[proofplan] We first show that a chamber determines a well-defined positive system: the sign of each root functional is constant on a chamber, and positivity is stable under addition of roots. For surjectivity, we start with an abstract positive system $\Phi^+$ and prove that the cone of vectors pairing positively with every root in $\Phi^+$ is non-empty. A vector in this cone lies in a unique chamber, and that chamber induces exactly $\Phi^+$. Finally, injectivity follows because a chamber is precisely a connected component of the complement of the root hyperplanes, so the full list of root signs determines the component. [/proofplan] [step:Show that each chamber assigns a constant sign to every root] Let $C$ be a Weyl chamber. For each $\alpha \in \Phi$, define the linear functional \begin{align*} \ell_\alpha:E &\to \mathbb{R} \\ v &\mapsto (\alpha,v). \end{align*} Since $C$ is contained in \begin{align*} E \setminus \bigcup_{\beta \in \Phi} H_\beta, \end{align*} we have $\ell_\alpha(v) \neq 0$ for every $v \in C$. The sets \begin{align*} C_\alpha^+ &:= \{v \in C : \ell_\alpha(v)>0\}, \\ C_\alpha^- &:= \{v \in C : \ell_\alpha(v)<0\} \end{align*} are open in $C$, disjoint, and cover $C$. Since $C$ is connected, exactly one of them is empty. Hence the sign of $(\alpha,v)$ is constant as $v$ varies over $C$. Therefore $\Phi_C^+$ is well-defined, and for every pair $\{\alpha,-\alpha\}$ with $\alpha \in \Phi$, exactly one of $\alpha$ and $-\alpha$ belongs to $\Phi_C^+$. [guided] Fix a Weyl chamber $C$. For a root $\alpha \in \Phi$, introduce the linear functional \begin{align*} \ell_\alpha:E &\to \mathbb{R} \\ v &\mapsto (\alpha,v). \end{align*} The hyperplane $H_\alpha$ is exactly the zero set of $\ell_\alpha$. Since $C$ is a connected component of the complement of all root hyperplanes, no point of $C$ lies in $H_\alpha$. Thus $\ell_\alpha(v)$ is never zero on $C$. Now split $C$ according to the sign of $\ell_\alpha$: \begin{align*} C_\alpha^+ &:= \{v \in C : \ell_\alpha(v)>0\}, \\ C_\alpha^- &:= \{v \in C : \ell_\alpha(v)<0\}. \end{align*} Both sets are open in the [subspace topology](/page/Subspace%20Topology) on $C$, because $\ell_\alpha$ is continuous. They are disjoint, and their union is all of $C$, because $\ell_\alpha$ never vanishes on $C$. If both were non-empty, they would disconnect $C$, contradicting that $C$ is connected. Hence one of the two sets is all of $C$. This proves that each root has a constant sign on a chamber. Since $-\alpha$ has the opposite functional, \begin{align*} \ell_{-\alpha}(v)=(-\alpha,v)=-(\alpha,v)=-\ell_\alpha(v), \end{align*} exactly one of $\alpha$ and $-\alpha$ is positive on $C$. [/guided] [/step] [step:Verify that the chamber signs form a positive system] Let $\alpha,\beta \in \Phi_C^+$, and suppose $\alpha+\beta \in \Phi$. For any $v \in C$, \begin{align*} (\alpha+\beta,v)=(\alpha,v)+(\beta,v)>0+0=0. \end{align*} Thus $\alpha+\beta \in \Phi_C^+$. Together with the previous step, this proves that $\Phi_C^+$ is a positive system. [/step] [step:Construct a vector positive on all roots in the given positive system] Let $\Phi^+ \subset \Phi$ be a positive system. Define the open cone \begin{align*} U := \{v \in E : (\alpha,v)>0 \text{ for every } \alpha \in \Phi^+\}. \end{align*} We prove that $U \neq \varnothing$. Let $\Delta \subset \Phi^+$ be the base of simple roots associated to $\Phi^+$, whose existence is given by the [Simple Roots of a Positive System](/theorems/???). This theorem says that $\Delta$ is linearly independent and every root $\alpha \in \Phi^+$ has a unique expansion \begin{align*} \alpha = \sum_{\delta \in \Delta} n_\delta(\alpha)\,\delta, \end{align*} where each coefficient $n_\delta(\alpha)$ is a non-negative integer and at least one coefficient is positive. Let $V := \operatorname{span}_{\mathbb{R}}(\Delta) \subset E$. Since $\Delta$ is linearly independent, there is a unique linear functional $L:V \to \mathbb{R}$ satisfying \begin{align*} L(\delta)=1 \qquad \text{for every } \delta \in \Delta. \end{align*} By the [Riesz representation theorem](/theorems/221) for finite-dimensional inner product spaces, applied to the inner product on $V$, there exists a unique vector $v_0 \in V$ such that \begin{align*} L(x)=(x,v_0) \qquad \text{for every } x \in V. \end{align*} Regard $v_0$ as a vector of $E$ through the inclusion $V \subset E$. For every $\alpha \in \Phi^+$, the expansion above gives \begin{align*} (\alpha,v_0) = \left(\sum_{\delta \in \Delta} n_\delta(\alpha)\,\delta,v_0\right) = \sum_{\delta \in \Delta} n_\delta(\alpha)(\delta,v_0) = \sum_{\delta \in \Delta} n_\delta(\alpha) >0. \end{align*} The last inequality holds because the coefficients are non-negative integers and at least one is positive. Hence $v_0 \in U$, so $U \neq \varnothing$. [guided] The goal is to find a vector $v \in E$ that pairs positively with every root in $\Phi^+$. Equivalently, we want to prove that the open cone \begin{align*} U := \{v \in E : (\alpha,v)>0 \text{ for every } \alpha \in \Phi^+\} \end{align*} is non-empty. We use the [Simple Roots of a Positive System](/theorems/???). Its hypotheses apply because $\Phi$ is a root system and $\Phi^+$ is a positive system in $\Phi$. The theorem gives a subset $\Delta \subset \Phi^+$, called the base of simple roots, with two properties: $\Delta$ is linearly independent, and every positive root $\alpha \in \Phi^+$ has a unique expansion \begin{align*} \alpha = \sum_{\delta \in \Delta} n_\delta(\alpha)\,\delta, \end{align*} where each $n_\delta(\alpha)$ is a non-negative integer and at least one coefficient is positive. This is the structural input that replaces the invalid strategy of minimizing supports of non-negative relations. Define the real vector subspace $V \subset E$ by \begin{align*} V := \operatorname{span}_{\mathbb{R}}(\Delta). \end{align*} Because $\Delta$ is linearly independent, prescribing values on $\Delta$ defines a unique linear functional $L:V \to \mathbb{R}$ by \begin{align*} L(\delta)=1 \qquad \text{for every } \delta \in \Delta. \end{align*} The restriction of the inner product of $E$ to $V$ is an inner product on the finite-dimensional space $V$. Therefore the finite-dimensional [Riesz representation theorem](/theorems/218) applies: there exists a unique vector $v_0 \in V$ such that \begin{align*} L(x)=(x,v_0) \qquad \text{for every } x \in V. \end{align*} We view this same $v_0$ as an element of $E$ using the inclusion $V \subset E$. Now take any positive root $\alpha \in \Phi^+$. Since $\alpha$ is a non-negative integer combination of simple roots, we compute \begin{align*} (\alpha,v_0) = \left(\sum_{\delta \in \Delta} n_\delta(\alpha)\,\delta,v_0\right) = \sum_{\delta \in \Delta} n_\delta(\alpha)(\delta,v_0) = \sum_{\delta \in \Delta} n_\delta(\alpha). \end{align*} Each summand is a non-negative integer, and at least one summand is positive because $\alpha$ is not the zero vector. Hence \begin{align*} (\alpha,v_0)>0. \end{align*} This holds for every $\alpha \in \Phi^+$, so $v_0 \in U$. Therefore $U \neq \varnothing$. [/guided] [/step] [step:Recover the prescribed positive system from the chamber containing the cone] Choose $v \in U$. Since $\Phi^+$ contains exactly one of $\alpha$ and $-\alpha$ for each root pair, for every $\beta \in \Phi$ exactly one of $\beta$ and $-\beta$ lies in $\Phi^+$. Thus $(\beta,v) \neq 0$ for every $\beta \in \Phi$, so \begin{align*} v \in E \setminus \bigcup_{\beta \in \Phi} H_\beta. \end{align*} Let $C$ be the Weyl chamber containing $v$. If $\alpha \in \Phi^+$, then $(\alpha,v)>0$, and the sign of $(\alpha,\cdot)$ is constant on $C$, so $\alpha \in \Phi_C^+$. Hence $\Phi^+ \subset \Phi_C^+$. Both $\Phi^+$ and $\Phi_C^+$ contain exactly one root from each pair $\{\alpha,-\alpha\}$, so the inclusion forces \begin{align*} \Phi^+=\Phi_C^+. \end{align*} Therefore every positive system is induced by some Weyl chamber. [/step] [step:Use root signs to distinguish different chambers] Let $C$ and $D$ be Weyl chambers with $\Phi_C^+=\Phi_D^+$. Choose $v \in C$ and $w \in D$. For every $\alpha \in \Phi$, the equality of positive systems implies that $(\alpha,v)$ and $(\alpha,w)$ have the same sign. Consider the line segment map \begin{align*} \sigma:[0,1] &\to E \\ t &\mapsto (1-t)v+tw. \end{align*} For each $\alpha \in \Phi$, the function $t \mapsto (\alpha,\sigma(t))$ is affine and has endpoint values with the same non-zero sign. Hence it never vanishes on $[0,1]$. Therefore \begin{align*} \sigma([0,1]) \subset E \setminus \bigcup_{\alpha \in \Phi} H_\alpha. \end{align*} The image $\sigma([0,1])$ is connected and intersects both $C$ and $D$, so $C=D$ because connected components are maximal connected subsets. Thus the assignment $C \mapsto \Phi_C^+$ is injective. Combining injectivity with the surjectivity proved above, the assignment is a bijection from Weyl chambers to positive systems. [/step]