I’ve been betraying the title of this blog and doing some abstract nonsense lately, mostly to relearn algebraic geometry for the -th time for some embarrassingly large . With the memory capabilities of a muffin, I am practically starting over each time. Luckily, each adventure feels slightly more beatable (exp points?), since I have a few more examples I can use for myself. Masnevets and I had a good discussion about a few basic examples of adjoint functors (recall the definition here: basically, we need a pair of functors and such that we have a natural isomorphism ), and thus we have a new Concrete Nonsense post.

Before we start, I want to state that I’m trying something new. This post is not intended to be an introduction to adjoints as I originally envisioned – I realized that there are many better sources for that. Instead, I’ll try to do a free association that juxtoposes a few elementary concepts. You don’t even have to know the definitions of adjoints to start seeing what I’m getting to, since I’ll be namedropping algebraic structures like Kanye West.

My first introduction to adjoints was from algebraic topology, where you naturally bump into the functors and . It was unnecessary at the time (for the scope of the course, at least) to know that they were adjoints, but now I know them as the “tensor-hom adjunction pair” (saying “tensor-hom” a lot helps me remember tensor as the left adjoint and hom as the right). Furthermore,  knowing this relationship allows me to remember some other things – in particular, knowing the left- and right- exactness of these functors, which I used to always mix up. Left adjoints are always right-exact, and right adjoints are always left-exact. Combined with knowing that tensoring is a left-adjoint, I now know that tensoring is right-exact and adjoints are left-exact.

I’ll now make a digression pertaining to something which I believe is important but seldom discussed. By now, some readers are probably complaining that my previous statement doesn’t make it easier to remember anything, because I have to remember an equally arbitrary fact – and what if our brains find it more intuitive that “left adjoints would be left-exact,” which is wrong? In my experiences, the sets of things that are easy to remember for mathematicians are extremely different from one person to the next. Thus, it may help (especially for “elementary” topics such as this one) to just chain lots of little ideas together, so if someone links concepts A, B, and C, and Alice finds it easier to remember B and C from A, she’ll benefit just as much as Bob, who likes to remember A and C from B (and Chris is sad that nobody likes C except him). As math presentation tends to be fairly “structured” – no surprise, knowing mathematicians – I wouldn’t mind seeing more “free associations” in print or the blogosphere, because this is the form of communication which reminds me most of chatting about math with other people while drinking coffee in the common room, an experience which has often taught me a lot more than going to most lectures.

(since this is a free association, before we go back to adjoints, we might as well see another way to remember the exactness assignments from Masnevets. A good “toy exact sequence” to use here is , where the second map is multiplication by and the third is taking mods . Tensoring by turns the second map into the zero map, so tensoring is not left-exact and thus must be right-exact.)

Many of you may be surprised by tensor-hom as my first choice of adjoint functors – that’s only because I learned that one first (this is a lie, and we’ll come back to later). There’s a much easier example using the forgetful functor. Consider the forgetful functor . It happens to be the right adjoint to the functor that sends a set to a free group generated by the set. The intuition here is this: by the definition of adjunction, we want “the number of maps out of the image of F” to be “the same” as “the number of maps into the image of G.” When forgets so much structure, we’ll get lots of maps into the set . To get lots of maps out of , we want a lot of freedom on the structure, and hence the free group. Of course, there’s nothing particularly binding here about groups; we can do this in general to get a “free functor” when we have a forgetful functor from a category of other algebraic structures to sets (we can’t do this *all* the time, but most of the time we are fine. I don’t understand the conditions here too well, which involves an overloading of the word “variety,”  so I won’t expound).

In fact, let’s take it a step further. We don’t only have to forget into sets. We can forget, for example, from abelian groups into groups, or from associative algebras into Lie algebras. What did we forget in these two cases? Commuting relations and the product, respectively. So intuitively, to “match the complexity” of the two ’s, we want to get just as much structure back: we want to make things commute in an arbitrary group; we want to be able to multiply things in a Lie algebra. It was quite a joyful “but of course” when I realized that the left adjoints of the two functors above came out to be abelianization and the universal enveloping algebra, respectively.

Finally, in this forgetful context I’ll return to the actual first example of adjoint functors I’ve seen (though I definitely did not know it in that context at the time). When we restrict a representation of , or equivalently a -module, to a sub-representation of , we’re “forgetting” how the other elements of act on our vector space. So is there a natural way to get them back? For each representation of of we can induce the representation . This ends up being another adjunction pair, of course. Here’s an immediate consequence: in the case that these groups are finite, note that the dimension of the two ’s must match; this just gives us , which is the Frobenius reciprocity formula from second-semester abstract algebra.

