 So, last time we saw the basic motivations for why we have to treat gauge field theories carefully. So, as we know the quantization goes through canonical structure and we saw that pi 0 a are all equal to 0 for any gauge field. So, right in the notation in the notation that the gauge potentials are written as a mu a T a and T a are generators I think no. So, this pi 0 just means that for the conjugate to a 0 are all 0 all the internal group index in the adjoint representation. We also saw that in many a sense the a 0 are also 0. So, a so called Coulomb gauge the Lorentz gauge is almost always imposed and it is a class of gauges which is simply that d mu a mu equal to 0 and here we would have to say a covariant derivative that is right. So, in the electromagnetism we said this equal to 0 and in non-Abelian case. So, this is a covariant gauge and within it one further puts the restriction that a 0 are equal to 0. So, the Coulomb gauge which is a subclass further imposes that a 0 a are all equal to 0 equivalently a 0 equal to 0 which leaves behind d i of a i equal to 0. Now, this we can see is a essentially a space derivative there is no time involved except that it is a differential condition it involves derivatives of fields it is there only space derivatives. So, it is at a particular time. So, if you actually think of taking derivatives as taking difference of fields at nearby points it is just a condition on values of fields at a on a particular time like surface right. So, if t is going this way we have some domain in the space like part and we are just saying that this has to be imposed here on a particular time like slice. So, you code in principle invert this and find out and just replay. So, you can solve for this and put it permanently in the dynamics and not have to worry about it later. So, in principle can be solved not that you will do it in practice well actually it is done also I should not say. So, in, but I do not want to get into too much of the detail formalism can be solved for a i in terms of space like Green's function for the Laplacian this is of course, a covariant gauge covariant Laplacian thus retain only physical degrees of freedom. So, I will give this as exercise that is one way, but it is a bit cumbersome and ultimately it has to do with very specific gauge. So, the general approach is to recall what we were doing last time that there exists. So, what is the general lesson we had that there are some coordinates that can be superfluous further there can be also space like differential conditions on the fields. So, we note that there are constraints in the phase space which can be taken care of by imposing gauge conditions. So, we say that C A of q p equal to 0 are conditions are constraints and some gamma A of q p are gauge conditions. The requirement that the imposition gamma A takes care of the condition C A and leaves behind only canonical independent set. So, we went through this last time. So, recall from last time things like what did we say C A new canonical variables which are you treat the C A as some coordinates and then q star q n minus n star as the coordinates and some p 1 p 2 p n and then p 1 star p n minus n star as the conjugate momenta where p i are found by inverting gamma A of written in this language right new variables. So, you have to invert these and find from them the p 1 2 p n there will of course, be A equal to 1 to n there will be as many conditions as there are constraints and that will be that many constraints. So, A runs from 1 to n here not to be confused with the A of the gauge fields, but they are going to become the same in the end. This is more general discussion right. So, this is sort of formal we do not know how exactly you are going to do it, but we assume that it is doable. Then we know that then the true we can construct physical vacuum to vacuum amplitude as a path integral only over the true variables well. So, put a j there and then we can say i s plus integral of all the dynamical variables right q star times j is keep it simple. So, that would be the vacuum to vacuum amplitude. So, you would modify it appropriately by introducing sources and so on. So, but this does not all I mean because the detail is not my point what we are going to do is in practice this is not what we do. So, what we do is right. So, I think we were writing the external current language only for phi version without the pies having been integrated out. So, that is why got concerned about the notation, but here we want to written both phi and pi. So, what we do is we. So, let me return to phi pi language or well we can continue to write this. So, we write omega plus omega minus as integral over all the d q's and all the d p's, but what we do is introduce a delta function of the gauge conditions and we also need to restrict the p's. So, we want to restrict the p's such that the gamma a are to 0 times x i s of some q's and p's. Now, there is a simple argument that says that the imposition of gamma a is as good as saying. So, yeah. So, we in fact, write this is integral over the q star p star then d of c if you like. So, we pretend that we made this canonical transformation from the old q p to the the symbolic ones and the the formulae p minus f is whatever you get from inverting the statement gamma a equal to 0. So, call this thing as we already used star somewhere, but so this is so this equation f a equal to 0 is a rather sorry. So, this equal yeah. So, p a equal to f a that is what this p a equal to f a of all these fields there is c and there is gamma, but gamma is equal is being set equal to 0. So, think of this as the formal inversion of the conditions gamma a equal to 0 I think this is enough to write. And so, we are imposing that here. The thing to remember now was that in order that this transformation does actually distangle the constrained and the unconstrained things correctly is that the Poisson bracket of C a and gamma a has to be non-zero. So, in a sense they remain they would after due rearrangement of the gammas into p's into this p equal to f a the f a would become conjugate to the C a. So, that requirement was that the gamma a and C a have a non-zero Poisson bracket. So, we are equal to 0 in non-equal to 0. So, that they remain the as a matrix treated when treated as a matrix, treated as an array labeled by a and