 I now want to make some particular remarks. One is that if you try to do perturbation theory using this. So, that is how this non-linearity becomes important in the diagrammatics. The perturbation theory shows that the coupling G is asymptotically free as it is called G grows weaker at. So, firstly from the F mu nu F mu nu terms such as d mu A mu times F right. So, and there is a G in front of them somewhere we did not put a G here I am completely confused about G because I have never really dealt with this theory in any detail. I know the general properties. So, there is G somewhere there in the non-linear terms that has to do with proper definitions of things yes. So, you remember we did d mu d nu and we got I G for the first terms and that got G square. So, there is a G multiplying this. So, I will tell you very frankly several books will not display a G when they just want to discuss the geometry and the underlying structure. If you want to do perturbation theory then you have to restore G here. So, if you do this then there will be a G here and a G square here. So, if you want to do perturbation theory then you have to introduce this G because what even the so called free field theory or the simplest possible theory you can possibly write for the field strengths or for these group algebra valued potentials A mu the kinetic terms come packaged with the interaction terms you cannot. So, by kinetic term we mean things that are just square of derivative right x dot square. So, but along with d mu A mu d nu A mu will you will automatically also get stuck with stuff like this which is all non-linear in A the only possible gauge invariant Lagrangian you can write is packaged with interactions but fully determined this is not left to imagination it has to be exactly G and it has to be exactly G squared. So, you do not have one derivative and two undirivative with some one coefficient and four A's with some other coefficient. This coefficient is exactly G square and the whatever the values that F A B C give you. So, it is a big package deal and a little unusual from our conventional approach of taking free Lagrangians and adding interactions to them. It is a package deal in which interactions if you so if you want to think of the A mu as the elementary potentials which you quantize by saying A and pi A commutator equal to i h cross then those quanta are automatically forced to be interacting you do not have a theory of free quanta free non-ibelian quanta. So, there is no such thing as dot dot dot however if you start if you assume applicability of perturbation theory and treat these as diagrams. So, we can treat this as a diagram with three gluons where one of them because there is a derivative there is a P mu on it and G will be the strength ok. So, some you have to put some A and mu sorry and F A B C and then there is a four gluon diagram with that G square mu that has to be mu. So, P mu with an A mu and then in two so one A mu sorry so this is A mu ok. So, new A new C and then mu B that ties to that P mu etcetera ok. Now, if you use these as elements of perturbation theory then you find that the coupling G gets renormalized in a way that it grows weaker at higher momenta at higher momenta scattering amplitude. So, this is the second biggest discovery of and asymptotically goes to 0 asymptotically in momenta goes to 0. So, the first biggest discovery is to discover that gauge symmetry describes all the interactions, but the second biggest discovery is that the couplings actually go to 0 asymptotically. So, this led to coining the very mysterious term asymptotic freedom asymptotically the gauge field theory is a free field theory. So, the dot dot dot meant that no such thing is the little too strong if you can live asymptotically at infinite momentum then you have a free field theory, but then you have to live asymptotically and live here. And, but there is another clever thing that was discovered by Toft that so wait in parts. So, we draw this by saying that alpha strong you know which is equal to G square over 4 pi for SU, but it is SU 3 not SU 2 it runs like this. So, it is so put 1 over here and it is something small and of course, going to 0 in the so square in S you know S is the S parameter of scattering is a function of S it goes to 0, but then you have to you can only put dotted lines here because here perturbation theory begins to fail if it grows towards 1 then the perturbation theory fails. So, we do not really know what happens here ok what once it approaches 1 we do not know whether it becomes 1 in fact, it becomes meaningless because now you cannot separate you. So, the main conceptual problem here is not that it is a just that it is a strongly coupled theory it is not as if you had the same excitations M U which are now fighting with each other more strongly. It is that you do not even know what the degrees of freedom are if this theory is asymptotically free then you can see a free quantum streaming out and you can say oh this pin 1 particle or helicity 1 particle is what in the interaction region interacts, but when it is here you cannot even see what the ingredient of the theory is. So, really whether one can then separately quantize the a's and then think of interaction among them all that fails. You remember in the beginning of the semester we went through this Fock Dirac quantization where we said you can identify whether there are bosons or fermions and then if there are bosons the states get can be labeled simply by the number of bosons. All that logic was applicable to the free theory or very weakly interacting theory. So, that the interactions were local, but by and large the system was non-interacting. That entire picture fails completely and we do not actually know how to well. So, quantizing the system here is a dicey affair. Some light on this was thrown by Adhoff's clever observation that if you use large n. So, in