Bar complex

Let R be an algebra over a field k, let M_1 be a right R-module, and let M_2 be a left R-module. Then, one can form the bar complex \operatorname{Bar}_R(M_1,M_2) given by :\cdots\rightarrow M_1 \otimes_k R \otimes_k R \otimes_k M_2 \rightarrow M_1 \otimes_k R \otimes_k M_2 \rightarrow M_1 \otimes_k M_2 \rightarrow 0\,, with the differential :\begin{align} d(m_1 \otimes r_1 \otimes \cdots \otimes r_n \otimes m_2) &= m_1 r_1 \otimes \cdots \otimes r_n \otimes m_2 \\ &+ \sum_{i=1}^{n-1} (-1)^i m_1 \otimes r_1 \otimes \cdots \otimes r_i r_{i+1} \otimes \cdots \otimes r_n \otimes m_2 + (-1)^n m_1 \otimes r_1 \otimes \cdots \otimes r_n m_2 \end{align} ==Resolutions==

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. Free Resolution of a Module Let M be a left R-module, with R a unital k-algebra. Then, the bar complex \operatorname{Bar}_R(R,M) gives a resolution of M by free left R-modules. Explicitly, the complex is :\cdots\rightarrow R \otimes_k R \otimes_k R \otimes_k M \rightarrow R \otimes_k R \otimes_k M \rightarrow R \otimes_k M \rightarrow 0\,, This complex is composed of free left R-modules, since each subsequent term is obtained by taking the free left R-module on the underlying vector space of the previous term. To see that this gives a resolution of M, consider the modified complex :\cdots\rightarrow R \otimes_k R \otimes_k R \otimes_k M \rightarrow R \otimes_k R \otimes_k M \rightarrow R \otimes_k M \rightarrow M \rightarrow 0\,, Then, the above bar complex being a resolution of M is equivalent to this extended complex having trivial homology. One can show this by constructing an explicit homotopy 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 :\begin{align} h_n(r_1 \otimes \cdots \otimes r_n \otimes m) &= \sum_{i=1}^{n-1} (-1)^{i+1} r_1 \otimes \cdots \otimes r_{i-1} \otimes 1 \otimes r_i \otimes \cdots \otimes r_n \otimes m \end{align} One can similarly construct a resolution of a right R-module N by free right modules with the complex \operatorname{Bar}_R(N,R). Notice that, in the case one wants to resolve R as a module over itself, the above two complexes are the same, and actually give a resolution of R by R-R-bimodules. This provides one with a slightly smaller resolution of R by free R-R-bimodules than the naive option \operatorname{Bar}_{R^e}(R^e,M). Here we are using the equivalence between R-R-bimodules and R^e-modules, where R^e = R \otimes R^\operatorname{op}, see bimodules for more details. ==The Normalized Bar Complex==

The normalized (or reduced) standard complex replaces A\otimes A\otimes \cdots \otimes A\otimes A with A\otimes(A/K) \otimes \cdots \otimes (A/K)\otimes A. ==See also==