Let $x_1, x_2\in \widetilde W$. Then each of the following three sets contains a unique maximum (with respect to the Bruhat order), and the maxima agree:

\begin{align*}
\{x_1x_2'\mid x_2'\leq x_2\},\quad \{x_1' x_2\mid x_1'\leq x_1\},\quad \{x_1' x_2'\mid x_1'\leq x_1,~x_2'\leq x_2\}.
\end{align*}

The common maximum is denoted $x_1\ast x_2$. If we write $x_1 \ast x_2 = x_1x_2' = x_1' x_2$, then

\begin{align*}
\ell(x_1\ast x_2) = \ell(x_1) + \ell(x_2') = \ell(x_1') + \ell(x_2).&\rightqed
\end{align*}

Demazure products have recently been studied in the context of affine Deligne-Lusztig varieties <cit.>. While the Demazure product is a somewhat simple Coxeter-theoretic notion, it is connected to the question of generic Newton points of elements in $\widetilde W$. He <cit.> shows how to compute generic Newton points in terms of iterated Demazure products, a method that we will review in Section <ref>. Conversely, He and Nie <cit.> use the Milićević's formula for generic Newton points <cit.> to show new properties of the Demazure product.

In this chapter, we prove a new description of Demazure products in $\widetilde W$, generalizing the aforementioned results of <cit.>. As applications, we obtain new results on the quantum Bruhat graph that shed some light on our previous results on the Bruhat order. Moreover, we give a new description of generic Newton points.

§.§ Computation of Demazure products

If one plays a bit with our Theorem <ref> or <cit.>, one will soon get an idea of how Demazure products should roughly look like. We capture the occurring formulas as follows.

Let $x_1 = w_1 \varepsilon^\mu_1, x_2 = w_2 \varepsilon^\mu_2\in \widetilde W$. Let $v_1, v_2\in W$ and define

\begin{align*}
x_1':=& w_1'\varepsilon^{\mu_1'} := (w_1 v_1) (w_2 v_2)^{-1}\varepsilon^{w_2 v_2 v_1^{-1}\mu_1 - w_2 v_2 \wt(v_1\Rightarrow w_2 v_2)},\\
x_2':=& w_2'\varepsilon^{\mu_2'} := v_1 v_2^{-1}\varepsilon^{\mu_2 - v_2 \wt(v_1\Rightarrow w_2 v_2)},\\
x_\ast :=& w_\ast \varepsilon^{\mu_\ast} := w_1 v_1 v_2^{-1}\varepsilon^{v_2 v_1^{-1}\mu_1 + \mu_2 - v_2\wt(v_1\Rightarrow w_2 v_2)} = x_1' x_2 = x_1 x_2'.
\end{align*}

In this situation, we want to compute the Demazure product $x_1\ast x_2$, knowing that $x_1\ast x_2$ can be written as $\tilde x_1 x_2 = x_1\tilde x_2$ for some $\tilde x_1\leq x_1$ and $\tilde x_2\leq x_2$. If $x_1$ is in a shrunken Weyl chamber with $\LP(x_1) = v_1$, and $x_2$ is shrunken with $\LP(x_2) = \{v_2\}$, then $x_\ast = x_1\ast x_2$ by <cit.>, so $\tilde x_1 = x_1'$ and $\tilde x_2 = x_2'$.

In the general case, our goal is to find conditions on $v_1, v_2\in W$ to ensure that $x_\ast = x_1\ast x_2$.

Before examining this situation further, it will be very convenient for our proofs to see that the property

\begin{align*}
(x_1\ast x_2)^{-1} = x_2^{-1}\ast x_1^{-1}
\end{align*}

is reflected by our construction in Situation <ref>.

Suppose we are in Situation <ref>. Let us write $y_1 := x_2^{-1}$ and $y_2 := x_1^{-1}$. Define $v_1' :=w_2  v_2 w_0$ resp. $v_2' :=w_1  v_1 w_0$.

Construct $y_1', y_2', y_\ast$ associated with $(y_1, y_2, v_1', v_2')$ as in Situation <ref>. Then

\begin{align*}
y_1' = (x_2')^{-1},\quad y_2' = (x_1')^{-1},\quad y_\ast = x_\ast^{-1}.
\end{align*}

* $v_1 \in \LP(x_1)$ iff $v_2' \in \LP(y_1)$.

* $v_2 \in \LP(x_2)$ iff $v_1' \in \LP(y_2)$.

* $d_{\QB(W)}(v_1\Rightarrow w_2 v_2) = d_{\QB(W)}(v_1'\Rightarrow w_1^{-1} v_2')$ and
$\wt(v_1\Rightarrow w_2 v_2) = -w_0\wt(v_1'\Rightarrow w_1^{-1} v_2)$.

\begin{align*}
y_1 = w_2^{-1}\varepsilon^{-w_2 \mu_2},\quad y_2 = w_1^{-1}\varepsilon^{-w_1 \mu_1}
\end{align*}

and compute

\begin{align*}
y_2' =& (w_2v_2  w_0)( w_1v_1 w_0)^{-1}\varepsilon^{-w_1 \mu_1 - w_1v_1  w_0\wt(w_2 v_2 w_0\Rightarrow (w_1)^{-1} w_1 v_1 w_0)}
\\=&(w_2 v_2) (w_1 v_1)^{-1} \varepsilon^{-w_1 \mu_1 + w_1 v_1\wt(v_1\Rightarrow w_2 v_2)} = (x_1')^{-1}.
\end{align*}

A similar computation, or a repetition of this argument for $x_1 = (y_2)^{-1}, x_2 = (y_1)^{-1}$, shows that $y_1' = (x_2')^{-1}$. Then the conclusion $y_\ast = x_\ast^{-1}$ is immediate.

For the Moreover statements, recall that

\begin{align*}
\LP(y_1) = \LP(x_2^{-1})\underset{\text{\crossRef{Lemma~}{lem:lengthFunctionalForProducts}}}=w_2 \LP(x_2) w_0.
\end{align*}

The same holds for $y_2 = x_1^{-1}$. The final statement is due to the fact that $v_1' = w_2 v_2 w_0$ and $w_1^{-1} v_2' = v_1 w_0$ using the duality anti-automorphism of the quantum Bruhat graph, cf. Lemma <ref>.

The first step towards proving $x_1\ast x_2 = x_\ast$ is the following estimate:

Let $x_1, x_2\in \widetilde W$ and $v_2\in \LP(x_1\ast x_2)$. There exists $v_1\in \LP(x_1)$ such that

\begin{align*}
\ell(x_1 \ast x_2) \leq \ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2 v_2).
\end{align*}

Write $x_1\ast x_2 = y x_2$ for some element $y = w' \varepsilon^{\mu'}\leq x_1$. Observe that $\ell(y x_2) = \ell(y) + \ell(x_2)$, so that $v_2$ must be length positive for $x_2$ and $w_2 v_2$ must be length positive for $y$.

