In Jantzen's AMS text 'Lectures on Quantum Groups' he makes the following remark (p.187, preface to Chapter 9):


"For general (complex semisimple f.d. Lie algebra) $\frak{g}$ we can consider for each simple root $\alpha$... a Lie subalgebra (of $\frak{g}$) isomorphic to $\frak{sl}\_{2}$ (ie, the Lie subalgebra $\frak{s}\_{\alpha}$ generated by suitable $X\_{\alpha}\in \frak{g}\_{\alpha}$, $Y\_{\alpha}\in\frak{g}\_{-\alpha}$). So, if $M$ is a f.d. $\frak{g}$-module, then one can find (for fixed $\alpha$) a basis $v\_{1},\ldots,v\_{n}$ such that $Y\_{\alpha}v\_{i}$ is either $0$ or a nonzero multiple of another $v\_{j}$ and such that also each $X\_{\alpha}v\_{h}$ is either $0$ or a nonzero multiple of another $v\_{l}$. However, in general, there does not exist a basis that works simultaneously for all simple $\alpha$. (There are exceptions, such as the adjoint representations...)..."


The first part of this statement is standard (we are applying complete reducibility of $M$ as an $\frak{s}\_{\alpha}$-module). However, I hope that someone can illuminate the last sentence on 'exceptions' - is this a typo/mis-statement? Or am I missing something here? 


It is not possible to obtain a simultaneous basis for the adjoint representation of $\frak{sl}\_{3}$: indeed, if $\alpha\_{1},\alpha\_{2}$ are the simple roots and $\frak{s}\_{1},\frak{s}\_{2}$ the corresponding $\frak{sl}\_{2}$-triples then we can decompose $\frak{sl}\_{3}$ as 


$\frak{sl}\_{3}$$\cong L(1)\oplus L(2)\oplus L(0)\oplus L(1)$


when we consider $\frak{sl}\_{3}$ as either a $\frak{s}\_{1}$-`or $\frak{s}_{2}$-`module. Here, $L(n)$ is the irreducible $\frak{sl}\_{2}$-module of dimension $n+1$. Also, in both decompositions we have $L(0)$ appears as a subspace of $\frak{h}$ (the $0$-weight space).


If we were to have a simultaneous basis as described above we would need a basis vector $u\in\frak{h}\subset\frak{sl}\_{3}$ corresponding to the copy of $L(0)$ appearing in the $\frak{s}\_{1}$- and $\frak{s}\_{2}$-decompositions of $\frak{sl}\_{3}$. This would imply that $\ker ad \;X\_{\alpha\_{1}}\cap \ker ad \; X\_{\alpha\_{2}}\cap \frak{h}$ is nonzero, which is impossible (as can be seen by a basic calculation) since we are in characteristic $0$.


Thanks in advance for your comments.



A remark in Jantzen's 'Lectures on Quantum Groups'
Your argument in the second to last paragraph is wrong. The basis of the adjoint representation given by any basis vector in each root space, and the basis of $\mathfrak{h}$ given by the simple coroots $H\_i=[E\_i,F\_i]$ has this property for any semi-simple Lie algebra.


The mistake you made was thinking that the basis had to be compatible with each $\mathfrak{sl}\_2$ decomposition, and in particular that all but one basis vector in $\mathfrak{h}$ must be sent to zero by bracket with $E\_i$ or $F\_i$. This is not what Jantzen said (you're right that no basis with that property can exist except in products of $\mathfrak{sl}\_2$'s); he only said that the bracket of one basis vector with $E\_i$ or $F\_i$ must be a multiple of another basis vector, and any basis of $\mathfrak{h}$ has that property. In fact, the only place where a basis vector from each root space and a completely arbitrary basis of $\mathfrak{h}$ will fail is that $[E\_i,F\_i]$ must be a multiple of a basis vector, which forces us to take the coroots.


