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Definition

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

⋯ → M 1 ⊗ k R ⊗ k R ⊗ k M 2 → M 1 ⊗ k R ⊗ k M 2 → M 1 ⊗ k M 2 → 0 , {\displaystyle \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

d ( m 1 ⊗ r 1 ⊗ ⋯ ⊗ r n ⊗ m 2 ) = m 1 r 1 ⊗ ⋯ ⊗ r n ⊗ m 2 + ∑ i = 1 n − 1 ( − 1 ) i m 1 ⊗ r 1 ⊗ ⋯ ⊗ r i r i + 1 ⊗ ⋯ ⊗ r n ⊗ m 2 + ( − 1 ) n m 1 ⊗ r 1 ⊗ ⋯ ⊗ r n m 2 {\displaystyle {\begin{aligned}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{aligned}}}

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 {\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

⋯ → R ⊗ k R ⊗ k R ⊗ k M → R ⊗ k R ⊗ k M → R ⊗ k M → 0 , {\displaystyle \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 {\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

⋯ → R ⊗ k R ⊗ k R ⊗ k M → R ⊗ k R ⊗ k M → R ⊗ k M → M → 0 , {\displaystyle \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 {\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

h n ( r 1 ⊗ ⋯ ⊗ r n ⊗ m ) = ∑ i = 1 n − 1 ( − 1 ) i + 1 r 1 ⊗ ⋯ ⊗ r i − 1 ⊗ 1 ⊗ r i ⊗ ⋯ ⊗ r n ⊗ m {\displaystyle {\begin{aligned}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{aligned}}}

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 Bar Complex

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} .

See also

• Koszul complex

Notes

• Cartan, Henri; Eilenberg, Samuel (1956), Homological algebra, Princeton Mathematical Series, vol. 19, Princeton University Press, ISBN 978-0-691-04991-5, MR 0077480{{citation}}: CS1 maint: ignored ISBN errors (link)
• Eilenberg, Samuel; Mac Lane, Saunders (1953), "On the groups of H ( Π , n ) {\displaystyle H(\Pi ,n)} . I", Annals of Mathematics, Second Series, 58: 55–106, doi:10.2307/1969820, ISSN 0003-486X, JSTOR 1969820, MR 0056295
• Ginzburg, Victor (2005). "Lectures on Noncommutative Geometry". arXiv:math.AG/0506603.
• Weibel, Charles (1994), An Introduction to Homological Algebra, Cambridge Studies in Advanced Mathematics, vol. 38, Cambridge: Cambridge University Press, ISBN 0-521-43500-5

References

1. 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 ↩