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in the last video we defined a local inverse for the exponential map so remember the exponential map takes any matrix and gives us an invertible matrix so it defines a map from little glnr to big glnr and what we did was we found a neighborhood of the origin called u inside a little glnr and a neighborhood of the identity v inside big g lnr and we found a map called log that goes back from v to u such that x composed with log is the identity log composed with x was the identity and the useful thing that we wrote down at the end of the last lecture was a power series expansion of log so it's log of the identity plus x equals x minus a half x squared plus a third x cubed etc so the point is anything inside v can be written as identity plus a little bit because v is a neighborhood of the identity okay so in an earlier lecture we proved this nice fact that if a and b commute uh which i'll write as uh a b equals b a then we have the law of logarithms that x a times x b equals x a plus b and what i want to discuss today is what happens if a and b don't commute well it turns out we can use this formula for the logarithm to give a formula for the logarithm of x by x b so let's let's just try and compute right if if if we look at this formula this is saying log of x by x b equals a plus b so what in general is log of x a x b well let's compute so it's log of um identity plus a plus a squared over two plus dot dot times identity plus b plus b squared over two plus dot dot dot so let's multiply these two power series together this is log of the identity plus a plus b um plus a b plus a squared over 2 plus b squared over 2 and then we get terms like a b squared over 2 plus a squared b over two and then we get things like a squared b squared etc higher order terms okay but the point is this is log of one plus something and if a and b is sufficiently small then this power series for log makes sense converges and we can apply it [Music] so log of 1 plus all of this stuff is um well let's let's give this stuff a name let's call it x all this stuff in brackets so a plus b plus a b plus a squared over two etc this infinite sum is x so this is x minus a half x squared plus a third x cubed minus dot dot dot and now we can just substitute in x so we're going to get a plus b plus a b plus a half a squared plus a half b squared and so what i'm going to do is in in the following i'm just going to keep the terms up to second order i'm going to ignore everything of order 3 and higher and just write it as dot dot dot so this is x plus dot dot and then okay minus a half x squared what do i get if i square this whole mess in brackets well you know i get a lot of cross terms i get things like a squared a b i get a b a i get a b squared and i get a lot of stuff that's cubic or higher order and then one third x cubed well everything inside x cubed is at least cubic so i get dot dot okay so we're for now ignoring the dot dot dots so what is this this is a plus b so that's the sort of first order term so if a and b were to commute this would be the full answer and everything else would cancel but you know that that's not what happens what do we get well there's a half a squared here and there's minus a half a squared here so they'll cancel there's a half b squared here and minus a half b squared here and they'll cancel so all that's left is a b minus a half a b plus b a and then of course there's the plus dot dot dot so all the cubic and higher order terms okay so that's a plus b plus well let's see if we evaluate this sum here we get a half a b minus ba plus dot dot dot so this quantity here a b minus b a is going to turn out to be very very important in this course so it's got a name it's called the commutator or the bracket of a and b it's written like this a b so in square brackets with a comma between them this is a b minus b a let me put a box around this because it's so important and you can see this vanishes if a and b commute which is why it's called the commutator so what we're saying is to first order x bay xb is x a plus b but then there are correction terms coming from the commutator and actually the cubic and higher order terms miraculously can all be written in terms of commutators of a and b so iterated commutators so for example if we kept all these cubic terms cubic order terms like a cubed and a b squared over two things the next term in this expansion would be um one over 12 a bracket a bracket b so minus 1 over 12 b bracket a bracket b so when i say a bracket b i mean square bracket a comma b close brackets so these are called the brackets after this guy surface li but they're also known as commutator brackets right so the miraculous thing about this formula is every term can be written as stuff involving a and b and brackets and that's it there's no a b right you never multiply the things together you only ever take brackets and this is a miracle it's not at all obvious that this is true and there's you can write a formula for the general term it's not particularly illuminating so it's going to be an exercise for you to er to check this cubic term is is correct and also to figure out the uh the next order term the fourth order term um but even having done the exercise it's i think not going to be clear to you that the general term should be written in this form as sort of a bracket b bracket blah blah blah blah blah it is a theorem and it's called the baker campbell hausdorff formula so this is the this is the formula that says x a x b equals x of a plus b plus a half a bracket b plus higher order terms where the higher order terms can all be written in terms of the bracket alone okay so there is an explicit formula it's actually not due to baker campbell or house stuff i think it's due to dinking but it's usually called the baker campbell house store formula um so we're not going to prove this um one of the projects if you wish to do it is um to prove this this formula but somehow while it's a very exciting formula it won't get used very much in what follows apart from this first correction term this half a bracket b term so the main point of this formula is that the group multiplication between these invertible matrices x bay and x b in big g lnr is determined entirely by the brackets operation between a and b as matrices in little g and then are so this is going to be a general theme that the multiplication in the group is determined by the brackets inside the x so on what's called the lee algebra