-Y

Last time I gave the definition of -rings and tried to motivate them from the perspective of K-theory. I’d like to continue with how to define abstract Chern classes for -rings and explain Riemann–Roch formalism. We’ll assume all -rings R have an augmentation, are equipped with a positive structure, and have an involution. I’ll once again use K-theory as motivation for the definitions.

In topological K-theory, if we have a space X with a vector bundle E, there exists a map such that the pullback decomposes into a direct sum of line bundles, and the map on cohomology rings is injective. The corresponding statement for -rings is that we can embed any -ring R into a larger one such that every element can be written as a sum of elements with augmentation .

Let be a graded ring. We let be the group under multiplication of formal power series of the form where . A map of Abelian groups (we use addition for R), written is a Chern class homomorphism if

• For each line element u, for i>1.
• For two line elements u and v, .

We will also assume that each is a polynomial. Chern class homomorphisms are compatible with the splitting principle in the following way. If we want to split an element X in R, then there exists a -ring R’ and a graded ring A’ which extends A, together with a Chern class homomorphism which restricts to the one for R. Hence, Chern classes are completely determined by what they do to line elements.

So we get “factorizations” . This allows us to define the Chern character via and the Todd class .

Now we move on to Riemann–Roch formalism. Let C be a category. A Riemann–Roch functor is a triple such that

• For all X in C, we have commutative rings K(X) and A(X). Let UK(X) and UA(X) be the underlying additive groups.
• For every morphism , we have ring homomorphisms (pullbacks) and which make K and A into contravariant functors from C to commutative rings.
• For every morphism , we have group homomorphisms (pushforwards) and which make UK and UA into covariant functors from C to Abelian groups.
• The projection formula where and and holds for H=K and H=A.
• is a natural transformation from K to A.

The statement that Riemann–Roch holds for a morphism f in C means that there exists an element such that for all x in K(X). The element measures the failure for to be a natural transformation of covariant functors.

If Riemann–Roch holds for two morphisms, then it also holds for their composition.

To bring in -rings, we focus more specifically on Chern class functors. This is a Riemann–Roch functor where we further assume that

• The image of K is the category of -rings with involution.
• The image of A is the category of graded rings whose morphisms are degree 0 maps.
• For every X in C, is a Chern class homomorphism.

Fix this Chern class functor. Since this post is already getting too technical, let me just mention that one defines what it means for a morphism in C to be an elementary embedding and elementary projection, and shows that Riemann–Roch holds for such morphisms. The elements in both cases are defined in terms of Todd classes.

Now we relate this to the Grothendieck–Riemann–Roch theorem. For this, we fix some Noetherian commutative ring R and let C be the category whose objects are schemes over Spec R that are quasi-projective and connected. The morphisms in our category are those which are projective local complete intersection morphisms: these are all morphisms of the form where E is a vector bundle on Y, and is the corresponding projective bundle, and i is a closed embedding whose image is locally a complete intersection (i.e., there is an open affine cover where in each open affine, the ideal of X is generated by a regular sequence). Here K is the K-theory of algebraic vector bundles over X, and A is a certain associated graded ring of K(X). We have pullbacks of vector bundles which makes K and A into contravariant functors. The pushforward on K is given by an alternating sum of higher direct images. In this case, i is an elementary embedding, and p is an elementary projection in the language used above.

What comes out of all of this (and the stuff I skipped on elementary embeddings and elementary projections), is that Riemann–Roch holds for all morphisms in this category, and the element can be written as the Todd class where is the virtual tangent bundle of f defined as
using the notation above. here denotes the relative tangent bundle, and is the normal bundle of X inside of (this is a bundle because X is a local complete intersection).

