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!

  • $\begingroup$ 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. $\endgroup$ Commented Mar 5, 2019 at 17:12
  • $\begingroup$ @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? $\endgroup$
    – Johnny T.
    Commented Mar 5, 2019 at 18:05
  • $\begingroup$ 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. $\endgroup$ Commented Mar 5, 2019 at 21:26
  • $\begingroup$ 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 $0$? $\endgroup$
    – Johnny T.
    Commented Mar 5, 2019 at 22:05
  • $\begingroup$ $0 \notin R$. -- $\endgroup$ Commented Mar 5, 2019 at 22:42

1 Answer 1


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.


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