 with an identity. And in fact, it's an associative algebra. So let me give you an example. So if you take V to be one-dimensional, so R1, then what we see is that the Clifford algebra of R1 is generated by two elements, one and E, and modularity relations, and the relation is E squared equals minus one, but this is clearly just the algebra of complex numbers. So something that you know and love. Okay, so maybe one more example. If you take V to be R2, now you will have three generators, right? One, E1, and E2, but you will have also the set one, which is a product of E1 and E2, so all in all, you will have three generators, and what this gives you is the algebra of paternions. So again, this is something that we know very well. Okay, with this at hand, I will use the following definition. So a vector bundle E, together with a connection A, is called a derived bundle. If the following conditions are satisfied, so first of all, E and nabla are impledient. That is, I have a scalar product on each fiber of E, and nabla respects the scalar product, and I also have a homomorphism from TM into the endomorphisms of E, which extends to the homomorphism of algebras, so cleaver TM into the endomorphisms of M. Now what this actually means is that whenever I have a tangent vector V in TM, so a map says into say rho of V in the endomorphism of E, and the requirement is that rho of V squared is minus norm of V squared times the identity. And the set condition is that this structure of the Clifford module is preserved by the connections that we have. That is nabla of V dot S is nabla V times S plus V times nabla S. All right, so what I mean by this, so here V is a local or a vector field, so this is a section of TM, and nabla of V denotes the lavish-Veta connection, so if you wish, this is the lavish-Veta connection applied to V. Okay, what is it good for? This is good for the following thing. So if I have a structure of a Dirac module, I can define the Dirac operator, so here is the definition. Now I choose a local frame U1 in the end, this is a local frame for TM, or let's say some open subset U, then the Dirac operator applied to S is just a sum. Well, I take the covariant derivative of S with respect to this vector, so say EI, I take EI and Clifford multiply with this covariant derivative, and I do this for all frames from one to n. Now let me give you one example of this. We could take E to be the exterior power of the cotangent bundle to M, I have a map from TM into, well, and the morphism of this bundle, so if I take a one form, so not necessarily one form, there's a differential form of any degree, and I have a tangent vector V, so I can map this into V, so if I Clifford multiplies this with omega, this is, so I first utilize a single V as a one form, and I take the batch product with omega minus I substitute V into omega. So you can easily compute that I will, that if I will apply Clifford multiplication in this form to my given form omega twice, I will have this property. And so we can compute the corresponding Dirac operator, and this happens to be, so in this particular case, D is just D plus D star. Again, something that you know very well, right? So this demonstrates that we have got something reasonable to study. And the basic property of a Dirac operator is a so-called Weizenberg formula, so here is theorem, and it tells us that if you squared the Dirac operator, this is the connection Laplacian, it will define this in a minute, plus some algebraic operator, R, so there, what is R of S? This is just some EI, EJ, Clifford multiplied with the curvature of nubla, evaluated at EI, EJ, and applied to S. So I and J is from one to N. Now the formula itself maybe is not very important, what is important that R is a zero-sword operator, this is in order to break scene. And what is this one? So nubla star nubla is just some nubla EI, nubla EI applied to S, minus nubla nubla EI, EI applied to S, when I is from one to N. In other words, so if you ignore this term, so this is the first order, the main term is this one of the second order, where you clearly see this is just a Laplacian here. Right now with this formula, the Weizenberg formula tells you essentially is that the square of the Dirac operator is a Laplacian plus an algebraic term. So something that physicists were looking for for many years. Are there any questions to that? In here, in here, of this bundle. Okay, in the remaining time, I wanted to discuss a bit, a particular example of the Dirac operator that we will need in the sequel. And this is related to the notion of a spin group and a spin structure. So what you know or what can be computed is that the fundamental group of the group S, O, N is in fact Z mod 2 Z. So let's just fact if you wish. But this implies that there is a group with double powers S, O, N, so there exists a group spin N. Such that this is a double power of S, O, N when, well, this is two to one. And I have to require here that N is at least three because for N equals two, this works a little different. In any case, and this group spin N is defined by this property uniquely. So if this covering is non-trivial. So let me give you an example. Now let us take N to be three, right? We can consider the group SP1, which is just a group of quaternions of unit lengths. And I claims there is a homomorphism into SO3, which practice follows. So if I have Q here, I can map it into the map X goes to QX, Q bar. Where X is an imaginary quaternion. And so I can think of this as a vector in R3, right? You can easily check that this is an orthonormal transformation of R3. And this gives us a map from SP1 into SO3. So topologically, this is the three sphere. And so we know that this is simply connected. And we also see that minus one here goes to the identity transformation. So it has a kernel plus minus one. And then you can check that this is a non-trivial, well it must be a non-trivial curve. And also in dimension four, the spin four group can be constructed explicitly. So again, we have SP1 times SP1 into SO4. So here to distinguish two copies of SP1, I will endow one with plus subscript and minus subscript. Now if you have two quaternions, Q plus and Q minus, I can send this to a map X goes to Q plus X, Q minus bar, where X is now a quaternion. Again, you can easily check that this is an orthonormal