w_1s_\beta\Rightarrow w_2)\leq~\,& {}^J{}{\mathrm{wt}}(w_1s_\alpha\Rightarrow w_1) + {}^J{}{\mathrm{wt}}(w_1\Rightarrow w_2) \\\underset{\text{L}3.25}\leq&{\mathrm{wt}}(w_1s_\alpha\Rightarrow w_1) + {}^J{}{\mathrm{wt}}(w_1\Rightarrow w_2) \\\leq~\,&\Phi^+(w\alpha)\alpha^\vee + {}^J{}{\mathrm{wt}}(w_1\Rightarrow w_2) \\=~\,&\chi_J(w\alpha)\alpha^\vee + {}^J{}{\mathrm{wt}}(w_1\Rightarrow w_2). \end{align*} $$

This finishes the proof.

Proof. We show the claim via induction on ${}^J{}\ell (w_1)$ . If $w_1\in {}^J{}W$ , then the claim follows from Lemma 3.25.

Otherwise, we find some $\alpha \in {\mathrm {cl}}(J)$ with $w_1^{-1}\alpha \in \Phi ^-$ . By assumption, also $w_2^{-1}\alpha \in \Phi ^-$ . Using Lemma 3.7, we get

$$ \begin{align*} {\mathrm{wt}}(w_1\Rightarrow w_2) &={\mathrm{wt}}(s_\alpha w_1\Rightarrow s_\alpha w_2) + \chi_J(\alpha)w_1^{-1}\alpha^\vee - \chi_J(\alpha)w_2^{-1}\alpha^\vee. \end{align*} $$

Since ${}^J{}\ell (s_\alpha w_1)<{}^J{}\ell (w_1)$ by Lemma 3.23, we want to show that $(s_\alpha w_1,s_\alpha w_2)$ also satisfy the condition stated in the lemma.

For this, let $\beta \in \Phi _J^+$ . If $\beta = \alpha $ , then $(s_\alpha w_1)^{-1}\alpha = -w_1^{-1}\alpha \in \Phi ^+$ by choice of $\alpha $ . Now, assume that $\beta \neq \alpha $ so that $s_\alpha \beta \in \Phi ^+_J$ . By the assumption on $w_1$ and $w_2$ , we must have $w_1^{-1}s_\alpha (\beta )\in \Phi ^+$ or $w_2^{-1} s_\alpha (\beta )\in \Phi ^-$ . In other words, we have

$$ \begin{align*} (s_\alpha w_1)^{-1}\beta\in \Phi^+\text{ or }(s_\alpha w_2)^{-1}\beta \in \Phi^-. \end{align*} $$

This shows that $(s_\alpha w_1, s_\alpha w_2)$ satisfy the desired properties.

By the inductive hypothesis and Lemma 3.28, we get

$$ \begin{align*} &{\mathrm{wt}}(s_\alpha w_1\Rightarrow s_\alpha w_2) + \chi_J(\alpha) w_1^{-1}\alpha^\vee - \chi_J(\alpha)w_2^{-1}\alpha^\vee\\ &= {}^J{}{\mathrm{wt}}(s_\alpha w_1\Rightarrow s_\alpha w_2) + \chi_J(\alpha) w_1^{-1}\alpha^\vee - \chi_J(\alpha)w_2^{-1}\alpha^\vee \\ &={}^J{}{\mathrm{wt}}(w_1\Rightarrow w_2). \end{align*} $$

This completes the induction and the proof.

Proof.

1. (a) We have

   $$ \begin{align*}(v')^{-1}\mu'&\geq(vs_\alpha)^{-1}\mu + \mathrm{wt}(v'\Rightarrow vs_\alpha) + \mathrm{wt}(wvs_\alpha\Rightarrow w'v')\\ &\geq v^{-1}\mu - \langle v^{-1}\mu,\alpha\rangle \alpha^\vee + \mathrm{wt}(v'\Rightarrow v) - \mathrm{wt}(vs_\alpha\Rightarrow v) \\&\quad + \mathrm{wt}(wv\Rightarrow w'v') - \mathrm{wt}(wv\Rightarrow wvs_\alpha)\\ &\underset{(\ast)}\geq v^{-1}\mu - \langle v^{-1}\mu,\alpha\rangle \alpha^\vee + \mathrm{wt}(v'\Rightarrow v) - \Phi^+(v\alpha)\alpha^\vee\\&\quad + \mathrm{wt}(wv\Rightarrow w'v') - \Phi^+(-wv\alpha)\alpha^\vee\\ &=v^{-1}\mu + \mathrm{wt}(v'\Rightarrow v) + \mathrm{wt}(wv\Rightarrow w'v') - (\ell(x,v\alpha)+1)\alpha^\vee\\ &\geq v^{-1}\mu + \mathrm{wt}(v'\Rightarrow v) + \mathrm{wt}(wv\Rightarrow w'v')\quad \pmod{\Phi_J^\vee}. \end{align*} $$
   The inequality $(\ast )$ is [Reference Schremmer28, Lemma 4.3].

