# 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 $\frak{sl}_{3}$. 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 $0$.

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To me it is not clear what the exceptions are meant to be exceptions to. So it seems best to ignore the sentence entirely. You understand perfectly what is happening. –  Wilberd van der Kallen Sep 6 '12 at 9:07

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.

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Ahh, thanks for your comment. I was certain that I must have been misreading this and glad for a fresh set of eyes to confirm. Cheers –  George Melvin Sep 6 '12 at 16:46