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

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    $\begingroup$ 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 $d^0(a)(g) = ga - a$. $\endgroup$ – peter a g Aug 10 at 3:17
  • $\begingroup$ @peterag thank you! $\endgroup$ – Shine On You Crazy Diamond Aug 10 at 3:18
  • $\begingroup$ @peterag to check my understanding, since we're a $\Bbb{Z}[G]$-module, that expression is the same as $(g-1)a$? $\endgroup$ – Shine On You Crazy Diamond Aug 10 at 3:19
  • $\begingroup$ yes, if you want. $\endgroup$ – peter a g Aug 10 at 3:20
  • $\begingroup$ @peterag I currently think it's $d^0(a)(g) = g - a$ since in the first term $f(g_1, \dots, g_{i})$, with $i = 0$. $\endgroup$ – Shine On You Crazy Diamond Aug 18 at 16:47

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