 So, in this nucleon wave function ok, this is the hypothesis due to Heisenberg and this basically assume that nuclear forces do not care whether you have a neutron or a proton and the charge it the and that they are independent of charge. The fact that proton has some charge does not matter to the nuclear force. Now, if that is true then the two things together should look like the same thing for nuclear forces and this is quite a far reaching hypothesis. So, what Young and Mills did was to say that therefore, just as electromagnetism involves rescaling you know the classical gauge invariance of electromagnetism where you shift a mu requires that you should simultaneously shift the phase of electron wave function by e raise to i g lambda x. Suppose here I firstly promote the overall phase to be a function of x. So, these theta's to be function of x and then I introduced the correct a mu field which absorbs the derivatives of that theta which enter this ok. So, this is Young and Mills and somebody called G Shaw. I think he became a professor also somewhere. This G Shaw did what Young and Mills became famous for in a in his PhD thesis. He submitted the thesis got his PhD and forgot all about it and the whole world was celebrating Young and Mills for having invented the thing. So, but somehow that name is never mentioned, but it is actually Young Mills and Shaw because G Shaw also observed the same things and proposed the same thing. What is also interesting is that in mathematics Aresman also proposed the same thing at almost the same time and he called it fiber bundles. All of this happened in 1956. So, the ideas emerged at about the all the same idea emerged from several sources about the same time. Aresman's motivation was probably just to generalize the covariant derivative which was too much tied to space time, tangent spaces and so on. So, in differential geometry. So, here what we say is look for transforming this big psi u of x. So, instead of global SU 2 we have u of x times psi of x and nu fields whose inhomogeneous transformation compensates derivatives of theta of x. So, here we observe and if you have I have told about SU 2 so many times. So, many of you may get bored of this, but let me tell you once again that SU 2 can be thought of as a solid sphere of dimension 3 a homogen you try to think of sphere in arbitrarily large number of dimensions. What is the main property of a sphere? It is homogeneous wherever. So, you first think of the spherical shell which is an S 2 its properties that wherever you are on it the space looks the same. So, it is a space of constant curvature or it is a space which is completely homogeneous it looks the same and it is compact it is not indefinite. So, what happens for SU 2 is you can parameterize this object in this particular way you set up three directions. So, totally there are three degrees of freedom because there is one over here and there is a unit vector in three dimensions there is one for each of the tau there is a component, but it is unit vector. So, it has only 2 degrees of freedom and the third one is this. So, suppose we set up the tau x tau y tau z axis not that that is the label, but it is actually theta x theta y theta z axis and. So, maybe so, yeah theta x theta y theta z I set up a unit vector theta cap in this space and then I rotate by some amount. So, this theta means a rotation of some amount. So, we say whatever amount by which you are going to rotate you put it on this ok. So, this is theta. So, theta cap exhausted 2 degrees of freedom to choose the direction and then how much you rotate is set up along that. So, any element like this can be thought up as some vector like this right in three abstract dimensions whose axis are labeled some theta x theta y theta z. But we know that SU 2 because this has expansion cos theta by 2 times identity plus i sin theta by 2 times theta cap dot tau. Therefore, we see that if we reach theta equal to 2 pi then we get cos of pi times 1. So, we get a minus 1 whereas, sin of pi will be 0. So, this term goes to 0. So, if I reach theta equal to 2 pi I get minus 1 outside right minus identity at the surface in any direction. So, if I go from minus 2 pi to plus 2 pi along any direction I always start with minus 1 go through plus 1 when everything is 0 u becomes equal to plus 1 and then return to minus 1. So, this is the kind of space it is and with different theta cap I should exhaust going from minus 2 pi to plus 2 pi. But you do not have to struggle too much about this open thing because outside it is exactly equal to minus 1 when you reach that it is actually just one point. Although it is drawn as a solid ball the outermost surface of this ball is abstractly speaking only one point ok. So, we are actually looking at a genuinely a four dimensional object in its three dimensional projection. So, since a few people do look puzzled I will spend a little more time to explain what this is. We do a one dimension lower analogy. So, suppose I was an ant which can only move around in two dimensions, but wanted to visualize a compact two dimensional sphere ok, but I do not have the eyes to see a sphere I do not have a third dimension in which to erect that sphere. Then what I would have is the plane and suppose in the plane I set up a disk. So, here I will have only theta x and theta I will just call it x and y because we do not have the meaning of group theory there, but this is a compact two dimensional sphere as visualized having only x and y. Normally you would think that a shell would require you to at least live in three dimensions to visualize it, but it is only a two dimensional space it is intrinsic dimensional it is only two. So, we can make it even out of a flat surface like this what you do is you