Definition of bar resolution I am reading Weibel's book about homological algebra. In section 6.5, the bar resolution, he uses some notation I really do not understand. So given G, a group, what is $[g_1\otimes...\otimes g_n]$? And what is $[g_1|...|g_n]$? 
 A: As Qiaochu mentions, Weibel literally means that they are symbols, and defines the standard resolution and normalized standard resolution resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ by choosing these as bases for the modules under discussion. However, my guess is that you need some motivation for these symbols/notations, and I'll summarize the ideas in Section I.5 of Cohomolgoy of Groups by Brown.
The topological motivation is to make a simplex (possibly an infinite-dimensional one) with $G$ acting freely on every facets. Then the simplicial chain complex of this simplex would be a free acyclic chain, since a simplex (even infinite-dimensional) is contractible. Taking the augmentation of this complex gives the free resolution of $\mathbb{Z}$ over $\mathbb{Z}G$.
To this end, we make such a simplex out of $G$. That is, a "vertex" is an element of $G$, and an $n$-facet (which has $(n+1)$ vertices) is an ordered $(n+1)$-tuple of elements of $G$. An element $g\in G$ acts on the $n$-facet $(g_0,...,g_n)$ by
$$ g . (g_0,...,g_n) = (g g_0,...,g g_n) . $$
So the $n$-th chain group $B_n^u$ is simply the free abelian group on the $n$-facets.
As $\mathbb{Z}G$-modules, however, $B_n^u$ does not have as basis the $n$-facets due to the equation above. So instead we take as basis the $G$-orbits of such facets, and denote the orbit of
$$ (g_0,g_0 g_1,...,g_0 \cdots g_n) = g_0. (1,g_1,g_1 g_2,...,g_1\cdots g_n) $$
by
$$ [g_1 \otimes \cdots \otimes g_n] . $$
And you will notice that Definition 6.5.1 in Weibel follows from the formulas of face maps on an $n$-facet (by omitting the $i$-th vertex).
Finally, the motivation of using the reduced $n$-th chain groups $B_n$ is to "discard" the degenerate $n$-simplices, i.e., those with repeated vertices. Note that an $n$-facet $(g_0,...,g_n)$ is degenerate if and only if $g_i=g_{i+1}$ for some $i$, so $$ [g_1\otimes\cdots\otimes g_n] $$ is degenerate if and only if $g_i=1$ for some $i$. That is why Weibel requires $[g_1|\cdots|g_n]$ satisfies $g_i\ne1$ for all $i$, and declares that $B_n$ is the quotient of $B_n^u$ by the submodule generated by the degenerate simplices.
