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Why are there at most two root lengths for a finite, simple, complex lie algebra? I know it is from the constraint that the $2(\alpha,\beta)/(\alpha,\alpha)$ is integer, but what is the argument?

Also, if $\alpha$ is a root, $k\alpha$ is also a root only when k is 1 or -1? How to prove this result?

Shadumu
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1 Answers1

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1.)The length of a root $\alpha$ is given by $\sqrt{(\alpha,\alpha)}$. The condition $2(\alpha,\beta)/(\alpha,\alpha)\in \mathbb{Z}$ enforces that the angle $\theta$ between two roots $\alpha$ and $\beta$ is one of the following seven possibilities: $\frac{\pi}{2}, \frac{\pi}{3},\frac{2\pi}{3},\frac{\pi}{4},\frac{3\pi}{4}, \frac{\pi}{6},\frac{5\pi}{6}$. To see this study the equation $(\alpha,\beta)=\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}\cos (\theta)$, so that $$ 4\cos^2(\theta)=2\frac{(\alpha,\beta)}{(\alpha,\alpha)}\cdot 2\frac{(\alpha,\beta)}{(\beta,\beta)}\in \{0,1,2,3\}, $$ because $0\le \cos^2(\theta)\le 4$, and $\alpha\neq \pm \beta$. Now for $\theta=\frac{\pi}{3},\frac{2\pi}{3}$ we see that the roots have equal length, and for the other angles there are two possible length. For $\theta=\frac{\pi}{4},\frac{3\pi}{4}$ the different roots have length proportional to $\sqrt{2}$, and for $\theta=\frac{\pi}{6},\frac{5\pi}{6}$ the factor is $\sqrt{3}$.

2.) By definition of an abstract root system, twice a root is not a root. So let $\beta=k\alpha$. We may assume that $|k|<1$, if not already $k=\pm 1$. By definition, $2\langle \beta,\alpha\rangle /\langle \alpha,\alpha\rangle$ is integral (this is again an axiom). This is only possible for $k=0$ or $k=\pm \frac{1}{2}$. The latter is impossible, because twice a root is not a root. It follows $k=0$, a contradiction. It follows that $k=\pm 1$.

If you are not assuming that the root system of a simple complex Lie algebra satisfies the axioms of an abstract root system, you have to show that the weight spaces $L_{\alpha}$ are $1$-dimensional, and $\dim L_{k\alpha}=0$ for all $k\ge 2$. This will use the Killing form of $L$, and a Diophantine argument for a dimension equation over $\mathbb{Z}$ concerning the weight spaces.

Dietrich Burde
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  • The constraint just tells us the ratios between any two non-orthogonal roots are of ${1, sqrt(2), sqrt(3)}$, it does not rule out the possibility that there might be three roots have different lengths, right? Why twice a root is not a root in abstract root system? – Shadumu Nov 25 '15 at 15:46
  • No, the angles show that either all length are equal, or at most two different lengths exist - compute all $(\alpha,\alpha)$ in all seven cases. Twice a root is never a root by definition of an abstract root system (like twice a vector is always again a vector in the vector space, by definition of a vector space). – Dietrich Burde Nov 25 '15 at 16:07
  • what about if the angle is $π/2$ then the ratio between lengths is undetermined? – Shadumu Nov 25 '15 at 17:01
  • Yes, exactly. Then $(\alpha,\alpha)/(\beta,\beta)$ is undetermined. – Dietrich Burde Nov 25 '15 at 20:31
  • @DietrichBurde How did you get the equation $(\alpha,\alpha)=\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}\cos (\theta)$? Shouldn't it be $(\alpha,\beta)=\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}\cos (\theta)$ – John Doe Nov 30 '18 at 03:03
  • @JohnDoe Yes, it was a typo, thank you. – Dietrich Burde Nov 30 '18 at 09:08
  • Hi @Dietrich Burde where in your proof (to that there are only two lenghts) we use that the root system is irreducible? – user652838 Jun 19 '22 at 19:09
  • @user726608 The point is that if you fix any root alpha and consider the set of roots not perpendicular to alpha (including alpha itself), then this set can only have roots of at most 2 distinct lengths. The reason is that if you have 3 vectors of different lengths, then some pair must have angle pi/2 (the product of the ratios of the lengths between the shorter 2 and the longer 2 must be the ratio between the shortest and the longest), but by Euclidean geometry it's not possible to find 2 perpendicular vectors each of whose angles away from alpha are one of the 6 allowed ones. – alphacapture Jun 20 '22 at 16:02
  • @DietrichBurde If you take any root alpha and iteratively add all roots not perpendicular to it until you can't anymore, then you only get the whole root system if it's irreducible (and otherwise you get the irreducible component containing alpha) – alphacapture Jun 20 '22 at 16:04
  • So the above answer actually answers my question? https://math.stackexchange.com/questions/4475423/possible-lenghts-in-a-root-system , though just did not really realize why irreducibility is necessary! @alphacapture – user652838 Jun 20 '22 at 16:52
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    @user726608 If it were me, I would have included the details I wrote out. But perhaps to someone who has studied the field, the details I filled in are obvious enough to omit. (Actually, I just had a look at Humphrey's book to see how he did it. I feel like my approach is a bit ad hoc and Humphrey's approach replaces my ad hoc computation that the other two roots can't be perpendicular with the observation that there must be two non-perpendicular roots of those lengths by irreducibility, and this approach maybe obvious to anyone who has studied the field; I may write this up later.) – alphacapture Jun 20 '22 at 17:39
  • Yes, I'll appreciate that:) – user652838 Jun 20 '22 at 18:01