 So, what we learnt about these is that we have matter fields in some representation of group G you know Lee group G and we introduce gauge potentials which belong to algebra of G often it is written with some kind of gothic character. So, and I think I have to add a G here A belongs to U sorry belongs to G no sorry not here because we have already defined A belonging to the algebra. So, you exponentiate you get the group elements, but then what we say is that the so called local gauge transformations are that your matter multiplied psi of x is equal to and now comes the main difference U of x times psi of x where U is in the appropriate representation. So, this is the simplest cases when psi is just in the fundamental or in the vector, but and then we said that the covariant derivative which we define to be d mu of psi and this is where I have to worry about the where the G goes probably from here itself I think I have to normalize the A correctly we will just see. So, there is the geometric point of view where you do not worry about this coupling G, but in physics we do need this G. So, better put it this is good because this is certainly very physical. Now, this has the property that same way as psi itself this same way as is what is meant by covariant. So, it transforms the same way psi provided a tilde of x is equal to U of x A of x U dagger x and this is minus I think d mu of U U dagger, but there is a 1 over G or something like this here which we can quickly try to fix and as you have the notes with you minus I over G, but there I took the psi to be e raised to I G times T. So, this is not quite like this, but it is. So, I do like this particular definition of covariant derivative with the I and the G there that makes everything physical and Hermitian and so, one has to put some minus I over G over here. Yeah I think this is correct. So, those three things go together the package deal that if psi goes like this and if A tilde A goes like this then this derivative will be covariant it will transform exactly like application of U to it. Now, we come to the equations of motion which were. So, we learn that d mu of f mu nu is equal to J mu J nu where f mu nu is written d mu d nu, but it has the detailed expression f mu nu A. So, this f mu nu is in this notation. So, it is also algebra valued f is also algebra valued and well this will also have some algebra valued currents. So, if we then look at component wise then it becomes d mu A nu A minus d nu A mu A and then I G times f A B C A mu B A nu C. So, the other part is Faraday's law which is not going to be all that important right now which is this where you permute these cyclically is equal to 0 that is sort of trivially satisfied because of the way f is defined. So, that is not a new input because we are treating the gauge fields as the fundamental thing the gauge potentials as the fundamental dynamical degrees of freedom a minus I by G. So, the thing is that for the purpose of doing all the important manipulations it is useful to use the gauge potentials. However, the gauge potentials have a huge ambiguity in them which people call the gauge degree of freedom, but at the same time it is actually lot of people call it redundancy of the description ok. So, gauge both describe the same thing we call it freedom because we want to make a virtue out of it because it allows us to make a prescription that all of physics is determined by. So, the symmetry freedom etcetera are meant to emphasize the restrictive power that L can power of the principle which is that the Lagrangian can only be a functional of f and of psi and of d psi these are the only things that it can consist of. So, the fact so that is the power of the gauge the covariant derivative only the covariant derivative can occur not ordinary derivative and that brings up the most important physical thing that somebody asked last time which is that remember the psi is a multiplet it is either a vector or a matrix in some representation of the group G for all of those there is only one coupling constant small G. So, this is called universality. So, whether you have neutrino which has a mass 0 as far as weak interactions are concerned or you have the top quark which has mass 176 G E V their coupling to the gauge bosons W plus minus and Z is exactly the same is the same number ok. So, the entire generation of fermions all have exactly same G. So, that is the power of this principle and that is when people try to think of positive words like it is a symmetry it is a freedom in the gauge fields and so on. But if we now begin to do the dynamics then we see the nuisance that this creates and that is where we observe that it is actually a big redundancy in the description. So, we begin with that now. So, note however, in the dynamical description firstly. So, the thing is that it allows you to define the Lagrangian very elegantly, but once you have defined the Lagrangian. So, all kinds of superfluous terms that you would be worried whether to include or not are gone not only that lot of the terms have exactly same coefficients. So, it is a very powerful restriction, but now you come to the dynamics and begin to look at it. The first thing is that the zeroth component of this equation nu equal to 0 requires that that d mu of f mu 0 is equal to 0, but you are or whatever the current is. So, it does not matter if it is non-zero. So, source free case just to keep things simple, but the assertions we are making about freedom versus non-freedom are all the same because this is some given current. So, what this means is because f is anti-symmetric if this index is already fixed to 0 this mu can only be space indices. So, this is actually the same as d i f i 0 equal to 0. Now, this is essentially an identity. So, it contains no dynamical information all the dynamical information was supposed to be here, but now you notice that the a 0 never has a second time derivative a 0 a have no sec in fact, from here we see that they do not have no second time derivative. In fact, there is no time derivative here at all it is all space derivatives. So, a 0 now this equation is nothing, but Gauss law because you know that f i 0 are the electric field