We’ll do one last thing: specialize to the case that R=k is a field and we consider only connected nonsingular quasiprojective varieties over k and proper morphisms between them. Given X, we can embed it inside some projective space via a map i. A map is proper if and only if the graph morphism induced by the maps i and f is a closed embedding. Whenever a nonsingular variety is embedded as a closed subvariety of another nonsingular variety, it is automatically a local complete intersection, so we see that this case is really a special case of the preceding paragraph (here E is a trivial rank n+1 vector bundle over Y). Also, on a nonsingular variety, every coherent sheaf has a finite resolution by vector bundles, so K(X) is in fact the Grothendieck group of coherent sheaves (though the multiplication for coherent sheaves which aren’t vector bundles involves sums and Tor functors). Furthermore, we can let A be the Chow ring functor (which shows that it coincides with the associated graded of K in this case), and our natural transformation is the usual Chern character ch. We’ll finish with what Grothendieck–Riemann–Roch says in this case:

Let be a proper morphism between nonsingular quasiprojective varieties, and let [F] be an element in K(X). Then

.

-Steven

In this post, I want to discuss Grothendieck’s -rings and how they provide an abstract setting for Riemann–Roch formalism. The references I’ll be using are

• Donald Knutson, Lambda-Rings and the Representation Theory of the Symmetric Group
• William Fulton and Serge Lang, Riemann–Roch Algebra

The definition of a -ring is a bit technical, but it starts with a commutative ring R together with operations for all nonnegative integers i such that and for all r in R together with some axioms. In particular, we should say what these lambda operations do to sums and products, and we might also want to know what compositions of them look like. To motivate these axioms, we’ll look at K-theory (where it originates).

Let X be a topological space, and consider the set of all vector bundles over X (topological, smooth, holomorphic, whatever you want). We define lambda operations using exterior powers: . Of course, the set of vector bundles on X isn’t a ring, but the free Abelian group of isomorphism classes of vector bundles is a ring if we use tensor product as the multiplication. But we have to define exterior products on “negatives” of isomorphism classes of vector bundles. For actual vector bundles, we have the identity

,

so we can use this to extend to “negatives” and we make this an axiom for a general -ring:

(L1) for all r and s in R.

In fact, the above identity holds if we pass to the Grothendieck group of vector bundles over X (add the relations whenever we have a short exact sequence of the form ) because in general, if E’ is an extension of E and F, then has a filtration whose associated graded is the direct sum on the RHS above.

What about products? i.e., what should be? The exterior power of a tensor product of two vector bundles has a rather complicated expression. Nonetheless, there exist integer valued polynomials in 2n variables such that

.

How do we get these polynomials? Let be algebraically independent variables, and let denote the ith elementary symmetric function in the Xs and Ys, respectively. Then we define the polynomials via the identity

.

So we have some complicated family of polynomials, and the axiom

(L2) for all r and s in R.

There are also some integer-valued polynomials of degree nm for expressing the compositions . I won’t get into that, but this gives the third axiom

(L3) for all r in R.

A morphism of -rings is a ring homomorphism which commutes with the lambda-operations.

For a simple combinatorial example, take X to be a point. In this case, vector bundles are just vector spaces, and . Identifying vector spaces with their dimension, the lambda operations become , which is just the binomial coefficient when n is nonnegative. A morphism is called an augmentation. For vector bundles, this map is given by sending a vector bundle to its rank and extending linearly to virtual vector bundles. We’ll assume our -rings are augmented.

We can place some further requirements and operations on our -rings. First, in , we naturally have a notion of what it means to be “positive”: any class which represents an actual vector bundle. The set of positive elements has the property that it’s closed under addition and multiplication, and every element of can be expressed as a difference of two positive elements. Furthermore, whenever x is positive, for sufficiently large i, and all positive elements of augmentation 1 (line bundles) have multiplicative inverses. We’ll take all of these features to be an axiom system for a “positive subset” of a -ring. Motivated by the K-theory, we’ll call positive elements of augmentation 1 line elements. We’ll assume that our -rings are equipped with a positive structure.

K-theory also has a nice involution: send a vector bundle to its dual bundle. In general, we’ll say that is an involution of our -ring if it satisfies

• for all x,
• for all x,
• for all line elements u.

We’ll further assume that our -rings are equipped with an involution.

Another example of -rings comes from representation rings. Given a group G and a representation V, we define to be the representation with the diagonal action of G. The augmentation here sends a representation to its dimension (over the ground field), the positive elements are the representations, and the involution sends a representation to its dual. One particular example is when G is the general linear group and we consider only rational representations, so that the representation ring is the ring of symmetric functions in n variables (together with a multiplicative inverse for the product of the n variables). In this case, the lambda operations are plethysm: where is the nth elementary symmetric function, and positive means Schur positive. [I first saw -rings in context of symmetric functions, so the definitions seemed a bit mysterious to me.]

