# How to show this $L$-module is simple? (related to the root space decomposition of semisimle Lie algebras)

Given a semisimple Lie algebra (finite dimensional over a field $$K$$ characteristic $$0$$ and algebraically closed), there exists a root space decomposition $$L = H \oplus \oplus_{\alpha \in R} L_{\alpha},$$ where $$H$$ is a maximal toral subalgebra, $$R = \{\alpha \in H^* : L_{\alpha} \not = 0 , \alpha \not = 0 \}$$ and $$L_{\alpha} = \{ x \in L : ad h(x) = \alpha(h) x \ \forall h \in H \}.$$

I want to prove that $$(L_{\alpha} + L_{-\alpha} + [L_{\alpha},L_{-\alpha}])$$-module $$\sum_{j \in \mathbb{Z}} L_{\beta+ j \alpha}$$ is simple when $$\alpha$$ and $$\beta$$ are linearly independent roots.
I know that each $$L_{\alpha}$$ is one dimensional for $$\alpha \in R$$.

Any comments would be appreciated. Thank you!

• Do you know that for any roots $\gamma, \delta$ whose sum is $\in R$, we have $[L_\gamma, L_\delta] =L_{\gamma+\delta}$? That should be quite helpful. Commented Mar 5, 2019 at 17:12
• @TorstenSchoeneberg Yes, but I couldn't really make use of it because $0 \not = x \in L_{\gamma}, 0 \not = y \in L_{\delta}$ implies $[x,y] \in L_{\gamma + \delta}$ but $[x,y]$ could be $0$ when $L_{\gamma + \delta} \not = 0$. Can I rule this situation out somehow? Commented Mar 5, 2019 at 18:05
• So you do not (yet) have the equality I stated, just the easy inclusion "$\subseteq$". The proof for the other inclusion which I know relies on basic representation theory of $\mathfrak{sl}_2$ (which taken a bit further would prove your claim as well). Maybe there are simpler arguments. Commented Mar 5, 2019 at 21:26
• oops I missed that, but I thought it can't be true for all cases: we know that $[L_{\alpha}, L_{-\alpha}]$ is one dimensional when $\alpha$ is a root but $L_0 = H$ which is not necessarily $1$ dimensional. I guess it holds when the sum is not $0$? Commented Mar 5, 2019 at 22:05
• $0 \notin R$. -- Commented Mar 5, 2019 at 22:42

This is easier than I thought when writing those comments.

Let $$k$$ be a field of characteristic $$0$$ and $$\mathfrak{sl}_2(k)$$ be the Lie algebra with $$k$$-basis $$X=\pmatrix{0&1\\0&0}, Y=\pmatrix{0&0\\-1&0}, H=\pmatrix{1&0\\0&-1}$$.

Proposition: If $$V$$ is a simple representation of $$\mathfrak{sl}_2(k)$$ of dimension $$n$$, then with $$\lambda=n-1$$, $$V$$ has a basis $$(e_{\lambda}, e_{\lambda-2}, ... , e_{-\lambda+2}, e_{-\lambda})$$ such that for all $$r \in \lbrace -\lambda, -\lambda+2, ..., \lambda-2, \lambda \rbrace$$, we have $$H e_r= r e_r$$, and $$X e_r = e_{r+2}$$ (resp. $$=0$$ for $$r=\lambda$$).

Proof: For this and more precise statements compare Bourbaki, Groupes et algèbres de Lie ch. VIII §1 no. 2 prop. 1/2 and corollaire, as well as no. 3 theorem 1. Basically, show that $$\ker(X) \neq 0$$, pick $$0 \neq e_\lambda \in \ker(X)$$, then define the other $$e_r$$ as cleverly normed scalar multiples of $$Y^r e_\lambda$$, which exhibits exactly how $$X,Y,H$$ act on them, particularly implying that they are linearly independent. $$\square$$

Corollary: For a finite dimensional representation $$V$$ of $$\mathfrak{sl}_2(k)$$ to be simple, it is necessary and sufficent that:

• all eigenvalues of $$H$$ acting on $$V$$ have the same parity (i.e. they are either all odd or all even; it follows from the proposition that they are integers), and
• for each such eigenvalue $$r$$ of $$H$$, the eigenspace $$V_r$$ is is one-dimensional.

Proof: This follows just from the well-known fact that $$\mathfrak{sl}_2(k)$$ and hence its finite-dimensional representations are semisimple, i.e. they are direct sums of simple modules as described in the proposition. $$\square$$

Now, take $$0 \neq x_\alpha \in L_\alpha$$ and extend this to an $$\mathfrak{sl}_2$$-triple with $$y_\alpha \in L_{-\alpha}, h_\alpha \in [L_\alpha, L_{-\alpha}]$$ satisfying the exact relations as $$X, Y, H$$ above. Then it's clear that

$$V:= \sum_{j\in \mathbb Z}L_{\beta+j\alpha}$$

is a finite-dimensional $$\mathfrak{sl}_2$$-representation. Each $$L_{\beta+j\alpha}$$ is the eigenspace for $$h_\alpha$$ to the eigenvalue $$\beta(h_\alpha)+j\alpha(h_\alpha) = \beta(h) +2j$$, so all these have the same parity; you say you already know they are one-dimensional, so by the above criterion, you're done.

Note that one can actually show that 1) all $$L_\alpha$$ are one-dimensional as well as 2) for roots $$\alpha, \beta$$ with $$\alpha+\beta \in R$$, we have $$[L_\alpha, L_\beta] = L_{\alpha+\beta}$$ (what I wanted to use in the comments) from the proposition by applying

Corollary 2: If $$V$$ is any finite-dimensional representation of $$\mathfrak{sl}_2(k)$$, then there is a unique $$\ell \in \mathbb Z$$ such that

$$V = \displaystyle\bigoplus_{r=-\ell}^\ell V_i$$ with $$H v_r =rv_r$$ for all $$v_r \in V_r$$;

also, the map $$V_{r} \twoheadrightarrow V_{r+2}$$ induced by $$X$$ is surjective for $$-1 \le r \le \ell-2$$.

Proof: See loc.cit., corollaire to prop. 2; basically, add several $$V$$'s of potentially different $$n$$'s as in the proposition, and see that these properties stay true. $$\square$$

Namely, to get that an arbitrary root-space $$L_\alpha$$ is one-dimensional, pick $$x_\alpha, y_\alpha, h_\alpha$$ as above and apply the last statement to $$r=0$$ (note $$L_0 = H$$, the Cartan subalgebra (not the element $$H$$ from above)) in a split semisimple Lie algebra) in $$V= \bigoplus_{j \in \mathbb Z} L_{j\alpha}$$; the other statement follows immediately from yours.