 This lecture is part of a series of lectures on homological algebra. And this lecture will be the introductory lecture, and what we will be discussing is the front of Tor, which for two abelian groups A and B gives you another abelian group by Tor A and B. And I'll start with a sort of historical introduction about where these groups originally come from. So the earliest appearance of these was probably introduced by Czech, who was looking at the following problem. Suppose you've got some sort of manifold M. Then it's got homology groups, which are usually denoted by H, I of M with coefficients in Z. And more generally, you can define homology groups of M with coefficients in a group G. Well, we don't really need to worry too much about what these were. And Czech was looking at the following problem, which is express H, I of M with coefficients in G in terms of H, I of M with coefficients in Z. In other words, if you know the integer homology groups of a manifold, can you find the homology groups with coefficients in some other group? And Czech found an answer as follows. There's an exact sequence where you take the integer homology and tensor with G. And then here you get the homology of M with coefficients in G, which is what you're trying to calculate. And then here we get this mysterious group, Chor, H, I minus 1, M with coefficients in Z and G. So that goes to nought. And this actually splits in a non-chronical way. So this tells you what the homology of M with coefficients in G is in terms of the homology of M with coefficients in Z, provided you know what this rather mysterious group is. Actually, this is kind of fake history because Czech didn't write down an exact sequence like this because most of this notation was actually invented after Czech was working. So this is the way history ought to have developed, not the way to actually develop. So the second example of where these groups, Chor, turn up is, suppose you take an exact sequence, nought goes to A, goes to B, goes to C, goes to zero of a billion groups and you tensor it with some other group M. So we get A tensor M goes to B tensor M goes to C tensor M goes to zero and this is exact. And the problem is that this map here need not be injective. So tension with M doesn't preserve exactness. And I just quickly remind everybody the standard example of this is where you take nought goes to Z, goes to Z, goes to Z over 2Z, goes to zero where this is multiplication by two. And if we tensor with Z over 2Z, you get Z over 2Z, goes to Z over 2Z, goes to Z over 2Z, goes to zero and this is multiplication by two and is not injective. And the fact that this isn't injective is sort of major headache because it makes calculating everything rather a pain. And it turns out this Chor group kind of fixes this problem because we get the following long exact sequence. We get nought goes to Chor A M, goes to Chor B M, goes to Chor C M, goes to A tensor M, goes to B tensor M, goes to C tensor M, goes to zero and this is now an exact sequence. So this group Chor C with M kind of controls the fact that this map here is not necessarily injective. In particular, if this group here happens to be zero, then this map here is injective. And we can control the kernel of this map by this group here and we can control the kernel of this map by this group here. And finally, this map here is injective. Actually, it's only injective because we're working over the integers over more general, for modules over more general rings. This map here need not be injective and things get more interesting or more complicated or whatever. So that's two places these Chor groups turn up. They turn up in this universal coefficient theorem in algebraic topology and they turn up when you try and control lack of exactness of tensor products. So the next question is how do you define these groups? So let's give the definition of Chor A and B for Abellion groups. And what we do is we choose a free resolution of A. What this means is we choose an exact sequence, nought goes to Z to the M, goes to Z to the N, goes to A, goes to zero. Since we're working over the integers, the kernel of this map is also free, which makes life very easy. And then we tensor this with B and we get B to the M, goes to B to the N, goes to A, goes to zero. As we commented, this map need not be injective in general. So it's got a kernel and the kernel is Chor A, B. So this definition gives various problems. First of all, where does the definition come from? So what was the motivation for it? This seems a rather odd construction, if you first see it. The second problem is it seems to depend on the resolution of this group A. So we need to do something about that. The third problem is why is it called Chor? And fourth problem is how do we compute it? Because this definition looks rather cumbersome, but as we'll see fairly shortly, it's actually reasonably easy to compute. And so we'll spend the rest of this talk answering these four questions. So first of all, let's explain where the definition comes from. Well, the motivation for this definition comes from algebraic topology. What we do is we choose a manifold M and suppose we can triangulate it. So M is sort of covered with these simplexes. And what we can do is we can form a chain complex. C naught goes to C1, goes to C2 and so on, where C i has a basis of the i-dimensional simplexes. And this is a complex that's got a boundary map where you take each simplex to its boundary as an algebraic topology. And we can compute the homology of M to be the homology groups of this complex. For instance, more precisely, H i of M is equal to the kernel of the map from C i to C i minus 1 divided by the image of C i plus 1 to C i. More generally, we can compute the homology of M with coefficients in a group G. And to do this, we take the complex C2 tenths of G, goes to C1 tenths of G, goes to C naught tenths of G, and so on. And this is defined to be the homology of this complex. So it was a common operation in topology to take a chain complex of free modules, tensor it with a group, and then take the homology of the resulting of the result. And this is exactly what we were doing in the definition of tour. We were taking some sort of complex, tensioning it with a group B and then taking the homology of that. You see tour is really the homology of this complex of this rather small complex of modules B, M and B to the N. So the definition of tour was possibly motivated by the definition of homology of a complex. So the second question is, why does the definition of tour not depend on which chain complex we use? Well, again, the