In the next post, I’ll discuss abstract Chern classes in the context of -rings and Riemann–Roch formalism, and say how this relates to Grothendieck–Riemann–Roch for proper maps between nonsingular varieties.

-Steven

In this post, I want to give an example of a Lie group which is not isomorphic to a subgroup of a matrix group. This contrasts with the algebraic picture, where every affine algebraic group can be realized as a subgroup of a matrix group. I’ll just sketch the details.

The group we’ll use is the universal cover G of . One can realize as a fiber bundle with fiber over via the map which sends a matrix to its first row, so its fundamental group is . Similarly, we can use this map to show that is a fiber bundle with fiber over and hence is simply connected.

Now let be any smooth map. It induces a map on Lie algebras . We can use this to define a map by . It’s straightforward to check that f preserves the Lie bracket. Since is simply connected, f is the derivative of some smooth map .

To show that r cannot be injective, we will show that the composition is not injective, where the second map is the usual inclusion. I claim that this composition is the same map as the composition , where is the covering space map. By construction, these two compositions induce the same differentials on their Lie algebras. Now using the fact that both G and are connected, this implies that the map on the level of Lie groups must be equal as well. But is not injective: we saw above that the fundamental group for is nontrivial, so we are done.

In this post, I want to discuss Pfaffians, a topic which I wish I had learned about as an undergraduate. I’m very interested in syzygies of ideals and such, and every now and then Pfaffians come up, so if only I knew what they were! Now that I know, I want to explain what they are and how they’re related to Plücker ideals.

Everything will be over the field K. If an n x n matrix has rank < r, then this can be checked by showing that all of the r x r submatrices of it have determinant 0. In particular, since these r x r minors are polynomials in the entries of the matrix, this says that the set of all matrices of rank < r is an algebraic subset Y of the space of all matrices. That it's irreducible can be seen by the following argument: let X be the space of n x n matrices, and let Gr(r-1, n) be the Grassmannian of r-1 planes in n-dimensional affine space. Then consider the subset Z of Gr(r-1, n) x X given by . If R is the tautological subbundle on Gr(r-1, n), then is a vector bundle over Gr(r-1, n) and hence is irreducible. But the image of Z under the projection is Y, so Y is also irreducible.

It's not so clear that the ideal generated by the r x r minors of a generic (= entries are algebraically independent variables over K) n x n matrix is radical, but this turns out to be true (one way to show that this is true is to find an explicit Gröbner basis for it).

But what if we only care about skew-symmetric matrices? To check if a matrix has rank < r, we could do the same as above, but the ideal generated by the r x r minors of a generic skew-symmetric matrix will NOT be radical. One problem already is that the determinant of any skew-symmetric matrix is always a perfect square in the field K, and hence our ideal should contain these square roots, which are called Pfaffians.

First I’d like to prove this last statement. Say M is a skew-symmetric n x n matrix. We want to find a number r such that M can be transformed isometrically into the matrix where is the r x r identity matrix (r could be 0, in which case these rows don’t appear), and the bottom row consists of n-2r rows of zeroes. First, M represents a bilinear form given by , where is the standard basis for . The skew-symmetry of M means that this form is skew-symmetric. Let L be the kernel of this form, i.e., the set of all x such that for all y. Then the form restricted to any complementary subspace L’ of L inside of will be nondegenerate. So for any nonzero x in L’, we can find y in L’ such that . Repeating this on the orthogonal complement of L’ inside of L, we can find a basis for L’ such that and is the Kronecker delta. Representing M in this basis gives the desired form. Note that if B is the matrix whose columns are , then the matrix representing the bilinear form with respect to this new basis is instead of .

This says three things: (1) the rank of M is always even, and (2) is either 0 or 1, so is the square of something, which we call the Pfaffian of M, and (3) if we want to test the condition that rank M < r, we need only check if the principal r x r minors (i.e., when the row indices and column indices are the same) are 0. Since these submatrices are themselves skew-symmetric, they also have Pfaffians, so it's equivalent to check if they are 0.

Do these give polynomial equations? Let M now be a generic skew-symmetric matrix with algebraically independent entries for , so that , and let K be the function field . By the above, the determinant of M is a square of something in , which we denote by Pf(M). But the determinant of M is a polynomial in , so by the fact that is a UFD and Gauss' lemma, this square root actually lives in . There's some ambiguity up to sign regarding which root to take, so we just pick one. So we see that the Pfaffians are polynomials in the entries of M, and hence the locus of skew-symmetric matrices of rank < r is also a subvariety of the space of skew-symmetric matrices. Since the rank of M must always be even, we should assume that r is odd. It turns out that the ideal generated by the Pfaffians of the (r+1) x (r+1) principal minors of M (just call these the (r+1) x (r+1) Pfaffians of M) will be radical.

