THE GEOMETRY OF THE EXCEPTIONAL LIE GROUPS

Subsection 5.2.1 Cartan subalgebra

First of all, there is a maximal subalgebra generated by non-null commuting elements. For \(\su(2)\text{,}\) this subalgebra is generated by a single element; we chose \(\sigma_0\text{.}\) For \(\su(3)\text{,}\) this subalgebra has dimension two; we chose the Gell-Mann matrices \(\lambda_3\) and \(\lambda_8\) as generators.

A Cartan subalgebra \(\hh\) of a Lie algebra \(\gg\) is a maximal subalgebra of simultaneously diagonalizable elements. Equivalently, the Cartan subalgebra of a simple Lie algebra is a maximal subalgebra generated by non-null commuting elements. The dimension of the Cartan subalgebra is independent of which of the many possible subalgebras is chosen.

The requirement that the elements be diagonalizable (in the adjoint representation) is nontrivial, as can be seen by considering the element

of \(\su(3)\) (really \(\sl(3,\RR)\)), which is null—and not diagonalizable. In higher dimensions, the inadvertent use of such elements can cause one to miscount the size of the Cartan subalgebra. For instance, in \(\su(4)\text{,}\) it is possible to find four mutually commuting elements that are linearly independent, although all of them are null, and none are diagonalizable. At most three elements of \(\su(4)\) can be simultaneously diagonalized; its Cartan subalgebra is 3-dimensional.

For \(\su(2)\text{,}\) the Cartan subalgebra is given by all multiples of \(\sigma_0\text{,}\) and for \(\su(3)\) we have

where the angled brackets denote the span of the given elements.

Subsection 5.2.2 Basis of Eigenvectors

Since the elements of \(\hh\) can be simultaneously diagonalized, by definition there is a basis consisting entirely of simultaneous eigenvectors of the elements of \(\hh\text{.}\) For \(\su(3)\text{,}\) this basis consists of \(\{\lambda_3, \lambda_8, \frac12(\lambda_1\mp\mu_2), \frac12(\lambda_4\mp\mu_5),\frac12(\lambda_5\mp\mu_7)\text{.}\) For instance,

Notice that any element of \(\hh\) has each basis element as an eigenvector, not just \(\lambda_3\) and \(\lambda_8\text{.}\) So, for instance,

Rather than identify the eigenspaces containing \(\lambda_4\mp\mu_5\) by their eigenvalues under \(\lambda_3\) and \(\lambda_8\text{,}\) we can instead label these eigenspaces by operators that give the eigenvalue for any element of \(\hh\text{.}\)

What are the properties of these operators? Let's call them \(\pm\alpha_2\text{,}\) since the eigenvalues for these two subspaces are equal and opposite for any element of \(\hh\text{.}\) We know that

and in fact

Thus, \(\alpha_2\) is a linear operator on \(\hh\text{,}\) that is, \(\alpha_2\) is in the dual space \(\hh^*\text{.}\) We can repeat this process with the remaining eigenspaces, resulting in a total of six elements of \(\hh^*\)—for this purpose, we don't count \(\hh\) itself as an eigenspace, although of course it is, with all eigenvalues zero.

Subsection 5.2.3 Roots

More generally, we can decompose any simple Lie algebra as

for some finite collection \(R\) of \(\alpha\in\hh^*\text{,}\) where

The \(\alpha\) are called the roots of \(\gg\text{.}\)

Subsection 5.2.4 Orthogonality

Suppose that \(X\in\gg_\alpha\) and \(Y\in\gg_\beta\text{.}\) Then

for any \(H\in\hh\text{.}\) We compute

so that

Thus, \(X\in\gg_\alpha\) maps \(\gg_\beta\) to \(\gg_{\alpha+\beta}\)—although the latter might be the trivial vector space \(\{0\}\text{.}\) In other words, each \(X\not\in\hh\) changes at least some eigenvalues. Similarly, \(X\) followed by \(Y\) (or vice versa) changes at least some eigenvalues—unless \(\alpha+\beta=0\text{.}\) Since our basis consists of eigenvectors, we have shown that

since a nonzero trace requires there to be elements whose eigenvalues do not change. Since we do not count \(\beta=0\) as a root, we write that case separately, namely

but the argument is the same. A more formal argument is given in Section 5.3

What about \(\hh\text{?}\) If \(H,K\in\hh\) and \(X\in\gg_\alpha\text{,}\) then

Thus, the inner product on \(\hh\) is

as can also be seen by writing \(H\) and \(K\) as diagonal matrices and taking the trace of the product. This inner product is nondegenerate, and so \(\hh\) admits an orthonormal basis.