2. (b) The calculation is completely analogous.

Proof of Theorem 1.1 using Theorem 4.2.

We use the notation of Theorem 1.1. In view of Lemma 4.3 and [Reference Schremmer28, Lemma 2.3], the condition

(*) $$ \begin{align} \exists v_2\in W:~ v_1^{-1}\mu_1 + {\mathrm{wt}}(v_2\Rightarrow v_1) + {\mathrm{wt}}(w_1 v_1\Rightarrow w_2 v_2)\leq v_2^{-1}\mu_2 \end{align} $$

is true for all $v_1\in {\mathrm {LP}}(x)$ iff it is true for all $v_1\in W$ . We see that asking condition $(\ast )$ for all $v_1\in W$ is equivalent to asking condition (2) of Theorem 4.2 for each Bruhat-deciding datum of the form $(v_1,\emptyset )$ with $v_1\in {\mathrm {LP}}(x_1)$ . In this sense, Theorem 4.2 implies Theorem 1.1.

Proof. $(v,\emptyset )$ is a Bruhat-deciding datum for x. By Lemma 4.3 and [Reference Schremmer28, Corollary 2.4], the inequality in Theorem 4.2 (2) is satisfied by some $v'\in W$ iff it is satisfied by the unique length positive element $v'$ for $x'$ .

Proof. (a) $\iff $ (b): We start with a key calculation for $v'\in {\mathrm {LP}}(x')$ :

$$ \begin{align*} &\langle (v')^{-1}\mu' - {\mathrm{wt}}(v'\Rightarrow v) - {\mathrm{wt}}(wv\Rightarrow w'v') - v^{-1}\mu,2\rho\rangle \\ \underset{\text{L}3.5} =&\ \langle (v')^{-1}\mu,2\rho\rangle - d(v'\Rightarrow v) -\ell(v') + \ell(v) \\ &\qquad-d(wv\Rightarrow w'v') - \ell(wv) + \ell(w'v') - \langle v^{-1}\mu,2\rho\rangle \\ &\underset{\text{L}2.3} = \ell(x') - \ell(x) - d(v'\Rightarrow v) - d(wv\Rightarrow w'v'). \end{align*} $$

First, assume that (a) holds, that is, $x\lessdot x'$ . By Theorem 4.2 and Lemma 4.3, we find $v'\in {\mathrm {LP}}(x')$ such that

$$ \begin{align*} (v')^{-1}\mu' - {\mathrm{wt}}(v'\Rightarrow v) - {\mathrm{wt}}(wv\Rightarrow w'v') - v^{-1}\mu\geq 0. \end{align*} $$

By the above key calculation, we see that

$$ \begin{align*} \ell(x') \geq \ell(x) + d(v'\Rightarrow v) + d(wv\Rightarrow w'v'), \end{align*} $$

where equality holds if and only if (b.1) is satisfied. Note that $x\lessdot x'$ implies that $x^{-1}x'$ must be an affine reflection, thus $w\neq w'$ . We see that $v\neq v'$ or $wv\neq w'v'$ , thus in particular

$$ \begin{align*} \ell(x)+1= \ell(x') \geq \ell(x) + d(v'\Rightarrow v) + d(wv\Rightarrow w'v')\geq \ell(x)+1. \end{align*} $$

Since equality must hold, we get (b.1) and (b.2).

Now, assume conversely that (b) holds. By (b.1) and Theorem 4.2, we see that $x<x'$ . Now, using the key calculation and (b.2), we get $\ell (x') =\ell (x)+1$ .

(b) $\iff $ (c): The condition (b.2) means that either $v=v'$ and $wv\rightarrow w'v'$ is an edge in ${\mathrm {QB}}(W)$ , or $wv = w'v'$ and $v'\rightarrow v$ is an edge. If we now distinguish between Bruhat and quantum edges, we get the explicit conditions of (c) (or (d)).

Let us first assume that (b) holds. We distinguish the following cases:

1. (1) $wv = w'v'$ and $v'\rightarrow v$ is a Bruhat edge: Then we can write $v' = s_\alpha v$ for some $\alpha \in \Phi ^+$ with $v^{-1}\alpha \in \Phi ^-$ . Now, the condition $wv = w'v'$ implies $w' = ws_\alpha $ . Condition (b.1) implies $v^{-1}\mu = (v')^{-1}\mu '$ , so $\mu ' = s_\alpha (\mu )$ . We get (c.1).