Since $y\leq x_1$, using Lemma <ref>, we find a length positive element $v_1$ for $x_1$ such that

\begin{align*}
(w_2 v_2)^{-1}\mu' + \wt(v_1\Rightarrow w_2 v_2) + \wt(w'w_2 v_2\Rightarrow w_1 v_1)\leq (v_1)^{-1}\mu_1.
\end{align*}

Pairing with $2\rho$ and using Lemma <ref>, we compute

\begin{align*}
&\langle 2\rho, (w_2 v_2)^{-1}\mu'\rangle + \ell(v_1) - \ell(w_2 v_2) \\&\qquad+ d(v_1\Rightarrow w_2 v_2) + \ell(w'w_2 v_2) - \ell(w_1 v_1) + d(w' w_2 v_2\Rightarrow w_1 v_1)\\&\leq \langle 2\rho,(v_1)^{-1}\mu_1\rangle.
\end{align*}

Using the length positivity of $w_2 v_2$ for $y$ and $v_1$ for $x_1$ (), we conclude

\begin{align*}
\ell(y) + d(v_1\Rightarrow w_2 v_2) + d(w'w_2 v_2\Rightarrow w_1v_1)\leq \ell(x_2).
\end{align*}

\begin{align*}
\ell(x_1\ast x_2) = \ell(y) + \ell(x_2) \leq \ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2v_2) - d(w' w_2 v_2\Rightarrow w_1 v_1).
\end{align*}

We obtain the desired conclusion.

We now study the Situation <ref> further.

Consider Situation <ref>, and assume that $v_1\in \LP(x_1)$. Then we always have the estimate

\begin{align*}
\ell(x_1') \geq \ell(x_1) - d_{\QB(W)}(v_1\Rightarrow w_2 v_2).
\end{align*}

The following are equivalent:

* Equality holds above:

\begin{align*}
\ell(x_1') = \ell(x_1) - d_{\QB(W)}(v_1\Rightarrow w_2 v_2).
\end{align*}

* $w_2 v_2$ is length positive for $x_1'$.

* For any positive root $\alpha$, we have

\begin{align*}
\ell(x_1,v_1\alpha) - \langle \wt(v_1\Rightarrow w_2 v_2),\alpha\rangle+ \Phi^+(w_2 v_2\alpha) - \Phi^+(v_1\alpha) \geq 0
\end{align*}

In that case, $x_1'\leq x_1$, so that $x_\ast \leq x_1\ast x_2$.

Consider the calculation

\begin{align*}
\ell(x_1') \underset{\text{\positiveLengthFormulaShort}}\geq& \left\langle (w_2 v_2)^{-1}\left(w_2 v_2 v_1^{-1}\mu_1 - w_2 v_2 \wt(v_1\Rightarrow w_2 v_2)\right),2\rho\right\rangle - \ell(w_2 v_2) + \ell(w_1 v_1)
\\\underset{\text{L\ref{lem:weight2rho}}}=&\langle v_1^{-1}\mu,2\rho\rangle - \ell(v_1) + \ell(w_2 v_2) - d(v_1\Rightarrow w_2 v_2)- \ell(w_2 v_2) + \ell(w_1 v_1)
\\\underset{\text{\positiveLengthFormulaShort}}=&\ell(x_1) - d(v_1\Rightarrow w_2 v_2).
\end{align*}

This shows the estimate and (i) $\iff$ (ii). In order to show (ii) $\iff$ (iii), we compute

\begin{align*}
\ell(x_1', w_2 v_2\alpha)=&\langle w_2 v_2\alpha, w_2 v_2 v_1^{-1}\mu_1 - w_2 v_2\wt(v_1\Rightarrow w_2 v_2),\alpha\rangle + \Phi^+(w_2 v_2\alpha) - \wt(w_1 v_1\alpha)
\\=&\ell(x_1,v_1\alpha) - \Phi^+(v_1\alpha) - \langle \wt(v_1\Rightarrow w_2 v_2),\alpha\rangle + \Phi^+(w_2 v_2\alpha).
\end{align*}

Finally, assume that (i) – (iii) are satisfied. We have to show $x_1'\leq x_1$. For this, we calculate

\begin{align*}
&(w_2 v_2)^{-1}\left(w_2 v_2 v_1^{-1}\mu_1 - w_2 v_2 \wt(v_1\Rightarrow w_2 v_2)\right) + \wt(v_1\Rightarrow w_2 v_2) \\&\qquad+ \wt(w_1 v_1\Rightarrow w_1 v_1) \\=&~ v_1^{-1}\mu_1.
\end{align*}

Since we assumed $w_2 v_2\in \LP(x_1')$, we conclude $x_1'\leq x_1$ by Theorem <ref>. Now by definition of the Demazure product, we get $x_\ast = x_1' x_2\leq x_1\ast x_2$.

By the duality presented in Lemma <ref>, we obtain the following:

Consider Situation <ref>, and assume that $v_2\in \LP(x_2)$. Then we always have the estimate

\begin{align*}
\ell(x_2') \geq \ell(x_2) - d_{\QB(W)}(v_1\Rightarrow w_2 v_2).
\end{align*}

The following are equivalent:

* Equality holds above:

\begin{align*}
\ell(x_2') = \ell(x_2) - d_{\QB(W)}(v_1\Rightarrow w_2 v_2).
\end{align*}

* $v_2$ is length positive for $x_2'$.

* For any positive root $\alpha$, we have

\begin{align*}
\ell(x_2, v_2\alpha) - \langle\wt(v_1\Rightarrow w_2 v_2),\alpha\rangle + \Phi^+(w_2 v_2\alpha) - \Phi^+(v_1\alpha)\geq 0.
\end{align*}

In that case, $x_2'\leq x_2$, so that $x_\ast \leq x_1\ast x_2$.

Under Lemma <ref>, this is precisely Lemma <ref>.

Suppose we are given Situation <ref>, and that $v_1\in \LP(x_1)$ and $v_2\in \LP(x_2)$. We have the estimate

\begin{align*}
\ell(x_\ast) \geq \ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2 v_2).
\end{align*}

Equality holds if and only if $v_2\in \LP(x_\ast)$.

Using again and Lemma <ref>, we calculate

\begin{align*}
\ell(x_\ast)\geq&\left\langle v_2^{-1}\left( v_2 v_1^{-1}\mu_1 + \mu_2 - v_2\wt(v_1\Rightarrow w_2 v_2)\right),2\rho\right\rangle - \ell(v_2) + \ell(w_1 v_1)
\\=&\langle v_1^{-1}\mu_1,2\rho\rangle + \langle v_2^{-1}\mu_2,2\rho\rangle - d(v_1\Rightarrow w_2 v_2) - \ell(v_1) + \ell(w_2 v_2) + \ell(v_2) + \ell(w_1 v_1)
\\=&\ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2 v_2)
\end{align*}

Both claims follow from this calculation.

Let $x = w\varepsilon^\mu\in \widetilde W$ and $u\in W$. Among all $v\in \LP(x)$, there is a unique one such that $d(v\Rightarrow u)$ becomes minimal. For this particular $v$, we have

\begin{align*}
\forall \alpha\in \Phi^+:~
\ell(x,v\alpha) - \langle \wt(v\Rightarrow u),\alpha\rangle+ \Phi^+(u\alpha) - \Phi^+(v\alpha) \geq 0.
\end{align*}

Let $x_2 = t^{u\lambda}$ with $\lambda\in \Xast$ superregular and dominant.
Let $v = v_1\in \LP(x)$ such that $d(v\Rightarrow u)$ becomes minimal. Set $v_2 = u$.

Consider Situation <ref> for $x_1 = x$ and $x_2$ as above. Now the condition (iii) of Lemma <ref> is satisfied by superregularity of $\lambda$. We conclude that $x_2'\leq x_2$, so that $x_\ast \leq x\ast x_2$.

Combining Lemma <ref> with Lemma <ref> shows

\begin{align*}
\ell(x) + \ell(x_2) - d(v\Rightarrow u) \geq \ell(x_1\ast x_2) \geq \ell(x_\ast)\geq
\ell(x) + \ell(x_2) - d(v\Rightarrow u).
\end{align*}

In particular, we get $x_1\ast x_2 = x_\ast$.