b or in other words the determinant of this should not be equal to 0. So, this is same as yeah. So, being the same as determinant of well you can work out this Poisson bracket it is same as d gamma by d p equal to 0 what is happening to me yeah right. So, that for the invertibility the inversion requires that this has to this determinant has to be non-zero and which ties in with our need that they become canonically conjugate to each that some combinations of them can be made into mutually canonical. We now argue that the d d p d delta function is the same as an equivalent delta function integral over the gammas well this is not what we. So, we stipulate that that this that because the p's are found from just inverting the statements about gammas. So, this would be true up to this is in the functional space it would be true up to some overall constants right which are not important. So, you can take it as a stipulation where we assume that the gammas can be normalized correctly or some overall norm so that this can work out correctly ok. So, up to a the point is that whatever that adjustable factor is not dependent on the phase space ok. So, let me write it here. I should tell you though that all this is all this is a way of justifying to oneself what one is doing and some of these things may have strange flaws these arguments are slick, but they may have flaws and in fact, there is this thing called gribo ambiguity which you should try to read on your own. So, they may not always be quite correct, but turns out that so people do identify them then fix them later and so on, but for the time being we assume that this works. The gribo problem by the way has never been completely resolved, but people just live with it yeah. So, this is if we assume this stipulation then we claim that this delta of P a minus F a can then be taken to be times this determinant d gamma by d p the Jacobian right transferring the measure here. So, we can equivalently think of this was some measure times some distribution. So, equivalently we take this to be the Jacobian of the transformation. So, this logic is correct right that I have C a equal to 0 emerging from analyzing the equations and I realize some of it is not dynamical. So, I say that oh that means, have to put this constraint and then to say that. So, the gauge condition is something independent of it. So, this is the picture I tried to draw last time and failed I do not know that I will succeed this time. So, there is some constraint surface and what we want is that that genuine path integral does not end up traversing this surface because if you in integral because these are all equivalent only one of them is a representative member. So, you somehow want to impose a gauge condition that picks out at trajectory that cuts this only once. So, at least in this way where although I have fashionably drawn many axes restricted to three-dimensional. So, it would have to be the co-dimension of the C a. So, this is the condition gamma a. So, if the condition gamma a such that it cuts this C a only once, but does cut it which is what makes the determinant non-zero then they are not independent then it will be. So, this is the geometric picture. Now, that you have the picture you do this clever set of arguments to convert these into canonical variables mutually canonical by finding F a which invert the gamma a and then declare them to be the conjugates of this q's yes. So, the identical copies are perpendicular to this. So, there are lot of these which are the redundant ones. So, that equal you could have set instead of 0 you could have set 5 or something which would all be equivalent and one has to write. So, let us think of specifically we said. So, if I had the amuse let us just say there is a 0 a 1 and a 3 a 2 and we said a 0 equal to 0 it means that I am restricted to remain in the a 1 a 2 plane. To do that I well in this case of course, it is trivial, but in principle what it requires to do is to impose the gauge condition that sorry because this is not a constraint this already is the gauge condition. We have to say divergence a equal to 0 gauge transformations take you within a particular set of that is what I meant here that gauge transformations would leave you within this. I think this is correct picture, but we can discuss it later because right now I mean the another flow of thought. So, but I just want to answer at least algebraically without the picture which I right now do not remember how it works, but that we are trying to convert the existence of constraints and the gauge conditions we place to repair them. Yeah. So, the gauge condition was divergence p equal to 0 pi or which is same as divergence e equal to 0 and to take care of that the gauge condition was divergence a was equal to 0 within the Lorentz gauge. So, these are the two things that are proposed to be made mutually canonical. So, there is a geometric picture of what the constraints and the gauge conditions are and then there is the canonical picture where you try to move from the geometric picture to the canonical picture and in the process you claim that imposing the delta function for these formal p's that were meant to invert gamma. So, the great advantage of this is that we actually get rid of the p's completely in the end. So, we now answer that this is same as the determinant of the Poisson bracket and we thus get rid of the superfluous p's which were placeholders to think through this thing. We need the matrix m a b which is variation of the gauge condition with respect to the gauge transformation that is what this boils down to. So, for example, what is the idea? For example, we had the gauge condition divergence e equal to 0 and, but under gauge transformation e tilde equal to old e minus da dt we are talking about transformations of a. So, a tilde are equal to old a minus d lambda by dt. I can only remember the covariance statement. So, a tilde equal to a mu minus d mu lambda. And for the non-Abelian case it is d mu. So, let me write the Abelian case and then the non-Abelian one as we struggled and realized one day delta a mu is equal to d mu of to the leading order is same as d mu of lambda a. So, it has d mu plus f a b c. So, even the infinitesimal one has the imprint of non-Abelian gauge transformation. Actually, this is why this gribov ambiguity arises.