 This lecture is part of an online course on Lie groups and will be about Engel's theorem. So Engel's theorem will help to explain the following mystery. Why is a nilpotent group called nilpotent? So if you recall a nilpotent group G0 is one such that if you kill off the centre and get a new group G1 and then kill off the centre of that and get a new group G2 and continue like this, we eventually get the trivial group. On the other hand the word nilpotent, well nilpotent means zero and potent means power. So something should be nilpotent if some power of it is equal to zero. And at first sight this concept of nilpotent seems to be very little to do with the definition of nilpotent for a group. And the explanation of it goes via Lie algebra. So we say Lie algebra L0 is nilpotent. If when you keep killing off the centre of L0 to get L1 then you kill off the centre of L1 to get L2 and so on. A Lie algebra is called nilpotent if that eventually ends up as the zero Lie algebra. Well again the definition of nilpotent for groups is obviously defined by analogy with the definition of nilpotent for Lie algebras. Well at first sight the definition of nilpotent for Lie algebras doesn't seem to have much to do with nilpotents either. So this difference is explained by Engel's theorem. So Engel's theorem says the following. We have a Lie algebra contained in the Lie algebra of n by n matrices. And so L is going to be a Lie algebra. Then if all elements of L are nilpotent L fixes some non-zero vector v of r to the n. This is assuming n is greater than zero of course. Well at first I'd better say what we mean by fixing a vector. So if g is a group then we say that a group element would fix a vector if g of v is equal to v. This is not the definition we use for Lie algebras. For Lie algebras the definition of fixing a vector says that L of v is equal to zero where L is in the Lie algebra. The reason for this is that we think of an element of a Lie algebra as being a sort of difference between an element of a group and the identity. So we might think of g as being approximately the identity element plus an element of a Lie algebra. So this is why the definition of a Lie algebra element fixing a vector is given like that. Well this still doesn't seem to have much to do with nilpotent matrices. But now we observe that we can repeat this with r to the n modulo v. Let's call this v zero or let's call it v1. And we can find a vector v2 in r to the n over v1 which is fixed by the Lie algebra L. And we can continue like this. We get v1, v2 and so on up to vn. And if we use these vectors as a basis for r to the n then all elements of L are strictly opera triangular. So L is now contained in the subspace of these matrices. And the set of these strictly opera triangular matrices is formerly algebra. And on the one hand the elements are obviously nilpotent. And on the other hand they satisfy the definition of being nilpotent Lie algebra that I mentioned earlier. For example if we take four dimensional space and look at this Lie algebra. We can see the center of it is more or less these matrices here. And if we're quotient out by those then the center of the resulting algebra is given by matrices where these are allowed to be none zero. And if we're quotient out by the center of that then we're quotient out everything and we get zero. So Engel's theorem shows that the condition for being a nilpotent Lie algebra is closely related to the condition that all matrices of the Lie algebra are nilpotent. I should have a sort of warning that a Lie algebra L inside the n by n matrices can be nilpotent as a Lie algebra but not nilpotent as matrices. For example we just take gl1 of r and we just take the matrix one in gl1 of r. Well this is obviously a nilpotent Lie algebra because it's abelian. On the other hand this is obviously not a nilpotent matrix. So the connection between nilpotent Lie algebra and nilpotent matrices isn't quite as clean as you would really like. So let's sketch the idea of the proof. This is a proof of Engel's theorem. The main idea is we repeatedly use induction on the dimension of the Lie algebra and the dimension of the vector space r to the n that is acting on. So we can sort of assume if we've got any smaller Lie algebra or any smaller space then we can apply Engel's theorem to that. So the proof really involves three steps. So step one is that L acts on itself by nilpotent operators. Well what do we mean by L acting on itself? Well any Lie algebra L acts on itself because if you've got an element L of the Lie algebra it acts as the map from L to L taking M to the bracket of L and M. So we can think of L as mapping to the Lie algebra of linear map from L to itself. This isn't necessarily injected because L might have a centre, in other words something that commutes with everything. And what we want