# What is the Lie subalgebra generated by $\mathfrak g_{\pm \alpha}\$?

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?

• 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 $f_\alpha$ ? Commented Feb 15, 2023 at 21:24

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).
• @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. Commented Feb 19, 2023 at 13:02