The above argument works whenever $v\in \LP(x)$ is chosen such that $d(v\Rightarrow u)$ becomes minimal. Since the value of $x_1\ast x_2$ does not depend on the choice of such an element $v$, nor does $x_\ast = x_1\ast x_2$. In particular, the classical part $\mathrm{cl}(x_\ast) = wvu^{-1}$ does not depend on $v$, hence $v$ is uniquely determined.

The formula $x_\ast = x_1\ast x_2 = x_1' x_2$ implies that $\ell(x_\ast) = \ell(x_1') + \ell(x_2)$. Using the previously computed length of $x_\ast$, we conclude $\ell(x_1') = \ell(x_1) - d(v\Rightarrow u)$. Now the estimate follows from Lemma <ref>.

Considering Lemma <ref> for the inverse $x^{-1}$, we obtain the following:

Let $x = w\varepsilon^\mu\in \widetilde W$ and $u\in W$. Among all $v\in \LP(x)$, there is a unique one such that $d(u\Rightarrow wv)$ becomes minimal. For this particular $v$, we have

\begin{align*}
&\forall \alpha\in \Phi^+:~
\ell(x,v\alpha) - \langle\wt(u\Rightarrow wv),\alpha\rangle- \Phi^+(u\alpha) + \Phi^+(wv\alpha) \geq 0.\rightqed
\end{align*}

Let $x \in \widetilde W$ and $u\in W$. The uniquely determined $v\in \LP(x)$ such that $d(v\Rightarrow u)$ is minimal will be denoted by $v = \rho^\vee_x(u)$. The uniquely determined $v\in \LP(x)$ such that $d(u\Rightarrow wv)$ is minimal will be denoted by $v = \rho_x(u) = w^{-1}\rho^\vee_{x^{-1}}(uw_0)w_0$.

The functions $\rho_x$ and $\rho_x^\vee$ will be studied in Section <ref>. For now, we state our announced description of Demazure products in $\widetilde W$.

Let $x_1 = w_1\varepsilon^{\mu_1}, x_2 = w_2\varepsilon^{\mu_2}\in \widetilde W$. Among all pairs $(v_1, v_2)\in \LP(x_1)\times \LP(x_2)$, pick one such that the distance $d(v_1\Rightarrow w_2 v_2)$ becomes minimal.

Construct $x_\ast$ as in Situation <ref>. Then

\begin{align*}
&x_1\ast x_2 = x_\ast = w_1 v_1 \varepsilon^{v_1^{-1}\mu_1 + v_2^{-1}\mu_2 - \wt(v_1\Rightarrow w_2 v_2)}v_2^{-1},\\& \ell(x_1\ast x_2) = \ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2 v_2),\\&v_2\in \LP(x_1\ast x_2).
\end{align*}

We have $x_\ast \leq x_1\ast x_2$ by Lemmas <ref> and <ref>. By Lemma <ref>, we find $(v_1', v_2')\in \LP(x_1)\times \LP(x_2)$ such that

\begin{align*}
\ell(x_1) + \ell(x_2) - d(v_1'\Rightarrow w_2 v_2') \geq \ell(x_1\ast x_2) \geq \ell(x_\ast)\geq
\ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2 v_2).
\end{align*}

By choice of $(v_1, v_2)$, the result follows.

We note the following consequences of Theorem <ref>.

Let $x_1 = w_1 \varepsilon^{\mu_1}, x_2 = w_2 \varepsilon^{\mu_2}\in \widetilde W$. Write

\begin{align*}
M = M(x_1, x_2) := \{&(v_1, v_2)\in \LP(x_1)\times \LP(x_2)\mid \\&\forall (v_1', v_2')\in \LP(x_1)\times \LP(x_2):~d(v_1\Rightarrow w_2 v_2)\leq d(v_1'\Rightarrow w_2 v_2')\}
\end{align*}

for the set of all pairs $(v_1, v_2)$ such that the theorem's condition is satisfied.

* The following two functions on $M$ are both constant:

\begin{align*}
&\varphi_1 : M\rightarrow W,\quad (v_1, v_2)\mapsto v_1 v_2^{-1},
\\&\varphi_2 : M\rightarrow \mathbb Z\Phi^\vee,\quad (v_1, v_2)\mapsto v_2 \wt(v_1\Rightarrow w_2 v_2).
\end{align*}

* The following is a well-defined bijective map:

\begin{align*}
M\rightarrow \LP(x_1\ast x_2),\quad (v_1, v_2)\mapsto v_2.
\end{align*}

* From the Theorem, we get that the function

\begin{align*}
M\rightarrow\widetilde W,\quad (v_1, v_2)\mapsto &w_1 v_1 v_2^{-1}\varepsilon^{v_2 v_1^{-1}\mu_1+\mu_2 - v_2\wt(v_1\Rightarrow w_2 v_2)}
\\=&w_1 \varphi_1(v_1, v_2) \varepsilon^{\varphi_1(v_1, v_2)^{-1}\mu_1 + \mu_2 - \varphi_2(v_1, v_2)}
\end{align*}

is constant with image $\{x_1\ast x_2\}$. This proves that $\varphi_1$ and $\varphi_2$ are constant.

* Injectivity follows from (a). Well-definedness follows from the theorem. For surjectivity, let $v_2\in \LP(x_1\ast x_2)$. Then certainly $v_2\in \LP(x_2)$. By Lemma <ref>, we find $v_1\in W$ such that $\ell(x_1\ast x_2)\leq \ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2 v_2)$. By the theorem, we find $(v_1', v_2')\in M$ with $\ell(x_1\ast x_2) = \ell(x_1) + \ell(x_2) - d(v_1'\Rightarrow w_2 v_2')$, such that $d(v_1\Rightarrow w_2 v_2) \leq d(v_1'\Rightarrow w_2 v_2')$. It follows that $(v_1, v_2)\in M$, finishing the proof of surjectivity.

In case $\ell(x_1x_2) =\ell(x_1) + \ell(x_2)$, we get $x_1x_2= x_1\ast x_2$. In this case, we recover Lemma lem:lengthAdditivity.

§.§ Generic action

Studying the Demazure product where one of the factors is superregular induces actions of $(\widetilde W, \ast)$ on $W$, that we denoted by $\rho_x$ resp. $\rho^\vee_x$ in Definition <ref>. In this section, we study these actions and the consequences for the quantum Bruhat graph.

Let $x_1 = w_1\varepsilon^{\mu_1}, x_2 = w_2\varepsilon^{\mu_2}\in \widetilde W$. Then

\begin{align*}
\rho_{x_1\ast x_2} = \rho_{x_2}\circ \rho_{x_1}.
\end{align*}

Note that if $z\in \widetilde W$ is in a shrunken Weyl chamber with $\LP(z) = \{u\}$ and $x\in \widetilde W$, then by Proposition <ref>,

\begin{align*}
\LP(z\ast x) = \{\rho_x(u)\}.
\end{align*}

Hence we have

\begin{align*}
\{\rho_{x_2}(\rho_{x_1}(u))\} = \LP\left((z\ast x_1)\ast x_2\right) = \LP\left(z\ast(x_1\ast x_2)\right) = \{\rho_{x_1\ast x_2}(u)\}.
\end{align*}

This shows the desired claim.

* There is a dual, albeit more complicated statement for the dual generic action $\rho^\vee$.

