Let R {\displaystyle R} be an algebra over a field k {\displaystyle k} , let M 1 {\displaystyle M_{1}} be a right R {\displaystyle R} -module, and let M 2 {\displaystyle M_{2}} be a left R {\displaystyle R} -module. Then, one can form the bar complex Bar R ( M 1 , M 2 ) {\displaystyle \operatorname {Bar} _{R}(M_{1},M_{2})} given by
with the differential
The bar complex is useful because it provides a canonical way of producing (free) resolutions of modules over a ring. However, often these resolutions are very large, and can be prohibitively difficult to use for performing actual computations.
Let M {\displaystyle M} be a left R {\displaystyle R} -module, with R {\displaystyle R} a unital k {\displaystyle k} -algebra. Then, the bar complex Bar R ( R , M ) {\displaystyle \operatorname {Bar} _{R}(R,M)} gives a resolution of M {\displaystyle M} by free left R {\displaystyle R} -modules. Explicitly, the complex is1
This complex is composed of free left R {\displaystyle R} -modules, since each subsequent term is obtained by taking the free left R {\displaystyle R} -module on the underlying vector space of the previous term.
To see that this gives a resolution of M {\displaystyle M} , consider the modified complex
Then, the above bar complex being a resolution of M {\displaystyle M} is equivalent to this extended complex having trivial homology. One can show this by constructing an explicit homotopy h n : R ⊗ k n ⊗ k M → R ⊗ k ( n + 1 ) ⊗ k M {\displaystyle h_{n}:R^{\otimes _{k}n}\otimes _{k}M\to R^{\otimes _{k}(n+1)}\otimes _{k}M} between the identity and 0. This homotopy is given by
One can similarly construct a resolution of a right R {\displaystyle R} -module N {\displaystyle N} by free right modules with the complex Bar R ( N , R ) {\displaystyle \operatorname {Bar} _{R}(N,R)} .
Notice that, in the case one wants to resolve R {\displaystyle R} as a module over itself, the above two complexes are the same, and actually give a resolution of R {\displaystyle R} by R {\displaystyle R} - R {\displaystyle R} -bimodules. This provides one with a slightly smaller resolution of R {\displaystyle R} by free R {\displaystyle R} - R {\displaystyle R} -bimodules than the naive option Bar R e ( R e , M ) {\displaystyle \operatorname {Bar} _{R^{e}}(R^{e},M)} . Here we are using the equivalence between R {\displaystyle R} - R {\displaystyle R} -bimodules and R e {\displaystyle R^{e}} -modules, where R e = R ⊗ R op {\displaystyle R^{e}=R\otimes R^{\operatorname {op} }} , see bimodules for more details.
The normalized (or reduced) standard complex replaces A ⊗ A ⊗ ⋯ ⊗ A ⊗ A {\displaystyle A\otimes A\otimes \cdots \otimes A\otimes A} with A ⊗ ( A / K ) ⊗ ⋯ ⊗ ( A / K ) ⊗ A {\displaystyle A\otimes (A/K)\otimes \cdots \otimes (A/K)\otimes A} .
Weibel 1994, p. 283. - Weibel, Charles (1994), An Introduction to Homological Algebra, Cambridge Studies in Advanced Mathematics, vol. 38, Cambridge: Cambridge University Press, ISBN 0-521-43500-5 ↩