 So, today we do the really fun stuff of Yang-Mills theories. This is what led Sidney Coleman the great expositor of field theories to say that this 70s brought us unimaginable wonders from like a different planet. So, the origin of this particular topic that we are going to do today has to do with the non-linear nature of Yang-Mills theory, but as we saw that non-linearity is introduced in a very specific geometric way and that is what leads to a very deep significance to this theory and leads to these effects that we are going to talk about today. So, let us begin with the idea of gauge transformations. So, recall that in the Abelian case a mu tilde equal to a mu minus d mu of some lambda of x. Now, if this a mu is 0 to begin with if this is equal to 0 then a tilde mu is called pure gauge. This is so, it has no it will not turn on any electroelectric or magnetic fields, but you might see this superfluous somebody hands you an a tilde mu and you worry whether it has any gauge fields or not. Well two ways to check one is just to take its anti-symmetrize derivative and you should find f mu nu is to be 0 I am sorry I am using this. So, I will not put any tilde. So, just calculate the f mu nu of this. So, this equal to 0 implies that a is pure gauge. So, if you find this so, this is the differential method, but there is an integral method. The integral method says try to solve that is look for lambda that can satisfy this. This then you read the other way round and you ask whether. So, these are first order differential equations. So, d 0 lambda equal to a 0 d 3 lambda equal to a 3 are just a set of partial differential equations which are first order. So, you can even attempt to solve them by this well known method that you say therefore, lambda must be equal to integral of a 0 and d x 0 and plus some residual f of x 1, x 2, x 3 right. And so on you do for the second one you write it is integral of this, but with another and then you again take the derivative. So, by deriving a consistency conditions added you know in constants of integration such as we can check if the integral if a unique lambda exists. So, we want to we want lambda of course, to be sensible. So, the whole idea is look for a unique lambda you know genuine space time function which is not multiple valued or etcetera. So, thus looking for now. So, you understand this mutual consider right you take derivative now you take the a 2 derivative of this lambda. So, there will be d f by d x 2 then you check that it is consistent with the d 2 equation of that and so on. So, now reason why I am writing out all this rather easy points is that there are cases in which this may not. So, firstly if you if method 1 has worked then method 2 is guaranteed to work because you know that it actually is secretly d mu of some lambda and then the curl of this would be 0 because curl of gradient is 0 and if the curl is 0 then the path integral of a is uniquely defined right essentially this is a path this is a path integral of a I mean line integral of a. So, the line integral of a is uniquely defined. So, if that integral can be done uniquely in the whole plane then you get a unique lambda. So, it should be independent of the path along which you do this integration integral. So, that is ensured of course, if it is like this now there are there are cases in which this may not be so obvious is that unique lambda may not exist in the sense of so the point is 2 is an integral method and 1 is differential method and 2 is the stronger method because it actually checks the whole plane. This may be locally true, but when you check the whole plane it may not come out to be true. So, in some way. So, unique method may not unique lambda may not exist if the space if the domain on which it is defined is not just r 4, but something with holes in it. So, for example, if you take the and we can only draw 3 D I mean we can visualize 3 axis x 1 x 2 x 3, but suppose I exclude a cylinder along the x 3 axis I remove the x 3 axis. So, I exclude this. So, I make a very non-trivial space so that the loops around this are not shrinkable. If there is the domain x is excluding this then you can get away with lambdas that are not actually unique, but say differ by 2 pi and then its effect you may not detect. So, firstly it is not simply connected and say 2 pi n shifts in lambda are not detectable. So, the lesson here was that because the integral of a along any path. So, this is c let us say and you are well integrating only in the x 0 plane, but this is x 1 and x 2 plane suppose this is x 0 and you are integrating from point 1 to point 2 this is the c. So, the point was that whether you took this path or this path or that path did not matter because you and you are in fact, the fact that it turns out unique was that regardless of what path you took it should give the same answer in the end that is not guaranteed provided there are some regions of space that are excluded then you cannot go around that or if you go around you may get a different answer and then you cannot complain. But if the domain is simply connected then you can shrink all loops to 0 and then single valuedness would require that lambda cannot have any discontinuity. In a simply connected space all loops can be shrunk to 0 shrunk to a point. So, lambda cannot have any approach a unique value as the loop shrinks to 0, but in a non-simply connected space you may be able to circumvent test number 2 because. So, to be specific let us only deal with 2 plus 1 dimensional space. So, that we do not have to draw a cylinder and 2 we have charges whose way matter wave functions undergo gauge transformation and then 3 we exclude the origin, origin is excluded ok. We also need a fourth condition which is that all charges are have the same charge. So, there are no wave function all wave function all the possible matter has g x equal to n times g ok, where g is the basic unit and n is integer. In that case you can get away with a non-unique lambda by allowing lambda to change by 2 pi as you wind around this loop and if you change it by 2 pi a 2 pi by g. So, under these assumptions under these conditions plus 2 pi by g times n and therefore, lambda is multi valued, but