[proofplan] We first record the root-space decomposition of the complexified [Lie algebra](/page/Lie%20Algebra) under the adjoint action of the maximal torus and use [finite dimensionality](/theorems/1534) to obtain finiteness. An invariant [inner product](/page/Inner%20Product) on $\mathfrak g$ identifies the real roots with vectors in a Euclidean space, so the usual root reflections are defined. Reflection invariance is supplied by the rank-one Weyl reflection theorem, while integrality and reducedness come from the root-string theorem. The only delicate point is reducedness: if $2\alpha$ were a root, the $\alpha$-string through $2\alpha$ would be forced to pass through $0$, contradicting the convention that roots are non-zero weights. [/proofplan] [step:Decompose the complexified Lie algebra into toral and root spaces] Let $GL(\mathfrak g_{\mathbb C})$ denote the group of complex-linear automorphisms of $\mathfrak g_{\mathbb C}$. Let $\operatorname{Ad}|_T:T\to GL(\mathfrak g_{\mathbb C})$ denote the restriction of the complexified adjoint representation to $T$. Since $T$ is a compact torus and $\mathfrak g_{\mathbb C}$ is finite-dimensional, the [[Weight Space Decomposition for Finite-Dimensional Representations of Compact Tori](/theorems/9726)][citetheorem:9726] applies to this representation and gives \begin{align*} \mathfrak g_{\mathbb C}=\bigoplus_{\lambda\in X^*(T)}\mathfrak g_\lambda. \end{align*} The zero-weight space is \begin{align*} \mathfrak g_1=\{X\in\mathfrak g_{\mathbb C}:\operatorname{Ad}(t)X=X\text{ for every }t\in T\}. \end{align*} Define the centralizer of $T$ in $G$ by \begin{align*} C_G(T):=\{g\in G:gt=tg\text{ for every }t\in T\}. \end{align*} The real part of $\mathfrak g_1$ is the Lie algebra of $C_G(T)$. Since $T$ is a maximal torus in the compact connected Lie group $G$, [citetheorem:9722] gives $C_G(T)=T$, and therefore \begin{align*} \mathfrak g_1=\mathfrak t_{\mathbb C}. \end{align*} Thus \begin{align*} \mathfrak g_{\mathbb C}=\mathfrak t_{\mathbb C}\oplus\bigoplus_{\lambda\in R}\mathfrak g_\lambda. \end{align*} Because $\mathfrak g_{\mathbb C}$ is finite-dimensional, only finitely many weight spaces can be non-zero. Hence $R$ and $\Phi$ are finite. By definition, $E=\operatorname{span}_{\mathbb R}(\Phi)$, so $\Phi$ spans its ambient real [vector space](/page/Vector%20Space). [/step] [step:Choose an invariant Euclidean structure and define the root reflections] Let $B_0:\mathfrak g\times\mathfrak g\to\mathbb R$ be any real inner product. Let $\mu_G$ denote the normalized Haar probability measure on $G$. Define \begin{align*} B:\mathfrak g\times\mathfrak g\to\mathbb R,\qquad B(X,Y):=\int_G B_0(\operatorname{Ad}(g)X,\operatorname{Ad}(g)Y)\,d\mu_G(g). \end{align*} The integrand is continuous because $\operatorname{Ad}:G\to GL(\mathfrak g)$ is smooth, so the integral is well-defined. Positivity of $B$ follows from positivity of $B_0$ and invertibility of each $\operatorname{Ad}(g)$. Right-invariance of $\mu_G$ gives, for every $h\in G$, \begin{align*} B(\operatorname{Ad}(h)X,\operatorname{Ad}(h)Y)=B(X,Y). \end{align*} Thus $B$ is an $\operatorname{Ad}(G)$-invariant inner product on $\mathfrak g$. Restrict $B$ to $\mathfrak t$ and let $(\cdot,\cdot)$ denote the induced dual inner product on $\mathfrak t^*$. Its restriction to $E\subset\mathfrak t^*$ is positive