Why this particular definition for the derivative on a group ring?

Question

I am reading Introduction to Knot Theory by Crowell and Fox, and I am a bit confused at the way they define a derivative on a group ring (and on a group). I understand a derivative (or derivation maybe more properly) to be a linear map which satisfies the Leibniz rule. However, they first define the derivative on a group to be $D:G\to JG$ $$D(g_1g_2)=Dg_1+g_1Dg_2$$ which can be uniquely extended to the map on the group ring $D:JG\to JG$ as $$D(v_1v_2)=(Dv_1)\mathfrak{t}(v_2)+v_1Dv_2$$ where $\mathfrak{t}$ is the trivializer $$\mathfrak{t}(\sum_i n_ig_i)=\sum n_i.$$ So I understand how this is all well-defined, and I am convinced it is a derivative (that's rather clear I guess). But why choose this, instead of just $$D(g_1g_2)=(Dg_1)g_2+g_1(Dg_2)$$ With a natural (unique, I think) extension to $$D(v_1v_2)=(Dv_1)v_2+v_1(Dv_2)$$ Perhaps they are going after some specific structure here, but does anyone have a reason to choose one over the other?

Answer 1

Suppose $G$ is a finite cyclic group generated by $g$ of order $n$ so that $g^{-1} = g^{n-1}$. Let $\Bbb{Z}[G]$ denote its group ring. The obvious choice of derivative would be $D = \frac{d}{dg}$ (meaning $Dg = 1$). We would like to have that $\frac{d}{dg} g^{-1} = \frac{d}{dg} g^{n-1}$ since these two elements are technically the same. (This says that $D$ is well-defined in loose terms.) If we use your proposed definition

$$D(g_1g_2) = D g_1\cdot g_2 + g_1\cdot Dg_2$$

applied to this situation, induction gives for $m < n$

$$D (g^m) = mg^{m-1}.$$

This is to be expected from what we know about $\Bbb R$. Particularly, we have that

$$D(g^{n-1}) = (n-1)g^{n-2} = (n-1)g^{-2}.$$

Additionally,

$$D(1) = D(1\cdot 1) = D(1)\cdot 1 + 1\cdot D(1).$$

That is to say that $D(1) = D(1) + D(1)$ so $D(1) = 0$ in the ring. Thus

$$0 = D(1) = D(g^{-1}g) = D(g^{-1})\cdot g + g^{-1}\cdot D(g) = D(g^{-1})\cdot g + g^{-1}.$$

That is to say that

$$ - g^{-1} = g\cdot D(g^{-1}) \Longrightarrow -g^{-2} = D(g^{-1}). $$

The above formula agrees with our intuition from $\Bbb R$, however, as one often finds in harmonic analysis, the nicety of $\Bbb R$ can really lead you astray into thinking things are simpler than they actually are. The two expressions for $D(g^{-1})$ do not match so blindly applying the Leibniz rule doesn't work in this setting. It seems that the rigidity of the group ring forces you somewhat into their "unnatural" definition for the derivative.

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If we go by their definition, the picture changes a bit:

$$D(g^2) = D(g\cdot g) = D(g) + g\cdot D(g) = 1 + g.$$

Likewise

$$D(g^3) = D(g\cdot g^2) = D(g) + g\cdot D(g^2) = 1 + g + g^2.$$

Repeating we get (for $m < n$)

$$D(g^m) = 1 + g + \cdots + g^{m-1}.$$

Thus we have

$$D(g^{n-1}) = 1 + g + \cdots + g^{n-2}.$$

By the formula in their text

$$D(g^{-1}) = -g^{-1}\cdot D(g) = -g^{-1}.$$

This doesn't seem to agree with $D(g^{n-1})$ but it does! By their (very reasonable and suggestive) shorthand,

$$\sum_{i=0}^k g^i = \frac{g^{k+1}-1}{g-1}.$$

Thus

$$\sum_{i=0}^{n-2} g^i = \frac{g^{n-1}-1}{g-1} = g^{-1}\frac{g^n - g}{g-1} = g^{-1}\frac{1 - g}{g-1} = -g^{-1}.$$

So we get that $D(g^{-1}) = D(g^{n-1})$ as we should.