2. (2) $wv = w'v'$ and $v'\rightarrow v$ is a quantum edge: Then we can write $v' = s_\alpha v$ for some $\alpha \in \Phi ^+$ with $v^{-1}\alpha \in \Phi ^+$ . Now, the condition $wv = w'v'$ implies $w' = ws_\alpha $ . Condition (b.1) implies $v^{-1}\mu +v^{-1}\alpha ^\vee = (v')^{-1}\mu '$ , so $\mu ' = s_\alpha (\mu )-\alpha ^\vee $ . We get (c.2).

3. (3) $v = v'$ and $wv\rightarrow w'v'$ is a Bruhat edge: Then we can write $w'v' = s_\alpha wv$ for some $\alpha \in \Phi ^+$ with $(wv)^{-1}\alpha \in \Phi ^-$ . Now, the condition $v=v'$ implies $w' = s_\alpha w$ . Condition (b.1) implies $v^{-1}\mu = (v')^{-1}\mu $ , so $\mu ' = \mu $ . We get (c.3).

4. (4) $v=v'$ and $wv\rightarrow w'v'$ is a quantum edge: Then we can write $w'v' = s_\alpha wv$ for some $\alpha \in \Phi ^+$ with $(wv)^{-1}\alpha \in \Phi ^-$ . Now, the condition $v=v'$ implies $w' = s_\alpha w$ . Condition (b.1) implies $v^{-1}\mu -(wv)^{-1}\alpha ^\vee = (v')^{-1}\mu $ , so $\mu ' = \mu -w^{-1}\alpha ^\vee $ . We get (c.4).

Reversing the calculations above shows that (c) $\implies $ (b).

For (c) $\iff $ (d), we just explicitly rewrite the conditions for length positivity of $v'$ , and the definition of edges in the quantum Bruhat graph.

Proof. Since $x(\alpha ,k)\in \Phi ^+$ by [Reference Schremmer28, Lemma 2.9], we have $x<xr_a$ . Since a is a simple affine root, we must have $x\lessdot xr_a$ . So one of the four possibilities (c.1)–(c.4) of Proposition 4.5 must be satisfied.

If (c.3) or (c.4) are satisfied, we get $v\in {\mathrm {LP}}(x')$ . Since $x' = x r_a$ is a length additive product, [Reference Schremmer28, Lemma 2.13] shows $s_\alpha v\in {\mathrm {LP}}(x)$ , finishing the proof.

Now, assume that (c.1) is satisfied. Then $x' = xs_\beta $ for some $\beta \in \Phi ^+$ means $k=0$ and $\alpha =\beta $ . Now, $v^{-1}\alpha \in \Phi ^+$ means that $\ell (s_\alpha v)> \ell (v)$ , so $s_\alpha v\rightarrow v$ cannot be a Bruhat edge.

Finally, assume that (c.2) is satisfied. Then $x' = xr_{(-\beta ,1)}$ for some $\beta \in \Phi ^+$ means that $k=1$ and $\alpha =-\beta \in \Phi ^-$ . Then $s_\alpha v\rightarrow v$ cannot be a quantum edge, as $\ell (s_\alpha v)<\ell (v)$ .

We get the desired claim or a contradiction, finishing the proof.

Proof. Let $\lambda $ be as in Proposition 4.9. Choosing $\lambda $ sufficiently large, we may assume that $x_1\varepsilon ^{\lambda }$ and $x_2\varepsilon ^{\lambda }$ are superregular with ${\mathrm {LP}}(x_1\varepsilon ^\lambda ) = {\mathrm {LP}}(x_2\varepsilon ^\lambda )=\{1\}$ . Now, $x_1\varepsilon ^\lambda \leq x_2\varepsilon ^\lambda $ if and only if

$$ \begin{align*} \mu_1 + {\mathrm{wt}}(w_1\Rightarrow w_2)\leq \mu_2, \end{align*} $$

by Corollary 4.4.

Proof. (1) $\implies $ (2): Suppose that $x\in {\mathrm {Adm}}(\lambda )$ , so $x\leq \varepsilon ^{u\lambda }$ for some $u\in W$ . Let also $v\in W$ . By Lemma 4.15, we find $\tilde u\in W$ such that

$$ \begin{align*} v^{-1}\mu + {\mathrm{wt}}(\tilde u\Rightarrow v) + {\mathrm{wt}}(wv\Rightarrow \tilde u)\leq \tilde u^{-1}u\lambda. \end{align*} $$

Thus,

$$ \begin{align*}v^{-1}\mu+\mathrm{wt}(wv\Rightarrow v)\leq&v^{-1}\mu + \mathrm{wt}(\tilde u\Rightarrow v) + \mathrm{wt}(wv\Rightarrow \tilde u)\\\leq &\tilde u^{-1}u\lambda\\\leq&(\tilde u^{-1}u\lambda)^{\mathrm{dom}} = \lambda. \end{align*} $$

Since (2) $\implies $ (3) is trivial, it remains to show (3) $\implies $ (1). So let $v\in {\mathrm {LP}}(x)$ satisfy $v^{-1}\mu + {\mathrm {wt}}(wv\Rightarrow v)\leq \lambda $ . By Theorem 4.2, we immediately get $x\leq \varepsilon ^{v\lambda }$ , showing (1).

