 This lecture is part of an online course about Lie groups, and we'll be discussing the question of the relation between Lie groups and Lie algebras. So in the previous lecture, we show that if you're given a Lie group, you can get a Lie algebra. And the question is how much of the Lie group does its Lie algebra actually capture? So for example, we can ask the following questions. If two groups G and H have the same Lie algebra, or at least isomorphically algebras, are they isomorphic? In other words, does the Lie algebra determine the Lie group? And we will see the answer to this as no. Secondly, if we've got two groups G and H, suppose you've got a homomorphism from G to H. Then it's easy to check that you get a homomorphism from the Lie algebra of G to the Lie algebra of H. I mean, it maps the tangent space of the identity of G to the tangent space of the identity of H, and it's not difficult to check this preserves the Lie bracket. And we can ask, is the converse true? So let's put a question mark there. If you've got a homomorphism between Lie algebras of Lie groups, does this give us a homomorphism between the Lie groups? And the answer is in general, no. Or suppose a special case of this, suppose G is a subgroup of H. Then the Lie algebra of G is a subalgebra of the Lie algebra of H. We can ask, is the converse true? So if we have a Lie algebra that's contained in the Lie algebra of H, does it come from some closed subgroup? I guess I should have said here G is a closed subgroup of H. None closed subgroups can be very weird. And generally, when you talk about a subgroup, you're very often talking about a closed subgroup, not always. And the fourth question we can ask, does any Lie algebra come from a Lie group? So in this lecture, we're implicitly assuming that all Lie groups and Lie algebras are finite dimensional. And for a dimensional case, it gets a lot trickier. I think I forgot to say the answer. This question three is no. And the answer to question four actually turns out to be yes for once. So what we're going to do is going to go through and give some examples to sort of illustrate what's going on. So the first question, you remember, said that if two Lie groups have the same Lie algebra, are they isomorphic? This is obviously false because G has the same Lie algebra as is connected component. So in general, if a group isn't connected, there will be a connected group with the same Lie algebra and the groups aren't isomorphic. But we should ask the question for connected groups. So suppose G and H are connected and have the same Lie algebra, which is a sloppy way of saying their Lie algebras are isomorphic. Well, are they the same? Well, no, because we can do things like we can take the reals, maps onto the circle group S1. And these both of Lie algebras isomorphic to the reals. Another example might be the group SL2 of R maps onto the group PSL2 of R. And the kernel is, consists of the elements plus or minus one. And again, these two groups have the same Lie algebra, but they're not isomorphic. For instance, this one has a non-trivial center and this one doesn't. And as I mentioned in an earlier lecture, what's going on is that these pairs of groups are locally isomorphic. So that there's a neighborhood of the identity of this group that looks like a neighborhood of the identity of this group. And if two groups are locally isomorphic, they have the same Lie algebra. So locally isomorphic implies the same Lie algebra. In fact, the converse is also true. The two groups have the same Lie algebra. They are at least locally isomorphic, but this takes a bit more work, so I'm not going to cover it this lecture. And you can get lots of examples of locally isomorphic groups by taking G modulo, a discrete closed subgroup. So that's what's going on in these two examples here. The discrete closed subgroup is the integers, and here it's just plus or minus one. And if you mod out by discrete closed subgroup, you're obviously going to get a locally isomorphic group because this doesn't affect what's going on near the identity. So in fact, discrete closed subgroups of G are quite restricted. So suppose G is connected, then any normal discrete subgroup is in the center of G. So in particular, it has to be abelian. And you can see this as follows. So let's fix some gamma in this closed normal subgroup. Let's call this closed normal subgroup gamma. And we look at the map taking G to the commutator, G to the minus one gamma to the minus one G gamma. And you notice the image is in gamma because we can write this as G to the minus one gamma to the minus one G. And this is in gamma because gamma is normal and we'll multiply it by gamma, which is also in gamma. So here we've got a map from G to gamma. And G is connected and gamma is discrete. So the image is a single point. On the other hand, if we take G to be the identity, then this is the identity. So the image must actually be the identity of G. Well, that just means that little G commutes with gamma for all G. So any normal discrete subgroup must actually be in the center. So the normal discrete subgroups we're quotienting out by here, I guess that should have been normal, are actually quite restricted. Well, so let's have a look at the next question, which says that you remember if you've got a homomorphism of groups, then this gives rise to a homomorphism from the Li algebra of G to the Li algebra of H. And we want to know, is the converse true? And again, the answer is no. So here we've got a homomorphism of groups from R to S1. And if we look at the Li algebra, it's just R to R, and this is actually an isomorphism. So there's an isomorphism of Li