to do is to show the action of L on L these things are all nilpotent. So we want to know are these nilpotent? Well we think of this as taking M to L of M minus M of L where we now think of these as matrices. And we notice that M goes to L times M is nilpotent because we assumed all the elements of L when nilpotent is matrices and similarly L goes to M times L is also nilpotent. And we notice that these two commute as linear transformations because right multiplication commutes with left multiplication. And now if we've got two commuting nilpotent objects so if x and y two commuting nilpotent things then x plus y is nilpotent. And the reason for this is that if we write if x to the n is equal to zero and y to the n is equal to zero then we notice that x plus y to the two n equals zero because if you expand this by the binomial theorem every term involves either x to the n or y to the n. So the adjoint action of L on itself which is this also acts by nilpotent matrices. So the second step of the proof is that L contains an ideal of co-dimension one. So we just recall what an ideal is. An ideal of a Lie algebra is something just such the bracket of anything in L and M is contained in M. So this is the analog for Lie algebras of a normal subgroup of groups. You can easily check that if we've got an ideal of a Lie algebra then we can take a quotient and that still has a Lie algebra structure. And this is easy, we just pick M to be a maximal sub-algebra that's not equal to L assuming L has dimension greater than zero. And now we're going to look at M acting on the space L modulo M. So here we've got a Lie algebra acting on a vector space which is just the vector space quotient of L by M. And now we're going to apply induction and by the induction hypothesis since M has lower dimension than L and we've just seen it acts by nilpotent matrices on this we see that there's a vector V in L over M with V not equal to zero and V is fixed by M. Well this means that if we take R, V and add it to M this is also a sub-algebra. So by maximality of M it's equal to L. So M has co-dimension one in L and you can also see that V M is contained in M because M acts trivially on V modulo M so M is an ideal. So this is step two we've shown that L always contains an ideal of co-dimension one assuming L is non-zero. And now the third step is very easy. So step three we've got L contains an M. M is an ideal of co-dimension equal to one. And now these both act on our vector space V and let's put V zero equals vectors fixed by M. And by induction V zero is not just zero because M has to fix some vectors on V because M has smaller dimension than L. And now L is generated by X and M so now let X act on V zero. Well this acts null-potently. This is where we finally use the condition that the matrices of L all act null-potently. So fixes a non-zero vector of V zero because any null-potent endomorphism of a non-zero vector space must have a vector that it kills or fixes if we're talking about the algebras. And this proves the theorem so V naught is fixed by the Lie algebra L which is what we wanted to prove. So that proves Engel's theorem. There's a sort of analog of this theorem due to Coulson for groups. It says that if G is an algebraic group contained in the general linear group over a field, an algebraic group just means it's a group that can be defined by polynomial equations. Then it says if G is unipotent then G is conjugate to a subgroup of the group with ones down the diagonal and zeroes here and something there. So what does unipotent mean? Well unipotent means all eigenvalues are equal to one. So null-potent means all eigenvalues are equal to zero. So this is the sort of group theoretic analog of a matrix being null-potent. If a matrix is null-potent then one plus that matrix is unipotent. And this is obviously very similar to the condition that the Lie algebra should be conjugate to something that is strictly oppertriangular. In some of the lectures I've made several rather disparaging remarks about null-potent Lie algebras and null-potent groups and said how horrible they were. And I would like to point out that although they're kind of evil, they're not completely evil and there is actually one nice property of null-potent Lie algebras that doesn't hold for general Lie algebras. So suppose we have a Lie algebra L contained in a group of strictly oppertriangular matrices. So in particular L is null-potent as a Lie algebra. Then the exponential map is an isomorphism from an L to the corresponding Lie group. Well what is the corresponding Lie group? Well if we like we can even define it to be the set of exponentials of matrices of L. I should say here we're working over the reals or more generally a field of characteristic zero because otherwise you can't define the exponential map. And this is very easy to see