you will never have a way of telling unless you have observables that can check this that you have gone around a loop. It turns out to be true in quantum mechanics because you have so called Bohm-Aharano effect. So, the Bohm-Aharano effect is in quantum mechanics is remembers the presence of a mu even at the of non-trivial magnetic fields at the origin. So, the point is that the wave function would then the Bohm-Aharano effect will be able to measure whether there was a, but note that that happens because you actually have a non-trivial magnetic field ok. So, but this is because non-trivial BZ exists. If it did not then you could get away with shifting by 2 pi n and you would not know. So, the example of that is in superconductors magnetic flux lines exist lambda is defined only 2 pi by e at any point. So, the summary of all this is that there are things lurking around in this gauge invariance business that have to do with the global properties of the space on which you are living and their connectivity ok. So, the overall moral of all this is that thus the notion of pure gauge cannot be is not cannot be tested easily if the space the domain of if the domain is not simply connected. Now, when we go to the Yang-Mills case even when the space is simply connected because the gauge group is non-trivial we can again have non-trivial gauge fields. So, non-Abelian case another class of exceptions due to group valuedness gauge transformations even on a simply connected domain even when dealing with a simply connected. So, that is the preamble and now we go to the non-Abelian case. So, in the superconductivity case the flux lines then the magnetic flux is quantized because the discontinued because the integral a dot dl which would measure the magnetic flux it can only change in units of 2 pi by E times n. So, the flux is quantized often times people say that superconductors display quantum mechanics in a bulk system, but I think that is not true. So, it has more to do with you may say that it shows that a quantum field A does have a classical limit and then obeys classical gauge invariance that is what it does show, but the observable is really not really microscopic ok. So, I mean the reasoning does not really involve quantum mechanics it involves just classical arguments and connect it in the simply connected argument. So, so long as the A mu field has a classical limit which is true for Abelian fields this is not a quantum mechanical result although there is a quantization ok. So, it is just a classical result, but there are other reasons why you may say it has to do with, but not this one ok. So, what we will do next is something quite interesting and let me write the title as the Jackie Rabi-Bacchua. So, now we go to the non-Abelian case and recall that we have that A mu and now I will write a u here instead of putting tilde we will say gauge transform by u is equal to u A mu u dagger and minus i over g u d mu u dagger. Therefore, configurations of the form are all pure gauge. So, here I should go and correct thus the notion of pure gauge can be tested, but with we well. So, let us remember that by pure gauge we mean a single valued lambda right, but here things are going to be more subtle. So, these are so if the original A was 0 then clearly the transform one is this and we drop this u and just say that if I am handed a configuration which is constructed out of a Lee group valued map. Now, we come to the so called Jackie Rabi observation. Let us simplify to the case where we will deal only with R 3 the space like R 3 and we will look at time independent case. So, the origin of the argument is that take S u 2 example topologically this is same as S 3. Everybody remembers why? You can construct a sequence of two spheres that go from north pole of the S 3 to south pole of S 3 and which are isomorphic to. So, which is and if you want I will quickly recapitulate the argument by going one dimension lower which is that disk in R 2 is isomorphic to S 2. So, since a disk in R 2 what is a disk it is a circular region including its boundary. So, that is disk and that is that can be shown to be equivalent to S 2 provided you identify the outermost circle with the south pole of S 2. So, what you do is you put the north pole at the center and then consider a sequence of circles going outward and at the same time on the S 2 you start with north pole and start drawing this circles. When you reach the south pole so, when you reach the boundary of the disk you map it into the south pole. So, this is so, it is not really a disk. So, it is disk with boundary shrunk to a point. You can think of it in reverse you take a sphere tear it at S 2 and then flatten it out. And just remember that where you tore although it became many points now you should think of it as one point. So, then you can visualize a two sphere which is intrinsically usually we embedded in 3D space can be visualized by an insect living in the disk. All you has to remember is the rule that the boundary is identified with a point. And for the same reason S u 2 has which has the form cos theta by 2 times identity plus i sin theta by 2 times theta cap dot sigma or dot tau. You can do the same map. So, treat theta as definition of map theta is the radius from origin and theta cap directions. Then you can see that the ball in 3D which is essentially the map by these two right the value of the radius theta and the direction theta cap value of theta and theta cap they together map out all the points in this ball. Then the ball in 3D with outermost point identified with spherical surface identified with one point is same as S 3 foliated as shells and both and this construction is isomorphic to both S u 2 as well as to S 3. S 3 because if I continued to if I continued to travel outward, but at some point shrunk the outermost circle to a point it maps a homogeneous space of dimension 2. If I have a sequence of two spheres and I take the outermost S 2 and shrink it to a point. So, outermost spherical surface S 2 identified with one point then it is S 3. In the case of S u 2 it is the point minus 1 minus identity ok. So, at theta equal to 2 pi regardless of which direction you reach it is all equal to minus 1.