 This lecture is part of an online course on commutative algebra and we'll be about Hilbert's theorem on finite generation of algebras of invariance. So what we're going to do is we're going to suppose g is a finite group acting on a finite dimensional vector space over the field which we'll call k of characteristic 0. Then we will show that the algebra of invariance of g is finitely generated and as we said there were several different meanings of finitely generated. So this means as a k algebra and we just recall the invariance are just the polynomials on k fixed by the action of g. So let's look at the proof of this. First of all, suppose the vector space is dimension n then we've got the algebra of all polynomials is just a polynomial ring kx1 up to xn. n is just the dimension of our vector space v and the key point of this is that this is graded. What this means is we can give each xi degree one and then every monomial has some degree so x1 to the n1, x2 to the n2 and so on has degree n1 plus n2 and so on. So let's call this ring r. We can write r as a direct sum r0 plus r1 plus r2 and so on where r i are the things of degree i. So r0 is just k and r1 is spanned by x1 up to xn and graded rings are generally easy to work with because you can often prove things by induction. You start at degree 0 and work your way up by induction and this is what we will do in order to prove Hilbert's theorem. So let's write i for the sub-algebra of invariants and i is also graded so we can write i is equal to i0 plus i1 plus i2 and so on and then we're going to put j to be the ideal of r generated by i1, i2 and so on and you should notice here that there is no i0. In fact i0 contains the element 1 so the ideal generated by i0 would just be the whole polynomial ring. So we're only taking these elements in order to define the ideal j and then j can then be written as j1 plus j2 plus j3 and so on so it's graded as well and of course there's no j0 or rather j0 is equal to zero. So now we notice that j is finitely generated ideal by the Hilbert's basis theorem and suppose it's generated by elements say a1, a2 up to ak and we can take the ais to be homogenous. So we want to show that a1 and so on generate the algebra i and now you need to distinguish very carefully between generating something as an algebra and generating something as an ideal so they generate the module or ideal i sorry not j. So that means every element in j is a linear combination of the ai with coefficients in r whereas generating i as an algebra means everything in i is a polynomial in the a's with coefficients in k so here a is an algebra over k and this is an ideal of r and in general there's no reason why if you've got a subring i and why the generators of j should be generators of i so let's look at the following example which we've actually had before so let's write out the basis for the elements as usual so we've got yy squared y cubed xy xx squared y so what I'm going to do now is I'm going to take j, no let's define i first so i is going to be the sub algebra generated by these things here and now you can see the ideal j is generated by all the things of degree greater than one so this is the ideal j and the ideal j is generated by x but does not generate the algebra i in fact the algebra i of course is not even finitely generated as we saw earlier so for general rings we can't prove that the ring is finitely generated as an algebra by fiddling around with some finitely generated ideal however we're not working with an arbitrary subring of polynomials we're working with a ring of invariants so we need to know what extra property does the ring of invariants have so and the answer is it has a Reynolds operator which is indicated by row this is a Greek letter row for Reynolds not a not a not a letter p and the Reynolds operator is very easy Reynolds operator applied to a polynomial is the average of f under group g and we can define the average to be sum over all elements of g of g applied to f and then we have to divide by the order of g and now this is where we are using the two assumptions we made in order to sum over g we need to use the fact that g is finite and in order to divide by the order of g we need k to be of characteristic zero or at least the characteristic of k does not divide the order of g so this proof would also work for a if the field had characteristic p greater than nor provided p didn't divide the order of g so let's see what properties this Reynolds operator has so properties of row first of all row of one is equal to one kind of completely trivial next row of f g is not equal to row of f row of g in general row is not a homomorphism of rings i mean there's no reason why it should be you can check this easily for g being of order two however row of f g is equal to f times row of g if f is fixed by the group g again that's almost trivial from this i see i've been using g for a polynomial as well as a group element but i guess you'll just have to cope with that and from this it follows that row of f is equal to f if f is fixed by g and these are the two properties of the Reynolds operator we're going to use you can rephrase them as follows so we've got a map from the ring i to the ring r so i is a sub ring of r so we have this exact sequence and you can think of all of these as being i modules and the Reynolds operator kind of goes back the other way so row is a homomorphism of i modules from r to i and furthermore it splits this exact sequence so we can write r is equal to i plus the kernel of row again the where we're considering both sides as i modules so it's the fact that this sequence splits which is that is the key to proving Hilbert's theorem well now that we've done introduced the Reynolds operator and we've got Hilbert's basis theorem we can now finish off the proof of Hilbert's theorem very easily so we're now proved by induction on the degree that a1 a2 generates i as a k algebra so we pick f in i with degree of f is greater than nought and we're going to use induction on the degree of f and we can put f is equal to a1 c1 plus a2 c2 and so on where ci is in r because f is in the ideal j and the ai generates the ideal j well there's no particular reason why the ci's should be invariant yet but what we do is we now apply row so we get f equals row of f because f is fixed by i sorry it's fixed by g and this is equal to row of a1 c1 plus row of a2 c2 and so on which is equal to a1 row of c1 plus a2 row of c2 and so on because all the ai are fixed by g and now all the elements row of ci are in i because if you make anything invariant under g then it's obviously fixed by g so the degree of ci is less than the degree of f because the degree of ai is greater than zero so by induction ci is in the algebra generated by a1 a2 and so on well now all the ci's are in the algebra generation by the a's and the sorry the row of ci's are in the algebra generated by the a's