Now that we've done that, how are Pfaffians related to Plücker ideals? First I'll say what those are: given the Grassmannian Gr(r, n), we can embed it inside of dimensional projective space by representing a subspace by an r x n matrix and then sending it to the -tuple of its r x r minors. The homogeneous ideal which defines the image is a Plücker ideal . Of course, we could forget that this ideal is homogeneous, and then it would define a subvariety (the affine cone) of dimensional affine space. In the case that r=2, the elements of dimensional affine space can be identified with skew-symmetric n x n matrices (there are elements above the diagonal), and the image of the Grassmannian corresponds to those which have rank at most 2 (because they are minors of a 2-dimensional subspaces). Hence the Plücker ideal is generated by the 4 x 4 Pfaffians of a generic n x n skew-symmetric matrix.

Why is this relevant? It's known how to write down the minimal free resolutions for the Pfaffians of a generic skew-symmetric matrix (you can find this in Section 6.4 of Jerzy Weyman's book Cohomology of Vector Bundles and Syzygies), so this gives for free the case of the Plücker ideals when r=2. The general case r>2 is not known, however. The case r=3 and n=6 can be done in Macaulay 2 in a few seconds (use the command res Grassmannian(2,5)), but I think any larger examples are computationally too expensive. In principal, they can be written down by hand using the techniques of Section 7.3 of the above book. There, the difficulties arise in trying to solve some plethysm problems for the general linear group. So it’s likely that a closed form solution will never be attainable.

-Steven

A not often-mentioned skill that I have been trying to develop is the evaluation of the “usefulness” of statements. Given my horrible memory and degrading mental RAM, I must prioritize which facts to consciously absorb and which ones to skim, since I’m sure I forget at least one theorem every time I learn a new one. Luckily, books aid to an extent by labelling the important things “Theorems” and “Lemmas,” though sometimes the latter are just as important as (if not much more than) the former, and sometimes even more important items appear in the Exercises section. Something about Exercise 7.70 in Richard Stanley’s Enumerative Combinatorics 2 really struck me as “useful.” The statement is:

where the are sets of variables, the product of the hook-lengths, gives the cycle structure, and are permutations of size .

Two things make the intuition of “useful” more concrete:

One, the statement has a lot of flexibility; By the symmetry of the symmetric group and the fact that cycles are preserved under inverses, you can sum over, say, products of all ordered -tuples of by rewriting the sum on the right as over , for example, or sum over pairs which give the same product, say . Furthermore, having the choice over the variables makes a lot of specialization techniques useful. For example, you can “pick out” cycles by specializing , where is an -th root of unity, because the power sums then give unless .

Two, the statement has a lot of power; the proof of this fact has a “Stanley difficulty” of [3] (out of 5) means it contains some nontrivial usage of the symmetric function machinery already. Thus, by using it we know we are backed with some complex stuff in the background. Note when , we get the amazing Hook-length formula directly; when , by observing that summing over is really summing over a single (a special case of our discussion above on “flexibility”), we get the familiar “inner product” identity. These are two nontrivial statements without an obvious connection with each other, a pleasant surprise.

This warm “finding a hidden gem” feeling is very nice.

-Y

P.S. Well, there is a third (and less glamorous) cue, which is that at least two problems in the problem sets came with the hint: “Look at Exercise 7.70.” The power of hindsight in all its glory, I suppose. However, I liked this statement too much that I needed an excuse to write it down, so shhh.

I wanted to present a proof of Lagrange’s four square theorem that I had seen a few years ago that I really like. I think there’s also some approaches using analytic number theory and algebraic number theory, but this one uses convex geometry (!) First, let me state the theorem (my convention here is that denotes the nonnegative integers):

Theorem (Lagrange). Every nonnegative integer can be written as a sum of four squares, i.e., the function given by is surjective.

In order to make this (reasonably) self-contained, let’s give some definitions we’ll use.

A lattice is a discrete subgroup of which spans in the sense of vector spaces. Basically, a lattice will be some subgroup of which is isomorphic to . If is a basis for , define to be the determinant of . Define the fundamental parallelepiped of to be the set .

