[proofplan] We use the $\mathfrak{sl}_2$-subalgebra associated to the root $\alpha$ and let it act on the direct sum of the root spaces occurring in the $\alpha$-string through $\beta$. The element $\alpha^\vee$ acts diagonally on these root spaces, with eigenvalues $\langle \beta,\alpha^\vee\rangle+2n$. Since this direct sum is invariant under the raising and lowering operators, $\alpha^\vee$ acts as a commutator on a finite-dimensional [vector space](/page/Vector%20Space), hence has trace zero. Computing the same trace from the root-space decomposition gives the desired identity. [/proofplan]

[step:Build the finite-dimensional string module preserved by the $\alpha$-operators]For each root $\gamma \in \Phi$, define the root space \begin{align*} \mathfrak g_\gamma := \{x \in \mathfrak g : [h,x] = \gamma(h)x \text{ for every } h \in \mathfrak h\}. \end{align*} Choose nonzero elements $e_\alpha \in \mathfrak g_\alpha$ and $f_\alpha \in \mathfrak g_{-\alpha}$ normalized so that \begin{align*} [e_\alpha,f_\alpha]=\alpha^\vee. \end{align*} Define the finite-dimensional complex vector space \begin{align*} M := \bigoplus_{n=-p}^{q} \mathfrak g_{\beta+n\alpha}. \end{align*} We claim that $M$ is preserved by the three endomorphisms \begin{align*} E:M &\to M, & E(x)&:=[e_\alpha,x],\\ F:M &\to M, & F(x)&:=[f_\alpha,x],\\ H:M &\to M, & H(x)&:=[\alpha^\vee,x]. \end{align*} Indeed, for $x \in \mathfrak g_{\beta+n\alpha}$ and $h \in \mathfrak h$, the Jacobi identity gives \begin{align*} [h,[e_\alpha,x]] &= [[h,e_\alpha],x]+[e_\alpha,[h,x]]\\ &= \alpha(h)[e_\alpha,x]+(\beta+n\alpha)(h)[e_\alpha,x]\\ &= (\beta+(n+1)\alpha)(h)[e_\alpha,x]. \end{align*} Thus $[e_\alpha,x]$ lies in $\mathfrak g_{\beta+(n+1)\alpha}$ if $\beta+(n+1)\alpha \in \Phi$, and is $0$ otherwise. Since the string has no roots beyond $\beta+q\alpha$, this shows $E(M)\subset M$. The same calculation with $f_\alpha \in \mathfrak g_{-\alpha}$ gives $F(M)\subset M$. Finally, since $\alpha^\vee \in \mathfrak h$, each root space $\mathfrak g_{\beta+n\alpha}$ is preserved by $H$, so $H(M)\subset M$.[/step]

[guided]The purpose of $M$ is to isolate exactly the root spaces lying on the line through $\beta$ in the $\alpha$-direction. We define \begin{align*} M := \bigoplus_{n=-p}^{q} \mathfrak g_{\beta+n\alpha}, \end{align*} where \begin{align*} \mathfrak g_\gamma := \{x \in \mathfrak g : [h,x] = \gamma(h)x \text{ for every } h \in \mathfrak h\} \end{align*} is the root space of a root $\gamma \in \Phi$. Choose nonzero root vectors $e_\alpha \in \mathfrak g_\alpha$ and $f_\alpha \in \mathfrak g_{-\alpha}$ normalized by \begin{align*} [e_\alpha,f_\alpha]=\alpha^\vee. \end{align*} We need to check that bracketing with these elements does not leave $M$. Let $x \in \mathfrak g_{\beta+n\alpha}$. For every $h \in \mathfrak h$, the Jacobi identity gives \begin{align*} [h,[e_\alpha,x]] &= [[h,e_\alpha],x]+[e_\alpha,[h,x]]\\ &= \alpha(h)[e_\alpha,x]+(\beta+n\alpha)(h)[e_\alpha,x]\\ &= (\beta+(n+1)\alpha)(h)[e_\alpha,x]. \end{align*} This means that $[e_\alpha,x]$ is either $0$ or a vector in the root space $\mathfrak g_{\beta+(n+1)\alpha}$. Since the displayed string contains all roots of the form $\beta+m\alpha$, there is no root past $\beta+q\alpha$, so the bracket is $0$ at the right endpoint and otherwise remains inside $M$. The same argument with $f_\alpha \in \mathfrak g_{-\alpha}$ gives \begin{align*} [h,[f_\alpha,x]] = (\beta+(n-1)\alpha)(h)[f_\alpha,x], \end{align*} so $[f_\alpha,x]$ lies in $\mathfrak g_{\beta+(n-1)\alpha}$ or is $0$. It is therefore also inside $M$. Finally, $\alpha^\vee \in \mathfrak h$, so by the definition of root spaces the map $x \mapsto [\alpha^\vee,x]$ preserves every summand of $M$. Thus $M$ is invariant under the three operators $E,F,H$.[/guided]

[step:Express the Cartan action as a commutator on the string module]Because $M$ is invariant under $E$ and $F$, their commutator is an endomorphism of $M$. For $x \in M$, \begin{align*} (EF-FE)(x) &= [e_\alpha,[f_\alpha,x]]-[f_\alpha,[e_\alpha,x]]\\ &= [[e_\alpha,f_\alpha],x]\\ &= [\alpha^\vee,x]\\ &= H(x), \end{align*} where the second equality is the Jacobi identity. Hence \begin{align*} H = EF-FE \end{align*} as endomorphisms of the finite-dimensional vector space $M$. Taking traces and using $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for endomorphisms of a finite-dimensional vector space, we obtain \begin{align*} \operatorname{tr}(H) = \operatorname{tr}(EF)-\operatorname{tr}(FE) = 0. \end{align*}[/step]