so this expression here is in i so we've shown that f sorry not in i in the algebra generated by a1 but by the a's so we've shown by induction that anything in i is in the algebra generated by the generators ai so this completes the proof of Hilbert's theorem that the algebra of invariance of a finite group in characteristic zero is finitely generated i'll say a little bit more about the the background for this first of all you can ask the question who was Reynolds and there was if you try looking for some guy Reynolds working in invariant theory you won't find anybody that's because Reynolds didn't work in invariant theory he actually worked in fluid dynamics of all things you may have heard of the Reynolds number of a fluid's flow that's the same Reynolds he invented Reynolds number as well as Reynolds operator his Reynolds operator was originally the following suppose you take a fluid flow varying in time well that's incredibly complicated so what Reynolds did was he said well let's replace the fluid flow at each point by its average over time so he was just looking at the average flow of a fluid over time and you see what's happening is you've got a group of time translations and what you're really doing is replacing the fluid flow by the average fluid flow over over time so so what Reynolds was doing was he was taking the average over the group of time translations and the group that the term Reynolds operator has been also used to mean taking the average over any other group well Hilbert didn't actually just do the case of finite groups he also proved it for groups like SLN over the complex numbers so I just want to say how you can extend Hilbert's result to this more general case first of all the proof above works equally well for compact groups and the point about compact groups is we can integrate over them well you can try integrating over non-compact groups like the reels but there's actually no guarantee your integrals converge in fact Reynolds operation of integrating over over time isn't really it's not entirely clear that it's well defined and whether or not you can take the average of a bounded continuous function over the reels is a kind of tricky question and for groups like SLN or C taking the average of a bounded function is an extremely dubious operation anyway for compact groups there's no problem because you can just integrate continuous functions over them so we can define row of f to be the integral over all g of g of f and here we divide by the this will be the volume of g and the point is that compact groups of finite volume non-compact groups you even if you do integrate over them the volume is infinite so if you divide by the volume you just get zero um so well that's fine but the problem is SLN of C is not compact so that doesn't work directly but now we can use a rather cunning idea known as vials unitarian trick so the point is the group SUN the special unitary group is compact so Hilbert's theorem applies to the special unitary group and now we look at the following picture we've got the special unitary group and this is contained in the special linear group over the complex numbers which also contains the special linear group over the reels and these groups all have le algebra so this has the special unitary group of le algebras and this is contained in the le algebra SLN of C which are just matrices of trace equal zero and this contains the le algebra SLN of r which again is just matrices of trace zero and if you go to a course on le groups you know that finite dimensional representations of a special unitary group are very similar to finite dimensional representations of the le algebra so here we have the representations similar in finite dimensions and again here the representations of these are more or less the same as the representations of these in finite dimensions um that's generally true representation finite dimensional representations of the le algebra correspond to finite dimensional representations of the le group if the the Lie group is simply connected. Now, actually the special linear group over the reals isn't simply connected, but it's close enough that finite dimensional representations of these two basically correspond. And when we're doing representation theory, we're sort of fitting around with finite dimensional representations of these groups here. And now, although the Lie algebras aren't the same, they're actually very similar because SLN of C is isomorphic to SLN of R tensored with the complex numbers. And it's also isomorphic to SUN of R tensored with the complex numbers. So although these two real Lie algebras are not the same, they do become the same if you tensed them over with the complex numbers. And what this means is that the finite dimensional representations, in fact, the complex representations of these two algebras are the same because their complexifications happen to be isomorphic. So we see that finite dimensional representations of this group are really the same as finite dimensional representations of this group because we can just go from there and then jump across to there and then go up to there. Now, this group has a Reynolds operator and what this means is that finite dimensional representations of this group do in fact have a Reynolds operator. So we can take the Reynolds operator of this group here and just transfer it to then and transfer it to then, transfer it to then. In fact, this group does have a Reynolds operator for finite dimensional representations. And so we can Hilbert's theorem goes through. I should have a warning that for infinite dimensional representations, this thing breaks down completely and the infinite dimensional representations of these two groups really have very little to do with each other. The reason why it breaks down is that to go from groups to the algebras, you need to use an exponential map. You can just take the exponential series applied to a matrix and that works fine. But the trouble is if you try the exponential map for an infinite dimensional linear transformation, it just doesn't converge in general. And there is still a relation between representations of the lead group and the lead algebra, but it's much, much more subtle. And in particular, for example, if you look at unitary representations of these groups, all irreducible unitary representations of this finite, whereas almost all irreducible unitary representations of this group are infinite. So in infinite dimensions, these two groups have very little to do with each other. So this proof we've given, as we've said, works for fields of characteristic zero. Next lecture, we will discuss what happens for fields of characteristic greater than zero. And Emmy Nerter managed to show that quite a lot of Hilbert's proof would go through for fields of characteristic greater than zero, only we need to use more commutative algebra because we can't make use of the Reynolds operator.