Coordinatization

First off, a few words on what we mean by coordinatization: as in algebraic geometry, we say that a coordinate function is some or possibly some , with all the niceness properties we’d expect to see in the context we’re working.

A particularly good example is Principal Component Analysis which yields a split linear automorphism on the ambient space that maximizes spread of the data points in the initial coordinates.

Topological coordinatization

The core question we’re working with right now is this:
Given a space (point cloud) X, and a (persistent) view of , can we use some map to generate a map inducing that map?

In topology there are ways around to compute spaces of maps directly from available homological information. The key phrase here is the Eilenberg-Moore spectral sequence, by which we compute using . We note that with the compact open topology.

We fix a field k of coefficients. We can find a spectral sequence that computes

Variations of this spectral sequence allow us to compute . A homotopy class of maps thus corresponds to a cohomology element from .

Monad approximation

Define
(unordered pairs)
and similarily unordered tuples

we can get a map, for pointed spaces

In the limit, we get the union .

The Dold-Thom theorem tells us:

Suppose we are really looking for maps .

We pick chain complexes . There is a hom chain complex consisting of graded maps with the homotopy/induced boundary

For this complex, we get
is the family of chain homotopy classes of chain maps .

Recall: are chain homotopic if there is a family such that .

Homotopic space maps yield homotopic chain complex maps. And thus

How, though, do we get from to ?

Cosimplicial spaces

A simplicial space is a contravariant . It has a total space or realization given by

In the dual setting of cosimplicial spaces, the total space is a subspace of .

There is a cosimplicial space

whose total space is .

Furthermore, there is a spectral sequence starting at (well understood space!)
going through

which recovers for us.

The mapping space maps to , which sends to chain maps. Differentials give us conditions on chain maps to come from actual maps on the topological level .

Suppose now that is a simplicial complex, and is known, say isomorphic to .

We pick as a model space, and work in . We pick out conditions form the spectral sequence above to pick out chain maps that might come from topological maps.

Suppose we had some chain map that sends .

We look for a chain homotopic map such that is a simplex (this will be WAY too optimistic, usually) or at least for small and close together in graph distances.

Yi Ding: there are likely many simplicial maps inducing any given . This makes optimizations on quality of maps (as measured in smallness properties of preimages of given simplices) troublesome, since we have a large search space and are likely to get stuck in local minima.

The topological approach creates very many maps, but we just want ONE – albeit with quality assurances. The big question is how to get there:

Shrinking the space? Homotopy collapses to reduce the number of generators?

Ask for more side conditions? If so, which?

Minimize the size of preimages and kernels – smoothing the resulting map?

Run with one condition until we reach a minimum, then change conditions to get out of that particular sink, repeat until it all looks stable?

Thus we need to search for strategies to improve convergence to actually good maps.

The questions we pose are similar to:

Jeff Ericksson, et al.: Given X, , how do we find a good representative ?

Finding good maps is really about finding good representatives in . Given simplicial bases we can pick out a first basis
(Kronecker ) for .

Finally: many problems in data analysis call for analyzing contractible spaces. So if we pick out the notion of a boundary, doing all of this relative to a boundary would give us increased power to use the resulting methods in data analysis.

He pointed out that the underlying principle inducing the rule

goes under many names, among those \emph{Inclusion-Exclusion}, favoured among computer scientists (and combinatoricists). He also pointed out that the origin of this principle is the Mayer-Vietoris long exact sequence

In this blog post, I’d like to give more meat to this assertion as well as point out how the general principle of Inclusion-Exclusion for finite sets follows immediately from Mayer-Vietoris.

Inclusion-Exclusion, and the passage from two sets to many

The basic principle of Inclusion-Exclusion says that if we have two sets, and , then the following relationship of cardinalities holds:

A first proof would be performed by referring to an appropriate Venn
diagram, or by pointing out that if we try to count all the elements
of by adding the counts for and , then we’ve overcounted the elements of , and thus need to subtract these.

Now, this is good and convincing for the situation where we have just two sets, but what if we had a whole family? What if we had ? Well, our old friend the complete induction rides to our rescue. Suppose we had already proven that for sets , the following formula holds:

If you think this formula is imposing, let me show you the first few ways it works out, so that you can get a feeling for it:

Here it starts becoming apparent why it is called the Inclusion-Exclusion formula: we include all elements from all sets, but then we’ve gotten too many elements from the pairwise intersections, so we exclude those, but then we’ve gotten too few elements from the triple intersections, so we re-include those, but then we’ve gotten too few elements from the quadruple intersections – and on it goes.

If our assumption holds, we can consider what the appropriate size of the union of different sets should be. We can pick one set, say , and treat it separately. Then, certainly, the union of the rest has size given by the formula, by our induction assumption. So, consider what happens when we add in the elements in . We then overcount all elements in any of the intersections , so we have to subtract those. But we then have undercounted all elements in any intersection on the form , so we need to add those back in. And so we go on, and for each intersection that occurred in the formula for sets, we find ourselves forced to add one new term, of the opposite sign, that counts the intersection of that old term with .

Thus, the new formula takes on the exact guise of the general formula we wanted to prove, since the terms involving are precisely indexed, in that formula, by the terms excluding , but with opposite signs.

Encoding set sizes homologically

So, now that we know that pairwise Inclusion-Exclusion implies the general Inclusion-Exclusion, let us take a look on how we can express Inclusion-Exclusion with homology. We note first of all that counts the number of path-connected components of a space . This follows easily from a few properties. First, a basis for the -chains of is given by the points in . Second, two -chains are homologous precisely if there is a -chain whose boundary is the difference between those -chains. The boundary of a path is the difference of its endpoints, so we end up with two basis elements being homologous precisely if there is a path connecting them.

Thus, two -chain elements are homologous precisely if they’re in the same path-component of . So, the homology classes of -chains are precisely the path-connected components.