* If $x = \omega r_{a_1}\cdots r_{a_n}$ is a reduced decomposition with simple affine roots $a_1,\dotsc,a_n\in \Delta_\af$ and $\omega\in \Omega$ of length zero, then

\begin{align*}
\rho_x &= \rho_{\omega\ast r_{a_1}\ast\cdots\ast r_{a_n}} = \rho_{r_{a_n}}\circ\cdots\circ \rho_{r_{a_1}}\circ \rho_\omega.
\end{align*}

The map $\rho_\omega$ is simply given by $\rho_\omega(v) = \cl(\omega)v$, as $\LP(\omega) = W$.
We now describe the $\rho_{r_{a_i}}$ as follows:

For a simple affine root $(\alpha,k)\in \Delta_\af$, we have

\begin{align*}
\ell(r_{(\alpha,k)},\beta) = \begin{cases}1,&\beta =\alpha,\\
\end{align*}

\begin{align*}
\LP(r_{(\alpha,k)}) = \{v\in W\mid v^{-1}\alpha\in \Phi^+\}.
\end{align*}

Let $v\in W$. If $v^{-1}\alpha\in \Phi^-$, then $s_\alpha v\in \LP(r_{(\alpha,k)})$ with $d(v\Rightarrow s_\alpha(s_\alpha v))=0$. Hence $\rho_{r_{(\alpha,k)}}(v) = s_\alpha v$.

If $v^{-1}\alpha\in \Phi^+$, then $v\in \LP(r_{(\alpha,k)})$ with $d(v\Rightarrow s_\alpha v)=1$ by Lemma <ref>. Since there exists no $u\in \LP(r_{(\alpha,k)})$ with $d(v\Rightarrow s_\alpha u)=0$, a distance of $1$ is already minimal. We see that $\rho_{r_{(\alpha,k)}}(v) = v$. Summarizing:

\begin{align*}
\rho_{r_{(\alpha,k)}}(v) = \begin{cases}v,&v^{-1}\alpha\in \Phi^+,\\
s_\alpha v,&v^{-1}\alpha\in \Phi^-.\end{cases}
\end{align*}

This gives an alternative method to compute $\rho_x$. One easily obtains a dual method to compute $\rho_x^\vee$ in a similar fashion.

Let $x \in \widetilde W$ and $v, v'\in \LP(x)$ be two length positive elements. There exists a shortest path $p$ from $v$ to $v'$ in the quantum Bruhat graph such that each vertex in $p$ lies in $\LP(x)$.

Let us first study the case $v' = 1$.

We do induction on $\ell(v)$. If $\ell(v) = 0$, the statement is clear.

Otherwise, there exists a quantum edge $v\rightarrow vs_\alpha$ for some quantum root $\alpha\in \Phi^+$ such that $d(v\Rightarrow v') = d(vs_\alpha\Rightarrow v')+1$ (Lemma <ref>). In this case, it suffices to show that $vs_\alpha\in \LP(x)$.

The quantum edge condition means that $\ell(vs_\alpha) = \ell(v) - \ell(s_\alpha)$. In other words, every positive root $\beta\in \Phi^+$ with $s_\alpha(\beta)\in \Phi^-$ satisfies $v(\beta)\in \Phi^-$.

Let $\beta\in \Phi^+$, we want to show that $\ell(x,vs_\alpha(\beta))\geq 0$. This follows from length positivity of $v$ if $s_\alpha(\beta)\in \Phi^+$. So let us assume that $s_\alpha(\beta)\in \Phi^-$. Then $vs_\alpha(\beta)\in \Phi^+$, applying the above observation to $-s_\alpha(\beta)$. Hence $\ell(x,vs_\alpha(\beta))\geq 0$, as $1\in \LP(x)$. This finishes the induction, so the claim is established whenever $v'=1$.

For the general case, we do induction on $\ell(v')$. If $v'=1$, we have proved the claim, so let us assume that $\ell(v')>0$. Then we find a simple root $\alpha\in \Delta$ with $s_\alpha v' < v'$. In particular, $(v')^{-1}\alpha\in \Phi^-$ so that $\ell(x,\alpha)\leq 0$. Consider the element $x' := xs_\alpha \gtrdot x$. We observe that for any $u\in W$ and $\beta\in \Phi$,

\begin{align*}
\ell(x',s_\alpha u\beta) = \ell(x,u\beta) + \ell(s_\alpha, -u\beta) = \begin{cases}\ell(x,u\beta),&u\beta \neq \pm \alpha,\\
-\ell(x,\alpha)+1>0,&u\beta = -\alpha,\\
\ell(x,\alpha)-1<0,&u\beta = \alpha.\end{cases}
\end{align*}

It follows that

\begin{align*}
\LP(x') = \{s_\alpha u \mid u\in \LP(x)\text{ and }u^{-1}\alpha \in \Phi^-\}.
\end{align*}

In particular, $s_\alpha v'\in \LP(x')$. Now suppose that $v^{-1}\alpha \in \Phi^-$. Then also $s_\alpha v\in \LP(x')$. We may apply the inductive assumption to get a path $p'$ from $s_\alpha v$ to $s_\alpha v'$ in $\LP(x')$. Multiplying each vertex by $s_\alpha$ on the left, we obtain the desired path $p$ in $\LP(x)$.

Finally assume that $v^{-1}\alpha \in \Phi^+$. Then $s_\alpha v\in \LP(x)$ by Corollary <ref>.

By Lemma <ref>, $v\rightarrow s_\alpha v$ is an edge in $\QB(W)$ and
\begin{align*}
d_{\QB(W)}(v\Rightarrow v') = d_{\QB(W)}(v\Rightarrow s_\alpha  v') = d_{\QB(W)}(s_\alpha v\Rightarrow v')+1.
\end{align*}
We get a path from $s_\alpha v$ to $v'$ in $\LP(x)$ by repeating the above argument, then concatenate it with $v\rightarrow s_\alpha v$.

This finishes the induction and the proof.

Let $x = w\varepsilon^\mu\in \widetilde W$ and $v, v'\in \LP(x)$. Then

\begin{align*}
v^{-1}\mu - (v')^{-1}\mu - \wt(v\Rightarrow v') + \wt(wv\Rightarrow wv')=0.
\end{align*}

In particular, $d(v\Rightarrow v') = d(wv\Rightarrow wv')$.

\begin{align*}
p : v = v_1\xrightarrow{\alpha_1} v_2\xrightarrow{\alpha_2} \cdots\xrightarrow{\alpha_{n-1}} v_n = v'
\end{align*}

be a path in $\LP(x)$ of weight $\wt(v\Rightarrow v')$. Now for $i=1,\dotsc,n-1$, observe that both $v_i$ and $v_i s_{\alpha_i}$ are in $\LP(x)$. Thus $\ell(x, v_i \alpha_i)=0$. We conclude that

\begin{align*}
&(v_i)^{-1}\mu - (v_{i+1})^{-1}\mu - \wt(v_i \Rightarrow v_{i+1}) + \wt(wv_i\Rightarrow wv_{i+1})
\\=&\langle v_i \alpha_i, \mu\rangle \alpha_i^\vee - \Phi^+(-v_i \alpha_i)\alpha_i^\vee + \wt(wv_i\Rightarrow wv_i s_{\alpha_i})
\\\leq& \langle v_i\alpha_i, \mu\rangle\alpha_i^\vee -  \Phi^+(-v_i \alpha_i)\alpha_i^\vee + \Phi^+(w v_i\alpha_i)\alpha_i^\vee
\\=&\ell(x,v_i\alpha_i)\alpha_i^\vee = 0.
\end{align*}

Summing these estimates for $i=1,\dotsc,n-1$, we conclude

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

Considering the same argument for $x^{-1}, wvw_0, wv'w_0$, we get the other inequality.

