 The first thing I wanted to advertise was this book which I think I had seen some time ago, but somehow never noticed it. But now I find that it is an exceptionally well written at least it contains a lot of the things that would go with what we call modern field theory. So, it is called by V. Permeshwar and Nair in Springer 2005. So, not that much has changed in this level of field theory in 14 years. So, do not have to worry. So, I think that is also good supplementary text and in future I would include some topics from there. Now, let us continue with gauge fields. So, we have some matter multiplet. This is as I said the original quite a bit of a revelation that you can actually think of nucleon as essentially two spin projections of the same code particle. This is the innovation due to Heisenberg and so, matter multiplet means in some group representation for example, proton and neutron as spin 2 of SU 2. And this is what we will deal with this is the simplest case. You can deal with higher representations by a simple generalization of group theory rules how the representation transforms. So, then we say gauge fields for example, for SU 2. So, introduce gauge fields as many as the generators of the group as you know SU 2 has representation 2, it has 3, it has 5 whichever representation that is the number of gauge fields needed is always the same it is just the number of generators. So, for example, for SU 2 we need triplet tau over the tau a are generators of a. So, there are as many a mu fields as the 4 vector fields as there are as many as the generators of the group. Then we introduce covariant derivative and if you like there is an identity matrix multiplying this d mu. Now, this has to do with how the psi transforms under gauge group. So, here it is because psi just transforms as u times psi, but a higher representation say higher tensor representation well let us not put this ok equal to u psi because psi is fundamental. For higher representations the action of d mu has to be suitably defined. The simplest mathematically the simplest way is that the generators belong to the adjoint representation and you can always write adjoint representation in the basis that is of any dimensions which is irreducible representation of the group. So, you have to write. So, write the irrep as d dimensional column vector and write generators T a as d by d matrices. This is more easily said than done, but this is at least mathematically a correct way of saying it. More generally people use you know tensorial representation that is they write the higher rep actually as a cross product of the fundamental you know like you write third rank anti symmetric made out of the lowest psi and so on. Then you have to remember to apply the T correctly to that anti symmetric representation, but I think this is the simplest recipe for a d dimensional irrep write a d by d representation of the generators of. So, since T a. So, you can do this for any representation of the group, but this recipe works therefore, writing a column vector and hitting it on the left trick works. The fourth thing then is this was 1, 2, 3 and the fourth thing is that with the introduction of covariant derivative we now note that the Lagrangian should be constructed to we build in the as the dynamical symmetry by insisting that the Lagrangian enjoys is invariant under transformations can someone just check what it was u a mu u dagger is that right. So, the above requirements then entail that the Lagrangian is made up entirely of gauge invariant combinations of psi it is covariant derivative and a field strength tensor to be made up out of the a mu. Here I just want to quickly comment that note the infinitesimal version, version is delta a mu a equal to d mu of epsilon a and then plus f a b c a mu b epsilon c. We next need a field strength corresponding to this a mu and this simplest choice turns out to to be to write d mu a mu just put covariant derivatives. So, the correct choice turns out to be I am just trying to check the science. Note that this can also be written as so, consider. So, we try to check we can see that d mu d mu we already know is a covariant quantity it will transform like d d tilde will be right it will transform like this you can take the u out of the well u inverse u d mu u inverse u dagger. So, if you take d mu commutator d mu then it is just equal to you can see that when I insert u u dagger u u dagger I will simply get, but now this d mu already contains a mu as well as derivatives. So, it contains i g a mu times d mu. So, I am writing this term first plus well d mu d mu is there which is not so important it will cancel and then d mu plus i g times d mu a mu and then minus g square times a mu a mu and let us do the same thing in reverse here. So, d mu d mu minus these terms cancel and we have and then minus g square. So, there is this factor of g 1 s 2 take care of and then remains the action of the derivatives on the next things. So, in this f mu nu definition we have to right. So, if I remove a g factor i g 1 over i g factor I should come out with. So, take out i g and put it here I will get this term and then with i g removed there will be an i times g left here and this minus sign will become plus