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2 corrected notation

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 M. \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. 1 # A remark in Jantzen's 'Lectures on Quantum Groups' 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 $M$. 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 $0$.