Since $B$ is nondegenerate, for any $X_\alpha\in\gg_\alpha$, there must be some $Y_\alpha\in\gg$ such that \begin{equation} B(X_\alpha,Y_\alpha) \ne 0 , \end{equation} and (\ref{ortho}) now implies that $Y_\alpha\in\gg_{-\alpha}$. Thus, if $\alpha$ is a root, so is $-\alpha$, and we have \begin{equation} [H,Y_\alpha] = -\alpha(H) Y_\alpha . \end{equation} Furthermore, since $\alpha\ne0$, there must be some $H\in\hh$ such that, using (\ref{Btrace}, \begin{equation} B([X_\alpha,Y_\alpha],H) = B(X_\alpha,[Y_\alpha,H]) = B(X_\alpha,\alpha(H)Y) = \alpha(H) B(X_\alpha,Y_\alpha) \ne 0 \label{Bid} \end{equation} so that \begin{equation} 2 H_\alpha = [X_\alpha,Y_\alpha] \ne 0 . \label{Hdef} \end{equation}

We now claim that $\alpha(H_\alpha)\ne0$. If not, then \begin{equation} [H_\alpha,X_\alpha] = 0 = [H_\alpha,Y_\alpha] . \end{equation} Consider now any representation of $\hh$, and suppose that the image of $H_\alpha$, which we will also call $H_\alpha$, has an eigenspace \begin{equation} V = \{v: H_\alpha v = \lambda v\} \end{equation} for some fixed eigenvalue $\lambda$. Since $X_\alpha$ and $Y_\alpha$ commute with $H_\alpha$, they both take eigenvectors to eigenvectors, that is, they take $V$ to itself. So $V$ itself is a representation of the subalgebra generated by $\{H_\alpha,X_\alpha,Y_\alpha\}$. But this means that, as matrices acting on $V$, we must have \begin{equation} [X_\alpha,Y_\alpha] = 2 H_\alpha = 2 \lambda \end{equation} since representations preserve commutators (by definition), and $H_\alpha$ is a multiple of the identity matrix when acting on $V$. Taking the trace of both sides immediately forces $\lambda=0$. Thus, the only eigenvalue of $H_\alpha$ is $0$, which means that $H_\alpha$ is nilpotent. But the only nilpotent element of $\hh$ is $0$, and $H_\alpha\ne0$. This contradiction establishes the claim.

We can now rescale $X_\alpha$ and $Y_\alpha$ if necessary to obtain \[ \alpha(H_\alpha) = 1 . \] Although this construction does not determine $X_\alpha$ and $Y_\alpha$ uniquely, $H_\alpha$ is uniquely determined. These special elements of $\hh$ will be referred to as preferred Cartan elements.

As discussed in the previous section, $\{H_\alpha,X_\alpha,Y_\alpha\}$ form a standard basis for $\sl(2,\RR)$, the split real form of $\su(2)$. Thus, all representations of this Lie subalgebra of $\gg$ have half-integer eigenvalues, and in particular, $\beta(H_\alpha)\in\frac12\ZZ$ for all roots $\beta$. The restriction of $\gg$ to real linear combinations of $H_\alpha$, $X_\alpha$, $Y_\alpha$ for all $\alpha\in R$ is therefore a real subalgebra of $\gg$, and is in fact the split real form of $\gg$.

As before, choose $T_\alpha\in\hh$ to be the element determined by \begin{equation} \alpha(H) = B(T_\alpha,H) \end{equation} for all $H\in\hh$. We need to verify that $T_\alpha$ is a multiple of $H_\alpha$.

Using (\ref{Bid}) and (\ref{Hdef}, we have \begin{equation} B(2H_\alpha,H) = \alpha(H) B(X_\alpha,Y_\alpha) \end{equation} so that in particular \begin{equation} B(2H_\alpha,H_\alpha) = \alpha(H_\alpha) B(X_\alpha,Y_\alpha) = B(X_\alpha,Y_\alpha) \end{equation} which is furthermore nonzero by assumption. Thus, \begin{equation} B(H_\alpha,H) = B(T_\alpha,H) B(H_\alpha,H_\alpha) = B\bigl(B(H_\alpha,H_\alpha)T_\alpha,H\bigr) \end{equation} for all $H\in\hh$, and we conclude that \[ T_\alpha = \frac{H_\alpha}{B(H_\alpha,H_\alpha)} . \] as claimed previously. It now follows immediately that \[ \alpha(H_\beta) = B(T_\alpha,H_\beta) = \frac{B(H_\alpha,H_\beta)}{B(H_\alpha,H_\alpha)} \in\frac12\ZZ \] leading to the angles and ratios discussed in the previous section.

It remains to show that the only multiples of a root $\alpha$ that are roots are $\pm\alpha$, and that each root only occurs once, that is, that $|\gg_\alpha|=1$.

We have shown that $\{H_\alpha,X_\alpha,Y_\alpha\}$ is a standard basis for $\sl(2,\RR)$, so that $X_\alpha$, $Y_\alpha$ act as raising and lowering operators, respectively, for $H_\alpha$. Suppose that $Z\in\gg_\alpha$, so that \begin{equation} [H_\alpha,Z] = \alpha(H_\alpha) Z = Z . \end{equation} We know that all representations of $\sl(2,\RR)$ consist of integer or half-integer “ladders”. So how does $\sl(2,\RR)$ act on $Z$? Moving down the ladder, \begin{equation} Z_0 = [Y_\alpha,Z] \end{equation} is an element of $\hh$, since $Y_\alpha$ decreases all the eigenvalues of $Z$ by $\alpha$, that is \begin{equation} [H,Z_0] = \bigl[H,[Y_\alpha,Z]\bigr] = \bigl[[H,Y_\alpha],Z\bigr] + = \bigl[Y_\alpha,[H,Z]\bigr] = (-\alpha+\alpha) [Y_\alpha,Z] = 0 \end{equation} for any $H\in\hh$. Since $B$ is positive-definite on $\hh$, we can expand $Z_0$ as \begin{equation} Z_0 = z H_\alpha + H_\perp \end{equation} where \begin{equation} B(H_\alpha,H_\perp) = 0 . \end{equation} But \begin{equation} B(H_\alpha,H_\perp) = 0 \Longrightarrow B(T_\alpha,H_\perp) = 0 \Longrightarrow \alpha(H_\perp) = 0 \end{equation} so that \begin{equation} \alpha(Z_0) = z\alpha(H_\alpha)+0 = z. \end{equation} Putting this all together, we have \begin{equation} [X_\alpha,Z_0] = -[Z_0,X_\alpha] = -\alpha(Z_0) X_\alpha = -z X_\alpha . \end{equation} But moving back up the ladder has to yield a multiple of $Z$, and we conclude that $Z$ is itself a multiple of $X_\alpha$, thus confirming that $|\gg_\alpha|=1$.

The argument against any other multiples of $\alpha$ other than $\pm\alpha$ being roots is similar. If $c\alpha$ is a root, then choosing $Z\in\gg_{c\alpha}$ leads to \begin{equation} [H_\alpha,Z] = (c\alpha)(H_\alpha) Z = c Z \end{equation} which forces $c$ to be half-integer. Suppose $c\in\ZZ$. Then we can move up or down the ladder from $Z$ to some element $Z_1\in\gg_\alpha$. By the argument above, $Z_1$ must be a multiple of $X_\alpha$. But then the ladder collapses, and $c=\pm1$. In particular, $2\alpha$ can not be a root. But now $\frac12\alpha$ also can not be a root, which rules out half-integer values of $c$.