2. In the second case, let $\tilde{u} \in W_J$ be such that $vs_\beta u\tilde{u} = (vs_\beta u)^J = v^{vs_\beta(\lambda+\beta^\vee)}$, then $\ell_{v^{vs_\beta(\lambda+\beta^\vee)}}(s_{v(\beta)}w) - \ell_v(w)$ rewrites as $\ell((u\tilde{u})^{-1}v^{-1}w) - \ell(vs_\beta u\tilde{u}) - \ell(v^{-1}w) + \ell(v)$. By Proposition 3.3, $\ell((u\tilde{u})^{-1}v^{-1}w) = \ell(v^{-1}w) - \ell(u\tilde{u})$. By Equation (1.3), $\ell(u\tilde{u}) = \ell(u) + \ell(\tilde{u})$ and $\ell(vs_\beta u\tilde{u}) = \ell vs_\beta u - \ell(\tilde{u})$. By Lemma 3.7 and 3.6, $\ell vs_\beta u = \ell(v) + \ell(s_\beta u) = \ell(v) + \ell(s_\beta) + \ell(u)$.

Thus in this case:

$$ 
\begin{aligned}
\ell_{v^{vs_\beta(\lambda+\beta^\vee)}}(s_{v(\beta)}w) - \ell_v(w) &= \ell((u\tilde{u})^{-1}v^{-1}w) - \ell vs_\beta u\tilde{u} - \ell(v^{-1}w) + \ell(v) \\
&= \ell(v^{-1}w) - \ell(u\tilde{u}) - (\ell vs_\beta u - \ell(\tilde{u})) - \ell(v^{-1}w) + \ell(v) \\
&= -\ell(s_\beta) - 2\ell(u).
\end{aligned} 
\quad (3.19) $$

3. By Lemma (1.11) we have

$$ 
\begin{aligned}
2\mathrm{ht}((\lambda + \beta^\vee)^{++}) &= 2(\mathrm{ht}(\lambda + \beta^\vee) - \sum_{\tau \in \mathrm{Inv}(u^{-1})} \langle \lambda + \beta^\vee, \tau \rangle) \\
&= 2(\mathrm{ht}(\lambda) + \mathrm{ht}(\beta^\vee) - \sum_{\tau \in \mathrm{Inv}(u^{-1})} \langle \lambda + \beta^\vee, \tau \rangle).
\end{aligned} 
\quad (3.20) $$

By reinjecting Expressions (3.18), (3.20) in Formula (3.16), and Expressions (3.19), (3.20) in Formula (3.17) we obtain either way:

$$ 
\begin{aligned}
\ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) &= 2\mathrm{ht}(\lambda) + 2\mathrm{ht}(\beta^\vee) - 2 \sum_{\tau \in \mathrm{Inv}(u^{-1})} \langle \lambda + \beta^\vee, \tau \rangle - 2\mathrm{ht}(\lambda) - \ell(s_\beta) - 2\ell(u) \\
&= \left( 2\mathrm{ht}(\beta^\vee) - \ell(s_\beta) \right) - 2 \left( \ell(u) + \sum_{\tau \in \mathrm{Inv}(u^{-1})} \langle \lambda + \beta^\vee, \tau \rangle \right).
\end{aligned} 
\quad \square $$

Using Lemma 1.10, it is easy to see that $2\mathrm{ht}(\beta^\vee) - \ell(s_\beta)$ is always positive and that, on the contrary, $\ell(u) + \sum_{\tau \in \mathrm{Inv}(u^{-1})} \langle \lambda + \beta^\vee, \tau \rangle$ is always non-positive. Therefore, the length difference is equal to 1 if and only if in the right-hand side of Equation (3.15), the first term is equal to 1 and the second term cancels out. This motivates the following definitions.

**Definition 3.9.** A coweight $\mu \in Y^+$ is **almost dominant** if and only if

$$ \forall \tau \in \Phi_+, \langle \mu, \tau \rangle \geq -1. \quad (3.21) $$

A root $\beta \in \Phi_+$ is **a quantum root** if and only if

$$ \ell(s_\beta) = 2\mathrm{ht}(\beta^\vee) - 1. \quad (3.22) $$
The notion of quantum roots comes from the definition of quantum Bruhat graphs, (see [LNS+12, §4.1]). With Notation 3.5, in Subsection 3.3 we prove that, if $\mathbf{y}$ covers $\mathbf{x}$ then $\lambda + \beta^\vee$ is almost dominant and we prove in Subsection 3.4 that $\beta$ needs to be a quantum root.

**Remark 3.10.** Note that if $\lambda + \beta^\vee$ is dominant, then the second term in the right-hand side of Expression (3.15) immediately cancels out, since in this case $u = 1_W$. In the reductive case, $\Phi$ is finite and therefore if $\lambda$ is far enough in the fundamental chamber (meaning that $\langle \lambda, \alpha_i \rangle$ is large for all $i \in I$, we say that $\lambda$ is superregular), then $\lambda + \beta^\vee$ is always dominant. Accordingly, covers of $\pi^{v(\lambda)}w$ for $\lambda$ superregular are easier to classify (see [LS10, Proposition 4.4], [Mil21, Proposition 4.4]).

## 3.3 Almost-dominance in properly affine covers

In this section, we prove that the second term of the right-hand side of Expression (3.15) need to be zero when $\mathbf{y}$ covers $\mathbf{x}$ (with Notation 3.5), through the following proposition:

**Proposition 3.11.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $\beta \in \Phi_+$ be a positive root and suppose that $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ or $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ covers $\pi^{v(\lambda)}w$. Then $\lambda + \beta^\vee$ is almost dominant, that is to say:

$$ \forall \tau \in \Phi_+, \langle \lambda + \beta^\vee, \tau \rangle \geq -1. \qquad (3.23) $$

It is deduced from the following two technical lemmas, we give their proofs after the proof of Proposition 3.11.

**Lemma 3.12.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$, $w \in W$, $\beta \in \Phi_+$. Suppose that there exists a pair $(\tau, n) \in \Phi_+ \times \mathbb{Z}$ such that:

$$ (i) \ n > 0 $$

$$ (ii) \ \langle \lambda + n\tau^\vee, \beta \rangle \geq -1 $$

$$ (iii) \ n < -\langle \lambda + \beta^\vee, \tau \rangle. $$

Then, $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ and $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$.

**Lemma 3.13.** Let $\lambda \in Y^{++}$ and $\beta \in \Phi_+$ be such that $\lambda + \beta^\vee$ lies in $Y^+$. Let $\tau \in \Phi_+$ be such that $\langle \lambda + \beta^\vee, \tau \rangle \leq -2$ and suppose that $\langle \tau^\vee, \beta \rangle \leq -2$. Then $\langle \lambda + \beta^\vee, s_\tau(\beta) \rangle \geq -1$.

*Proof of Proposition 3.11.* We prove the contrapositive: Let $\tau \in \Phi_+$ be a positive root such that $\langle \lambda + \beta^\vee, \tau \rangle \leq -2$, we will produce non-trivial chain from $\pi^{v(\lambda)}w$ to $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ and $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$. In particular since $\lambda$ is dominant, $\langle \beta^\vee, \tau \rangle \leq -2$.

The numbers $\langle \tau^\vee, \beta \rangle$ and $\langle \beta^\vee, \tau \rangle$ have the same sign ([Bar96, Lemma 1.1.10]), therefore we have that $\langle \tau^\vee, \beta \rangle \leq -1$.

Suppose first that $\langle \tau^\vee, \beta \rangle \leq -2$, then $(\tau, -(\langle \lambda + \beta^\vee, \tau \rangle + 1))$ is a pair which satisfy the conditions of Lemma 3.12:
The notion of quantum roots comes from the definition of quantum Bruhat graphs, (see [LNS+12, §4.1]). With Notation 3.5, in Subsection 3.3 we prove that, if $\mathbf{y}$ covers $\mathbf{x}$ then $\lambda + \beta^\vee$ is almost dominant and we prove in Subsection 3.4 that $\beta$ needs to be a quantum root.

**Remark 3.10.** Note that if $\lambda + \beta^\vee$ is dominant, then the second term in the right-hand side of Expression (3.15) immediately cancels out, since in this case $u = 1_W$. In the reductive case, $\Phi$ is finite and therefore if $\lambda$ is far enough in the fundamental chamber (meaning that $\langle \lambda, \alpha_i \rangle$ is large for all $i \in I$, we say that $\lambda$ is superregular), then $\lambda + \beta^\vee$ is always dominant. Accordingly, covers of $\pi^{v(\lambda)}w$ for $\lambda$ superregular are easier to classify (see [LS10, Proposition 4.4], [Mil21, Proposition 4.4]).

## 3.3 Almost-dominance in properly affine covers

In this section, we prove that the second term of the right-hand side of Expression (3.15) need to be zero when $\mathbf{y}$ covers $\mathbf{x}$ (with Notation 3.5), through the following proposition:

**Proposition 3.11.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $\beta \in \Phi_+$ be a positive root and suppose that $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ or $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ covers $\pi^{v(\lambda)}w$. Then $\lambda + \beta^\vee$ is almost dominant, that is to say:

$$ \forall \tau \in \Phi_+, \quad \langle \lambda + \beta^\vee, \tau \rangle \ge -1. \qquad (3.23) $$

It is deduced from the following two technical lemmas, we give their proofs after the proof of Proposition 3.11.

**Lemma 3.12.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$, $w \in W$, $\beta \in \Phi_+$. Suppose that there exists a pair $(\tau, n) \in \Phi_+ \times \mathbb{Z}$ such that:

$$ (i) \ n > 0 $$

$$ (ii) \ \langle \lambda + n\tau^\vee, \beta \rangle \ge -1 $$

$$ (iii) \ n < -\langle \lambda + \beta^\vee, \tau \rangle. $$

Then, $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ and $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$.

**Lemma 3.13.** Let $\lambda \in Y^{++}$ and $\beta \in \Phi_+$ be such that $\lambda + \beta^\vee$ lies in $Y^+$. Let $\tau \in \Phi_+$ be such that $\langle \lambda + \beta^\vee, \tau \rangle \le -2$ and suppose that $\langle \tau^\vee, \beta \rangle \le -2$. Then $\langle \lambda + \beta^\vee, s_\tau(\beta) \rangle \ge -1$.

*Proof of Proposition 3.11.* We prove the contrapositive: Let $\tau \in \Phi_+$ be a positive root such that $\langle \lambda + \beta^\vee, \tau \rangle \le -2$, we will produce non-trivial chain from $\pi^{v(\lambda)}w$ to $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ and $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$. In particular since $\lambda$ is dominant, $\langle \beta^\vee, \tau \rangle \le -2$.

The numbers $\langle \tau^\vee, \beta \rangle$ and $\langle \beta^\vee, \tau \rangle$ have the same sign ([Bar96, Lemma 1.1.10]), therefore we have that $\langle \tau^\vee, \beta \rangle \le -1$.

Suppose first that $\langle \tau^\vee, \beta \rangle \le -2$, then $(\tau, -(\langle \lambda + \beta^\vee, \tau \rangle + 1))$ is a pair which satisfy the conditions of Lemma 3.12:
(i) Since $\langle \lambda + \beta^{\vee}, \tau \rangle \le -2$, and $-(\langle \lambda + \beta^{\vee}, \tau \rangle + 1) \ge 1 > 0$.

(ii) By Lemma 3.13, $\langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle \ge -1$, thus

$$ \begin{aligned} \langle \lambda - (\langle \lambda + \beta^{\vee}, \tau \rangle + 1)\tau^{\vee}, \beta \rangle &= \langle s_{\tau}(\lambda + \beta^{\vee}) - \beta^{\vee} - \tau^{\vee}, \beta \rangle \\ &= \langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle - 2 - \langle \tau^{\vee}, \beta \rangle \\ &\ge \langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle \ge -1. \end{aligned} $$

(iii) Clearly $-(\langle \beta^{\vee}, \tau \rangle + 1) < -\langle \beta^{\vee}, \tau \rangle$.

Suppose now that $\langle \tau^{\vee}, \beta \rangle = -1$, we show that $(\tau, 1)$ is a pair satisfying the conditions of Lemma 3.12:

(i) The first point is trivially verified.

(ii) Since $\langle \tau^{\vee}, \beta \rangle = -1$ and $\lambda$ is dominant, $\langle \lambda + \tau^{\vee}, \beta \rangle \ge -1$.

(iii) Since $\langle \lambda + \beta^{\vee}, \tau \rangle \le -2$, $1 < -\langle \lambda + \beta^{\vee}, \tau \rangle$.

Hence, either way, if such a $\tau \in \Phi_+$ exists, then by Lemma 3.12 $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ and $\pi^{vs_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$. $\square$

*Proof of Lemma 3.12.* We use conditions (i), (ii), (iii) in the statement to produce chains from $\pi^{v(\lambda)}w$ to $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ and $\pi^{vs_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)}w$.

Suppose first that Inequality (ii) is strict, then we show that we have the following chains:

$$ \pi^{v(\lambda)}w < \pi^{vs_{\tau}(\lambda+n\tau^{\vee})} s_{v(\tau)}w < \pi^{vs_{\tau}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\tau)}s_{v(\beta)}w < \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w \quad (3.24) $$

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda+n\tau^{\vee})} s_{v(\tau)}w < \pi^{vs_{\beta}(\lambda+\beta^{\vee}+n\tau)} s_{v(\beta)}s_{v(\tau)}w < \pi^{vs_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)}w. \quad (3.25) $$

a) By Condition (i), since $\lambda$ is dominant and $\tau$ is a positive root, using Formula (3.8) with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \tau, v, w, n)$, we have

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda+n\tau^{\vee})} s_{v(\tau)}w. \quad (3.26) $$

Using Formula (3.9) with the same parameters gives

$$ \pi^{v(\lambda)}w < \pi^{vs_{\tau}(\lambda+n\tau^{\vee})} s_{v(\tau)}w. \quad (3.27) $$

b) We supposed that Inequality (ii) is strict, so $\langle \lambda + n\tau^{\vee}, \beta \rangle \ge 0$. Then by Formula (3.8) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + n\tau^{\vee}, \beta, vs_{\tau}, s_{v(\tau)}w, 1)$, we get

$$ \pi^{vs_{\tau}(\lambda+n\tau^{\vee})} s_{v(\tau)}w < \pi^{vs_{\tau}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\tau)}s_{v(\beta)}w. \quad (3.28) $$

Moreover by Formula (3.9) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + n\tau^{\vee}, \beta, v, s_{v(\tau)}w, 1)$,

$$ \pi^{v(\lambda+n\tau^{\vee})} s_{v(\tau)}w < \pi^{vs_{\beta}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\beta)}s_{v(\tau)}w. \quad (3.29) $$
(i) Since $\langle \lambda + \beta^{\vee}, \tau \rangle \le -2$, and $-(\langle \lambda + \beta^{\vee}, \tau \rangle + 1) \ge 1 > 0$.

(ii) By Lemma 3.13, $\langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle \ge -1$, thus

$$ \begin{aligned} \langle \lambda - (\langle \lambda + \beta^{\vee}, \tau \rangle + 1) \tau^{\vee}, \beta \rangle &= \langle s_{\tau}(\lambda + \beta^{\vee}) - \beta^{\vee} - \tau^{\vee}, \beta \rangle \\ &= \langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle - 2 - \langle \tau^{\vee}, \beta \rangle \\ &\ge \langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle \ge -1. \end{aligned} $$

(iii) Clearly $-(\langle \beta^{\vee}, \tau \rangle + 1) < -\langle \beta^{\vee}, \tau \rangle$.

