In the book "Lie Groups: Beyond and Introduction" by Anthony Knapp, the author describes roots of a semisimple (A Lie algebra without any non-zero solvable ideal) Lie algebra as elements of the dual space of a Cartan subalgebra, and the decomposition $\mathfrak{g} = \mathfrak{h} \oplus \bigoplus\limits_{\alpha \in h^*\setminus \left\lbrace 0 \right\rbrace} \mathfrak{g}_{\alpha}$. Those $\alpha \in h^*\setminus \left\lbrace 0 \right\rbrace$ for which $\mathfrak{g}_{\alpha} \neq \left\lbrace 0 \right\rbrace$, are called roots. Here, $\mathfrak{g}_{\alpha}$ is the generalized common eigenspace for $ad \ H$, $H \in \mathfrak{h}$ with eigenvalue $\alpha \left( H \right)$.
In developing the theory, the author uses the notation $\Delta$ to denote the set of all roots of a semisimple Lie algebra $\mathfrak{g}$ relative to a Cartan subalgebra $\mathfrak{h}$ and proves the following result:
Theorem: Let $\alpha \in \Delta$ and $\beta \in \Delta \cup \left\lbrace 0 \right\rbrace$. Then, we have:
- The $\alpha$-string containing $\beta$ has the form $\beta + n \alpha$, where $-p \leq n \leq q$ with $p, q \geq 0$; there are no gaps. Moreover, $2 \dfrac{\langle \beta, \alpha \rangle}{\langle \alpha, \alpha \rangle} = p – q \in \mathbb{Z}$.
- If $\beta + n \alpha \neq 0$ for all $n \in \mathbb{Z}$, then the subalgebra $\mathfrak{sl}_{\alpha} = \left\lbrace H_{\alpha}, E_{\alpha}, E_{- \alpha} \right\rbrace$, which is isomorphic to $\mathfrak{sl} \left( 2, \mathbb{C} \right)$, acts irreducibly on $\mathfrak{g}' =\bigoplus\limits_{n \in \mathbb{Z}} \mathfrak{g}_{\beta + n \alpha}$.
Here, $H_{\alpha} \in \mathfrak{h}$, $E_{\alpha} \in \mathfrak{g}_{\alpha}$ and $E_{- \alpha} \in \mathfrak{g}_{- \alpha}$ are chosen such that they satisfy the bracket relations of $\mathfrak{sl} \left( 2, \mathbb{C} \right)$. Also, $\langle \cdot, \cdot \rangle$ is a bilinear form on $\mathfrak{h}^*$ given by $\langle \phi, \psi \rangle = B \left( H_{\phi}, H_{\psi} \right)$.
Immediately after proving this result, the author proves the following two corollaries:
Corollary 1: If $\alpha, \beta \in \Delta \cup \left\lbrace 0 \right\rbrace$ and $\alpha + \beta \neq 0$, then $\left[ \mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta} \right] = \mathfrak{g}_{\alpha + \beta}$.
Corollary 2: Let $\alpha, \beta \in \Delta$ such that $\beta + n \alpha \neq 0$ for all $n \in \mathbb{Z}$. Let $E_{\alpha} \in \mathfrak{g}_{\alpha}$, $E_{- \alpha} \in \mathfrak{g}_{- \alpha}$ and $E_{\beta} \in \mathfrak{g}_{\beta}$. Then,
$$\left[ E_{- \alpha}, \left[ E{\alpha}, E_{\beta} \right] \right] = \dfrac{q \left( p + 1 \right)}{2} B \left( E_{\alpha}, E_{- \alpha} \right) E_{\beta}.$$
Now, in the proof of corollary 1, the author says that whenever $\beta + n \alpha \neq 0$ for all $n \in \mathbb{Z}$, we can use point (2) of the theorem. So, we look at $\mathfrak{sl}_{\alpha}$ which acts reducibly on $\mathfrak{g'}$ and hence has a highest weight and a corresponding highest weight vector. Since $E_{\alpha}$ increases the index, using the isomorphism of $\mathfrak{sl}_{\alpha}$ with $\mathfrak{sl} \left( 2, \mathbb{C} \right)$, we get that the highest weight is $\beta + q \alpha$ and the highest weight vector is $E_{\beta + q \alpha}$. Now, the author says (and I quote), "Thus $\left[ \mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta} \right] = \left\lbrace 0 \right\rbrace$ forces $q$ to be zero."
My question is why should $\left[ \mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta} \right]$ be zero? There is no direct way to say this. Even the irreducibility does not seem to be helping here. Also, if it is true that $\left[ \mathfrak{g}_{\alpha}, \mathfrak{g}_{\beta} \right] = \left\lbrace 0 \right\rbrace$, then in Corollary 2, the left hand side is the zero vector and the right hand side is a non-zero vector.
I will be thankful for any insights into this!
Best Answer
$[\mathfrak{g}_\alpha, \mathfrak{g}_\beta]$ doesn't have to be zero of course. But if $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \ne \{0\}$, then since $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset \mathfrak{g}_{\alpha+\beta}$ and $\mathfrak{g}_\gamma$ is at most 1-dimensional (Proposition 2.21), the equality $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] = \mathfrak{g}_{\alpha+\beta}$ must hold in this case. So Knapp is just considering the alternative case that $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] = \{0\}$, and showing the equality holds in this case also (i.e. that $\mathfrak{g}_{\alpha+\beta} = \{0\}$ or equivalently $\alpha+\beta \notin \Delta$).