algebras back the other way, but there's no homomorphism of groups from S1 to R corresponding to this. Well, what's going on here is that this group here, S1, is not simply connected. So we recall that simply connected means that any loop inside the group can be sort of contracted down to a single point if it provided you keep a sort of point fixed. In other words, we can take a loop and sort of pull it down like that. And obviously the real line is simply connected. If you've got any loop, in other words, the image of a circle in the real line, it sort of might look something like this. And you can obviously just contract it down. On the other hand, if you've got the circle group S1, there's a pretty obvious loop that just goes round S1 that you can't contract down. And it turns out that if G is simply connected, then a map from Li algebra of G to the Li algebra of H induces a homomorphism from G to H. Again, we don't really have enough technology to prove this quite yet. But as I said, this lecture is just giving a few examples. So the third question says that suppose G, suppose you've got a Li algebra contained in the Li algebra of a subgroup of a group H. Is there a closed subgroup corresponding to L? And the answer is not quite. We have the following funny example. Let's take G to be a product of two copies of S1, which is just R. You can think of as R2 modulo the lattice Z2. And what I mean to do is to take L to be the multiples of X alpha X beta for X in R, where alpha over beta is irrational. And we can think of G as being a sort of square with the top and bottom identified. So what you should do is you should think of the top of this as being identified with the bottom and the left hand side identified with the right hand side. And now if we tried finding a subgroup corresponding to L, we can find such a subgroup. It's just a subgroup of all multiples of alpha and beta. And the subgroup sort of looks like this when it comes up there and then it comes up there. And it sort of goes on like this. And the problem is the subgroup of all points of the form X alpha X beta is dense in G. So it's not closed. And so in some sense we do get a subgroup, but if a subgroup isn't closed inside a group, then you run into sort of funny problems trying to take its Lie algebra. So there's sort of a subgroup corresponding to this subalgebra, but you have to be a little bit careful about it. Finally, we can ask the fourth question says that does any Lie algebra come from a Lie group? Again we're doing the finite dimensional case. And here the answer is yes, but it's actually a little bit tricky to prove. Let me say a bit about how to prove the easy case where the center of the Lie algebra L is just zero. So what's the center of a Lie algebra? The center is the set of a group. It's the set of group elements that commute with everything else. For Lie algebras, commuting with something corresponds to the Lie bracket being zero. So this would say that x, y equals naught for all y or rather the set of x such that x, y equals zero for all y. And if the center of L is none zero, then L can be then written as a subly algebra with the n by n matrices over r where n is equal to the dimension of L. And we do this by actually identifying L with n by n matrices. So linear transformations of L correspond to elements of this Lie algebra. And what we want to do is to get a map taking from each element of L to some linear transformation of L. This will give us a map from L to the Lie algebra of gln. Let's call this map rho. And rho of an element a is the linear transformation such that rho a of b is equal to ab. And we need to check that this is a homomorphism of Lie algebras. So we want rho a b to be rho a rho b minus rho b rho a. Well in order to check this we need to see that abc is equal to abc minus bac. And this sort of looks more or less like the Jacobi identity with possibly a few signs. And these signs come from the fact that you sometimes switch the orders of a and b. So this follows from the Jacobi identity and anti-symmetry. So this gives one interpretation of what the Jacobi identity is. It's saying the natural action of L on itself is actually a homomorphism of L to the Lie algebra of n by n matrices. Now we've got L as a sub-algebra of n by n matrices. And you can show from this that L actually corresponds to a subgroup. Although doing that, well you need to use something like the exponential map that we'll be covering in a later lecture. So I'll postpone this for the moment. You notice the fact that L is a sub-algebra of m and L follows from the fact that the centre is equal to zero because anything in the centre of L would have imaged zero under this map here. So that's why we need the centre of L to be zero for this easy proof. If the centre of L isn't zero then it's still true that L is a sub-algebra of some of the algebra of n by n matrices but it's rather more difficult to prove. Again we might do that in a later lecture. So let's have a summary. So we've seen that Lie algebras don't quite correspond to Lie groups because there are the various problems. However, Lie algebras are very similar to simply connected Lie groups. So we saw that most of the problems that arose that Lie groups not quite correspond to Lie algebras were related to the fact that the group was either not connected or not simply connected. So what I'm going to do now is to say a little bit about the relation between Lie groups and simply connected Lie groups. Well, if the group isn't connected then the group of components can be any discrete group and we can't really say much about that. So what we're really going to do is compare connected