transformation of a four-dimensional space. So we have a map from this group into SO4 and minus one, minus one, so the diagonal minus one goes clearly to the identity element. So again, this is a map two to one. And therefore, and this is again simply connected, right? So this is just topologically S3 cross S3. In other words, so what we have shown is that the group spin three is SP1 and this is just the same as SU2. And the group spin four is SP1 times SP1. All this fairly concrete. In general, what you have is the group spin n can be defined to be the group generated by elements of the form Vi, Vj inside the Clifford algebra of Rn. Well, you know, these are vectors in Rn of length one. So okay, but you know, the general dimension won't concern us too much. We will focus more on dimension four, where we have extremely explicit construction of the group spin n. Now, for this group, I have two representations, row plus and row minus into SU2. So if you wish plus and SU2 minus, this is just, well, forgetting one of those factors and again, forgetting the other factor. And this gives me two vector bundles, well, not yet. Now, let me say that m is called spin if there exists a bundle P. So this is a spin n bundle over m, such that we have a two to one covering P of the principle bundle of orthonormal frames of m. So a spin structure, so such a spin bundle P may or may not exist. So for instance, in dimension three, any manifold is spin. In dimension four, CP2 is not a spin manifold, right? So it may or may not exist, as I said, but let us assume it does exist. Then these two representations, row plus and row minus give me two vector bundles, S plus minus. So this is just P times row plus minus C2. So these are permission rank two vector bundles. But more importantly, what you can check is that we have a map from GM into the space of homomorphism from S plus to S minus. And this, in fact, extends to a homomorphism of Gleifert algebra. In fact, what I'm saying is that if you take the bundle S, which is S plus plus S minus, this is a Gleifert bundle. So to have a Gleifert bundle, I need to have a connection on this bundle. So where do I get a connection here? This is very simple. If I have a spin structure, I have a two to one covering over the principal bundle of orthonormal frames. Here I have the Levy-Chivita connection, so I can pull it back to the bundle P and this is again a connection on this bundle, thanks to the fact that the le algebra of spin n and S o n are just the same. If I have this, now the corresponding Dirac operator is of the form 0, 0, d plus, d minus, we'll put slashes everywhere. And the Weissenberg formula in this case is d minus applied after d plus applied to some spin or psi. So this is now a section of S plus is nubla star nubla applied to psi plus 1 fourth of the scalar character of the metric G times psi. So as I said, this is a scalar curvature. So in particular, if you have a metric of a positive scalar curvature, this formula implies that there are no harmonic spinors. Something that has certain consequences. Right, so this is a question about your presentations. So what I'm saying is that if you view S plus as a representation of spin four and S minus as a representation of spin four, then the homomorphism is just a is a complexification of the canonical representation of spin four through S o four. And this actually tells you that there is such an isomorphism, there is such a homomorphism. I mean, you can do this also explicitly, but this is a sort of a very quick way to see that there is such a essentially unique homomorphism. Okay, let me discuss quickly one variant of this. So if you have a spin structure, what you can do, you can twist S plus by a Hermitian line bundle and also S minus by Hermitian line bundle. And this gives you the twisted Dirac operator, which is defined as by the same formula, meaning that if you have a connection A on L, the Hermitian connection on L, this gives you a connection together with the Levy-Tvita connection on these bundles. And again, you have a Dirac bundle and you can apply the same rule. And in this case, the Weisenberg formula is dA minus dA plus, applied to psi is nabla A star nabla A applied to psi plus one-fourth of the scalar curvature applied to psi. And one extra term, which is the self-dual part of the curvature, Clifford multiplied with psi. Now, what is important to realize here is that we have here something which is known as a spin C structure, which is essentially just a pair of vector bundles. I will still denote them by S plus and S minus. So these are Hermitian rank two bundles, such that the second exterior power of S plus is the same as the second exterior power of S minus. Right, in the case of the spin manifold, this would be trivial, but we can generalize this to a more general setting, but this is non-trivial bundle. And we have the bundle homomorphism into home S plus, S minus. Again, this extends in the usual way to the homomorphism of Clifford algebras and we have the corresponding Diracuprate. So, but the point is that, so here is a fact, any oriented, well, Romanian, any closed, oriented Romanian form manifold admits a spin C structure. Right, so not any form manifold is spin, but there is always at least one spin C structure and in fact, what you can show is that the set of all spin C structure so let me denote this by S of M is and H2MZ torso. So what this means is the following. Once I have a pair of Hermitian rank two bundles with this property, I can always take a Hermitian line bundle and twist my bundles with this Hermitian line bundle and this will give me another spin C structure. And if I will apply any line bundles that I have to my disposal, I will get all spin C structures on my manifold M, right? This means that I have an action of H2 on the set and this action is first for free and transitive, right? This is what torso means. Okay, so this will be a basic input for the Zeibar-Quitten theory, something that we will discuss in the last two lectures. The next two lectures will be essentially about analysis of something which is not necessarily connected to the material of the last two lectures as I have said in the beginning. This will give us sort of a tool by which we will study the Zeibar-Quitten equations. Okay, but that's all I wanted to tell you today.