# nLab derivation on a group

For convenience we assume below that $M$ is a $G$-module, it does not in general have to be abelian and it suffices to have it a $G$-group.

Suppose $G$ is a group and $M$ a $G$-module and let $\delta :G\to M$ be a derivation. This means $\delta \left({g}_{1}{g}_{2}\right)=\delta \left({g}_{1}\right)+{g}_{1}\delta \left({g}_{2}\right)$ for all ${g}_{1},{g}_{2}\in G$. (Note: not $\delta \left({g}_{1}\right){g}_{2}+{g}_{1}\delta \left({g}_{2}\right)$ as for the other notion of derivation.)

For calculations, the following lemma is very valuable, although very simple to prove.

###### Lemma

If $\delta :G\to M$ is a derivation, then

1. $\delta \left({1}_{G}\right)=0$;

2. $\delta \left({g}^{-1}\right)=-{g}^{-1}\delta \left(g\right)$ for all $g\in G$;

3. for any $g\in G$ and $n\ge 1$,

$\delta \left({g}^{n}\right)=\left(\sum _{k=0}^{n-1}{g}^{k}\right)\delta \left(g\right).$\delta(g^n) = (\sum^{n-1}_{k=0}g^k)\delta(g).
###### Proof

As was said, these are easy to prove.

$\delta \left(g\right)=\delta \left(1g\right)+1\delta \left(g\right)$, so $\delta \left(1\right)=0$, and hence (1); then

$\delta \left(1\right)=\delta \left({g}^{-1}g\right)=\delta \left({g}^{-1}\right)+{g}^{-1}\delta \left(g\right)$\delta(1) = \delta(g^{-1}g) = \delta(g^{-1}) + g^{-1}\delta(g)

to get (2), and finally induction to get (3).

### Remark:

The Fox derivatives are examples. It is worth noting that this lemma allows a simplification of the conditions given there (as noted there).

### Example (using the Fox derivative w.r.t a generator)

Let $X=\left\{u,v\right\}$, with $r\equiv uvu{v}^{-1}{u}^{-1}{v}^{-1}\in F=F\left(u,v\right),$ then

$\frac{\partial r}{\partial u}=1+uv-uvu{v}^{-1}{u}^{-1},$\frac{\partial r}{\partial u} = 1 + u v - u v u v^{-1} u^{-1},
$\frac{\partial r}{\partial v}=u-uvu{v}^{-1}-uvu{v}^{-1}{u}^{-1}{v}^{-1}.$\frac{\partial r}{\partial v} = u - u v u v^{-1} - u v u v^{-1} u^{-1} v^{-1}.

This relation, $r$, is the typical braid group relation, here in ${\mathrm{Br}}_{3}$.

Revised on October 9, 2012 13:36:47 by Tim Porter (95.147.237.26)