[guided]The reason for introducing the $\mathfrak{sl}_2$-operators is that the Cartan element $\alpha^\vee$ is their commutator. Since $E$ and $F$ are endomorphisms of $M$, the compositions $EF$ and $FE$ are also endomorphisms of $M$. For every $x \in M$, the Jacobi identity gives \begin{align*} [e_\alpha,[f_\alpha,x]]-[f_\alpha,[e_\alpha,x]] = [[e_\alpha,f_\alpha],x]. \end{align*} By our normalization $[e_\alpha,f_\alpha]=\alpha^\vee$, so \begin{align*} (EF-FE)(x) = [\alpha^\vee,x]=H(x). \end{align*} Thus \begin{align*} H=EF-FE. \end{align*} Now we use a linear-algebra fact: on a finite-dimensional vector space, the trace of a commutator is zero because $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. Applying this to $E$ and $F$ gives \begin{align*} \operatorname{tr}(H) = \operatorname{tr}(EF-FE) = \operatorname{tr}(EF)-\operatorname{tr}(FE) = 0. \end{align*} This is the key global constraint on the string: the sum of all $\alpha^\vee$-weights occurring in $M$ must vanish.[/guided]

[step:Compute the same trace from the root-space decomposition]For each integer $n$ with $-p \le n \le q$, the operator $H$ acts on $\mathfrak g_{\beta+n\alpha}$ by the scalar \begin{align*} (\beta+n\alpha)(\alpha^\vee) = \beta(\alpha^\vee)+n\alpha(\alpha^\vee) = \langle \beta,\alpha^\vee\rangle+2n. \end{align*} Since $\mathfrak g$ is complex semisimple, each root space $\mathfrak g_{\beta+n\alpha}$ is one-dimensional. Therefore \begin{align*} \operatorname{tr}(H) &= \sum_{n=-p}^{q}\left(\langle \beta,\alpha^\vee\rangle+2n\right)\\ &= (p+q+1)\langle \beta,\alpha^\vee\rangle +2\sum_{n=-p}^{q} n. \end{align*} The finite arithmetic sum is \begin{align*} \sum_{n=-p}^{q} n = -\sum_{m=1}^{p}m+\sum_{m=1}^{q}m = -\frac{p(p+1)}{2}+\frac{q(q+1)}{2}. \end{align*} Hence \begin{align*} \operatorname{tr}(H) &= (p+q+1)\langle \beta,\alpha^\vee\rangle -p(p+1)+q(q+1)\\ &= (p+q+1)\langle \beta,\alpha^\vee\rangle +(q-p)(p+q+1)\\ &= (p+q+1)\left(\langle \beta,\alpha^\vee\rangle+q-p\right). \end{align*}[/step]

[guided]We now compute $\operatorname{tr}(H)$ using the direct-sum decomposition of $M$. Fix an integer $n$ with $-p \le n \le q$. If $x \in \mathfrak g_{\beta+n\alpha}$, then the definition of the root space gives \begin{align*} H(x) = [\alpha^\vee,x] = (\beta+n\alpha)(\alpha^\vee)x. \end{align*} Since $\alpha(\alpha^\vee)=2$, this scalar is \begin{align*} (\beta+n\alpha)(\alpha^\vee) = \beta(\alpha^\vee)+n\alpha(\alpha^\vee) = \langle \beta,\alpha^\vee\rangle+2n. \end{align*} For a finite-dimensional complex semisimple Lie algebra, every root space is one-dimensional. Therefore each summand $\mathfrak g_{\beta+n\alpha}$ contributes exactly the scalar $\langle \beta,\alpha^\vee\rangle+2n$ to the trace. Summing over all roots in the string gives \begin{align*} \operatorname{tr}(H) &= \sum_{n=-p}^{q}\left(\langle \beta,\alpha^\vee\rangle+2n\right)\\ &= (p+q+1)\langle \beta,\alpha^\vee\rangle +2\sum_{n=-p}^{q} n. \end{align*} The remaining sum is an elementary arithmetic sum: \begin{align*} \sum_{n=-p}^{q} n = -\sum_{m=1}^{p}m+\sum_{m=1}^{q}m = -\frac{p(p+1)}{2}+\frac{q(q+1)}{2}. \end{align*} Substituting this into the trace formula gives \begin{align*} \operatorname{tr}(H) &= (p+q+1)\langle \beta,\alpha^\vee\rangle -p(p+1)+q(q+1)\\ &= (p+q+1)\langle \beta,\alpha^\vee\rangle +(q-p)(p+q+1)\\ &= (p+q+1)\left(\langle \beta,\alpha^\vee\rangle+q-p\right). \end{align*} This is the same trace as before, now expressed entirely in terms of the two endpoints of the string.[/guided]

[step:Equate the two trace computations to obtain the string formula] From the commutator computation, \begin{align*} \operatorname{tr}(H)=0. \end{align*} From the root-space computation, \begin{align*} \operatorname{tr}(H) = (p+q+1)\left(\langle \beta,\alpha^\vee\rangle+q-p\right). \end{align*} Since $p,q \in \mathbb N \cup \{0\}$, we have $p+q+1>0$. Therefore \begin{align*} \langle \beta,\alpha^\vee\rangle+q-p=0, \end{align*} and hence \begin{align*} \langle \beta,\alpha^\vee\rangle=p-q. \end{align*} This is the desired root string identity. [/step]