 This lecture will be a continuation of the previous lecture on the Bernstein-Sato polynomial. So what we'll be doing in this lecture and the next lecture is showing how to prove the existence of this polynomial. In fact, in the rest of this lecture we will hardly mention the Bernstein-Sato polynomial at all. Instead, we'll be proving an inequality called Bernstein's inequality. And this is an inequality involving modules over the vial algebra. So I better explain what the vial algebra is. Well, the vial algebra A is just the ring of polynomials over the complex numbers in variables x1 up to xn. So we've got a polynomial ring, but then we also introduce the partial derivatives of x1 up to xn. So I'm writing this as an abbreviation for differentiation with respect to x1 and differentiation with respect to xn. So this can be thought of as the ring of differential operators with polynomial coefficients. And you've got to be a little bit careful here because this is non-commutative. It's pretty close to being commutative because xi xj is equal to xj xi. And partial differentiation commutes, at least if you're applying it to polynomials, so these all hold. And delta i xj is equal to xj delta i if i is not equal to j. However, delta i xi is equal to xi delta i plus 1. This is the Leibniz rule. So it's pretty close to being a commutative algebra. The only thing that stops it being a commutative algebra is this term 1 here. So it's pretty close to being a commutative ring. And we can in fact apply a lot of the techniques of commutative algebra to it. We can also think of modules. So if M is a module over the vial algebra, I suppose it's the vial algebra modulo sum ideal. And since it's non-commutative, you need to worry about whether it's a left ideal or a right ideal or whatever. But suppose you've got a cyclic module, we can think of this as being a system of differential equations because each element of this ideal is going to be a differential equation. So an ideal can be thought of as a collection of differential operators. And in particular, a homomorphism of modules from a over i to some ring of functions, say smooth functions, so the ring of smooth functions is of course a module over differential operators. So this can be thought of as solutions to the differential equations in i. And if you think about it, a homomorphism from a to this, you just look at the image of one, and it's going to be a smooth function which is a solution to all the differential operators in i. So the vial algebra can be sort of used to turn the problem of solving differential equations into a problem of linear algebra over this non-commutative ring. Of course this doesn't make the theory of differential equations trivial because this is actually quite a complicated ring and working out this is of course no easier than solving the differential equation. But it gives you some new ideas for handling differential equations. So the first thing we want to do is to work out what is the centre of A? And the answer is the centre of A is just the complex number. So you remember A was equal to the sort of non-commutative polynomial ring over these so it certainly contains C as a as a subring and the main point here is that it doesn't actually contain anything else. And the proof of this is quite easy. What we do is if we've got a differential operator, we can map it to xd minus dx. And this is a C linear map from A to A and it's actually really easier to work out what it is in an obvious basis of A where the basis consists of a monomial in x times a monomial in these differential operators. So it should be x1. And you can work out the kernel of all the maps taking d to xi, d minus dx i for all i. And the kernel of all is just the polynomials in x1 up to xn. And similarly the kernel of d goes to delta i, d minus d delta i for all i is just polynomials in delta 1 up to delta n. Notice there's actually a sort of symmetry between the partial derivatives in the x. In other words, there's actually an automorphism of this vial algebra which just switches the xi's with the di's up to sine or something. Anyway, we see from this that anything commuting with all the xi and delta i must be a polynomial in the xi and also a polynomial in the di so it must be in C. It should have a slight warning here. When you work out these linear maps and work out their kernels, we are actually using the fact that C has characteristic zero. So it should warn you in characteristic P greater than zero that the center of the vial algebra is no longer trivial. So the center contains, for example, delta i to the P because you actually find that x delta i to the P is equal to delta. It actually commutes with x because the commutator turns out to be P times something. And if you work in characteristic P, that's actually zero. So the center is actually bigger in characteristic P than in characteristic naught. In some sense the reason for this is that in characteristic P greater than naught there are other differential operators. So in characteristic zero this is all differential operators with polynomial coefficients. In characteristic greater than naught there are other differential operators. You can think of these informally as being something like d to the P over P factorial. And you have to be a bit careful here because we're in characteristic P so we can't actually define division by P factorial but we can sort of do this over the integers and then reduce it mod P so we get some extra differential operators. So anyway in characteristic P there are additional complications but we're not going to worry about these. We're just going to work over the complex numbers. So we found the center of the vial algebra. Now what we want to do is to convert the vial algebra into a commutative algebra. So we have a Bernstein filtration on A. So we have a naught contained in a1 contained in a2 and a naught is going to be the complex numbers. And a i is spanned by all monomials in x1 to xn delta 1 to delta n of degree less than or equal to i. So you're allowed to multiply most i of these together and then take the span of those. And remember these are non commutatives so you have to worry a little bit about the order but we allow all monomers in any order. And then it's pretty obvious that a i times a j is contained in a i plus j. The other thing you notice is that if P is in a i and q is in a j then pq minus qp is actually in a to the i plus j minus 1. So that sort of follows because it's true for monomials of degree 1. You remember xi times delta j