 I want to talk today about the Zeibach-Widding invariant. More or less the whole point why we did all this stuff in the previous lectures. Anyway, so what we had, recall, we had the gauge group. This was a space of maps from 4 to m to u1. And this acts on c, which was w32s plus times a32, all that. So as I discussed the last time, it is not really important which degrees we take here. So one of the arguments today will require a somewhat higher degree. So we will take here 5 and 4. But as you know, you could have fixed any k sufficiently large and back with any k here. In any case, so this action is not free, right? So why is this the case? Now, if you consider the stabilizer of a point psi a, you will immediately see that whenever, let's say, is that way, if psi is 0, so everywhere is 0, I have always known trivial stabilizer, which is just a circle given by constant maps. And so that's how this is if and only if. And this is just an easy observation. In any case, so if you have a solution with c identically 0, this is called reducible. So these are points where the action of the group is not free. So we will just discard those. And I will denote by c star the space of all those points psi a, where psi is not identically 0. OK, now recall from the last lecture. So what we wanted to have, we wanted to have the modulate space to be a compact space. And we proved this the last time. We wanted to have this a manifold. And this required to have local slices at each point. And we wanted to have an orientation. So here is one of the properties that we need. And the claim is that for any psi a in c star, there exists local slice. So local slice through this point psi a. And this doesn't have anything to do with the cyberquitney equations. This is an absolutely general claim. So here is a sketch of the proof. So what do we have? You remember, we have the infinitesimal action. This was r psi a on psi. It was minus psi psi dA d psi. So it was infinitesimal action. And in general, if you have any manifold and you have an action here of some group, so as in the example of the last time, but now if we fix a point, we have the image of the infinitesimal action. So let me do this infinitesimally. This is this line. And if you want to have a slice, we have to pick a space which is sort of orthogonal or at least complementary to this space. And we can try to choose the orthogonal complement to that. But this is the slice. Now what are the image of the infinitesimal action? This is nothing else but just the kernel of the formal adjoint operator. And this one is easy to compute. So we have r psi a adjoint on psi dot, say, a dot, is star a dot plus i times the imaginary part of psi dot. And as a slice, we can declare s to be an open neighborhood of the 0 in the kernel of r psi a u. There's a map from w 4, 2 into w 3, 2. That's all this means. OK, and so it is very easy to compute. We have now a natural embedding of s into the configuration space. And what you can easily prove is that the tangent space at the point psi a to s plus the image of r psi a is a whole space, that is t psi a to c. And which means, at least at infinitesimal, at the point psi a, we have a slice. But then this also holds in the small neighborhood of the origin. So that we get a slice, at least locally. OK, so this is a good news. Now we have slices through each point, at least through irreducible points. Now we need also a perturbation. So in general, if we just take the pre-image of some fixed point, there is no reason to expect that the modular space will be smooth. So what we need, we need some family of maps. And this is how we do. So we take s w again. I will denote this by the same letters from c times w 3, 2, lambda 2 plus t star m, as well as imaginary numbers, into 3 to s minus times w 3, 2, lambda 2 plus. And so what is this map? I will take psi a, a point in c, and eta. This is a self-dual 2 form on m, purely imaginary one. I will send this to the, let me write this explicitly, so d a plus of psi and f a plus minus mu of psi minus eta. So I just added one term, eta, to the second equation, and that's it. And the point why we are doing this, because the following proposition can be proved, is that 0 is a regular value for this, the tube map cyber written. I don't want to give you the full proof, but at least an idea how this works in sketch of proof. Well, what you can easily see is that if you differentiate this map with respect to eta, it is surjective on the second component. So here is nothing to prove. Now what we want to show is that the map, we need to show that the map from t psi a c into w3 2 s minus. So this is just the differential of the cyber written map. What we have is psi dot a dot maps to d a plus psi dot plus 1 half a dot times psi is surjective. You want to establish this. Now what does this mean? What does this mean that this map is not surjective? It means that I have an element, so I know that the image of this map is a closed subspace. So I can choose if it's not the whole space, I can always choose an element which is orthogonal in the L2 sense to the image of this map. So if not, then there exists no. Let me say phi in gamma s minus. So if you wish w3 2 such that the scalar product of d a plus psi dot plus 1 half dot psi was phi in L2 sense 0. Now if I put a dot equals 0, what I will have is that ga minus of phi equals 0. So this implies for a dot equals 0. So I have this equality. Anyway, in the same way, I can also see that the L2 scalar product of a hat times psi with phi is 0, and this holds for any a dot. And also the point is this. So I have chosen the exponent here so that we have an embedding in c0 s minus. And if you assume that phi doesn't vanish at some point, then it doesn't vanish in the neighborhood of this point. And I can always choose a dot so that this will never be 0. And this actually implies that phi vanishes everywhere. OK, and this proves that the image of this map is actually surjective. So maybe it is true what I said, but so here is one more step. So what do you see from this argument that phi actually vanishes on an open subset of your manifold m, but then you can invoke the unique continuation principle for elliptic problems. And this actually tells you that phi must vanish everywhere. So that's the argument. Technically, what you have here is a so-called yonus 0. I don't know quite how to pronounce this, certainly. In