 This lecture is part of an online course on Leigh groups and will be about the Baker-Campbell House Store formula. So what does this formula say? Well, I suppose A and B are N by N matrices over, say, the reals. Then, as we saw in the last lecture, you can try writing X with A times X with B to be X of A plus B, except that because A and B don't commute in general, this formula doesn't actually hold. You need to add in a correction term, and then you need to add in further higher-order correction terms. Oops, I forgot the A there. And so on. And what you notice is that all the terms I've written down here can be expressed using the matrices A, B we started with and just the Leigh bracket. It would be quite trivial to write down a formula like this if we're allowed to multiply matrices together. However, the key point of the Baker-Campbell House Store formula is that this expression here only involves the Leigh bracket. It converges for A and B small in some sense, in other words, near in some neighbourhood of zero, but it does not converge in general. In fact, we saw a consequence of this non-convergence in the previous lecture that the exponential map wasn't in general on to, that if the Baker-Campbell House Store formula always converged, then we would be able to prove that the exponential map is on to, but it just isn't. So now before proving the Baker-Campbell House Store formula, let's see some applications. So the first application is that a Leigh group is determined up to local isomorphism by its Leigh algebra. And this follows fairly easily. So you remember if we have a Leigh algebra, which is some sort of vector space and we had the corresponding Leigh group, which is some sort of curved manifold, then you remember the exponential map was an isomorphism between some neighbourhood of zero in the Leigh algebra and some neighbourhood of the identity E in the Leigh group. So we had an exponential map and we had its inverse, which was a logarithm map, which doesn't actually converge for everything on the Leigh group, but at least converges in some neighbourhood. And what this means that the Baker-Campbell House Store formula means the product here is determined by the Leigh bracket and the Baker-Campbell House Store formula. Because the Baker-Campbell House Store formula simply tells you how to multiply two elements in the image of Xp. Incidentally, this means there's no need for higher-order terms when you try and approximate a Leigh group. You remember the Leigh bracket is a sort of second-order term and you couldn't imagine that if you tried writing down this formula here the terms here might be, you know, maybe you'd get a fourth-order term that wouldn't be expressible in terms of the Leigh bracket and then you would have to add some sort of complicated fourth-order expression in terms of A and B in order to recover the group. Well, that doesn't happen over the Reels. The Baker-Campbell House Store formula says that the Leigh bracket is really all you need to determine the Leigh group at least near the identity. Incidentally, this fails over fields of characteristic p greater than zero. There you find that you can have two algebraic groups and characteristic zero with the same Leigh algebra that are not locally isomorphic. In fact, you can see this is going to fail in non-zero characteristic because you have to divide by various integers and you can't do that in characteristic greater than zero. So the reason why the Leigh algebra works well over the Reels is because of the Baker-Campbell House Store formula and in cases when it fails then the connection between the Leigh algebra and the group is much vaguer. So next application. Suppose we've got, suppose G and H are Leigh groups suppose we have a homomorphism from the Leigh algebra of G to the Leigh algebra of H. If G is simply connected we get a homomorphism from G to H induced by this. Well, obviously if G isn't connected we've got little hope of finding a homomorphism in general because the Leigh algebra can only see the connected component of G. So however, we also need to use the fact that G is simply connected and to explain why, let's recall what happens in analytic continuation. So in analytic continuation in the complex numbers suppose you've got some region and suppose that you've got an analytic function at some point and suppose you can extend it to some little neighbourhood and then maybe at some point here we can extend it to a little neighbourhood of that point and suppose we can keep on extending it. So for each path in the region if you've got a path from here to a point over here we can find, we can analytically continue our function at the origin to this point over here. And if we have another path then we can analytically continue along this other path and these two analytic continuations aren't necessarily the same. For example, if we take the function log of x it's not defined at zero but if we define it at one and analytic continues along this path to minus one we find the logarithm should be pi i but if we analytically continue along this path we get minus pi i. Now the problem here is that these two paths are not homotopic you can't slide one to the other. You notice if we can slide one path into the other by a sequence of paths like this then the value that you get by analytic continuation would be the same because along any path or nearby paths give the same value so if you can sort of slide one path to the other you get the same analytic continuation. So in a simply connected region it doesn't matter which path you use to define analytic continuation. Now for defining maps between Leigh groups the argument is kind of rather similar. So if we've got a Leigh group here let's call this Leigh