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If a root system $R$ is irreducible (not a product of two root systems) then $R$ does not contain three vectors of pairwise different lengths.

To show this do we need just to compute all the angles and the ratios between the root lenghts as explained (1) in the following post:

Two questions on roots of finite, simple, complex lie algebra

Howeber how the fact that $R$ is irreducible is involved here?

user652838
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    The root system of a simple Lie algebra is irreducible - see here. – Dietrich Burde Jun 18 '22 at 16:45
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    I think OP asks where exactly in the proof do we use irreducibility, which is a good question. (I hope it's clear to OP that we do need it somewhere, as we can realize something like $A_1 \times A_1 \times A_1$ as mutually perpendicular pairs of vectors with arbitrary lengths; but I think the question really is at what point of the proof do we use the condition.) – Torsten Schoeneberg Jun 18 '22 at 16:48
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    A very similar question was comment-answered with a reference: https://math.stackexchange.com/q/2615781/96384 – Torsten Schoeneberg Jun 18 '22 at 16:51
  • Is the proof/answer in the link I added enough, i.e. proves my claim..? (The answer in the book was weird to me) – user652838 Jun 18 '22 at 18:03
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    I am not sure to be honest. I think one might want to add an extra argument to what is proven in the linked answer, as it seems to consider only the ratio of two different root lengths to begin with. The way Humphreys proceeds in the book seems pretty straightforward to me, and I feel like something like that should be used. What do you find "weird" about it? – Torsten Schoeneberg Jun 20 '22 at 03:45
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    @Torsten Schoeneberg since I did not study deeply about how the weyl group works on the root system..so I am trying to show this in a standard way. What the extra argument is? – user652838 Jun 20 '22 at 14:09
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    I cannot think of an argument which does not involve the Weyl group in one way or another. One should familiarize oneself with the Weyl group at some point anyway. – Torsten Schoeneberg Jun 20 '22 at 23:53

1 Answers1

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This is proven e.g. in Bourbaki, Lie Groups and Algebras vol. VI §1 no.4 Proposition 12; or in Humphreys, Introduction to Lie Algebras and Representation Theory, §10.4, Lemma C.

We have (see e.g. Dietrich Burde's answer to Two questions on roots of finite, simple, complex lie algebra):

(*) If $\alpha, \beta$ are two roots which are not perpendicular to each other, and w.l.o.g. $(\alpha, \alpha) \le (\beta, \beta)$, then

$$\dfrac{(\beta, \beta)}{(\alpha, \alpha)} \in \{1,2,3\}.$$

Now assume an irreducible root system $\Phi$ (in its Euclidean space $E$) contains three roots $\alpha, \beta, \gamma$ of mutually different lengths. W.l.o.g. $(\alpha, \alpha)=1$. Now the obvious idea is that if e.g. $(\beta, \beta) = 2$ and $(\gamma, \gamma)=3$, we would have $\dfrac{(\gamma, \gamma)}{(\beta, \beta)} =\frac32 \notin \{1,2,3\}$ in contradiction to the above. The problem is that a priori, some or all of our three roots could be perpendicular to each other. And indeed one can easily construct such examples if the root system is not irreducible, e.g. in $A_1 \times A_1 \times A_1$, which is made up of three mutually orthogonal roots (and their negatives), each of them can have any positive length you want.

To exclude that, we need to move our roots around a bit until they are not orthogonal to each other. We do that with the Weyl group $W$ of our root system

Fact: $\{w(\alpha): w \in W\}$ spans $E$. (And for this we need irreducibility.)

This implies that for any root $\phi$, there is some $w \in W$ such that $w(\alpha)$ is not perpendicular to $\phi$ (because if they all were, so would be their span $E$, but that contains $\phi \in \Phi$).

In particular, there is a $w_1\in W$ such that $\alpha':=w_1(\alpha)$ is not perpendicular to $\beta$, and a $w_2\in W$ such that $\alpha'' := w_2(\alpha)$ is not perpendicular to $\gamma$. Since it's clear or easily proven that for any $w \in W$, $(w(\phi),w(\phi))= (\phi, \phi)$, this allows us to use (*) for the first time and infer w.l.o.g. that $(\beta, \beta) = 2$ and $(\gamma, \gamma)=3$. But then we apply the fact again to find $w_3$ with $w_3(\beta)$ not perpendicular to $\gamma$, and can conclude as wanted.

Torsten Schoeneberg
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