any case, this proves the surjectivity of the differential of this Ibraquitin map. So this 0 is a regular value for this pertupt map. I mean, this allows us to prove the surjectivity. So I used here this perturbation to show that the second component of the differential satisfaction to this factor is surjective. I didn't write this. I just said we need to prove that the first component is surjective, because the second one is sort of romantic. OK. Now, so what we have obtained so far is the following corollary. And the statement is that for generic eta in W32 lambda 2 plus, the moduli space m eta star, that as we consider, the space of all solutions to the perturbs are Ibraquitin equations psi a. Such that dA plus of psi is 0. And fA plus minus, let me write, equals mu of psi plus eta. And we additionally require that psi is not everywhere vanishing. Of course, we divide this by the gauge group. It is a smooth manifold of dimension t equals 1 fourth of c1 of l that squared minus 2 times the Euler characteristic of m minus 3 times the signature. OK, so what you have already seen is that this is actually a manifold for a generic eta. And the only thing which is perhaps unclear about this dimension comes from. And this dimension comes from the idea Zinger index theorem. So this is. So I don't want to give you the precise statement. But the point is that whenever you have a linear elliptic operator, the idea Zinger index theorem tells you that the index of the corresponding operator can be computed in terms of the characteristic classes. And if you plug this into the corresponding formula, the output is this formula. So that's how it goes. But now, that's a very good news. But we have created here a little problem named is that this is not anymore a compact space. Because we have removed certain points. And this sort of causes, again, a problem which we have to worry about. And this is what we will do next. So let me now discuss reducible solutions. So here is lemma. So assume the first-gen class of the determinant line bundle is non-zero. So here, I think of the first-gen class as a class in the second-geron cohomology group. But if you think of this as a class in an integer-valued cohomology, so I'm saying that C1 is not a torsion class. Then for a generic metric G on M, the unperturbed cyberquitten equations have no reducible solutions. And this holds provided B2 plus of M is at least 1. So again, I don't want to prove the theorem. But let me just tell you why this is the case. All right, so let us assume that we have, indeed, a reducible solution. So assume 0A is a solution, which just means that FA plus is 0. But what this means is that if I take the first-gen class of the determinant line bundle, this is represented by I over 2 pi FA. And this is minus I over 2 pi. Sorry, this is, again, I over 2 pi FA minus. So now I have, so for any harmonic self-dual 2 form on M, if I take the wedge product of omega with FA, this is always 0, because whenever you have self-dual and anti-self-dual forms, they wedge to 0. So what we get is from this, well, we have FA wedge omega is 0. But in other words, this tells us that C1 L that cup product with any form omega is 0. And now I have assumed that this is not 0. So let me take a basis of self-dual harmonic 2 forms. So omega 1, omega 2, omega k. So this is a basis of H2 plus. And you can see this relation as k equations for a metric g. So I have picked here a basis. But this basis depends implicitly on the choice of the metric. And this gives you k equations for g. And if g is not in the set, g such set, C1 L that cup H omega k is 0. So now what I have here, I have a codimension k subset in the set of all metrics. And if I stay away from the subset, then there is no reducible solutions. k is B2 plus, yes. And so if k is at least one, I have at least one equation so that I can always move away from this sort of bad set. If B2 plus is 1, then generically in the one parameter family, you will have a reducible solution. Exactly, yeah. That's precisely correct. You can't say really much. I mean, that's really a bad case. So you have to, if B2 plus is at least one, you have the wall crossing phenomenon. But if B2 plus is 0, then you are more or less lost. That's precisely the reason why this doesn't work. Or at least why Poincare conjecture is not tackled in this way. So what we have now, we have now a smooth manifold which is compact and non-singular. The only remaining bit is an orientation. And let me discuss this now. So the basic lemma, which is more or less just a trivial statement, but let me state it anyway, is the following. So if you have two families of home operators, let me write this as TP0, where P is in some parameter space, P, which is assumed just to be a topological space. If TP0 and TP1, where P is again in the same parameter space, two homotopic families of, well, let me say, linear Fredholm maps, then the determinant bundle of T0, or the family T0, is isomorphic to the determinant of T1. And what I actually mean is a little more stronger statement. I mean, if I have a trivialization of the determinant bundle of T0, this gives me a trivialization of the determinant bundle of T1. Now, what is approved? The proof is just a simple observation. And here, by the way, I mean a homotopic within the space of linear Fredholm maps. But if you have a homotopy within Fredholm maps, you have the determinant bundle for T0. So this is a homotopy on P cross 0, 1. But the base here is homotopy equivalent to both P cross 0 and P cross 1. So whenever you have a trivialization on one of the ends, you will have a trivialization on the whole space. This will give you a trivialization on the other end. OK. Now, so why do they state this? Now, so let me recall the deformation complex. So what we had was 0 into omega 0 into gamma s plus plus omega 1. So all forms here in this complex are imaginary valued, but they don't prices anymore. So into gamma s minus plus omega 2 plus plus 0. And we have computed here the maps explicitly. This was r psi a. And this was a differential of the Cyberg-Witton map in at the point psi a. And so what I have shown was that r psi a was of the form 0d plus some r0. And this is the 0's order operator. And similarly, the differential of the Cyberg-Witton map at the point psi a. Now, this is a 2 by 2 matrix. So it can be written as dA plus 0, 0, and d plus plus 0's order.