this you take SU n group. So, the n of SU n you make it large then it turns out that that n enters somewhere here I should have written it in that form. So, the coupling itself becomes. So, in the diagrams all any loop of with any graph of n loops has a 1 over n in front of it. Therefore, in the limit of large n all the loops become subdominant and only tree diagrams survive. So, for large n only that the tree diagrams are dominant which is the free field theory. Well not exactly free, but because the tree diagrams are still there, but for a quantum field theory tree is supposed to be the free part of it. People think of it as essentially the classical thing. There is no real quantum mechanics itself that things can break up and combine, but there are no in a tree diagram there is no internal momentum to be integrated over right because the energy momentum conservation fixes all the momenta in the. So, all you have is a matrix element. So, tree diagram essentially is free fields although things can break up and recombine, but there are no quantum corrections. So, SU n theory is for large n you get free fields. In fact, I think only the planar diagram survive or something like this. So, this was a interesting simplification. The reason I am telling you this is that although we have a differential equation you know from that F minu F minu Lagrangian we can write equations of motion which I did not write I will come back to it. So, we can write the equations of motion for F, but it is meaningless to solve them because there is nothing called a classical Yang-Mills field. It is meaningless there is no state of the system. So, for photons we have plenty of classical looking states. In fact, they are fully quantum, but we can think of free streaming photons, we can count individual photons, we have you know children attach meters and measure potentials. You cannot do any such thing to gauge fields they actually do not have a classical existence at all. Anybody who is trying to solve a differential equation find an exact solution of it and try to think that he has actually found something physical is mistaken because it has no such interpretation. There is no such thing. So, you might try to say some state psi right in which I find this as an expectation value. There are no such states available in which you can take expectation value of your field and then say oh this is my classical value. In other words we cannot write any effective potential which is you know whatever. There is no such thing because Z of a classical it has no meaning in no way that you can derive such an effective potential out of the full path integral the functional of the theory. However, classical solutions do play a very interesting role and so, I will come back to it later. So, first thing is there is no such thing as free non-nibelian gauge fields and the second thing is that there is in fact, no classical gauge field because the only gauge invariant Lagrangian you can write for it comes package with interactions and if you put in the interaction then it is strongly coupled and there is nothing like that is classical. You can say oh, but I am going to observe it at only very high moment well good luck because you have to first get one to give it to give it high momentum, but you can never get one out. So, in reality the we only find them. So, this asymptotically asymptotic freedom implies confinement which is a funny word because then you at least presume that you know what is confined, but you do not know what is really confined. Well, just say quarks and gluons neither quarks nor gluons can be pulled out indefinitely because if you do it then you are going to infrared limit. If the point is to separate two things out and to see them separately you will have to go to large distances, but large distances usually mean low wavelengths which means the infrared this is the S parameter. So, at large collision energies you can think of it as weak, but when you are trying to pull it apart you are in this region and then you can never really pull it apart. So, this required a proof by lattice gauge theory. So, the only proof we know what Wilson did was to introduce. So, the lattice is primarily a regulator you introduce some separation you to not to avoid the ultraviolet parts. So, you have integrated ultraviolet parts you just have points in space and what he proved was that any two quarks the force between them grows as the area bounded by the flux lines that connect them. So, right. So, the statement is that the area of this energy of this grows linearly as the area. So, if you try to take contributions of all the possible loops then you get infinite answer. So, this indicative proof that so, and you have to you can think of some quarks sitting on this loop. So, I have already put the quarks, but he just proves it for gluons at between the gauge fields. Any gauge loop gauge loop its energy grows as the linearly as the area, but nobody has the full picture in the same language. So, the problem with lattice is a very peculiar one. So, once lattice gauge theory was invented people actually started doing numerical calculations because now you take the functional integral and instead of doing the dA you simply start doing product over. So, here you remember you had to do product over all space time points, but now you can do it over the lattice points x i right you can do this. So, people put it on a numerically on a lattice and then try to calculate. They get various answers, but eventually they have to extrapolate them to lattice spacing going to 0 to recover the continuum. What happens then is a very interesting thing because lattice is the real lattice is in real life. They themselves show phase transitions as you change the lattice spacing depending on what interaction you have put between the nearby between the members of the lattice. There are phase transitions. So, you do