THE GEOMETRY OF THE EXCEPTIONAL LIE GROUPS

Subsection 5.3.1 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

and (5.3.3) now implies that \(Y_\alpha\in\gg_{-\alpha}\text{.}\) Thus, if \(\alpha\) is a root, so is \(-\alpha\text{,}\) and we have

Furthermore, since \(\alpha\ne0\text{,}\) there must be some \(H\in\hh\) such that \(\alpha(H)\ne0\text{.}\) Using (5.3.7) now leads to

which implies that

where the first equality is a definition.

We now claim that \(\alpha(H_\alpha)\ne0\text{.}\) If instead \(\alpha(H_\alpha)=0\text{,}\) then (5.3.2) and (5.3.10) imply that

so both \(X_\alpha\) and \(Y_\alpha\) commute with \(H_\alpha\text{.}\) Consider now any representation \(V\) of \(\gg\text{,}\) and suppose that the image of \(H_\alpha\text{,}\) which we will also call \(H_\alpha\text{,}\) has an eigenspace

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

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 \(\lambda=0\text{.}\) ¹ Thus, the only eigenvalue of \(H_\alpha\) is \(0\text{,}\) which means that \(\beta(H_\alpha)=0\) for any root \(\beta\text{,}\) which in turn forces \(H_\alpha=0\text{.}\) This contradiction establishes the claim, namely that \(\alpha(H_\alpha)\ne0\text{.}\)

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

thus demonstrating that the subalgebra generated by \(\{H_\alpha,X_\alpha,Y_\alpha\}\) is a copy of \(\sl(2,\RR)\text{.}\) 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 5.3.2 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

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

Using (5.3.11) and (5.3.12) we have

so that in particular

which is furthermore nonzero by assumption. Thus,

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

as claimed previously. It now follows immediately that

leading to the angles and ratios discussed in Section 5.2.

Subsection 5.3.3 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

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,

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

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

where

But

so that

Putting this all together, we have

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 \(|\gg_\alpha|=1\) (and also that \(|\hh_\alpha|=1\)).

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

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 \(c\text{.}\)