Now we just need some convex geometry. The main tool will be Minkowski’s theorem. The proof is short, so I’ll include it.

Lemma (Blichfeldt). Let be a lattice and be a measurable set. If , then there exist distinct such that .

Proof. Let be the fundamental parallelepiped of . Then (disjoint union), and hence . Define . Then

.

Since each , there must exist distinct such that . Take and set and .

Lemma (Minkowski). Let be a lattice and be a centrally symmetric (i.e., implies ) convex set such that . Then contains a nonzero element of .

Proof. Set , so . Then there exist distinct such that . Since , we have .

Now we can write down a proof of Lagrange’s four square theorem.

Proof. Pick . We break the proof up into three steps.

Step 1: Reduce to the case that is prime. The sum of four squares is the square of the norm of a quaternion . Since norm is multiplicative, the product of two sums of four squares is itself a sum of four squares. This can be checked directly, but it’s not really that interesting.

Step 2: Find such that . If , take , . Otherwise, assume is odd. Define
. Choose with . Then , and since , which implies that , and hence . So has elements.

Similarly, define . Pick such that . Then as before, , so also has elements. Hence cannot be empty by the pigeonhole principle, so we can find and such that .

Step 3: Construct a suitable lattice and centrally symmetric convex body such that a lattice point in will correspond to an expression of as a sum of four squares. Define . Being a subgroup of , it is clear that is discrete. Also, the set surjects onto under the projection, so has finite index in , and hence has full rank and . Define . Since

,

we can apply Minkowski’s theorem to find such that . Now implies , and hence is a multiple of . From , this multiple must be 1.

And there you have it!

-Steven

Arts and crafts and mathematics have met before. For instance, mathematical knitting isn’t new. I’d like to talk about my humble (read: shoddy) contribution.

An important structure in homological algebra, especially when talking about things up to (co)chain homotopy,  is that of a triangulated category.  If you haven’t seen triangulated categories before, Wikipedia is decent, but Chapter 10 of Weibel’s Introduction to Homological Algebra is better. Here, I’m just going to be concerned with the octahedral axiom, whose statement is notoriously hard to picture. (Throughout this post, I’ll be using the formulations and notations of Weibel.)

Let be an additive category and let be an automorphism of this category. A triangle in is an ordered triple of objects along with three morphisms: , , and . We write for short. A triangulated cateogry is an additive category equipped with an automorphism and a distinguished family of triangles which are called the exact triangles. We usually draw an exact triangle as shown in the diagram to the right. These data must satisfy four axioms, the last and most confusing of which is the following.

Octahedral Axiom: Let be a category equipped with an automorphism . Suppose we have three exact triangles: , , and . Then there is a fourth exact triangle such that , , , and .

This is usually pictured as an octahedron not unlike the image on the right, which I’ve shamelessly re-LaTexed from a diagram in section 10.2 of Weibel. (The picture looks better there.) In this view, the equalities of morphisms in the Axiom take on geometric significance: the equalities and mean that the four non-exact faces all commute (two did automatically), and the equalities and say that the two shortest paths between and in either direction agree. (There is an asymmetry here: the other two pairs of antipodal vertices do not have any such paths to compare.)

There is another common visualization, a planar one involving line segments consisting of two morphisms each. See Weibel or Wikipedia.

There are a few annoying drawbacks to the octahedral diagram above:

• At a glance it’s not immediately clear which triangles commute and which are exact.
• At a glance it’s not immediately clear which three exact triangles were given and which one was implied by the Axiom.
• Some arrows involve the objects , , and , while some involve , , and . But in the diagram, we only have room for one or the other, so we omit the . Since the convention for drawing an exact triangle already takes this into account, this no problem when considering exact triangles; but when we view the commuting triangles, we have to keep in mind this convention of writing in place of .

I like to see things clearly, immediately. I don’t to have to think about which triangles commute and which are exact, or whether an arrow is really hitting or . All the problems above are fixed by my small pet project for this past weekend. Check out the eye candy (in the second picture, you can kind of see six of the eight sides):

Other than the shoddy workmanship, here are some things to notice about the model and why it improves on the diagram above:

• It’s 3D! You can hold it (but please, be gentle)!
• Blue sides are exact triangles; yellow sides are commutative triangles. The “output” exact triangle has been labeled (though not in the pictures above). So the meanings of the triangles are obvious at a glance.
• If a map goes to or from instead of , then the vertex corresponding to is labeled as when part of an exact triangle (following the usual convention), but as on the yellow, commuting faces. This is my favorite feature of the model.

