[proofplan] We first record that each chamber determines a positive root system by the signs of the linear forms $x \mapsto (x,\alpha)_E$, and that a wall reflection carries a chamber to the unique adjacent chamber across that wall. Transitivity is proved by joining a point of one chamber to a point of another by a generic line segment and reflecting successively across the hyperplanes crossed by that segment. Freeness is proved by showing that an element preserving a chamber preserves its positive roots, while every non-identity Weyl group element sends at least one positive root to a negative root. [/proofplan] [step:Attach a positive root system to a chamber] Let $C$ be a Weyl chamber. For each $\alpha \in \Phi$, the continuous function \begin{align*} \ell_\alpha : E &\to \mathbb{R} \\ x &\mapsto (x,\alpha)_E \end{align*} does not vanish on $C$, because $C$ is contained in the complement of the hyperplane arrangement. Since $C$ is connected, $\ell_\alpha$ has constant sign on $C$. Define \begin{align*} \Phi_C^+ := \{\alpha \in \Phi : (x,\alpha)_E > 0 \text{ for every } x \in C\}. \end{align*} Then exactly one of $\alpha$ and $-\alpha$ lies in $\Phi_C^+$ for each pair $\{\alpha,-\alpha\} \subset \Phi$, because \begin{align*} (x,-\alpha)_E = -(x,\alpha)_E. \end{align*} Moreover \begin{align*} C = \{x \in E : (x,\alpha)_E > 0 \text{ for every } \alpha \in \Phi_C^+\}. \end{align*} Indeed, the right-hand side is one connected component of the complement of the hyperplane arrangement with the same sign pattern as $C$, and therefore equals $C$. [guided] Fix a Weyl chamber $C$. The hyperplanes $H_\alpha$ are precisely the zero sets of the linear maps \begin{align*} \ell_\alpha : E &\to \mathbb{R} \\ x &\mapsto (x,\alpha)_E . \end{align*} Because $C$ lies in the complement of all these hyperplanes, $\ell_\alpha(x) \neq 0$ for every $x \in C$. Since $C$ is connected and $\ell_\alpha$ is continuous, the sign of $\ell_\alpha$ cannot change on $C$. Thus every root is either positive on all of $C$ or negative on all of $C$. We define the positive roots determined by $C$ by \begin{align*} \Phi_C^+ := \{\alpha \in \Phi : (x,\alpha)_E > 0 \text{ for every } x \in C\}. \end{align*} For each root pair $\{\alpha,-\alpha\}$, exactly one member lies in $\Phi_C^+$, because replacing $\alpha$ by $-\alpha$ changes the sign of the linear form. The chamber itself is recovered from this sign data: \begin{align*} C = \{x \in E : (x,\alpha)_E > 0 \text{ for every } \alpha \in \Phi_C^+\}. \end{align*} The inclusion from left to right is the definition of $\Phi_C^+$. Conversely, the right-hand side is the region of the arrangement with exactly the same signs as $C$. A chamber is a connected component of the complement of the arrangement, so two points with the same strict sign pattern lie in the same component. Hence the displayed set is precisely $C$. [/guided] [/step] [step:Reflect a chamber across one of its walls] Let $D$ be a Weyl chamber, and suppose $H_\beta$ is a wall of $D$ for some $\beta \in \Phi$. The root reflection \begin{align*} s_\beta : E &\to E \\ x &\mapsto x - 2\frac{(x,\beta)_E}{(\beta,\beta)_E}\beta \end{align*} fixes $H_\beta$ pointwise and sends the half-space $(x,\beta)_E>0$ to the half-space $(x,\beta)_E<0$. For every $\alpha \in \Phi$, the defining property of a root system gives $s_\beta(\alpha) \in \Phi$. Consequently \begin{align*} s_\beta(H_\alpha)=H_{s_\beta(\alpha)}. \end{align*} Thus $s_\beta$ preserves the hyperplane arrangement and maps chambers to chambers. Since it fixes the wall $H_\beta$ and reverses the $\beta$-sign, $s_\beta(D)$ is the chamber adjacent to $D$ across the wall $H_\beta$. [guided] The reflection across the hyperplane $H_\beta$ is the map \begin{align*} s_\beta : E &\to E \\ x &\mapsto x - 2\frac{(x,\beta)_E}{(\beta,\beta)_E}\beta . \end{align*} If $x \in H_\beta$, then $(x,\beta)_E=0$, so $s_\beta(x)=x$. Thus $s_\beta$ fixes $H_\beta$ pointwise. Also, \begin{align*} (s_\beta x,\beta)_E &= \left(x - 2\frac{(x,\beta)_E}{(\beta,\beta)_E}\beta,\beta\right)_E \\ &= (x,\beta)_E - 2(x,\beta)_E \\ &= -(x,\beta)_E. \end{align*} Therefore the reflection exchanges the two open half-spaces bounded by $H_\beta$. We also need to know that $s_\beta$ sends chambers to chambers, not merely open sets. This follows from the root-system axiom that $s_\beta(\Phi)=\Phi$. For any root $\alpha \in \Phi$, \begin{align*} x \in H_\alpha &\iff (x,\alpha)_E=0 \\ &\iff (s_\beta x,s_\beta\alpha)_E=0 \\ &\iff s_\beta x \in H_{s_\beta\alpha}. \end{align*} Hence $s_\beta(H_\alpha)=H_{s_\beta\alpha}$, and the full hyperplane arrangement is preserved. A homeomorphism preserving the arrangement sends connected components of its complement to connected components of its complement. Therefore $s_\beta(D)$ is a Weyl chamber. Finally, because $s_\beta$ fixes the common wall $H_\beta$ and reverses the sign of $(x,\beta)_E$, the chamber $s_\beta(D)$ lies on the opposite side of $H_\beta$ from $D$. It is therefore the unique chamber adjacent to $D$ across that wall. [/guided] [/step] [step:Move from one chamber to another by crossing hyperplanes one at a time] Let $C$ and $C_1$ be Weyl chambers. Choose points $x \in C$ and $y \in C_1$ such that the segment \begin{align*} \gamma : [0,1] &\to E \\ t &\mapsto (1-t)x + ty \end{align*} meets at most one root hyperplane at each time. Such a choice is possible because the excluded choices are contained in a finite union of proper affine subspaces determined by pairwise intersections $H_\alpha \cap H_\beta$. Let \begin{align*} 0 < t_1 < \cdots < t_m < 1 \end{align*} be the times at which $\gamma$ meets a root hyperplane. For each $j \in \{1,\dots,m\}$, choose $\alpha_j \in \Phi$ such that \begin{align*} \gamma(t_j) \in H_{\alpha_j}. \end{align*} By genericity, no other root hyperplane contains $\gamma(t_j)$ except $H_{\alpha_j}=H_{-\alpha_j}$. The segment therefore passes from one chamber to the adjacent chamber across $H_{\alpha_j}$ at time $t_j$. Starting with $D_0=C$, define recursively \begin{align*} D_j := s_{\alpha_j}(D_{j-1}) \end{align*} for $j=1,\dots,m$. By the previous step, $D_j$ is exactly the chamber containing $\gamma(t)$ for $t_j<t<t_{j+1}$, with the convention $t_{m+1}=1$. Hence $D_m=C_1$. Therefore \begin{align*} w := s_{\alpha_m}\cdots s_{\alpha_1} \in W \end{align*} satisfies $w(C)=C_1$. This proves transitivity. [guided] We now prove that every chamber can be reached from every other chamber by Weyl group reflections. Choose $x \in C$ and $y \in C_1$. We perturb them inside their chambers if necessary so that the straight segment \begin{align*} \gamma : [0,1] &\to E \\ t &\mapsto (1-t)x + ty \end{align*} does not pass through an intersection of two distinct root hyperplanes. This is possible because there are only finitely