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Let $\Gamma$ be the set of all simple roots of a simple Lie algebra $\mathfrak g.$ Let $\Gamma_1 \subset \Gamma$ be an arbitrary subset. Then what is Lie subalgebra generated by $\mathfrak g_{\pm \alpha}\ $? My intuition suggests that it would be $\mathfrak n_i^{+} \mathfrak n_i^{-} : = \left \{\sum\limits_j \left [x_j^i, y_j^i \right ]\ \bigg |\ x_j^i \in \mathfrak n_i^{+}, y_j^i \in \mathfrak n_i^{-} \right \},$ where $\mathfrak n_i^{+} : = \bigoplus\limits_{\alpha \in \Gamma_i} \mathfrak g_{\alpha}$ and $\mathfrak n_i^{-} : = \bigoplus\limits_{\alpha \in \Gamma_i} \mathfrak g_{-\alpha}.$ In the book I am following it is mentioned that such a subalgebra is generated by the Chevalley generators $e_{\alpha}$ and $f_{\alpha}.$ I don't understand how is it the case. Could anyone please shed some light on this?

Thanks for your time.

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    $\begingroup$ As to your last doubt, is not $\mathfrak g_\alpha$ the one-dimensional space spanned by $e_\alpha$, and $\mathfrak g_{-\alpha}$ the one-dimensional one spanned by $f_\alpha$? Isn't it clear then that the subalgebra generated by the $\mathfrak g_{\pm \alpha}$ is the subalgebra generated by those $e_\alpha$ and $f_\alpha$ ? $\endgroup$ Commented Feb 15, 2023 at 21:24

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The notation here is a little unclear. You define $\Gamma_1$ but not $\Gamma_i$. I would also note there is no "set of all simple roots". A set of simple roots is effectively a choice of basis for the root system. Every root can be a simple root depending on our choice.

If you mean that $\Gamma_i = \Gamma_1$, then you should see that what you have constructed is simply a subspace of the Cartan subalgebra since for simple roots $\alpha,\beta$, we have $\alpha-\beta$ is never a root. Thus $[\mathfrak{g}_\alpha, \mathfrak{g}_{-\beta}] = 0$ unless $\alpha = \beta$ when it is the span of the coroot $h_\alpha$ of $\alpha$.

Instead you want to build outwards as well. Recall also the subalgebra generated by a subspace $V \leq \mathfrak{g}$ is the subalgebra spanned by $V + [V,V] + [V,[V,V]] + \cdots$. So you want to consider $[\mathfrak{n}^+,\mathfrak{n}^+]$ and $[\mathfrak{n}^-,\mathfrak{n}^-]$ for example, and you don't want to stop after just one bracket (or you wont get a subalgebra in general).

It should become pretty clear which root spaces must lie in this generated subalgebra as you consider these brackets.

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    $\begingroup$ The "every root can be a simple root" deserves to be a question here for reference. Last time I tried to find it in print I was not lucky. $\endgroup$ Commented Feb 15, 2023 at 22:10
  • $\begingroup$ @MarianoSuárez-Álvarez Theorem 10.3 in Humphreys does the job. The idea is to choose a vector $\gamma$ perpendicular to $\pm\alpha$ but not any other root and then shift it slightly to $\gamma'$ so that $(\gamma',\alpha) = \epsilon> 0$ but $|(\gamma',\beta) | > \epsilon$. Then the hyperplane orthogonal to $\gamma'$ gives an partial order where $\alpha$ is a simple root. I could make this into a community wiki question+answer if you think it warrant it. $\endgroup$
    – Callum
    Commented Feb 19, 2023 at 13:02

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