sit at the center. So, let me draw the real thing that we know. So, suppose this is what we call S 2 a spherical shell it will have a north pole and a south pole. So, let us map the north pole to this point and then I start drawing circles which grow in size. So, correspondingly I draw the only thing I forego is that the length of this circle does not match the length here. Here the length keeps growing the circumference of circle keeps growing here as I come closer to south pole the circles begin to grow smaller until the shrink to a point ok. So, if I was an ant confined to this and I just make a new rule although it looks like I have this I ascribe it a smaller length and smaller length I change my unit as I go further and further out you know it is a sin theta factor. So, I make it smaller and smaller until when I reach the outermost circle of the disk I declare it to be size 0. So, I am confined to a disk the confined to a plane, but I make my distance rules or length measurement rules in such a way that when I reach the circumference of the disk I declare it to be size 0 then I have effectively visualized an S 2 sitting in a plane ok. Something that normally would require three dimensional to visualize I have done using a two dimensional space. So, this thing where you have a sequence of two spheres right each one is a two sphere each value of theta allowing theta cap to vary over everything is a sequence of two spheres which eventually gives you only one point. So, this is like the north pole of the S 3 and finally, this is the south pole of the S 3 embedded in three dimensions ok. So, SU 2 is basically a is as a point set it is isomorphic to S 3 and in this particular way you can visualize it like this. So, what we are going to do is we will look for. So, there are three degrees of freedom that describes this group three parameters. If we are going to do this kind of transformations we will need at least three compensating gauge fields right. So, we expect to get. So, suppose we propose a covariant derivative I am just jumping a little bit yeah we can say notionally we will expect that for. So, note that we could have always written theta into theta cap dot tau to be equal to theta a tau a where each unit vector component will get multiplied by theta to produce theta a and it can be written like this. So, that is what we are writing. So, just as we in the abelian case we had a mu going to all a mu minus derivative of some lambda here it will be the we will make the theta as a function of x and then this will this is how we expect that this will happen, but we can do it in a. So, I now jump a little bit. So, propose that d mu of psi is equal to d mu plus i g a mu a tau a psi just as we had in that case except that instead of single a mu I have a triplet and the if you want there is an identity matrix multiplying this d mu times i. So, this is a 2 by 2 matrix acting on a 2 vector right because a mu a tau a what is a mu a tau a it is equal to a mu 3 times tau 3. So, a mu 3 and tau 3 is 1 minus 1. So, it is a 2 by 2 matrix and it will act on this Heisenberg's 2 component nucleon wave function. So, now we ask if we let then what a mu tilde do we need. So, that d mu tilde of psi tilde is equal to u of x same story as before right there we had taken out e raised to i. So, the point is this is a space time dependent object a derivative acts on it will differentiate this as well, but somehow by magic we want to end up with this only multiplying this without any derivatives. So, the derivatives have to be compensated by the transformation of a mu. So, right everyone is clear about this. So, all we need to do is to carry this out and see what happens. So, we say d mu of. So, d mu tilde psi tilde is equal to and we will use a shorthand notation for m a tau a together we will call it it is a matrix. So, we just write that. So, i g a tilde on u times psi. So, that is equal to d mu u times psi plus i g a tilde times u psi and plus when this genuinely acts on this. So, what we should do, but we want this to look like u times something right. So, I just pull out a u because I want the answer to look like u times something. So, I put everything else inside. However, simple trick it looks today that we can just teach like this was the greatest invention of the last century. So, this then is to be redefined as. So, this is equal to u of x times d mu of psi as expected and plus i g times u dagger a u plus 1 over g. So, we pulled out i g. So, 1 over i g right acting on psi to put back tilde on the a. So, far we did not know what it was. But we want this to read the old derivative. The old derivative was just a there. So, it follows that this is nothing, but a. So, need. So, the transformation law can now be written out. So, I went the other way, but usually people write a tilde in terms of the nu. So, if you want to do that we can quickly do that because it does not take much time multiply by u and u dagger. So, thus well let us first write this. So, is to not make mistake thus a must be equal to u dagger a tilde u plus or minus i over g times u dagger d mu u equivalently a dagger a tilde is equal to u a u dagger right. If I do this then it will transfer them here and then I will get a minus and I put this on this side. So, plus i over g I have to do the same thing here this u u dagger is 1, but recall that u is a unitary matrix. So, u u dagger is 1. So, often people write it in this form right because d mu of u u dagger is equal to 0. So, d mu u u dagger is same as minus u times d mu u dagger. So, this is our great transformation law the greatest the greatest applicable theoretical discovery of last century may have been many others, but this is certainly the all the forces of all the nuclear forces obey this law.