 This lecture is part of a graduate course on Lie groups, and we'll be about the Poincaré-Berkov-Witt theorem. So the Poincaré-Berkov-Witt theorem tells you about the structure of the universal enveloping algebra UL of a Lie algebra L. So I'd better stop by saying what the universal enveloping algebra is. This is informally the associative algebra generated by the Lie algebra L. In other words, what this means is there's a map from L to U of L, and if this takes, I'm going to denote the image of an element A by A in U of L by a sort of abusive notation. So if these are the images of the elements A and B, then the commutator of A and B and L must be mapped to A, B minus B, A in the universal enveloping algebra. So this is a Lie algebra, and this is an associative algebra, and we can think of this as being in some sense the universal associative algebra generated by L. So it's generated by the vector space L subject to these relations. For example, I suppose L is a bilion. So in other words, A, B equals 0 for all A and B. And I suppose L has a basis, A1 up to A, N. Then the universal enveloping algebra is just a polynomial algebra on these elements A1 up to A, N, because they just all commute and the universal algebra generated by commuting elements is just a polynomial algebra. Another example, I suppose L is spanned by A and B with the relation A, B equals B. And the universal enveloping algebra is generated by elements A and B, subject to the relation A, B minus B, A equals B. Another example is suppose L is the Lie algebra of a group, Lie group G. So in other words, it's the left invariant first-order differential operators on G. And in my task, so L accounts for the first-order differential operators, what about higher-order differential operators? Well, it turns out that the universal enveloping algebra of L is just the left invariant differential operators. So we will see this later during the course of the Poincare-Berkel-Fitt theorem for Lie groups. And the basic question is, it's kind of a bit difficult to visualize U of L, so we can ask how big is the universal enveloping algebra of L? And what we're going to do is we're going to find upper and lower bounds for the size of U of L. It's not quite clear what we mean by the size yet because U of L is just an infinite-dimensional vector space. So what we want to do is to show that in some sense U of L is the same size as a polynomial algebra on n generators where n is the dimension of L. And what we need to do is to explain what I mean by saying it's the same size. So let's start by finding an upper bound for the size of U of L. Well, this is quite easy. What we do is we pick a basis A1 up to An of L. And now we can note that U of L is certainly spanned by elements A1, A2, up to Aik for all sequences I1, up to Ik of integers. However, we can do better than this. This is really too big. So we want to cut this down a bit. And what we notice is that Aij minus AjAi is equal to AjAj, which is in L. So we can swap the orders of all the AIs at the cost of having something of lower degree. So we can swap Aij at the following cost. We add terms of lower degree. Well, I haven't. I mean, the degree doesn't actually make sense, but it's obvious what it's meant. So Ui is actually spanned by elements A1 up to Aik such that I1 is less than or equal to I2 is less than or equal to I3 and so on. So we can assume that all these subscripts are in increasing order. And now we can see that this is really the same size as a polynomial algebra. And to make this more precise, we're going to filter the universally enveloping algebra of L. This means we choose subspaces F0 contained in F1 contained in F2 and so on. And Fi is going to be spanned by all, let's call that a k, by all A i1, A i2 with at most k a's in them. And now we notice that and we can form a graded algebra F0 plus F1 over F0 plus F2 over F1 and so on. And you can see that this is in some sense the same size as the universal enveloping algebra. And this is a graded algebra because Fi, Fj is contained in Fi plus j. But we can do better than that because Fi, Fj commute up to terms in Fi plus j minus 1. So in fact, this is a commutative algebra. And furthermore, it's generated by L or rather the image of L in F1 over F0. So we've actually got a map from the symmetric algebra of L onto this algebra here. And this is just a polynomial algebra. So this gives a map from a polynomial algebra onto a graded algebra that's in some sense the same size as the universal enveloping algebra. Now the Poincare-Berkov-Vitt theorem says that this is an isomorphism. When I say this, I mean this map here. Well, we've sort of proved the easy half of the Poincare-Berkov-Vitt theorem, which is to find an upper bound for this algebra here that we've shown there's a map from the polynomial algebra onto it. What we've got to do is to show that it's defined a lower bound for this to show it's the same size as this. I mean for all we know this algebra could just collapse completely and maybe all these spaces here are equal to zero. For example, we can do this construction for any bilinear bracket whatsoever on an algebra. And we will still be able to get this. And for most, if we choose any old bracket, there's no reason why this should be an isomorphism. The proof that this is an isomorphism needs to use the fact that the bracket on a Lie algebra is anti-symmetric and satisfies the Jacobi identity. And if it doesn't, then this map is generally not an isomorphism and does collapse. So let's have a look at some easy cases. So here we're finding a lower bound for the universal enveloping algebra of L. Well, some easy