Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!)

share|improve this question
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

3 Answers 3

up vote 6 down vote accepted

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.

share|improve this answer
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.

share|improve this answer

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 (proof below). In particular, every automorphism of an irreducible root system is an isometry. 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.

This proof is in "Álgebras de Lie" by Luiz San Martin, Editora da UNICAMP but I will use Humphreys's notation instead.

Lemma: In an irreducible root system $\Phi$, given any pair of roots $\alpha$, $\beta$ there exists a sequence of roots $\alpha = \alpha_1, \, \alpha_2,\, \ldots,\, \alpha_n = \beta$ which are consecutively non-orthogonal, that is, such that $(\alpha_i, \alpha_{i+1}) \neq 0$, for $i=1,2,\ldots,n-1$.

Proof: Write $\alpha \sim \beta$ if there is such a sequence of roots connecting $\alpha$ and $\beta$ and $\alpha \not\sim \beta$ on the contrary. Consider $\Phi_\alpha = \{ x \in \Phi:\, \alpha \sim x \} $ and $\Phi^0_\alpha = \{ y \in \Phi:\, \alpha \not\sim y \} $. These subsets are orthogonal since if a $y \Phi^0_\alpha $ was not orthogonal to a $x \in \Phi_\alpha$ then $\alpha \sim x$ and $x \sim y$ would imply $\alpha \sim y$. Since $\alpha \in \Phi_\alpha$ and $\Phi$ is irreducible it follows that $\Phi^0_\alpha$ must be empty, thus $\alpha$ connects to every other root.

Proposition: an isomorphism between two irreducible root systems must be conformal, that is, it must scale the metrics by a constant factor.

Proof: To prove that a linear map $\phi$ is conformal, it is enough to check this on a generating set of vectors. By definition, an isomorphism $\phi$ preserves the Killing number between any two roots $\alpha$ and $\beta$: $< \phi(\alpha),\, \phi(\beta) > \,=\, < \alpha,\, \beta >$. In terms of the inner product $(\cdot,\cdot)$ this is

$\displaystyle \frac{2(\phi(\alpha),\phi(\beta))}{(\phi(\beta),\phi(\beta))} = \frac{2(\alpha,\beta)}{(\beta,\beta)} $

and thus

$ (\phi(\alpha),\phi(\beta)) = c_\beta\, (\alpha,\beta) \qquad$ where $ c_\beta = \displaystyle \frac{(\beta,\beta)}{(\phi(\beta),\phi(\beta))} $

We can exchange $\alpha$ and $\beta$ in the argument above to get $ (\phi(\beta),\phi(\alpha)) = c_\alpha\, (\beta,\alpha) $. Since the inner product is symmetric we get

$ (\phi(\alpha),\phi(\beta)) = c_\alpha\, (\alpha,\beta) \qquad$ where $ c_\alpha = \displaystyle \frac{(\alpha,\alpha)}{(\phi(\alpha),\phi(\alpha))} $

From the two previous equations we have that $c_\alpha = c_\beta$ whenever $(\alpha,\beta) \neq 0$.

By the irreducibility and the previous Lemma we get that $c_\alpha$ is the same for every root, and thus this is the constant scaling factor.

Corollary: an automorphism of an irreducible root system must be an isometry.

Proof: By the previous Proposition, this automorphism $\phi$ must be conformal. Let $c>0$ be the conformal scaling factor and let $L$ be the norm of the longest root in $\Phi$. Then $\sqrt{c}L$ is the norm of the longest root in $\phi(\Phi)$. Since $\phi(\Phi) = \Phi$, we have that $\sqrt{c}L = L$ so that $\sqrt{c}=1$ and thus $c=1$ and $\phi$ is an isometry.

share|improve this answer
Can you give me the proof of the claim: "an isomorphism between two irreducible root systems must be conformal: it must scale the metrics by a constant factor "? Thank you very much. –  SnowAngel6147 Jul 21 at 16:21
@SnowAngel6147, just edited my post to include this proof. Please see if it helps. –  Lucas Seco Jul 23 at 16:29
That's what I'm looking for. Thank you very much! –  SnowAngel6147 Jul 24 at 2:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.