Let $L$ be a Lie algebra over a field of characteristic $p>0$ and $D$ a derivation of $L$. For every $x\in L$ denote by $\mathrm{ad} x$ the adjoint map $\mathrm{ad}x: L \rightarrow L, a\mapsto [x,a]$. Is the following relation true?

$$D^{p-1}((\mathrm{ad} x)^{p-1}(D(x)))=(\mathrm{ad} x)^{p-1}(D^p(x))$$

I proved it for small values of $p$, but I am not able to find a general argument. In the case the result is well-known, a reference will be very welcome.


Yes, this identity holds, and both sides are equal to $-D^p(x^p)$.

We need the following facts, all of which are straightforward to verify, where $D$ is an arbitrary derivation, still working over characteristic $p>0$: $$(ad_x)^k(y)=\sum_{m=0}^{k}(-1)^{m+1} {k\choose m}x^m y x^{k-m}$$ $$D(x^k)=\sum_{m=0}^{k-1}x^m D(x)x^{k-1-m}$$ $$D^p(xy)=D^p(x)y+xD^p(y)$$

$$(-1)^m {p-1\choose m}=1 \text{mod p}$$

Then we have:

\begin{align*} D^{p-1}((ad_x)^{p-1}(D(x))&=D^{p-1}\bigg(\sum_{m=0}^{p-1}(-1)^{m+1}{p-1\choose m}x^mD(x)x^{p-1-m}\bigg)\\ &=-D^{p-1}\bigg(\sum_{m=0}^{p-1}x^mD(x)x^{p-1-m}\bigg)\\ &=-D^{p}(x^p)\\ &=-\sum_{m=0}^{p-1}x^mD^p(x)x^{p-1-m}\\ &=\sum_{m=0}^{p-1}(-1)^{m+1}{p-1\choose m}x^mD^p(x)x^{p-1-m}\\ &=(ad_x)^{p-1}(D^p(x)) \end{align*}

  • $\begingroup$ Great! Thank you very much. $\endgroup$ – Rocky Smith May 28 '19 at 7:43
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    $\begingroup$ Sorry if this is dumb, but if $L$ is just a Lie algebra, what does it mean to multiply elements the way you are doing? Are you working in the universal enveloping algebra? $\endgroup$ – Kevin Casto Feb 21 '20 at 22:54
  • $\begingroup$ Yep, this is all taking place in the universal enveloping algebra, noting that the canonical map from $L$ to $U(L)$ is injective, so we can verify stuff there. $\endgroup$ – Chris H Feb 22 '20 at 3:26

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