Suppose now that $\langle \tau^{\vee}, \beta \rangle = -1$, we show that $(\tau, 1)$ is a pair satisfying the conditions of Lemma 3.12:

(i) The first point is trivially verified.

(ii) Since $\langle \tau^{\vee}, \beta \rangle = -1$ and $\lambda$ is dominant, $\langle \lambda + \tau^{\vee}, \beta \rangle \ge -1$.

(iii) Since $\langle \lambda + \beta^{\vee}, \tau \rangle \le -2$, $1 < -\langle \lambda + \beta^{\vee}, \tau \rangle$.

Hence, either way, if such a $\tau \in \Phi_+$ exists, then by Lemma 3.12 $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ and $\pi^{v s_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$. $\square$

*Proof of Lemma 3.12.* We use conditions (i), (ii), (iii) in the statement to produce chains from $\pi^{v(\lambda)}w$ to $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ and $\pi^{v s_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)}w$.

Suppose first that Inequality (ii) is strict, then we show that we have the following chains:

$$ \pi^{v(\lambda)}w < \pi^{v s_{\tau}(\lambda + n \tau^{\vee})} s_{v(\tau)}w < \pi^{v s_{\tau}(\lambda + \beta^{\vee} + n \tau^{\vee})} s_{v(\tau)}s_{v(\beta)}w < \pi^{v(\lambda + \beta^{\vee})} s_{v(\beta)}w \quad (3.24) $$

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda + n \tau^{\vee})} s_{v(\tau)}w < \pi^{v s_{\beta}(\lambda + \beta^{\vee} + n \tau)} s_{v(\beta)}s_{v(\tau)}w < \pi^{v s_{\beta}(\lambda + \beta^{\vee})} s_{v(\beta)}w. \quad (3.25) $$

a) By Condition (i), since $\lambda$ is dominant and $\tau$ is a positive root, using Formula (3.8) with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \tau, v, w, n)$, we have

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda + n \tau^{\vee})} s_{v(\tau)}w. \quad (3.26) $$

Using Formula (3.9) with the same parameters gives

$$ \pi^{v(\lambda)}w < \pi^{v s_{\tau}(\lambda + n \tau^{\vee})} s_{v(\tau)}w. \quad (3.27) $$

b) We supposed that Inequality (ii) is strict, so $\langle \lambda + n \tau^{\vee}, \beta \rangle \ge 0$. Then by Formula (3.8) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + n \tau^{\vee}, \beta, v s_{\tau}, s_{v(\tau)}w, 1)$, we get

$$ \pi^{v s_{\tau}(\lambda + n \tau^{\vee})} s_{v(\tau)}w < \pi^{v s_{\tau}(\lambda + \beta^{\vee} + n \tau^{\vee})} s_{v(\tau)}s_{v(\beta)}w. \quad (3.28) $$

Moreover by Formula (3.9) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + n \tau^{\vee}, \beta, v, s_{v(\tau)}w, 1)$,

$$ \pi^{v(\lambda + n \tau^{\vee})} s_{v(\tau)}w < \pi^{v s_{\beta}(\lambda + \beta^{\vee} + n \tau^{\vee})} s_{v(\beta)}s_{v(\tau)}w. \quad (3.29) $$
c) Since $\langle \tau^\vee, \tau \rangle = 2$, Condition (iii) is equivalent to $-n < -\langle \lambda + \beta^\vee + n\tau^\vee, \tau \rangle$, so, using Formula (3.9) for $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \beta^\vee + n\tau^\vee, \tau, vs_\tau, s_{v(\tau)}s_{v(\beta)}w, -n)$, we get

$$ \pi^{vs_\tau(\lambda+\beta^\vee+n\tau^\vee)} s_{v(\tau)} s_{v(\beta)} w < \pi^{v(\lambda+\beta^\vee)} s_{v(\beta)} w. \quad (3.30) $$

Moreover using Formula (3.8) for $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda+\beta^\vee+n\tau^\vee, \tau, vs_\beta, s_{v(\beta)}s_{v(\tau)}w, -n)$,

$$ \pi^{vs_\beta(\lambda+\beta^\vee+n\tau)} s_{v(\beta)} s_{v(\tau)} w < \pi^{vs_\beta(\lambda+\beta^\vee)} s_{v(\beta)} w. \quad (3.31) $$

Thus, if Inequality (ii) is strict, combining (3.26)(3.29) and (3.31) we obtain the chain (3.24). Moreover combining (3.27), (3.28) and (3.30) we obtain the chain (3.25). This proves Lemma 3.12 in this case.

Suppose now that $\langle \lambda + n\tau^\vee, \beta \rangle = -1$. Then note that $\lambda + n\tau^\vee + \beta^\vee = s_\beta(\lambda + n\tau^\vee)$, and Formulas (3.8), (3.9) can not be used for the middle inequalities of chains (3.24) and (3.25) anymore.

1. Case $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ if $\langle \lambda + n\tau^\vee, \beta \rangle = -1$.

a) If $vs_\tau(\beta) \in \Phi_+$ then we can apply Formula (1.12) to the element $\pi^{vs_\tau(\lambda+\tau^\vee)}s_{v(\tau)}w$ and the positive affinized root $vs_\tau(\beta)[0]$, and since $\langle \lambda + n\tau^\vee, \beta \rangle = -1 < 0$, we still have

$$ \pi^{vs_\tau(\lambda+n\tau^\vee)} s_{v(\tau)} w < s_{vs_\tau(\beta)[0]} \pi^{vs_\tau(\lambda+\tau^\vee)} s_{v(\tau)} w = \pi^{vs_\tau(\lambda+\beta^\vee+n\tau^\vee)} s_{v(\tau)} s_{v(\beta)} w $$

and the chain (3.24) still holds.

b) If $vs_\tau(\beta) \in \Phi_-$, that is to say $s_\tau(\beta) \in \text{Inv}(v)$, then by Proposition 2.3

$$ \pi^{v(\lambda)} w < \pi^{vs_{s_\tau(\beta)}(\lambda)} s_{v_{s_\tau(\beta)}(\lambda)} w \quad (3.32) $$

and by Formula (3.9) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \tau, vs_{s_\tau(\beta)}, s_{v_{s_\tau(\beta)}(\lambda)}w, 1)$ we get

$$ \pi^{vs_{s_\tau(\beta)}(\lambda)} s_{v_{s_\tau(\beta)}(\lambda)} w < \pi^{vs_\tau s_\beta(\lambda+n\tau^\vee)} s_{v(\tau)} s_{v(\beta)} w = \pi^{vs_\tau(\lambda+\beta^\vee+n\tau^\vee)} s_{v(\tau)} s_{v(\beta)} w. \quad (3.33) $$

For the vectorial elements, we used the fact that $s_{vs_\tau(\beta)} = s_{v(\tau)}s_{v(\beta)}s_{v(\tau)} = vs_\tau s_\beta s_\tau v^{-1}$ and $s_{vs_{s_\tau(\beta)}(\tau)} = vs_\tau s_\beta s_\tau s_\beta s_\tau v^{-1}$, so $s_{vs_{s_\tau(\beta)}(\tau)}s_{vs_\tau(\beta)} = vs_\tau s_\beta v^{-1} = s_{v(\tau)}s_{v(\beta)}$.

Combining Inequalities (3.32) and (3.33) we obtain the chain

$$ \pi^{v(\lambda)} w < \pi^{vs_\tau s_\beta s_\tau(\lambda)} s_{v(\tau)} s_{v(\beta)} s_{v(\tau)} w < \pi^{v(s_\tau(\lambda+\beta^\vee+n\tau^\vee))} s_{v(\tau)} s_{v(\beta)} w < \pi^{v(\lambda+\beta^\vee)} s_{v(\beta)} w \quad (3.34) $$

which replaces chain (3.24).

2. Case $\pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ if $\langle \lambda + n\tau^\vee, \beta \rangle = -1$.
c) Since $\langle \tau^{\vee}, \tau \rangle = 2$, Condition (iii) is equivalent to $-n < -\langle \lambda + \beta^{\vee} + n\tau^{\vee}, \tau \rangle$, so, using Formula (3.9) for $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \beta^{\vee} + n\tau^{\vee}, \tau, vs_{\tau}, s_{v(\tau)}s_{v(\beta)}w, -n)$, we get

$$ \pi^{vs_{\tau}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\tau)} s_{v(\beta)} w < \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w. \quad (3.30) $$

Moreover using Formula (3.8) for $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda+\beta^{\vee}+n\tau^{\vee}, \tau, vs_{\beta}, s_{v(\beta)}s_{v(\tau)}w, -n)$,

$$ \pi^{vs_{\beta}(\lambda+\beta^{\vee}+n\tau)} s_{v(\beta)} s_{v(\tau)} w < \pi^{vs_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)} w. \quad (3.31) $$

Thus, if Inequality (ii) is strict, combining (3.26)(3.29) and (3.31) we obtain the chain (3.24). Moreover combining (3.27), (3.28) and (3.30) we obtain the chain (3.25). This proves Lemma 3.12 in this case.

Suppose now that $\langle \lambda + n\tau^{\vee}, \beta \rangle = -1$. Then note that $\lambda + n\tau^{\vee} + \beta^{\vee} = s_{\beta}(\lambda + n\tau^{\vee})$, and Formulas (3.8), (3.9) can not be used for the middle inequalities of chains (3.24) and (3.25) anymore.

1. Case $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w$ if $\langle \lambda + n\tau^{\vee}, \beta \rangle = -1$.

a) If $vs_{\tau}(\beta) \in \Phi_+$ then we can apply Formula (1.12) to the element $\pi^{vs_{\tau}(\lambda+\tau^{\vee})} s_{v(\tau)} w$ and the positive affinized root $vs_{\tau}(\beta)[0]$, and since $\langle \lambda + n\tau^{\vee}, \beta \rangle = -1 < 0$, we still have

$$ \pi^{vs_{\tau}(\lambda+n\tau^{\vee})} s_{v(\tau)} w < s_{vs_{\tau}(\beta)[0]} \pi^{vs_{\tau}(\lambda+\tau^{\vee})} s_{v(\tau)} w = \pi^{vs_{\tau}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\tau)} s_{v(\beta)} w $$

and the chain (3.24) still holds.

b) If $vs_{\tau}(\beta) \in \Phi_-$, that is to say $s_{\tau}(\beta) \in \text{Inv}(v)$, then by Proposition 2.3

$$ \pi^{v(\lambda)} w < \pi^{vs_{\tau}(\beta)(\lambda)} s_{v(\tau)} w \quad (3.32) $$

and by Formula (3.9) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \tau, vs_{s_{\tau}(\beta)}, s_{vs_{\tau}(\beta)}w, 1)$ we get

$$ \pi^{vs_{s_{\tau}(\beta)}(\lambda)} s_{v(\tau)} s_{v(\beta)} w < \pi^{vs_{\tau}s_{\beta}(\lambda+n\tau^{\vee})} s_{v(\tau)} s_{v(\beta)} w = \pi^{vs_{\tau}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\tau)} s_{v(\beta)} w. \quad (3.33) $$

For the vectorial elements, we used the fact that $s_{vs_{\tau}(\beta)} = s_{v(\tau)}s_{v(\beta)}s_{v(\tau)} = vs_{\tau}s_{\beta}s_{\tau}v^{-1}$ and $s_{vs_{s_{\tau}(\beta)}(\tau)} = vs_{\tau}s_{\beta}s_{\tau}s_{\beta}s_{\tau}v^{-1}$, so $s_{vs_{s_{\tau}(\beta)}(\tau)}s_{vs_{\tau}(\beta)} = vs_{\tau}s_{\beta}v^{-1} = s_{v(\tau)}s_{v(\beta)}$.

Combining Inequalities (3.32) and (3.33) we obtain the chain

$$ \pi^{v(\lambda)} w < \pi^{vs_{\tau}s_{\beta}s_{\tau}(\lambda)} s_{v(\tau)} s_{v(\beta)} s_{v(\tau)} w < \pi^{v(s_{\tau}(\lambda+\beta^{\vee}+n\tau^{\vee}))} s_{v(\tau)} s_{v(\beta)} w < \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w \quad (3.34) $$

which replaces chain (3.24).

2. Case $\pi^{vs_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)} w$ if $\langle \lambda + n\tau^{\vee}, \beta \rangle = -1$.
a) If $w^{-1}vs_{\tau}(\beta) \in \Phi_{-}$, by Formula (1.12) applied to $\pi^{v(\lambda+n\tau^{\vee})}s_{v(\tau)}w$ and the affinized root $v(\beta)[\langle\lambda+n\tau^{\vee}, \beta\rangle]$, since $\langle\lambda+n\tau^{\vee}, \beta\rangle < 0$,

$$ \pi^{v(\lambda+n\tau^{\vee})} s_{v(\tau)} w < \pi^{v(\lambda+n\tau^{\vee})} s_{v(\beta)} s_{v(\tau)} w = \pi^{vs_{\beta}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\beta)} s_{v(\tau)} w $$

and the chain (3.25) still holds.

b) If $w^{-1}vs_{\tau}(\beta) \in \Phi_{+}$, then using Formula (1.12) with $\pi^{v(\lambda)}w$ and the affinized reflection $vs_{\tau}(\beta)[\langle\lambda, s_{\tau}(\beta)\rangle]$ (which is always possible because, if $\langle\lambda, s_{\tau}(\beta)\rangle = 0$ then by minimality of $v$, $vs_{\tau}(\beta) \in \Phi_{+}$), we obtain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\tau)}s_{v(\beta)}s_{v(\tau)}w. \qquad (3.35) $$

Moreover, by Formula (3.8) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \tau, v, s_{vs_{\tau}(\beta)}w, n)$, we get

$$ \pi^{v(\lambda)}s_{v(\tau)}s_{v(\beta)}s_{v(\tau)}w < \pi^{v(\lambda+n\tau^{\vee})}s_{v(\beta)}s_{v(\tau)}w. \qquad (3.36) $$

Hence combining Inequalities (3.35) and (3.36) we obtain a chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\tau)}s_{v(\beta)}s_{v(\tau)}w < \pi^{vs_{\beta}(\lambda+\beta^{\vee}+n\tau^{\vee})}s_{v(\beta)}s_{v(\tau)}w < \pi^{vs_{\beta}(\lambda+\beta^{\vee})}s_{v(\beta)}w. \qquad (3.37) $$

Therefore, in all cases, if such a pair $(\tau, n)$ exists, then $\pi^{v(\lambda+\beta^{\vee})}s_{v(\beta)}w$ and $\pi^{vs_{\beta}(\lambda+\beta^{\vee})}s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$. $\square$

*Proof of Lemma 3.13.* The proof relies on the assumption that $\lambda + \beta^{\vee}$ lies in the Tits cone, which is equivalent to saying that there is only a finite number of positive roots $\alpha$ such that $\langle\lambda + \beta^{\vee}, \alpha\rangle < 0$.

