[For convenience we assume below that is a -module, it does not in general have to be abelian and it suffices to have it a -group.]
Suppose $G$ is a group and $M$ a $G$-module and let $\delta : G \to M$ be a derivation. This means $\delta(g_1g_2) = \delta(g_1) +g_1\delta(g_2)$ for all $g_1, g_2 \in G$. (Note: not $\delta(g_1)g_2 + g_1\delta(g_2)$ as for the other notion of derivation.)
For calculations, the following lemma is very valuable, although very simple to prove.
If $\delta : G \to M$ is a derivation, then
$\delta(1_G) = 0$;
$\delta(g^{-1}) = -g^{-1}\delta(g)$ for all $g \in G$;
for any $g \in G$ and $n\geq 1$,
As was said, these are easy to prove.
$\delta(g) = \delta(1g) + 1\delta(g)$, so $\delta(1)= 0$, and hence (1); then
to get (2), and finally induction to get (3).
The Fox derivatives are examples. It is worth noting that this lemma allows a simplification of the conditions given there (as noted there).
Let $X = \{u,v\}$, with $r \equiv u v u v^{-1} u^{-1} v^{-1} \in F = F(u,v),$ then
This relation, $r$, is the typical braid group relation, here in $Br_3$.
Last revised on October 9, 2012 at 13:36:47. See the history of this page for a list of all contributions to it.