Nilpotent matrices related to Lie algebras of special orthogonal groups in characteristic 0

In terms of matrix theory, the question I'm led to is the following: Start with an $n$-dimensional vector space over an algebraically closed field of characteristic 0 such as $\mathbb{C}$, which has a non-degenerate symmetric bilinear form. Consider all $n \times n$ skew-adjoint matrices $A$ relative to this form.

Given $n\geq 5$, what is the least power $k$ for which all nilpotent matrices of this type satisfy $A^k=0$?

For any nilpotent $n \times n$ matrix $A$, it follows from the Cayley-Hamilton Theorem that $A^n=0$. But in the special case here it seems plausible to expect a slightly smaller minimum: namely, $n-1$ if $n$ is odd and $n-2$ if $n$ is even. I also wonder what is written down in the literature along this line.

There is of course a hidden agenda, relative to simple Lie algebras attached to special orthogonal groups over a field like $\mathbb{C}$. In the classification of simple types $A_\ell-D_\ell$ of rank $\ell$, the respective Coxeter numbers are $\ell+1, 2\ell, 2\ell, 2(\ell-1)$. (Types $B, C$ share the same Weyl group.) Types $A, C$ are realized naturally as $n\times n$ matrices with $n=h$, but the other Lie algebras of orthogonal type yield $n=2\ell+1$ and $n=2\ell$. So my question for the latter types means: does every nilpotent element $e$ of this natural matrix Lie algebra satisfy $e^h =0$ as in types $A,C$?

Behind this question is a related prime characteristic question for restricted Lie algebras, motivated in part by Kostant's classical 1959 paper in Amer. J. Math. (Corollary 5.4). In the general setting of simple Lie algebras he showed that regular (= principal) "nilpotent" elements $e$ are characterized by a condition on their adjoint operators: $(\mathrm{ad} \:e)^{2q}\neq 0$ where $q$ is the sum of coefficients of the highest root expressed relative to simple roots. Moreover, the next power annihilates all regular nilpotents. Earlier he showed that $q+1 = h$ is the Coxeter number of the Weyl group. (But there is a misprint in that corollary.)

ADDED: As Victor points out, except for a small decrease in type $D$ the four classical families of simple Lie algebras have index of nilpotence in their natural representations given by the Cayley-Hamilton approach in type $A$. My question arose from passing to characteristic $p>0$ via a Chevalley basis over $\mathbb{Z}$, then extending scalars. For $p \geq h$, results from the mid-1980s on cohomology of restricted Lie algebras and support varieties of modules (Jantzen, Friedlander-Parshall, ... ) reveal that for the built-in $[p]$ operation on such a Lie algebra one has $\text{ad}\: e^{[p]} = 0 = (\text{ad} \:e)^p$ for all nilpotents $e$. But in natural matrix representations like those for types $A-D$, the $[p]$ operation is the usual matrix power. Here the slightly modified Cayley-Hamilton power needed for vanishing agrees.

At the extreme of $E_8$ there is more divergence: Here the "natural" smallest faithful representation is given by the adjoint module with $n=248$, whereas $h=30$. For $p = 31$ this power of each $\text{ad} \:e$ vanishes, contrasting with Kostant's characteristic 0 result which requires a power at least $59$ when $e$ is regular.

• For odd n there is n-dimensional orthogonal irreducible representation of sl(2). Then $e\in sl(2)$ would be represented by skew-adjoint operator with $e^{n-1}\ne 0$. This seems to contradict your statement.. Mar 3, 2011 at 4:27

Let $\lambda$ be a partition of $n.$ Then there exists a skew-symmetric nilpotent matrix whose Jordan blocks sizes are $\lambda_i$ if and only if every even parts has even multiplicity. This follows easily from the representation theory of $\mathfrak{sl}_2$ and is duly recorded in standard sources, e.g. Collingwood and McGovern. It follows that the maximum "nilpotence index" of a skew-symmetric $n\times n$ matrix is $n$ for odd $n$ and $n-1$ for even $n.$