cases. First of all, if L is a bilion, well in this case U of L we saw earlier is just a polynomial algebra. So it's sort of obvious that the previous map was an isomorphism. Another case that is fairly easy to do is when L is a Lie algebra of a Lie group, finite dimensional one. Now in fact, any finite dimensional Lie algebra is the Lie algebra of a Lie group. So this does in some sense cover all cases. However, it turns out to be quite tricky to prove that all the algebras in finite dimensions are Lie algebras of Lie groups. But if you can, then we can prove the Poincare book of Victorimus follows. What we do is we pick local coordinates. So the identity is 0, 0, 0, and so on. And then we can pick a basis for L to consist of left invariant vector fields. So we have a, let's call them a11d by dx1 plus a12d by dx2 and so on. And a21d by dx2, sorry, d by dx1 plus a22d by dx2 and so on. And now we can choose coordinates and we can just choose these so that these numbers here are all 1 at the identity. And everything else is just 0 at the identity. So what our basis for L looks like, it looks like d by dx1 plus terms vanishing at the identity plus anything yet, plus terms vanishing at the identity and d by dx2 plus terms vanishing at the identity and so on. And now if we call this a1 and this a2 and so on, now if we take ai1 times ai2 times ai3 and so on, this is equal to d by dx1 d by d, sorry, i1 d by di2 and so on. Plus terms vanishing at the identity plus lower order terms. And by looking at the leading term of this, we see that these are linearly independent for i1 less than i2 is less than i3 and so on. So we can show that all these terms in the universal enveloping algebra are actually linearly independent by examining their action on the smooth functions on the lead group. So informally what we've done is we've found a representation of u of L on a vector space v. In this case the vector space v is functions on the group G. So what do we mean by a representation of u of L? Well a representation of u of L just means a module acted on by the ring u of L. And we notice this is the same as a representation of the lead algebra. So a representation of the lead algebra, if we call it rho, is just a vector space v together with an action rho of a of elements of a on v such that rho of a, rho of b, minus rho of b, rho of a, is equal to rho of a b. And almost by definition of the universal enveloping algebra, a representation of the lead algebra is much the same as a module over the universal enveloping algebra. So the problem is what we want to do now is to find a representation of the universal enveloping algebra u of L so that we are able to show that all these elements a i1, a i2 and so on with i1, listen to i2 and so on are linearly independent. And the question is what are we going to take the representation v to b? Well we could copy the previous example and take v to be something like functions on the group. Well it's not quite clear what that means but we might take a ring of formal power series in the dual of the lead algebra or something. We're not going to do that. We're going to take v to be the universal enveloping algebra of L. Well we can't actually do this yet because the whole point of this is to construct the universal enveloping algebra of L and in particular show it doesn't collapse. And so we haven't actually yet constructed u of L in a way that we can use it so that's why I put it in inverted commas. So what we're going to do is to try and build up something that will eventually turn out to be u of L. So this doesn't quite work because we don't yet know that u of L doesn't collapse. So we take v to be spanned by expressions a i1, a i2, up to a ik for i1 is less than or equal to i2 is less than or equal to so on. And at the moment we're just taking these to be formal expressions. We can't yet say that this is the product of these elements in the universal enveloping algebra because we haven't yet constructed the universal enveloping algebra. So we're just going to take these to be formal expressions for the moment and try and build up an action of L on this so that this will eventually turn out to be the product of these elements. So we've got a vector space with this basis and the next step is to find an action of L on v. And so if we call this action rho, we have an operator rho of ai has to have mapped v to v. So we want to know what is rho ai of ai1, ai2 and so on, a up to aik. And we'll eventually want this action to be just multiplication of this element by ai in the universal enveloping algebra. So what we do is we define this action the way that it has to be if this is the product in the universal enveloping algebra. So first of all, if i is less than or equal to i1, we can just define this to be ai, ai1, ai2 and so on. If i is less than or equal to i1, if i is greater than i1, we have to do something because we only defined these elements when the subscripts are increasing. So what we do is we define it to be rho of ai, ai1, ai2, ai3 and so on, plus rho of ai1, rho of ai times ai2 and so on. This is for i greater than i1. And you notice that ai, ai1 should act, we want it to act as ai, ai1 minus ai1, ai, so this is the only way we could define it. And what we need to check is that this is well defined. So both of these expressions here are well defined by induction. So first of all, this is well defined because we're multiplying an element of l by a smaller monomial in the ai1, which we can assume is defined by induction. And here again, we're multiplying rho of ai by some smaller number of monomials. And here we're multiplying it by rho of ai1 for i1 less