(Realistically, this post assumes familiarity with derived functors, chain complexes, and their homology. Ideally, the reader has played around with and a bit, as well as a few examples such as singular cohomology, group homology, etc. A lot of this material is taken from Weibel’s Introduction to Homological Algebra.)

I’ve been putting a lot of energy into understanding homological algebra recently (following Weibel’s book). And if there’s one thing you do all the time in homological algebra, it’s resolve things (a resolution of a module by X objects, where X is some adjective, is an exact sequence or with each an X object). Resolutions help you to compute derived functors (e.g. the cohomology of something), which is a common goal. So I want to talk about how you can compute the derived functors by resolving either or , and why we should care.

This may or not really count as part of the “Shoulda Series” (1), since I’m pretty sure someone did tell me this at some point. But either way, I had to re-discover it for myself “in practice” before I “got it.”

Throughout, let be a ring with unit (not necessarily commutative, but you can take it to be if you want) and let and similar symbols be -modules. We will use (resp., ) for projective (resp., injective) -modules. (Take all modules to be, say, right.) The first thing to recall is that for fixed and , the functors and are right exact, while the functors and are left exact. The first three are functors , but the fourth is a functor ; this important point will become central in just a bit, so make sure you understand why! (Basically, a contravariant functor is the same as a covariant functor . By , the opposite cateogry of , we simply mean the category obtained from by keeping the same objects and reversing all the arrows.)

We may now define the and functors:

Definition: Define:
1. to be the left derived functors of ,
2. to be the left derived functors of ,
3. to be the right derived functors of , and
4. to be the right derived functors of .

The good news is:

Theorem: There are natural isomorphisms and .

We denote the common values and . The basic idea of the proof of the half of this theorem is to take projective resolutions of and of , take the tensor product bicomplex formed by these two resolutions, and then show that a certain chain complex is acyclic. (This chain complex is closely related to , the total direct sum complex associated to the bicomplex .) One then shows that is naturally isomorphic to each of the two derived functors. For the proof is similar, using injective resolutions, , and instead. See Weibel, section 2.7 for the details.

In many situations, our goal is to compute (or at least gain knowledge about) and . Recall that to compute left derived functors we resolve by projective objects, and to compute right derived functors we resolve by injective objects. Projective objects in the category are great: a module is projective if and only if it is a direct summand of a free module. In particular, all free modules are projective. In practice, one can often use finite-rank free resolutions, which are comparatively easy to compute with (and can lead to finiteness results on the derived functors, automatically). One great example of this is the bar resolution (this is the chain complex described here), whose existence immediately tells you that the group homology of a finite group has finite rank whenever the representation does.

But injective objects are not as nice to work with. The only decent general fact I am aware of is the following.

Baer’s Criterion: Let be an -module. Then is injective if and only if for every ideal of and every module homomorphism , there is a homomorphism extending .

This isn’t bad, but it isn’t nearly as helpful as in the projective case. And in fact, most injective modules turn out to be huge and/or nasty in some sense. So it appears that in general, will be harder to compute than . This is a real shame, since usually has more interesting structure! (Think of as cohomology, where there is usually an interesting product, e.g., cup product on the singular cohomology of topological spaces.)

But all is not lost: Remember that the functor is contravariant; equivalently, it is a (covariant) functor . Since the universal property defining projective objects is dual to the universal property defining injective objects, it follows that the injectives of are precisely the projectives of ! So when computing , we can either resolve by injective -modules (usually messy and/or difficult) or resolve by projective -modules (usually much nicer). For instance, the bar resolution mentioned earlier, which is a resolution of the trivial -module by free -modules for a group , can be used to compute group cohomology, i.e., the groups . Hence as long as we are content to always resolve the first variable, is just as easy to compute in general as .

Finally, I want to discuss something which puzzles me. The tensor product functor is left adjoint to the functor; that is, we an isomorphism

valid whenever is a right -module, is an -bimodule, and is a right -module; this isomorphism is natural in all three modules. And one can show that this adjunction holds for the corresponding derived functors as well. So there is a very fundamental symmetry between the bifunctors and . Simplifying to the case where is commutative, we have

In this most important of adjunctions, why is there an opposite-category variable in one bifunctor but not in the other? Life would seem to make more sense if each of the two had one ordinary- and one opposite-category variable. I suspect that this may have to do with the fact that things are not as symmetric as they seem: even if is commutative so that left and right are equivalent, we are still talking about algebras (rings) and modules, while dually we could also talk about coalgebras and comodules. See the questions below, and enlighten me, please.

