Properties of Roots

Section 1.13 Properties of Roots

In this section, we fill in the missing details when deriving the properties of the roots of a simple Lie algebra \(\gg\text{.}\)

We assume that a Cartan algebra \(\hh\subset\gg\) of simultaneously diagonalizable elements has been chosen, and that \(\gg\) has been decomposed into eigenspaces

\begin{equation} \gg = \hh \mathop{\oplus}_{\alpha\in R} \gg_\alpha ,\tag{1.13.1} \end{equation}

so that, for any \(H\in\hh\text{,}\) we have

\begin{equation} [H,X_\alpha] = \alpha(H) X_\alpha\tag{1.13.2} \end{equation}

for a finite set of roots \(R\subset\hh^*\text{.}\) Since \(\gg\) is simple, the Killing form \(B\) on \(\gg\) is nondegenerate. As argued in the previous section,

\begin{equation} B(\gg_\alpha,\gg_\beta) = 0 = B(\gg_\alpha,\hh)\label{ortho}\tag{1.13.3} \end{equation}

for \(\alpha+\beta\ne0\) (and \(\alpha\ne0\text{,}\) since \(0\not\in R\)). Recall further that the Jordan–Chevalley decomposition shows that any linear operator can be divided uniquely into the sum of a diagonalizable operator and a nilpotent operator. Thus, nonzero elements of \(\hh\) can not be nilpotent.

We begin with a simple lemma about traces. Since

\begin{equation} \tr(XY) = \tr(YX) ,\tag{1.13.4} \end{equation}

the trace is cyclic, so that, for instance,

\begin{equation} \tr(XYZ) = \tr(ZXY) .\tag{1.13.5} \end{equation}

Thus,

\begin{equation} \tr([X,Y]Z) = \tr(XYZ-YXZ) = \tr(XYZ-XZY) = \tr(X[Y,Z])\tag{1.13.6} \end{equation}

which implies that

\begin{equation} B([X,Y],Z) = B(X,[Y,Z]) .\label{Btrace}\tag{1.13.7} \end{equation}

Subsection 1.13.1 Preferred Cartan Elements

Since \(B\) is nondegenerate, for any \(X_\alpha\in\gg_\alpha\text{,}\) there must be some \(Y_\alpha\in\gg\) such that

\begin{equation} B(X_\alpha,Y_\alpha) \ne 0 ,\tag{1.13.8} \end{equation}

and (1.13.3) now implies that \(Y_\alpha\in\gg_{-\alpha}\text{.}\) Thus, if \(\alpha\) is a root, so is \(-\alpha\text{,}\) and we have

\begin{equation} [H,Y_\alpha] = -\alpha(H) Y_\alpha .\tag{1.13.9} \end{equation}

Furthermore, since \(\alpha\ne0\text{,}\) there must be some \(H\in\hh\) such that, using (1.13.7)

\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}\tag{1.13.10} \end{equation}

so that

\begin{equation} 2 H_\alpha = [X_\alpha,Y_\alpha] \ne 0 .\label{Hdef}\tag{1.13.11} \end{equation}

We now claim that \(\alpha(H_\alpha)\ne0\text{.}\) If not, then

\begin{equation} [H_\alpha,X_\alpha] = 0 = [H_\alpha,Y_\alpha] .\tag{1.13.12} \end{equation}

Consider now any representation of \(\hh\text{,}\) and suppose that the image of \(H_\alpha\text{,}\) which we will also call \(H_\alpha\text{,}\) has an eigenspace

\begin{equation} V = \{v: H_\alpha v = \lambda v\}\tag{1.13.13} \end{equation}

for some fixed eigenvalue \(\lambda\text{.}\) Since \(X_\alpha\) and \(Y_\alpha\) commute with \(H_\alpha\text{,}\) they both take eigenvectors to eigenvectors, that is, they take \(V\) to itself. So \(V\) itself is a representation of the subalgebra generated by \(\{H_\alpha,X_\alpha,Y_\alpha\}\text{.}\) But this means that, as matrices acting on \(V\text{,}\) we must have

\begin{equation} [X_\alpha,Y_\alpha] = 2 H_\alpha = 2 \lambda\tag{1.13.14} \end{equation}

since representations preserve commutators (by definition), and \(H_\alpha\) is a multiple of the identity matrix when acting on \(V\text{.}\) Taking the trace of both sides immediately forces \(\lambda=0\text{.}\) Thus, the only eigenvalue of \(H_\alpha\) is \(0\text{,}\) which means that \(H_\alpha\) is nilpotent. But the only nilpotent element of \(\hh\) is \(0\text{,}\) and \(H_\alpha\ne0\text{.}\) This contradiction establishes the claim.