The in particular part follows from inspecting the argument given. Alternatively, pair the identity just proved with $2\rho$, then apply Lemma <ref> and .

The corollary can be shown directly by evaluating the Demazure product

\begin{align*}
\varepsilon^{w v' 2\rho} \ast x \ast \varepsilon^{v 2\rho}
\end{align*}

in two different ways, using the associativity property of Demazure products.

Let $x = w\varepsilon^\mu\in \widetilde W$, $v\in \LP(x)$ and $u\in W$. Then

\begin{align*}
d(u\Rightarrow wv) = d(u\Rightarrow w\rho_x(u)) + d(w\rho_x(u)\Rightarrow wv).
\end{align*}

Let $\lambda$ be superregular and $y := \varepsilon^{u\lambda}$. Define the element

\begin{align*}
z := y \ast x = u \rho_x(u)^{-1} \varepsilon^{\rho_x(u) \lambda + \mu - \rho_x(u) \wt(u\Rightarrow w\rho_x(u))}.
\end{align*}

Then $z$ is superregular with $\LP(z) = \{\rho_x(u)\}$. Consider the element

\begin{align*}
\tilde y' := u(wv)^{-1}\varepsilon^{wv\lambda - wv \wt(u\Rightarrow wv)}.
\end{align*}

This is superregular with $\LP(\tilde y') = \{wv\}$. Note that Theorem <ref> implies $\tilde y'\leq y$, as

\begin{align*}
(wv)^{-1}(wv \lambda - wv\wt(u\Rightarrow wv)) + \wt(u\Rightarrow wv) + \wt(u\Rightarrow u) = \lambda.
\end{align*}

Thus $\tilde z\leq z$, where

\begin{align*}
\tilde z = \tilde y x = u v^{-1} \varepsilon^{v \lambda + \mu - v \wt(u\Rightarrow wv)}.
\end{align*}

Note that $\tilde z$ is superregular with $\LP(\tilde z) = \{v\}$. In light of Theorem <ref>, the inequality $\tilde z\leq z$ means

\begin{align*}
v^{-1}(v \lambda + \mu - v \wt(u\Rightarrow wv)) +& \wt(\rho_x(u)\Rightarrow v) + \wt(u\Rightarrow u)\\&\leq \rho_x(u)^{-1}(\rho_x(u) \lambda + \mu - \rho_x(u) \wt(u\Rightarrow w\rho_x(u))).
\end{align*}

Rewriting this, we get

\begin{align*}
v^{-1}\mu - \wt(u\Rightarrow wv) + \wt(\rho_x(u)\Rightarrow v) \leq \rho_x(u)^{-1}\mu - \wt(u\Rightarrow w\rho_x(u)).
\end{align*}

Corollary <ref> yields the equation

\begin{align*}
v^{-1}\mu -\rho_x(u)^{-1}\mu +\wt(\rho_x(u)\Rightarrow v) = \wt(w\rho_x(u) \Rightarrow wv).
\end{align*}

We conclude

\begin{align*}
\wt(u\Rightarrow wv) \geq \wt(u\Rightarrow w\rho_x(u)) + \wt(w\rho_x(u)\Rightarrow wv).
\end{align*}

This implies the desired claim.

By the duality from Lemma <ref>, we obtain the following.

Let $x = w\varepsilon^\mu\in \widetilde W$, $v\in \LP(x)$ and $u\in W$. Then

\begin{align*}
&d(v\Rightarrow u) = d(v\Rightarrow \rho^\vee_x(u)) + d(\rho^\vee_x(u)\Rightarrow u).\rightqed
\end{align*}

In the language of <cit.>, this means that the set $w\LP(x)$ contains a unique minimal element with respect to the tilted Bruhat order $\preceq_u$. Since $w \LP(x) = \LP(x^{-1})w_0$, it follows that the set $\LP(x)$ contains a unique maximal element with respect to $\preceq_u$. If $x = \varepsilon^\mu$ is a pure translation element, this recovers <cit.>.

The converse statements are generally false, i.e. $\LP(x)$ will in general not contain tilted Bruhat minima, and $w\LP(x)$ will not contain maxima. For a concrete example, choose $x$ to be a simple affine reflection of type $A_2$.

The set $\LP(x)$ satisfies a number of interesting structural properties with respect to the quantum Bruhat graph, namely containing shortest paths for any pair of elements (Lemma <ref>) and the existence of tilted Bruhat maxima. One may ask the question which subsets of $W$ occur as the set $\LP(x)$ for some $x\in \widetilde W$.

Let $x =w\varepsilon^\mu \in \widetilde W$ and $u_1, u_2\in W$. Then the function

\begin{align*}
\varphi: W\rightarrow \Xast,~ v\mapsto v^{-1}\mu - \wt(u_1\Rightarrow wv) - \wt(v\Rightarrow u_2)
\end{align*}

has a global maximum at $\rho_x(u_1)$, and another global maximum at $\rho^\vee_x(u_2)$.

If $v\in W$ is not length positive for $x$, and $vs_\alpha$ is an adjustment, it is easy to see that $\varphi(v)\leq\varphi(vs_\alpha)$. So we may focus on $\varphi\vert_{\LP(x)}$.

Let $v\in \LP(x)$ and $v' = \rho_x(u_1)$, so that

\begin{align*}
\varphi(v) =~\,& v^{-1}\mu - \wt(u_1\Rightarrow wv) - \wt(v\Rightarrow u_2) \\=~\,& v^{-1}\mu - \wt(u_1\Rightarrow w v') - \wt(wv'\Rightarrow wv) - \wt(v\Rightarrow u_2)
\\\underset{\text{C\ref{cor:lengthPositiveComparison}}}=&(v')^{-1}\mu - \wt(v'\Rightarrow v) - \wt(u_1\Rightarrow w v') - \wt(v\Rightarrow u_2)
\\\leq~\,&(v')^{-1}\mu - \wt(u_1\Rightarrow wv') - \wt(v'\Rightarrow u_2) = \varphi(v').
\end{align*}

This shows the first maximality claim. The second one follows from the duality of Lemma <ref>.

Let $x_1 = w_1 \varepsilon^{\mu_1}, x_2 = w_2 \varepsilon^{\mu_2}\in \widetilde W$ and $v_1\in \LP(x_1)$. Theorem <ref> states that $x_1\leq x_2$ in the Bruhat order if and only if there is some $v_2\in W$ with

\begin{align*}
v_1^{-1}\mu_1 + \wt(v_2\Rightarrow v_1) + \wt(w_1v_1\Rightarrow w_2v_2) \leq v_2^{-1}\mu_2.
\end{align*}

By the above corollary, it is equivalent to require this inequality for $v_2 = \rho_{x_2}(w_1 v_1)$. One can alternatively require it for $v_2 = \rho_{x_2}^\vee(v_1)$.

Let $x_1 = w_1\varepsilon^{\mu_1}, x_2 = w_2 \varepsilon^{\mu_2}\in \widetilde W$ and $v_1 \in \LP(x_1), v_2\in \LP(x_2)$. The following are equivalent:

* The distance $d(v_1\Rightarrow w_2 v_2)$ is minimal for all pairs in $\LP(x_1)\times \LP(x_2)$, i.e. $(v_1, v_2)\in M(x_1, x_2)$.

* $v_1 = \rho^\vee_{x_1}(w_2 v_2)$ and $v_2 = \rho_{x_2}(v_1)$.

