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Jacobson's formula ($ p$ -powers and Lie brackets)

The treatment here essentially follows [1].

Let $ p$ be a prime number. Let $ A$ be a (not necessarily commutative, but unital, associative) algebra over $ \mathbb{F}_p$ . We may also regard $ A$ as a Lie algebra over $ \mathbb{F}_p$ , the bracket product being the ordinary commutator. We would like to obtain a formula for

$\displaystyle (a+b)^p
$

for $ a,b\in A$ with the help of Lie brackets. To do that, we first introduce an transcendent element (=variable) $ T$ which commutes with any element of $ A$ . Then we expand $ (T a+b)^p$ in terms of $ T$ . Namely,

(1) $\displaystyle (T a+b)^p=T^p a^p+ b^p +\sum_{j=1}^{p-1} T^j s_j(a,b)$

where $ s_j(a,b)$ is a non-commutative polynomial in $ a,b$ . Our task then is to find a nice formula for $ s_j$ .

A first thing to do is to differentiate the equation (1) by $ T$ .

(2) $\displaystyle \sum_{k=0}^{p-1} (T a + b)^k a (T a +b)^{p-1-k} =\sum_{j=1}^{p-1} j T^{j-1} s_j(a,b).$

To compute the left hand side, we use a nice trick. For any element $ x\in A$ , we denote by $ \lambda(x)$ (respectively, $ \rho(x)$ ) an operator defined by the left multiplication (respectively, the right multiplication) of $ x$ . That is,

      $\displaystyle \lambda(x): A \ni f \mapsto x f \in A,$
      $\displaystyle \rho(x): A \ni f \mapsto f x \in A.$

It should be noted that for all $ x,y\in A$ , $ \lambda(x)$ and $ \rho(y)$ always commute. Indeed, we have

$\displaystyle \lambda(x)\rho(y)z=x(zy)=(xz)y=\rho(y)\lambda(x).
$

So from an ordinary result on commutative algebra, we have

$\displaystyle (\lambda(a)-\rho(a))^p=\lambda(a)^p -\rho(a)^p=\lambda(a^p)-\rho(a^p).
$

We define $ \operatorname{ad}(a)$ to be

$\displaystyle \operatorname{ad}(a)=\lambda(a)-\rho(a).
$

Then the equation above may be written as

$\displaystyle (\operatorname{ad}(a))^p=\operatorname{ad}(a^p).
$

Another interesting formula is the following.

$\displaystyle (\lambda(a)-\rho(a))^{p-1}=\sum_{j=0}^{p-1} \lambda(a)^j \rho(a)^{p-1-j}
$

(To verify that it holds, we notice that for any commutative variable $ T,U$ , an identity

$\displaystyle (T-U)^p= T^p-U^p=(T-U)\sum_{j=0}^{p-1} T^j U^{p-1-j}
$

holds.) The above formula may then be rewritten as

$\displaystyle (\operatorname{ad}(a))^{p-1}(b)=\sum_{j=0}^{p-1} a^j b a^{p-1-j}.
$

By suitable substitutions, we thus have

(3) $\displaystyle (\operatorname{ad}(T a+b ))^{p-1}(a)=\sum_{j=0}^{p-1} (T a+ b)^j a (T a+b)^{p-1-j}.$

comparing the equations (2) and (3), we see that each $ s_j(a,b)$ belongs to the Lie sub algebra of $ A$ generated by $ a,b$ .

To sum up, we have obtained the following proposition.

PROPOSITION 6.2 (Jacobson's formula)   Let $ p$ be a prime number. Let $ A$ be an algebra over $ \mathbb{F}_p$ (which is not necessarily commutative, but unital, associative as we always assume.) Then for any elements $ a,b\in A$ , we have

$\displaystyle (a+b)^p=a^p+b^p+\sum_{j=1}^{p-1} s_j(a,b)
$

where $ s_j$ is a universal polynomial in $ a,b$ given in the following manner.

$\displaystyle s_j(a,b)=(1/j)\operatorname{coeff}((\operatorname{ad}(T a+b))^{p-1}a,T^{j-1}).
$

(Here, $ \operatorname{coeff}(\bullet, T^j) $ denotes the coefficient of $ T^j$ in $ \bullet$ .) In particular, $ s_j(a,b)$ belongs to the Lie algebra generated by $ a,b$ .

An important corollary is the following.

COROLLARY 6.3   Let $ p$ be a prime number. Let $ A,B$ be algebras over a ring $ R$ of characteristic $ p$ . Let $ L$ be a Lie subalgebra of $ A$ (that means, $ R$ -submodule which is closed under commutators.) Let $ \phi: A\to B$ be a $ R$ -linear map and assume that

$\displaystyle [\phi(x),\phi(y)]=\phi([x,y])
$

holds for any $ x,y\in L$ . Let us define a map $ \psi:L\to B$ by

$\displaystyle \psi(x)=\phi(x)^p-\phi(x^p).
$

Then the map $ \psi$ is $ p$ -linear. That means, $ \psi$ satisfies

$\displaystyle \psi(f x+g y)=f^p \psi(x)+g^p \psi(y) \qquad
(\forall x,y \in L, \forall f,g \in R).
$

PROOF.. The only thing which needs to be verified is the additivity of $ \psi$ .

      $\displaystyle \psi(x+y)$
    $\displaystyle =$ $\displaystyle (\phi(x)+\phi(y))^p-\phi((x+y)^p)$
    $\displaystyle =$ $\displaystyle \left(\phi(x)^p+\phi(y)^p+\sum_j s_j(\phi(x),\phi(y))\right)$
      $\displaystyle -\left (\phi(x^p)+\phi(y)^p+\sum_j \phi(s_j(x,y))\right)$
    $\displaystyle =$ $\displaystyle \psi(x)+\psi(y)$
      $\displaystyle + \left (\sum_j (s_j(\phi(x),\phi(y))-\phi(s_j(x,y))\right)$
    $\displaystyle =$ $\displaystyle \psi(x)+\psi(y)$

(Note that each $ s_j$ commutes with $ \phi$ because it is built only with commutators.) $ \qedsymbol$


next up previous
Next: appendix Up: a formula for -powers Previous: -powers of derivations
2012-02-29