We can now rescale \(X_\alpha\) and \(Y_\alpha\) if necessary to obtain

\begin{equation} \alpha(H_\alpha) = 1 .\tag{1.13.15} \end{equation}

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.

Subsection 1.13.2 Root Angles

As discussed in the previous section, \(\{H_\alpha,X_\alpha,Y_\alpha\}\) form a standard basis for \(\sl(2,\RR)\text{,}\) the split real form of \(\su(2)\text{.}\) 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 \(\beta\text{.}\) The restriction of \(\gg\) to real linear combinations of \(H_\alpha\text{,}\) \(X_\alpha\text{,}\) \(Y_\alpha\) for all \(\alpha\in R\) is therefore a real subalgebra of \(\gg\text{,}\) and is in fact the split real form of \(\gg\text{.}\)

As before, choose \(T_\alpha\in\hh\) to be the element determined by

\begin{equation} \alpha(H) = B(T_\alpha,H)\tag{1.13.16} \end{equation}

for all \(H\in\hh\text{.}\) We need to verify that \(T_\alpha\) is a multiple of \(H_\alpha\text{.}\)

Using (1.13.10) and (1.13.11) we have

\begin{equation} B(2H_\alpha,H) = \alpha(H) B(X_\alpha,Y_\alpha)\tag{1.13.17} \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)\tag{1.13.18} \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)\tag{1.13.19} \end{equation}

for all \(H\in\hh\text{,}\) and we conclude that

\begin{equation} T_\alpha = \frac{H_\alpha}{B(H_\alpha,H_\alpha)} .\tag{1.13.20} \end{equation}

as claimed previously. It now follows immediately that

\begin{equation} \alpha(H_\beta) = B(T_\alpha,H_\beta) = \frac{B(H_\alpha,H_\beta)}{B(H_\alpha,H_\alpha)} \in\frac12\ZZ\tag{1.13.21} \end{equation}

leading to the angles and ratios discussed in the previous section.

Subsection 1.13.3 Multiples of Roots, and Multiplicity

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

We have shown that \(\{H_\alpha,X_\alpha,Y_\alpha\}\) is a standard basis for \(\sl(2,\RR)\text{,}\) so that \(X_\alpha\text{,}\) \(Y_\alpha\) act as raising and lowering operators, respectively, for \(H_\alpha\text{.}\) Suppose that \(Z\in\gg_\alpha\text{,}\) so that

\begin{equation} [H_\alpha,Z] = \alpha(H_\alpha) Z = Z .\tag{1.13.22} \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 \(Z\text{?}\) Moving down the ladder,

\begin{equation} Z_0 = [Y_\alpha,Z]\tag{1.13.23} \end{equation}

is an element of \(\hh\text{,}\) since \(Y_\alpha\) decreases all the eigenvalues of \(Z\) by \(\alpha\text{,}\) 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\tag{1.13.24} \end{equation}

for any \(H\in\hh\text{.}\) Since \(B\) is positive-definite on \(\hh\text{,}\) we can expand \(Z_0\) as

\begin{equation} Z_0 = z H_\alpha + H_\perp\tag{1.13.25} \end{equation}

where

\begin{equation} B(H_\alpha,H_\perp) = 0 .\tag{1.13.26} \end{equation}

But

\begin{equation} B(H_\alpha,H_\perp) = 0 \Longrightarrow B(T_\alpha,H_\perp) = 0 \Longrightarrow \alpha(H_\perp) = 0\tag{1.13.27} \end{equation}

so that

\begin{equation} \alpha(Z_0) = z\alpha(H_\alpha)+0 = z.\tag{1.13.28} \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 .\tag{1.13.29} \end{equation}

But moving back up the ladder has to yield a multiple of \(Z\text{,}\) and we conclude that \(Z\) is itself a multiple of \(X_\alpha\text{,}\) thus confirming that \(|\gg_\alpha|=1\text{.}\)

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\tag{1.13.30} \end{equation}

which forces \(c\) to be half-integer. Suppose \(c\in\ZZ\text{.}\) Then we can move up or down the ladder from \(Z\) to some element \(Z_1\in\gg_\alpha\text{.}\) By the argument above, \(Z_1\) must be a multiple of \(X_\alpha\text{.}\) But then the ladder collapses, and \(c=\pm1\text{.}\) 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 \(c\text{.}\)