(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 **co**homology, 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 **co**algebras and **co**modules. 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.]

About question 4.

If you take the first Weyl algebra, then it’s actually isomorphic to its opposite algebra.

By:

Grétar Amazeenon September 30, 2008at 11:20 AM

Another example of a noncommutative algebra thats isomorphic to its opposite is the enveloping algebra of sl_2. But if you look at sl_3 then the enveloping algebra is not isomorphic to its opposite algebra, but they are Morita equivalent because a left module over the opposite algebra is just a right module over the algebra itself, and since the enveloping algebra of sl_3 is a Hopf algebra then the categories of left modules and the category of right modules are equivelent. This argument actually holds for all lie algebras, g. The enveloping algebra of a lie algebra is Morita equivalent to its opposite.

By:

Grétar Amazeenon September 30, 2008at 11:33 AM

Grétar: thanks for the examples! Coincidentally, we proved that universal enveloping algebras are isomorphic to their opposite in my representation theory class just yesterday.

By:

Alexander Ellison October 2, 2008at 3:18 PM

Does that hold for all universal enveloping algebras? Somehow I thought the enveloping algebra of sl_3 was not isomorphic to its opposite, but certainly Morita equivalent. Maybe I’ve made a mistake somewhere in my calculations.

By:

Grétar Amazeenon October 2, 2008at 3:53 PM

If L is a Lie algebra, then x \mapsto -x for all x in L gives an anti-automorphism of U(L), hence an isomorphism from U(L) to U(L)^op.

By:

Alexander Ellison October 2, 2008at 5:08 PM

… which of course raises the question: Can we find a ring A which is Morita equivalent but not isomorphic to its opposite?

By:

Alexander Ellison October 2, 2008at 10:06 PM

Well I guess the obvious thing to do is to try to find a Hopf algebra thats not isomorphic to its opposite.

By:

Grétar Amazeenon October 3, 2008at 4:44 PM

Probably the simplest example of a ring that is Morita equivalent but not isomorphic to its opposite ring is the ring of 3×3 matrices (over some field, say) with first row having zero second and third entries. I.e., matrices of the following shape:

(*00)

(***)

(***)

By:

Jeremy Rickardon March 18, 2009at 10:35 AM

Hey Jeremy, I’ve been looking for such an example for a while now, how do you explain (a) your ring not being isomorphic, and (b) your rings being Morita equivalent to its opposite?

By:

Jubbly Jargonon March 24, 2009at 6:11 PM

The ring and its opposite are Morita equivalent because they are both Morita equivalent to the ring U of upper triangular 2×2 matrices (which *is* isomorphic to its opposite).

U is the direct sum of two projective (right) modules, one of which, call it P, is 2-dimensional (the right ideal of U consisting of the matrices whose bottom row is zero) and the other of which, call it Q, is 1-dimensional (the right ideal consisting of the matrices whose top row is zero).

Then the ring I described is the endomorphism ring of the direct sum P+P+Q, whereas its opposite is the endomorphism ring of P+Q+Q, so by Morita theory both are Morita equivalent to U.

The ring is not isomorphic to its opposite because it has a 1-dimensional projective right module but not a 1-dimensional projective left module.

By:

Jeremy Rickardon March 26, 2009at 5:01 PM

That’s a nice compact proof, I managed to get a proof independently, but it was very similar to yours. Thanks for the cool example.

By:

Jubbly Jargonon March 31, 2009at 4:01 PM

Nice example! Two questions:

Q1: Is there a name for these algebras which are Morita equivalent to their opposites but not necessarily isomorphic?

Q2: What if the algebra is separable? Does being Morita equivalent to its opposite imply it is isomorphic?

By:

Chris Schommer-Prieson April 5, 2009at 8:39 PM

Doesn’t seem like such algebras can be seperable dude:

An associative k-algebra A is seperable when for any field extension k/L we get that A tensor_k B is semisimple.

k/k is a valid field extension and A tensor_k k is just the k-algebra A.

A is Morita equivalent to the upper-triangular matrices over k which (I’m asserting; and I’ve a proof somewhere) won’t be semi-simple.

[+] Maybe you mean matrix algebras bigger than just the 3×3 case?

By:

Jubbly Jargonon April 8, 2009at 5:46 PM

By the way, when I start talking about A being Morita equivalent to the triangular fellas, I am refering to Jeremy’s example from thenon.

By:

Jubbly Jargonon April 8, 2009at 5:47 PM

Chris,

(1) As far as I know, nobody has given this kind of algebra a name.

(2) For a separable example (over a suitable field k), you could take a division algebra A over k that is not isomorphic to its opposite algebra B. Then A x M_2(B) is Morita equivalent to its opposite algebra B x M_2(A), but not isomorphic.

By:

Jeremy Rickardon April 9, 2009at 11:54 AM

By the way, both the examples I’ve given start from the same idea:

Start with a ring that is isomorphic to its opposite ring (upper triangular 2×2 matrices in the first example, and AxB in the second) and then take a Morita equivalent ring in a way that breaks the symmetry.

I haven’t thought about it much, so the answer may be easy, but a harder question might be whether there’s an example that isn’t Morita equivalent to any ring that is isomorphic to its opposite. That’s not possible for finite-dimensional algebras over a field, but for general rings I don’t know.

By:

Jeremy Rickardon April 9, 2009at 3:40 PM

It’s a very nice construction method actually, didn’t see the whole picture initially.

If a ring is isomorphic to its opposite then it must be Morita equivalent. This follows from one result by Lam (Lectures On Modules And Rings; 1998) saying that the rings Morita equivalent to a ring R are exactly rings isomorphic to eMn(R)e where Mn is the n-n matrix ring and e is some full idempotent. Now take n=1 and e = I and you’re done.

By:

Jubbly Jargonon April 9, 2009at 4:36 PM

I just noticed a more basic way you could reason this out:

S is isomorphic to R.

R is a finitely generated projective generator as an R-module.

End(R) = Hom(R,R) ~ R ~ S.

(With ~ an isomorphism.)

Now the Morita theorems give R morita equivalent to S.

By:

Jubbly Jargonon April 9, 2009at 4:43 PM

Hi, just a quick comment: the hom and tensor bifunctors land in the category of abelian groups in general, if your ring isn’t necessarily commutative.

By:

bstonekon June 19, 2012at 10:56 PM