group G and draw a picture of Leigh group H and they've got identity elements and suppose we've got a homomorphism between their Leigh algebras taking G to H. Well using the Baker-Cambell House Store formula we can use that to define a map from a small neighbourhood of G to a small neighbourhood of the identity in G to a small neighbourhood of the identity in H. And now if we've got a path in G we can sort of continue this map along this path by using the fact that we've got the map to find a small neighbourhood and using the group operation. I mean the map from G to H has to be a homomorphism of groups so we can sort of continue it along like this. But we could also choose another path to continue it along. And there's no particular reason in general why the continuations along these two paths should be the same. For example if we've got a map from the circle to the line well we can certainly have a local homomorphism from the circle to the line but if we extend it along this path then the image sort of looks like that and if we extend it along this orange path then the image sort of looks like that and you see the image of this point is not well defined. However if the Lie group G is simply connected then we see that if you just slide a path slightly you're going to get the same continuation so if you can slide one path into the other which you can always do if G is simply connected then the image is going to be well defined and independent of the path. So this shows that Lie algebras are very similar to simply connected Lie groups because homomorphisms between Lie algebras are almost the same as homomorphisms between the corresponding simply connected Lie groups. So now we'll start with explaining how to prove the Campbell Baker House Store formula. First of all if we look at the formula we see it's not really a formula about any particularly algebra we can think of this as a formula about the free Lie algebra on A and B which is a sort of algebra generated by A and B where the only relations you put on them are the axioms for a Lie algebra. So if we can prove this for a free Lie algebra on two generators A and B then it will automatically be true for all the algebras apart from convergence problems that I'm not going to worry about. So let's start by letting R be the free associative algebra on two elements X and Y. So we can think of this as being non-commutative polynomials. So a typical element might look like something plus something times X plus something times Y plus something times X squared plus something times X Y plus something times Y X plus something times Y squared plus maybe higher order terms. And you see the key difference between this and polynomial algebras is that X and Y need not commute so we need a term for X, Y and we also need a term for Y, X in general. And just as with a polynomial algebra you can take a completion that's a ring of power series we can do the same thing we can take a completion of this which is a sort of non-commutative power series and you can check that if we allow these elements are going to be sort of infinite sums like this and you can check that the sum and product of these is well defined so we get a ring. And what I'm going to do now is to define a co-product on R hat. So the co-product will be well first we're going to define it on R so it would be a ring homomorphism from R to R tensor R. So the so ring homomorphism so you can see R is really the free algebra generated by X and Y so we can define any ring homomorphism by giving the images of X which is going to be X tensor 1 plus 1 tensor X and Y which is going to be Y tensor 1 plus 1 tensor Y so this defines a homomorphism from R to R tensor R and you can easily see that it induces a homomorphism from in the same way to the completion. Now I want to explain what this co-product is well what it is really is it's a sort of co-product of a Hopf algebra so I'll just briefly explain what a Hopf algebra is this isn't really needed for the proof of the Baker-Campbell House Store formula it's just background information so suppose we take the group algebra let's say RG of a group G so this is an associative algebra where the product is the same as the product of a group G so it has a product going from RG tensor RG to RG but it also has a co-product from RG to RG tensor with RG and this co-product is defined as follows delta of G is equal to G tensor G whenever G is in the is in the group and it's defined by linearity and you can check that this is a homomorphism of algebras and these are part of the structure of something called a Hopf algebra a Hopf algebra also has a unit and a co-unit and an antipode as I'm not going to bother to write down because I don't need them but you can look them up on Wikipedia or something so a group algebra is a Hopf algebra and conversely if you've got a Hopf algebra it behaves very much like a group algebra it has many other properties that many things you can do with groups you can do with Hopf algebras the group algebras are rather special property that it's co-product is co-commutative you see if you flip these two round it doesn't change it Hopf algebras that are where the co-product isn't co-commutative are sometimes called quantum groups so that's essentially what a quantum group is it's more or less the same as a Hopf algebra of a special type anyway it turns out that from a any Lie algebra you can construct a Hopf algebra called its universal enveloping algebra you can never remember how many p's that are in enveloping which we will discuss a little bit later and if you take the freely algebra on two generators it turns out that its universal enveloping algebra is just the free associative algebra on two generators that we've been talking