Questions

Here are a few questions which are bothering me, mostly related to the above. Comments, suggestions, examples, problems, etc. are more than welcome!

1. Philosphically/fuzzily/whateverly, why is there this weird asymmetry between / and /? Maybe the only answer is that “there happens to be an interesting adjunction of bifunctors where one side is covariant-covariant and the other side is contravariant-covariant.” But this is really unsatisfying.

2. Is the answer to question 1 related to the fact that we are talking about algebras and modules, rather than coalgebras and comodules? If this is the case, then what do these bifunctors and adjunctions look like in the case of bialgebras?

3. Cohomology/ has an interesting product structure. Does homology/ have a coproduct structure? If so, when is it interesting?

4. Less related, but recently bothering me: Does anyone know of an example of a non-commutative ring which is Morita equivalent to its opposite?

[Background for 4: We say rings and are Morita equivalent if the categories are and of left modules are isomorphic. So in this case, I am asking for a ring whose left and right modules agree in some reasonable natural way, but which is not commutative.]

Inspired by Tim Gowers’s illuminating informal discussions of mathematical topics, and in particular this one on vector spaces, I am starting my own “Shoulda Series.” That is, a series of notes on things about which I might have said: “Someone should have told me this a long time ago!” In particular, I don’t want to use this series to “give away” the sorts of things you should really figure out for yourself.

Question: Why do we avoid choosing bases for vector spaces whenever possible? In particular, why is an isomorphism defined independent of bases “better” than one which uses bases?

One of thousands of possible answers: Since I’m a geometer at heart, I think the following is a great reason. In short, the moral is: basis-free isomorphisms generalize to vector bundles and basis-dependent ones usually do not. This is because “choosing a basis” of a vector space is the point-wise analogue of the “choosing a local trivialization” of a vector bundle. To do anything globally, one generally needs to use several different trivializations; a bona fide isomorphism of vector bundles requires a family of vector space isomorphisms compatible with the transition functions between the various trivializations. (For those who know what this means, the precise check is that the Cech 1-cocycles corresponding to the bundles in question differ by a Cech 1-coboundary, where the coefficient sheaf is the constant sheaf associated to .)

For a wealth of examples, consider the following. Any two finite-dimensional vector spaces (over the same field) are isomorphic, but not naturally so. But there are usually several different isomorphism classes of (real or complex) vector bundles of a given dimensionon on a given space.

A more specific example will also bring out a slight improvement of our moral above. Let be a real vector space and be a real vector bundle. Recall that the isomorphism between and its dual is not natural. Hence is not always isomorphic to its dual . (Here, is the trivial vector bundle .) But if is given an inner product , the isomorphism becomes natural: the map is simply . So if a vector bundle admits an inner product (in the sense of vector bundles), then we should expect an isomorphism . Indeed this is the case, and in fact this is a common phenomenon: using a partition of unity, you can check that every real vector bundle on a paracompact base space admits an inner product.

However, the situation is different with complex vector bundles. Let be a complex vector bundle. While every complex vector bundle paracompact base space admits a Hermitian inner product, this product induces an isomorphism between the dual bundle and the complex conjugate bundle ; neither of these need be isomorphic to itself.

For example, consider the tangent bundle to the Riemann sphere and its dual, . Considered as real 2-plane bundles, these are isomorphic since is paracompact. But they are non-isomorphic as complex line bundles. For readers familiar with characteristic classes, there is an easy way to see this algebraically. The Chern classes of a complex vector bundle take values in the integral cohomology of the base space. They obey the relation , so if for any odd , then is not isomorphic to . The Stiefel-Whitney classes of a real vector bundle obey the same law. But since they take values in cohomology, and the relation is an equality!

Aside: Unoriented phenomena (e.g. Stiefel-Whitney classes) tend to use relations, while oriented phenomena tend to use relations. Complex vector spaces have a canonical orientation on their underlying real vector bundles, so complex phenomena (e.g. Chern classes) fall under oriented phenomena. Examples are mod 2 versus oriented intersections (see Guillemin & Pollack, Differential Topology) and the oriented versus unoriented versions of Poincaré duality (see Hatcher, Algebraic Topology, freely available online). While I’m on the subject of references, two great references for vector bundles and characteristic classes are Milnor & Stasheff’s Characteristic Classes and Hatcher’s Vector Bundles and K-Theory (the latter is unfinished and freely available online).

We conclude with our improved moral statement.

Improved moral: A natural isomorphism of vector spaces generalizes to vector bundles. An isomorphism of vector spaces making use of a structure which “globalizes” well (e.g. inner products, when the base space is paracompact) will also generalize to vector bundles. Other isomorphisms often will not.