definite. For each $\alpha\in\Phi$, define the [linear map](/page/Linear%20Map) \begin{align*} s_\alpha:E\to E,\qquad s_\alpha(\beta):=\beta-\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\alpha. \end{align*} Since $\alpha\ne 0$, the denominator $(\alpha,\alpha)$ is positive, so $s_\alpha$ is well-defined. [/step] [step:Use Weyl group elements to prove reflection invariance] Let $\alpha\in\Phi$. Choose $\lambda_\alpha\in R$ such that $\alpha=\alpha_{\lambda_\alpha}$. Define the normalizer of $T$ in $G$ by \begin{align*} N_G(T):=\{g\in G:gTg^{-1}=T\}. \end{align*} The theorem [citetheorem:9727] applies because $G$ is compact and connected, $T$ is a maximal torus, $\lambda_\alpha$ is a root character of the complexified adjoint representation, and our real root is normalized as $\alpha_{\lambda_\alpha}=-i\,d(\lambda_\alpha)_e$ with the Euclidean structure induced from an $\operatorname{Ad}(G)$-invariant inner product. It states that the root reflection attached to $\alpha$ is induced by an element of the Weyl group $N_G(T)/T$. Hence there exists $n\in N_G(T)$ such that the action of conjugation by $n$ on $E$ is $s_\alpha$. For every root character $\lambda\in R$, the conjugated weight space is \begin{align*} \operatorname{Ad}(n)(\mathfrak g_\lambda)=\mathfrak g_{n\lambda n^{-1}}, \end{align*} where the character $n\lambda n^{-1}:T\to S^1$ is defined by \begin{align*} (n\lambda n^{-1})(t):=\lambda(n^{-1}tn). \end{align*} Since $\operatorname{Ad}(n)$ is an invertible complex-linear map on $\mathfrak g_{\mathbb C}$, the target weight space is non-zero whenever $\mathfrak g_\lambda$ is non-zero. To identify the induced real root, let $\operatorname{Ad}(n)^{-1}|_{\mathfrak t}:\mathfrak t\to\mathfrak t$ denote the restriction of $\operatorname{Ad}(n)^{-1}$ to $\mathfrak t$, which is well-defined because $n\in N_G(T)$. For $H\in\mathfrak t$, \begin{align*} d(n\lambda n^{-1})_e(H)=d\lambda_e(\operatorname{Ad}(n)^{-1}H). \end{align*} Multiplying by $-i$ gives \begin{align*} \alpha_{n\lambda n^{-1}}(H)=\alpha_\lambda(\operatorname{Ad}(n)^{-1}H). \end{align*} Thus the Weyl [group action](/page/Group%20Action) preserves $R$, and under the real differential identification it preserves $\Phi$. Thus \begin{align*} s_\alpha(\Phi)=\Phi. \end{align*} [guided] Fix a root $\alpha\in\Phi$, and choose $\lambda_\alpha\in R$ with $\alpha=\alpha_{\lambda_\alpha}$. We need to prove that reflecting every root across the hyperplane perpendicular to $\alpha$ again gives roots. Define \begin{align*} N_G(T):=\{g\in G:gTg^{-1}=T\} \end{align*} to be the normalizer of $T$ in $G$. The rank-one structure theorem for compact Lie groups is exactly designed for this: [citetheorem:9727] applies because $G$ is compact and connected, $T$ is a maximal torus, $\lambda_\alpha$ is a root character of the complexified adjoint representation, and our real root is the functional $\alpha_{\lambda_\alpha}=-i\,d(\lambda_\alpha)_e$ equipped with the dual inner product induced by an $\operatorname{Ad}(G)$-invariant inner product. The conclusion is that the reflection \begin{align*} s_\alpha:E\to E,\qquad s_\alpha(\beta)=\beta-\frac{2(\beta,\alpha)}{(\alpha,\alpha)}\alpha \end{align*} is realized by conjugation by some element $n\in N_G(T)$. Now let $\lambda\in R$ be a root character and let $X\in\mathfrak g_\lambda$ be non-zero. Define the