# restricted lie algebras definition

Jacobson (Lie algebras, p.187) defines what is meant by a restricted Lie algebra: Def4: A restricted Lie algebra, $L$, of characteristic $p\not = 0$ is a Lie algebra of characteristic $p$ in which there is defined a mapping $a\rightarrow a^{[p]}$ such that

1. $(\alpha a)^{[p]}=\alpha^p a^{[p]}$
2. $(a+b)^{[p]}=a^{[p]}+b^{[p]}+\sum_{i=1}^{p-1}s_{i}(a,b)$, where $is_{i}(a,b)$ is the coefficient of $\lambda^{i-1}$ in $a(ad(\lambda a+b))^{p-1}$ and
3. [$ab^{[p]}$]$=a(ad b)^p$

My questions lie in the understanding of how identity 2 comes about. On page 187 Jacobson introduces the polynomial ring $\mathfrak U(\lambda)$ and writes $(\lambda a+b)^p= \lambda ^{p} a^p+b^p +\sum_{i=1}^{p-1}s_{i}(a,b)\lambda^{i}$ where $s_{i}(a,b)$ is a polynomial in $a,b$ of total degree $p$. So in particular, for $p=2$ we have $(\lambda a+ b)^2=\lambda^2a^2+b^2+\lambda(ab+ba)$ so I conclude that $s_{1}(a,b)=ab+ba$

I have a two questions. First, at the bottom of page 187, he says, that if p=2, then $s_{1}(a,b)=$[$a,b$], but [$a,b$]$=ab-ba$, not $ab+ba$ like I got above. Where is my mistake?

Second, he differentiates $(\lambda a+b)^p= \lambda ^{p} a^p+b^p +\sum_{i=1}^{p-1}s_{i}(a,b)\lambda^{i}$ with respect to $\lambda$ and gets $\sum^{p-1}_{i=0}(\lambda a+b)^ia(\lambda a+b)^{p-i-1}=\sum_{i=1}^{p-1}is_{i}(a,b)\lambda^{i-1}$. I do not understand how the left hand side came about. I would like to see, or at least have it explained to me how to differentiate the left hand side. Thank you for your time.

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When it comes to motivation, you might already know it, but if $G$ is an algebraic group over a field of characteristic $p\neq 0$, then its Lie algebra is a restricted Lie algebra. So they do come up naturally. – M Turgeon Jul 17 '12 at 17:04
if $p=2$ then $ab+ba=ab-ba$, so you made no mistake there – user8268 Jul 17 '12 at 19:31
for the second question - this is just Leibniz rule – user8268 Jul 17 '12 at 19:42

The motivation for the second axiom of a restricted lie algebra as they're listed here is the binomial theorem. Given any $k$-algbera $A$ with $k$ a field of characteristic $p$, we can define a bracket on $A$ by $[a,b]:=ab-ba$ and a $[p]$-power map by $a^{[p]}:=a^p$. It is actually a fairly difficult exercise to show that with this structure, $A$ is indeed a restricted lie algebra (if $A$ were commutative, the exercise would be much easier).

In the case that $A$ is commutative, the second axiom applied to $A$ is exactly the binomial theorem in characteristic $p$: $(a+b)^p=a^p+b^p$. In the general case that $A$ is not commutative, the middle terms are nonzero, and they are given by the $s_i(a,b)$.

I encourage you to check this for small values of $p$. For example, write out $(a+b)^5$ not assuming that $ab=ba$ and see the middle terms that arise.

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