# A question on the root systems of simple Lie algebras in the 90 degree case

I've been taking a look at simple Lie algebras for particle physics and I've found myself wondering about the following question. It can be shown that the adjacent roots in a root diagram corresponding to a simple Lie algebra must always lie at either 0 ,30, 45, 60 or 90 degrees (if the rank $\geq2$). In all cases but the 90 degree case (forgetting the trivial 0 degree case), the relative lengths of the roots are uniquely determined given the angle between them. But it turns out that in the 90 degree case there are no constraints whatsoever on their relative lengths. So my first question is: are there infinitely many Lie algebras corresponding to the 90 degree case, one for every possible ratio in the lengths of the roots? Or is there only one algebra, which just doesn't care about the relative lengths of its roots?

Any help would be greatly appreciated! (Ps I am an extremely mediocre graduate physicist, so please go easy on me!)

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I assume 69 is a typo, meaning 60? –  Jim Humphreys Apr 7 '11 at 13:19
yes - i'm so sorry! –  K McKenzie Apr 7 '11 at 15:12

If you look only at a simple Lie algebra, no two "adjacent" simple roots in the Dynkin diagram can form a right angle: being joined by at least one edge forces a different angle. In the simple case there is no ambiguity about relative lengths of roots, but of course in a direct sum of simple Lie algebras the different simple ideals involved are not directly related.

By the way, the nontrivial angles between adjacent simple roots are actually obtuse rather than acute angles.

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Thank you very much for this! (And yes, I was dimly aware that the angle between the simple roots of a simple Lie algebra is always obtuse, but I had in my mind the whole root diagram here - sorry if that was confusing.) May I ask if it follows from what you're saying that the case in which the roots (i.e. any adjacent roots in the root diagram) are at 90 degrees to one another never corresponds to the root diagram of a simple Lie algebra, whereas those at angles 30, 45 and 60 degrees always do? That would be really useful for me to know as it happens... Thanks again! –  K McKenzie Apr 7 '11 at 15:11
Yes, if you refer just to adjacent simple roots; but of course a simple Lie algebra such as $\mathfrak{sl}_4$ can have two orthogonal simple roots corresponding to end nodes of the Dynkin diagram, each connected to a third simple root. What's important is that the relative lengths of roots are well determined, though of course the unit of length you choose can be arbitrary. (In this particular example all roots must have equal length, but ratios of 2 and 3 for squared lengths occur elsewhere in the classification.) –  Jim Humphreys Apr 7 '11 at 17:20

The length of a root is not a notion attached to a Lie algebra on its own, but to a nondegenerate quadratic Lie algebra: a Lie algebra with a choice of invariant inner product. If instead of the question you did asked, you had said

are there infinitely many quadratic Lie algebras corresponding to the 90 degree case, one for every possible ratio in the lengths of the roots?

the answer would have been "Yes!" In fact, there is a quadratic Lie algebra for each choice of non-zero length for the two roots. The underlying Lie algebras are, of course, isomorphic, but the forms are different.

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Ben, I also asked myself that same question and the notion that made it clear for me was that of an isomorphism between root systems: it is a linear map that sends all the roots of one system to all the roots of the other system, preserving the Cartan-Killing numbers of the corresponding roots (see Humphreys' book, Section 9.2)

The beauty is that such an isomorphism does not need to be an isometry! So the the infinitely many two-dimensional root systems of the 90$^o$ case are all isomorphic between themselves, in particular all isomorphic to $A_1 \times A_1$.

In terms of the corresponding Lie algebras, you have also infinitely many, all of them isomorphic to ${\frak sl}(2) \times {\frak sl}(2)$.

PS - It can be shown that an isomorphism between two irreducible root systems must be conformal: it must scale the metrics by a constant factor. In particular, every automorphism of an irreducible root system is an isometry. (Ask me if you need a proof of this, it is very simple but it is not in Humphreys' book, I think.)
The irreducibility here is crucial since an $A_1 \times A_1$ root system with distinct root lengths admit an automorphism which is not an isometry.

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