 Now, if I put here a parameter t, which runs between 0 and 1, so for t, I can also put t in front of here. This will be a homotopy through for whole maps between, you know, for t equals 1, I have exactly the complex that I have, but for t equals 0, I will have just the terms that I have written down on the blackboard. So what we have actually for t equals 0 is a sum of two complexes, which we have already seen before. So what was that? So the first complex was 0 goes to 0, goes to gamma s plus, gamma s minus 0, and here is dA plus, and the other complex was 0 goes to omega 0, omega 1, omega 2 plus, 0, and we have just d and d plus. Now what you can easily see is that the determinant of dA plus, so what is that? This is the determinant of the kernel of dA plus, well, you know, lambda top, tensor, lambda top, the core kernel of dA plus, but we can identify this with the, you know, the kernel of dA minus. Now the point is that both dA plus and dA minus are complex linear operators, so in particular the kernels are complex linear vector spaces, and they always have a preferred orientation, at least if you fix your convention, right? So conventions are not really uniform, you have to be aware of that, but once we fix the convention, this is always R, okay? Now it's also easy to handle this complex, right? So the corresponding operator here is d plus plus d star, and if you take the, if you want to consider the determinant, so we have the kernel of this is just h1mR, well, we have this always purely imaginary, and the core kernel is also easy to compute, this is just h0. Let me just write h0 plus h2 plus, and these are spaces which are defined, which are determined by the topology of your form manifold, so you may ask to, or you may orient these spaces, and this doesn't depend on the smooth structure. So the upshot of this discussion is that an orientation of those homology spaces actually determines you an orientation, or a trivialization of the determinant line bundle. So let me write this down, so very good. So we now have all the ingredients that we need, and let me state this formally, so we have a right at our main theorem, and the theorem is this, so assume v2 plus of m is at least 2, then mg eta is a smooth compact oriented. So I write oriented, but I should have written probably orientable, and with choice of orientations of homology groups we have oriented manifold of dimension d, and d is a number which already disappeared from the blackboard, but you remember. And this holds for generic g and eta. So any two choices, g0, eta0, g1, eta1, m, g0, eta0, and m, g1, eta1 are oriented. So by now you actually know already the proof of the theorem, maybe the only thing what I need to say is, why do we require v2 plus at least 2? This is something that we have already discussed. This is because if you choose g0 eta, so two pairs of parameters, we want to connect them by a path, and we want to consider the module as a parameterized modular space, and we will avoid reducibles whenever v2 plus is at least 2. This is where it appears. Okay. Now let me come to the point. So this hybrid variant, here is what we have. So I will pick any m0 in my base manifold m, then what I have, I can take any map as values in u1, and I can evaluate this at the point m0. So I have a map into u1. This is clearly surjective. And the kernel of this is denoted by g0. So this is just the subset of all those gauge transformations which are identity at the point m0, and we have the following exact sequence. Okay. Now what we have is I can take m hat eta to be the space of all solutions psi eta such that sw eta psi eta is 0, but I will divide this by the base gauge group g0. And now what you see here is the rest of the actions. So we have an action here by u1. And if I quotient by u1, I will get my original modular space back. In other words, what I have is m eta hat fibers over m eta, and the fiber of that is u1. In other words, we have a principle u1 bundle over the modular space. So this is principle u1 bundle. And let me denote by mu its first 10 class. Okay. Now, so it was not explicit in the notations here, but implicitly just to define those modular spaces, we have chosen a spin C structure. So I have here made a choice of sigma in S m. And this is the set of all spin C structures. Okay. Now I can define the cyberquidding variant relative to this choice of a spin C structure. So sw of sigma is defined to be an integral over m eta of mu to the power d by 2 whenever d is even, or it's zero otherwise. And so with the work that we have done so far, you know that this number that I have defined does not really depend on the choices, so that we have an orient, we have a number. Well, if you want to be strict here, this the cyberquidding variant is only the absolute value of the cyberquidding variant is well defined, right? Because we have made here the choice of orientation, but if you can fix your underlying topological form manifold, then you can make sense of this as an integer. So if you combine all this, this is just a map from S of m into Z. And the basic property that, so this is the cyberquidding variant. The basic property that this variant have is the following, sw vanishes on all, but finally spin C structures. And the proof is not really hard. This follows from the prior estimates that we have had already before, but this is an exercise for you, so you will solve this today in the afternoon. Okay, so let me finish this lecture with a sort of collection of results that you can obtain with the help of the cyberquidding variant. So there is no time left for proofs anymore, but I thought it may be interesting to see where this could be applied to. In any case, one of the first theorems in this area was by Witton. So if m admits a metric of positive scalar curvature and b to plus of m is at least two, then the cyberquidding variant vanishes identically for any spin C structure. And again, the proof is sort of, just follows from the Weissenberg formula. So that's something that you can prove on your own. Now the next result, this would be all sort of, the whole theory could be absolutely boring. It could have happened that we would end it up with a very complicated description of the zeroes map, right? And that would be very disappointing. But Taub's proved very general statement which tells you the following. So assume m is symplactic, and as always we assume that b to plus is at least two, then the claim is that the cyberquidding invariant of m does not vanish. The statement is a little more precise, but I don't want to go into details here. Should I explain what a symplactic manifold is? Okay, now, so this gives you lots and lots of examples where the cyberquidding invariant doesn't vanish, so you know that your invariant is an interesting one. Now, so by the way, you can play here and you can combine those two results. So if you know that b to plus of your manifold is at least two, then either, so if it admits a symplactic structure, it doesn't admit a metric of positive scalar curvature and vice versa. This is the only way I know, which shows you that a priori there is no reason to expect that symplactic structure is related in any way to metrics of positive curvature, but this actually tells you there is set in relation. Okay, so yes, and CP2 is also not a count example. So another result is the following theorem, usually attributed to Whitton. So who gave us a sketch of the proof is, the statement is as follows. So assume M1 and M2 have both b to plus at least one, then the cyberquidding invariant of the connected sum M1 with M2 vanishes. Do I need to explain what the connected sum is? Okay, so what this result tells you is that you can interpret this in the following way. So assume you know that the cyberquidding invariant is nonzero and you try to decompose your manifold into simpler building blocks, but here simpler building blocks would mean manifolds with b to plus at least one and you will establish this as impossible. And one more result, so I wanted to show you is the theorem of Donaldson, originally proved by Donaldson much earlier by other methods, but you can prove this with the help of the cyberquidding theory is that there are many topological closed four manifold which do not admit a smooth structure. So I don't want to specify what many means, but this is really a huge sort of family of four manifolds which are, you know, where you have an atlas where transition functions are continuous but can't be made smooth or, you know, ck for k say two or something. Okay, and one more result which is sort of complementary to the last theorem is the following. So again the results that I will state in a minute was proven earlier by other means but still it's interesting to state I think. So what I mean by fintechel and Stern is that there are infinitely many manifolds over here I should say well let me just say manifolds of dimension four say parameterized by some number which are homeomorphic pairwise non-defilmorphic. So in other words what so what you can say is that if you take four manifolds and there may be a sort of topological four manifold then there may or may not be a smooth structure and if there is one it may happen that there are infinitely many of those smooth structures. Okay, thank you very much for your attention it was pleasure for me and if you wish please give me the evaluation forms back.