We will show that $\langle\lambda + \beta^{\vee}, (s_{\tau}s_{\beta})^{n}(\tau)\rangle \ge 0$ for $n$ large enough implies $\langle\lambda + \beta^{\vee}, s_{\tau}(\beta)\rangle \ge -1$, which implies the lemma. To shorten the computation, let us write $a = -\langle\beta^{\vee}, \tau\rangle$ and $a^{\vee} = -\langle\tau^{\vee}, \beta\rangle$. So the assumptions $\langle\lambda + \beta^{\vee}, \tau\rangle \le -2$ and $\langle\tau^{\vee}, \beta\rangle \le -2$ imply that $a \ge 2 + \langle\lambda, \tau\rangle$ and $a^{\vee} \ge 2$. In the basis $(\beta, \tau)$ of $\mathbb{R}\beta \oplus \mathbb{R}\tau$, the matrix of $s_{\tau}s_{\beta}$ is $M = \begin{pmatrix} -1 & a \\ -a^{\vee} & aa^{\vee} - 1 \end{pmatrix}$. We have $\chi_M = X^2 + (2 - aa^{\vee})X + 1$ thus, since $aa^{\vee} \ge 4$, $M^2 = (aa^{\vee} - 2)M - I_2$. Write $M^n = \mu_n M + \nu_n I_2$ for $n \in \mathbb{N}$, then an easy computation shows that $\nu_n = -\mu_{n-1}$ and $\mu_{n+1} = (aa^{\vee} - 2)\mu_n - \mu_{n-1}$. In particular since $aa^{\vee} - 2 \ge 2$ and $\mu_0 = 0 < \mu_1$, by iteration $(\mu_n)_{n \in \mathbb{N}}$ is strictly increasing.

Let $x = \langle\lambda, \beta\rangle \ge 0$ and $y = \langle\lambda, \tau\rangle \in [0, a - 2]$. Then

$$ \begin{aligned} \langle\lambda + \beta^{\vee}, (s_{\tau}s_{\beta})^{n}(\tau)\rangle &= \langle\lambda + \beta^{\vee}, a\mu_{n}\beta + ((aa^{\vee} - 1)\mu_{n} - \mu_{n-1})\tau\rangle \quad (3.38) \\ &= (x + 2)\mu_{n}a + ((aa^{\vee} - 1)\mu_{n} - \mu_{n-1})(y - a). \end{aligned} $$

Since $\lambda + \beta^{\vee}$ lies in the Tits cone, $\langle\lambda + \beta^{\vee}, (s_{\tau}s_{\beta})^{n}(\tau)\rangle$ is non-negative for $n$ large enough. Moreover, $\mu_{n-1} < \mu_n$ for all $n \in \mathbb{N}$ and $a - y > 0$. Therefore we deduce from Equation (3.38) that, for $n$ large enough,

$$ (x + 2)\mu_n a \ge (a - y)((aa^{\vee} - 1)\mu_n - \mu_{n-1}) > (a - y)\mu_n(aa^{\vee} - 2). $$
a) If $w^{-1}vs_{\tau}(\beta) \in \Phi_{-}$, by Formula (1.12) applied to $\pi^{v(\lambda+n\tau^{\vee})}s_{v(\tau)}w$ and the affinized root $v(\beta)[\langle\lambda+n\tau^{\vee}, \beta\rangle]$, since $\langle\lambda+n\tau^{\vee}, \beta\rangle < 0$,

$$ \pi^{v(\lambda+n\tau^{\vee})} s_{v(\tau)} w < \pi^{v(\lambda+n\tau^{\vee})} s_{v(\beta)} s_{v(\tau)} w = \pi^{v s_{\beta}(\lambda+\beta^{\vee}+n\tau^{\vee})} s_{v(\beta)} s_{v(\tau)} w $$

and the chain (3.25) still holds.

b) If $w^{-1}vs_{\tau}(\beta) \in \Phi_{+}$, then using Formula (1.12) with $\pi^{v(\lambda)}w$ and the affinized reflection $vs_{\tau}(\beta)[\langle\lambda, s_{\tau}(\beta)\rangle]$ (which is always possible because, if $\langle\lambda, s_{\tau}(\beta)\rangle = 0$ then by minimality of $v$, $vs_{\tau}(\beta) \in \Phi_{+}$), we obtain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\tau)}s_{v(\beta)}s_{v(\tau)}w. \qquad (3.35) $$

Moreover, by Formula (3.8) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \tau, v, s_{v\tau}(\beta)w, n)$, we get

$$ \pi^{v(\lambda)}s_{v(\tau)}s_{v(\beta)}s_{v(\tau)}w < \pi^{v(\lambda+n\tau^{\vee})}s_{v(\beta)}s_{v(\tau)}w. \qquad (3.36) $$

Hence combining Inequalities (3.35) and (3.36) we obtain a chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\tau)}s_{v(\beta)}s_{v(\tau)}w < \pi^{v s_{\beta}(\lambda+\beta^{\vee}+n\tau^{\vee})}s_{v(\beta)}s_{v(\tau)}w < \pi^{v s_{\beta}(\lambda+\beta^{\vee})}s_{v(\beta)}w. \qquad (3.37) $$

Therefore, in all cases, if such a pair $(\tau, n)$ exists, then $\pi^{v(\lambda+\beta^{\vee})}s_{v(\beta)}w$ and $\pi^{v s_{\beta}(\lambda+\beta^{\vee})}s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$. $\square$

*Proof of Lemma 3.13.* The proof relies on the assumption that $\lambda+\beta^{\vee}$ lies in the Tits cone, which is equivalent to saying that there is only a finite number of positive roots $\alpha$ such that $\langle\lambda+\beta^{\vee}, \alpha\rangle < 0$.

We will show that $\langle\lambda+\beta^{\vee}, (s_{\tau}s_{\beta})^{n}(\tau)\rangle \ge 0$ for $n$ large enough implies $\langle\lambda+\beta^{\vee}, s_{\tau}(\beta)\rangle \ge -1$, which implies the lemma. To shorten the computation, let us write $a = -\langle\beta^{\vee}, \tau\rangle$ and $a^{\vee} = -\langle\tau^{\vee}, \beta\rangle$. So the assumptions $\langle\lambda+\beta^{\vee}, \tau\rangle \le -2$ and $\langle\tau^{\vee}, \beta\rangle \le -2$ imply that $a \ge 2 + \langle\lambda, \tau\rangle$ and $a^{\vee} \ge 2$. In the basis $(\beta, \tau)$ of $\mathbb{R}\beta \oplus \mathbb{R}\tau$, the matrix of $s_{\tau}s_{\beta}$ is $M = \begin{pmatrix} -1 & a \\ -a^{\vee} & aa^{\vee}-1 \end{pmatrix}$. We have $\chi_M = X^2 + (2 - aa^{\vee})X + 1$ thus, since $aa^{\vee} \ge 4$, $M^2 = (aa^{\vee}-2)M - I_2$. Write $M^n = \mu_n M + \nu_n I_2$ for $n \in \mathbb{N}$, then an easy computation shows that $\nu_n = -\mu_{n-1}$ and $\mu_{n+1} = (aa^{\vee}-2)\mu_n - \mu_{n-1}$. In particular since $aa^{\vee}-2 \ge 2$ and $\mu_0 = 0 < \mu_1$, by iteration $(\mu_n)_{n \in \mathbb{N}}$ is strictly increasing.

Let $x = \langle\lambda, \beta\rangle \ge 0$ and $y = \langle\lambda, \tau\rangle \in [0, a-2]$. Then

$$ \begin{aligned} \langle\lambda + \beta^{\vee}, (s_{\tau}s_{\beta})^{n}(\tau)\rangle &= \langle\lambda + \beta^{\vee}, a\mu_{n}\beta + ((aa^{\vee}-1)\mu_{n} - \mu_{n-1})\tau\rangle \quad (3.38) \\ &= (x+2)\mu_{n}a + ((aa^{\vee}-1)\mu_{n} - \mu_{n-1})(y-a). \end{aligned} $$

Since $\lambda + \beta^{\vee}$ lies in the Tits cone, $\langle\lambda + \beta^{\vee}, (s_{\tau}s_{\beta})^{n}(\tau)\rangle$ is non-negative for $n$ large enough. Moreover, $\mu_{n-1} < \mu_n$ for all $n \in \mathbb{N}$ and $a-y > 0$. Therefore we deduce from Equation (3.38) that, for $n$ large enough,

$$ (x+2)\mu_n a \ge (a-y)((aa^{\vee}-1)\mu_n - \mu_{n-1}) > (a-y)\mu_n(aa^{\vee}-2). $$
Hence

$$ (x + 2) > (a - y)(a^{\vee} - \frac{2}{a}) = aa^{\vee} - a^{\vee}y - 2 + 2\frac{y}{a}. $$

Therefore $\langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle = x + 2 + a^{\vee}y - aa^{\vee} > -2 + 2\frac{y}{a}$ and, since it is an integer, we deduce $\langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle \geq -1 \geq 1 - a^{\vee}$, which proves the result. $\square$

**Corollary 3.14.** Let $\lambda \in Y^{++}$, $v \in W^{\lambda}$, $w \in W$. Let $\beta \in \Phi_{+}$ be a positive root such that $\lambda + \beta^{\vee} \in Y^{+}$. Suppose that $\mathbf{y} \in \{\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w, \pi^{v s_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)}w\}$ covers $\mathbf{x} = \pi^{v(\lambda)}w$. Then

$$ \ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 2\mathrm{ht}(\beta^{\vee}) - \ell(s_{\beta}). \qquad (3.39) $$

*Proof.* Let $u = v^{\lambda+\beta^{\vee}} \in W$. Then for any $\tau \in \mathrm{Inv}(u^{-1})$, by Lemma 2.1, $\langle \lambda + \beta^{\vee}, \tau \rangle$ is negative. We deduce from Proposition 3.11 that $\langle \lambda + \beta^{\vee}, \tau \rangle = -1$ for any such $\tau$. Therefore

$$ \sum_{\tau \in \mathrm{Inv}(u^{-1})} \langle \lambda + \beta^{\vee}, \tau \rangle = -|\mathrm{Inv}(u^{-1})| = -\ell(u). \qquad (3.40) $$

We then directly obtain Expression (3.39) from Expressions (3.15) and (3.40). $\square$

## 3.4 Properly affine covers and quantum roots

We now prove in Proposition 3.19 that, with Notation 3.5, if $\beta$ is not a quantum root, then $\mathbf{y}$ do not cover $\mathbf{x}$. This is enough, together with Corollary 3.14, to conclude that $\ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 1$. There is a subtlety if the root $\beta$ lies in a subsystem of $\Phi$ of type $G_2$, we suppose that this is not the case in Lemma 3.16 and Lemma 3.17, and we deal with the $G_2$ case in Lemma 3.18. Let us first give another characterization of quantum roots.

**Lemma 3.15.** A root $\beta \in \Phi_{+}$ is a quantum root if and only if $\langle \beta^{\vee}, \gamma \rangle = 1$ for all $\gamma \in \mathrm{Inv}(s_{\beta}) \setminus \{\beta\}$.

*Proof.* Recall that a quantum root is a root $\beta \in \Phi_{+}$ such that $2\mathrm{ht}(\beta^{\vee}) = \ell(s_{\beta}) + 1$. By Corollary 1.10, this is equivalent to

$$ \sum_{\gamma \in \mathrm{Inv}(s_{\beta})} \langle \beta^{\vee}, \gamma \rangle = \ell(s_{\beta}) + 1. \qquad (3.41) $$

For any $\gamma \in \mathrm{Inv}(s_{\beta})$, $\gamma$ is a positive root and $s_{\beta}(\gamma) = \gamma - \langle \beta^{\vee}, \gamma \rangle \beta$ is a negative root, therefore $\langle \beta^{\vee}, \gamma \rangle \geq 1$. Moreover, $\langle \beta^{\vee}, \beta \rangle = 2$ and $|\mathrm{Inv}(s_{\beta})| = \ell(s_{\beta})$. Therefore Equation (3.41) is satisfied if and only if $\langle \beta^{\vee}, \gamma \rangle$ is exactly one for all $\gamma \in \mathrm{Inv}(s_{\beta}) \setminus \{\beta\}$. $\square$

**Lemma 3.16.** Let $\lambda \in Y^{++}$, $v \in W^{\lambda}$, $w \in W$ and $\beta \in \Phi_{+}$. Let $\gamma \in \mathrm{Inv}(s_{\beta}) \setminus \{\beta\}$ be such that $\langle \beta^{\vee}, \gamma \rangle \geq 2$ and suppose that $\beta \notin \mathrm{Inv}(s_{\gamma})$. Then $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ and $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ do not cover $\pi^{v s_{\beta}(\lambda)}w$.
Hence

$$ (x + 2) > (a - y)(a^{\vee} - \frac{2}{a}) = aa^{\vee} - a^{\vee}y - 2 + 2\frac{y}{a}. $$

Therefore $\langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle = x + 2 + a^{\vee}y - aa^{\vee} > -2 + 2\frac{y}{a}$ and, since it is an integer, we deduce $\langle \lambda + \beta^{\vee}, s_{\tau}(\beta) \rangle \geq -1 \geq 1 - a^{\vee}$, which proves the result. $\square$

**Corollary 3.14.** Let $\lambda \in Y^{++}$, $v \in W^{\lambda}$, $w \in W$. Let $\beta \in \Phi_{+}$ be a positive root such that $\lambda + \beta^{\vee} \in Y^{+}$. Suppose that $\mathbf{y} \in \{\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w, \pi^{v s_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)}w\}$ covers $\mathbf{x} = \pi^{v(\lambda)}w$. Then

$$ \ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 2\mathrm{ht}(\beta^{\vee}) - \ell(s_{\beta}). \qquad (3.39) $$

*Proof.* Let $u = v^{\lambda+\beta^{\vee}} \in W$. Then for any $\tau \in \mathrm{Inv}(u^{-1})$, by Lemma 2.1, $\langle \lambda + \beta^{\vee}, \tau \rangle$ is negative. We deduce from Proposition 3.11 that $\langle \lambda + \beta^{\vee}, \tau \rangle = -1$ for any such $\tau$. Therefore

$$ \sum_{\tau \in \mathrm{Inv}(u^{-1})} \langle \lambda + \beta^{\vee}, \tau \rangle = -|\mathrm{Inv}(u^{-1})| = -\ell(u). \qquad (3.40) $$

We then directly obtain Expression (3.39) from Expressions (3.15) and (3.40). $\square$

## 3.4 Properly affine covers and quantum roots

We now prove in Proposition 3.19 that, with Notation 3.5, if $\beta$ is not a quantum root, then $\mathbf{y}$ do not cover $\mathbf{x}$. This is enough, together with Corollary 3.14, to conclude that $\ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 1$. There is a subtlety if the root $\beta$ lies in a subsystem of $\Phi$ of type $G_2$, we suppose that this is not the case in Lemma 3.16 and Lemma 3.17, and we deal with the $G_2$ case in Lemma 3.18. Let us first give another characterization of quantum roots.

**Lemma 3.15.** A root $\beta \in \Phi_{+}$ is a quantum root if and only if $\langle \beta^{\vee}, \gamma \rangle = 1$ for all $\gamma \in \mathrm{Inv}(s_{\beta}) \setminus \{\beta\}$.