not know whether you are actually approaching a lattice phase transition or you have reached the continuum limit. So, conceptually there are some issues with lattice gauge theory. Now, so it is an unproven mathematical fact, but it is a well established empirical fact that the scattering the effective coupling grows weaker at high. So, this now in LHC of course, which is such a high energy they are mostly weakly coupled quarks and things that they did in early days can now be applied very freely by extrapolation. The energy dependent running is exactly as expected and so on. So, there is a lot of confirmation of QCD in this regime. We know that it is exactly like this. We also have never pulled out any free quark or glue on. So, the confinement hypothesis looks correct. Whether it mathematically follows from this particular annual symmetry is not approved, but it would be too surprising if there was something uglier that give the same answer because this is very elegant and has all the ingredients needed to explain it. The last thing I can do is we get to this topological part and the way to motivate it is that well. So, before we go on just the equations of motion are that no surprise in guessing. Renew is constructed out of the quark. So, it is psi bar gamma mu tau a right. So, it is. So, this is all like this. So, that is what the currents are. And the other associated equation is the Faraday's law and the and dive B equal to zero law is simply written by mu nu rho and then we put this on them. So, this means cyclic permutations. This notation is for mu putting this box means mu nu rho plus nu rho mu plus etcetera. So, this is exactly for the as in electromagnetism. So, structurally it looks very similar, but as I told you there are subtleties because there will be a mu a mu terms here. In the f there will be small f a b c a mu b nu c and so, there is no point trying to put solve this equation. But we will see some clever equations clever solutions that do exist. One other point is that the Lagrangian also contains another candidate term. So, recall in Maxwell case we had, but we also have e dot b equal to epsilon mu nu rho sigma f mu nu f rho sigma is also a Lorentz invariant. This we do not put in the Lagrangian by arguing that this the this is not parity invariant. If you as you remember in the poor man's language e is a true vector and b is a pseudo vector. So, if you do a space inversion this term will change sign ok. So, you do not want to include such terms in the Lagrangian. In the more sophisticated language here you what you do know is that the epsilon tensor is invariant under the flip of all the four f mu nu. It is the it is the volume element of that number of space time dimensions. The volume element does not change sign. So, it will not change sign whereas, the other things will change sign together. So, it is the same reason that so, it is a pseudo scalar it is not a genuine scalar it is a scalar, but it is pseudo scalar. Therefore, there is a determinant that comes in the flipping. So, we may not want to include such a term. Also in the Abelian case it turns out to be a total derivative. So, you can throw away the term by saying well it is something to do with things at infinity. But in the non-Abelian case that term on infinity also matters ok. So, it turns out that and actually you can choose gauge field configurations that are pure gauge, but they are laid out in space time in such a way that they cannot be contracted to 0 because you have a u as a function of x. So, even though the function even if also we will see next time we have space R 3 or we could have R 4 as well space time. From this we are mapping into this SU 2 space right you make some map from here to there. Now, this is a compact as we had argued last time it is like a compact ball which is an S 3. If you take this R 3 and treat all of its boundary S 2 as one point the boundary is S 2 right. So, if you treat S 2 as one point then this R 3 also becomes a three sphere. For the same argument we had here remember you start from the origin and you come out the outermost shall you identify with one point because it is e raised to 2 pi i right. So, it is just equal to by 2. So, it is just minus 1. So, outermost surface is equal to minus identity of SU 2. For the same reason if you identify the outermost in other words you map the outermost SU 2 exactly into this minus 1 then structurally this has become topologically same as this. And then the number of ways you can map an S 3 into an S 3 is classified by integers. If you start mapping you reach halfway point here you map cover the whole of this space. So, suppose some sphere here you map into minus 1 and then continue going out such that you begin to go inward. So, that when you reach out at infinity you have map back to 1. So, you start with one go there and map back there that map is distinct from when you start from here and end only here the entire thing maps here. And then you can play this game many times you can reach minus 1 then reach 1 then minus 1 then 1 and so on. So, you can map this into this in many number of ways which are essentially indexed by integers with positive and negative winding number. So, because of that fact even if you give me a pure. So, suppose I construct a gauge field a mu which is simply equal to minus i what had we written d mu u d u dagger or whatever right. This is a pure gauge field because the u a u part is 0 you did not start with any gauge field to begin with. Suppose you construct a gauge field that is entirely of this form. So, it is a gauge transform of nothing, but this u could be a non-trivial map. In that case you are stuck with a non-trivial mu which although it is pure gauge it has no physical field strengths it is not identical to it is not the same as the vacuum you started with earlier. So, this kind of things arise in gauge fields which we will do next time.