So, the -degree homology classes count path-components. How do we use that? Well … if we consider a finite set with the discrete topology – i.e. open sets are single points and any unions of single point sets, then a path is a continuous function , i.e. one where the preimage of any open set is open. Thus, the preimage of any point is an open set. Suppose a path contains more than one point. Then we can partition the interval into two disjoint open sets, by taking one set to be the preimage of one of the points, and the other set to be the preimage of everything else. However, the interval is connected, which means that if there are two such open sets, one of them has to be empty.

Thus, the path-connected components of a discrete topological space are the points themselves. So if is discrete, then .

Long exact sequences in homology and the Mayer-Vietoris sequence

It is a theorem from homological algebra that if we have a short exact sequence of chain complexes

then there is a long exact sequence in homology given by

Almost all interesting long exact sequences in topology are special cases of this particular fact. We will derive one interesting sequence: the Mayer-Vietoris sequence from this.

Take a topological space . The singular chain complex forms a chain complex with the free abelian group of -chains and the differential given by the usual formula.

Suppose we have two spaces: and . Then any -chain on is an -chain on and an -chain on . So we get an inclusion map by . We choose this sign instead of the image to make the next step cleaner. It is an injection because is a free abelian group, and the injections
are set maps on bases for the corresponding chain groups.

Compatibility of this map with the differential is almost as easy. If we first apply the differential and then map into the direct sum, we get the map . If we instead first map into the direct sum, and then apply the differentials, we get the map . These are equal, and thus the map we’ve constructed is a chain map.

So, is an injective map of chain complexes.

If we now have a pair of chains, we can construct a chain on the union of the two spaces. An -chain on is automatically an -chain on . Similarly, an -chain on is automatically an -chain on . So both and inject into by the same kind of argument as with the injection .

The injections we get this way can be put together to a map by, for instance, . I am going to fudge over the issue of whether the -chains on and form a generating set for the -chains on , and just take it on faith that this works. The details are not particularly difficult – they are just obnoxiously long and annoying.

Now, given that we believe we can get a generating set by taking the union of the bases for the -chains on and , then this means that the map I gave above is surjective. Indeed, a basis element is hit by the element , and a basis element is hit by the element .

This argument proves exactness at the first and last term of the short exact sequence

and it remains to prove exactness at the middle term. So we need to prove that

Now, since , the inclusion of the image in the kernel is obvious. So, suppose now that , and . Then, in , we know that . But then, since and similarly since . So and . But then this kernel element really is in the image of the inclusion we gave.

So this is a short exact sequence of chain complexes. It follows from the theorem of long exact sequences in homology that there is an associated long exact sequence, called the Mayer-Vietoris sequence:

Putting it all together

Suppose is a discrete topological space. Then any continuous map from a -simplex to is by necessity constant at one point. So all the singular simplices in are constant maps, and thus a basis for any -chain group is given by just the points in as such. The differential acts on such a constant map by sending it to an alternating sum of identical maps the cardinality of which is one more than the dimension of the simplex. So the differential vanishes at every other point in the chain complex, and is an isomorphism on the rest. But then we have exactly two possibilities for a homology computation:

Case 1:

Here, the kernel is the entire chain group. But so is the image. Thus the homology group is trivial.

Case 2:

Here, the kernel is trivial, but then again, so is the image. Thus, again, so is the homology group.

The only part not really covered by this is the homology group , since this is computed from a diagram on the shape . So behaves like we showed above, and is the only non-trivial homology group of the discrete space.

But then, the only bit that remains non-zero in the Mayer-Vietoris sequence is

If we have a short exact sequence of free abelian groups, then this relates the number of basis elements in each of the groups with the structure of the sequence. Let our short exact sequence be given by . then, by injectivity of the map , the cardinality of a basis of the image in is equal to the cardinality of a basis for . By surjectivity of the map , and by Noether’s theorem, the cardinality of a basis of is equal to the cardinality of a basis of the kernel of this map, plus the cardinality of a basis of . By exactness in the middle, this kernel basis is equal to the image basis from A. So, we have, writing for the cardinality of a basis of the free abelian group , that . Thus, .

And if we apply this to the Mayer-Vietoris sequence fragment we acquired, it follows that

and thus

I’ve picked out the examples I have in order to have two spaces with the same Betti numbers, but with different cohomological ring structure.

Sphere with two handles

I choose a triangulation of the sphere with two handles given the boundary of a tetrahedron spanned by the nodes a,b,c,d and the edges be, ef, bf and cg, ch, gh spanning two triangles.

We get a cochain complex on the form

with the codifferential given as

and

Computing nullspaces and images, we get a one-dimensional , generated by , a two-dimensional with one basis given by and , and a one-dimensional with the linear dual of any of the four 2-cells as an acceptable basis choice, for instance . Throughout, given a simplex I write for the functional that takes and for all other simplices .

Now, the Encyclopedia of Topology, volume II has a paper by Viro and Fuchs on homology and cohomology. They state a direct construction of the cup product as defined by the following:
where the q:s are the degrees of the c:s. They also state that (at least if X is connected) any degree 0 coclass acts as identity under the cup product.

So, any product of something in with anything else is already clear. And any product of a degree 1 or 2 coclass with a degree 2 coclass will have to vanish – since there isn’t anything for it to possibly hit. Remains the three possible products of two coclasses both of degree 1.

Now, by the definition of the coclasses, we need something that is an equivalence class of linear duals of 2-cells to split into things that only operate on the two handles. However, these are geometrically disjoint – so any cell we could feed into such a product would vanish on the components. For instance,

So all the higher degree cohomology products have to vanish.

Torus

We pick a triangulation of the torus with 9 vertices, 27 one-cells and 18 2-cells, given as the identification space of a square, in a usual manner. It’s going to be obnoxious to write down boundary maps, and list all cells, so I’ll just refer you to the following picture instead:

Now, setting up the same computations to get the cohomology classes, we arrive at a one-dimensional class in degree zero, represented by the sum of all duals of all vertices, two classes in degree one, with representatives given, for instance, by and , and one class in degree two, represented by the dual of any 2-cell – for instance .

Now is where I find things get tricky. Again, we get the 0-degree class acting as, essentially, an identity. And the only products that could possibly be nontrivial now would be products of two classes of degree 1. So let’s call the classes and , and lets call the degree 2 class .
Using the Viro-Fuchs construction of the cup product, I should be able to say that if we consider and therefore (there is some reference to the codifferential matrices hidden in the signs here), and similarily, by considering the simplex idg (= dgi with a cyclic permutation) twice we would get .

This, somehow, feels fishy. Could anyone check my reasoning for me, please?

Edited to add: indeed this was fishy. The kind of blatant reordering I did to compute with dgi isn’t permissible. Also, we know that the square of a degree 1 coclass has to vanish, since the cohomology ring is graded commutative. However, we’re not far from the truth: If we want to compute , we should look at all pairs of coclasses in the expression for each, and look for pairs that “connect” to form a valid 2-cell. So, from u we can immediately discard bc, ci, di, fh, fi and gi, since nothing in v starts with c,h or i. Remains to check which of the cells ce and ef connect to anything at all. We get, in the end, the pairings
ce,eg
ce,eh
ef,fh
ef,fi
corresponding to the resulting potential 2-coclasses , , and . However, looking over our sketch again, we see that the cells ceg, ceh and efi are absent from our triangulation of the torus. Hence, the only potential 2-coclass representative surviving is , which represents the actual cup product, yielding

We’ll start by generating a dataset. Essentially, I’ll take the trefolium, sample points on the curve, and then perturb each point ever so slightly.

As a result, we get a dataset that looks like this:

So, let’s pick a sample from the dataset. What I’d really want to do now would be to do the witness complex construction, but I haven’t figured enough out about how R ticks to do quite that. So we’ll pick a sample and then build the 1-skeleton of the Rips-Vietoris complex using Euclidean distance between points. This means, we’ll draw a graph on the dataset with an edge between two sample points whenever they are within ε from each other.

So we pick a sample from this sample. Every 31 points might be good. (number arrived at by guessing wildly, and drawing the resulting images until they looked pretty enough)

We’d get, from this, the following result:

Now, we’ll want to build the corresponding skeleton. Let’s do it for a few different εs to demonstrate the difference.

The result is:

We notice that as the radius we observe grows, we connect all the loops, but by the time the loops are completely connected, there are also cross connections forming toward the middle. However, with any luck, these cross connections will be so short-lived, in terms of the radii we use, so that the homology classes we extract from the Rips-Vietoris complexes are noticably more persistent.

Doing the corresponding computation relies on me figuring enough out to use the Plex software suite, or writing my own, and thus will be subject of a much later blog post. This one was mainly “Look – I use R” and “Look – pretty pictures”. Enjoy.

Just got accepted for publication in the Journal of Homotopy and Related Structures.

Damn, this feels good!

The next couple of weeks will be spent doing homology of simplicial complexes, singular homology, equivalence of the two, neat things you can do with them; and then we’ll start moving towards a Borsuk-Ulam-y topological combinatorics direction.

I might end up pulling combinatorics papers from my old “gang” in Stockholm on graph complexes, and graph property complexes, and poke around those with them.

I find Weibel very readable – once the interest is there. It’s a good reference, and not as opaque as, for instance, the MacLane + Hilton-Stammbach couplet can be at points.

The interest, however, is something I blame my alma mater for. Once upon a time, Jan-Erik Roos went to Paris and studied with Grothendieck. When he got back, he got a professorship at Stockholm University without having finished his PhD. He promptly made sure that nowadays (when he’s an Emeritus stalking the halls) there is not a single algebraist at Stockholm University without some sort of intuition for homological algebra.

So, my MSc advisor, Jörgen Backelin, gave me a subject building on from things that he touched in his PhD thesis, since I was obviously interested in combinatorics. And as such, nothing fits me better than looking at Ext and Tor over monomial rings (corresponding to coordinate hyperplane varieties…)

The other Very Interesting teacher, Jan-Erik Björk, at that university held a course in homological algebra that I attended. It was very handwavy, but with enough of deep understanding underneath that some things just clicked into place.

The story goes on. All in all, out of my 5 years at Stockholm University, at least 3 was spent doing homological algebra in addition to whatever else I was doing, and they were spent in a tight clique of undergrads and early grad students that all shared a high interest in the subject matter. In my transcript, I have an imposing distribution of courses:

where one credit corresponds to one week of fulltime study, roughly. The +20 for homological stuff is for my thesis project, which was on homological algebra, but wasn’t a lecture course. I also have some 5 or 10 points of homological stuff I never got exams done for. So all in all, I spent about a year fulltime with only homological algebra (slightly more distributed in time), and a year fulltime with only algebra of sorts that were not explicitly homological in nature.

And as they say, practice does make perfect. I have, from the various lecture courses I took, an intuition for the category of chain complexes, and for the derived category. I have an understanding for the basics of model categories. I have studied Operads and PROPs with Sergei Merkulov, and seen what happens when you take the basic stance that “We want to equate a structure with its free resolution”, and then run for the hills with it.

The other week I was discussing my graduation plans with my advisor, and he asked what my Rigorosum was going to be about. I told him I expected to do it in homological algebra, and he answered that he wasn’t certain that there was anyone available who’d be capable to accurately test my knowledge of the field. He is a group cohomologist, and he outblazes me when it comes to intuition for that – and especially when it comes to the topological notions in the field.

I, however, am comfortable thinking about differential graded modules and dg-algebras, and doing homological algebra with these. And this seems to place me, possibly, as the single person in my state with a good working knowledge of modern homological algebra.

The things I talked about in the post this follows up on are current knowledge. None of it is particularily original, but the presentation is a result of my personal history. You might very well get similar presentations if you ask my course mates from Stockholm University – but then, that school has a very special athmosphere when it comes to homological algebra.

Our beloved Dr. Mathochist just gave me the task of taking care of any readers prematurely interested in it while telling us all just a tad too little for satisfaction about Khovanov homology.

And I received a letter from the Haskellite crowd – more specifically from alpheccar, who keeps on reading me writing about homological algebra, but doesn’t know where to begin with it, or why.

I have already a few times written about homological algebra, algebraic topology and what it is I do, on various levels of difficulty, but I guess – especially with the carnival dry-out I’ve been having – that it never hurts writing more about it, and even trying to get it so that the non-converts understand what’s so great about it.