*Proof.* Recall that a quantum root is a root $\beta \in \Phi_{+}$ such that $2\mathrm{ht}(\beta^{\vee}) = \ell(s_{\beta}) + 1$. By Corollary 1.10, this is equivalent to

$$ \sum_{\gamma \in \mathrm{Inv}(s_{\beta})} \langle \beta^{\vee}, \gamma \rangle = \ell(s_{\beta}) + 1. \qquad (3.41) $$

For any $\gamma \in \mathrm{Inv}(s_{\beta})$, $\gamma$ is a positive root and $s_{\beta}(\gamma) = \gamma - \langle \beta^{\vee}, \gamma \rangle \beta$ is a negative root, therefore $\langle \beta^{\vee}, \gamma \rangle \geq 1$. Moreover, $\langle \beta^{\vee}, \beta \rangle = 2$ and $|\mathrm{Inv}(s_{\beta})| = \ell(s_{\beta})$. Therefore Equation (3.41) is satisfied if and only if $\langle \beta^{\vee}, \gamma \rangle$ is exactly one for all $\gamma \in \mathrm{Inv}(s_{\beta}) \setminus \{\beta\}$. $\square$

**Lemma 3.16.** Let $\lambda \in Y^{++}$, $v \in W^{\lambda}$, $w \in W$ and $\beta \in \Phi_{+}$. Let $\gamma \in \mathrm{Inv}(s_{\beta}) \setminus \{\beta\}$ be such that $\langle \beta^{\vee}, \gamma \rangle \geq 2$ and suppose that $\beta \notin \mathrm{Inv}(s_{\gamma})$. Then $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ and $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)}w$ do not cover $\pi^{v s_{\beta}(\lambda)}w$.
Proof. By the contrapositive of Proposition 3.11, we can suppose that $\langle \lambda + \beta^{\vee}, \tau \rangle \ge -1$ for any $\tau \in \Phi_+$. Let $\gamma$ be as in the statement and write $\alpha = s_\gamma(\beta) \in \Phi_+$. We will construct non-trivial chains in the same fashion as in the proof of Lemma 3.12. Beforehand, we show by computation that $\langle \lambda + \gamma^{\vee}, \alpha \rangle \ge -1$. If $\langle \gamma^{\vee}, \beta \rangle = 1 = -\langle \gamma^{\vee}, \alpha \rangle$ it is clear since $\lambda$ is dominant, else if $\langle \gamma^{\vee}, \beta \rangle \ge 2$,

$$ \begin{aligned} \langle \lambda + \gamma^{\vee}, \alpha \rangle &= \langle \lambda + \beta^{\vee} - \alpha^{\vee} + (1 - \langle \beta^{\vee}, \gamma \rangle) \gamma^{\vee}, \alpha \rangle \\ &= \langle \lambda + \beta^{\vee}, \alpha \rangle + (1 - \langle \beta^{\vee}, \gamma \rangle) \langle \gamma^{\vee}, \alpha \rangle - 2 \\ &= \langle \lambda + \beta^{\vee}, \alpha \rangle + (\langle \beta^{\vee}, \gamma \rangle - 1) \langle \gamma^{\vee}, \beta \rangle - 2. \end{aligned} $$

Since $\langle \beta^{\vee}, \gamma \rangle \ge 2$ and $\langle \gamma^{\vee}, \beta \rangle \ge 2$, $(\langle \beta^{\vee}, \gamma \rangle - 1)\langle \gamma^{\vee}, \beta \rangle \ge 2$, and by assumption $\langle \lambda + \beta^{\vee}, \alpha \rangle \ge -1$. Thus, $\langle \lambda + \gamma^{\vee}, \alpha \rangle \ge -1$ either way.

We construct chains which are slight modifications of the ones constructed in the proof of Lemma 3.12.

1. Suppose first that $\langle \lambda + \gamma^{\vee}, \alpha \rangle \ge 0$. Then we show that we have the following chains:

$$ \pi^{v(\lambda)} w < \pi^{v(\lambda+\gamma^{\vee})} s_{v(\gamma)} w < \pi^{v(\lambda+\gamma^{\vee}+\alpha^{\vee})} s_{v(\alpha)} s_{v(\gamma)} w < \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w \quad (3.42) $$

$$ \pi^{v(\lambda)} w < \pi^{v s_{\gamma}(\lambda + \gamma^{\vee})} s_{v(\gamma)} w < \pi^{v s_{\gamma} s_{\alpha}(\lambda + \gamma^{\vee} + \alpha^{\vee})} s_{v(\gamma)} s_{v(\alpha)} w < \pi^{v s_{\beta}(\lambda + \beta^{\vee})} s_{v(\beta)} w. \quad (3.43) $$

Indeed:

(a) The coweight $\lambda$ is dominant and $\gamma \in \Phi_+$, so $\langle \lambda, \gamma \rangle \ge 0$ and Formula (3.8) (resp. (3.9)) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \gamma, v, w, 1)$ proves the leftmost inequality in the chain (3.42) (resp. (3.43)).

(b) Since $\langle \lambda+\gamma^{\vee}, \alpha \rangle \ge 0$, Formula (3.8) (resp. (3.9)) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \gamma^{\vee}, \alpha, v, s_{v(\gamma)}w, 1)$ (resp. $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \gamma^{\vee}, \alpha, vs_{\gamma}, s_{v(\gamma)}w, 1)$) proves the second inequality in the chain (3.42) (resp. (3.43)).

(c) Note that $\lambda + \beta^{\vee} = (\lambda + \gamma^{\vee} + \alpha^{\vee}) + (\langle \beta^{\vee}, \gamma \rangle - 1)\gamma^{\vee}$. Moreover $0 < \langle \beta^{\vee}, \gamma \rangle - 1$ and $-\langle \lambda + \gamma^{\vee} + \alpha^{\vee}, \gamma \rangle = \langle \beta^{\vee}, \gamma \rangle - \langle \lambda, \gamma \rangle - 2 < \langle \beta^{\vee}, \gamma \rangle - 1$. Therefore by applying Formula (3.8) (resp. (3.9)) to $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \gamma^{\vee} + \alpha^{\vee}, \gamma, v, s_{v(\alpha)}s_{v(\gamma)}w, \langle \beta^{\vee}, \gamma \rangle - 1)$ (resp. $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda+\gamma^{\vee}+\alpha^{\vee}, \gamma, vs_{\gamma}s_{\alpha}, s_{v(\gamma)}s_{v(\alpha)}w, \langle \beta^{\vee}, \gamma \rangle - 1)$) we obtain the rightmost inequality in the chain (3.42) (resp. (3.43)).

2. We now suppose that $\langle \lambda + \gamma^{\vee}, \alpha \rangle = -1$. Then $\lambda + \gamma^{\vee} + \alpha^{\vee} = s_{\alpha}(\lambda + \gamma^{\vee})$ and the above chains do not always hold. We focus here on the case of $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w$.

(a) If $v(\alpha) \in \Phi_+$, since $\langle \lambda + \gamma^{\vee}, \alpha \rangle < 0$, the inequality $\pi^{v s_{\alpha}(\lambda+\gamma^{\vee})} s_{v(\alpha)} s_{v(\gamma)} w > \pi^{v(\lambda+\gamma^{\vee})} s_{v(\gamma)} w$ still holds, by Formula (1.12) applied with $s_{v(\alpha)}[0]$. The rest of the chain (3.42) still holds and the whole chain remains correct.
*Proof.* By the contrapositive of Proposition 3.11, we can suppose that $\langle \lambda + \beta^{\vee}, \tau \rangle \ge -1$ for any $\tau \in \Phi_+$. Let $\gamma$ be as in the statement and write $\alpha = s_{\gamma}(\beta) \in \Phi_+$. We will construct non-trivial chains in the same fashion as in the proof of Lemma 3.12. Beforehand, we show by computation that $\langle \lambda + \gamma^{\vee}, \alpha \rangle \ge -1$. If $\langle \gamma^{\vee}, \beta \rangle = 1 = -\langle \gamma^{\vee}, \alpha \rangle$ it is clear since $\lambda$ is dominant, else if $\langle \gamma^{\vee}, \beta \rangle \ge 2$,

$$ \begin{aligned} \langle \lambda + \gamma^{\vee}, \alpha \rangle &= \langle \lambda + \beta^{\vee} - \alpha^{\vee} + (1 - \langle \beta^{\vee}, \gamma \rangle) \gamma^{\vee}, \alpha \rangle \\ &= \langle \lambda + \beta^{\vee}, \alpha \rangle + (1 - \langle \beta^{\vee}, \gamma \rangle) \langle \gamma^{\vee}, \alpha \rangle - 2 \\ &= \langle \lambda + \beta^{\vee}, \alpha \rangle + (\langle \beta^{\vee}, \gamma \rangle - 1) \langle \gamma^{\vee}, \beta \rangle - 2. \end{aligned} $$

Since $\langle \beta^{\vee}, \gamma \rangle \ge 2$ and $\langle \gamma^{\vee}, \beta \rangle \ge 2$, $(\langle \beta^{\vee}, \gamma \rangle - 1)\langle \gamma^{\vee}, \beta \rangle \ge 2$, and by assumption $\langle \lambda + \beta^{\vee}, \alpha \rangle \ge -1$. Thus, $\langle \lambda + \gamma^{\vee}, \alpha \rangle \ge -1$ either way.

We construct chains which are slight modifications of the ones constructed in the proof of Lemma 3.12.

1. Suppose first that $\langle \lambda + \gamma^{\vee}, \alpha \rangle \ge 0$. Then we show that we have the following chains:

$$ \pi^{v(\lambda)} w < \pi^{v(\lambda+\gamma^{\vee})} s_{v(\gamma)} w < \pi^{v(\lambda+\gamma^{\vee}+\alpha^{\vee})} s_{v(\alpha)} s_{v(\gamma)} w < \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w \quad (3.42) $$

$$ \pi^{v(\lambda)} w < \pi^{v s_{\gamma}(\lambda+\gamma^{\vee})} s_{v(\gamma)} w < \pi^{v s_{\gamma} s_{\alpha}(\lambda+\gamma^{\vee}+\alpha^{\vee})} s_{v(\gamma)} s_{v(\alpha)} w < \pi^{v s_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)} w. \quad (3.43) $$

Indeed:

(a) The coweight $\lambda$ is dominant and $\gamma \in \Phi_+$, so $\langle \lambda, \gamma \rangle \ge 0$ and Formula (3.8) (resp. (3.9)) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \gamma, v, w, 1)$ proves the leftmost inequality in the chain (3.42) (resp. (3.43)).

(b) Since $\langle \lambda+\gamma^{\vee}, \alpha \rangle \ge 0$, Formula (3.8) (resp. (3.9)) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \gamma^{\vee}, \alpha, v, s_{v(\gamma)}w, 1)$ (resp. $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \gamma^{\vee}, \alpha, v s_{\gamma}, s_{v(\gamma)}w, 1)$) proves the second inequality in the chain (3.42) (resp. (3.43)).

(c) Note that $\lambda + \beta^{\vee} = (\lambda + \gamma^{\vee} + \alpha^{\vee}) + (\langle \beta^{\vee}, \gamma \rangle - 1)\gamma^{\vee}$. Moreover $0 < \langle \beta^{\vee}, \gamma \rangle - 1$ and $-\langle \lambda + \gamma^{\vee} + \alpha^{\vee}, \gamma \rangle = \langle \beta^{\vee}, \gamma \rangle - \langle \lambda, \gamma \rangle - 2 < \langle \beta^{\vee}, \gamma \rangle - 1$. Therefore by applying Formula (3.8) (resp. (3.9)) to $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda + \gamma^{\vee} + \alpha^{\vee}, \gamma, v, s_{v(\alpha)}s_{v(\gamma)}w, \langle \beta^{\vee}, \gamma \rangle - 1)$ (resp. $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda+\gamma^{\vee}+\alpha^{\vee}, \gamma, v s_{\gamma} s_{\alpha}, s_{v(\gamma)}s_{v(\alpha)}w, \langle \beta^{\vee}, \gamma \rangle - 1)$) we obtain the rightmost inequality in the chain (3.42) (resp. (3.43)).

2. We now suppose that $\langle \lambda + \gamma^{\vee}, \alpha \rangle = -1$. Then $\lambda + \gamma^{\vee} + \alpha^{\vee} = s_{\alpha}(\lambda + \gamma^{\vee})$ and the above chains do not always hold. We focus here on the case of $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w$.

(a) If $v(\alpha) \in \Phi_+$, since $\langle \lambda + \gamma^{\vee}, \alpha \rangle < 0$, the inequality $\pi^{v s_{\alpha}(\lambda+\gamma^{\vee})} s_{v(\alpha)} s_{v(\gamma)} w > \pi^{v(\lambda+\gamma^{\vee})} s_{v(\gamma)} w$ still holds, by Formula (1.12) applied with $s_{v(\alpha)}[0]$. The rest of the chain (3.42) still holds and the whole chain remains correct.
(b) If $v(\alpha) \in \Phi_-$, then $vs_\alpha < v$, and we have a chain

$$ \pi^{v(\lambda)}w < \pi^{vs_\alpha(\lambda)}s_{v(\alpha)}w < \pi^{vs_\alpha(\lambda+\gamma^\vee)}s_{v(\alpha)}s_{v(\tau)}w = \pi^{v(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\alpha)}s_{v(\tau)}w < \pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w. \quad (3.44) $$

The reflection used for the first inequality is $s_{-v(\alpha)[0]}$, and it holds by Formula (1.12) because $\langle v(\lambda), -v(\alpha) \rangle = -\langle \lambda, \alpha \rangle < 0$. Note that this is non-zero because $v$ is the minimal representative of $vW_\lambda$ and thus $vs_\alpha < v$ implies $s_\alpha \notin W_\lambda$ so $\langle \lambda, \alpha \rangle \neq 0$. For the second and the third inequalities we use Formula (3.8) with $(\mu_0, \alpha_0, v_0, w_0, m)$ equal to $(\lambda, \gamma, vs_\alpha, s_{v(\alpha)}w, 1)$ and $(\lambda + \alpha^\vee + \gamma^\vee, \gamma, v, s_{v(\alpha)}s_{v(\gamma)}w, \langle \beta^\vee, \gamma \rangle - 1)$ respectively.

3. We suppose that $\langle \lambda + \gamma^\vee, \alpha \rangle = -1$ and we deal with the case of $\pi^{vs_{v(\beta)}(\lambda+\beta^\vee)}s_{v(\beta)}w$. Then

$$ \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\gamma)}s_{v(\alpha)}w = \pi^{vs_\gamma(\lambda+\gamma^\vee)}s_{v\gamma(\alpha)}s_{v(\gamma)}w = s_{v\gamma(\alpha)}[\langle \lambda+\gamma^\vee, \alpha \rangle]\pi^{vs_\gamma(\lambda+\gamma^\vee)}s_{v(\gamma)}w. \quad (3.45) $$

Moreover $(s_{v(\gamma)}w)^{-1}(vs_\gamma(\alpha)) = w^{-1}v(\alpha)$. Thus, since $\langle \lambda + \gamma^\vee, \alpha \rangle < 0$,

(a) If $w^{-1}v(\alpha) \in \Phi_-$, by Formula (1.12), $\pi^{vs_\gamma(\lambda+\gamma^\vee)}s_{v(\gamma)}w < \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\gamma)}s_{v(\alpha)}w$ and the chain (3.43) still holds.

(b) If $w^{-1}v(\alpha) \in \Phi_+$, then, since $\langle \lambda, \alpha \rangle = \langle \gamma^\vee, \beta \rangle - 1 > 0$, by Formula (1.12), $\pi^{v(\lambda)}w < s_{v(\alpha)}[\langle \lambda, \alpha \rangle]\pi^{v(\lambda)}w = \pi^{v(\lambda)}s_{v(\alpha)}w$. Then by Formula (3.9) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \gamma, v, s_{v(\alpha)}w, 1)$ we have

$$ \pi^{v(\lambda)}s_{v(\gamma)}s_{v(\beta)}s_{v(\gamma)}w < \pi^{vs_\gamma(\lambda+\gamma^\vee)}s_{v(\beta)}s_{v(\gamma)}w = \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\gamma)}s_{v(\alpha)}w $$

and we have a chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\alpha)}w < \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\gamma)}s_{v(\alpha)}w < \pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w. \quad (3.46) $$

<figure><img src="image_6.png" /></figure>

**Lemma 3.17.** Let $\beta \in \Phi_+$ be a positive root and suppose that there exists $\gamma \in \text{Inv}(s_\beta) \setminus \{\beta\}$ such that $\langle \beta^\vee, \gamma \rangle \ge 2$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ne 3$. Then $\gamma$ can be chosen such that $\beta \notin \text{Inv}(s_\gamma)$.

