[proofplan] We prove the contrapositive in both directions. If the Dynkin diagram is disconnected, its vertices split into two nonempty sets with all cross inner products equal to zero; the spans of these two sets give an [orthogonal decomposition](/theorems/436), and the standard connected-support property of roots shows that every root lies entirely in one side. Conversely, if the root system is reducible, the two orthogonal components force the simple roots to split into two nonempty subsets, and no Dynkin edge can cross between them. Thus diagram connectedness is exactly the same condition as irreducibility of the root system. [/proofplan] [step:Split a disconnected Dynkin diagram into orthogonal simple-root sets] Assume that $\Gamma(\Delta)$ is disconnected. Then there exist nonempty disjoint subsets $\Delta_1,\Delta_2 \subset \Delta$ such that \begin{align*} \Delta &= \Delta_1 \sqcup \Delta_2 \end{align*} and no edge of $\Gamma(\Delta)$ joins a vertex of $\Delta_1$ to a vertex of $\Delta_2$. By the definition of the Dynkin diagram, this means \begin{align*} (\alpha,\beta)_E = 0 \end{align*} for every $\alpha \in \Delta_1$ and every $\beta \in \Delta_2$. Define linear subspaces $E_1,E_2 \subset E$ by \begin{align*} E_1 &:= \operatorname{span}_{\mathbb{R}}(\Delta_1), & E_2 &:= \operatorname{span}_{\mathbb{R}}(\Delta_2). \end{align*} The displayed orthogonality of all generators implies \begin{align*} (x,y)_E = 0 \end{align*} for every $x \in E_1$ and every $y \in E_2$. Hence $E_1 \perp E_2$. [/step] [step:Use connected support to place each root in one orthogonal summand] For a root $\gamma \in \Phi$, write its simple-root expansion \begin{align*} \gamma = \sum_{\alpha \in \Delta} n_\alpha \alpha, \end{align*} where each $n_\alpha \in \mathbb{Z}$, and either all nonzero coefficients are nonnegative or all nonzero coefficients are nonpositive. Define the support of $\gamma$ by \begin{align*} \operatorname{supp}_{\Delta}(\gamma) := \{\alpha \in \Delta : n_\alpha \ne 0\}. \end{align*} The standard connected-support lemma for roots says that $\operatorname{supp}_{\Delta}(\gamma)$ is connected as a subgraph of $\Gamma(\Delta)$. Since $\Gamma(\Delta)$ has no edge between $\Delta_1$ and $\Delta_2$, a connected subset of vertices cannot meet both $\Delta_1$ and $\Delta_2$. Therefore \begin{align*} \operatorname{supp}_{\Delta}(\gamma) \subseteq \Delta_1 \quad\text{or}\quad \operatorname{supp}_{\Delta}(\gamma) \subseteq \Delta_2. \end{align*} Equivalently, every root $\gamma \in \Phi$ lies in $E_1$ or in $E_2$. Define \begin{align*} \Phi_1 &:= \Phi \cap E_1, & \Phi_2 &:= \Phi \cap E_2. \end{align*} Then $\Phi = \Phi_1 \sqcup \Phi_2$, because every root lies in one of the two subspaces and $E_1 \cap E_2 = \{0\}$: if $v \in E_1 \cap E_2$, then $v \in E_1$ and $v \in E_2$, so $(v,v)_E = 0$, hence $v = 0$. Since $0 \notin \Phi$, the union is disjoint. Moreover $\Phi_1$ and $\Phi_2$ are nonempty, since $\Delta_1 \subset \Phi_1$ and $\Delta_2 \subset \Phi_2$. [claim:Each $\Phi_i$ is a root subsystem in its span] For $i \in \{1,2\}$, the set $\Phi_i$ is a root system in $\operatorname{span}_{\mathbb{R}}(\Phi_i)$. [/claim] [proof] Fix $i \in \{1,2\}$ and let $j \in \{1,2\}$ be the other index. The set $\Phi_i$ is finite and does not contain $0$, since $\Phi_i \subset \Phi$ and $\Phi$ is a finite root system. It spans $\operatorname{span}_{\mathbb{R}}(\Phi_i)$ by definition. If $\alpha \in \Phi_i$, then $-\alpha \in \Phi$ because $\Phi$ is a root system, and $-\alpha \in E_i$ because $E_i$ is a linear subspace; hence $-\alpha \in \Phi_i$. Let $\alpha,\beta \in \Phi_i$. Define the reflection $s_\alpha: E \to E$ by \begin{align*} s_\alpha(x) := x - \frac{2(x,\alpha)_E}{(\alpha,\alpha)_E}\alpha. \end{align*} Since $\Phi$ is a root system, $s_\alpha(\beta) \in \Phi$. Since both $\alpha$ and $\beta$ lie in $E_i$, the displayed formula gives $s_\alpha(\beta) \in E_i$. Therefore $s_\alpha(\beta) \in \Phi \cap E_i = \Phi_i$, so $\Phi_i$ is stable under reflections by its roots. The crystallographic integrality condition is inherited from $\Phi$, because the integers $2(\beta,\alpha)_E/(\alpha,\alpha)_E$ are the same whether $\alpha,\beta$ are regarded as roots of $\Phi$ or of $\Phi_i$. Reducedness is also inherited from $\Phi$: if $\alpha \in \Phi_i$ and $c\alpha \in \Phi_i$ for a real scalar $c$, then $c\alpha \in \Phi$, so reducedness of $\Phi$ forces $c = \pm 1$. Thus $\Phi_i$ is a reduced crystallographic root system in $\operatorname{span}_{\mathbb{R}}(\Phi_i)$. [/proof] The spans of $\Phi_1$ and $\Phi_2$ are contained in $E_1$ and $E_2$, respectively, and $E_1 \perp E_2$. Hence $\Phi$ is the disjoint union of two nonempty mutually orthogonal root subsystems. Thus $\Phi$ is reducible. [guided] Let $\gamma \in \Phi$ be arbitrary. Since $\Delta$ is a simple root basis, $\gamma$ has a unique expansion \begin{align*} \gamma = \sum_{\alpha \in \Delta} n_\alpha \alpha \end{align*} with $n_\alpha \in \mathbb{Z}$, and the coefficients are all $\ge 0$ or all $\le 0$. We define the support of $\gamma$ relative to $\Delta$ by \begin{align*} \operatorname{supp}_{\Delta}(\gamma) := \{\alpha \in \Delta : n_\alpha \ne 0\}. \end{align*} The key structural fact about simple-root expansions is the connected-support lemma: the support of any root is connected in the Dynkin diagram. Applying this lemma to the root $\gamma$ and the simple root basis $\Delta$, the vertex set $\operatorname{supp}_{\Delta}(\gamma)$ is connected as a subgraph of $\Gamma(\Delta)$. The partition $\Delta = \Delta_1 \sqcup \Delta_2$ has no edge crossing from $\Delta_1$ to $\Delta_2$. If a connected set of vertices met both $\Delta_1$ and $\Delta_2$, then a path inside that connected set would have to contain an edge crossing from one side of the partition to the other, contradicting the absence of such edges. Hence \begin{align*} \operatorname{supp}_{\Delta}(\gamma) \subseteq \Delta_1 \quad\text{or}\quad \operatorname{supp}_{\Delta}(\gamma) \subseteq \Delta_2. \end{align*} If the support is contained in $\Delta_1$, then the expansion of $\gamma$ uses only vectors in $\Delta_1$, so $\gamma \in E_1$. If the support is contained in $\Delta_2$, then $\gamma \in E_2$. Thus every root lies in at least one of the two subspaces. Now define \begin{align*} \Phi_1 &:= \Phi \cap E_1, & \Phi_2 &:= \Phi \cap E_2. \end{align*} Both sets are nonempty because $\Delta_1 \subset \Phi_1$ and $\Delta_2 \subset \Phi_2$. The preceding paragraph gives $\Phi = \Phi_1 \cup \Phi_2$. The union is disjoint: if $v \in E_1 \cap E_2$, then $v$ is orthogonal to itself because $E_1 \perp E_2$, so $(v,v)_E = 0$ and