conjugated character $n\lambda n^{-1}:T\to S^1$ by \begin{align*} (n\lambda n^{-1})(t)=\lambda(n^{-1}tn). \end{align*} For $t\in T$, using $n^{-1}tn\in T$ because $n\in N_G(T)$, we compute \begin{align*} \operatorname{Ad}(t)\operatorname{Ad}(n)X=\operatorname{Ad}(n)\operatorname{Ad}(n^{-1}tn)X. \end{align*} Since $X\in\mathfrak g_\lambda$, the last expression equals \begin{align*} \lambda(n^{-1}tn)\operatorname{Ad}(n)X=(n\lambda n^{-1})(t)\operatorname{Ad}(n)X. \end{align*} Thus $\operatorname{Ad}(n)X$ lies in $\mathfrak g_{n\lambda n^{-1}}$. Because $\operatorname{Ad}(n)$ is invertible and $X\ne 0$, this vector is non-zero, so $n\lambda n^{-1}$ is again a root character. We now compute the differential action explicitly. Since $n\in N_G(T)$, the linear map $\operatorname{Ad}(n)^{-1}|_{\mathfrak t}:\mathfrak t\to\mathfrak t$ is well-defined. For $H\in\mathfrak t$, \begin{align*} d(n\lambda n^{-1})_e(H)=d\lambda_e(\operatorname{Ad}(n)^{-1}H). \end{align*} Multiplying both sides by $-i$ gives \begin{align*} \alpha_{n\lambda n^{-1}}(H)=\alpha_\lambda(\operatorname{Ad}(n)^{-1}H). \end{align*} Thus passing to real differentials is exactly the Weyl group action on real roots. Since the chosen element $n$ acts as $s_\alpha$ on $E$, this gives $s_\alpha(\Phi)\subset\Phi$. Applying the same argument to $n^{-1}$ gives the reverse inclusion, so $s_\alpha(\Phi)=\Phi$. [/guided] [/step] [step:Apply the root-string theorem to obtain crystallographic integrality] We use the [[Root String Theorem](/theorems/4693)][citetheorem:9728] for the compact Lie algebra $\mathfrak g$ relative to the maximal toral subalgebra $\mathfrak t$. Its hypotheses apply because $G$ is compact and connected, $T$ is a maximal torus with Lie algebra $\mathfrak t$, the roots in $\Phi$ are precisely the real roots $\alpha_\lambda=-i\,d\lambda_e$ obtained from the non-zero weight spaces of the complexified adjoint representation, and the inner product on $E$ is induced from an $\operatorname{Ad}(G)$-invariant inner product. Therefore, for every $\alpha,\beta\in\Phi$, the set \begin{align*} \{\beta+k\alpha:k\in\mathbb Z\}\cap\Phi \end{align*} has the form \begin{align*} \{\beta-p\alpha,\beta-(p-1)\alpha,\dots,\beta+q\alpha\} \end{align*} for some integers $p,q\ge 0$, and these integers satisfy \begin{align*} p-q=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}. \end{align*} Therefore \begin{align*} \frac{2(\beta,\alpha)}{(\alpha,\alpha)}\in\mathbb Z \end{align*} for every $\alpha,\beta\in\Phi$. This is the crystallographic Cartan-integer condition. [/step] [step:Exclude nontrivial scalar multiples of a root] It remains to prove reducedness. First suppose that $\alpha\in\Phi$ and $2\alpha\in\Phi$. Apply the root-string theorem to the $\alpha$-string through $\beta:=2\alpha$. Since $\beta-\alpha=\alpha$ is a root, the downward length $p$ is at least $1$. The root-string formula gives \begin{align*} p-q=\frac{2(2\alpha,\alpha)}{(\alpha,\alpha)}=4. \end{align*} Since $q\ge 0$, this implies $p\ge 4$. Hence the string contains $\beta-2\alpha=0$. This contradicts the definition of $\Phi$, whose elements are non-zero roots. Therefore $2\alpha\notin\Phi$ for every $\alpha\in\Phi$. Now let $\alpha,\beta\in\Phi$ and suppose $\beta=c\alpha$ for some real number $c>0$. Applying the Cartan-integer condition to the ordered pairs $(\alpha,\beta)$ and $(\beta,\alpha)$ gives \begin{align*} 2c\in\mathbb Z \end{align*} and \begin{align*} \frac{2}{c}\in\mathbb Z. \end{align*} Thus $c\in\{1/2,1,2\}$. The cases $c=2$ and $c=1/2$ would say that one of $\alpha,\beta$ is twice the other, which has just been excluded. Hence $c=1$. Allowing negative scalar multiples, the only roots on the line $\mathbb R\alpha$ are $\alpha$ and $-\alpha$. Therefore $\Phi$ is reduced. [guided] Reducedness means that a root line contains no roots except the two opposite roots. The main obstruction to exclude is the possible pair $\alpha,2\alpha$, because the integrality condition will reduce all other scalar-multiple possibilities to that one. Assume first that $\alpha\in\Phi$ and $2\alpha\in\Phi$. We apply the [Root String Theorem][citetheorem:9728] to the string in the $\alpha$-direction through the root $\beta:=2\alpha$. The theorem applies here for the same reason as in the integrality step: $\alpha$ and $\beta$ are real roots obtained from the compact torus weight decomposition of the complexified adjoint representation, and the inner product is the invariant one used in the theorem statement. It gives integers $p,q\ge 0$ such that the roots on this line through $\beta$ are exactly \begin{align*} \beta-p\alpha,\beta-(p-1)\alpha,\dots,\beta+q\alpha \end{align*} and such that \begin{align*} p-q=\frac{2(\beta,\alpha)}{(\alpha,\alpha)}. \end{align*} Substituting $\beta=2\alpha$ gives \begin{align*} p-q=\frac{2(2\alpha,\alpha)}{(\alpha,\alpha)}=4. \end{align*} Because $q\ge 0$, we get $p=q+4\ge 4$. The meaning of $p\ge 4$ is that the string contains all roots from $\beta-p\alpha$ up to $\beta+q\alpha$, in particular it contains $\beta-2\alpha$. But \begin{align*} \beta-2\alpha=2\alpha-2\alpha=0. \end{align*} This is impossible, since $\Phi$ was defined using only non-zero weights. Hence no root has its double as a root. Now suppose two roots lie on the same real line. Choose $\alpha,\beta\in\Phi$ and write $\beta=c\alpha$ with $c>0$. The Cartan-integer condition applied to $(\beta,\alpha)$ gives \begin{align*} \frac{2(\beta,\alpha)}{(\alpha,\alpha)}=2c\in\mathbb Z. \end{align*} The same condition applied to $(\alpha,\beta)$ gives \begin{align*} \frac{2(\alpha,\beta)}{(\beta,\beta)}=\frac{2c(\alpha,\alpha)}{c^2(\alpha,\alpha)}=\frac{2}{c}\in\mathbb Z. \end{align*} So both $2c$ and $2/c$ are positive integers. This forces $c\in\{1/2,1,2\}$. If $c=2$, then $\beta=2\alpha$, which has been excluded. If $c=1/2$, then $\alpha=2\beta$, also excluded. Therefore $c=1$. Including the negative direction, the only roots proportional to $\alpha$ are $\alpha$ and $-\alpha$. This is exactly the reducedness axiom. [/guided] [/step] [step:Assemble the root-system axioms] We have proved that $\Phi$ is finite, spans $E$, contains no zero vector, is invariant under every reflection $s_\alpha$ for $\alpha\in\Phi$, satisfies the Cartan-integrality condition \begin{align*} \frac{2(\beta,\alpha)}{(\alpha,\alpha)}\in\mathbb Z \end{align*} for all $\alpha,\beta\in\Phi$, and is reduced. These are precisely the axioms for a reduced crystallographic root system in the Euclidean vector space $E$. If $\Phi=\varnothing$, then $E=\operatorname{span}_{\mathbb R}(\varnothing)=\{0\}$, and the same verification is vacuous. Thus $\Phi$ is a reduced crystallographic root system in its real span. [/step]