groups to simply connected groups. So in other words, suppose we had somehow managed to understand all the simply connected groups. How can we find all the connected groups from that? Well, as we said, if we've got a simply connected group then we can form another group by taking the quotient of G by gamma where gamma is discrete and normal and as we saw earlier this implies that it's actually in the centre of G and this is now a connected Lie group. And we can do the converse. If we've got any connected Lie group we can actually construct a simply connected Lie group such that our connected group is a quotient of this. And I'll just recall how to do this. So suppose H is a connected group. We put G to be the universal cover of H. And let's recall from algebraic topology what this consists of so we fix the identity E is a base point of H and G is the set of all homotopy classes of paths in H starting at E. This means a map F from the unit interval to H with F of 0 equals E. So if we've got a group H like this and point E then a typical path would be some path like that. As I said, these are homotopy classes of paths. These are homotopy classes fixing F0 and F1. So what we do is we take the path and we fix the beginning and end of the path and we're allowed to move the path provided we keep the start and end points fixed. So this path here is considered to be homotopic to this path here and these are all going to represent the same element of the universal cover. And the basic result of algebraic topology that this universal cover is simply connected. And it's also a group. So in order to see it's a group let's just take H and if we've got some path starting at E going to some point here and some other path going to some other point. So this goes to a point G and this orange path might go to a point H. We can sort of just shift the path going to H by multiplying it by G everywhere. We take every point of this path and multiply it by G. So this now goes to a point GH. And you can easily check that this is a well-defined operation and in fact gives you a group structure on the universal covering space. And furthermore this group is locally isomorphic to H sorry to G. Again this is a general result from algebraic topology and works for any space. It's nothing to do with being a group. So we've got a map from a simply connected group on to our group G which is a local isomorphism. So these have the same least algebra but this group here is simply connected. Incidentally the group we quotient out by is actually you can check it's actually isomorphic to the fundamental group Pi 1 of the group G we started with. I mentioned earlier that a discrete normal subgroup like this here has to be abelian so this sort of suggests the slightly surprising result that the fundamental group of any Lie group is always abelian. In fact you can check this directly as follows. So suppose we've got two elements of the fundamental group of G. So an element of the fundamental group of G will be a path from the unit interval to the group G such that the image of beginning and end of the path are both the identity of G. So suppose we take one element of the fundamental group and we take another element of the fundamental group. So this will now give us a map H mapping to G and again the images of the beginning and end of the path will be in G. Now we can compose these two paths in two different ways. So we can multiply this element of the fundamental group by this element or we can do them in the other order. And in order to show that the fundamental group is abelian we've got to show these two maps a homotopic. In other words we've got to kind of fill in this square here which will give us a homotopy between this path in G and this path in G. And that's quite easy because we can just map any point x, y in this square to H of x times f of y where we multiply these two elements in the group. In fact you can see in order for this argument to work we don't actually need G to be a group we just need it to have a nice continuous product with an identity element. So fundamental groups of Lie groups are particularly easy to deal with because they're abelian. So anyway let's see some examples of this. Well the simple example is one we've had before if we take G to be the circle group S1 then it's universal covering space. It's just the reals which sort of sits over in a sort of helical pattern like this and so on. So this would be a copy of R mapping to S1 and the fundamental group of a circle is of course just C and you can see the kernel of this map from the reals to the group S1 is again just isomorphic to Z. Well for a more interesting example let's look at the general linear group in two variables over the reals and we want to know what is the group of components and what is the fundamental group Pi1. And to do this we recall the Gram-Schmidt process which is sometimes called the Iwasawa decomposition. Iwasawa decomposition is something that works for more general groups. So the Gram-Schmidt process you remember is a way of turning any base of R2 into an orthonormal base and the Iwasawa decomposition writes the group as a product of a certain compact subgroup and a certain abelian subgroup and a certain nilpotent subgroup and in these three cases the nilpotent subgroup is going to be these matrices and the abelian subgroup is going to be these matrices with diagonal entries greater than zero and the compact subgroup is just going to be the group of orthogonal matrices. And you can see that writing gl2 of R as a product of these three groups is really the Gram-Schmidt process because multiplying something by this nilpotent subgroup is