(i) $\Rightarrow$ (ii): Certainly, $v_1$ minimizes the function $d(\cdot\Rightarrow w_2 v_2)$ on $\LP(x_1)$, showing the first claim. The second claim is analogous.

(ii) $\Rightarrow$ (i): Consider Situation <ref>. By Lemmas <ref> and <ref>, we conclude that $w_2 v_2$ must be length positive for $x_1'$. It follows that $x_\ast \leq x_1\ast x_2$ and
\begin{align*}
\ell(x_\ast) = \ell(x_1') + \ell(x_2) = \ell(x_1) + \ell(x_2) - d(v_1\Rightarrow w_2 v_2).
\end{align*}
By Lemma <ref>, $v_2$ is length positive for $x_\ast$.
Write $x_1\ast x_2$ as $\tilde w \varepsilon^{\tilde \mu}$.
Using Lemma <ref> with Lemma <ref>, the condition $x_\ast \leq x_1\ast x_2$ yields some $v_2'\in \LP(x_1 \ast x_2)$ with

\begin{align*}
v_1^{-1}\mu_1 + v_2^{-1}\mu_2 - \wt(v_1\Rightarrow w_2 v_2) + \wt(v_2'\Rightarrow v_2) + \wt(w_1 v_1\Rightarrow \tilde w v_2')\leq (v_2')^{-1}\tilde\mu.
\end{align*}

By Proposition <ref>, we find $v_1'$ such that $(v_1', v_2')\in M(x_1, x_2)$. By Theorem <ref>, we can express $x_1\ast x_2$ in terms of $(v_1', v_2')$. Then the above inequality becomes

\begin{align*}
&v_1^{-1}\mu_1 + v_2^{-1}\mu_2 - \wt(v_1\Rightarrow w_2 v_2) + \wt(v_2'\Rightarrow v_2) + \wt(w_1 v_1\Rightarrow w_1 v_1')\\&\leq (v_1')^{-1}\mu_1 + (v_2')^{-1}\mu_2 - \wt(v_1'\Rightarrow w_2 v_2').
\end{align*}

Since $v_1, v_1'\in \LP(x_1)$ and $v_2, v_2'\in \LP(x_2)$, we can apply Corollary <ref> twice to obtain

\begin{align*}
\wt(v_1\Rightarrow v_1') + \wt(w_2 v_2'\Rightarrow w_2 v_2) - \wt(v_1\Rightarrow w_2 v_2)\leq -\wt(v_1'\Rightarrow w_2 v_2').
\end{align*}

Rewriting, we get

\begin{align*}
\wt(v_1\Rightarrow v_1') +\wt(v_1'\Rightarrow w_2 v_2') + \wt(w_2 v_2'\Rightarrow w_2 v_2) \leq \wt(v_1\Rightarrow w_2 v_2).
\end{align*}

In other words, there is a shortest path from $v_1$ to $w_2 v_2$ that passes through $v_1'$ and $w_2 v_2'$. By condition (ii), this is only possible if $v_1 = v_1'$ and $v_2 = v_2'$, showing (i).

Consider Situation <ref> with $v_1 \in \LP(x_1), v_2\in \LP(x_2)$. There exists $(v_1', v_2')\in M(x_1, x_2)$ such that
\begin{align*}d(v_1\Rightarrow w_2 v_2) = d(v_1\Rightarrow v_1') + d(v_1'\Rightarrow w_2 v_2') + d(w_2v_2'\Rightarrow w_2 v_2).\end{align*}

For convenience, we define a set of admissible pairs by

\begin{align*}
&A := \{(v_1', v_2')\in \LP(x_1)\times \LP(x_2)\mid \\&\qquad d(v_1\Rightarrow w_2 v_2) = d(v_1\Rightarrow v_1') + d(v_1'\Rightarrow w_2 v_2') + d(w_2v_2'\Rightarrow w_2 v_2)\}.
\end{align*}

Then $(v_1, v_2)\in A$, so that $A$ is non-empty. Choose $(v'_1, v_2')\in A$ such that $d(v_1'\Rightarrow w_2 v_2')$ becomes minimal among all pairs in $A$. We claim that $(v_1', v_2')\in M(x_1, x_2)$. For this, we use Lemma <ref>. It remains to show that $v_1' = \rho^\vee_{x_1}(w_2 v_2')$ and $v_2' = \rho_{x_2}(v_1)$. By Proposition <ref> and Corollary <ref>, we obtain

\begin{align*}
&d(v_1'\Rightarrow w_2 v_2') = d(v_1'\Rightarrow \rho^\vee_{x_1}(w_2 v_2')) + d(\rho^\vee_{x_1}(w_2 v_2')\Rightarrow w_2 v_2'),
\\&d(v_1'\Rightarrow w_2 v_2') = d(v_1'\Rightarrow w_2 \rho_{x_2}(v_1)) + d(w_2 \rho_{x_2}(v_1)\Rightarrow w_2 v_2').
\end{align*}

It follows that $(\rho^\vee_{x_1}(w_2 v_2'), v_2') \in A$ and $(v_1', \rho_{x_2}(v_1')) \in A$. By choice of $(v_1', v_2')$ and the above computation, we get that $v_1' = \rho^\vee_{x_1}(w_2 v_2')$ and $v_2' = \rho_{x_2}(v_1')$. This finishes the proof.

For $x_1, x_2\in \widetilde W$, we have $\LP(x_1\ast x_2)= \rho_{x_2}(\LP(x_1)) = \rho_{x_1}^\vee(w_2 \LP(x_2))$, where $w_2\in W$ is the classical part of $x_2$.

We only show $\LP(x_1\ast x_2) = \rho_{x_2}(\LP(x_1))$, the other claim is completely dual.

If $v_2\in \LP(x_1\ast x_2)$, we find $v_1\in \LP(x_1)$ such that $(v_1, v_2)\in M(x_1, x_2)$. By Lemma <ref>, $v_2 = \rho_{x_2}(v_1)\in \rho_{x_2}(\LP(x_1))$.

Now let $v_2\in \rho_{x_2}(\LP(x_1))$ and write $v_2 = \rho_{x_2}(v_1)$ for some $\widetilde{v_1}\in \LP(x_1)$. By Corollary <ref>, we find $(v_1', v_2')\in M(x_1, x_2)$ such that

\begin{align*}d(v_1\Rightarrow w_2 v_2) =& d(v_1\Rightarrow w_2 v_2') + d(w_2v_2'\Rightarrow w_2 v_2).\end{align*}

Since $v_2 = \rho_{x_2}(v_1)$, we use Proposition <ref> to obtain

\begin{align*}d(v_1\Rightarrow w_2 v_2') =& d(v_1\Rightarrow w_2 v_2) + d(w_2v_2\Rightarrow w_2 v_2').\end{align*}

This is only possible if $v_2 = v_2'$. Since $v_2'\in \LP(x_1\ast x_2)$ by Proposition <ref>, we obtain the desired claim $v_2\in \LP(x_1\ast x_2)$.

§.§ Generic $\sigma$-conjugacy class

To conclude the paper, we apply our results to the notion of generic $\sigma$-conjugacy classes. For this, we have to assume that our affine Weyl group actually comes from a quasi-split reductive group $G$ over a non-archimedian local field $F$, as described in Section sec:notation. This means that $W$ is the finite Weyl group of $G$, and $X_\ast$ are the $\Gal(\overline{\breve F}/\breve F)$-coinvariants of the cocharacter group of a maximal torus. Denote by $B(G)$ the set of $\sigma$-conjugacy classes in $G(\breve F)$. For $x\in \widetilde W$, we write $[x]\in B(G)$ for the $\sigma$-conjugacy classes associated with any representative of $x$ in $G(\breve F)$, and $[b_x]$ for the generic $\sigma$-conjugacy class of the Iwahori double coset indexed by $x$.

