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Let $L$ be a lie algebra over finite field, for $ x,y$ in $L$ I want to solve the following bracket: $[yx^k,x]=?$ How can we describe that in the format of $[...[y,x],x],...,x]=[y,x]_i$ ($i-times$)

Nil
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  • I suspect you'll need to pursue an inductive approach: compute the first few cases of integer $k$, locate a pattern, then prove it by induction on $k$. – Semiclassical Sep 07 '14 at 14:23
  • What do you mean by the powers of an element of a Lie algebra? – Qiaochu Yuan Sep 07 '14 at 16:21
  • I mean $[yx^k,x]$. In fact I am looking for a proper formula to describe the $[x^k,y^n]$ in general. For special case I want to know how can compute $[yx^k,x]$. – Nil Sep 07 '14 at 16:28
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    Well powers of $ x_\alpha $ and $ y_\alpha $ where $\alpha $ is weight ( or for the adj root) act as annihilators, through the braket,of a vector space. This is due to finite dimensionality and strings for the y and the concept of maximal weighy vector for x. You can find these formulas online. I am not sure if this is what you are looking for, but I am just putting it out there. Also powers of lie algebra elements only exist in a universal enveloping algebra and representations. – dylan7 Sep 07 '14 at 17:25
  • My main aim is for lie algebras over finite fields, Please accept my apologize, I should explain it initially. – Nil Sep 07 '14 at 18:01
  • The problem is that $x^k$ makes no sense in a Lie algebra. – Mariano Suárez-Álvarez Sep 07 '14 at 18:35
  • So, what about $[yx^k,x]$ ? Is it meaningless? – Nil Sep 07 '14 at 18:38

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The product $yx^k$ is not yet defined, because the product of two elements $x,y$ in a Lie algebra is written $[x,y]$. Of course we might consider another bilinear product $(x,y)\mapsto xy$ on the underlying vector space (e.g., Poisson algebra). The notion $ad(x)^k (y)=[x,[x,[x,\cdots [x,y]]]\cdots ]$ makes sense, of course. For Lie algebra representations, sometimes the product $x^ky$ appears in the universal enveloping algebra $U(\mathfrak{g})$ of the Lie algebra $\mathfrak{g}$. This is an associative product.

Dietrich Burde
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