So here goes.

Alpheccar writes that to his understanding, the idea is to build topological spaces out of algebraic gadgets, and then do topology on them. This is a part of the story, and certainly historically very important, but it is far from all of it.

Motivations

The revolution for homological algebra pretty much started with Eilenberg-MacLane – who wrote an articlebook that did the constructions necessary for the very topological versions of homological algebra – but without ever involving the actual topological spaces.

The point is that the way you do algebraic topology is that you tend to set up a functor Top → R-ChMod that assigns a chain complex of R-modules to each (nice enough) topological space, and then you add functors R-ChMod → R-Mod that extract informations from these. Typical examples are cellular chain complexes with coefficients somewhere nice for the first functor, and then homology or cohomology for the second functor – depending on what viewpoint is the most obvious.

The revolution was that we simply throw out that first functor.

In order to study (co)homology, we don’t really need to care that there was a topological space somewhere to begin with. We only, really, need a nice enough category of chain complexes (if it’s abelian, then that’s fine – we get the long exact sequences in homology and other niftiness easily then, but if it’s not, triangulated will do…) and we study certain types of functors from these to module categories.

Homological algebra as a tool for algebraic topology

Since, in the viewpoint introduced above, homological algebra is a part of the process used in algebraic topology, it turns out to be really neat to sit down and just prove a lot of neat results in homological algebra – with the background that at some later point, these might be useful once you sit down with the topology. I got hold of this particular point early – I had started my MSc thesis work in homological algebra before I took my first real topology course, and during that course, the less pointset topology and the more algebraic topology we did, the easier everything got. The fundamental results we needed to grasp to do algebraic topology in any amount of seriosity were basically just applications of all the cornerstone results in homological algebra, and thus perfectly obvious to my clique of arrogant undergrads.

This particular piece goes far. Why don’t we need to worry about whether we’re doing homology or cohomology? Answer: since Ext and Tor are dual in certain specific ways, which ends up meaning that although internal algebraic structures might be finicky, the module structure is very neat, and in k-Mod = Vect_k, we end up with no worries anywhere.

Testing grounds

The viewpoint of homological algebra as a tool for algebraic topology goes pretty deep. When I ask my advisor what to put in texts where I motivate why our field is important, in the standard answer he gives me the following pops up:

Group cohomology is important, since it is a field where topological methods can be tested reasonably safely, since we have the group theoretic attack vectors in addition to the purely topological.

On the other hand, group cohomology also turns out to be important, since we get important information for the study of groups out of the homological algebra side of things.

Low order Ext

The area where this is most notable is in representation theory. This field comes in several flavours: group representations, where we study kG-Mod for some (sometimes finite) group G; Lie algebra representations, where we study g-Mod for some Lie algebra g; quiver representations, where we study kQ-Mod for some (finite) quiver algebra kQ – and so on. One question that tends to crop up here, and with a high degree of importance for the non-homological algebraists around me – is what happens if we know only parts of our group? Can we say something about the entire group based on that?

It turns out that we can. There are very neat correspondences between the lower order Ext groups over kG and the behaviour of G itself. I’m going to stick to group representations here, since that’s the area I know best – however, this is something that pops up analogously all over the place.

Extensions

Suppose you have some R-module K that you know embeds, in some specific way, into some larger R-module M. And suppose you find the quotient L=M/K in some manner. What could, then, M be? One obvious answer is , but is this enough? This ends up depending on Ext¹_R(L,K), with each element of this particular Ext group indexing precisely one such extension.

This is at the core of Maschke’s theorem, by the way, which says that if the characteristic of the field k doesn’t divide the group order |G|, then by a specific construction, the only extensions possible for any kG-modules are the split extension – the one where we just take the direct sum.

This all leads to a wealth of useful information and ideas in representation theory. For instance, there is a way to describe modules proposed by Dave Benson and some co-authors, where you draw diagrams with each vertex being a simple module, occupying that spot in a composition series, and the edges being taken from the relevant Ext¹.

Invariants and coinvariants

Suppose you have a group acting on a vector space. This can be taken extremely physical – quantum mechanics is all about this kind of situation, or so I’ve heard. Then it might be interesting to figure out the invariant subspace: {a|ga=a for all g in G}. This is Ext⁰. Or we might want a basis for the complement: representatives for every way that things can move. This is the coinvariant vector space, defined as A/(ga-a), and this is just Tor⁰.

Simples, projectives and the stable module category

Simple modules are nice. They don’t have invariant subspaces. In the best of all worlds – which is when Maschke holds – simple modules are precisely the irreducible modules. However, when Maschke doesn’t hold, we can have non-trivial Ext¹, and thus we can build larger modules out of simples by a kind of gluing: they aren’t just a nice direct sum of simples, but something ickier.

Thus, unless Maschke holds, there will be weird things happening in the module category.

These weird things, though, are controllable. To be specific, we can consider the smallest possible irreducible modules. These will end up being building blocks, and for nice enough worlds, these will also end up corresponding closely to the simples – in the way that we can allocate a simple to an irreducible projective in a bijective manner.

So … what is this projective I keep throwing around? Take a free module. This is a direct sum of a finite number of copies of the ring R. This will have direct summands. By picking apart all summands into further direct summands, at some point we hit bottom: we cannot pick anything apart any longer. This is, by the theorem of Krull-Schmidt, a well-defined state of being. We can permute things, but in essence, a module is just its decomposition into irreducibles.

So, anything that is a direct sum of a free module is a projective. We can lose projectivity by taking quotients – so if we add relations, we may well get lost. But as long as we just look for direct summands, we’re pretty much home free. Now, the irreducible projectives have to be summands of the ring R itself, so they end up actually being (left) ideals in the ring. And each of them corresponds intimately to a simple module.

One trick that’s very beloved among the people who worry about these things is to get rid of anything projective, and look at the stable module category. In this, we just quotient away anything projective – morphism sets are taken modulo morphisms that factor through a projective… This way, we only have the “essential”, or as it is known to the experts of the field “difficult” information left. Then Extⁿ(M,N)=Hom(ΩⁿM,N), where Ωⁿ is the nth syzygy – see below for more on this.

