 Let us make connections between connections so these two notions obviously are very closely related as a name suggests so in fact we can even provide Dictionary so on the one hand we have Vector bundles on the other hand we have principle bundles now So as we have discussed already if I have a vector bundle I can construct a principle bundle So let's do a name. So e I can construct a principle bundle of frames of e and In the other direction if I have a principle G bundle P I can construct an associated bundle P times G row Say Rk this is e so this gives Way ways to go back and forth between principle and vector bundles but Let us perhaps go a little further. So if you have a section s So this is a section What does this correspond on the right-hand side? So it turns out that this corresponds to a equivariant map from P into a Representation say Rk and this is required to be G Equivariant and so the the correspondence is very explicit right, so if if you have an and a G equivalent map as hats and you can define as head of the class say P X just to be So I can define s of P X to be as head of P If you wish the class This guy now if you have a connection Nubla On a vector bundle this gives us a connection on the principle bundle. So let me denote this by a in omega 1 P G in the standard way, so something that I've already mentioned if you have a Local trivialization of your vector bundle the connection becomes a one form with values in the Lie algebra And this is exactly what you have if you look at this globally What you get is a one form on the total space of P rather than just a bunch of local one forms and As we discuss here, we have the gauge group Of a vector bundle and we have the corresponding notion of the gauge group on the right-hand side So what is that? G of P Let's just say Set if you wish of bundle automorphins. These are the maps from pi from P to P such that pi P times G is She by times G But in other words what we see from here is that the value of phi at the point P must be in the same Orbit of the Lie group action so that they say can represent any map phi as a map Psi from P to G and the condition is that Psi of P times G is Psi of P G G inverse Okay Since we have we have here also discuss the notion of the curvature. So for a connection nublar. I have the curvature F nublar Which was defined as a composition you remember of the differentials the nublar and you have a corresponding notion of the curvature here For a connection a so now F a becomes Da plus one-half a wedge a Well here the brackets together with wedge means simultaneous Brackets in the Lie algebra and taking the wedge product in the space of one forms. And so what you have is That of course these notions are related so you know if If you have a connection nublar which induces your connection A on the principle bundle you take its curvature And this is more or less the same as a curvature of this connection F nublar. All right now the most important property of the curvature is the so-called Pianchi identity the theorem is That if you take the nublar and apply this to the curvature So the curvature is a two form if you apply the nublar to this you will get a three form on a manifold There was values in the endomorphism bundle. This stands out to be zero Or if you think of the connection in terms of the principle bundle or Locally, this is the same as saying that D of F a equals a wedge F a up to a sign So here is a minus sign. Okay. Maybe let me give you One example of that So here is an example. So let me assume that P over M is a principle U1 bundle So a particular feature is that you one is an abliently group. All right. So what we have is the following so the gauge group of P as we have discussed earlier. It's just a set of maps psi from P to U1 So I said in this case G P times G is just Psi of P Right. In other words psi is an invariant map with respect to the U1 action That is this is just the same as the space of maps from the base manifold Into U1 Now if you have a connection A on P The gauge equivalent connection to that. So let us pick also a gauge transformation. So this is a smooth map Right. It's just an element in this space So then the corresponding One form of G is a plus G minus 1 DG Let us try to compute the curvature of this connection. So F a G is Because of a Lee Algebra is ablient This term over here is trivial. So we have we have just the differential of a and this is D a plus T of G inverse DG But this is close. So this vanishes and we have F a Yes, you do. Yeah, so what I mean is that they cover Indeed the identity map here on the base manifold M Okay, so in other words What we see is that the in the ablient case a curvature actually It's independent of the gauge transformation. Okay, are there any questions to that? All right, then Let us move a little bit so Let me consider now elements of the Tran wild theory So first we consider the notion of an invariant polynomial. So what we have is We take a map P from the lee algebra G into Say R or C. So let me take to the complex numbers and I Say that P is an invariant Or maybe G invariant polynomial If the following whole so P is G invariant as a map that is P Had G XI is P XI right for any G and then XI in the lee algebra of G and P is a polynomial. So P is Polynomial meanings that you know if you choose Say a basis XI one up to XI n so basis Full G then P of X1 XI one plus X and XI and is a Polynomial in X1 Xn and Sometimes it will be convenient to write this homogeneous polynomials, which just means so piece homogeneous Of degrees a D F P of lambda XI as lambda to the power D P of XI so Here are Some examples So we will take Here as an example we take G to be UK and So the corresponding D algebra is UK and we could take Say P of XI to be Trace of XI to the power D and this is obviously a homogeneous polynomial of degree D more importantly we can take so P of XI to be the determinant of lambda one plus I over to pi XI and so Now we have a parameter lambda here. So if we expand this in lambda you will get an expression like lambda to the power K Plus C1 of XI lambda to the power K minus one plus and so on CK of XI Now these are homogeneous. So each CJ is a homogeneous polynomial of degree J Now so we can compute easily certain terms right if you put if you put lambda equals zero what we get is CK of XI is a determinant of XI essentially. So C K XI yes, and for instance C1 of XI is just I over to pi Trace of XI Alright now if you have a principle bundle together with a connection So P M is a Principle bundle If you pick a connection nabla on P We have the character of nabla, right? This is now a two-form We can think of this at least as a two-form on the total space of G with values in the Lie algebra G We can apply P to F nabla And this will give us some form on the total space of P was values Say in complex numbers But now it's easy to check that this form is actually G invariant and basic so basic here means that whenever you take you know Particle so V is a Particle tangent Factor to the principle bundle P is that that is it is tangent to the fiber of the projection map if you Inserts this into P of F nabla you will have zero and This implies that actually there is a unique form Maybe F nabla So I would just you know by the same symbol but when saying is that this is a form Essentially on the base manifold M was values in C so let me Temporarily denotes this with Twiddle Such that the pullback of this form nabla Twiddle is by Nubble so it drops a symbol Twiddle in the sequel and now the Yankee identity Tells us actually that this form Whatever way you think either this form on the Total space of the principle bundle or the the form on the base is closed. So P Nubla Twiddle is zero In other words, this means that we can Consider the the Ramko homology class of P F nabla So this is H The Ram see and this is Well defined Independent of the choice of connection now In other words would be a constant is an invariant of a vector bundle or a principle bundle itself Now here is an Example which will be of Importance to us So if you have a you want a UK bundle as we have had over there then C J of P. So, you know P is a K bundle then C J of P So this is now an element in the two J's in the Ramko homology group of M Was values in C. So this is called the J's turn class maybe Let me just finish by saying that for the Classes for the polynomial C K that we have constructed here. They have the properties that If you take C J bar, this is C J So in fact, they are real polynomials and this means that we have constructed classes in the Real co homology group rather than the complex one Okay, let me stop here and we will continue our discussion today in the afternoon