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I want to determine the root-system of the lie algebra $sl(3,\mathbb C)$. Does someone know a good (and complete) reference for this problem?

I know that the root-system is $A_2$ but I want to see a complete proof (calculation) for this.

Thanks in advance.

Marc
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  • Compute $DA-AD$ when $A,D\in\mathfrak{sl}_3$ with $D$ diagonal, and deduce the result. – YCor Feb 12 '16 at 21:21

1 Answers1

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The root system of $A_2$ consists of $\Phi=\{ \pm \alpha,\pm \beta , \pm (\alpha+\beta) \}$, which can be constructed as follows. With the Cartan subalgebra $\mathfrak{h}_{\mathbb{R}}\cong \mathbb{R}^2$ and the canonical scalar product on $\mathbb{R}^2$ we can realize the simple roots as $$ \alpha=\begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix}, \quad \beta=\begin{pmatrix} 1 \\ 0 \end{pmatrix}. $$ Then we obviously have $(\alpha,\alpha)=(\beta,\beta)=1$, and in addition \begin{align*} \langle \alpha, \alpha^{\vee}\rangle & = \frac{2(\alpha,\alpha)}{(\alpha,\alpha)}= 2 ,\\ \langle \alpha, \beta^{\vee}\rangle & = \frac{2(\alpha,\beta)}{(\alpha,\alpha)}= -1,\\ \langle \beta, \beta^{\vee}\rangle & =2, \end{align*} and we obtain the Cartan numbers for type $A_2$. This shows that the root system of $A_2$ consists of $\Phi=\{ \pm \alpha,\pm \beta , \pm (\alpha+\beta) \}$. For additional information also see the MSE question here, for the case of $B_2$. Of course, the same discussion is valid for type $A_2$.

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Dietrich Burde
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  • Thanks for the answer! But how I know that this belongs to $sl(3,\mathbb C)$? – Marc Feb 12 '16 at 21:48
  • The Cartan numbers characterise the Lie algebra. And $A_2$ is $\mathfrak{sl}_3(\mathbb{C})$. – Dietrich Burde Feb 12 '16 at 22:08
  • I think your last sentence is the main thing which I dont understand. I thought that $A_2$ is the name of the root system. But you said "$A_2$ is $sl(3,\mathbb C)$", then what is the definition of $A_2$? (Sorry for this "stupid" question.. I am a beginner in this Topic, I hope you can answer this question for me) – Marc Feb 12 '16 at 22:53
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    Who downvoted this post ? Incredible ! –  Feb 13 '16 at 20:24