The Frobenius action on $W$ and $\widetilde W$ will be denoted $\presig{(\cdot)}$, so the Frobenius image of $x$ is $\presig x$.

Throughout this section, we fix an element $x=w\varepsilon^\mu\in \widetilde W$. Following He <cit.>, we consider twisted Demazure powers of $x$.

Let $n\geq 1$. We define the $n$-th $\sigma$-twisted Demazure power of $x$ as

\begin{align*}
x^{\ast,\sigma,n} := x\ast (\presig x)\ast\cdots \ast\left(\prescript{\sigma^{n-1}}{}x\right)\in \widetilde W.
\end{align*}

For $n\geq 2$, let us write

\begin{align*}
x_n := \prescript{\sigma^{1-n}}{}{}\left(\left(x^{\ast,\sigma,n-1}\right)^{-1}x^{\ast,\sigma,n}\right),
\end{align*}

such that

\begin{align*}
x^{\ast,\sigma,n} = x^{\ast,\sigma,n-1}\ast \left(\prescript{\sigma^{n-1}}{}x\right) = x^{\ast,\sigma,n-1}\cdot \left(\prescript{\sigma^{n-1}}{}x_n\right).
\end{align*}

We can calculate $x_n$ in terms of $x$ and $\prescript{\sigma^{1-n}}{}\LP(x^{\ast,\sigma,n-1})$ using Theorem <ref>. By Corollary <ref>, we have

\begin{align*}
\LP(x^{\ast,\sigma,n}) = \rho_{\prescript{\sigma^{n-1}}{}x}\left(\LP(x^{\ast,\sigma,n-1})\right)
=\cdots=\rho_{\prescript{\sigma^{n-1}}{}x}\circ\cdots\circ \rho_{\presig x}\left(\LP(x)\right).
\end{align*}

Observe that by definition of the generic action $\rho_x$, we may write

\begin{align*}
\rho_{\prescript{\sigma^n}{}x}(\prescript{\sigma^n}{}(u)) = \prescript{\sigma^n}{}(\rho_x(u)).
\end{align*}

Let us define the map $\rho_{x,\sigma} := \rho_x\circ \prescript{\sigma^{-1}}{}(\cdot):W\rightarrow W$ by

\begin{align*}
\rho_{x,\sigma}(u) := \rho_x(\prescript{\sigma^{-1}}{}(u)).
\end{align*}

\begin{align*}
\LP(x^{\ast,\sigma,n}) =&\rho_{\prescript{\sigma^{n-1}}{}x}\circ\cdots\circ \rho_{\presig x}\left(\LP(x)\right).
\\=&\left(\prescript{\sigma^{n-1}}{}(\cdot)\circ \rho_x\circ \prescript{\sigma^{1-n}}{}(\cdot)\right)\circ\cdots\circ\left(\prescript{\sigma^{1}}{}(\cdot)\circ \rho_x\circ \prescript{\sigma^{-1}}{}(\cdot)\right)(\LP(x))
\\=&\prescript{\sigma^{n-1}}{}(\cdot)\circ \rho_{x,\sigma}\circ\cdots\circ \rho_{x,\sigma}(\LP(x))
\\=&\prescript{\sigma^{n-1}}{}{}\left(\rho_{x,\sigma}^{n-1}(\LP(x))\right).
\end{align*}

* There exists an integer $N>1$ such that for each $n\geq N$,
\begin{align*}
x_N = x_n\text{ and }\rho_{x,\sigma}^N(\LP(x)) = \rho_{x,\sigma}^n(\LP(x)).
\end{align*}
Denote the eventual values by $x_\infty:= x_N$ resp. $\rho_{x,\sigma}^\infty(\LP(x)):= \rho_{x,\sigma}^N(\LP(x))$.

* We have

\begin{align*}
\rho_{x,\sigma}^\infty(\LP(x)) =&\{v\in \LP(x)\mid\exists n\geq 1:~v = \rho_{x,\sigma}^n(v)\}.
\\\lim_{n\to\infty}\frac{\ell(x^{\ast,\sigma,n})}n = &\ell(x_\infty).
\end{align*}

* The element $x_\infty$ is fundamental. For each $v\in \rho_{x,\sigma}^\infty(\LP(x))$, it can be written as

\begin{align*}
x_\infty = (\prescript{\sigma^{-1}}{}v)\rho_{x,\sigma}(v)^{-1}\varepsilon^{\mu - \rho_{x,\sigma}(v)\wt\left(\prescript{\sigma^{-1}}{}v\Rightarrow w\rho_{x,\sigma}(v)\right)}.
\end{align*}

* Observe that $\rho^n_{x,\sigma}$ induces an endomorphism $\LP(x)\rightarrow \LP(x)$. We obtain a weakly decreasing sequence of subsets of $W$

\begin{align*}
\LP(x)\supseteq \rho_{x,\sigma}(\LP(x))\supseteq\rho_{x,\sigma}^2(\LP(x))\supseteq\cdots.
\end{align*}
Since $W$ is finite, this sequence must stabilize eventually.

Because $x_n$ only depends on the values of $\rho_{x,\sigma}^{n-1}(\LP(x))$ and $x$, the result follows.

* Both claims follow immediately from (a).

* Let $N$ be as in (a), and let $n\geq 1$. Then

\begin{align*}
x^{\ast,\sigma,N+n} = x^{\ast,\sigma,N}\cdot \prescript{\sigma^N}{}x_\infty\cdots \prescript{\sigma^{N+n-1}}{}x_\infty
\end{align*}

is a length additive product. In particular,

\begin{align*}
\ell(x_\infty\cdots\prescript{\sigma^{n-1}}{}x_\infty) = n\ell(x_\infty).
\end{align*}

By <cit.> or Proposition prop:fundamental, $x_\infty$ is fundamental.

Next let $v\in \rho^\infty_{x,\sigma}(\LP(x))$. Then also $\rho_{x,\sigma}(v)\in \rho^\infty_{x,\sigma}(\LP(x))$, and we get

\begin{align*}
\prescript{\sigma^{N}}{}\rho_{x,\sigma}(v)\in \LP(x^{\ast,\sigma,N+1}) = \LP(x^{\ast,\sigma,N}\ast \prescript{\sigma^N}{}x) = \LP(x^{\ast,\sigma,N}\cdot\prescript{\sigma^N}{}(x_\infty)).
\end{align*}

In view of Proposition <ref>, we find a uniquely determined element $\prescript{\sigma^{N}}{}v'\in \LP(x^{\ast,\sigma,N})$ such that

\begin{align*}
(\prescript{\sigma^{N}}{}v', \prescript{\sigma^{N}}{}\rho_{x,\sigma}(v))\in M(x^{\ast,\sigma,N}, \prescript{\sigma^N}{}x).
\end{align*}

Then by Theorem <ref>,

\begin{align*}
x_\infty = v'\rho_{x,\sigma}(v)^{-1}\varepsilon^{\mu - \rho_{x,\sigma}(v)\wt(v'\Rightarrow w\rho_{x,\sigma}(v))}.
\end{align*}

Note that $\presig v'\in\prescript{\sigma^{1-N}}{}\LP(x^{\ast,\sigma,N}) = \rho_{x,\sigma}^\infty(\LP(x))$. The minimality condition on the tuple
$(\prescript{\sigma^{N}}{}v', \prescript{\sigma^{N}}{}\rho_{x,\sigma}(v))$ moreover implies that $\rho_x(v') = \rho_{x,\sigma}(\presig v') = \rho_{x,\sigma}(v)$ (Lemma <ref>).

