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I understand the geometric interpretation of derived functors, as well as their usefulness in giving a simple, purely algebraic description of cohomology.

I also understand how resolutions are used to calculate derived functors. However, I don't have a geometric interpretation of these calculations. I'm able to digest and play with the diagram-chasing arguments, but I have no geometric intuition.

I read that the theory of model categories provides a geometric explanation for the use of resolutions in homological algebra, by establishing a parallel between homological algebra and abstract homotopy theory. How is that? Could one of the more experienced geometers here paint for me a simple picture of the basic ideas?

Many thanks!

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I think a relevant keyword here is "(co)fibrant replacement." – Qiaochu Yuan Mar 29 '12 at 19:34
Dear @QiaochuYuan, could you please elaborate? Thanks! – Bruno Joyal Mar 29 '12 at 19:35
@Bruno: An injective resolution for $A$ is a cofibrant replacement for $A$, regarded as a cochain complex concentrated in degree $0$. This is part of a much bigger picture which establishes homological algebra as a special case of homotopical algebra. You could start reading here. – Zhen Lin Mar 30 '12 at 0:10
Thank you, @ZhenLin! I will take a look at the reference. – Bruno Joyal Mar 30 '12 at 6:01
up vote 4 down vote accepted

Pardon the sketchy details to follow here: there are many details that I'm burying under the rug for sake of brevity (and giving some sort of answer!). I'd recommend the first chapter of Hovey's book to get an idea for how these arguments go.

A more geometric example of this kind of argument is the construction of homotopy colimits, which are a "derived" version of colimit. Lets look at diagrams of the form $B \gets A \to C$.

Consider the two diagrams $D^n \gets S^{n-1} \to D^n$. This has colimit $S^n$, seen as gluing two disks along their boundary spheres.

Here is a weakly homotopic diagram: $* \gets S^{n-1} \to *$. This has a map into our first diagram which is object-wise a homotopy equivalence. But the colimit of this diagram is $*$, which is homotopically different from $S^n$. This means that colimit, seen as a functor from this three-space diagram category to spaces, is not a homotopy-invariant functor.

Geometrically, this can be a bit undesirable. The discrepancy arises from the fact that the maps $S^{n-1} \to *$ are not cofibrations; we're not really gluing together two points along their "common $S^{n-1}$" like we did for the pair of maps $S^{n-1} \to D^n$. Geometrically, we want the outcome to be the more interesting $S^n$.

So, we make a derived version, called homotopy colimit. It's a version of colimit that is homotopy invariant in the diagram category, but also agrees with colimit if the diagram is made up of these "nice inclusions" (cofibrations) (Again, we're sticking with these 3 space diagrams for now).

There are two ways to think about homotopy colimit that I've bumped into. One is an explicit construction: the homotopy colimit of $B \gets A \to C$ is a quotient of the coproduct $B \coprod (A \times I) \coprod C$ where we glue in $A \times 0$ to $B$ and $A \times 1$ to $C$.

If we check that with a diagram $* \gets A \to *$, we find that this gives the suspension of $A$. So in our example of $* \gets S^{n-1} \to *$, we do indeed compute $S^n$ as the homotopy colimit.

Another way to think about things is that we really wanted $D^n \gets S^{n-1} \to D^n$ to be the "correct" diagram in which to take this colimit. All the maps in this diagram are "nice" inclusions. In general, I can replace the diagram $B \gets A \to C$ with a homotopy equivalent diagram $B' \gets A' \to C'$ where:

1) $A'$ is a CW-complex (A "cofibrant" space)

2) $A' \to B'$ is a CW-inclusion

3) $A' \to C'$ is a CW-inclusion.

Then, I take homotopy colimit. So, for example, $* \gets S^{n-1} \to *$ can be replaced with $D^n \gets S^{n-1} \to D^n$ in this way.

It turns out that if I replace $B \gets A \to C$ with $B' \gets A' \to C'$, I will not only get a colimit that does not depend on the choices I made, but also one that is homotopy invariant in the category of diagrams. A homotopy equivalent diagram $E \gets D \to F$ will produce a homotopy equivalent homotopy colimit. (Compare this to projective resolutions, where no matter which resolution I choose, I get the same derived functor value).

The replacement $B' \gets A' \to C'$ is itself a "cofibrant" object in a model category structure on this diagram category. The model structure is constructed in such a way that colimit and the diagonal functor $A \mapsto (A \gets A \to A)$ form a Quillen adjunction -- which implies, among other things, that weakly equivalent cofibrant diagrams have weakly equivalent colimits.

Whenever you have a Quillen adjunction, as a pair of adjoint functors $(L,R)$, you get a pair of derived functors $(L',R')$ which are homotopy invariant versions of $L$ and $R$ in a similar way: $L'(X)$ is given by evaluating $L$ on a cofibrant replacement of $X$. $R'(Y)$ is gotten by evaluating $R$ on a fibrant replacement of $Y$.

And I've rambled for long enough -- hopefully this helps a bit :).

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It helps a lot! Thank you very much Thomas. – Bruno Joyal Mar 30 '12 at 6:04

This isn't a geometric approach, but for intuition, I prefer the free resolution to the projective resolution. I tend to think of "projective" as being essentially, "All the properties we need in this category to ensure that we get the same results as we would from a free resolution." :)

If $...F_3\rightarrow F_2\rightarrow F_1\rightarrow M$ is your free resolution, then $F_1$ essentially represents information about a set of generators for $M$, stripped of all relations. $F_2$ then represents a set of relationships between those generators that generate all the relationships (but again stripped of all information about their relationships.) Essentially, the free resolutions give you a way to measure the complexity of the relationships between elements of $M$. We are listing a set of generators, then listing a (complete) set of relationships between those generators (from which all other relationships can be determined,) then listing a (complete) set of meta-relationships between the relationships, etc.

Oddly, I recall having this occur to me when taking a Combinatorics course with Richard Stanley using his book, Combinatorics and Commutative Algebra.

In there, certain counting problems that require complicated "inclusion-exclusion" arguments can be best realized by taking free resolutions of certain structures. I'm flaking on the details, but I recall one of the problems was "how many $n\times n$ matrices of non-negative integers have all rows and columns sum up to $d$?"

The set of solutions forms a graded something-or-other with $d$ being the grade, and the obvious generators being permutation matrices, yielding the first $F_1\rightarrow M$. But if you just count the number of elements of $F_1$ which go to the grade $d$, then you are over-counting because of the relationships in $M$ between those generators. So $F_2$ is used to count your over-count, and $F_3$ is used to then count your under-count, etc.

(Obviously, for combinatorics, you need a finite free resolution for this counting strategy to ever terminate.)

I feel like there is a geometric explanation implicit in this, but I always think of it in terms of relationships and relationships-between-relationships, and a measurement of the underlying complexity of the relationships inside the algebraic object $M$.

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Thank you Thomas for your insight! – Bruno Joyal Mar 30 '12 at 6:03

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