about and the co-product on this free associative algebra turns out to be the co-product of this considered as a universal enveloping algebra so that's sort of trying to explain why we're considering the free associative algebra on x and y and messing around with this funny co-product anyway let's get back to r we say that an element little r in r is called primitive if delta r is equal to r tensor 1 plus 1 tensor r so in particular x and y are primitive and now we notice that the primitive elements form a Lie algebra and for this we just need to work out so suppose r and s are primitive we need to work out delta of the bracket of r and s and this is quite easy because it's delta r delta s minus delta s delta r which is equal to r tensor 1 plus 1 tensor r times s tensor 1 plus 1 tensor s plus something similar which I'm feeling too lazy to write out or the minus something similar and if you multiply this all out this is equal to rs tensor 1 plus 1 tensor rs plus r tensor tensor s plus s tensor r minus a lot of junk and this turns out to be r s minus s r tensor 1 plus 1 tensor rs minus s r which is equal to rs tensor 1 plus 1 tensor rs s so that's a rather nice result because we can get a Lie algebra inside the associative algebra just by taking the primitive elements now let's look at what happens to the exponential map so you remember we've got an exponential map which goes from r to r it goes from the elements of r hat with constant term 0 to the elements of r hat with constant term 1 so this is just taking anything so we're just defining x of some non-commutative power series to be 1 plus a plus a squared over 2 and so on and you can see that this converges if a has 0 constant term and there's an inverse map given by the logarithm map where the logarithm of 1 plus something is defined by the usual power series so these maps are actually bijections because we're working with formal power series there's no problem with convergence and primitive elements under x are more or less the same as group-like elements so what's a group-like element? well it has to satisfy delta of g is equal to g tends to g so you remember for group rings the actual group elements in the group rings satisfied this identity here there's also another condition saying that the co-unit on this must be 1 in other words it must have constant term so primitive means delta of r equals r tends to 1 plus 1 tends to r and that's quite easy to check for instance suppose r is primitive then we can work out delta of x of r turns out to be x of delta of r because since r is a ring homomorphism it's not difficult to check it kind of commutes with delta and this is going to be x of r tends to 1 plus 1 tends to r which is going to be x of r tends to 1 times x of 1 tends to r which is just equal to x of r tends to with x of r so x of r satisfies the group like property and conversely a very similar calculation shows that if something is group like then it's logarithm is primitive so we actually get a bijection between these two objects here and the primitive elements as we've just seen form a Lie algebra and the group like elements as you might guess form a group which is very easy to check you can think of this as being an infinite dimensional Lie group and we now have an exponential that which is actually a bijection between the Lie algebra and this group so now we can try and prove the Baker-Kamble-Hausdorff formula well what we do is we just take x and y in our hat and we know that these are primitive so x of x and x of y is group like you remember this says that delta of a is equal to a tensor a and this means that x with x times x with y is also group like because it's easy to check the product of two group like elements as group like and finally the log of x with x times x with y is now primitive so this is exactly a term that appears in the Campbell-Baker-Hausdorff formula remember we were saying x with x times x with y is equal to x of something and you notice this something is exactly this expression here so to finish off we just have to prove that any primitive element of our hat is in the completion of the Lie algebra generated by x and y well what I'm going to do is I'm going to leave this to next lecture because it's actually something we'd quite like to prove for all Lie algebras and the universal enveloping algebras and not just this particular special case this is, this fact here is actually a theorem due to Friedrich's theory so we're going to postpone this part of the proof of the Baker-Campbell-Hausdorff lecture I'll just finish off by pointing out that there is an explicit form of the Baker-Campbell-Hausdorff formula you notice this is just a sort of abstract existence proof it shows that there is a formula you can write down in terms of the Lie brackets but doesn't quite say what it is of course you could work at any finite number of terms that you want just by brute force and there is an explicit formula for it and I'm not going to write it down for reasons that will become very obvious in a moment I'm just going to show you a picture of it and it looks like this and it's pretty obvious that if I try to write it down I would be almost certain to make a mistake somewhere and so I'm not actually going to so there is an explicit formula but it's rather hard to apply because it's so complicated as you've seen most of the applications don't actually need the explicit formula if we want to show that Lie groups are determined up to local isomorphism by the Lie algebra we don't need the explicit formula all we need to know is there is some formula like this okay so that's all for the Baker-Campbell-Hausdorff formula and next lecture we will be discussing universal enveloping algebras and Friedrich's theorem