Question I had whilst finding the root decomposion of $sl(n,\mathbb{C})$ by a nonmaximal (and $1$-dimensional) semisimple abelian subalgebra. Consider $L=sl(n,\mathbb{C})$. Let $h = e_{1,1}-e_{2,2}$ and let $H = \operatorname{span}\{h\}$.
$sl(n,\mathbb{C})$ has the basis: $\{e_{i,j} i \neq j\} \cup \{e_{i,i} - e_{i+1,i+1}\}$.
So, I've come to the conclusion that the weight space decomposion $sl(n,\mathbb{C})$ will compose of $H$ and then 4 weight spaces:
$$\alpha_+(c(e_{1,1}-e_{2,2})=c$$
$$\alpha_{+2}(c(e_{1,1}-e_{2,2})=2c$$
$$\alpha_-(c(e_{1,1}-e_{2,2})=-c$$
$$\alpha_{-2}(c(e_{1,1}-e_{2,2})=-2c$$
where in general $\alpha: H \rightarrow \mathbb{C}$ is a weight.
Lets look at the root space $L_{\alpha_+} = \{x \in sl(2,\mathbb{C}): [h,x]=\alpha_+(h)x$ for all $h \in H \}$.
Then we have that, for example, both of the basis vectors $e_{1,6}$ and $e_{1,5}$ are in $L_\alpha$:
$$[h,e_{1,6}] = [c(e_{1,1}-e_{2,2}),e_{1,6}]$$
$$= ce_{1,1}e_{1,6}-ce_{1,6}e_{1,1} - ce_{2,2}e_{1,6} + ce_{1,6}e_{2,2}$$
$$=ce_{1,6}$$
And similarly for $e_{1,5}$. This would mean that the root space of $L_\alpha$ is at least $2$-dimensional. Is this okay in this situation or did I do something wrong? I'm pretty sure if $H$ was a Cartan subalgebra that would mean that $L_{\alpha}$ would be 1 dimensional, right?
 A: Well one would and should not call these weight spaces of an arbitrary abelian (semisimple) subalgebra "root" spaces because the weights do not, in general, form a root system. For more striking examples see https://math.stackexchange.com/a/2112543/96384.
And here, yes, the weight space to the weight you call $\alpha_{+1}$ is spanned by the $e_{1,j}$ with $j \ge 3$ and $e_{i,2}$ with $i \ge 3$, hence has dimension $2n-4$; likewise the weight space to $\alpha_{-1}$ is $(2n-4)$-dimensional. The weight spaces to the weights $\alpha_{\pm2}$ have only dimension $1$ each, and I leave it to you to find a basis for the weight space to the $0$ weight, a.k.a. centraliser of $H$, which must fill up the missing dimensions to $\mathrm{dim}(\mathfrak{sl}_n) =n^2-1$. (A big part of it is made up of $e_{i,j}$ with both $i \neq j \ge 3$, and then see what you can still do on the diagonal.)
By the way, you are sort of describing the decomposition of $\mathfrak{sl}_n$ as an $\mathfrak{sl}_2$-module here for the embedding of $\mathfrak{sl}_2$ into the top left corner of $\mathfrak{sl}_n$. The case $n=3$ of that was recently asked about here, with matching results (note that what's called $V_2$ there contains the weights $-2,0,2$ each with multiplicity $1$; then each of the $V_1$'s contains weights $\pm 1$ each with multiplicity $1$; and one more $V_0$ gives another $1$-dimensional space of weight $0$. Adding up, we have the weights $\pm2$ with dimensions $1$ each, the weights $\pm1$ with dimension $2$ each, and the weight $0$ with dimension $2$).