Proof. We have

$$ \begin{align*} \text{(1)}\iff&\forall a\in \Delta_{\mathrm{af}}:~\left(\mu +\widetilde\omega_a - w^{-1}\widetilde\omega_a\right)^{{\mathrm{dom}}}\leq\lambda \\\iff&\forall a\in \Delta_{\mathrm{af}}, v\in W:~v^{-1}\left(\mu + \widetilde\omega_a - w^{-1}\widetilde\omega_a\right)\leq\lambda \\\iff&\forall v\in W:\sup_{a\in \Delta_{\mathrm{af}}}v^{-1}\left(\mu + \widetilde\omega_a- w^{-1}\widetilde\omega_a\right)\leq\lambda \\\iff&\text{(2)}. \end{align*} $$

Now, assume that x is in a shrunken Weyl chamber, ${\mathrm {LP}}(x) = \{v\}$ and $a\in \Delta _{\mathrm {af}}$ . We claim that

$$ \begin{align*} \left(\mu + \widetilde\omega_a - w^{-1}\widetilde\omega_a\right)^{{\mathrm{dom}}} = v^{-1}\left(\mu + \widetilde\omega_a - w^{-1}\widetilde\omega_a\right). \end{align*} $$

Once this claim is proved, the equivalence (1) $\iff $ (3) follows.

It remains to show that $v^{-1}\left (\mu + \widetilde \omega _a - w^{-1}\widetilde \omega _a\right )$ is dominant. Hence, let $\alpha \in \Phi ^+$ . We obtain

$$ \begin{align*} \left\langle v^{-1}\left(\mu + \widetilde\omega_a - w^{-1}\widetilde\omega_a\right),\alpha\right\rangle &=\langle \mu,v\alpha\rangle + \langle \widetilde\omega_a,v\alpha\rangle - \langle \widetilde\omega_a,wv\alpha\rangle \\ &\geq\langle \mu,v\alpha\rangle - \Phi^+(-v\alpha) - \Phi^+(wv\alpha) \\ &=\ell(x,v\alpha)-1\geq 0.\\[-36pt] \end{align*} $$

Proof. (1) $\implies $ (2): Comparing condition (3) of Proposition 4.12 with condition (3) of Lemma 4.13 for superregular elements $x\in \widetilde W$ yields the desired claim.

(2) $\implies $ (1): We can directly compare condition (2) of Proposition 4.12 with condition (2) of Lemma 4.13.

(2) $\iff $ (3): Call an irreducible root system $\Phi '$ good if condition (2) is satisfied for $\Phi '$ , and bad otherwise. Certainly, $\Phi $ is good iff each irreducible component of $\Phi $ is good. Moreover, root systems of type $A_n$ are good, we saw this in formula (3.1).

If $\Phi _J\subseteq \Phi $ is bad for some $J\subseteq \Delta $ , then certainly $\Phi $ is bad as well. It remains to show that root systems of types $C_3$ and $G_2$ are good and that root systems of types $B_3, C_4$ and $D_4$ are bad. Each of these claims is easily verified using the Sagemath computer algebra system [30, 29].

Proof. First, note that the relation

$$ \begin{align*} x\preceq x':\iff \forall v\exists v':~ v^{-1}\mu + {\mathrm{wt}}(v'\Rightarrow v) + {\mathrm{wt}}(wv\Rightarrow w'v')\leq (v')^{-1}\mu \end{align*} $$

is transitive. Thus, it suffices to show the implication $x\leq x'\implies x\preceq x'$ for generators $(x, x')$ of the Bruhat order.

In other words, we may assume that $x' = xr_{\mathbf a}$ for an affine root $\mathbf a = (\alpha ,k)\in \Phi _{\mathrm {af}}^+$ with

$$ \begin{align*} x\mathbf a = (w\alpha,k-\langle \mu,\alpha\rangle)\in \Phi_{\mathrm{af}}^+. \end{align*} $$

This means that $w' = ws_\alpha $ and $\mu ' = \mu + (k-\langle \mu ,\alpha \rangle )\alpha ^\vee $ , where $k-\langle \mu ,\alpha \rangle \geq \Phi ^+(-w\alpha )$ . We now do a case distinction depending on whether the root $v^{-1}\alpha $ is positive or negative.