*Proof.* Note that, by [Bar96, Lemma 1.1.10], for any $\beta, \gamma \in \Phi$, $\langle \beta^\vee, \gamma \rangle$ and $\langle \gamma^\vee, \beta \rangle$ have the same sign, so if $\langle \beta^\vee, \gamma \rangle \ge 2$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ne 3$, either $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee \rangle \ge 4$, either $\langle \beta^\vee, \gamma \rangle = 2$ and $\langle \gamma^\vee, \beta \rangle = 1$. We treat separately these cases:

1. Let us first suppose that there exists $\gamma \in \text{Inv}(s_\beta)$ such that $\langle \beta^\vee, \gamma \rangle = 2$ and $\langle \gamma^\vee, \beta \rangle = 1$. Suppose that $\beta \in \text{Inv}(s_\gamma)$, so $s_\gamma(\beta) = \beta - \gamma < 0$, and $s_\beta(\gamma) = \gamma - 2\beta < 0$. Then we show that $\beta \notin \text{Inv}(s_{\tilde{\gamma}})$ for $\tilde{\gamma} = -s_\beta(\gamma)$:

$$ s_{\tilde{\gamma}}(\beta) = s_\beta s_\gamma s_\beta(\beta) = -s_\beta(\beta - \gamma) = \gamma - \beta = -s_\gamma(\beta) > 0. $$

Moreover $s_\beta(\tilde{\gamma}) = -\gamma < 0$ and $\langle \beta^\vee, \tilde{\gamma} \rangle = \langle \beta^\vee, \gamma \rangle = 2$ therefore, $\gamma$ can be chosen such that $\beta \notin \text{Inv}(s_\gamma)$.
(b) If $v(\alpha) \in \Phi_-$, then $vs_\alpha < v$, and we have a chain

$$ \pi^{v(\lambda)}w < \pi^{vs_\alpha(\lambda)}s_{v(\alpha)}w < \pi^{vs_\alpha(\lambda+\gamma^\vee)}s_{v(\alpha)}s_{v(\tau)}w = \pi^{v(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\alpha)}s_{v(\tau)}w < \pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w. \quad (3.44) $$

The reflection used for the first inequality is $s_{-v(\alpha)[0]}$, and it holds by Formula (1.12) because $\langle v(\lambda), -v(\alpha) \rangle = -\langle \lambda, \alpha \rangle < 0$. Note that this is non-zero because $v$ is the minimal representative of $vW_\lambda$ and thus $vs_\alpha < v$ implies $s_\alpha \notin W_\lambda$ so $\langle \lambda, \alpha \rangle \neq 0$. For the second and the third inequalities we use Formula (3.8) with $(\mu_0, \alpha_0, v_0, w_0, m)$ equal to $(\lambda, \gamma, vs_\alpha, s_{v(\alpha)}w, 1)$ and $(\lambda + \alpha^\vee + \gamma^\vee, \gamma, v, s_{v(\alpha)}s_{v(\gamma)}w, \langle \beta^\vee, \gamma \rangle - 1)$ respectively.

3. We suppose that $\langle \lambda + \gamma^\vee, \alpha \rangle = -1$ and we deal with the case of $\pi^{vs_{v(\beta)}(\lambda+\beta^\vee)}s_{v(\beta)}w$. Then

$$ \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)} s_{v(\gamma)} s_{v(\alpha)} w = \pi^{vs_\gamma(\lambda+\gamma^\vee)} s_{v\gamma(\alpha)} s_{v(\gamma)} w = s_{v\gamma(\alpha)}[\langle \lambda+\gamma^\vee, \alpha \rangle] \pi^{vs_\gamma(\lambda+\gamma^\vee)} s_{v(\gamma)} w. \quad (3.45) $$

Moreover $(s_{v(\gamma)}w)^{-1}(vs_\gamma(\alpha)) = w^{-1}v(\alpha)$. Thus, since $\langle \lambda + \gamma^\vee, \alpha \rangle < 0$,

(a) If $w^{-1}v(\alpha) \in \Phi_-$, by Formula (1.12), $\pi^{vs_\gamma(\lambda+\gamma^\vee)}s_{v(\gamma)}w < \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\gamma)}s_{v(\alpha)}w$ and the chain (3.43) still holds.

(b) If $w^{-1}v(\alpha) \in \Phi_+$, then, since $\langle \lambda, \alpha \rangle = \langle \gamma^\vee, \beta \rangle - 1 > 0$, by Formula (1.12), $\pi^{v(\lambda)}w < s_{v(\alpha)}[\langle \lambda, \alpha \rangle] \pi^{v(\lambda)}w = \pi^{v(\lambda)}s_{v(\alpha)}w$. Then by Formula (3.9) applied with $(\mu_0, \alpha_0, v_0, w_0, m) = (\lambda, \gamma, v, s_{v(\alpha)}w, 1)$ we have

$$ \pi^{v(\lambda)}s_{v(\gamma)}s_{v(\beta)}s_{v(\gamma)}w < \pi^{vs_\gamma(\lambda+\gamma^\vee)}s_{v(\beta)}s_{v(\gamma)}w = \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\gamma)}s_{v(\alpha)}w $$

and we have a chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\alpha)}w < \pi^{vs_\gamma s_\alpha(\lambda+\gamma^\vee+\alpha^\vee)}s_{v(\gamma)}s_{v(\alpha)}w < \pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w. \quad (3.46) $$

<figure><img src="image_7.png" /></figure>

**Lemma 3.17.** Let $\beta \in \Phi_+$ be a positive root and suppose that there exists $\gamma \in \text{Inv}(s_\beta) \setminus \{\beta\}$ such that $\langle \beta^\vee, \gamma \rangle \ge 2$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ne 3$. Then $\gamma$ can be chosen such that $\beta \notin \text{Inv}(s_\gamma)$.

*Proof.* Note that, by [Bar96, Lemma 1.1.10], for any $\beta, \gamma \in \Phi$, $\langle \beta^\vee, \gamma \rangle$ and $\langle \gamma^\vee, \beta \rangle$ have the same sign, so if $\langle \beta^\vee, \gamma \rangle \ge 2$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ne 3$, either $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee \rangle \ge 4$, either $\langle \beta^\vee, \gamma \rangle = 2$ and $\langle \gamma^\vee, \beta \rangle = 1$. We treat separately these cases:

1. Let us first suppose that there exists $\gamma \in \text{Inv}(s_\beta)$ such that $\langle \beta^\vee, \gamma \rangle = 2$ and $\langle \gamma^\vee, \beta \rangle = 1$. Suppose that $\beta \in \text{Inv}(s_\gamma)$, so $s_\gamma(\beta) = \beta - \gamma < 0$, and $s_\beta(\gamma) = \gamma - 2\beta < 0$. Then we show that $\beta \notin \text{Inv}(s_{\tilde{\gamma}})$ for $\tilde{\gamma} = -s_\beta(\gamma)$:

$$ s_{\tilde{\gamma}}(\beta) = s_\beta s_\gamma s_\beta(\beta) = -s_\beta(\beta - \gamma) = \gamma - \beta = -s_\gamma(\beta) > 0. $$

Moreover $s_\beta(\tilde{\gamma}) = -\gamma < 0$ and $\langle \beta^\vee, \tilde{\gamma} \rangle = \langle \beta^\vee, \gamma \rangle = 2$ therefore, $\gamma$ can be chosen such that $\beta \notin \text{Inv}(s_\gamma)$.
2. Let us now suppose that there exists $\gamma \in \text{Inv}(s_\beta)$ such that $\langle \beta^\vee, \gamma \rangle \ge 2$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ge 4$. Write $\beta = v_\beta(\beta_0) = s_{\alpha_1} \dots s_{\alpha_n}(\beta_0)$ where the $\alpha_i$ and $\beta_0$ are simple roots, and suppose that $n$ is of minimal length amongst possible expressions of $\beta$. Therefore $s_{\alpha_1} \dots s_{\alpha_n} s_{\beta_0} s_{\alpha_n} \dots s_{\alpha_1}$ is a reduced expression of $s_\beta$ and

$$ \mathrm{Inv}(s_\beta) = \{s_{\alpha_1} \dots s_{\alpha_{p-1}}(\alpha_p) \mid p \le n\} \sqcup \{\beta\} \sqcup \{s_{\alpha_1} \dots s_{\alpha_n} s_{\beta_0} s_{\alpha_n} \dots s_{\alpha_{n+1-p}}(\alpha_{n-p}) \mid p \le n\}. \quad (3.47) $$

Let $k$ be the smallest such that $\gamma_k = s_{\alpha_1} \dots s_{\alpha_{k-1}}(\alpha_k)$ verifies $\langle \beta^\vee, \gamma_k \rangle \ge 2$ and $\langle \beta^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \beta \rangle \ge 4$.

The expression $s_{\alpha_1} \dots s_{\alpha_{k-1}} s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_1}$ is an expression of $s_{\gamma_k}$, thus

$$ \mathrm{Inv}(s_{\gamma_k}) \subset \{s_{\alpha_1} \dots s_{\alpha_{p-1}}(\alpha_p) \mid p \le k-1\} \sqcup \{\gamma_k\} \sqcup \{s_{\alpha_1} \dots s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_{k+1-p}}(\alpha_{k-p}) \mid p \le k-1\}. \quad (3.48) $$

Suppose by contradiction that $\beta \in \mathrm{Inv}(s_{\gamma_k})$. Since $v_\beta$ is of minimal length, $\beta$ is not in the first set, hence there is $p \in [1, k-1]$ such that $\beta = s_{\alpha_1} \dots s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_{k+1-p}}(\alpha_{k-p})$.

We show that $\gamma_{k-p} = s_{\alpha_1} \dots s_{\alpha_{k-p-1}}(\alpha_{k-p}) \in \mathrm{Inv}(s_\beta)$ verifies $\langle \beta^\vee, \gamma_{k-p} \rangle \ge 2$, which contradicts the minimality of $k$. Note that $\beta = -s_{\gamma_k}(\gamma_{k-p})$. We compute:

$$ \begin{aligned} \langle \beta^\vee, \gamma_{k-p} \rangle &= \langle -s_{\gamma_k}(\gamma_{k-p}^\vee), \gamma_{k-p} \rangle \\ &= -(2 - \langle \gamma_{k-p}^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \gamma_{k-p} \rangle) \\ &= \langle \beta^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \beta \rangle - 2. \end{aligned} $$

So since $\langle \beta^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \beta \rangle \ge 4$, we get $\langle \beta^\vee, \gamma_{k-p} \rangle \ge 2$, and with a similar computation, we find that $\langle \gamma_{k-p}^\vee, \beta \rangle = \langle \beta^\vee, \gamma_{k-p} \rangle \ge 2$ as well, so $\langle \beta^\vee, \gamma_{k-p} \rangle \langle \gamma_{k-p}^\vee, \beta \rangle \ge 4$. This contradicts the minimality of $k$ and thus $\beta \notin \mathrm{Inv}(s_{\gamma_k})$.

□

**Lemma 3.18.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $\beta \in \Phi_+$ and let $\gamma \in \mathrm{Inv}(s_\beta)$ be such that $\beta \in \mathrm{Inv}(s_\gamma)$ and $\langle \beta^\vee, \gamma \rangle = 3$, $\langle \gamma^\vee, \beta \rangle = 1$.

Then $\pi^{v(\lambda+\beta^\vee)} s_{v(\beta)} w$ and $\pi^{v s_\beta(\lambda+\beta^\vee)} s_{v(\beta)} w$ do not cover $\pi^{v(\lambda)} w$.

*Proof.* We show that, with the assumptions of the statement, $\beta$ and $\gamma$ appear as positive roots of a root subsytem of $\Phi$ isomorphic to $G_2$, and we use this system to construct chains replacing the ones in the proof of Lemma 3.16.

First, note that $-s_\gamma(\beta)$ lies in $\mathrm{Inv}(s_\beta)$ (so $s_\beta s_\gamma(\beta)$ is positive). Indeed, we can write, as in the proof of Lemma 3.17, $\beta = s_{\alpha_1} \dots s_{\alpha_n}(\beta_0)$ for a minimal $n$, and $\gamma = s_{\alpha_1} \dots s_{\alpha_{k-1}}(\alpha_k)$ for some $k \le n$. Then, since $\beta \in \mathrm{Inv}(s_\gamma)$, $\beta$ is also of the form $s_{\alpha_1} \dots s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_{k-p+1}}(\alpha_{k-p})$ for some $p \le k-1$, and thus $-s_\gamma(\beta) = s_{\alpha_1} \dots s_{\alpha_{k-p-1}}(\alpha_{k-p}) \in \mathrm{Inv}(s_\beta)$. Therefore we have the following positive roots, and their associated coroots (the notation will become clear afterwards):

$$ 1. \theta_1 := -s_\gamma(\beta) = \gamma - \beta \in \Phi_+, \text{ with associated coroot } \theta_1^\vee = -s_\gamma(\beta^\vee) = 3\gamma^\vee - \beta^\vee. $$
2. Let us now suppose that there exists $\gamma \in \text{Inv}(s_\beta)$ such that $\langle \beta^\vee, \gamma \rangle \ge 2$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ge 4$. Write $\beta = v_\beta(\beta_0) = s_{\alpha_1} \dots s_{\alpha_n}(\beta_0)$ where the $\alpha_i$ and $\beta_0$ are simple roots, and suppose that $n$ is of minimal length amongst possible expressions of $\beta$. Therefore $s_{\alpha_1} \dots s_{\alpha_n} s_{\beta_0} s_{\alpha_n} \dots s_{\alpha_1}$ is a reduced expression of $s_\beta$ and

$$ \mathrm{Inv}(s_\beta) = \{s_{\alpha_1} \dots s_{\alpha_{p-1}}(\alpha_p) \mid p \le n\} \sqcup \{\beta\} \sqcup \{s_{\alpha_1} \dots s_{\alpha_n} s_{\beta_0} s_{\alpha_n} \dots s_{\alpha_{n+1-p}}(\alpha_{n-p}) \mid p \le n\}. \quad (3.47) $$

Let $k$ be the smallest such that $\gamma_k = s_{\alpha_1} \dots s_{\alpha_{k-1}}(\alpha_k)$ verifies $\langle \beta^\vee, \gamma_k \rangle \ge 2$ and $\langle \beta^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \beta \rangle \ge 4$.

The expression $s_{\alpha_1} \dots s_{\alpha_{k-1}} s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_1}$ is an expression of $s_{\gamma_k}$, thus

$$ \mathrm{Inv}(s_{\gamma_k}) \subset \{s_{\alpha_1} \dots s_{\alpha_{p-1}}(\alpha_p) \mid p \le k-1\} \sqcup \{\gamma_k\} \sqcup \{s_{\alpha_1} \dots s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_{k+1-p}}(\alpha_{k-p}) \mid p \le k-1\}. \quad (3.48) $$

Suppose by contradiction that $\beta \in \mathrm{Inv}(s_{\gamma_k})$. Since $v_\beta$ is of minimal length, $\beta$ is not in the first set, hence there is $p \in [1, k-1]$ such that $\beta = s_{\alpha_1} \dots s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_{k+1-p}}(\alpha_{k-p})$.

