The answers to [this question](https://mathoverflow.net/questions/37101/computing-the-index-of-a-lie-algebra-what-is-known-beyond-the-reductive-case) 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](http://en.wikipedia.org/wiki/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.



Inequality for the index of a Lie algebra using its Levi decomposition
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](http://www.ams.org//mathscinet-getitem?mr=2146652) (2005). Also, see A. Panasyuk, [Reduction by stages and the Raïs-type formula for the index of a Lie algebra with an ideal](http://www.ams.org//mathscinet-getitem?mr=2369183) (2008).


