# Derived functors and coboundary operator

I understand that one can define the cohomology of an object $A$ in terms of a complex (non-zero in positive degrees) in some Abelian category, together with differentials, such that the composition of any two composable differentials is zero.

When properly defined, this complex gives a cohomology theory $H^\ast$ for objects like $A$ and should satisfy the axioms of a cohomology theory (something like being a universal $\delta$-functor).

An equivalent construction is to start with a left-exact functor—call it $H^0$—and right-derive it. The right-derived functors of $H^0$ also satisfy the axioms of a cohomology theory and by universality of universal $\delta$-functors, this construction is equivalent to the above construction of complexes and explicit differentials.

I am wondering how to derive (in the non-technical sense) the formulas of the coboundary operators from a single coboundary operator (at level 0). That is, given a left-exact functor $H^0$, how do I construct differentials in some complex of objects associated to my object $A$ such that the cohomology of the resulting complex is given by the right-derived functors of $H^0$?

For example, the Hochschild cohomology functors $HH^i$ (cohomology of an associative algebra $A$ with values in a bimodule $M$ over $A$) are the right-derived functors of the functor $HH^0$ defined by $$HH^0=\ker \delta^0$$ where $$\delta^0(a)=a\cdot m-m\cdot a$$ for $m\in M$ and $a\in A$. When $HH^0$ is defined by a single equation (in this case $a\cdot m=m\cdot a$), then form of $\delta^0$ is obvious.

Now, $HH^1$ is the first right-derived functor of $HH^0$. $HH^1$ can be calculated as $\ker \delta^1/\operatorname{im} \delta^0$, where $$(\delta ^1 f)(a,b)=a\cdot f(b)-f(ab)+f(a)\cdot b$$ where $f\colon A\to M$. How did I construct the formula for $\delta^1$ from the formula of $\delta^0$?

I would be grateful for any references, which describe this for, say, group cohomology, Hochschild cohomology, Lie algebra cohomology, sheaf cohomology.

Edit. My motivation is the following. I am trying to understand how the formulas of coboundary operators (which is usually an alternating sum over "contractions") arise. The theory of derived functors seems to suggest that they should be a "derived" form of the formula for $\delta^0$. As far as I understand, constructing injective resolutions is not very easy. In the case of group cohomology, the bar resolution always works, but is practically uncomputable. There are special resolutions for cyclic groups, which are still big. Computing group cohomology is usually most convenient when one can find an Eilenberg-Mac Lane space $K(G,1)$ (preferably compact) and calculate the group cohomology as the singular cohomology of this space (even possible with sheaf cohomology in twisted coefficients).

For group cohomology, the composition $\delta^2\delta^1$ being 0 corresponds to the equality $g\cdot(g'\cdot m)=(gg')\cdot m$, so in some sense the "derived" formula for the invariants functor (which is $H^0$) suggests the characterizing equation of an action. To define the set of invariants, however, we didn't need to know that $g\cdot(g'\cdot m)=(gg')\cdot m$.

A decent cohomology theory for some object, where it is not clear which axioms to impose, could in principle suggests (some of) the most natural axioms for this object.

This is why I am interested in understanding how the general coboundary operator formulas are derived from a single formula in degree 0.

• Right derived functors are not differentials. What do you mean? – Zhen Lin Dec 12 '13 at 12:43
• @ZhenLin Edited. Is this more clear? – Earthliŋ Dec 12 '13 at 19:20
• Are you trying to avoid the standard construction via injective resolutions? – Zhen Lin Dec 12 '13 at 23:15
• @ZhenLin In principle, yes. I added more details in the question body. – Earthliŋ Dec 13 '13 at 12:09
• I'm not convinced they can be constructed like that in general. For instance, the bar complex for group cohomology comes from taking a projective resolution of the trivial module $\mathbb{Z}$, not an injective resolution. – Zhen Lin Dec 13 '13 at 13:04