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The answers to this question indicate that there is a fairly vast literature on the index of Lie algebras. Unfortunately I was not able to find in this literature (or maybe to extract from it) an answer to the following simple-looking question:

Is it true (and under which conditions) that for a Lie algebra $\mathfrak{g}$ with a Levi decomposition $\mathfrak{g}=\mathfrak{h}\triangleright\mathfrak{s}$, where $\mathfrak{s}$ is solvable and $\mathfrak{h}$ is semisimple, we have $$\mathrm{ind}\ \mathfrak{g}\leq \mathrm{ind}\ \mathfrak{h}+\dim\mathfrak{s},$$ and, if yes, when the above inequality is strict (i.e. has $<$ instead of $\leq$) or, on the other hand, when does it turn into an equality (except for the trivial case when $\mathfrak{g}$ is abelian)?

More broadly, are there any other inequalities estimating the index of a Lie algebra from above (or perhaps from below) using its Levi decomposition?

Many thanks in advance.

NOTE: The original question, to which Francois Ziegler provided the counterexample in his answer, concerned the opposite inequality $\mathrm{ind}\ \mathfrak{g}\geq \mathrm{ind}\ \mathfrak{h}+\dim\mathfrak{s}$, but this very answer has motivated me to modify the question.

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  • $\begingroup$ If someone gives a good answer, you should not modify the question to the opposite. It would be better to accept the answer, and then post a better question. $\endgroup$ – Dietrich Burde Dec 9 '14 at 11:10
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Your initially proposed inequality fails if $G$ is Euclid's group $\mathrm{SO}(3)\ltimes\mathbf R^3$. Indeed we have $\operatorname{ind}\mathfrak g = 2$ (generic coadjoint orbits of $G$ have dimension 4), $\operatorname{ind}\mathfrak{so}(3) = 1$ (generic coadjoint orbits of $\mathrm{SO}(3)$ have dimension 2), hence $ \operatorname{ind}\mathfrak g < \operatorname{ind}\mathfrak{so}(3) + \dim\mathbf R^3. $

Edit. The reverse inequality, on the other hand, is true (under the sole assumption that $\mathfrak s$ is an ideal: it needs not be the solvable radical, nor a semidirect factor). Indeed, dualizing $0\to\mathfrak s\to\mathfrak g\to\mathfrak h\to0$ gives an exact sequence $$ 0\to\mathfrak h^*\overset\iota\hookrightarrow\mathfrak g^*\to\mathfrak s^*\to0 $$ by means of which each coadjoint orbit $\mathcal O$ of $H=G/S$ becomes a coadjoint orbit $\iota(\mathcal O)$ of $G$ (on which $S$ acts trivially). Now take $\mathcal O$ to have maximal dimension among coadjoint orbits of $H$. Then we have \begin{align} \operatorname{ind}\mathfrak g &\leqslant \operatorname{codim}_{\mathfrak g^*}\iota(\mathcal O)\\ %&=\dim\mathfrak g^*-\dim\iota(\mathcal O)\\ &=\dim\mathfrak g^*-\dim\mathcal O\\ &=\dim\mathfrak g^*- (\dim\mathfrak h^*-\operatorname{codim}_{\mathfrak h^*}\mathcal O)\\ &=\dim\mathfrak s + \operatorname{ind}\mathfrak h. \end{align} References. You may be interested in §40.4 "Index and semi-direct products" in this book (2005). Also, see A. Panasyuk, Reduction by stages and the Raïs-type formula for the index of a Lie algebra with an ideal (2008).

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  • $\begingroup$ Many thanks! Your answer suggests that the opposite inequality is far more likely to hold in some generality, and is hence more interesting, so I changed the question accordingly (and restored the reference-request tag). $\endgroup$ – just-learning Dec 7 '14 at 19:48
  • $\begingroup$ Extra thanks for the added reference, will have a look. $\endgroup$ – just-learning Dec 7 '14 at 20:34

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