So, homological algebra lets us understand the stable module category, which in turn lets us understand the parts that are essential to the module category structure.

Resolutions

I just promised you I’d tell you about syzygies. First off, some personal information – because readers always love that!

If you find me on IRC, on EFNet or on Freenode, you’ll find me under the nick Syzygy-. The – is there because there is someone who’s been using Syzygy for years and years on EFNet and because I’m not deliberately trying to be a bastard if I can help it. The rest of the nick is there to a certain extent because I like the way I write it in longhand.

And to a certain extent because it is an epitome of why homological algebra is interesting in my eyes.

Suppose we are interested in a finitely presented module, which we might be for many reasons, including being interested in algebraic geometry and in solving systems of polynomial equations. We might then just figure out what relations hold within a set of generators, which gives us a generating set, and some relation set.

These, relations, though are far from guaranteed to be the whole story. It’s probable that there are non-trivial relations between the relations. What do we do? We figure out what these are. They span the first syzygy module of the module we started with, denoted by ΩM. But this is unlikely to be free, so we can keep on going.

This way, we get a sequence of modules, all of which are free – since we just choose a generating set in each step – and with maps between them adding all the extra relations. But this is nothing other than a free resolution of the starting module. And here comes the candy that hooked me for this discipline: studying modules over their resolutions is the same thing as studying what chain complexes are, deep down, which in turn is the same thing as studying homological algebra.

Want to figure out what a module map means for the family of syzygies? What you really want is a chain map in the chain complex category. But some of these maps – or even portions of maps – will not carry relevant information. So we factor those away, and we get a slightly weirder category. But here, equality doesn’t quite mean what it should, so we add in more equality relations. And suddenly, we live in a derived category – and in here, the Hom sets are Ext groups, and the tensor products are Tor groups.

Number theory, geometry, and computation!

To continue this tour de force, consider the theorems in vector calculus relating various triple and double integrals. (note – I never dealt with this. I rode on technicalities to root out calculus from my curriculum so it would fit more algebra….) These theorems, in the end, only state that , which is at the core of what homological algebra is all about.

If we formalize this particular recognition a bit, and tug at the corners, we end up with de Rham cohomology, which deals with what you can do with differential forms on manifolds (layman speak: things you can integrate. The f(x)dx after the integral sign is a typical differential form) – and this is one of the many many places where cohomology rather than homology ends up being the “right” way to view things just because you started out as a geometer instead of .. well .. topologist or algebraist.

The same kind of thing happens in algebraic geometry as well. You start out happily with your varieties, you conclude that as soon as things get interesting, the nice and pretty concepts of coordinate rings don’t hold up, and you’re forced to go to coordinate sheaves. And then you try to figure out what you can do with sheaves of functions on a variety – and before you know it, you reconstructed sheaf cohomology. This, by the way, a quick look at wikipedia told me, lets you define euler characteristics for varieties in a way consistent with the classical uses of it.

I am no geometer, and I’m not the person to tell you about the intricacies of these things. If you understand them, though, I’d love to figure them out at some point.

The discussion of Khovanov homology is a slightly (though not very) similar thing to this. Again, I have no real idea, and am treading on thin ice here.

So, alpheccar. Is this what you asked for? Please tell me what more you want covered, and I’ll write up some more! This was fun writing!

On Monday, I’ll give a talk on my research into -structures in group cohomology. If you’re curious, I already put the slides up on the web.

I’ll try to blog from T’bilisi, but I don’t know what connectivity I’ll have at all.

I will straight off assume that continuity is something everyone’s comfortable with, and build on top of that.

Homotopies and homotopy equivalences

We say that two continuous maps, f,g:X→Y between topological spaces are homotopical, and write , if there is a continuous map such that H(x,0)=f(x) and H(x,1)=g(x). This captures the intuitive idea of step by step nudging one map into the other in formal terms.

Two spaces X,Y are homeomorphic if there are maps , such that and .

Two spaces X,Y are homotopy equivalent if there are maps , such that and .

Now, if f,g are maps X→Y and f=g, then , since we can just set H(x,t)=f(x)=g(x) for all t, and get a continuous map out of it. Thus homeomorphic spaces are homotopy equivalent, since the relevant maps are equal, and thus homotopic.

There are a couple of more properties for homotopic maps we’ll want. It respects composition – so if and h:Y→Z and e:W→X then and . This can be seen by considering h(H(x,t)) and H(e(x),t) respectively.

Denote by D² the unit disc in , and by {*} the subset {(0,0)} in . Then . In one direction, the relevant map is just the embedding, and in the other direction, it collapses all of D² onto {*}. One of the two relevant compositions is trivially equal the identity map, and in the other direction, the linear homotopy H(x,t)=tx will do well. Thus the disc and the one point space are homotopy equivalent.

The fundamental groupoid

Let X be a topological space (most probably with a number of neat properties – I will not list just what properties are needed though), and consider for each pair x,y of points in X, the set [x,y] of homotopy classes of paths from the point x to the point y. A path, here, is a continuous map [0,1]→X. We can compose classes – if and , then we can consider the map
. This is a path from x to z, and so belongs to a class in [x,z]. This class is well defined from the choices of γ, γ' since homotopies and composition of maps work well together.

This gives us a composition. It is associative - on homotopy classes. What happens if we look at maps instead of homotopy classes is part of the subject of my own research. It has an identity at each point x - the constant path γ(t)=x, and for each class in [x,y] there is a class in [y,x] such that their composition is homotopic to the constant path in [x,x].

Thus, we get a groupoid. This is called the fundamental groupoid, and denoted by . If we fix a point, and consider [x,x], then this is a group, called the fundamental group with basepoint x, and denoted by .

For , a linear homotopy will make any two paths in [x,y] homotopic – and so |[x,y]|=1 in for any x,y.

