Let $A$ be a free associative ${\mathbb Z}$-algebra on generators $I = \{ x_i ~:~ I \in {\mathbb N} \}$. Setting the usual bracket $[a,b] = ab-ba$, it forms a free Lie ring $A^{(-)}$. The Dynkin operator $\delta : A \rightarrow A^{(-)}$ on associative monomials in the free generating set $I$ is defined as $\delta (x_{i_1} x_{i_2} \dotsc x_{i_m}) = [\dotsc [x_{i_1}, x_{i_2}], \dotsc, x_{i_m}]$ (left normed commutator). I am trying to understand a proof of the following lemma in book by E.I.Khukhro ($p$-automorphisms of finite $p$-groups) stating : $\delta(x_{k+1}[x_1, \dotsc, x_k]) = [x_{k+1}, [x_1, \dotsc, x_k]]$.

The step while we prove $\delta (x_{k+1} x_k [x_1, \dotsc, x_{k-1}]) = [[x_{k+1}, x_k], [x_1, \dotsc, x_{k-1}]]$, the argument says "we replace $x_{k+1}x_k$ by the commutator $[x_{k+1}, x_k]$ and then regard this commutator as a new variable, which implies this step.

The replacement seems to change the definition of Dynkin operator by changing the variables. It is unclear how to derive this step. On another thought, a direct approach seems to look like a modified version of the lemma, which require another proof.

Is there any general formula for $\delta(x_{k+l} \dotsc x_{k+1}[x_1, \dotsc, x_k])$ with a proof via induction on $k+l$?

  • $\begingroup$ I have no idea how you define $\delta$, and even how it is supposed to be an operator from which space to which other space (maybe the free Lie ring to itself). $\endgroup$
    – YCor
    Mar 31 '19 at 19:07
  • $\begingroup$ What do you mean by "associative Lie monomials"? $\endgroup$
    – YCor
    Mar 31 '19 at 19:17
  • $\begingroup$ Products of the form $x_{i_1} \dotsc x_{i_m}$ where the variables need not be commutative, but these are associative. $\endgroup$
    – Siddhartha
    Mar 31 '19 at 19:18
  • 1
    $\begingroup$ Maybe also recall the convention for multiple brackets: $[x_1,\dots,x_k]$ is defined as $[x_1,[x_2,\dots,x_k]]$ or as $[[x_1,\dots,x_{k-1}],x_k]$? $\endgroup$
    – YCor
    Mar 31 '19 at 19:24
  • 1
    $\begingroup$ "left-normed": $[x_1,\dots,x_k]=[[x_1,\dots,x_{k-1}],x_k]$. $\endgroup$
    – YCor
    Mar 31 '19 at 20:06

Lemma : If $\delta$ is the Dynkin operator defined as above, then

$$ \delta(x_{k+l} x_{k+l-1} \dotsc x_{k+1} [x_1, \dotsc, x_k]) = \left[ [x_{k+l}, x_{k+l-1}, \dotsc, x_{k+1}], [x_1, \dotsc, x_k] \right] $$

where the notation $[x_1, \dotsc, x_k] := [ [x_1, \dotsc, x_{k-1}], x_k]$ (left normed commutator).

Proof : Induction on $k$, where $l \in {\mathbb N}$ is arbitrary. For $k=1$, it follows from definition. Assume this holds up to some $k-1$ where $k \geq 2$. We use notations $A = [x_{k+l}, \dotsc, x_{k+1}], B = [x_1, \dotsc, x_{k-1}], C = x_{k+l} \dotsc x_{k+1} [x_1, \dotsc, x_k]$ for convenience. Then $$ \delta(C) = \delta(x_{k+l} \dotsc x_{k+1} B x_k) - \delta(x_{k+l} \dotsc x_{k+1} x_k B) \hspace{1in} (\ast) $$ Now write $B = \sum_{i_1, \dotsc, i_{k-1}} y_{i_1} \dotsc y_{i_{k-1}}$ as sum of the non-commutating monomials on $x_1, \dotsc, x_{k-1}$. By induction hypothesis, the second part of $(\ast)$ is equal to $-[[A, x_k], B]$. The first part is $$ = \delta( \sum_{i_1, \dotsc, i_{k-1}} x_{k+l} \dotsc x_{k+1} y_{i_1} \dotsc y_{i_{k-1}} x_k) = \sum_{i_1, \dotsc, i_{k-1}} [x_{k+l}, \dotsc, x_{k+1}, y_{i_1}, \dotsc, y_{i_{k-1}}, x_k] $$ using linearity and definition of Dynkin operator. Now this $$ = \sum_{i_1, \dotsc, i_{k-1}} \left( [A, y_{i_1}, \dotsc, y_{i_{k-1}}] x_k - x_k [A, y_{i_1}, \dotsc, y_{i_{k-1}}] \right) $$ $$ = \delta \left( x_{k+l} \dotsc x_{k+1} \sum_{i_1, \dotsc, i_{k-1}} y_{i_1} \dotsc y_{i_{k-1}} \right) x_k - x_k \delta \left( x_{k+l} \dotsc x_{k+1} \sum_{i_1, \dotsc, i_{k-1}} y_{i_1} \dotsc y_{i_{k-1}} \right) $$ $$ = \delta(x_{k+l} \dotsc x_{k+1} B) x_k - x_k \delta(x_{k+l} \dotsc x_{k+1} B) = [A, B]x_k - x_k[A, B] = [[A, B], x_k] $$ using induction hypothesis. Putting these in $(\ast)$ we get $$ \delta(C) = [[A, B], x_k] - [[A, x_k], B] = [A, [B, x_k]] $$ by Jacobi identity, completing the proof.


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