We show that $\gamma_{k-p} = s_{\alpha_1} \dots s_{\alpha_{k-p-1}}(\alpha_{k-p}) \in \mathrm{Inv}(s_\beta)$ verifies $\langle \beta^\vee, \gamma_{k-p} \rangle \ge 2$, which contradicts the minimality of $k$. Note that $\beta = -s_{\gamma_k}(\gamma_{k-p})$. We compute:

$$ \begin{aligned} \langle \beta^\vee, \gamma_{k-p} \rangle &= \langle -s_{\gamma_k}(\gamma_{k-p}^\vee), \gamma_{k-p} \rangle \\ &= -(2 - \langle \gamma_{k-p}^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \gamma_{k-p} \rangle) \\ &= \langle \beta^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \beta \rangle - 2. \end{aligned} $$

So since $\langle \beta^\vee, \gamma_k \rangle \langle \gamma_k^\vee, \beta \rangle \ge 4$, we get $\langle \beta^\vee, \gamma_{k-p} \rangle \ge 2$, and with a similar computation, we find that $\langle \gamma_{k-p}^\vee, \beta \rangle = \langle \beta^\vee, \gamma_{k-p} \rangle \ge 2$ as well, so $\langle \beta^\vee, \gamma_{k-p} \rangle \langle \gamma_{k-p}^\vee, \beta \rangle \ge 4$. This contradicts the minimality of $k$ and thus $\beta \notin \mathrm{Inv}(s_{\gamma_k})$.

<figure><img src="image_8.png" /></figure>

**Lemma 3.18.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $\beta \in \Phi_+$ and let $\gamma \in \mathrm{Inv}(s_\beta)$ be such that $\beta \in \mathrm{Inv}(s_\gamma)$ and $\langle \beta^\vee, \gamma \rangle = 3$, $\langle \gamma^\vee, \beta \rangle = 1$.

Then $\pi^{v(\lambda+\beta^\vee)} s_{v(\beta)} w$ and $\pi^{v s_\beta(\lambda+\beta^\vee)} s_{v(\beta)} w$ do not cover $\pi^{v(\lambda)} w$.

*Proof.* We show that, with the assumptions of the statement, $\beta$ and $\gamma$ appear as positive roots of a root subsytem of $\Phi$ isomorphic to $G_2$, and we use this system to construct chains replacing the ones in the proof of Lemma 3.16.

First, note that $-s_\gamma(\beta)$ lies in $\mathrm{Inv}(s_\beta)$ (so $s_\beta s_\gamma(\beta)$ is positive). Indeed, we can write, as in the proof of Lemma 3.17, $\beta = s_{\alpha_1} \dots s_{\alpha_n}(\beta_0)$ for a minimal $n$, and $\gamma = s_{\alpha_1} \dots s_{\alpha_{k-1}}(\alpha_k)$ for some $k \le n$. Then, since $\beta \in \mathrm{Inv}(s_\gamma)$, $\beta$ is also of the form $s_{\alpha_1} \dots s_{\alpha_k} s_{\alpha_{k-1}} \dots s_{\alpha_{k-p+1}}(\alpha_{k-p})$ for some $p \le k-1$, and thus $-s_\gamma(\beta) = s_{\alpha_1} \dots s_{\alpha_{k-p-1}}(\alpha_{k-p}) \in \mathrm{Inv}(s_\beta)$. Therefore we have the following positive roots, and their associated coroots (the notation will become clear afterwards):

$$ 1. \theta_1 := -s_\gamma(\beta) = \gamma - \beta \in \Phi_+, \text{ with associated coroot } \theta_1^\vee = -s_\gamma(\beta^\vee) = 3\gamma^\vee - \beta^\vee. $$
2. $\tilde{\beta} := -s_{\beta}(\gamma) = 3\beta - \gamma \in \Phi_+$, with associated coroot $\tilde{\beta}^{\vee} = -s_{\beta}(\gamma^{\vee}) = \beta^{\vee} - \gamma^{\vee}$.

3. $\tilde{\gamma} := s_{\beta}s_{\gamma}(\beta) = 2\beta - \gamma \in \Phi_+$, with associated coroot $\tilde{\gamma}^{\vee} = s_{\beta}s_{\gamma}(\beta^{\vee}) = 2\beta^{\vee} - 3\gamma^{\vee}$.

Let us also denote $\theta_2 = s_{\theta_1}(\gamma) = 3\beta - 2\gamma$, with associated coroot $\theta_2^{\vee} = \beta^{\vee} - 2\gamma^{\vee}$. Then one can check that $\{\theta_1, \theta_2\}$ form the positive simple roots of a $G_2$ root system (in the sense that $\langle \theta_1^{\vee}, \theta_2 \rangle = -3$ and $\langle \theta_2^{\vee}, \theta_1 \rangle = -1$), such that $\gamma = s_{\theta_1}(\theta_2)$, $\beta = s_{\theta_1}s_{\theta_2}(\theta_1)$, $\tilde{\gamma} = s_{\theta_2}(\theta_1)$ and $\tilde{\beta} = s_{\theta_2}s_{\theta_1}(\theta_2)$. However, $\theta_2$ may not be a positive root in $\Phi$, and we thus need to distinguish these two cases.

Let us first suppose that $\theta_2$ lies in $\Phi_+$. Notice that

$$ \theta_1^{\vee} + \tilde{\beta}^{\vee} + \theta_2^{\vee} = (3\gamma^{\vee} - \beta^{\vee}) + (\beta^{\vee} - \gamma^{\vee}) + (\beta^{\vee} - 2\gamma^{\vee}) = \beta^{\vee}, \quad (3.49) $$

and

$$ s_{\theta_1} s_{\tilde{\beta}} s_{\theta_2} = s_{\theta_1} (s_{\theta_2} s_{\theta_1} s_{\theta_2} s_{\theta_1} s_{\theta_2}) s_{\theta_2} = s_{\theta_1} s_{\theta_2} s_{\theta_1} s_{\theta_2} s_{\theta_1} = s_{\theta_2} s_{\tilde{\beta}} s_{\theta_1} = s_{\beta}. \quad (3.50) $$

Moreover, we have

$$ \langle \theta_2^{\vee}, \tilde{\beta} \rangle = \langle \beta^{\vee} - 2\gamma^{\vee}, 3\beta - \gamma \rangle = 1 > 0 \quad (3.51) $$

$$ \langle \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1 \rangle = \langle 2\beta^{\vee} - 3\gamma^{\vee}, \gamma - \beta \rangle = -1. \quad (3.52) $$

1. Suppose first that $\langle \lambda, \theta_1 \rangle > 0$. Since $\lambda$ is dominant and by Computation (3.51), (3.52), $-\langle \lambda + \theta_2^{\vee}, \tilde{\beta} \rangle < 0$ and $-\langle \lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1 \rangle \le 0$. Using Formula (3.8) (resp. (3.9)) with $(\mu_0, \alpha_0, m) = (\lambda, \theta_2, 1)$ for the first inequality, $(\lambda + \theta_2^{\vee}, \tilde{\beta}, 1)$ for the second and $(\lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1, 1)$ for the third (recall Formulas (3.49), (3.50)), we obtain the chain (3.53) (resp. (3.54))

$$ \pi^{v(\lambda)} w < \pi^{v(\lambda+\theta_2^{\vee})} s_{v(\theta_2)} w < \pi^{v(\lambda+\theta_2^{\vee}+\tilde{\beta}^{\vee})} s_{v(\tilde{\beta})} s_{v(\theta_2)} w < \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w \quad (3.53) $$

$$ \pi^{v(\lambda)} w < \pi^{v s_{\theta_2}(\lambda + \theta_2^{\vee})} s_{v(\theta_2)} w < \pi^{v s_{\theta_2} s_{\tilde{\beta}}(\lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee})} s_{v(\theta_2)} s_{v(\tilde{\beta})} w < \pi^{v s_{\beta}(\lambda + \beta^{\vee})} s_{v(\beta)} w. \quad (3.54) $$

2. If $\langle \lambda, \theta_1 \rangle = 0$, then $-\langle \lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1 \rangle = 1$ so the last inequality in the chains (3.53) and (3.54) do not always hold, we have the following case distinction, which we already encountered in Lemma 3.12 and Lemma 3.16:

(a) If $v(\theta_1) \in \Phi_+$, the chain (3.53) still holds, else $v s_{\theta_1} < v$, $\lambda + \beta^{\vee} = s_{\theta_1}(\lambda + \theta_2^{\vee} + \tilde{\beta})$ and we instead have the chain

$$ \pi^{v(\lambda)} w < \pi^{v s_{\theta_1}(\lambda)} s_{v(\theta_1)} w < \pi^{v s_{\theta_1}(\lambda + \theta_2^{\vee})} s_{v(\theta_1)} s_{v(\theta_2)} w < \pi^{v s_{\theta_1}(\lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee})} s_{v(\theta_1)} s_{v(\tilde{\beta})} s_{v(\theta_2)} w \quad (3.55) $$

where the last term is actually equal to $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w$.
2. $\tilde{\beta} := -s_{\beta}(\gamma) = 3\beta - \gamma \in \Phi_+$, with associated coroot $\tilde{\beta}^{\vee} = -s_{\beta}(\gamma^{\vee}) = \beta^{\vee} - \gamma^{\vee}$.

3. $\tilde{\gamma} := s_{\beta}s_{\gamma}(\beta) = 2\beta - \gamma \in \Phi_+$, with associated coroot $\tilde{\gamma}^{\vee} = s_{\beta}s_{\gamma}(\beta^{\vee}) = 2\beta^{\vee} - 3\gamma^{\vee}$.

Let us also denote $\theta_2 = s_{\theta_1}(\gamma) = 3\beta - 2\gamma$, with associated coroot $\theta_2^{\vee} = \beta^{\vee} - 2\gamma^{\vee}$. Then one can check that $\{\theta_1, \theta_2\}$ form the positive simple roots of a $G_2$ root system (in the sense that $\langle \theta_1^{\vee}, \theta_2 \rangle = -3$ and $\langle \theta_2^{\vee}, \theta_1 \rangle = -1$), such that $\gamma = s_{\theta_1}(\theta_2)$, $\beta = s_{\theta_1}s_{\theta_2}(\theta_1)$, $\tilde{\gamma} = s_{\theta_2}(\theta_1)$ and $\tilde{\beta} = s_{\theta_2}s_{\theta_1}(\theta_2)$. However, $\theta_2$ may not be a positive root in $\Phi$, and we thus need to distinguish these two cases.

Let us first suppose that $\theta_2$ lies in $\Phi_+$. Notice that

$$ \theta_1^{\vee} + \tilde{\beta}^{\vee} + \theta_2^{\vee} = (3\gamma^{\vee} - \beta^{\vee}) + (\beta^{\vee} - \gamma^{\vee}) + (\beta^{\vee} - 2\gamma^{\vee}) = \beta^{\vee}, \quad (3.49) $$

and

$$ s_{\theta_1} s_{\tilde{\beta}} s_{\theta_2} = s_{\theta_1} (s_{\theta_2} s_{\theta_1} s_{\theta_2} s_{\theta_1} s_{\theta_2}) s_{\theta_2} = s_{\theta_1} s_{\theta_2} s_{\theta_1} s_{\theta_2} s_{\theta_1} = s_{\theta_2} s_{\tilde{\beta}} s_{\theta_1} = s_{\beta}. \quad (3.50) $$

Moreover, we have

$$ \langle \theta_2^{\vee}, \tilde{\beta} \rangle = \langle \beta^{\vee} - 2\gamma^{\vee}, 3\beta - \gamma \rangle = 1 > 0 \quad (3.51) $$

$$ \langle \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1 \rangle = \langle 2\beta^{\vee} - 3\gamma^{\vee}, \gamma - \beta \rangle = -1. \quad (3.52) $$

1. Suppose first that $\langle \lambda, \theta_1 \rangle > 0$. Since $\lambda$ is dominant and by Computation (3.51), (3.52), $-\langle \lambda + \theta_2^{\vee}, \tilde{\beta} \rangle < 0$ and $-\langle \lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1 \rangle \le 0$. Using Formula (3.8) (resp. (3.9)) with $(\mu_0, \alpha_0, m) = (\lambda, \theta_2, 1)$ for the first inequality, $(\lambda + \theta_2^{\vee}, \tilde{\beta}, 1)$ for the second and $(\lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1, 1)$ for the third (recall Formulas (3.49), (3.50)), we obtain the chain (3.53) (resp. (3.54))

$$ \pi^{v(\lambda)} w < \pi^{v(\lambda+\theta_2^{\vee})} s_{v(\theta_2)} w < \pi^{v(\lambda+\theta_2^{\vee}+\tilde{\beta}^{\vee})} s_{v(\tilde{\beta})} s_{v(\theta_2)} w < \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w \quad (3.53) $$

$$ \pi^{v(\lambda)} w < \pi^{v s_{\theta_2}(\lambda + \theta_2^{\vee})} s_{v(\theta_2)} w < \pi^{v s_{\theta_2} s_{\tilde{\beta}}(\lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee})} s_{v(\theta_2)} s_{v(\tilde{\beta})} w < \pi^{v s_{\beta}(\lambda + \beta^{\vee})} s_{v(\beta)} w. \quad (3.54) $$

2. If $\langle \lambda, \theta_1 \rangle = 0$, then $-\langle \lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee}, \theta_1 \rangle = 1$ so the last inequality in the chains (3.53) and (3.54) do not always hold, we have the following case distinction, which we already encountered in Lemma 3.12 and Lemma 3.16:

(a) If $v(\theta_1) \in \Phi_+$, the chain (3.53) still holds, else $v s_{\theta_1} < v$, $\lambda + \beta^{\vee} = s_{\theta_1}(\lambda + \theta_2^{\vee} + \tilde{\beta})$ and we instead have the chain

$$ \pi^{v(\lambda)} w < \pi^{v s_{\theta_1}(\lambda)} s_{v(\theta_1)} w < \pi^{v s_{\theta_1}(\lambda + \theta_2^{\vee})} s_{v(\theta_1)} s_{v(\theta_2)} w < \pi^{v s_{\theta_1}(\lambda + \theta_2^{\vee} + \tilde{\beta}^{\vee})} s_{v(\theta_1)} s_{v(\tilde{\beta})} s_{v(\theta_2)} w \quad (3.55) $$

where the last term is actually equal to $\pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w$.
(b) If $w^{-1}v(\theta_1) \in \Phi_-$, then since $\langle \lambda+\theta_2^\vee+\tilde{\beta}^\vee, \theta_1 \rangle < 0$, by Formula (1.12) applied with the affinized root $vs_{\theta_2}s_{\tilde{\beta}}(\theta_1)[\langle \lambda+\theta_2^\vee+\tilde{\beta}^\vee, \theta_1 \rangle]$, the third inequality of Chain (3.54) still holds, and thus the whole chain remains correct. Otherwise if $w^{-1}v(\theta_1) \in \Phi_+$ we instead have the chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\theta_1)}w < \pi^{vs_{\theta_2}(\lambda+\theta_2^\vee)}s_{v(\theta_2)}s_{v(\theta_1)}w < \pi^{vs_{\theta_2}s_{\tilde{\beta}}(\lambda+\theta_2^\vee+\tilde{\beta}^\vee)}s_{v(\theta_2)}s_{v(\tilde{\beta})}s_{v(\theta_1)}w \quad (3.56) $$

where the last term is actually equal to $\pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ since $\lambda + \theta_2^\vee + \tilde{\beta}^\vee = s_{\theta_1}(\lambda + \beta^\vee)$.