hence $v = 0$; since $0 \notin \Phi$, no root can lie in both $\Phi_1$ and $\Phi_2$. It remains to check the point required by reducibility: the pieces are root subsystems, not just subsets. Fix $i \in \{1,2\}$. The set $\Phi_i$ is finite and contains no zero vector because $\Phi_i \subset \Phi$. It spans $\operatorname{span}_{\mathbb{R}}(\Phi_i)$ by definition. If $\alpha \in \Phi_i$, then $-\alpha \in \Phi$ because $\Phi$ is a root system, and $-\alpha \in E_i$ because $E_i$ is a linear subspace; therefore $-\alpha \in \Phi_i$. For reflection stability, let $\alpha,\beta \in \Phi_i$. Define the reflection $s_\alpha: E \to E$ by \begin{align*} s_\alpha(x) := x - \frac{2(x,\alpha)_E}{(\alpha,\alpha)_E}\alpha. \end{align*} Since $\Phi$ is a root system, $s_\alpha(\beta) \in \Phi$. Since $\alpha$ and $\beta$ both lie in the linear subspace $E_i$, the formula for $s_\alpha(\beta)$ shows that $s_\alpha(\beta) \in E_i$. Hence $s_\alpha(\beta) \in \Phi \cap E_i = \Phi_i$. The crystallographic condition is inherited because $2(\beta,\alpha)_E/(\alpha,\alpha)_E$ is already an integer in the ambient root system $\Phi$. Reducedness is inherited as well: if $\alpha \in \Phi_i$ and $c\alpha \in \Phi_i$ for a real scalar $c$, then $c\alpha \in \Phi$, so reducedness of $\Phi$ gives $c = \pm 1$. Thus each $\Phi_i$ is a reduced crystallographic root system in its own span. Finally, the spans of $\Phi_1$ and $\Phi_2$ are contained in $E_1$ and $E_2$, respectively, and $E_1 \perp E_2$. Hence $\Phi$ is the disjoint union of two nonempty mutually orthogonal root subsystems. This is exactly a nontrivial reducible decomposition of $\Phi$. [/guided] [/step] [step:Split a reducible root system into two nonempty sets of simple roots] Conversely, assume that $\Phi$ is reducible. Then there exist nonempty root subsystems $\Phi_1,\Phi_2 \subset \Phi$ such that \begin{align*} \Phi &= \Phi_1 \sqcup \Phi_2 \end{align*} and \begin{align*} (\rho_1,\rho_2)_E = 0 \end{align*} for every $\rho_1 \in \Phi_1$ and every $\rho_2 \in \Phi_2$. Define \begin{align*} \Delta_1 &:= \Delta \cap \Phi_1, & \Delta_2 &:= \Delta \cap \Phi_2. \end{align*} Since $\Phi = \Phi_1 \sqcup \Phi_2$, each simple root lies in exactly one of $\Phi_1$ and $\Phi_2$, so \begin{align*} \Delta = \Delta_1 \sqcup \Delta_2. \end{align*} Both $\Delta_1$ and $\Delta_2$ are nonempty: if, for example, $\Delta_1 = \varnothing$, then every simple root lies in $\Phi_2$, so $\operatorname{span}_{\mathbb{R}}(\Phi) = \operatorname{span}_{\mathbb{R}}(\Delta) \subseteq \operatorname{span}_{\mathbb{R}}(\Phi_2)$, contradicting the nonempty orthogonal component $\Phi_1$. For $\alpha \in \Delta_1$ and $\beta \in \Delta_2$, we have $\alpha \in \Phi_1$ and $\beta \in \Phi_2$, so the orthogonality of the decomposition gives \begin{align*} (\alpha,\beta)_E = 0. \end{align*} Thus no edge of $\Gamma(\Delta)$ joins a vertex of $\Delta_1$ to a vertex of $\Delta_2$. [/step] [step:Conclude the equivalence between connectedness and irreducibility] The first two steps prove that if $\Gamma(\Delta)$ is disconnected, then $\Phi$ is reducible. The third step proves that if $\Phi$ is reducible, then $\Gamma(\Delta)$ is disconnected. Taking contrapositives, $\Gamma(\Delta)$ is connected if and only if $\Phi$ is irreducible. This is the desired equivalence. [/step]