Nilpotent Lie algebras of vector fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T01:48:51Z http://mathoverflow.net/feeds/question/65115 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65115/nilpotent-lie-algebras-of-vector-fields Nilpotent Lie algebras of vector fields Pasha Zusmanovich 2011-05-16T07:35:25Z 2011-05-16T15:58:27Z <p>Let $L$ be a finite-dimensional nilpotent subalgebra of the Lie algebra $W_n$ of all vector fields in $n$ variables (I am interested both in polynomial and formal vector fields). Does there exist a bound in terms of $n$ on the index of nilpotency of $L$?</p> <p>For $n=1$ the answer is trivially "yes": every finite dimensional subalgebra is of dimension $\le 3$, concentrated in degrees $-1$, $0$, $1$. For $n>1$, I don't know.</p> <p><b>Motivation</b>: the question looks for me interesting in its own right, but also arises in control theory. There, one has a criterion for nonlinear systems to be so-called finitely discretizable (roughly, to admit a polynomial solution of some sort) in terms of nilpotency of the Lie algebra generated by the corresponding vector fields. So, when one applies this criterion on practice, one wants to be sure that it is enough to check the vanishing of the iterated Lie brackets up to a given degree. For the control theory application, the base field is $\mathbb R$, but the question does not seem to be dependent on the base field (as long as it is of characteristic zero at least).</p> http://mathoverflow.net/questions/65115/nilpotent-lie-algebras-of-vector-fields/65125#65125 Answer by Vladimir Dotsenko for Nilpotent Lie algebras of vector fields Vladimir Dotsenko 2011-05-16T09:54:10Z 2011-05-16T09:54:10Z <p>I am afraid not: the subalgebra of $W_2$ spanned by $\partial_x, \partial_y, x\partial_y, \ldots, x^m\partial_y$ has the nilpotency index $m+1$. </p>