2
$\begingroup$

Suppose we have a Lie algebra with structure constants

$$\mathrm{d}e^i=\sum_{j<k}a_{ijk}e^j\wedge e^k$$

for some coefficients $a_{ijk}$.

In this setting, how may be checked (perhaps computationally?) that our algebra is nilpotent? I wonder whether there is a somewhat nice algorithm involving the coefficients.

Thank you for your attention and answers.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ The structure constants are right there; take a dual basis and see that the structure constants are these $a_{ijk}$. Then just check as usual. $\endgroup$
    – Ben McKay
    Commented Feb 27, 2016 at 8:48
  • $\begingroup$ @BenMcKay Yes, of course I may use the standard definition, but I wonder whether there exists a specially refined (preferrably algebraic) condition for those coefficients. For instance, checking that a matrix is nilpotent is equivalent to the characteristic polynomial being $x^n$, which is an easy-to-use algebraic condition on the coefficients. $\endgroup$
    – Jjm
    Commented Feb 27, 2016 at 8:54
  • $\begingroup$ The lower central series can be computed "computationally". $\endgroup$
    – YCor
    Commented Feb 27, 2016 at 11:30
  • $\begingroup$ ... and the most obvious algorithm is to compute all the $n^n$ $n$-fold brackets ($n$ is the dimension) $[e_{i_1},\dots,e_{i_n}]$ and check whether they are all zero. It's not the most efficient: computing inductively the lower central series requires a polynomial number of operations in terms of $n$ (the coefficients can grow in the process, but this depends on how we compute in the ground field anyway). $\endgroup$
    – YCor
    Commented Feb 27, 2016 at 17:01

2 Answers 2

Reset to default
2
$\begingroup$

By Engel's theorem, the Lie algebra is nilpotent just when, for any fixed $k$, the matrix $A_k=(a_{ijk})$ is nilpotent.

Edit: By Engel's theorem, the Lie algebra is nilpotent just when every linear combination of the matrices $A_k=(a_{ijk})$ is nilpotent.

Improve this answer
$\endgroup$
5
  • 5
    $\begingroup$ Seems like saying that when there's a basis $(e_i)$ such that each $ad(e_i)$ is nilpotent, then the Lie algebra is nilpotent. This is false in $sl_2$, e.g. with a basis $(s,x,y)$ with law $[x,y]=s-x+y$, $[s,x]=s+x+y$, $[s,y]=s-x-y$. $\endgroup$
    – YCor
    Commented Feb 27, 2016 at 11:28
  • $\begingroup$ @YCor: Thanks. I should have required that every linear combination of the matrices $A_k$ is nilpotent. $\endgroup$
    – Ben McKay
    Commented Feb 27, 2016 at 15:20
  • $\begingroup$ @YCor: It used to be true (I don't know whether it is still true) that you couldn't delete an answer that had been accepted as correct by the OP. The best you could (still can?) do is edit the answer, pointing out the error and, if you can't correct it, then putting up a warning that the answer has been found to be in error. $\endgroup$
    – Robert Bryant
    Commented Feb 28, 2016 at 13:31
  • $\begingroup$ About the edited answer, while at first sight it does not seem to be an algorithm, it works at it just mean that $(t_1A_1+\dots+t_nA_n)^n=0$, where $t_i$ are formal variables. (Of course, it is only an algorithm relying on the assumption that we can compute in the field.) $\endgroup$
    – YCor
    Commented Feb 29, 2016 at 11:37
  • $\begingroup$ An idea in this direction: for a Lie algebra L to be nilpotent, it is sufficient to show that, for a given basis B of L, all the iterated brackets of elements of B have to be nilpotent, see: arxiv.org/abs/1010.0821 but this is far from being effective due to the number of nilpotent matrices to consider $\endgroup$
    – Bulois Michael
    Commented Mar 1, 2016 at 9:15
3
$\begingroup$

The book of Willem de Graaf "Lie Algebras: Theory and Algorithms" contains some algorithms for determining whether or not a Lie algebra $L$ is nilpotent. Of course, Engel's theorem is one of the main tools. We can very efficiently see that a given Lie algebra $L$ is not nilpotent, by checking first the trace condition, i.e., $tr(ad(x))=0$ for all $x\in L$, or to find a nonzero eigenvalue of some $ad(x)$. There are also algorithms for calculating the nilradical, even for fields of prime characteristic $p>0$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Note that an algorithm, in the implementation meaning, only makes sense if we start with a computable field. Talking of an algorithm for a Lie algebra with real or complex coefficients is senseless (how do you input a real number?). The efficiency of the algorithm, in particular, depends on two variables: the dimension, and the max of "lengths" of structure coefficients (though I'm not sure there a single notion of length, for say, coefficients in $\mathbb{Q}[t]$) $\endgroup$
    – YCor
    Commented Feb 28, 2016 at 12:46
  • $\begingroup$ I thought of algorithms given in the book, and implemented in "gap", which do work for real and complex Lie algebras. $\endgroup$
    – Dietrich Burde
    Commented Feb 28, 2016 at 13:12
  • $\begingroup$ They work in real/complex Lie algebra if you input algebraic constants, I guess... you have to compute some numbers in the subring generated by the constants and be able to determine whether these number are zero. There are some issues. Computations are made with some number of digits; if you can't predict beforehand the correct number of digits, you might compute a number to be zero by mistake because its first nonzero digit is too far. $\endgroup$
    – YCor
    Commented Feb 28, 2016 at 13:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .