 This lecture is part of an online commission of algebra course and will be about stably free modules So you remember that last lecture we had a long sequence of modules Which are minor variations of free modules and the ones of these that were closest to free modules at the stably free modules So a module M is stably free if M direct sum A finite number of copies of R to the M is free The key point here that we as we'll explain later is that M has to be finite So this is all over a ring R And the first thing we have to do is to give an example of a module that is stably free But not free. It's completely trivial that any free module is stably free because you can just take M equals 0 So here is a stably free module That is not free and The simplest one is the tangent space of a sphere S2 so we draw a sphere S2 and We take its tangent space, which means at every point we've got this little tangent plane of the sphere Well, there's a bit of a problem with that is it's not at all clear how the tangent space of a sphere is a module over a ring Well the way you turn it into a module over a ring is as follows We let the ring R be continuous functions on the sphere with values in the in the real numbers and We let the module M be the space of a continuous Vector fields So this is just a function from the sphere So the value at each point is just a tangent vector at that point So you can think of an element of M as just being some sort of vector field on the sphere. So you choose a Tangent vector field at each point of the sphere and this has to vary continuously And we notice that M is a module over the ring of continuous functions because it's a module under point wise multiplication If you've got a continuous function with some value at a point you just multiply the the tangent vector at that point by that continuous function So we've got to show first of all this module is Stably free and secondly, we've got to show that it's not free so let's first check that M is Stably free and M is stably free because M direct sum R is isomorphic to R cubed if we add one copy of the Free module over R we get to some of three copies of the free module over R And this is easy to see because M corresponds to the tangent Bundle Of S2 in other words for each point S2 We're assigning its tangent vector and R is going to correspond to the normal bundle so each point of S2 we assign a vector of the normal space at that point which is the Vectors and R3 or thumb or to the tangent space and you notice the normal bundle of Sphere S2 is is actually just isomorphic to the one-dimensional free bundle where you just take S2 times a copy of the reels as a vector bundle over it and this is because if you take a normal bundle over a sphere at each point, we're just Signing a vector in the normal space and you can just take a Point say at distance one from the sphere in some metric and that will give you a cross-section of the normal bundle showing its trivial on the other hand if you take the sum of the tangent bundle and the normal bundle of the sphere then to each point you're just assigning Something that's obviously isomorphic to three-dimensional Standard three-dimensional space you're working in so the sum of the tangent bundle and the normal bundle is canonically isomorphic to Sum of three copies of the reels So the tangent bundle plus the normal bundle is a free vector bundle So that's how you do it topologically. It's kind of obvious that M plus R is isomorphic to R cubed and we can also do this algebraically so So first of all we take the ring of all continuous functions on R Well, the ring of all continuous functions on R is a bit of a mess if you're doing it algebraically So let's just take the ring of all polynomial functions on on the sphere. So let's take all polynomials on Three-dimensional space and quotient out by the polynomials vanishing on the sphere So we get a ring which you can think of as being polynomials on S2 and Now we can define a module M so So we're going to take this to be a sub module of of some of three copies of R So R3 is going to be the set of all points a b c and now we don't want all points a b c a b c or an R. So this is a You can think of this as corresponding to choosing it For each point of the sphere you choose a Vector lying in R3, but we we only want the ones that are tangent to a given point of the sphere so We want a x plus by plus c z is equal to zero, which if you think about a bit corresponds to restricting to the tangent space So here M is obviously a sub module of R3 and you can write the decomposition of R3 as M plus R By mapping a point a b c To well, we've got to make it into a point with this property here. So we might as a minus x d B minus y d C minus Z d and Point D sets that so D is in R and this is in M and here D is just going to be the point A x plus by plus C z and you can easily check that this gives Shows that M plus R is isomorphic to R cubed So that shows that M is stably free and if we want to show that M is not free Well, this follows from the hairy sphere theorem So you remember the hairy sphere theorem says if you take a two-dimensional sphere and Have a nice vector field on it at each point then any any continuous vector fields On or on on on S2 vanishes somewhere at some point and This means that the tangent Bundle can't be trivial So the tangent bundle is Is not a free bundle because if it were a free bundle then we could take a Section of the free bundle that didn't vanish anywhere And so this implies that M is not a free module over over the ring R Actually, you can do this in other dimensions and there's nothing particularly special about a two-dimension Sphere, so let's take an n minus one dimensional sphere in n-dimensional Real vector space so as before we have module M over The ring R, which is either continuous functions or polynomial functions on the on the sphere I don't really really care and We see that M Direct sum with R is isomorphic to R to the n just as we did for n equals 3 so so This always gives a stably free module and we can ask when is M free and It's free When n equals 1 2 4 or 8 so this is over s 0 s 1 s 3 and s 7 It's fairly easy to see that it's free in these cases. So these three cases s 0 s 1 and s 3 are groups They correspond to the Elements of norm 1 inside the reels or the complex numbers or Hamiltonian quaternions And if you've got any group then it's tangent bundle is trivial because there's a canonical way To identify the tangent space at each point with the tangent space at the origin because you can just left Multiply by a group element. So the tangent space So the tangent space is trivial or rather free so S7 is a little bit tricky because that's not quite a group. It's a but it's got a a sort of non associative multiplication with inverses The key point is that the multiplication is a sort of inverse and this this sort of corresponds to the multiplication in the Octonians or Cayley numbers And it was an open question for a long time back whether when there were any other spheres For which the tangent bundle is trivial. This is this is the famous vector fields on a sphere problem Can you find n linearly independent tangent vector fields on an n dimensional sphere or guess n minus 1 dimensional sphere? And this was finally solved by JF Adams Which says no other N work and there seems