components. So, this just says d i equal to 0 or d i v constrained to be whatever source you have. So, this is not a dynamical equation constrained on the initial conditions of the PDE. You have a partial differential equation to solve. So, you want to set up fields and its derivatives on the initial surface. What this says is that you cannot do the special derivatives arbitrarily the x dependence if. So, you are going to be having have time evolution starting from some surface at t equal to 0 let us say, but the you have to set up e fields here and you also have to set up there you set up the full e field and the time derivative, but the gradient of the e field in this also is constrained it is not arbitrarily. Now, let us look at this from a canonical point of view because pi i a let us go to the index notation I mean individual algebra components would be equal to variation of the Lagrangian with respect to a i a dot right d l by d q dot. So, the q here is the ith component of gauge potential and internal index a, but then Lagrangian is equal to quarter a nu a minus d nu a mu a plus the f a b c term and the whole thing squared effectively right it is Lagrangian is that squared. So, if you try to vary this you have to search where the time derivatives are and if you set mu equal to 0 t 0 then you get the a i dots because right I mean. So, you get nonzero and so this is ok. So, let us look at e and m first f mu nu Lagrangian is quarter f mu nu f mu nu and this we can write as equal to quarter f mu nu sorry quarter d mu a nu minus d nu a mu and leave this untouched because it is actually the same thing. So, if we vary d l by d a mu dot then that comes only from here right. So, firstly it removes the this is square of f. So, it is same as saying one half of f mu nu times variation of right derivative of this with respect to anything is 2 times the derivative of this times derivative of that. So, I put this out and then derivative of this, but that I write out as this and then it basically says that it is either this or it is this depending on whether I choose the. So, we could have taken this as rho ok. So, if nu is equal to rho then this term contributes, but if nu is equal to mu then this term contributes and then this has to be the time derivative, but it is anti symmetric, but this is also anti symmetric. So, you get twice the contribution once from here once from there which removes that half and you are essentially left with f f mu 0. So, let me explain. So, this means that one this is same as d 0 of a rho right. So, now when varying you have to set mu equal to 0 and then mu equal to rho only. So, the answer will come out proportional to delta mu rho delta mu 0 delta nu rho minus delta nu 0 delta mu rho correct. That is what the this variation will be proportional to and when in fact, that is the that is the answer because when that tell is if mu is equal to 0 and nu is equal to rho I get 1 because I am varying this with respect to itself or minus that, but because the. So, that sets mu equal to 0 here I put nu, but it you take the other term. So, it sets mu equal to 0 and sets mu equal to rho. So, we have to put rho there. So, I should put 0 take the first term because that is the positive sign. So, that is right. So, that is what the answer will look and the first term gives exactly this and the second term is actually the same after you exchange the index here indices here. So, and now you see that this does not change if you go to the Yang-Mills case because in the Yang-Mills case this term is not does not contain derivatives of A. So, it is not going to contribute. So, you vary f squared. So, you are left with this. So, you add here f abc if you like, but times the whole f mu nu A and then the variation then enforces these indices. So, in the Yang-Mills case it will be f 0 rho A that is the only difference. So, we find that the canonical conjugates to the gauge potentials are the f 0 mu components, but that means that A 0 has no canonical conjugate pi mu yeah pi mu A are equal to f 0 mu A, but that automatically means that. So, that is why the A 0 is completely redundant it has no dynamics really it is derivatives do not time derivatives do not appear, but now we get the additional constraint we have the constraint that d i of f 0 i A equal to 0. So, there are as many equations as there are A range of A or which I meant to say really d i of the canonical language pi i. So, in the canonical dynamics the even the space like momenta are not really completely unconstrained. So, Dirac the great formalist tried to put this in a elegant language and then he classified that things of the type A 0 should be called second class constraint. Some part of British value system is here what is really superfluous we call second class. On the other hand when the constraint is implicit we call it first class in particular you have to check that in the course of dynamics you do not violate it. So, it is not only a matter of setting initial condition, but you have to maintain that at all times that because it is. So, that is why we call it a constraint it has to be carried along and this Dirac calls first class. Here we have yeah. So, a property of property of first class constraints is that dynamical independence C A C B Poisson bracket should be equal to 0 and which is true of this because you can think of each of these as a it is a bit unfortunate I should call it something else this is a more general statement. So, here A B etcetera equal to A comma B etcetera and we can see that it holds because canonical bracket of any particular pi with any other pi is nonzero. So, in the Poisson bracket you pull the derivatives out the derivative is a well there is a linear operation if it was ordinary derivative in the case of this ok. So, it is actually not all that trivial in the case of young males of course, it is not a linear operation, but one can check that it will work out ok. So, this is a non-trivial exercise because you will have to use Jacobi the Jacobi identity of the F A B C's to make sure. So, this equal to 0. So, this is an exercise. So, for electromagnetism it is obvious, but that there is only one constraint. So, nothing to check.