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I am working with the representation theory of complex simple Lie algebras, and have a question:

It is intuitively clear that the root systems $A_1\times A_1$, $A_2$, $B_2$, and $G_2$ comprise all the root systems of rank 2, and that the root systems of rank 3 are the direct sums of rank 1 and rank 2 systems along with the irreducibles $A_3$, $B_3$, and $C_3$. But how does one prove that all other root systems are isomorphic to the ones above?

I am thinking that we should use the properties of a root system, e.g. that

  1. the root system $R$ spans $\mathbb{R}^n$,
  2. $\pm \alpha\in R$,
  3. reflecting in the hyperplane $\alpha^\perp$ takes $R$ to itself,
  4. and that $2\frac{(\beta,\alpha)}{(\alpha,\alpha)}\in \mathbb{Z}$ for $\alpha,\beta$ being roots.

But, after some fiddling around, I am having trouble finding the conditions that give each root system. An explanation or partial explanation would be very helpful and much appreciated here.

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2 Answers 2

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To classify the root systems of rank 2 is useful to observe that, if $\alpha$ and $\beta$ are two non multiple root in $\Phi$ and $\theta$ is the angle between them, we have $$\langle \beta , \alpha \rangle = \frac{2(\beta,\alpha)}{(\alpha,\alpha)}=2\frac{||\beta||}{||\alpha||}\cos\theta$$ By condition 4, we obtain $$\langle \alpha, \beta \rangle \langle \beta, \alpha \rangle= 4 \cos^2\theta \in \mathbb{Z}$$ and then the only possible values for $\theta$ are $\pi/2 ,\, \pi/3 ,\,2\pi/3, \, \pi/4, \, 3\pi/4,\,\pi/6, \, 5\pi/6 $. By the first formula you can also obtain coditions on lenght of roots in $\Phi$ and then, with this information, you can easly check that the only rooth systems of rank 2 are $A_1\times A_1, \, A_2, \, B_2$ and $G_2 $.

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  • $\begingroup$ I have a question - this seems to show that a rank 2 root system must contain one of these 4 root systems, but how does it show that it actually is equal to one of them? $\endgroup$ Commented Jun 13, 2020 at 13:40
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    $\begingroup$ You can easily prove the equality looking at the rigidity conditions on the values of $\theta$. (Suppose there is another root vector $v$, then consider the possible values of the angles with other root vectors... ) $\endgroup$ Commented Jun 14, 2020 at 9:02
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In the following answer, I majorly use results in Humphrey's Introduction to Lie algebras and Representation Theory.

From section 10.1, we know that for any root system $\Phi\subset E$, $\Phi$ has a base, say $\Delta$. And the cardinality of $\Delta$ is just the rank of $\Phi$, from property (B1) in 10.1.

Firstly, let's consider the case when $Rank(\Phi)=2$. In this case, $\Delta=\{\alpha_1,\alpha_2\}$. Consider the Cartan integer $<\alpha_1,\alpha_2>$, it can only be 0,-1,-2,-3, from $⟨α,β⟩⟨β,α⟩=4cos^2θ\in\mathbb{Z}$, as Sabino Di Trani mentioned and from property (B2) in section 10.1. The major point is: it's not hard to recover $\Phi$ from $\Delta$, if we've already know the Cartan integers(this method works for arbitrary rank, c.f. Proposition 11.1). In fact, you can recover $\Phi$ using root strings. The result is: if $<\alpha_1,\alpha_2>=0$, then $\Phi\cong A_1\times A_1$; if $<\alpha_1,\alpha_2>=-1$, then $\Phi\cong A_2$; if $<\alpha_1,\alpha_2>=-2$, then $\Phi\cong B_2$; if $<\alpha_1,\alpha_2>=-3$, then $\Phi\cong G_2$.

Now, let's consider the case when $Rank(\Phi)=3$. In this case, $\Delta=\{\alpha_1,\alpha_2,\alpha_3\}$. By Proposition 11.1, $\Phi$ is determined by the Cartan integers. These integers cannot be too arbitrary. To make a clearer statement, we can use results from the classification of Dynkin diagrams(See Theorem 11.4). We can enumerate all of the possible cases for 3-point-Dynkin-diagrams. Then, we get all of the possible Cartan integers, which correspond to the root systems you've mentioned. Those are all the cases.

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