many roots, hence only finitely many hyperplanes, and the condition that a segment meet $H_\alpha \cap H_\beta$ for two non-proportional roots $\alpha,\beta$ imposes a proper affine condition on the endpoints. Let \begin{align*} 0 < t_1 < \cdots < t_m < 1 \end{align*} be the crossing times. At each time $t_j$, the point $\gamma(t_j)$ lies on exactly one hyperplane of the arrangement, namely $H_{\alpha_j}$ for some root $\alpha_j \in \Phi$, up to replacing $\alpha_j$ by $-\alpha_j$. Since no second hyperplane is crossed at the same time, the segment enters the chamber adjacent to the previous one across the wall $H_{\alpha_j}$. Define $D_0=C$, and after the $j$-th crossing define \begin{align*} D_j := s_{\alpha_j}(D_{j-1}). \end{align*} The preceding step says precisely that reflection in $H_{\alpha_j}$ sends the chamber before the crossing to the chamber after the crossing. Thus $D_j$ is the chamber containing $\gamma(t)$ for $t$ immediately after $t_j$. After the last crossing, the segment lies in $C_1$, so $D_m=C_1$. Consequently the Weyl group element \begin{align*} w := s_{\alpha_m}\cdots s_{\alpha_1} \end{align*} sends $C$ to $C_1$. Since $C$ and $C_1$ were arbitrary, the action is transitive. [/guided] [/step] [step:Show that a chamber stabilizer preserves its positive roots] Suppose $w \in W$ satisfies $w(C)=C$. For every $\alpha \in \Phi_C^+$ and every $x \in C$, we have $w^{-1}x \in C$, hence \begin{align*} (x,w\alpha)_E = (w^{-1}x,\alpha)_E > 0. \end{align*} Therefore $w\alpha \in \Phi_C^+$. Since $w$ is a bijection of $\Phi$, this gives \begin{align*} w(\Phi_C^+) = \Phi_C^+. \end{align*} [guided] Assume $w(C)=C$. We want to translate this geometric statement into a statement about roots. Let $\alpha \in \Phi_C^+$. By definition, this means \begin{align*} (z,\alpha)_E > 0 \end{align*} for every $z \in C$. Now take any $x \in C$. Since $w(C)=C$, also $w^{-1}x \in C$. Therefore \begin{align*} (w^{-1}x,\alpha)_E > 0. \end{align*} The Weyl group acts by orthogonal transformations, so \begin{align*} (w^{-1}x,\alpha)_E = (x,w\alpha)_E. \end{align*} Hence \begin{align*} (x,w\alpha)_E > 0 \end{align*} for every $x \in C$. This says exactly that $w\alpha \in \Phi_C^+$. Since $w$ permutes the finite set $\Phi$, the inclusion $w(\Phi_C^+) \subset \Phi_C^+$ is an equality. Thus every chamber stabilizer preserves the positive root system attached to that chamber. [/guided] [/step] [step:Prove that a non-identity Weyl group element sends some positive root to a negative root] Let $\Delta_C \subset \Phi_C^+$ denote the set of indecomposable positive roots, meaning those $\delta \in \Phi_C^+$ that cannot be written as $\delta=\alpha+\beta$ with $\alpha,\beta \in \Phi_C^+$. We use the standard simple-root theorem for finite reduced root systems: $\Delta_C$ is a basis of the root lattice, every element of $\Phi_C^+$ is a non-negative integer linear combination of elements of $\Delta_C$, and $W$ is generated by the simple reflections $s_\delta$ with $\delta \in \Delta_C$. We also use the standard inversion criterion for a finite Weyl group with respect to $\Delta_C$: if \begin{align*} w=s_{\beta_m}\cdots s_{\beta_1} \end{align*} is a reduced expression with $\beta_j \in \Delta_C$, then \begin{align*} \alpha_j := s_{\beta_1}\cdots s_{\beta_{j-1}}\beta_j \in \Phi_C^+ \end{align*} for each $j \in \{1,\dots,m\}$, and \begin{align*} w\alpha_m \in -\Phi_C^+. \end{align*} The hypotheses of this criterion are satisfied because the preceding simple-root theorem identifies $(W,\{s_\delta : \delta \in \Delta_C\})$ as the Coxeter system attached to the chamber $C$, and the word was chosen reduced. If $w\neq e$, choose such a reduced expression with $m\geq 1$. Define \begin{align*} \alpha := s_{\beta_1}\cdots s_{\beta_{m-1}}\beta_m. \end{align*} By the inversion criterion, $\alpha \in \Phi_C^+$ and $w\alpha \in -\Phi_C^+$. Hence every non-identity element of $W$ sends at least one positive root to a negative root. [guided] The goal is to prove the key algebraic fact behind freeness: a non-identity Weyl group element cannot preserve all positive roots. We first name exactly which standard structure theorem is being used, because this step is where the chamber geometry is converted into Coxeter combinatorics. Let $\Delta_C \subset \Phi_C^+$ be the set of indecomposable positive roots, so $\delta \in \Delta_C$ precisely when $\delta \in \Phi_C^+$ and $\delta$ cannot be written as $\delta=\alpha+\beta$ with $\alpha,\beta \in \Phi_C^+$. For a finite reduced root system, the simple-root theorem attached to the chamber $C$ says that $\Delta_C$ is a basis of the root lattice, every positive root is a non-negative integer linear combination of roots in $\Delta_C$, and the reflections $s_\delta$ with $\delta \in \Delta_C$ generate $W$. The reducedness hypothesis is used here to ensure that the chamber walls determine one primitive root direction each, so the indecomposable positive roots are exactly the simple roots defining those walls. Now take $w\in W$ with $w\neq e$. Since the simple reflections generate $W$, choose a reduced expression \begin{align*} w=s_{\beta_m}\cdots s_{\beta_1} \end{align*} with $m\geq 1$ and each $\beta_j \in \Delta_C$. We apply the standard inversion criterion for finite Weyl groups with respect to this reduced expression. It states that the roots \begin{align*} \alpha_j := s_{\beta_1}\cdots s_{\beta_{j-1}}\beta_j \end{align*} are positive roots in $\Phi_C^+$, and that the final one is sent by $w$ to a negative root. The criterion applies because $\Delta_C$ is the simple system determined by $C$ and the displayed expression is reduced. Set \begin{align*} \alpha := \alpha_m=s_{\beta_1}\cdots s_{\beta_{m-1}}\beta_m. \end{align*} By the inversion criterion, $\alpha \in \Phi_C^+$. The same criterion gives \begin{align*} w\alpha \in -\Phi_C^+. \end{align*} Thus we have found a positive root whose image under $w$ is negative. Therefore no non-identity Weyl group element preserves the whole positive root system $\Phi_C^+$. [/guided] [/step] [step:Conclude uniqueness from freeness] Suppose $w \in W$ satisfies $w(C)=C$. By the previous stabilizer step, \begin{align*} w(\Phi_C^+) = \Phi_C^+. \end{align*} If $w \neq e$, the preceding step gives some $\alpha \in \Phi_C^+$ such that \begin{align*} w\alpha \in -\Phi_C^+, \end{align*} contradicting $w(\Phi_C^+)=\Phi_C^+$. Hence $w=e$, so the stabilizer of $C$ is trivial. Now let $C$ and $C_1$ be chambers. Transitivity gives at least one $w \in W$ with $w(C)=C_1$. If $w_1(C)=C_1$ and $w_2(C)=C_1$, then \begin{align*} w_2^{-1}w_1(C)=C. \end{align*} The stabilizer of $C$ is trivial, so $w_2^{-1}w_1=e$, and therefore $w_1=w_2$. Thus there exists a unique Weyl group element carrying $C$ to $C_1$, which proves simple transitivity. [/step]