What is the basis of the root space $\mathfrak g_\theta$, where $\theta = \alpha_1+\cdots+\alpha_n$?

Question

Let $\mathfrak g = \mathfrak{sl}_{n+1}$ with its canonical basis $\{x^\pm_i, h_i\: i=1,\cdots, n\},$ where $x_i^+ = e_{i,i+1}, x_i^- = e_{i+1,i}, h_i = e_{i,i}-e_{i+1,i+1}$ and $e_{ij}$ denotes the canonical basis of the matrices $n+1\times n+1$.

Denoting by $\alpha_i, i=1,\cdots, n, $ the simple positive roots, I want to know what is the root space $\mathfrak g_\theta = \{x\in \mathfrak g: [h,x] = \theta(h) x \; \forall \; h\in \mathfrak h \}$, where $\theta = \alpha_1+\cdots+\alpha_n .$

I know that $\mathfrak g_\theta$ is unidimensional, so it suffices to find a element $x_\theta\in \mathfrak g$ such that $[h,x_\theta] = \theta(h) x_\theta$ for all $h$. Since $\theta$ is the sum of the positive roots, my first guess was to say that $x_\theta = \sum x_j^+$, but it didn't work, since

$[h,\sum x_j^+] = \sum[h,x_j^+] = \sum \alpha_j(h)x_j^+$

which does not give back the sum $\alpha_1(h)+\cdots+\alpha_n(h)$.

Any hint on how to find this element $x_\theta$? Thank you very much!

Answer 1

Just to turn my comment into a hint-answer: If $[h,x] = \alpha(h)\cdot x$ and $[h,y] =\beta(h)\cdot y$, then with the Jacobi identity one gets

$$[h, [x,y]] = \left(\alpha(h) + \beta(h) \right) \cdot [x,y].$$

In particular, a natural candidate for non-zero vector in a root space to $\sum_{i \in I} \alpha_i$ is some nested commutator (careful of the ordering here, cf. ladder operators of simple Lie algebra and simple roots) of the corresponding $x_i$, $i \in I$.

Note that in the case of $\mathfrak{sl}_n$, when written out as matrices, another natural set of basis vectors for the root spaces are all $e_{i,j}$ ($i\neq j$), and indeed up to $\pm$ these should be the same as above. It is quite intricate though to decide the $\pm$ here, depends on many choices of ordering.