For S¹ – the circle – we can choose to view it as [0,1]/(0=1). Then we can consider the paths f_m(t)=a(1-t)+bt+nt. This is a path from a to b, and it winds n times around the circle. Each path in [a,b] is homotopic to a f_m, by a linear homotopy, which just rescales the speeds through various bits and pieces, and possibly straightens out when you double back. Thus, . Furthermore, if you compose f_mf_n, you’ll get f_n+m.

If we pick out the fundamental group out of this groupoid, we’ll get the well known fundamental group .

Now, suppose we have two homotopy equivalent spaces X and Y, with the homotopy equivalence given by f:X→Y and g:Y→X. Then consider the map f_*:[x,y]_X→[f(x),f(y)]_Y given by f_*γ(t)=f(γ(t)). I claim
1) f_* is bijective.
2) f_* works well with composition of classes.

For bijectivity we start with injectivity in one direction. Consider two paths in [x,y]. We need to show . If , then . However, then

which contradicts . Thus , and so also .

The proof is symmetric in the choice of direction, and so we can just repeat the same argument to get that g_* is also an injection. Thus we can conclude that f_* is in fact a bijection.

Now, for the second part, we consider and . We need to show that . But is the path that first runs through in half the time, then runs through in the rest of the time, and just transports this path point by point to Y. And transports point by point to Y and transports point by point to Y, and just runs through the first of these in half the time, then the rest in the rest of the time.

Thus, homotopy equivalent spaces have the same fundamental groupoid.

We had, in my first representation theory post, a mention of Maschke’s theorem. This states that if the characteristic of our field doesn’t divide the group order, then simple and irreducible mean the same thing.

Now, obviously, the actual proof you normally see first deals with a construction that works for when the characteristic doesn’t divide the group order – which uses 1/|G| at one point. So, what happens when this is impossible to work with? When the conditions of Maschke simply do not hold?

The very simplest answer is that then we can get modules that are glued together by simple modules with some meshing. Such that they aren’t direct sums any more. The ways we can glue together modules are through extensions – i.e. we can glue together A,C by forming a short exact sequence
0 → C → B → A → 0
and we’ll have that B is a module such that B/C=A. Now, the typical such module is the direct sum of A and C – and if Maschke holds, this is indeed all there is.

If Maschke does not hold, then the set of all such B forms an abelian group, called Ext¹(A,C), with addition formed in a slightly messy way called Bair summation. This is actually the same Ext as in all my cohomology posts. So the dimension 1 cohomology actually measures what kind of modules you can have at all.

Suppose the field k has characteristic p and the group order is pⁿ for some n. Then an order argument, and a few properties of modules will yield that there is only one simple module: the trivial module k. So the only extensions really interesting will be in Ext¹(k,k).

And at this point I can bring in my knowledge of group cohomology: if the characteristic is 2, and the group is C₄, then the cohomology ring was mentioned in my post on the higher multiplication structure of the cohomology rings of groups of order 4. It is – so each degree has dimension 1, so there are precisely two extensions of the trivial module over the cyclic 4-group. One is the direct sum, and the other is an irreducible, but not simple module.

Precisely this example, however, I feel I won’t quite explore to the potential it is worth right now, so I’ll pick an example where I know that I won’t be telling anything wrong.

Consider the symmetric group S₃. This has order 6, so the interesting cases are mod 2 and mod 3. For the 3-case, we have the trivial representation and the sign representation – since these always occur when they have a chance to. These are both 1-dimensional, simple, and irreducible.

At this point, we’ll start needing various theorems for the analysis. First off, we can count simple modules over the group algebra by counting conjugacy classes in the group where the order of the elements isn’t divided by the field characteristics. This way, we’ll know how many simples we have.

Furthermore, there is a theorem that states that regardless of Maschke, we certainly have a bijection between simples and irreducibles. I’m not going to go into how this is established, but if you’re interested, the relevant keyword is the radical of a module. Furthermore, if we decompose the algebra itself into a sum of irreducible modules, then each module occurs with a multiplicity corresponding to the dimension of the simple module it belongs to.

I’ve tried three times now to start writing out in painful detail precisely what’s going on here – but I can’t really, today, get a single example worked through from start to end. For S₃, though, we know the conjugacy classes: the identity, the transpositions and the 3-cycles. For each of the modular cases one of these disappears, and so for the 2-case, we get the trivial module and one 2-dimensional simple; and for the 3-case, we get the trivial module and the sign representation. These correspond to irreducibles, that make up the group algebra – in both cases two kinds of irreducibles, and for the 2-case, one occuring once, and one occuring twice, and for the 3-case both occuring just once.

Writing down the irreducibles just doesn’t seem to work out for me today. I blame the flu.

Edit: Fixed a typo pointed out by Dylan Thurston in the comments. Again, I blame the flu.

A success here would be very good news indeed, since for one it would indicate that such an adaptation should be possible, and for another it would possibly give me a way to lend strength both to the previous calculation and to a conjecture I have in the calculation of group cohomology with A_∞ means.

So, we start. We recover, from the previous post, the structure of the cohomology ring as k[x,y,z]/(xy), with x,y in degree 1, and z in degree 2. Furthermore, we have a higher operation, m₄, with m₄(x,y,x,y)=m₄(y,x,y,x)=z.

Thus, with the theorem, stating that the maps

are actually the duals of the maps

embedding the relators of the original algebra into the tensor algebra over the generators.

So, for our case, we have the maps




and we look to dualizing them. Considering for a while what this all means, and fixing notation, with a,b the generators, dual to x,y, we end up with the maps



from which we may recover a presentation of the group algebra as

As such, this is an unqualified success. We recover our original group algebra presentation from the A_∞-structure, and thus should be able to do similarily as a test for completion of future calculations as well. This, of course, needs to be proven before relied upon, but it lends credence to my hopes.

To do this, I fix the group algebra

and the cohomology ring

with ,

Furthermore, we pick a canonical nice resolution P, continuing the one I calculated previously. This has the i:th component Λⁱ⁺¹, and the differentials looking like

for differentials starting in odd degree, and

for differentials starting in even degree. The first few you can see on the previous calculation, or if you don’t want to bother, they are

Now, armed with this, we can get cracking. By lifting, we get canonical representating chain maps for x,y,z described, loosly, by the following:

x takes an element in , keeps the first, third, et.c. elements and throws out the even ordered elements; so
For an element in , the last element gets extra treatment, so

For the lowest degrees, we also have

This describes the image of our quasi-isomorphism .