because in fact the series for X is finite. In fact you can see for any matrix in L, A to the N is equal to zero for some fixed N. So the exponential of A is just equal to 1 plus A plus A squared over 2 plus A to the N minus 1 over N minus 1 factorial and it's just a polynomial in A. And similarly log of 1 plus A is equal to A minus A squared over 2 plus A cubed over 3 plus or minus all the way up to plus or minus A to the N minus 1 over N minus 1. So the series for log of X and exponential of X now converge everywhere as long as you're just working with oppertriangular matrices. So in particular you can see that the set of matrices of the form X with A for A in the Lie algebra is the simply connected Lie group. And it's simply connected because it's isomorphic as a topological space to L which is just a vector space which is obviously simply connected. So the correspondence between simply connected Lie groups and Lie algebras is particularly easy because the exponential and logarithm map always converge. Now let's have an example of a nilpotent Lie group that is not simply connected and is not a matrix group. Meaning we can't identify it as finite matrices over the reals although we can identify it as an infinite matrix group. And for this we just take G to be the group of all these matrices. So this is a nilpotent Lie group of dimension 3 and we're quotient out by the set of matrices of the form of this form where N is in Z. So this is in the center of the group G so we're just quotient out by a subgroup in the center isomorphic to Z. So this is the famous Heisenberg group which appears a lot in quantum mechanics because it controls the position and momentum operators of a boson. And let's look at the Lie algebra. Well the Lie algebra is spanned by three elements. So we take X to be 0 1 0 0 0 0 0 0 0 and Y to be this matrix with a 1 there and Z to be the last one which is 0 0 1 0 0 0 0 0 0. And now we can work out the brackets between X, Y and Z and we find that the bracket of X and Y is Z or possibly minus Z I sometimes get this wrong. So Z bracketed with X and Z bracketed with Y is just 0 so Z is the center or spans the center of this Lie algebra. And now we want to show that we can't represent G in terms of finite matrices. Well we look at the image of Z. Here we're taking a map from G to the general linear group over N of the reals and the Lie algebra will map to something in the N by N matrices over the reals. And we want to look at the image of Z in N by N matrices over the reals. And now we notice that X of I Z sorry X of Z must be equal to 1 sorry we should be working over the complex numbers not the reals that I can diagonalize everything. We know that X of Z is equal to 1 and this implies that Z is diagonalizable. Because if a matrix isn't diagonalizable and you put into Jordan form you see that the X of it can't be 1 and we just get a one parameter subgroup isomorphic to the reals. So we know Z is diagonalizable and the other hand Z is in the center of the group or Lie algebra G. So we can pick some eigen space V of Z in C to the N and we notice that V is acted on by G because Z is in the center of the Lie algebra. So all eigen spaces are also preserved by G. So the Z looks like well it has to be diagonal and it's got only one eigen value on V so it must look like this in V for some A in C. And now we use the Lie brackets we know that X Y is equal to Z so trace of X Y is equal to the trace of Z. Well this is just the trace of X Y minus the trace of Y X which is equal to 0 because you know for matrices the trace of a product of two matrices doesn't depend on the order of the product. So the trace of Z must be 0 but the trace of Z is just equal to the dimension of V times A so A equals 0 as the characteristic is equal to 0. So Z acts trivially on any eigen space so the action of Z is trivial on V so the map from G to GLN over C is trivial on X of Z which is the set of group elements of this form. So there's no faithful finite dimensional representation of this group G. I'll just finish by mentioning the fact that here I sort of emphasized we were in characteristic 0. So in characteristic 0 if you've got matrices X Y such that X Y minus Y X is equal to the identity matrix. This implies if we're in characteristic 0 the matrices must be 0 dimensional so this is really not possible in characteristic equal to 0. I'll just point out that the condition of the characteristic 0 is actually essential because in characteristic P greater than 0 we can take X to be D by DX and Y to be multiplication by X on the space V which is spanned by 1 X X squared up to X to the P up to X to the P minus 1. We notice that D by DX the power of P is equal to 0 so this is why we can do this in characteristic P but not characteristic 0. Okay that would be all for the about nilpotent Lie algebras for the moment.