Subsection 5.2.5 Symmetry about the Origin

We now claim that if \(\alpha\in R\) then also \(-\alpha\in R\text{.}\) We have already seen that the roots of \(\su(3)\) have this property; we now prove it in general.

Suppose \(\alpha\) is a root, but not \(-\alpha\text{.}\) Then \(\gg_\alpha\perp\gg\text{,}\) since we showed above that \(\gg_\alpha\) is perpendicular to all elements not in \(\gg_{-\alpha}\text{!}\) By the assumed non-degeneracy of \(B\text{,}\) this can't happen, so \(\gg_{-\alpha}\) must contain a nonzero vector.

Thus, the roots always come in pairs, symmetric about the origin.

Subsection 5.2.6 \(\su(2)\) Subalgebras

Continuing along these lines, it turns out that each \(\{\gg_{\pm\alpha}\}\) pair generates an \(\su(2)\) subalgebra of \(\gg\text{.}\) We outline this construction here, but save most of the details for Section 5.3

For \(X_\alpha\in\gg_\alpha\text{,}\) \(Y_\alpha\in\gg_{-\alpha}\text{,}\) the argument used above to show orthogonality now shows that \([X,Y]\in\hh\text{.}\) Using non-degeneracy, we can show that \(H_\alpha=[X,Y]\ne0\text{.}\) Thus, \(\langle H_\alpha,X_\alpha,Y_\alpha \rangle\) is a subalgebra of \(\gg\text{;}\) by rescaling if necessary we can bring the commutators into the standard form for \(\su(2)\) (really \(\sl(2,\RR)\)). In particular, we can choose roots satisfying ¹

But we have already analyzed the representations of \(\su(2)\text{!}\) In particular, all eigenvalues of (standard) \(\su(2)\) in any representation must be in \(\frac12\ZZ\text{!}\) Thus, \(\alpha(H_\beta)\) must be a half-integer, and in particular must be real, for any combination of our preferred elements \(H_\beta\in\hh\) and preferred roots \(\alpha\in R\text{.}\)

From this point on, we work with the split real form of \(\gg\text{,}\) restricting if necessary to real linear combinations of the \(H_\alpha\text{,}\) \(X_\alpha\text{,}\) and \(Y_\alpha\text{,}\) and of the preferred roots satisfying (5.2.16). In particular, the Killing form (5.2.15) on \(\hh\) is now positive definite.

We illustrate this construction with \(\su(3)\text{.}\) An orthonormal basis for \(\hh\) is given by \(\{\lambda_3,\lambda_8\}\text{,}\) and direct computation shows that

so these three elements (with the factors of \(\frac12\)) form a standard basis for \(\sl(2,\RR)\text{.}\) Thus, we can choose

and define \(\alpha_1\in\hh^*\) by

Similarly,

yielding

and leading to

The roots of \(\su(3)\) are then \(\{\pm\alpha_1,\pm\alpha_2,\pm\alpha_3\}\text{.}\)

Subsection 5.2.7 Root Angles

The final piece of this construction is to determine the angles between the roots. Since \(\alpha\in\hh^*\text{,}\) there are unique elements \(T_\alpha\in\hh\) such that

for any \(H\in\hh\text{.}\) It is not hard to show that

Putting this all together, we have shown that

since, as argued above, \(\alpha(H_\beta)\) must be a half-integer.

There is a natural inner product on \(\hh^*\text{,}\) given by

The angle between two roots is thus given by

and we also have

From these two relations we can work out not only the possible angles between two roots \(\alpha\) and \(\beta\text{,}\) but also the ratio of their lengths. The results are collected in Table 5.2.1. ²

Returning to \(\su(3)\) and computing the Killing product in the adjoint representation, since the \(\{X_\alpha,Y_\alpha\}\) pairs are clearly related to each other by cyclic transformations, so are the \(H_\alpha\) (despite the asymmetric form of our basis element \(\lambda_8\)). Thus,

so that

which in turn forces

Similarly, all the cross terms will be equal, so it is enough to compute

(since \(\lambda_3\) and \(\lambda_8\) are orthogonal and have the same magnitude), so that \(\theta=\frac{2\pi}{3}\text{.}\)

Remarkably, one obtains the same answer in this case by treating the pairs of eigenvalues

as Euclidean coordinates. This alternative derivation works only when the roots all have the same magnitude, which, as can be seen from Table 5.2.1, is not always the case. Nonetheless, we do have