to be no particularly easy proof of this Frank Adams is original proof of the very hard piece using higher co homology operations and things like this And there are now slightly simpler proofs using k theory, but there seems to be no Particularly easy proof of this. I mean you can give easy proofs for some values of n And the hard case is when n is a power of 2 So that actually seems to give an example of a an algebraic problem for which the only known solution involves Algebraic topology Another example when stable Stably free modules turn up is in sares conjecture So sares conjecture which was finally proved by quillen and Suslyn What's the following he said is any projective module over Over a polynomial ring over a field free I guess this should be a finitely generated projective module and sares motivation for this was that Affine space kind of corresponds to A real vector space and over over a real vector space One can show in algebraic topology that all finite dimensional vector bundles of trivial I'm now finally generation projective modules over rings kind of correspond to vector bundles as we will see a little bit later So the analogous Statement would be that any finitely projective module over Over the coordinate ring of affine space is free So what does this have to do with stably free modules? Well, it turns out it's easy to show It is Stably free And we can do this as follows So first of all There's a very famous theorem of Hilbert that we may or may not get to later on in the course Which shows that if you've got a projective module or from that matter any module over a polynomial ring Let's call this projective module P zero then it has a finite free resolution by free modules Up to Fn Now if this module happens to be projective Then we can split this map So there's a map from P naught to F naught splitting it so F naught can be written as P1 plus P naught mapping to P naught and Similarly the map from F1 to F zero Maps F1 on to P1 and this also splits so we can write F1 as P2 plus P1 And so on and it goes all the way up to here. So We get naught goes to Pn plus one and All the all the free modules hit all the free modules here split So what this means is that we now get P0 plus P1 plus P2 plus P3 plus P4 All the way up to Pn plus one is isomorphic to P0 plus P1 plus P2 plus P3 and so on and all these pairs are Free So we've got P0 plus a finite rank free module is a finite rank free module. So P0 is stably free however So we can get to stably free quite easily but getting from a stably free module to a free module is really very difficult It turns out to be really quite hard to tell the difference between stably free modules and free modules because most Natural invariance you can think of for modules turn out to be much the same for stably free modules and free modules There's a There's actually an elementary way of phrasing says conjecture. So So says conjecture to be reduced to showing that every stably free module over the polynomial ring kx1 up to xn is free and This is the same This is you can easily reduce to the case where you just add just one copy of M So we want to show that if M plus R is isomorphic to R to the N Then this implies M is isomorphic to R to the N minus 1 Sorry, that's I don't want this. That's the same as that one that you call that M and and If we've got an identification of M plus R with R to the M then the A generation of R can be identified with the vector in R to the M So we have a vector F1 up to FM with Fi in R given by this Identification and you can easily check that F1 up to FM generate The unit ideal in R and furthermore This space is isomorphic to R to the M Turns it's quite easy to check that this is equivalent to the fact that the vector F1 FM Can be extended to an invertible Matrix in M by M matrices over the ring R So this is an elementary version of says conjecture that says nothing about modules or anything else it says that if you've got Vector of M polynomials in N variables that generate the unit ideal then you can extend it to an invertible M by M matrix. It's very easy to check this for M equals 1 rather harder for other N So We can also investigate Stably free modules of small rank in particular rank zero or one. So let's look at stably free Modules of Rank zero And so what this means is that M Plus R to the N is isomorphic to R to the N for some N the rank is obviously going to be the difference between this number and this number which is zero in this case So first of all if we look at M plus R1 This isomorphic to R1 It's quite easy to show that M equals zero more generally if M plus N Is isomorphic to R with M and N sub modules in other words M and N ideals Then you can immediately see that M N is equal to zero Because it's a direct sum and so So M and N not free If M or N is None zero so if a sum of two ideals is Isomorphic to R and one of them one of the ideals is isomorphic to R the other one must be zero So this implies that if M plus R is isomorphic to R then M is isomorphic Well M is just zero Um more generally if M plus R to the N is isomorphic to R to the N Then what you can do is You can take nth exterior powers of both sides Lambda to the N so we recall that lambda to the N of a plus b nth exterior power is lambda to the N of a Plus lambda to the N minus one of a Tensant with lambda to the one of b and so on So if we take the nth exterior power of R to the N We just get R And if we take the nth exterior power of this We get an R plus Next one will be M tensor with R to the N and then other things we don't care about So we've now reduced the case. You've got a module plus R is isomorphic to R. So M tensor with R to the N is isomorphic to zero. So M is isomorphic to zero So so the only stably free module of rank zero is just zero Next we can look at stably free of rank one Unfortunately, this turns out to be somewhat easier than the case of rank zero So if M plus R to the N is isomorphic to R to the N plus one Then all we can do is we take N plus one exterior powers and We find here We get R because the N plus one thick stereo power of R to the N plus one is just zero and here We take the N plus one thick stereo power of this which is zero and Then we take the nth exterior power of this which is R Tensant with the first exterior power of M, which is just M and then we've got various other things Involving the Second exterior power of M tends with something Which we don't really care about Anyway, so so here we've got M plus something is Isomorphic to R And now We can add R to the N to both sides So we find something or other plus M plus R to the N is isomorphic to And R to the N plus one So So we know that these two are equal by assumption here Um, so By the case of stably free modules of rank zero we see all this stuff here must vanish so We find that this then implies that M is isomorphic to R because all this stuff here is zero and So Stably free modules of rank zero and one up either zero or a one-dimensional free module Stably free modules of rank two are considerably more complicated In fact, we saw at the beginning of this lecture that we could have a stably free module of rank two that wasn't free Okay, next lecture. I'll be saying a little bit more about stably free modules and in particular Introducing the island berg McLean swindle in order to study them a little bit more