The other day I read this post over at Notational Notions. The main result is the following.

Proposition: Let be an infinite field and any -vector space. Then cannot be written as a union of finitely many proper subspaces.

can be either finite- or infinite-dimensional in the above. The very basic idea here is that in an algebraic setting, unions of “algebraically nice” subsets need not be “algebraically nice.” An immediate corollary of this Proposition is the following (somewhat silly) version of the Prime Avoidance Lemma.

Silly Prime Avoidance (SPA) Lemma: Let be a ring containing an infinite field. Let be ideals of such that . Then for some .

The goal of this post is to discuss the SPA Lemma and to see what it means in terms of “geometry over a finite field.”

Proof: Considering as a -vector space, every ideal of is a subspace. Now apply the Proposition with , and the subspaces in question being . Since there are finitely many , we can only have if for some . Q.E.D.

(Recall that the full Prime Avoidance Lemma allows the following modification of the hypotheses: an arbitrary ring, as long as at most two of the are non-prime. While neither of the two sets of hypotheses are stronger than the other, the “Silly” version seems a bit less “serious” to me. See Chapter 3 of Eisenbud’s Commutative Algebra with a View Toward Algebraic Geometry for a thorough discussion.)

Can we relax the condition that be infinite? Certainly not, since if is finite, then can be a finite set, which is easy to exhaust by a finite union: Consider . Then the ideal is in the union of the ideals , , and . but is not contained in any of them individually. (Note that none of the ideals , , or are prime.) So really, in order to ask whether or not we can strengthen the SPA Lemma, we should address the following problem.

Problem: Let be a finite field and let be a unital -algebra. Find an ideals of such that but not contained in for all . Let’s ask that be Noetherian, too. (Hints (highlight to view): 1. We will need at least three which are not prime, by the full Prime Avoidance Lemma. 2. Clearly, cannot be a principal ideal; in particular, cannot be a PID.)

While you think about the Problem, let’s consider what this means for geometry over finite fields. Let be a finitely presented -algebra and an ideal of . Suppose are closed subsets of with corresponding ideals . The SPA Lemma says that if on each particular not all the vanish, then there is some (with ) which vanishes on none of the . (When we ask if vanishes at a point of , we are asking whether the image of f in the residue field at the corresponding prime is zero; see Chapter 1 of Eisenbud and Harris’s The Geometry of Schemes.).

What, then, would a solution to the Problem be?  It would be an affine scheme (or affine algebraic set, if you don’t like schemes and don’t mind taking R to be nilpotent-free) over and a finite set of polynomials on , along with closed subsets such that every combination (with ) vanishes on one of the .  One consequence of this is that we have no hope of “simultaneously separating” the using a function from the (non-unital) subalgebra : i.e., there cannot be a function which takes different values on each of the .  One possible intuitive explanation for this is that the residue fields are extensions of ; in nice cases, then, they will all be finite themselves.  And as long as they don’t vary too wildly (e.g., if there is a such that all the residue fields are at the largest), then the set of possible values of functions is a finite set; by the Pigeonhole Principle, if we choose too large, then we’ll never separate them all simultaneously.  Our result is stronger: not only will some value repeat, but at least one of the values will always be zero.

Does anyone know a better geometric intuition for this, or a more precise statement of what sort of “non-separation” this entails?  I would love to understand this more clearly.  More generally, I’d like to understand how well we can separated closed subsets of arbitrary schemes.  Anyway, we conclude with a solution to our problem.

Solution: (Highlight to view.)  Let and .  Let , , and ; note that these may be prime.  For , let be the ideal generated by and all polynomials of degree 2 and higher with no linear or constant term.  Since R is Noetherian, these must be finitely generated (even though we don’t need this to be true, it’s nice to know it is).  It is easily verified that this is a solution to the problem. 

Remark 1: (Highlight to view.)  This solution shows that the hypothesis that “no more than two of the are non-prime” is sharp.

Remark 2: (Highlight to view.)  Two easy generalizations of the solution: you can do this over any and for any number of variables by just adding another ideal of the form for each possible homogeneous linear polynomial.  And the quotient of this by any ideal generated by functions with no linear or constant term will still be an example; in particular, this yields an example of Krull dimension one, e.g. .

(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.]