most recent 30 from http://mathoverflow.net 2013-05-22T18:09:34Z http://mathoverflow.net/feeds/question/58420 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58420/what-is-known-on-finite-dimensional-nilpotent-lie-algebras-with-maximal-index CLomp 2011-03-14T12:40:12Z 2013-03-30T17:14:41Z

<p>The index of a Lie algebra is $\mathrm{ind}(\mathfrak{g})=\mathrm{min}_{\lambda \in \mathfrak{g}^{*}} \mathrm{dim} \mathfrak{g}^{\lambda}$, where $\mathfrak{g}^{\lambda} = \lbrace x\in \mathfrak{g} \mid \lambda\circ \mathrm{ad}_{x} = 0 \rbrace$.</p> <p>Is there any way to classify all complex n-dimensional nilpotent Lie algebra $\mathfrak{g}$ whose index is $\mathrm{ind}\ \mathfrak{g} = n-2$ ?</p> <p>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 $2\leq i &lt; n$.</p>

http://mathoverflow.net/questions/58420/what-is-known-on-finite-dimensional-nilpotent-lie-algebras-with-maximal-index/126034#126034 Dietrich Burde 2013-03-30T17:14:41Z 2013-03-30T17:14:41Z

<p>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 $\alpha (L)\le (\dim (L)+ind (L))/2$.</p>