1
$\begingroup$

I found two different root systems of $sl(3,\mathbb C)$ in the internet.

The first one: enter image description here

And the second one:

enter image description here

I think they should be the same. So...how can I see that these root systems are isomorphic?

$\endgroup$
4
  • $\begingroup$ Edit: Actually the first one cant be a root system, right? The property that $2\frac{\langle \beta,\alpha\rangle}{\langle \alpha,\alpha\rangle}\in \mathbb Z$ is not satisfied...I am confused.. $\endgroup$ Feb 14, 2016 at 13:11
  • 1
    $\begingroup$ Right. Also the first one would have two different root lengths. You should rely on this root system for $A_2$. Compare it with type $B_2,C_2$ and $G_2$. $\endgroup$ Feb 14, 2016 at 13:12
  • $\begingroup$ Alright. I got the first one from math.umn.edu/~phamx352/docs/LieAlg-Review(revised).pdf . You know where the mistake is? Maybe one can derive $A_2$ from the results of that sheet. $\endgroup$ Feb 14, 2016 at 13:15
  • 1
    $\begingroup$ They have no scaling. I still recommend you the standard references. $\endgroup$ Feb 14, 2016 at 13:16

1 Answer 1

2
$\begingroup$

In $\mathfrak{sl}_3$, the standard torus $\mathfrak{h}$ is spanned by the coroots $H_1=E_{11}-E_{22}$ and $H_2=E_{22}-E_{33}$. The roots are $\alpha_1=\epsilon_1-\epsilon_2$ and $\alpha_2=\epsilon_2-\epsilon_3$, where the $$\epsilon_i:\{\mbox{Diagonal matrices}\}\to\mathbb{C}$$ are linear functionals given by $\epsilon_i(E_{jj})=\delta_{ij}$. Writing $$\epsilon_1=(1,0,0),\;\;\;\epsilon_2=(0,1,0),\;\;\;\epsilon_3=(0,0,1)$$ you get the second realization of the root system above: $$ \alpha_1=(1,-1,0),\;\;\;\alpha_2=(0,1,-1),\;\;\;\alpha_1+\alpha_2=(1,0,-1).$$

On the other hand, we can realize $\alpha:\mathfrak{h}\to\mathbb{C}$ by how it acts on $H_1$ and $H_2$. That is, we can write $\alpha=(\alpha(H_1),\alpha(H_2))$. We have \begin{align*} \alpha_1(H_1)&=2 &\alpha_2(H_1)&=-1\\ \alpha_1(H_2)&=-1&\alpha_2(H_2)&=2 \end{align*} In this way, we may write $\alpha_1=(2,-1)$, $\alpha_2=(-1,2)$ and $\alpha_1+\alpha_2=(1,1)$. This isn't exactly the root system above (it is a $90^\circ$ rotation of it). I haven't read through the associated notes too carefully, but the difference appears to boil down to a different choice of basis for $\mathfrak{h}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .