# What is the differential $d^0(f) : C^0(G, A) \to C^1(G, A)$ knowing the general formula?

Romyar Sharifi's Lecture Notes

It's from the first differential formula appearing on page 7 and that formula is:

$$d^i : C^i(G, A) \to C^{i+1}(G, A), \\ d^i(f)(g_0, \dots, g_i) = g_0 f(g_1, \dots, g_i) + \sum\limits_{j=1}^i (-1)^j f(g_0, \dots, g_{j-2}, g_{j-1}g_j, g_{j+1}, \dots, g_i) + \\ (-1)^{i+1} f(g_1, \dots, g_i)$$

where $$A$$ is a $$\Bbb{Z}[G]$$-module, $$G$$ is a group, and $$C^i(G, A) = \{ f: G^i \to A \vert f \text{ is a } \textbf{function} \}$$.

Two questions come to mind in proving that $$d^{i+1} \circ d^i = 0$$ (i.e. $$d$$ is indeed a differential of a cochain complex).

1. Does the author really mean "function", i.e. any set map even if it's not a group homomorphism?
2. Given the formula, what is $$d^0(f)(g_0)$$?

I am not sure about the first, but for the second my guess is:

$$d^0(f)(g_0) = g_0$$

Is that correct?

Note $$G^i = G \times \dots \times G \ (i \text{ times})$$.

• 1) yes - function. 2) no... your guess doesn't parse as the new function has to take values in $A$, not in $G$ as you have it: one can identify $C^0$ with elements of $A$ (the domain is the empty product), and $d^0(a)$ for $a\in A$ has to be element of $C^1$, i.e., a function from $G$ to $A$. Reading the formula you wrote above, one has $d^0(a)(g) = ga - a$. – peter a g Aug 10 at 3:17
• @peterag thank you! – Shine On You Crazy Diamond Aug 10 at 3:18
• @peterag to check my understanding, since we're a $\Bbb{Z}[G]$-module, that expression is the same as $(g-1)a$? – Shine On You Crazy Diamond Aug 10 at 3:19
• yes, if you want. – peter a g Aug 10 at 3:20
• @peterag I currently think it's $d^0(a)(g) = g - a$ since in the first term $f(g_1, \dots, g_{i})$, with $i = 0$. – Shine On You Crazy Diamond Aug 18 at 16:47