nLab
derivation on a group

[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 GG is a group and MM a GG-module and let δ:GM\delta : G \to M be a derivation. This means δ(g 1g 2)=δ(g 1)+g 1δ(g 2)\delta(g_1g_2) = \delta(g_1) +g_1\delta(g_2) for all g 1,g 2Gg_1, g_2 \in G. (Note: not δ(g 1)g 2+g 1δ(g 2)\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.

Lemma

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

  1. δ(1 G)=0 \delta(1_G) = 0;

  2. δ(g 1)=g 1δ(g)\delta(g^{-1}) = -g^{-1}\delta(g) for all gGg \in G;

  3. for any gGg \in G and n1n\geq 1,

    δ(g n)=( k=0 n1g k)δ(g).\delta(g^n) = (\sum^{n-1}_{k=0}g^k)\delta(g).
Proof

As was said, these are easy to prove.

δ(g)=δ(1g)+1δ(g)\delta(g) = \delta(1g) + 1\delta(g), so δ(1)=0\delta(1)= 0, and hence (1); then

δ(1)=δ(g 1g)=δ(g 1)+g 1δ(g)\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={u,v}X = \{u,v\}, with ruvuv 1u 1v 1F=F(u,v),r \equiv u v u v^{-1} u^{-1} v^{-1} \in F = F(u,v), then

ru=1+uvuvuv 1u 1, \frac{\partial r}{\partial u} = 1 + u v - u v u v^{-1} u^{-1},
rv=uuvuv 1uvuv 1u 1v 1. \frac{\partial r}{\partial v} = u - u v u v^{-1} - u v u v^{-1} u^{-1} v^{-1}.

This relation, rr, is the typical braid group relation, here in Br 3Br_3.

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