So good morning and welcome back. Today we continue our study of the special linear group

SL2C, so the Relativistic Spin Group, and in particular of its Lie algebra. And today

we would like to conduct the analysis that we did in theory in the second previous lecture.

We'd like to do it for the example of the special linear group and its algebra, namely the construction

of the Dynkin diagram. So last time we arrived at the point that we constructed from the

coordinate-induced tangent vectors d by dx i at the identity. So those are elements of the

tangent space at the identity element of the special linear group SL2C. So this is this guy.

We constructed from these and where i runs from 1 to 3 the left invariant vector fields on all of

SL2C by virtue of taking the left translation to a point ABCD and the push forward thereof. So we

apply this to the first, second, and third tangent vector at the identity and we found already last

time that this gives rise to the following vector field left invariant vector field on SL2C A minus

BC 0 A 0 B 0 and this is D but in the coordinate chart chosen the first coordinate chart we

constructed this is 1 plus BC over A but this is just a symmetry of numbers where M is the row and

i indicates the column and this is to be multiplied by D by dx i but now at the position at the point

ABCD of the underlying SL2C manifold. So from a tangent vector at a point from three tangent

vector at the point we construct three right, three left invariant vector fields of this form. So why

did we do this? Well we're working towards the following so we have the we have the theorem that

taking the tangent space of some Lie group at some point that tangent space can be equipped with a

Lie bracket which here for distinction from the differential geometric Lie bracket I indicate by

this double bracket symbol that as a Lie algebra so not only as a vector space but also as a Lie

algebra this is isomorphic to the set of left invariant vector fields L of G on a Lie group

equipped or which inherits so this is a the set is certainly a subset of all the vector fields on

the Lie group so a subset of the section of the tangent bundle of the Lie group which is an infinite

dimensional vector space over R or C whereas this is a as we saw a finite dimensional vector space

but this L of G the set of left invariant vector fields inherits a bracket from the differential

geometric Lie bracket here which is this one and it's of course much more handy to deal with the

tangent space at the identity and to equip it with a bracket in order to mirror this behavior here

but here this bracket as I said we get for free because the differential geometric Lie bracket

whereas on this side this bracket we need to induce from over there and so of course how that

works in this context is that we define the bracket here d by dx I which is supposed to be a vector at

the identity and d by dx J also vector at the identity with respect to say the first coordinate

chart and we define this bracket by its action on the function to be given by now here we have the

differential geometric Lie bracket as indicated above we push forward the d by dx I vector to a

point ABCD in the manifold so if you are really picky we indicate here that this is a vector at

the identity that's being pushed forward and again L ABCD push forward of d by dx J at the

identity here like this and now the differential geometric Lie bracket closes we decided to define

this by action on an arbitrary function f which is a smooth function on SL2C okay but now this of

course yields the derivative of f at the identity this yields the derivative of f everywhere last

time I indicated that one possibility to remedy this is that before we act on f we actually have

this pushed the opposite direction but an even easier way to obtain the relevant vector at the

identity is to evaluate the whole guy here so this bracket to evaluate it at the identity before we

apply it to the function f so we get a vector at the identity and in order so this fully defines

this lie bracket on this tension space but in order to study the corresponding structure here we

definitely need to to calculate the right-hand side so unfortunately at this point there's no

better method than to just carry through the calculation so we need to calculate the following

three brackets we need to calculate aha okay so yeah need to calculate the following brackets d

by da at the identity with d by DB at the identity where I now use the fact that in the coordinates

ABC the first coordinate chart of course the x1 stands for a the x2 stands for B the x3 stands

for C it's convenient to label it here in the general formula like this but once we start

calculating these brackets it's easier or more transparent to label them by the actual names