We now turn to the case of $\theta_2 \in \Phi_-$. Notice that $\beta^\vee = -\theta_2^\vee + \tilde{\gamma}^\vee + \gamma^\vee$ and $s_\beta = s_\gamma s_{\tilde{\gamma}} s_{\theta_2} = s_{\theta_2} s_{\tilde{\gamma}} s_\gamma$. Moreover, $\langle -\theta_2^\vee, \tilde{\gamma} \rangle = \langle 2\gamma^\vee - \beta^\vee, 2\beta - \gamma \rangle = -1$ and $\langle -\theta_2^\vee + \tilde{\gamma}^\vee, \gamma \rangle = \langle \beta^\vee - \gamma^\vee, \gamma \rangle = 1$. Therefore, since $\lambda$ is dominant and $-\theta_2$ is a positive root:

1. If $\langle \lambda, \tilde{\gamma} \rangle > 0$, then using Formula (3.8) (resp. (3.9)) with $(\mu_0, \alpha_0, m) = (\lambda, -\theta_2, 1)$ for the first inequality, $(\lambda - \theta_2^\vee, \tilde{\gamma}, 1)$ for the second and $(\lambda - \theta_2^\vee + \tilde{\gamma}^\vee, \gamma, 1)$ for the third, we obtain the chain (3.57) (resp. (3.58))

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda-\theta_2^\vee)}s_{v(\theta_2)}w < \pi^{v(\lambda-\theta_2^\vee+\tilde{\gamma}^\vee)}s_{v(\tilde{\gamma})}s_{v(\theta_2)}w < \pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w \quad (3.57) $$

$$ \pi^{v(\lambda)}w < \pi^{vs_{\theta_2}(\lambda-\theta_2^\vee)}s_{v(\theta_2)}w < \pi^{vs_{\theta_2}s_{\tilde{\gamma}}(\lambda-\theta_2^\vee+\tilde{\gamma}^\vee)}s_{v(\tilde{\gamma})}s_{v(\theta_2)}w < \pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w. \quad (3.58) $$

2. Suppose now that $\langle \lambda, \tilde{\gamma} \rangle = 0$, so $\lambda - \theta_2^\vee + \tilde{\gamma} = s_{\tilde{\gamma}}(\lambda - \theta_2^\vee)$. Then

(a) If $v(\tilde{\gamma}) \in \Phi_+$, the chain (3.57) still holds. Else, $vs_{\tilde{\gamma}} < v$ and we instead have the chain

$$ \pi^{v(\lambda)}w < \pi^{vs_{\tilde{\gamma}}(\lambda)}s_{v(\tilde{\gamma})}w < \pi^{vs_{\tilde{\gamma}}(\lambda-\theta_2^\vee)}s_{v(\tilde{\gamma})}s_{v(\theta_2)}w < \pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w \quad (3.59) $$

where the first inequality comes from Proposition 2.3 and the two others from Formula (3.8).

(b) If $w^{-1}v(\tilde{\gamma}) \in \Phi_-$, then the chain (3.58) still holds. Else $w^{-1}v(\tilde{\gamma}) \in \Phi_+$ and we instead have the chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\tilde{\gamma})}w < \pi^{vs_{\theta_2}(\lambda-\theta_2^\vee)}s_{v(\theta_2)}s_{v(\tilde{\gamma})}w < \pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w \quad (3.60) $$

where the first inequality is deduced from Formula (1.12) used with the affinized root $v(\tilde{\gamma})[\langle \lambda, \tilde{\gamma} \rangle]$, and the two others from Formula (3.9) as for the chain (3.58).

□

**Proposition 3.19.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $\beta \in \Phi_+$ and suppose that $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ or $\pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ cover $\pi^{v(\lambda)}w$. Then $\beta$ is a quantum root.

*Proof.* We prove the converse, suppose that $\beta$ is not a quantum root. By Lemma 3.15, there is $\gamma \in \text{Inv}(s_\beta) \setminus \{\beta\}$ such that $\langle \beta^\vee, \gamma \rangle \ge 2$. If $\beta \notin \text{Inv}(s_\gamma)$ we apply Lemma 3.16. We can also apply it in case $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ne 3$ by Lemma 3.17. Finally if $\beta \in \text{Inv}(s_\gamma)$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle = 3$ we apply Lemma 3.18. Therefore if $\beta$ is not a quantum root then $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ and $\pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$. □
(b) If $w^{-1}v(\theta_1) \in \Phi_-$, then since $\langle \lambda+\theta_2^\vee+\tilde{\beta}^\vee, \theta_1 \rangle < 0$, by Formula (1.12) applied with the affinized root $vs_{\theta_2}s_{\tilde{\beta}}(\theta_1)[\langle \lambda+\theta_2^\vee+\tilde{\beta}^\vee, \theta_1 \rangle]$, the third inequality of Chain (3.54) still holds, and thus the whole chain remains correct. Otherwise if $w^{-1}v(\theta_1) \in \Phi_+$ we instead have the chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\theta_1)}w < \pi^{vs_{\theta_2}(\lambda+\theta_2^\vee)}s_{v(\theta_2)}s_{v(\theta_1)}w < \pi^{vs_{\theta_2}s_{\tilde{\beta}}(\lambda+\theta_2^\vee+\tilde{\beta}^\vee)}s_{v(\theta_2)}s_{v(\tilde{\beta})}s_{v(\theta_1)}w \quad (3.56) $$

where the last term is actually equal to $\pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ since $\lambda + \theta_2^\vee + \tilde{\beta}^\vee = s_{\theta_1}(\lambda + \beta^\vee)$.

We now turn to the case of $\theta_2 \in \Phi_-$. Notice that $\beta^\vee = -\theta_2^\vee + \tilde{\gamma}^\vee + \gamma^\vee$ and $s_\beta = s_\gamma s_{\tilde{\gamma}} s_{\theta_2} = s_{\theta_2} s_{\tilde{\gamma}} s_\gamma$. Moreover, $\langle -\theta_2^\vee, \tilde{\gamma} \rangle = \langle 2\gamma^\vee - \beta^\vee, 2\beta - \gamma \rangle = -1$ and $\langle -\theta_2^\vee + \tilde{\gamma}^\vee, \gamma \rangle = \langle \beta^\vee - \gamma^\vee, \gamma \rangle = 1$. Therefore, since $\lambda$ is dominant and $-\theta_2$ is a positive root:

1. If $\langle \lambda, \tilde{\gamma} \rangle > 0$, then using Formula (3.8) (resp. (3.9)) with $(\mu_0, \alpha_0, m) = (\lambda, -\theta_2, 1)$ for the first inequality, $(\lambda - \theta_2^\vee, \tilde{\gamma}, 1)$ for the second and $(\lambda - \theta_2^\vee + \tilde{\gamma}^\vee, \gamma, 1)$ for the third, we obtain the chain (3.57) (resp. (3.58))

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda-\theta_2^\vee)}s_{v(\theta_2)}w < \pi^{v(\lambda-\theta_2^\vee+\tilde{\gamma}^\vee)}s_{v(\tilde{\gamma})}s_{v(\theta_2)}w < \pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w \quad (3.57) $$

$$ \pi^{v(\lambda)}w < \pi^{vs_{\theta_2}(\lambda-\theta_2^\vee)}s_{v(\theta_2)}w < \pi^{vs_{\theta_2}s_{\tilde{\gamma}}(\lambda-\theta_2^\vee+\tilde{\gamma}^\vee)}s_{v(\tilde{\gamma})}s_{v(\theta_2)}w < \pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w. \quad (3.58) $$

2. Suppose now that $\langle \lambda, \tilde{\gamma} \rangle = 0$, so $\lambda - \theta_2^\vee + \tilde{\gamma} = s_{\tilde{\gamma}}(\lambda - \theta_2^\vee)$. Then

(a) If $v(\tilde{\gamma}) \in \Phi_+$, the chain (3.57) still holds. Else, $vs_{\tilde{\gamma}} < v$ and we instead have the chain

$$ \pi^{v(\lambda)}w < \pi^{vs_{\tilde{\gamma}}(\lambda)}s_{v(\tilde{\gamma})}w < \pi^{vs_{\tilde{\gamma}}(\lambda-\theta_2^\vee)}s_{v(\tilde{\gamma})}s_{v(\theta_2)}w < \pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w \quad (3.59) $$

where the first inequality comes from Proposition 2.3 and the two others from Formula (3.8).

(b) If $w^{-1}v(\tilde{\gamma}) \in \Phi_-$, then the chain (3.58) still holds. Else $w^{-1}v(\tilde{\gamma}) \in \Phi_+$ and we instead have the chain

$$ \pi^{v(\lambda)}w < \pi^{v(\lambda)}s_{v(\tilde{\gamma})}w < \pi^{vs_{\theta_2}(\lambda-\theta_2^\vee)}s_{v(\theta_2)}s_{v(\tilde{\gamma})}w < \pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w \quad (3.60) $$

where the first inequality is deduced from Formula (1.12) used with the affinized root $v(\tilde{\gamma})[\langle \lambda, \tilde{\gamma} \rangle]$, and the two others from Formula (3.9) as for the chain (3.58).

□

**Proposition 3.19.** Let $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $\beta \in \Phi_+$ and suppose that $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ or $\pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ cover $\pi^{v(\lambda)}w$. Then $\beta$ is a quantum root.

*Proof.* We prove the converse, suppose that $\beta$ is not a quantum root. By Lemma 3.15, there is $\gamma \in \text{Inv}(s_\beta) \setminus \{\beta\}$ such that $\langle \beta^\vee, \gamma \rangle \ge 2$. If $\beta \notin \text{Inv}(s_\gamma)$ we apply Lemma 3.16. We can also apply it in case $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle \ne 3$ by Lemma 3.17. Finally if $\beta \in \text{Inv}(s_\gamma)$ and $\langle \beta^\vee, \gamma \rangle \langle \gamma^\vee, \beta \rangle = 3$ we apply Lemma 3.18. Therefore if $\beta$ is not a quantum root then $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ and $\pi^{vs_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$ do not cover $\pi^{v(\lambda)}w$. □
## 3.5 Conclusion

We now have everything to prove Theorem A:

**Theorem A.** Suppose that $\mathbf{y}, \mathbf{x} \in W_+^a$ are such that $\mathbf{x} \le \mathbf{y}$. Then

$$ \mathbf{x} \triangleleft \mathbf{y} \iff \ell^a(\mathbf{y}) = \ell^a(\mathbf{x}) + 1. \qquad (3.61) $$

*Proof.* If $\mathbf{x} \le \mathbf{y}$ with $\ell^a(\mathbf{y}) = \ell^a(\mathbf{x}) + 1$, then by strict compatibility of $\ell^a$ (Theorem 1.6), $\mathbf{y}$ covers $\mathbf{x}$. Conversely suppose that $\mathbf{y}$ covers $\mathbf{x}$. If $\mathbf{y}$ and $\mathbf{x}$ have same dominance class then Theorem 2.18 implies that $\ell^a(\mathbf{y}) = \ell^a(\mathbf{x}) + 1$. Else, if $\mathrm{proj}^{Y^+}(\mathbf{y}) \notin W \cdot \mathrm{proj}^{Y^+}(\mathbf{x})$, by Proposition 3.1, $\mathbf{y}$ is of the form $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ or $\pi^{v_{s_\beta}(\lambda+\beta^\vee)}s_{v(\beta)}w$, for $\mathbf{x} = \pi^{v(\lambda)}w$ with $\lambda \in Y^{++}$, $v \in W^\lambda$, $w \in W$ and $\beta \in \Phi_+$. Then, by Corollary 3.14, we have $\ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 2\mathrm{ht}(\beta^\vee) - \ell(s_\beta)$. Moreover, by Proposition 3.19, $\beta$ is a quantum root and therefore in this case as well:

$$ \ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 1. $$

□

Along the way, we have obtained a classification of covers, which we summarize in Proposition 3.20. This is to be compared with [Sch23, Proposition 4.5].

**Proposition 3.20.** Let $\mathbf{x} = \pi^{v(\lambda)}w \in W_+^a$ with $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $J \subset S$ be the set of simple reflections such that $W_\lambda = W_J$, recall Notation 1.2 and Definition 3.9. Then covers of $\mathbf{x}$ are the elements of the following form.

1. $\pi^{v(\lambda)}s_{v(\beta)}w = \mathbf{x}s_{w^{-1}v(\beta)[0]}$ for $\beta \in \Phi$ such that $\ell(s_\beta v^{-1}w) = \ell(v^{-1}w) + 1$.

2. $\pi^{v_{s_\beta}(\lambda)}s_{v(\beta)}w = s_{v(\beta)[0]}\mathbf{x}$ for $\beta \in \Phi_+$ such that:

(a) $\langle \lambda, \beta \rangle \neq 0$

(b) $\ell(\mathrm{vs}_\beta) = \ell(v) - 1$

(c) If $u$ denotes $\mathrm{vs}_\beta$ and $u_J$ the maximal $W_J$-suffix of $u$, then $v u_J^{-1}$ is on a minimal gallery from $v$ to $w$.

3. $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w = s_{v(\beta)[\langle \lambda, \beta \rangle+1]}\mathbf{x} = \mathbf{x}s_{w^{-1}v(\beta)[1]}$ for $\beta \in \Phi_+$ such that:

(a) $\beta$ is a quantum root

(b) $\lambda + \beta^\vee$ is an almost dominant coweight

(c) For $u = v^{\lambda+\beta^\vee}$, $v$ is on a minimal gallery from 1 to $vu$, that is to say $\ell(vu) = \ell(v) + \ell(u)$

(d) For $\tilde{v} = v^{v(\lambda+\beta^\vee)}$, $s_{v(\beta)}\tilde{v}$ is on a minimal gallery from $v$ to $w$.

4. $\pi^{v_{s_\beta}(\lambda+\beta^\vee)}s_{v(\beta)}w = s_{v(\beta)[-1]}\mathbf{x}$ for $\beta \in \Phi_+$ such that:

(a) $\beta$ is a quantum root
## 3.5 Conclusion

We now have everything to prove Theorem A:

**Theorem A.** Suppose that $\mathbf{y}, \mathbf{x} \in W_+^a$ are such that $\mathbf{x} \le \mathbf{y}$. Then

$$ \mathbf{x} \triangleleft \mathbf{y} \iff \ell^a(\mathbf{y}) = \ell^a(\mathbf{x}) + 1. \qquad (3.61) $$

*Proof.* If $\mathbf{x} \le \mathbf{y}$ with $\ell^a(\mathbf{y}) = \ell^a(\mathbf{x}) + 1$, then by strict compatibility of $\ell^a$ (Theorem 1.6), $\mathbf{y}$ covers $\mathbf{x}$. Conversely suppose that $\mathbf{y}$ covers $\mathbf{x}$. If $\mathbf{y}$ and $\mathbf{x}$ have same dominance class then Theorem 2.18 implies that $\ell^a(\mathbf{y}) = \ell^a(\mathbf{x}) + 1$. Else, if $\mathrm{proj}^{Y^+}(\mathbf{y}) \notin W \cdot \mathrm{proj}^{Y^+}(\mathbf{x})$, by Proposition 3.1, $\mathbf{y}$ is of the form $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w$ or $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w$, for $\mathbf{x} = \pi^{v(\lambda)}w$ with $\lambda \in Y^{++}$, $v \in W^\lambda$, $w \in W$ and $\beta \in \Phi_+$. Then, by Corollary 3.14, we have $\ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 2\mathrm{ht}(\beta^\vee) - \ell(s_\beta)$. Moreover, by Proposition 3.19, $\beta$ is a quantum root and therefore in this case as well:

$$ \ell^a(\mathbf{y}) - \ell^a(\mathbf{x}) = 1. $$

□

Along the way, we have obtained a classification of covers, which we summarize in Proposition 3.20. This is to be compared with [Sch23, Proposition 4.5].