For , we get the interleaved effect: every second element, but now with even indices, and some extra treatments at the end. So an element in will behave like

and for an element in we get

The last generator, z, is rather boring. The corresponding chain map only lifts elements to other degrees: shaving off the first two components of whatever it is applied to.

We define on products of the generators by simply composing the corresponding chain maps as long as the product is defined. The interesting stuff, from an A_∞ point of view occurs when the product vanishes, thus for x,y in the first line. A calculation shows us that xy is the only interesting element of the ideal , since , and we define and .

Thus, we’d be curious as to what happens with xy, and yx. Both products are zero in the cohomology ring; but the composition of the corresponding chain maps are not zero.

By calculation, we get given in the first few step by the matrices


and so on, with the lower right entry alternatingly ab and ba.

is the same thing, but with ab and ba interchanged.

We want and to be homotopies between the 0 chain maps, and these two respectively. By juggling the relevant matrices in Magma for a while, I conclude that has lower right entry a for all odd degree map components, and has the entry above that b, same components. All even components vanish.

Thus, for the first nonzero component, we have

for x,y and

for y,x.

Now, if we look at Φ₃, most of the possibilities vanish because of the way we defined f₂ for the non xy, yx cases. Thus, the only interesting entries remaining are xyx and yxy. These are, respectively,


and calculation of these expressions is a matter of composing the chain maps we have a tentative grasp of already. This gives us the case of xyx alternatingly between b in lower right corner for all even degrees and a just to the left of it for odd degrees, thus starting with the matrices


and for the case yxy, we have a similar pattern, but with a in the lower right for odd degrees and b above it for even degrees, giving the first two maps as

So, we can immediately conclude that our m₃ is going to be the zero element of Γ, since none of these chain maps are homotopic to any non-trivial coclass representatives. Thus we’ll need to find homotopies from these to the zero chain maps for our values of f₃.

A homotopy suitable for f₃(x,y,x) would be h with





It goes on, but I have not yet discerned a pattern clear enough to describe a generic element of this chain map. I probably should, at some point.

For f₃(y,x,y), we would need a homotopy for the other sequence, and calculations lead me to put down h with





and here we can discern a pattern to the matrices. They will have a sequence of 1 starting out at the third column, and going down skipping every second place.

Armed with these calculations, we may set out to calculate for our pleasure Φ₄. Due to the heuristic we use in defining the f_i for things “lifted” from the generators we’ve defined above, we shall discard anything except for xyxy and yxyx from study; all other cases will just give the relations we use in calculating f₂ or f₃ from the cases we’ve calculated.

Thus, we’ll be interested in

and

and thus we can, by using the already calculated matrices and a CAS (I use Magma right now) just calculate the matrices. In both these cases we get the same thing: the matrices



and so on; which are precisely the matrices we chose for our .

Thus, we’ll set m₄(x,y,x,y)=m₄(y,x,y,x)=z and f₄=0, and stop calculating right here.

We will find the A_∞-structures on the group cohomology ring by establishing an A_∞-quasi-isomorphism to the endomorphism dg-algebra of a resolution of the base field. We’ll write m_i for operations on the group cohomology, and μ_i for operations on the endomorphism dg-algebra. The endomorphism dg-algebra has μ₁=d and μ₂=composition of maps, and all higher operations vanishing, in all our cases.

Elementary abelian 2-group

Let’s start with the easy case. Following to a certain the notation used in Dag Madsen’s PhD thesis appendix (the Canonical Source of the A_∞-structures of cyclic group cohomology algebras), and the recipe given in A-infinity algebras in representation theory, we may start by stating what we know as we start:

our group algebra. We can resolve using a neat canonical resolution, derived from the tensor products of the resolutions of the field with modules over the cyclic 2-group. This gives us the resolution

where each differential is given by a matrix D with D_i,i=a and D_i+1,i=b.

Recall that . Thus is a dg-algebra, whose homology is precisely the group cohomology.

Now, by the minimality theorem (proven by Kadeishvili first, and reproven by a veritable host of mathematicians), there is a quasi-isomorphism of A_∞-algebras that lifts the identity in homology. This we can use to figure out the A_∞-structure for our cohomology ring: we know that the is an honest-to-glod dg-algebra, and thus has an A_∞-structure with all higher multiplications (by which I mean 3-ary and higher) vanishing. We can also pick representatives for our cohomology ring elements as representatives of homotopy classes of chain maps . This gives us a quasi-isomorphism of dg-algebras , lifting the identity, and which we can augment to an A_∞-quasi-isomorphism.

Which is what we’ll want to do now.

We’ll (at this stage) use the fact that we know what looks like: it’s the algebra . There are two 1-coclasses, both represented by a morphism , namely one composing the projection onto the first factor with the augmentation map, and one composing the projection onto the second factor. We’ll name the first of these x, and the second y, and note that they lift to chain maps that shave off the first and last summand of respectively in each degree.

So for our quasi-isomorphism , we now have sending each to the chain map shaving off the first i and last j components. We also know , namely 0, multiplication, differential and composition respectively. The first axiom we’ll investigate for A_∞-maps states that

But now m₁=0 and so this reduces to

so f₂ is a map such that its differential is equal to the “commutator” of f₁ and multiplication.

Now, pick some coclasses u,v. These will map to shaving maps as described above, and their product will map to a shaving map that does just the same as the composition of the individual shaving maps; so if u is xⁱy^j, and v is x^ky^l, then f₁(u) shaves off i components in the front and j components in the back, and f₁(v) shaves off k in the front and l in the back. So, the composition of these two maps is the map that drops i+k components from the front, and j+l components from the back. On the other hand f₁(uv)=f₁(x^i+ky^j+l) is the map that drops i+k components from the front, and j+l components from the back.

