MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

My reading of the passage is different. Take for example the adjoint representation. Here one can choose a Chevalley basis, whose properties meet Jantzen's requirements relative to each fixed simple root. In your set-up you need to keep in mind that there is no canonical direct sum decomposition of the type you indicate, but a Chevalley basis does give (up to sign choices) something unique.

One thing that makes this example (or some "natural" modules) work in Jantzen's discussion is the fact that weight spaces are 1-dimensional, apart from the zero weight space. In general, it's much trickier to fix an all-purpose basis in each separate weight space. This is what makes Lusztig's "canonical basis" ideas so striking, along with Kashiwara's development of "crystal bases" (the subject of Jantzen's Chapters 9-11.

Anyway, as Wilberd points out, the chapter introduction includes some informal language which by itself isn't so important compared to the rigorous material following.

It's usually a good idea to address a question like this to the author (when possible) and also check to see if errata are posted anywhere. For instance, there are some lists (mostly involving very minor corrections) posted at Jantzen's homepage in Aarhus:

share|cite|improve this answer
Thanks for the comment and helpful advice. Cheers – George Melvin Sep 6 '12 at 16:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.