**Proposition 3.20.** Let $\mathbf{x} = \pi^{v(\lambda)}w \in W_+^a$ with $\lambda \in Y^{++}$, $v \in W^\lambda$ and $w \in W$. Let $J \subset S$ be the set of simple reflections such that $W_\lambda = W_J$, recall Notation 1.2 and Definition 3.9. Then covers of $\mathbf{x}$ are the elements of the following form.

1. $\pi^{v(\lambda)}s_{v(\beta)}w = \mathbf{x}s_{w^{-1}v(\beta)[0]}$ for $\beta \in \Phi$ such that $\ell(s_\beta v^{-1}w) = \ell(v^{-1}w) + 1$.

2. $\pi^{v s_\beta(\lambda)}s_{v(\beta)}w = s_{v(\beta)[0]}\mathbf{x}$ for $\beta \in \Phi_+$ such that:

(a) $\langle \lambda, \beta \rangle \neq 0$

(b) $\ell(\mathrm{vs}_\beta) = \ell(v) - 1$

(c) If $u$ denotes $\mathrm{vs}_\beta$ and $u_J$ the maximal $W_J$-suffix of $u$, then $v u_J^{-1}$ is on a minimal gallery from $v$ to $w$.

3. $\pi^{v(\lambda+\beta^\vee)}s_{v(\beta)}w = s_{v(\beta)[\langle \lambda, \beta \rangle+1]}\mathbf{x} = \mathbf{x}s_{w^{-1}v(\beta)[1]}$ for $\beta \in \Phi_+$ such that:

(a) $\beta$ is a quantum root

(b) $\lambda + \beta^\vee$ is an almost dominant coweight

(c) For $u = v^{\lambda+\beta^\vee}$, $v$ is on a minimal gallery from 1 to $vu$, that is to say $\ell(vu) = \ell(v) + \ell(u)$

(d) For $\tilde{v} = v^{v(\lambda+\beta^\vee)}$, $s_{v(\beta)}\tilde{v}$ is on a minimal gallery from $v$ to $w$.

4. $\pi^{v s_\beta(\lambda+\beta^\vee)}s_{v(\beta)}w = s_{v(\beta)[-1]}\mathbf{x}$ for $\beta \in \Phi_+$ such that:

(a) $\beta$ is a quantum root
(b) $\lambda + \beta^{\vee}$ is an almost dominant coweight

(c) For $u = v^{\lambda+\beta^{\vee}}$, $v$ is on a minimal gallery from 1 to $v s_{\beta} u$

(d) For $\tilde{v} = v^{v_{\beta}(\lambda+\beta^{\vee})}$, $s_{v(\beta)}\tilde{v}$ is on a minimal gallery from $v$ to $w$.

In particular, suppose that $\lambda \in Y^{++}$ is regular and is such that $\lambda + \beta^{\vee}$ is also regular for any quantum root $\beta \in \Phi_+$, we then say that $\lambda$ is superregular. Proposition 3.20 can be simplified for superregular coweights. This is to be compared with [LS10, Proposition 4.4] and [Wel22, Theorem 2].

**Proposition 3.21.** Let $\mathbf{x} = \pi^{v(\lambda)}w \in W_+^a$ with $\lambda \in Y^{++}$ a superregular coweight and $v, w \in W$. Then covers of $\mathbf{x}$ are the elements of the following form.

1. $\mathbf{x} s_{\beta[0]} = \pi^{v(\lambda)} w s_{\beta}$ for $\beta \in \Phi_+$ such that $\ell(v^{-1} w s_{\beta}) = \ell(v^{-1} w) + 1$.

2. $s_{\beta[0]}\mathbf{x} = \pi^{s_{\beta}v(\lambda)}s_{\beta}w$ for $\beta \in \Phi_+$ such that $\ell(s_{\beta}v) = \ell(v) - 1$.

3. $\mathbf{x} s_{w^{-1}v(\beta)[1]} = \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w$ for $\beta \in \Phi_+$ a quantum root such that $\ell(v^{-1}w) = \ell(s_{\beta}) + \ell(s_{\beta}v^{-1}w)$ (otherwise said $s_{\beta}v^{-1}w \le_R v^{-1}w$).

4. $s_{v(\beta)[-1]}\mathbf{x} = \pi^{v s_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)} w$ for $\beta \in \Phi_+$ a quantum root such that $\ell(v s_{\beta}) = \ell(v) + \ell(s_{\beta})$ (otherwise said $s_{\beta} \le_R v s_{\beta}$).

For Kac-Moody root systems, the existence of superregular coweights is not clear a priori. However in an upcoming joint work with Hébert we prove that any Kac-Moody root system admits a finite number of quantum roots, which ensures the existence of superregular coweights. We also use this finiteness to deduce that any element of $W_+^a$ admits a finite number of covers; in particular intervals in $W_+^a$ are finite.

## References

[Bar96] Nicole Bardy. Systèmes de racines infinis. *Mém. Soc. Math. Fr.* (N.S.), (65):vi+188, 1996.

[BB05] Anders Björner and Francesco Brenti. *Combinatorics of Coxeter groups*, volume 231 of *Graduate Texts in Mathematics*. Springer, New York, 2005.

[BKP16] Alexander Braverman, David Kazhdan, and Manish M. Patnaik. Iwahori-Hecke algebras for $p$-adic loop groups. *Invent. Math.*, 204(2):347–442, 2016.

[BPGR16] Nicole Bardy-Panse, Stéphane Gaussent, and Guy Rousseau. Iwahori-Hecke algebras for Kac-Moody groups over local fields. *Pacific J. Math.*, 285(1):1–61, 2016.

[BPHR22] Nicole Bardy-Panse, Auguste Hebert, and Guy Rousseau. Twin masures associated with Kac-Moody groups over Laurent polynomials. *arXiv preprint arXiv:2210.07603*, 2022.
(b) $\lambda + \beta^{\vee}$ is an almost dominant coweight

(c) For $u = v^{\lambda+\beta^{\vee}}$, $v$ is on a minimal gallery from 1 to $v s_{\beta} u$

(d) For $\tilde{v} = v^{v_{\beta}(\lambda+\beta^{\vee})}$, $s_{v(\beta)}\tilde{v}$ is on a minimal gallery from $v$ to $w$.

In particular, suppose that $\lambda \in Y^{++}$ is regular and is such that $\lambda + \beta^{\vee}$ is also regular for any quantum root $\beta \in \Phi_+$, we then say that $\lambda$ is superregular. Proposition 3.20 can be simplified for superregular coweights. This is to be compared with [LS10, Proposition 4.4] and [Wel22, Theorem 2].

**Proposition 3.21.** Let $\mathbf{x} = \pi^{v(\lambda)}w \in W_+^a$ with $\lambda \in Y^{++}$ a superregular coweight and $v, w \in W$. Then covers of $\mathbf{x}$ are the elements of the following form.

1. $\mathbf{x} s_{\beta[0]} = \pi^{v(\lambda)} w s_{\beta}$ for $\beta \in \Phi_+$ such that $\ell(v^{-1} w s_{\beta}) = \ell(v^{-1} w) + 1$.

2. $s_{\beta[0]}\mathbf{x} = \pi^{s_{\beta}v(\lambda)}s_{\beta}w$ for $\beta \in \Phi_+$ such that $\ell(s_{\beta}v) = \ell(v) - 1$.

3. $\mathbf{x} s_{w^{-1}v(\beta)[1]} = \pi^{v(\lambda+\beta^{\vee})} s_{v(\beta)} w$ for $\beta \in \Phi_+$ a quantum root such that $\ell(v^{-1}w) = \ell(s_{\beta}) + \ell(s_{\beta}v^{-1}w)$ (otherwise said $s_{\beta}v^{-1}w \le_R v^{-1}w$).

4. $s_{v(\beta)[-1]}\mathbf{x} = \pi^{v s_{\beta}(\lambda+\beta^{\vee})} s_{v(\beta)} w$ for $\beta \in \Phi_+$ a quantum root such that $\ell(v s_{\beta}) = \ell(v) + \ell(s_{\beta})$ (otherwise said $s_{\beta} \le_R v s_{\beta}$).

For Kac-Moody root systems, the existence of superregular coweights is not clear a priori. However in an upcoming joint work with Hébert we prove that any Kac-Moody root system admits a finite number of quantum roots, which ensures the existence of superregular coweights. We also use this finiteness to deduce that any element of $W_+^a$ admits a finite number of covers; in particular intervals in $W_+^a$ are finite.

## References

[Bar96] Nicole Bardy. Systèmes de racines infinis. *Mém. Soc. Math. Fr.* (N.S.), (65):vi+188, 1996.

[BB05] Anders Björner and Francesco Brenti. *Combinatorics of Coxeter groups*, volume 231 of *Graduate Texts in Mathematics*. Springer, New York, 2005.

[BKP16] Alexander Braverman, David Kazhdan, and Manish M. Patnaik. Iwahori-Hecke algebras for $p$-adic loop groups. *Invent. Math.*, 204(2):347–442, 2016.

[BPGR16] Nicole Bardy-Panse, Stéphane Gaussent, and Guy Rousseau. Iwahori-Hecke algebras for Kac-Moody groups over local fields. *Pacific J. Math.*, 285(1):1–61, 2016.

[BPHR22] Nicole Bardy-Panse, Auguste Hebert, and Guy Rousseau. Twin masures associated with Kac-Moody groups over Laurent polynomials. *arXiv preprint arXiv:2210.07603*, 2022.
[CFG21] Pierre-Emmanuel Chaput, Lucas Fresse, and Thomas Gobet. Parametrization, structure and Bruhat order of certain spherical quotients. *Representation Theory. An Electronic Journal of the American Mathematical Society*, 25:935-974, October 2021.

[IM65] N. Iwahori and H. Matsumoto. On some Bruhat decomposition and the structure of the Hecke rings of *p*-adic Chevalley groups. *Inst. Hautes Études Sci. Publ. Math.*, (25):5-48, 1965.

[KL80] David Kazhdan and George Lusztig. Schubert varieties and Poincaré duality. In *Geometry of the Laplace operator* (Proc. Sympos. Pure Math., Univ. Hawaii, Honolulu, Hawaii, 1979), volume XXXVI of *Proc. Sympos. Pure Math.*, pages 185-203. Amer. Math. Soc., Providence, RI, 1980.

[Kum02] Shrawan Kumar. *Kac-Moody groups, their flag varieties and representation theory*, volume 204 of *Progress in Mathematics*. Birkhäuser Boston, Inc., Boston, MA, 2002.

[LNS+12] Cristian Lenart, Satoshi Naito, Daisuke Sagaki, Anne Schilling, and Mark Shimozono. A uniform model for kirillov-reshetikhin crystals i: Lifting the parabolic quantum bruhat graph. *International Mathematics Research Notices*, 2015, 112012.

[LS10] Thomas Lam and Mark Shimozono. Quantum cohomology of *g/p* and homology of affine grassmannian. *Acta Mathematica*, 204, 03 2010.

[Mil21] Elizabeth Milićević. Maximal Newton Points and the Quantum Bruhat Graph. *Michigan Mathematical Journal*, 70(3):451-502, 2021.

[MO18] Dinakar Muthiah and Daniel Orr. Walk algebras, distinguished subexpressions, and point counting in Kac-Moody flag varieties. In *Representations of Lie algebras, quantum groups and related topics*, volume 713 of *Contemp. Math.*, pages 187-203. Amer. Math. Soc., Providence, RI, 2018.

[MO19] Dinakar Muthiah and Daniel Orr. On the double-affine Bruhat order: the $\varepsilon = 1$ conjecture and classification of covers in ADE type. *Algebr. Comb.*, 2(2):197-216, 2019.

[Mut18] Dinakar Muthiah. On Iwahori-Hecke algebras for *p*-adic loop groups: double coset basis and Bruhat order. *Amer. J. Math.*, 140(1):221-244, 2018.

[Mut19] Dinakar Muthiah. Double-affine Kazhdan-Lusztig polynomials via masures. *arXiv preprint arXiv:1910.13694*, 2019.

[Rém02] Bertrand Rémy. Groupes de Kac-Moody déployés et presque déployés. *Astérisque*, (277):viii+348, 2002.
[CFG21] Pierre-Emmanuel Chaput, Lucas Fresse, and Thomas Gobet. Parametrization, structure and Bruhat order of certain spherical quotients. *Representation Theory. An Electronic Journal of the American Mathematical Society*, 25:935–974, October 2021.

[IM65] N. Iwahori and H. Matsumoto. On some Bruhat decomposition and the structure of the Hecke rings of *p*-adic Chevalley groups. *Inst. Hautes Études Sci. Publ. Math.*, (25):5–48, 1965.

[KL80] David Kazhdan and George Lusztig. Schubert varieties and Poincaré duality. In *Geometry of the Laplace operator* (Proc. Sympos. Pure Math., Univ. Hawaii, Honolulu, Hawaii, 1979), volume XXXVI of *Proc. Sympos. Pure Math.*, pages 185–203. Amer. Math. Soc., Providence, RI, 1980.

[Kum02] Shrawan Kumar. *Kac-Moody groups, their flag varieties and representation theory*, volume 204 of *Progress in Mathematics*. Birkhäuser Boston, Inc., Boston, MA, 2002.

[LNS+12] Cristian Lenart, Satoshi Naito, Daisuke Sagaki, Anne Schilling, and Mark Shimozono. A uniform model for kirillov-reshetikhin crystals i: Lifting the parabolic quantum bruhat graph. *International Mathematics Research Notices*, 2015, 1120–1146.

[LS10] Thomas Lam and Mark Shimozono. Quantum cohomology of *g/p* and homology of affine grassmannian. *Acta Mathematica*, 204, 03 2010.

[Mil21] Elizabeth Milićević. Maximal Newton Points and the Quantum Bruhat Graph. *Michigan Mathematical Journal*, 70(3):451–502, 2021.

[MO18] Dinakar Muthiah and Daniel Orr. Walk algebras, distinguished subexpressions, and point counting in Kac-Moody flag varieties. In *Representations of Lie algebras, quantum groups and related topics*, volume 713 of *Contemp. Math.*, pages 187–203. Amer. Math. Soc., Providence, RI, 2018.

[MO19] Dinakar Muthiah and Daniel Orr. On the double-affine Bruhat order: the $\varepsilon = 1$ conjecture and classification of covers in ADE type. *Algebr. Comb.*, 2(2):197–216, 2019.

[Mut18] Dinakar Muthiah. On Iwahori-Hecke algebras for *p*-adic loop groups: double coset basis and Bruhat order. *Amer. J. Math.*, 140(1):221–244, 2018.

[Mut19] Dinakar Muthiah. Double-affine Kazhdan-Lusztig polynomials via masures. *arXiv preprint arXiv:1910.13694*, 2019.

[Rém02] Bertrand Rémy. Groupes de Kac-Moody déployés et presque déployés. *Astérisque*, (277):viii+348, 2002.
[Ron89] Mark Ronan. Lectures on buildings, volume 7 of perspectives in mathematics. Academic Press Inc., Boston, MA, 11:12, 1989.

[Sch23] Felix Schremmer. Affine bruhat order and demazure products. 2023.

[Wel22] Amanda Welch. Classification of cocovers in the double affine Bruhat order. Electron. J. Comb., 29(4):research paper p4.7, 19, 2022.
[Ron89] Mark Ronan. Lectures on buildings, volume 7 of perspectives in mathematics. *Academic Press Inc., Boston, MA*, 11:12, 1989.

[Sch23] Felix Schremmer. Affine bruhat order and demazure products. 2023.

[Wel22] Amanda Welch. Classification of cocovers in the double affine Bruhat order. *Electron. J. Comb.*, 29(4):research paper p4.7, 19, 2022.