4
$\begingroup$

The index of a Lie algebra is $\mathrm{ind}(\mathfrak{g})=\mathrm{min}_{\lambda \in \mathfrak{g}^{*}} \mathrm{dim} \mathfrak{g}^{\lambda}$, where $\mathfrak{g}^{\lambda} = \lbrace x\in \mathfrak{g} \mid \lambda\circ \mathrm{ad}_{x} = 0 \rbrace$.

Is there any way to classify all complex n-dimensional nilpotent Lie algebra $\mathfrak{g}$ whose index is $\mathrm{ind}\ \mathfrak{g} = n-2$ ?

Examples would be the filiform Lie algebras, if I am not mistaken, e.g. $\mathfrak{g}$ generated by $\{x_1, \ldots, x_n\}$ subject to $[x_1,x_i]=x_{i+1}$ for $2\leq i < n$.

$\endgroup$
2
  • 1
    $\begingroup$ Just adding some words to the definition: the index of Lie algebra is the codimension of the generic coadjoint orbit. For semisimple Lie algebras the index agrees with the rank. So among the semisimple Lie algebras, only those of type $A_1$ have maximal index. $\endgroup$ Mar 14, 2011 at 13:46
  • $\begingroup$ There is an article you can find in <kirj.ee/public/proceedings_pdf/2010/issue_4/…>. In proposition 4 of this article, there is a formula you might find interesting. It relates the index of a Lie algebra with the rank of a matrix. There are some examples computed. $\endgroup$
    – dan232
    Jul 23, 2012 at 15:45

1 Answer 1

3
$\begingroup$

It is known, that the index of a Lie algebra is a semi-invariant for degenerations (by Ooms and Elashvili), i.e., if $L_1$ degenerates to $L_2$, then $ind(L_1)\le ind(L_2)$. This is very useful. For example, it follows that any filiform Lie algebra of dimension $n$ has index less or equal than $n-2$, where only the standard graded filiform $L(n)$, which you have defined above, has exactly index $n-2$. In general, there are many other Lie algebras of dimension $n$ and index $n-2$, e.g., also the quasi-filiform Lie algebras $L(n-1)\oplus \mathbb{C}$. See here also the work Adini and Makhlouf. The Hasse-diagram of complex nilpotent Lie algebras in dimension 6 gives explicit examples, e.g., we have degenerations from the top algebra $L_{6,20}$ as follows (notation of Magnin for the Lie algebras) $L_{6,20}\rightarrow L_{6,18}\rightarrow L_{6,17} \rightarrow L_{6,16} \rightarrow L_{5,5} \oplus \mathbb{C}\rightarrow \mathbb{C}^6$, with index numbers $2 \rightarrow 2 \rightarrow 2 \rightarrow 4 \rightarrow 4 \rightarrow 6$. See my paper arXiv:0911.2995 for this, and a discussion on the maximal dimension of an abelian subalgebra $\alpha (L)$, which is related to the index by $\alpha (L)\le (\dim (L)+ind (L))/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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