When is an acylic chain complex contractible

2012-07-25T00:07:23Z

When is an acyclic chain complex contractible?

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)?

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.

Thanks everybody!

Answer by Ralph for When is an acylic chain complex contractible

2012-07-25T05:49:35Z

There is a useful characterization in Brown: Cohomology of Groups, Prop. 0.3:

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.

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.

This observation can be axiomized as follows: A ring (with unit) is called hereditary, if each submodule of a projective module is again projective. As a corollary:

Each acyclic chain complex of projective modules over a hereditary ring is contractible.

An example of an non-commutative hereditary ring is given by the upper-triangular matrices over a field.

BTW: Tom's remark also follows easily from the criterion: Let all $C_n$ be free and $C_n=0$ for $n < 0$. 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.

When is an acylic chain complex contractible - MathOverflow

When is an acylic chain complex contractible

Answer by Ralph for When is an acylic chain complex contractible