sign that is also correct. So, we have to put a g over here. So, we write this as this is nothing, but d mu of acting on whatever is after a mu and. So, we write this as now we can write it out in a more convenient form by you know you invoking the. So, in the yeah. So, introducing and you do not have to write this that is just another way of thinking about it, but it is equal to d mu of nu a minus d nu of mu a and then. So, I forgot to write I meant to write introducing T a where T a T b equal to some f a b c times f a b c are the structure constants and we use the normalization for T's trace of T a T b equal to 2 times delta a b. This 2 is was introduced in early literature because of SU 2. So, yeah. So, but the so this is the normalization we can see that trace of T's are of course, 0. So, write like this in the case of SU 2 the structure constants are i times epsilon a b c. Now, what we have therefore constructed is a space time dependent gauge transformations I mean which is what we mean usually by non-ibillion or Young Mills transformations compared to global transformations right where u belongs to the league group and all the fields yeah. So, league group means that u u inverse is also u u inverse is also in the group and all the physical fields are algebra valued. Now, as I have been saying the greatest news of this century is that this mathematics suffices to describe all the fundamental forces. So, the Lagrangian for the system for gauge fields I will leave the front out first we said trace f mu nu f mu nu. This we would generalize from Maxwell's right because we have f mu nu f mu nu and there we put a one quarter because there is to get canonical one half for kinetic energy for the space like fields time like gauge time like gauge component a 0 does not have any kinetic energy anyway. So, only the space like ones have and we need minus sign because they come with minus sign because of this signature and we put a one fourth to make it one half in that case because there is a doubling here. Here we have to put a one eighth because there is an extra two coming from the trace of the TATB which are inside this. Additionally I remember seeing a 1 over g square in front of these. So, that happens if so, you will also see in textbooks yeah. So, in this f mu nu definition sometimes this g is not put in that case you have to supply a g here to recover the physical f mu nu's right. And so, this is also equivalently equal to minus a quarter f mu nu a f mu nu a where summation over a is understood. So, this correctly generalizes the Lagrangian from the Abelian case and for the matter it has to be d mu psi star d mu psi or d mu psi dagger because if you think that dagger is a matrix as well as a Hermitian conjugate. So, you have to put a dagger. So, it is d mu psi dagger the dagger of the whole thing which means that it will go and hit the i in the covariant derivative the i g d mu minus i g that i will become negative because of this dagger. So, that the whole thing remains Hermitian. So, here psi is of course, a scalar field for scalar field psi it will be like this or by that I mean yeah scalar and should be simply psi bar d mu gamma mu psi for spinor or spinors in psi let me put it like this scalars in psi. So, this psi is a multiplied some matter multiplied if you have several scalars space scalars space time scalars then you will have this this is like the Klein Gordon d mu phi star d mu phi except that now you have to put covariant derivative acting on the phi. So, this is how the Higgs kinetic energy will look like Higgs field which is a doublet under SU 2 its kinetic energy looks exactly like this whereas, all the fermions it will be like the Dirac case where now the bar has to apply to each component and the gamma mu is of course, attaching itself to each. So, it is a hybrid notation there is a group theory some group theory contraction and there is a gamma 4 contraction. So, there is a gamma 4 for each of the euro and column I think it is best to write it as and sorry because it is a bar you probably want to write like this this is clear right it is psi bar. So, this should be simply written as gamma mu psi 1 gamma mu psi n let the gamma mu stick to the psi then you would not go wrong. So, the group theory multiplication then will be psi 1 bar gamma mu psi 1 psi 2 bar gamma mu psi 2 etcetera. This is the group index left to right or up to down is group index and then individual it will be a Dirac spinner right. So, this is the easiest way to remember how to write it and of course, you can write mass term you know as per Lawrence properties, but no mass term for gauge fields. However, there is a big mischief because the gauge fields are now highly non-linear. The U 1 gauge theory was essentially a linear theory, but here the amuse coupled to themselves because of this structure of F mu nu where it has F contains not only F these two terms are linear in A, but this term is quadratic in A. So, the gauge field is non-linear in the field strength is non-linear in the potentials and this creates a very complicated geometric structure which when you are lucky can be delineated by some very clever geometric arguments, but if you are not lucky then you just have to suffer it.