Let $\mathfrak{g}$ be a semi-simple finite dimensional Lie algebra over $\mathbb C$.
Is it true that
$$
\dim \mathfrak h \ge \mathop{rank} \mathfrak g 
$$
for **any** maximal abelian subalgebra $\mathfrak h \subset \mathfrak g$ ? Here by maximal I mean that $\mathfrak h$ is not properly contained in any other abelian subalgebra of $\mathfrak g$.



Minimal dimension of maximal abelian subalgebras
I think the answer to my question is no. A search at Mathscinet revealed the following:



> 
> Gerstenhaber conjectured in [Ann. of
>  Math. (2) 73 (1961), 324--348;
>  MR0132079 (24 #A1926)] that a maximal
>  commutative subalgebra of the
>  associative ring of $n\times n$
>  matrices $M\_n(k)$ always has dimension
>  $\ge n$, and proved this is the case
>  for subalgebras generated by two
>  elements.
> 
> 
> The conjecture was disproved by 
>  Courter in [Duke Math. J. 32 (1965),
>  225--232; MR0174583 (30 #4784)], by
>  constructing a maximal commutative
>  subalgebra of $M\_{14}(k)$ having
>  dimension $13$. Tensor powers give further examples of dimension $13^m$ in $M\_{14^m}(k)$.
> 
> 
> Gustafson [J. Algebra 42 (1976), no.
>  2, 557--563; MR0422341 (54 #10331)]
>  proved the lower bound $\ge n^{2/3}$
>  for arbitrary maximal commutative
>  subalgebras of $M\_n(k)$.
> 
> 
> More recently, Laffey [Linear Algebra
>  Appl. 71 (1985), 199--212; MR0813045
>  (87a:15025)] improved the lower bound
>  to $(2n)^{2/3}$ and exhibited
>  infinitely many maximal commutative
>  subalgebras of $M\_n(k)$ having
>  dimension $3n^{2/3} - 4$.
> 
> 
> 


I haven't looked at the papers mentioned above yet but they seem to imply that there are
maximal abelian Lie subalgebras of $\mathfrak{gl}(n,k)$ ( or $\mathfrak{sl}(n,k)$ )
having dimension smaller than $n-1$. 


Indeed, start with a maximal commutative subalgebra $H \subset M\_n(k)$. Since, it is commutative the Lie bracket $[A,B]=AB-BA$ is trivial over it. Therefore it gives rise
to an abelian subalgebra $\mathfrak h \subset \mathfrak{gl}(n,k)$ which must be maximal. Otherwise the associative subalgebra generated by any abelian Lie subalgebra containing it would be a commutative subalgebra of $M\_n(k)$ strictly containg $H$.




---


I finally took a look at the papers mentioned above. 
Courter's example is the subalgebra of $M\_{14}(k)$
generated by the Identity and elements of the form
$$ 
\left(
 \begin{array}{cccccccccccccc}
 0&0&a&0&b&0&c&e&0&f&0&g&m&n\\
 0&0&0&a&0&b&d&0&e&0&f&h&p&q\\
 & & & & & & & & & & & &c&0\\
 & & & & & & & & & & & &d&0\\
 & & & & & & & & & & & &0&c\\
 & & & & & & & & & & & &0&d\\
 & & & & & & & & & & & &a&b\\
 & & & & & & & & & & & &g&0\\ 
 & & & & & &0 & & & & & &h&0\\ 
 & & & & & & & & & & & &0&g\\ 
 & & & & & & & & & & & &0&h\\ 
 & & & & & & & & & & & &e&f\\ 
 & & & & & & & & & & & &0&0\\
 & & & & & & & & & & & &0&0
 \end{array}
\right)
$$
where $a,b,c,d,e,f,g,h,m,n,p,q \in k$ are arbitrary. These very same elements
generate a maximal abelian subalgebra of $\mathfrak{sl}(14,k)$ of dimension $12$ as I, or rather my computer, checked.


