I've a very basic notational question on group cohomology. Let $G$ be a finite group and $M$ a $G$-module. For $i\geq 0$, let $P_i=\mathbb Z[G^{i+1}]$ be the free $\mathbb Z$-module on $G^{i+1}$, made into a $G$-module by left multiplication. Define $d_i: P_i \to P_{i-1}$ by $$d_i(g_0,...,g_i)=\sum_{j=0}^i(-1)^j(g_0,...,g_{j-1},g_{j+1},...,g_i)$$
My question is why does $d_0$ map $d_0(\sum_{g\in G}n_g g)$ to $\sum_{g\in G}n_g$?
I am confused about how to define $P_{-1}$.
Many thanks
The definition $P_i = \mathbb{Z}\left[G^{i+1}\right]$ should be made for all $i \geq -1$, not only for $i \geq 0$. Thus, $P_{-1} = \mathbb{Z}\left[G^{0}\right]$. But $G^0$ is the trivial group consisting of the one element $\left(\right)$ (the empty list, a.k.a. the $0$-tuple). If you plug in $i = 0$ into the definition of $d_i$, you obtain
(1) $d_0\left(\left(g_1\right)\right) = \left(\right)$ for every $g_1 \in G$.
Now, it is common to identify $G^1$ with $G$, so that the $1$-tuple $\left(g_1\right)$ becomes $g_1$. Furthermore, it is common to define $\mathbb{Z}\left[G^{0}\right]$ with $\mathbb{Z}$, so that the empty list $\left(\right)$ becomes the integer $1$. Hence, (1) rewrites as
(2) $d_0\left(g_1\right) = 1$ for every $g_1 \in G$.
Since $d_0$ is $\mathbb{Z}$-linear, this yields that $d_0\left(\sum_{g\in G} n_g g\right) = \sum_{g\in G} n_g$ for all $\left(n_g\right)_{g \in G} \in \mathbb{Z}^G$.