[proofplan] We argue by contradiction using the root string property for roots of a finite-dimensional semisimple Lie algebra. If $2\alpha$ were a root, then the $\alpha$-string through $2\alpha$ would contain both $\alpha$ and $-\alpha$ as roots in the same affine line. But root strings are contiguous with no gaps, while the missing middle element is $0$, which is not a root. This contradiction proves that no double of a root is again a root. [/proofplan] [step:Assume that a double root exists and set up the relevant root string] Suppose, toward a contradiction, that there exists $\alpha \in \Phi$ such that $2\alpha \in \Phi$. Define $\beta \in \Phi$ by \begin{align*} \beta := 2\alpha. \end{align*} Thus $\beta$ is the assumed double of $\alpha$. We apply the [Root String Theorem for semisimple Lie algebras](/theorems/???) to the pair of roots $\beta, \alpha \in \Phi$. Its hypotheses are satisfied: $\beta \in \Phi$ by the contradiction assumption, $\alpha \in \Phi$ by the theorem statement, $\beta \neq \alpha$ because $2\alpha = \alpha$ would imply $\alpha = 0$, and $\beta \neq -\alpha$ because $2\alpha = -\alpha$ would imply $3\alpha = 0$; in characteristic zero this again gives $\alpha = 0$, contradicting the fact that roots are nonzero. We use the version of the [root string theorem](/theorems/4693) for roots of finite-dimensional semisimple Lie algebras; this version proves contiguity of root strings directly and does not assume in advance that the root system is reduced. The theorem says that the set \begin{align*} \{m \in \mathbb{Z} : \beta + m\alpha \in \Phi\} \end{align*} is a finite interval of integers. Equivalently, if two integer translates $\beta + m_1\alpha$ and $\beta + m_2\alpha$ are roots with $m_1 < m_2$, then every intermediate translate $\beta + m\alpha$ for $m_1 \leq m \leq m_2$ is also a root. [/step] [step:Find two roots in the same $\alpha$-string separated by the zero weight] Since $\beta = 2\alpha$, we compute the following translates: \begin{align*} \beta - \alpha &= 2\alpha - \alpha = \alpha, \\ \beta - 3\alpha &= 2\alpha - 3\alpha = -\alpha, \\ \beta - 2\alpha &= 2\alpha - 2\alpha = 0. \end{align*} Both $\alpha$ and $-\alpha$ belong to $\Phi$: $\alpha \in \Phi$ by hypothesis, and root systems arising from semisimple Lie algebras are closed under negation. Hence $\beta-\alpha \in \Phi$ and $\beta-3\alpha \in \Phi$. By the contiguity of the $\alpha$-string through $\beta$, the intermediate translate $\beta-2\alpha$ must also belong to $\Phi$. Therefore $0 \in \Phi$. [guided] The contradiction comes from looking not only at $2\alpha$ and $\alpha$, but also at the opposite root $-\alpha$. By the symmetry axiom in the definition of a [root system](/page/Root%20System), every root occurs with its negative, so $-\alpha \in \Phi$. We now verify the hypotheses of the root string theorem in this concrete situation. We have defined $\beta := 2\alpha$, and the contradiction assumption gives $\beta \in \Phi$. The theorem statement gives $\alpha \in \Phi$. Since roots are nonzero, $\alpha \neq 0$. Therefore $\beta \neq \alpha$, because $2\alpha = \alpha$ would imply $\alpha = 0$, and $\beta \neq -\alpha$, because $2\alpha = -\alpha$ would imply $3\alpha = 0$; the ground field has characteristic zero, so this again implies $\alpha = 0$. Thus the [Root String Theorem for semisimple Lie algebras](/theorems/???) applies to the ordered pair $(\beta, \alpha)$. Define the integer string set $S \subset \mathbb{Z}$ by \begin{align*} S := \{m \in \mathbb{Z} : \beta + m\alpha \in \Phi\}. \end{align*} The root string theorem says that $S$ is a finite interval of integers. In particular, if $m_1, m_2 \in S$ and $m_1 < m < m_2$, then $m \in S$. With $\beta := 2\alpha$, the roots $\alpha$ and $-\alpha$ both lie on this same $\alpha$-string through $\beta$: \begin{align*} \beta - \alpha &= 2\alpha - \alpha = \alpha \in \Phi, \\ \beta - 3\alpha &= 2\alpha - 3\alpha = -\alpha \in \Phi. \end{align*} Equivalently, $-3 \in S$ and $-1 \in S$. Since $S$ is an interval of integers, the intermediate integer $-2$ also belongs to $S$. Hence \begin{align*} \beta - 2\alpha = 2\alpha - 2\alpha = 0 \end{align*} belongs to $\Phi$. Thus the assumption $2\alpha \in \Phi$ forces $0 \in \Phi$. [/guided] [/step] [step:Contradict the definition of a root system and conclude] By definition, a [root](/page/Root) in the [root system](/page/Root%20System) of a semisimple Lie algebra is a nonzero linear functional on $\mathfrak{h}$. Therefore $0 \notin \Phi$. The previous step forced $0 \in \Phi$, a contradiction. Hence the assumption that $2\alpha \in \Phi$ is false. Therefore, for every $\alpha \in \Phi$, one has $2\alpha \notin \Phi$. [/step]