really making a base orthogonal by adding a multiple of one vector to another. Then multiplying the base by this diagonal matrix is just multiplying each basis vector by a positive real which you can use to make the matrices all have norm one. And finally what we're left with is an orthonormal base and any two orthonormal bases are related by a unique orthogonal rotation so it's an element of the orthogonal group. So the Gram-Schmidt process is just a way of saying that every element of the gl2 of R can be written uniquely in this form. And now we can figure out what the fundamental group is because these two spaces are both contractable. So in fact this is isomorphic to the reals and this is isomorphic to the product of two copies of the positive reals and they're both contractable. So the components and the fundamental group of the general linear group are the same as the components and the fundamental group of the two-dimensional orthogonal group. And this is much easier to work out because the orthogonal group of R has two components one of which is SO2 of R, which is just the rotations and the other is the reflections of R2. So we've got two components. We can also work out the fundamental group. Let's just notice that the group of components of gl2 of R which is usually noted by pi zero is just isomorphic to z modulo 2z. So it's order 2. And pi1 of gl2 of R is isomorphic to pi1 of SO2 of R. Well the group of rotations of R2 is just isomorphic to a circle. So this is just the fundamental group of a circle which is isomorphic to z. So the fundamental group of gl2 of R is actually z. So we have a simply connected group. Let's call this gl2 of R plus and put a twiddle over it. So this means you take the identity component and this means you take the universal cover. And this maps to gl2 of R and the quotient is a group of order 2 and the kernel of this map is z. And it's kind of really quite difficult to describe gl2 of the universal cover of gl2 of R in terms of matrices. The problem is this is not isomorphic to a group of matrices. In fact, we will see later when we look at the representation theory of gl2 of R plus that any map of this group on a finite dimensional vector space must vanish on this kernel z. So it really factors through gl2 of z. So this group is kind of hard to get hold of. You can't write down explicit matrices for it. At least not in finite dimensions. You can if you allow infinite matrices. Similarly, the group sl2 of R also has a universal cover with an infinite center z. And these covers are also a bit mysterious. The only one that I've ever seen turn up anywhere is the so-called metaplectic group where you take the double cover of sl2z. So you can take a cover with fundamental group z but you can then quotient out by 2z and get a group that's just a double cover. And this double cover of sl2z actually turns up in the theory of modular forms. It's very closely related to modular forms of half integral weight. But again it has no faithful finite dimensional representation so it's a little bit tricky to handle. So that's looked at the fundamental group of gl2. Now let's take a look at the fundamental group of gl3 of R. Well as before it has the same group of components and the same fundamental group as O3 of R. We can again do the Gram-Schmidt process and see that gl3 of R is homotopy equivalent to the orthogonal group O3 of R. So O3 of R has a subgroup SO3 of R, the special orthogonal group of things of determinant, plus one of index two. So pi zero of gl3 of R and pi zero of O3 of R, which is a group of order two. So it is two components which are of course the matrices of positive determinant and the matrices of negative determinant. Pi one of gl3 of R is isomorphic to pi one of SO3 of R because we can forget about the other component. And this pi one of SO3 of R is not at all obvious. It turns out to be another group of order two. One way to see this is to use quaternions. So you recall that quaternions are the group of the ring of all matrices a plus bi plus cj plus dk. And it has the multiplicative group has a subgroup S3 which is the set of a plus bi plus cj plus dk such that a squared plus b squared plus c squared plus d squared equals one. So obviously a sphere. This is a group and it maps on to the group SO3 of R. And the way it does this is we can think of SO3 R as being the rotations of the vector space bi plus cj plus dk. And the way S3 acts as rotations is if we've got some quaternion g here it acts on a vector v as g of v is g v g to the minus one where this vector is g. So we can define a homomorphism from the group S3 to rotations of three dimensional space. You can easily check that its image does in fact preserve the obvious quadratic form. And the kernel of this map is just the matrices plus or minus one. So what we've got is we see that the group S3 is a double cover of SO3 of R. And the group S3 is of course simply connected. And so the fundamental group of SO3 is the kernel of the map from its simply connected cover which is just a group of order two. So you see that the fundamental groups of lead groups can sometimes be a little bit surprising. I mean you probably wouldn't guess if you didn't know that the fundamental group of GL3 was Z2 and the fundamental group of GL2 was the integers. So by the way, the fact that the fundamental group of orthogonal groups often has order two is a special case of the construction of spinor groups that we'll be doing later. And in fact S3 is one of the simplest examples of a spin group. Okay, so next lecture we will be discussing the exponential map that gives another relation between the lead algebra and the lead group.