Case $v^{-1}\alpha \in \boldsymbol{\Phi ^-}$ . Put $v' = s_\alpha v$ such that $wv = w'v'$ . Then using [Reference Schremmer28, Lemma 4.3],

$$ \begin{align*} &v^{-1}\mu + {\mathrm{wt}}(v'\Rightarrow v) + {\mathrm{wt}}(wv\Rightarrow w'v') \\=\,&v^{-1}\mu + {\mathrm{wt}}(v s_{-v^{-1}\alpha}\Rightarrow v) + 0 \\\leq\,&v^{-1}\mu -\Phi^+(-\alpha)v^{-1}\alpha^\vee \\\leq\,&v^{-1}\mu - kv^{-1}\alpha^\vee \\=\,&(s_\alpha v)^{-1}(s_\alpha(\mu) + k\alpha^\vee) = (v')^{-1}\mu'. \end{align*} $$

Case $v^{-1}\alpha \in \boldsymbol{\Phi ^+}$ . Put $v' = v$ such that $w'v' = wvs_{v^{-1}\alpha }$ . Then using [Reference Schremmer28, Lemma 4.3],

$$ \begin{align*} &v^{-1}\mu + {\mathrm{wt}}(v'\Rightarrow v) + {\mathrm{wt}}(wv\Rightarrow w'v') \\=\,&v^{-1}\mu + {\mathrm{wt}}(wv\Rightarrow wvs_{v^{-1}\alpha}) \\\leq\,&v^{-1}\mu + \Phi^+(-w\alpha)v^{-1}\alpha^\vee \\\leq\,&v^{-1}\mu + (k-\langle \mu,\alpha\rangle)\alpha^\vee=(v')^{-1}\mu'. \end{align*} $$

This finishes the proof.

Proof. Let $J = J_1\cap \cdots \cap J_m$ . Then we get

$$ \begin{align*} \mu + {\mathrm{wt}}(w\Rightarrow w')\leq \mu'\quad \pmod{\Phi_J^\vee}. \end{align*} $$

Let $C_1 := \ell (x^{-1} x')$ and pick $C_2>0$ such that the conclusion of Corollary 3.12 holds true. We can find an element $\lambda \in \mathbb Z\Phi ^\vee $ such that $\langle \lambda ,\alpha \rangle =0$ for all $\alpha \in J$ and

$$ \begin{align*} \langle \lambda,\alpha\rangle\geq C_2 \end{align*} $$

for all $\alpha \in \Phi ^+\setminus \Phi _J$ . Since $1\in W$ is length positive for both x and $x'$ , it follows from [Reference Schremmer28, Lemma 2.13] that

$$ \begin{align*} \ell(x\varepsilon^\lambda) = \ell(x) + \ell(\varepsilon^\lambda),\qquad \ell(x'\varepsilon^\lambda) = \ell(x') + \ell(\varepsilon^\lambda). \end{align*} $$

So it suffices to show $x\varepsilon ^\lambda \leq x'\varepsilon ^\lambda $ . Note that $x\varepsilon ^\lambda , x'\varepsilon ^\lambda \in \Omega _J^{C_2}$ by choice of $\lambda $ . Moreover, we have

$$ \begin{align*} \mu + \lambda + {\mathrm{wt}}(w\Rightarrow w')\leq \mu'+\lambda\quad \pmod{\Phi_J^\vee} \end{align*} $$

by assumption. Therefore, the inequality $x\varepsilon ^\lambda \leq x'\varepsilon ^\lambda $ follows from Corollary 3.12.

Proof. Induction on $\ell (x')$ .

If $(1,J,\dotsc ,J_m)$ is also Bruhat-deciding for $x'$ , we are done by Lemma 4.16. Otherwise, we must have that $1\in W$ is not length positive for $x'$ or that $J := J_1\cap \cdots \cap J_m$ allows some $\alpha \in \Phi _J$ with $\ell (x',\alpha )\neq 0$ .

First, consider the case that $1\in W$ is not length positive for $x'$ . Then we find a positive root $\alpha \in \Phi ^+$ with $\ell (x',\alpha )<0$ . Hence, $a := (-\alpha ,1)\in \Phi _{\mathrm {af}}^+$ with $x' a \in \Phi ^-$ so that

$$ \begin{align*} x" := w"\varepsilon^{\mu"}:= x'r_{a} = w's_\alpha \varepsilon^{\mu'- (1+\langle\mu',\alpha\rangle)\alpha^\vee}<x'. \end{align*} $$

We calculate

$$ \begin{align*} \mu + {\mathrm{wt}}(w\Rightarrow w") &\leq \mu + {\mathrm{wt}}(w\Rightarrow w') + {\mathrm{wt}}(w'\Rightarrow w's_\alpha) \\ &\leq\mu' + \Phi^+(-w'\alpha)\alpha^\vee \\ &=\mu'- (1+\langle\mu',\alpha\rangle)\alpha^\vee + (\langle \mu',\alpha\rangle +1+\Phi^+(-w'\alpha))\alpha^\vee \\ &=\mu" + (\ell(x',\alpha)+1)\alpha^\vee\leq \mu"\quad \pmod{\Phi_J^\vee}. \end{align*} $$

By induction, $x\leq x"$ . Since $x"<x'$ , we conclude $x<x'$ and are done.