So they are the same. And thus we don’t need to bother with any homotopies, or any higher order operations or higher order maps. We set f² to be the zero map, and consider ourself finished and happy. The cohomology is a dg-algebra in its own right, and this is all there is to it in A_∞-terms. And we’re done.

This result implies, by the way, via a proposition from Keller, that the elementary abelian 2-groups have Koszul group algebras. An argument using restrictions of non-nilpotent coclasses to cyclic subgroups will tell you that these are the only finite groups that have Koszul group algebras.

With this example.

Cyclic 4-group

This is the one canonical example known beforehand in group cohomology. It was calculated by Dag Madsen in his PhD-thesis, and cited ever since. I will perform the same calculation, but in a blinding detail you won’t find in a thesis or a paper on the subject.

So, for starters, we find ourself with the group algebra

and the cohomology ring

quasi-isomorphic to the endomorphism dg-algebra of a neat resolution

where c in Γ represent the morphism m→cm.
Furthermore, as usual, the μ_i are all known, and m₁=0, and m₂ is the multiplication in Γ.

Once we’ve choosen representatives for the coclasses in , we can lift this choice to an A_∞-quasi-isomorphism. In this process, we’ll find and define the relevant higher multiplications for Γ, thus finding the A_∞-structure for that algebra.

Thus f₁ sends

and

and

Working through the coherence axioms, the first we encounter is the one defining . This is the one investigating the relationship between
and . So, we pick , and investigate the two expressions and . Writing it down in detail tells us that the only point where a difference occurs is if both u and v are odd, and this case we can figure out from the case where u=v=ξ since η only translates chainmaps higher up in degree.

Now, , so .
And we can read off the following diagram

and thus we can conclude that in degree 2. We shall be juggling maps a lot, so I will use the notation (x y z w)[i] for the map in that drops i degrees, and where the four last positions are multiplication by x, y, z and w respectively. So, with this notation, we have
f₁(η)=(1 1 1 1)[2]
f₁(ξ)=(1 x² 1 x²)[1]
f₁(ξ)f₁(ξ) = (x² x² x² x²)[2]

Composing the lowest degree component of the map (x² x² x² x²)[2] with the augmentation map, we see that in cohomology, it corresponds to the 0 element. So it is actually homotopic to the image of under f₁, and this particular homotopy is what we’ll want to be.

So we’ll want a homotopy h, defined by that
dh+hd = (x² x² x² x²)[2]
so we can immediately conclude that h needs to be of degree 1, since composition with d will add another degree step. So our h will look something like

and we can check what happens as we chase through the diagram. If we start at an odd-indexed position, we’ll get the two components
hd(o) = h(x o) = x h(o)
dh(o) = x³ h(o)
for o of odd degree. If we instead start in an even degree, we get
hd(e) = h(x³e) = x³h(e)
dh(e) = x h(e)
where e is an element of even degree. By close inspection in the diagram, we note that the expressions involving x³ all involve h applied on elements of odd degree, and so we’ll set those to vanish, and fill in the needed values by letting h(e)=x. Thus we get the chain map
h = (x 0 x 0) [1]
or in diagrammatic form

Thus , and f₂ on odd and odd elements are all translates of this, and all other parameters to f₂ give us a zero map. This brings us to a point where we can start investigating m₃.

We can extract f₁m₃ and m₁f₃ from the 3rd A_∞-morphism axiom, and put the rest into a map of its own. This will end up to be something supposed to be homotopic to the image of m₃, and so we can define m₃ and the homotopy once we have them.

The rest of the axiom is

(where I am using the fact that we’re in characteristic 2 extensively)
Applying this on elements x,y,z gives us a possibility to distinguish between cases.

If only one element is of odd degree, then every f₂ occurring will have at least one even argument, and so will vanish.

If two elements are of odd degree, we get the expressions
for x,y odd
for x,z odd
for y,z odd.
Each of these vanish if we take the behaviour of f₂ for higher odd coclasses into account: these are just translates of the behaviour defined in f₂(ξ,ξ), and so should vanish, since we defined f₂ on pairs of odd classes to just be translates of the value on ξ and ξ.

Remains the case with all three elements of odd degree. Again, higher odd elements behave by translating the behaviour of ξ, and so it is enough to study the behaviour on ξ, ξ, ξ.

In this case, we get

which is the sum of the maps given in the following two diagrams

and

By reading off the diagrams, we note that this sum is the chain map
(x³ x³ x³ x³)[2]
and we can further note that this corresponds to the cocycle 0 of degree 2 (since the augmentation composed with x³ vanishes), so we put m₃=0, to correspond to what this is homotopic to. And f₃(ξ,ξ,ξ) needs to be mapped precisely to this homotopy.

So, a homotopy h between 0 and (x³ x³ x³ x³)[2] is a map h with
dh+hd = (x³ x³ x³ x³)[2]
From the same considerations as above, we can conclude that h will have the two components
hd(o) = h(x o) = x h(o)
dh(o) = x³ h(o)
for o of odd degree. If we instead start in an even degree, we get
hd(e) = h(x³e) = x³h(e)
dh(e) = x h(e)
where e is an element of even degree.

Now, this time, since we want the x³ to occur, we can pick h to be 0 on even degree components and the identity on odd degree components, giving us
h = (0 1 0 1)[1]
and this being the image of f₃(ξ,ξ,ξ).

Going ever onwards, the next axiom we check is the one, that when we leave m₁f₄ and f₁m₄ out of the mix, we’ll get

and again, we can start looking at cases based on number of odd arguments.

For only one odd argument, all the f₂ and f₃ will vanish.

For two odd arguments, the same thing will happen.

For three odd arguments, we get the expressions
for x even
for y even
for z even
for w even
where the expressions vanish since all summands in each are the same translations of .

And for we get

Again, we calculate each summand by using a corresponding diagram, and get the three diagrams

for f₁f₃ and

for f₃f₁ and

for f₂f₂

Thus we can read off that f₂f₂ vanishes, and that the complete expression for is the one we’d write down as
(1 1 1 1)[2]
which we recognize as , so we set and .

And this concludes the calculation of the A_∞-structure on , and also gives a rather clear hint as to how to do it for in general.