is delta j times xi plus something in a0. And this property sort of propagates all the way through. What this means is that p and q, while they don't commute, they sort of commute a little bit in some sense. That pq times qp are both in a i plus j but the difference is actually in a i plus j minus 1. So what we can do is we can look at the following graded ring. It's a naught plus a1 over a naught plus a2 over a1 and so on. And this is a commutative ring. And you notice it's sort of about the same size as the vial algebra in some sense. And in fact it's a polynomial ring in the images of the xi and delta i. You remember the xi and delta i, all of images in a1 modulo a0. And you can easily check this is just a polynomial ring in these elements here. So the vial algebra isn't too far away from being a polynomial ring. We can turn it into a polynomial ring by doing this filtration and grading trick. Now we want to do the same thing for modules. So I suppose m is a module over a generated by a complex vector space m0. Then we put mi is just a i m0. So we have a i mj is contained in mi plus j. So we've got a filtration on m. And if we take m0 plus m1 over m0 plus m2 over m1 plus and so on. This is a module over the ring r which is a naught plus a1 over a0 and so on. And in some sense this module here is the same size as m. Because we're sort of quoted out by m1 there and we've got an m1 there and so on. So in some vague idea, in some vague sense you can think of this being the same size as m. Well now we can apply the theory of Hilbert polynomials to m. And let's assume that m0, the dimension of m0 is less than infinity. Here we're taking the dimension over the complex numbers. So it's finitely generated. So it's a finitely generated module over a notarian ring of polynomials. So the dimension over the complex numbers of mi is a polynomial in i for i sufficiently large. Because this is what the theory of Hilbert polynomials tells us. And we define the dimension of m to be the degree of the dimension over c of mi. Notice by the way there are two completely different notions of dimension here. Here we've got the sort of ring theoretic dimension. So you remember in ring theory we define the dimension of a ring or a module. And it doesn't really have all that much to do with this dimension which is the dimension of a vector space. So don't confuse these two notions of dimension. They're quite different. Well and we can define the multiplicity of m to be the dimension of m factorial times the leading coefficient. And you remember if we take the Hilbert polynomial and multiply its degree factorial times the leading coefficient this is an integer. Well you might think this is going to depend on m0. Well it doesn't. The dimension of m and the multiplicity of m are independent of m0. It's because changing the basis of m0 just changes the Hilbert polynomial by at most a finite shift. And in other words if you add an extra element to m0 the Hilbert polynomial of this new m0 and of the old one each is bounded by the other shifted by a constant. So they have the same degree and leading coefficient. The lower coefficients of the polynomial do indeed depend on the choice of m0. So they're more complicated to deal with and we usually just ignore them. It's the leading coefficient and the degree that are the most useful bits of a Hilbert polynomial because they tend to be much more stable under changing things. So in the commutative case where m is a module over the polynomial ring in 2n variables dimension of m can be any integer from 0 to 2n. For instance if m has finite length it's dimension 0. If it's the whole polynomial ring it's just dimension 2n. And if you take this ring and quotient up by some of these x i's you get a module with some intermediate dimension. So you might guess that the same thing happens over the vial algebra but there's a dramatic difference. Here we get Bernstein's inequality. It says that if m is not equal to 0 then the dimension of m is at least n. It's trivial to show the dimension of m is less than or equal to 2n. So what has happened is that dimensions from 0 to n-1 are not possible in the non commutative case. So what we're going to do in the rest of this lecture is just prove Bernstein's inequality. So let's have a proof. What we do is the key point is to show the map from a to hom over c from m i to m 2i is injective. So suppose that a times m i is equal to 0 where a is in a i. We want to show that a is equal to 0. Just change pens this one seems to be a bit temperamental. We'll do by induction on i. So i less than 0 is trivial as a i is equal to 0. Assume true for i-1. So we've got a is an a i and a times m i equals 0. Then we have a times delta j, the commutative of a with delta j is an a i minus 1. And a m i is equal to 0. So a delta j m i minus 1 is equal to 0. So a delta j is equal to 0 by induction. Similarly, a xj equals 0. I've sort of mentioned that the x's and the deltas or dels or whatever you call this letter are sort of symmetric. So anything you can prove for one you can prove from the other. So a is in the center of the vial algebra. And do you remember we worked out the center of the vial algebra earlier? It was just the complex numbers. So a is a complex number. And now we're just about done. So a is in c and a m i and a m equals 0. So a m is equal to 0. And as m is not equal to 0 this is where we use the fact the module m is non-trivial. So we find that a must actually be 0. So this shows that a i is contained in home of m i to m 2 i. And now we can finish easily because the dimension over c is a polynomial of degree 2n. And the dimension of these are both polies of degree given by the dimension of m. That's the dimension over c and this is the ring theoretic dimension. So this side here is a polynomial of degree at most 2 times the dimension of m. And if you compare these two we see the dimension of m must be at least 2n over 2 which is equal to n which is Bernstein's inequality. So the minimum possible dimension of a module over the vial algebra is therefore n rather than 0 as you might guess by analogy with the commutative case. If the dimension of m is equal to n or m is equal to 0, m is called a holonomic module. This is related to the notion of a holonomic system of differential equations in case you're wondering where the term comes from. Okay, so next lecture we're going to show how to use Bernstein's inequality to study holonomic modules and use properties of holonomic modules to prove the existence of the Bernstein polynomial.