motivation for the proof of this comes from algebraic topology. So M can have two different triangulations. And the problem is, do we get the same homology groups independent of the triangulation? And this is actually a very tricky question. The slightly easier question you can ask is suppose we've got a map from M to N. In fact, suppose you've got two maps from M to N, F and G. We can ask, do these induce the same map from the homology groups of M to the homology groups of N? And the answer is they do provide an F and G a homotopic. On the level of chain complexes, we can talk about two chain complexes being homotopic, or at least two maps between chain complexes being homotopic as follows. So suppose we've got two maps, F here and G. So we've got two maps, F and G, from a chain complex C to a chain complex D. And these F and G induce the same map on homology if they are homotopic. So what does homotopic mean? Well, homotopic looks a bit odd. What it means is there's a map going from CI to DI plus one. Let's call this map D such that, let's not call it, sorry, not call it D, it's called S. So D is the map going like that, such that SD minus DS is equal to F minus G. And this is the algebraic analog of two maps being homotopic in algebraic topology. And it turns out that this idea of using homotopic maps can be used to show that if you've got two different resolutions of an Abellion group, then they give the same tour groups. So we're going to go into this in more detail in a later lecture. But the answer to the question is how do you know that the definition of tour doesn't depend on the resolution is that we're going to use this idea of two maps between complexes being homotopic that comes from algebraic topology. So the third question is why is this group called tour? What is the name tour comes from? The answer is that tour A and B or finitely generated Abellion groups depends only on the torsion subgroups of A and B. So the torsion subgroups just consist of the elements of finite order. We'll quite figure out why the elements of finite order are called torsion, but anyway. So fourth question we want to do is how do we compute these groups? So what we're going to do is we're going to compute tour A and B for A and B finitely generated Abellion groups. Well, first of all, we note that tour of A plus B, C is isomorphic to tour of A, C plus tour B, C and tour of A, B plus C is obviously going to satisfy the same form as tour A, B plus tour A and C. And the proof of this is easy and completely uninteresting to watch. So I'm going to skip it. It basically depends on the fact that the same formula is true if you replace tour by tensor products and is just an easy exercise. So we just need to compute tour A, B for A and B cyclic because any finitely generated Abellion group is a direct sum of cyclic groups so we can reduce to this case. So let's compute tour of Z with G. Well, what we do is we choose a resolution of Z, naught goes to Z, naught goes to naught, goes to Z, goes to Z, goes to naught. So here we've got a resolution of Z by two free Abellion groups. So this would be Z to the M and this would be the group Z to the N and it's particularly trivial to do this because Z is already a free Abellion group. And now we tense it with G so we get naught, tense and G goes to Z, tense and G and this is equal to zero and this is equal to G and tour of Z of G is the kernel of this. Well, it's pretty obvious what the kernel of this is, it's just zero. So tour of Z with coefficient of tour of Z and G is always zero for any finitely generated in fact any Abellion group G whatsoever. So that was easy. Now let's look at tour Z over NZ with coefficients in G. Now the free resolution of this is a little bit more interesting because we get naught goes to Z, goes to Z, goes to Z over NZ, goes to naught where this is of course multiplication by N. Now we tense this with G and we get Z, tense and G goes to Z, tense and G goes to Z over NZ, tense and G goes to zero. And now what we want is to work out the kernel of this. So this map here is multiplication by N and Z, tense and G is of course just G. So we've really got a map G goes to G, which is multiplication by N. And then here we have tour Z over NZ of G. Well, so this is just equal to the elements of order N in G. So now we can work out tour of Z over NZ for G cyclic. So tour Z over NZ and Z over MZ is going to be the elements of order N in Z over MZ. And if you think about it, you'll see this is isomorph to Z over MNZ, where this is the greatest common divisor of M and N, and tour of Z over NZ, Z with coefficients in Z is in particular just zero. So this is calculated tour for all finitely generated Abelian groups. You notice that tour of A, B for finitely generated Abelian groups depends only on the torsion of A and B as we mentioned earlier. You can also note that in the examples we've calculated tour A and B seems to be isomorphic to tour B and A. This turns out to be true in general. In fact, it's also true over arbitrary rings, although this takes a little bit of work to prove. Note that this isn't at all obvious because although A tends to be isomorphic to B tends to A, we've defined tour A symmetrically. We take a free resolution of A and tense it with B. So for this one, we take a free resolution of B and we tense it with A. So it's actually quite remarkable we get the same answer, whichever we do. We can also note that for A and B finite, tour A, B happens to be isomorphic to A tends to B. But this is a rather bad isomorphism and you shouldn't really use it. The point is there's no natural isomorphism from tour A and B to A tends to B in general. This is a natural isomorphism. There's only one sensible isomorphism between these two groups. Although you can find an isomorphism of this group, it depends on the choice of generators of A and B. And if you change generators, you'll get a different isomorphism between these two groups. So I'll just finish by mentioning possible further reading. The ultimate source of homological algebra is the original book on homological algebra by Carton and Island Boat. Although it's many decades old, it's still one of the best introductions to homological algebra written by the founders of homological algebra, who are two of the greatest mathematicians of last century. So anyone interested in homological algebra should just get a copy of this book and try reading it. Okay, next lecture we will be doing more about the tensor product of Abellion groups. In particular, we will try and prove that the definition of tensor product doesn't actually depend on the choice of resolution.