When is an acylic chain complex contractible - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:59:11Z http://mathoverflow.net/feeds/question/103056 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible When is an acylic chain complex contractible Joseph Victor 2012-07-25T00:07:23Z 2012-07-25T05:56:21Z <p>When is an acyclic chain complex contractible? </p> <p>I know an acyclic chain complex of free modules over a PID (or field) are always contractible, but what about over a more complicated ring, like a graded algebra over Z/p (for instance the mod p Steenrod Algebra)? </p> <p>EDIT: I want to assume the chain complex is bounded below, but not necessarily that the ring is commutative. I suspect that it isn't always true for a non-commutative ring, but I don't really have a counter example. </p> <p>Thanks everybody! </p> http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible/103071#103071 Answer by Ralph for When is an acylic chain complex contractible Ralph 2012-07-25T05:49:35Z 2012-07-25T05:56:21Z <p>There is a useful characterization in Brown: Cohomology of Groups, Prop. 0.3: </p> <blockquote> <p>A chain complex $C$ over any ring is contractible iff it is acyclic and each short exact sequence $0 \to \ker(d_n) \to C_n \to \operatorname{im}(d_n) \to 0$ splits. </p> </blockquote> <p>This immediately explains the OP's PID example: If $C$ is a complex of free modules over a PID, then $\operatorname{im}(d_n) \le C_{n-1}$ is also free and thus the sequence splits. </p> <p>This observation can be axiomized as follows: A ring (with unit) is called <em>hereditary</em>, if each submodule of a projective module is again projective. As a corollary: </p> <blockquote> <p>Each acyclic chain complex of projective modules over a hereditary ring is contractible. </p> </blockquote> <p>An example of an non-commutative hereditary ring is given by the upper-triangular matrices over a field. </p> <p>BTW: Tom's remark also follows easily from the criterion: Let all $C_n$ be free and $C_n=0$ for $n &lt; 0$. Since $C_0$ is free, the short exact sequence $0 \to \ker(d_1) \to C_1 \to C_0 \to 0$ splits. Hence $\ker(d_1)=\operatorname{im}(d_2)$ is a direct summand of a free module and therefore projective. By induction then all of the short exact sequences split. </p>