Next, consider the case that $1\in W$ is indeed length positive for $x'$ , but we find some $\alpha \in \Phi _J$ with $\ell (x',\alpha )\neq 0$ . We may assume $\alpha \in \Phi ^+$ , and then $\ell (x',\alpha )>0$ by length positivity. Then $a = (\alpha ,0)\in \Phi _{\mathrm {af}}^+$ with $x' a\in \Phi ^-$ . We conclude that

$$ \begin{align*} x" := w"\varepsilon^{\mu"}:= x'r_{ a} = w's_\alpha \varepsilon^{\mu' -\langle \mu',\alpha\rangle\alpha^\vee}<x'. \end{align*} $$

We calculate

$$ \begin{align*} \mu + {\mathrm{wt}}(w\Rightarrow w")&\leq\mu + {\mathrm{wt}}(w\Rightarrow w') + {\mathrm{wt}}(w'\Rightarrow w's_\alpha) \\ &\leq\mu' + \Phi^+(-w'\alpha)\alpha^\vee \\ &=\mu" + (\Phi^+(-w'\alpha)+\langle \mu',\alpha\rangle)\alpha^\vee \\ &\equiv\mu"\quad \pmod{\Phi_J^\vee}, \end{align*} $$

as $\alpha ^\vee \in \Phi _J^\vee $ . So as in the previous case, we get $x\leq x"<x'$ and are done.

This completes the induction and the proof.

Proof. Among all $v'\in W$ satisfying the inequality

$$ \begin{align*} \mu + {\mathrm{wt}}(v'\Rightarrow 1) + {\mathrm{wt}}(w\Rightarrow w'v')\leq (v')^{-1}\mu'\quad \pmod{\Phi_J^\vee}, \end{align*} $$

pick one of minimal length in W. We prove that $\ell (x',\gamma )<0$ for all $\gamma \in \mathrm{max}\ \mathrm{inv}(v')$ .

Suppose that this was not the case, so $\ell (x',\gamma )\geq 0$ for some $\gamma \in \mathrm{max}\ \mathrm{inv}(v')$ . The condition $\gamma \in {\mathrm {inv}}(v')$ implies $\ell (s_\gamma v')<\ell (v')$ . Moreover, ${\mathrm {wt}}(v'\Rightarrow 1) = {\mathrm {wt}}(s_\gamma v'\Rightarrow 1)-(v')^{-1}\gamma ^\vee $ by Proposition 3.18. We calculate

$$ \begin{align*} \mu + {\mathrm{wt}}(s_\gamma v'\Rightarrow 1)& + {\mathrm{wt}}(w\Rightarrow w's_\gamma v') \\ &= \mu + {\mathrm{wt}}(v'\Rightarrow 1) + (v')^{-1}\gamma^\vee + {\mathrm{wt}}(w\Rightarrow w's_\gamma v')\\ &\leq \mu + {\mathrm{wt}}(v'\Rightarrow 1) + (v')^{-1}\gamma^\vee + {\mathrm{wt}}(w\Rightarrow w'v') + {\mathrm{wt}}(w'v'\Rightarrow w's_\gamma v') \\ &\leq(v')^{-1}\mu' + (v')^{-1}\gamma^\vee + {\mathrm{wt}}(w' v' \Rightarrow w' s_\gamma v') \\ &=(v')^{-1}\mu' + (v')^{-1}\gamma^\vee + {\mathrm{wt}}(w' s_\gamma v' s_{-(v')^{-1}(\gamma)}\Rightarrow w' s_\gamma v') \\ &\leq(v')^{-1}\mu' + (v')^{-1}\gamma^\vee - \Phi^+(w'\gamma) (v')^{-1}\gamma^\vee \\ &=(s_\gamma v')^{-1}\mu' + \langle \mu',\gamma\rangle (v')^{-1}\gamma^\vee + (v')^{-1}\gamma^\vee - \Phi^+(w'\gamma)(v')^{-1}\gamma^\vee \\ &=(s_\gamma v')^{-1}\mu' + \ell(x', \gamma)(v')^{-1}\gamma^\vee \leq (s_\gamma v')^{-1}\mu'\quad \pmod{\Phi_J^\vee}. \end{align*} $$

This is a contradiction to the choice of $v'$ , so we get the desired claim.