The map $\rho_{x,\sigma} : \rho_{x,\sigma}^\infty(\LP(x))\rightarrow \rho_{x,\sigma}^\infty(\LP(x))$ is a surjective, and the set $\rho_{x,\sigma}^\infty(\LP(x))$ is finite. It follows that the restriction of $\rho_{x,\sigma}$ to  $\rho_{x,\sigma}^\infty(\LP(x))$ is bijective. Recall that $v$ and $\presig v'$ are two elements of $\rho_{x,\sigma}^\infty(\LP(x))$ whose images under $\rho_{x,\sigma}$ coincide. Thus $v =\presig v'$, finishing the proof.

* The $\sigma$-conjugacy class $[x_\infty]\in B(G)$ is the generic $\sigma$-conjugacy class of $x$.

* For any $v\in \rho_{x,\sigma}^\infty(\LP(x))$, we have $\ell(x_\infty) = \ell(x)-d(v\Rightarrow\presig(w\rho_{x,\sigma}(v)))$.

* Fix $v\in \rho_{x,\sigma}^\infty(\LP(x))$ and define $J=\supp_\sigma(\rho_{x,\sigma}(v)^{-1}v)$, so $J\subseteq \Delta$ consists of all $\sigma$-orbits of simple roots whose corresponding simple reflections occur in some reduced decomposition of $\rho_{x,\sigma}(v)^{-1}v\in W$.

We can express the generic Newton point of $x$ as

\begin{align*}
\nu_x = \pi_J\left(v^{-1}\mu - \wt(v\Rightarrow \presig(w v))\right).
\end{align*}

Here, $\pi_J$ denotes the projection function as defined in <cit.>.

* By a result of Viehmann <cit.>, we can express the generic $\sigma$-conjugacy class of $x$ as

\begin{align*}
[b_x] = \max\{[y]\mid y\leq x\} = \max\{[y]\mid y\leq x\text{ and }y\text{ is fundamental}\}.
\end{align*}

In particular, $[b_x]\geq [x_\infty]$. For the converse inequality, pick some $y\leq x$ fundamental with $[b_x] = [y]\in B(G)$.

By definition of the Demazure product, we get

\begin{align*}
x^{\ast,\sigma,n} = x \ast \left(\presig x\right)\cdots\ast \left(\prescript{\sigma^{n-1}}{}x\right) \geq y \left(\presig y\right)\cdots \left(\prescript{\sigma^{n-1}}{}y\right).
\end{align*}

Thus, using the fact that $y$ and $x_\infty$ are fundamental, we get

\begin{align*}
\langle \nu(x_\infty),2\rho\rangle =& \ell(x_\infty) = \lim_{n\to \infty} \frac{\ell(x^{\ast,\sigma,n})}n \\\geq& \lim_{n\to \infty} \frac{\ell(y \presig y\cdots \prescript{\sigma^{n-1}}{}y)}n = \lim_{n\to \infty} \ell(y) = \langle \nu(y),2\rho\rangle = \langle \nu(b_x),2\rho\rangle.
\end{align*}

This estimate shows that $[x_\infty] = [b_x]$.

* This follows from the explicit description of $x_\infty$ in Lemma <ref> together with Lemma <ref>
and the simple observation $\rho_{x,\sigma}(v)\in \LP(x_\infty)$.

* Let us write $x_\infty = w_\infty\varepsilon^{\mu_\infty}$. The generic Newton point of $x$ is the Newton point of $x_\infty$, which we express using Lemma lem:newtonPointsAsAverages.

Let $N\geq 1$ such that the action of $(\sigma\circ w_\infty)$ on $X_\ast$ becomes trivial. We want to show for each $v\in \rho_{x,\sigma}^\infty(\LP(x))$ that

\begin{align*}
v^{-1}\sum_{k=1}^N(\sigma\circ w_\infty)^k \mu_\infty\in X_\ast\otimes\mathbb Q
\end{align*}

is dominant.

Note each $v\in \rho_{x,\sigma}^\infty(\LP(x))$ may be written as $v = \rho_{x,\sigma}(u)$ for some $u\in \rho_{x,\sigma}^\infty(\LP(x))$. By Lemma <ref>, it follows that $w_\infty = (\prescript{\sigma^{-1}}{}u)v^{-1}$. Thus $u = \presig(w_\infty v)\in \rho_{x,\sigma}^\infty(\LP(x))$. This shows $\presig(w_\infty v)\in \rho_{x,\sigma}^\infty(\LP(x))$ for each $v\in \rho_{x,\sigma}^\infty(\LP(x))$. It follows for each $\alpha\in \Phi^+$ that

\begin{align*}
&\left\langle v^{-1}\sum_{k=1}^N(\sigma\circ w_\infty)^k \mu_\infty,\alpha\right\rangle =\sum_{k=1}^N \langle \mu_\infty,(\sigma\circ w_\infty)^k v\alpha\rangle
\\=&\sum_{k=1}^N \left(\langle \mu_\infty,(\sigma\circ w_\infty)^k v\alpha\rangle + \Phi^+((\sigma\circ w_\infty)^k v\alpha) - \Phi^+((\sigma\circ w_\infty)^{k+1} v\alpha)\right)
\\=&\sum_{k=1}^N \ell(x_\infty,(\sigma\circ w_\infty)^k v\alpha)\geq 0.
\end{align*}

This shows the above dominance claim. As $v\in \rho_{x,\sigma}^\infty(\LP(x))$ was arbitrary, the same claim holds for $\rho_{x,\sigma}(v)$. With
\begin{align*}J:=\supp_\sigma(\rho_{x,\sigma}(v)^{-1}\presig(w_\infty \rho_{x,\sigma}(v))) = \supp_\sigma(\rho_{x,\sigma}(v)^{-1} v),\end{align*}
Lemma lem:newtonPointsAsAverages proves that

\begin{align*}
\nu(x_\infty) =~& \pi_J(\rho_{x,\sigma}(v)^{-1}\mu_\infty) \underset{\text{L\ref{lem:xInfConstruction}}}=\pi_J(\rho_{x,\sigma}(v)^{-1}\mu - \wt(\prescript{\sigma^{-1}}{} v\Rightarrow w\rho_{x,\sigma}(v)))
\\\underset{\text{Def.~}\pi_J}=&\pi_J(\rho_{x,\sigma}(v)^{-1}\mu - \wt(v\Rightarrow\presig(w\rho_{x,\sigma}(v)))).
\end{align*}

Now observe that

\begin{align*}
\rho_{x,\sigma}(v)^{-1}\mu\equiv \,&v^{-1}\mu\pmod{\mathbb Q\Phi_J^\vee},
\\
\wt(v\Rightarrow\presig(w\rho_{x,\sigma}(v)))\equiv\,& \wt(v\Rightarrow\presig(wv))\pmod{\mathbb Q\Phi_J^\vee}.\qedhere
\end{align*}

Part (a) of the above Theorem readily implies <cit.>. Our previous result Corollary cor:genericGKPMinDistance expresses the generic Newton point $\nu_x$ as a formula similar to part (c) of the above Theorem, but the allowed elements $v\in \LP(x)$ in the cited result are usually different ones. If $x$ is in a shrunken Weyl chamber, this formula for the generic Newton point coincides with <cit.>.