TITLE: Isomorphic root systems?
QUESTION [1 upvotes]: I found two different root systems of $sl(3,\mathbb C)$ in the internet.
The first one: 
And the second one: 

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

REPLY [2 votes]: 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 $\mathfrak{h}$.
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