Proof. Induction on $\ell (x')$ .

By Lemma 4.18, we may assume that for each $i\in \{1,\dotsc ,m\}$ and $\gamma \in \mathrm{max}\ \mathrm{inv}(v_i')$ , we have $\ell (x',\gamma )<0$ .

If $1\in W$ is length positive for $x'$ , that is, $\ell (x',\alpha )\geq 0$ for all $\alpha \in \Phi ^+$ , then we get $\mathrm{max}\ \mathrm{inv}(v^{\prime }_i)=\emptyset $ for all $i=1,\dotsc ,m$ , that is, $v_i'=1$ . Now, the claim follows from Lemma 4.17.

Thus, suppose that the set

$$ \begin{align*} \{\alpha\in \Phi^+\mid \ell(x',\alpha)<0\} \end{align*} $$

is nonempty. We fix a root $\alpha $ that is maximal within this set. Now, $\mathbf a = (-\alpha ,1)\in \Phi _{{\mathrm {af}}}^+$ satisfies $x'\mathbf a \in \Phi _{\mathrm {af}}^-$ , as $\ell (x',\alpha )<0$ . Consider

$$ \begin{align*} x" := w"\varepsilon^{\mu"}:= x'r_{\mathbf a} = w's_\alpha \varepsilon^{\mu'- (1+\langle\mu',\alpha\rangle)\alpha^\vee}<x'. \end{align*} $$

We want to show $x\leq x"$ using the inductive assumption. So pick an index $i\in \{1,\dotsc ,m\}$ . We do a case distinction based on whether the root $(v_i')^{-1}\alpha $ is positive or negative.

Case $(v_i')^{-1}\alpha \in \boldsymbol{\Phi ^-}$ . Then $\alpha \in {\mathrm {inv}}(v_i')$ , so there exists some $\gamma \in \mathrm{max}\ \mathrm{inv}(v_i')$ with $\alpha \leq \gamma $ . By choice of $v_i'$ , we get $\ell (x',\gamma )<0$ . By maximality of $\alpha $ and $\alpha \leq \gamma $ , we get $\alpha =\gamma $ . In other words, $\alpha \in \mathrm{max}\ \mathrm{inv}(v_i')$ .

Define $v_i" := s_\alpha v_i'$ . Then by Proposition 3.18, ${\mathrm {wt}}(v_i'\Rightarrow 1) = {\mathrm {wt}}(v_i"\Rightarrow 1)-(v_i')^{-1}\alpha ^\vee $ . We compute

$$ \begin{align*}&\mu + \mathrm{wt}(v'^{\prime}_i\Rightarrow 1) + \mathrm{wt}(w\Rightarrow w" v'^{\prime}_i) \\=\,&\mu + \mathrm{wt}(v_i'\Rightarrow 1) + (v_i')^{-1}\alpha^\vee + \mathrm{wt}(w\Rightarrow w'v^{\prime}_i)\\\leq\,&(v^{\prime}_i)^{-1}\mu' + (v^{\prime}_i)^{-1}\alpha^\vee\\=\,&(s_\alpha v^{\prime}_i)^{-1}(\mu' - (1+\langle \mu',\alpha\rangle)\alpha^\vee) = ( v'^{\prime}_i)^{-1}\mu"\quad \pmod{\Phi_{J_i}^\vee}. \end{align*} $$

Case $(v_i')^{-1}\alpha \in \boldsymbol{\Phi ^+}$ . We define $v_i" := v_i'$ and use [Reference Schremmer28, Lemma 4.3] to compute

$$ \begin{align*}&\mu + \mathrm{wt}(v'^{\prime}_i\Rightarrow 1) + \mathrm{wt}(w\Rightarrow w"v'^{\prime}_i) \\\leq\,&\mu + \mathrm{wt}(v^{\prime}_i\Rightarrow 1) + \mathrm{wt}(w\Rightarrow w'v^{\prime}_i) +\mathrm{wt}(w'v^{\prime}_i\Rightarrow w'v^{\prime}_is_{(v_i')^{-1}\alpha})\\\leq\,&(v^{\prime}_i)^{-1}\mu' + \Phi^+(-w'\alpha)(v_i')^{-1}\alpha^\vee\\=\,&(v^{\prime}_i)^{-1}(\mu' - (1+\langle \mu', \alpha\rangle)\alpha^\vee) + (\langle \mu',\alpha\rangle + 1 + \Phi^+(-w'\alpha))(v_i')^{-1}\alpha^\vee\\=\,&(v^{\prime}_i)^{-1}\mu" + (\ell(x',\alpha)+1)(v_i')^{-1}\alpha^\vee \leq (v'^{\prime}_i)^{-1}\mu"\quad \pmod{\Phi_{J_i}^\vee}. \end{align*} $$

In any case, we get the desired inequality

$$ \begin{align*}\mu + \mathrm{wt}(v'^{\prime}_i\Rightarrow 1) + \mathrm{wt}(w\Rightarrow w"v'^{\prime}_i)\leq (v_i")^{-1}\mu"\quad \pmod{\Phi_{J_i}^\vee}. \end{align*} $$

By induction, $x\leq x"<x'$ , completing the induction and the proof.

Proof. Induction on $\ell (v)$ . If $v=1$ , this follows from Lemma 4.19.

Let $J := J_1\cap \cdots \cap J_m$ . If $\alpha \in J$ , then $vs_\alpha $ trivially satisfies the same condition as v. So we may assume that $v\in W^J$ .

Since $v\neq 1$ , we find a simple root $\alpha \in \Delta $ with $v^{-1}\alpha \in \Phi ^-$ . In particular, $\ell (x,\alpha )\leq 0$ such that $x<xs_\alpha $ .

We claim that $(s_\alpha v, J_1,\dotsc ,J_m)$ is a Bruhat-deciding datum for $xs_\alpha $ . Indeed, for $\beta \in \Phi $ , we use [Reference Schremmer28, Lemma 2.12] to compute

$$ \begin{align*} \ell(xs_\alpha, s_\alpha v\beta) &= \ell(x,v\beta) + \ell(s_\alpha,s_\alpha v\beta) \\ &=\ell(x,v\beta)+\begin{cases}1,&v\beta = -\alpha,\\-1,&v\beta=\alpha,\\0,&v\beta\neq \pm \alpha.\end{cases} \end{align*} $$

If $\beta \in \Phi ^+$ , the condition $v^{-1}\alpha \in \Phi ^-$ forces $v\beta \neq \alpha $ , showing

$$ \begin{align*} \ell(xs_\alpha, s_\alpha v\beta) \geq \ell(x,v\beta)\geq 0. \end{align*} $$

Now, consider the case $\beta \in \Phi _J^+$ . Then $\ell (x,v\beta )=0$ by assumption. Moreover, $v\beta \in \Phi ^+$ as $v\in W^J$ so that $v\beta \neq -\alpha $ . We conclude $\ell (xs_\alpha ,s_\alpha v\beta )=\ell (x,v\beta )=0$ in this case.

This shows that $(s_\alpha v,J_1,\dotsc ,J_m)$ is Bruhat-deciding for $xs_\alpha $ . Since $\ell (s_\alpha v)<\ell (v)$ , we may apply the inductive hypothesis to $xs_\alpha $ to prove $xs_\alpha \leq \max (x', x's_\alpha )$ . We distinguish two cases.

Case ${\ell (x',\alpha )\leq 0}$ . This means $x'<x's_\alpha $ , so we wish to prove $xs_\alpha < x's_\alpha $ , using the inductive hypothesis. So let $i\in \{1,\dotsc ,m\}$ . By Lemma 4.3, we may assume that $v_i'$ is length positive for $x'$ .

First, assume that $(v_i')^{-1}\alpha \in \Phi ^-$ . By Lemma 3.7, we get

$$ \begin{align*} {\mathrm{wt}}(v_i'\Rightarrow v) = {\mathrm{wt}}(s_\alpha v_i'\Rightarrow s_\alpha v). \end{align*} $$

Define $v_i":= s_\alpha v_i'$ . Then

$$ \begin{align*} &(s_\alpha v)^{-1}(s_\alpha \mu) + {\mathrm{wt}}(v_i"\Rightarrow s_\alpha v) + {\mathrm{wt}}(ws_\alpha s_\alpha v\Rightarrow w's_\alpha v_i") \\& = v^{-1}\mu + {\mathrm{wt}}(v_i'\Rightarrow v) + {\mathrm{wt}}(wv\Rightarrow w' v_i') \\ &\leq(v_i')^{-1}\mu' = (v_i")(s_\alpha\mu')\quad \pmod{\Phi_{J_i}^\vee}. \end{align*} $$

Next, assume that $(v_i')^{-1}\alpha \in \Phi ^+$ . By length positivity, we must have $\ell (x',\alpha )=0$ . By Lemma 3.7, we get

$$ \begin{align*} {\mathrm{wt}}(v_i'\Rightarrow v) = {\mathrm{wt}}(v_i'\Rightarrow s_\alpha v). \end{align*} $$

Define $v_i" := v_i'$ . Then using [Reference Schremmer28, Lemma 4.3],