than i, so by induction on i, we can assume that this is well defined. So this does in fact give a well defined collection of operators mapping v to itself. Now we run into the following problem. Is this an action of l on v? So to do this, we need to check that rho ai bracketed with aj is equal to rho ai, rho aj minus rho aj, rho ai. And I'm not going to do this check because it's a straightforward but rather tedious calculation. It uses the Jacobi identity for l at several points plus anti-symmetry, of course. And you need to check several cases and use lots of induction on the length of monomials and so on. For example, if we've got ai, aj acting on ai1, ai2, then we get lots of different cases depending on the order of i, j and i1. For instance, if i and j are less than or equal to i1, we get one argument showing that this satisfies this identity. Some of i or j are greater than i1, then we get a more complicated argument and I'm not going to go through it because if I did, everyone would just fast forward to the next part of the lecture. So we can leave this as an exercise or something. So now I'm going to give some applications of the Poincare-Berkoff-Ripp theorem. So the first problem is how to recover the Lie algebra from the universal enveloping algebra. So suppose somebody gives you the universal enveloping algebra and you want to know what is the Lie algebra you started off with. Well, it's an easy way to construct this using the coproduct. So the universal enveloping algebra actually has some extra structure called a coproduct. It's a map from u of l to u of l tensor with u of l. And it's defined as a map of associative algebras that delta of a is a tensor 1 plus 1 tensor a. And by the universal property of the universal enveloping algebra, you can check that this does actually give a homomorphism of associative algebras because we remember we checked that this thing preserves the Lie brackets in some sense. So what we're going to do is to show the following theorem that l consists of the primitive elements of u of l. You remember primitive means delta of a is equal to a tensor 1 plus 1 tensor a. And we've seen that the elements of the Lie algebra all satisfy this. And before proving this, we just want to notice that this theorem is actually false. So let's write down a counter example. Let's take l to be one-dimensional Lie algebra, k, where the field k has characteristic p greater than 0. And if it has a basis a, we notice that delta a is equal to a tensor 1 plus 1 tensor a. But we notice that delta of a to the p is also equal to a to the p tensor 1 plus 1 tensor a to the p. So a to the p is also primitive. So this theorem actually fails in characteristic p. We're assuming that l is a Lie algebra over the reals, but this also works for any field of characteristic 0. So we'd better make a note here that we really are using the fact that we're working over characteristic 0 for this theorem to work. So let's show how to prove this. So the first case, which is easy, is we just take l to be abelian. And then the universal enveloping algebra of l, so I suppose l has a basis x1 up to xn. The universal enveloping algebra of l is just a polynomial algebra in variables x1 up to xn. And delta f equals f tensor 1 plus 1 tensor f turns out to be equivalent to saying that f of x plus y is equal to f of x plus f of y. Here x is equal to x1 up to xn and so on. So these are called additive polynomials. And all additive polynomials are easy to find because we may as well assume f is homogeneous of degree k by just splitting f into its homogeneous pieces. And then we notice that 2 to the k times f of x is equal to f of 2 to the x, which is equal to 2f of x. So 2 to the k minus 2f is equal to 0. And since the characteristic is equal to 0, this implies f equals 0 if the degree is greater than what. If f is characteristic p, then this actually does fail. For example, the polynomial f of x equals x to the p is additive in characteristic p greater than 0. So this is the point at which the proof fails in characteristic p, that there are additive polynomials that are nonlinear. In the general case, we can easily reduce to the Abelian case. What we do is we just filter u of l by f0 contained in f1 over f0, sorry, plus f2 over f1 and so on. And we notice that this is a polynomial algebra. And suppose that f in u of l is primitive and suppose f is in fn. Well, then we can think of this as being the universal enveloping algebra of an Abelian algebra. So any primitive element must just be of degree 1. And now we notice that the image of f in fn over fn minus 1 is primitive. So it must be in fn minus 1 if n is greater than 1, because the only primitive elements for an Abelian universal enveloping algebra of degree n are 0. And so we see that f is actually in f1, so it must be an element of l plus a constant term and you can easily eliminate the constant term. So f must in fact be in l, which is contained in f1. So in characteristic 0, the Lie algebra is just the collection of primitive elements of the universal enveloping algebra. And as an application of this, we can now prove a result we used in the previous lecture. So let's take l to be the free Lie algebra on elements x and y. Then u of l is just the free associative algebra on x and y, which is all non-commutative polynomials in x and y. So the primitive elements of u of l are just the elements of the Lie algebra l. In other words, primitive elements in the free associative algebra can all be written in terms of x and y in the Lie bracket. And this was the result we used in the proof of the Baker-Cambell-Hausdorff formula in the previous lecture.