 Last time we have finished or the whole lecture was devoted essentially to a linear gauge theory. And what we have established that linear gauge theories are not powerful enough, at least for the purposes the gauge theory is sort of valued. So today we want to consider a non-linear setup and I will try to do this a little bit abstract. Abstract, so here are Fred Horn maps. What we have is this, we have two Banach manifolds, say X and Y and the map F between X and Y, a smooth map is called Fred Horn. If you know the derivative at each point is a Fred Horn map between TXX and T. Whenever we have a Fred Horn map, we can associate to this an integer which is called an index. So the index of DXF that is a dimension of the kernel of DXF minus the dimension of the core kernel of DXF. This actually doesn't depend on X, so we never say X is connected and this is called the index of F. The basic theorem is if F is Fred Horn and Y is a regular value for F, then the pre-image of a regular value is a smooth sub-manifold of dimension D. That is the index of F. Right, this is something that you already know very well from the finite dimensional setup and Fred Horn maps are exactly the class of maps for which many facts from basic analysis goes through in almost unchanged form. So this is the place where we need that our manifolds are in fact Banach. So we want to apply here an implicit function theorem and the easiest setup that this actually does work is a setting of Banach spaces, Banach manifolds. So another theorem of great importance is the one called a theorem of Sartre and Smale and the statement is that almost any, well in quotes, almost any value is a regular value for a Fred Horn map. So let me be a little bit more precise. So the set of regular values for the whole map is of the second category. This means this is a countable intersection of open everywhere dense sets. In particular, this is dense. So I won't prove this theorem, but in fact, this is more or less an easy application of a finite dimensional Sartre theorem. Now, one more notion that I need is, I will say that F is proper if a pre-image of a compact set is compact. So now let us do something interesting. So let me assume that we have a regular value. So Y is a regular value. The index of F is zero. And so if F is proper, the fact that Y is a regular value means that the pre-image is a manifold. Index is zero, so this is a zero manifold and F is proper means that the pre-image is compact. So we have a finite set of points. So you can count the number of points in the pre-image of Y. And this is a well-defined number. And if I take this mode two, this is called the degree mode two of F. Now an obvious question here is, does this depend on Y? And of course, because the notation already says, this is independent. So more precisely, we have the following theorem. The mode two degree is well-defined. And secondly, if F zero is homotopic to F one, and so homotopic means here inside the space of red hole maps. So the homotopy should be through red hole maps. Then the degree mode two of F zero equals the degree mode two of F one. Again, the proof here doesn't actually differ from the finite dimensional case. So I gave a proof in my notes. If you are interested, you can have a look. But essentially, the infinite dimensions here do not show up. So the proof goes through as you know this from the finite dimension. But in any case, so this has an interesting corollary. If you know that the degree mode two of F is one, then F is subjective. And so this gives you, for instance, a topological condition in which you can ensure that your equation is solvable. Okay, so are there any questions to that? If not, let me complicate things a little bit. So we always want to complicate, play the whole lectures about complications. So here is an alternative approach. So I will assume that, so assume we have, instead of just one map, we have a family of maps. So let me denote this by Kali F from X times W. By the way, so here is one more condition that I should have stated. Is it as well defined provided Y is connected? All right, if Y is disconnected, we essentially, we sort of have just different maps. In any case, so let us assume that we have the whole family of maps. Well, this is again a Banach manifold. I will assume this is connected, Y, and I will assume that the following properties are satisfied. So the best property is that for any W and W, the map FW, so this is now from X to Y, just a restriction of Kali F to X times W, is for the whole of index zero, I will also assume that this is proper. The second hypothesis is that there exists W zero in W such that FW zero is our original map F. So that is we have just a deformation of a given map F. And the second condition is that Y is a regular value for Kali F. Right, this doesn't imply that Y is a regular value for straight F, it may not be the case. Okay, in any case what we have, because Y is a regular value for F, we can take the pre-image of Y, now this is a sub-manifold in X times W. Here I have a natural projection onto W, so this gives me a map Pi to W, and it's easy to show that, so here is, if you wish, the fact, lemma, that Pi is in fact for the whole, the index of Pi equals the index of FW. If you wish, this is just the same as the index of straight F. Say it again, yes. I mean, so this fact is more general. I do assume that this is zero, but even if you don't assume this, it still holds. And so by the Sartz-Mail theorem, we know that there exists W if you wish, close, so arbitrarily close to W zero, such that W is a regular value for Pi, which means that the pre-image of W, and this is just the same as FW, so the pre-image of Y with respect to FW is a sub-manifold of dimension D, so this is again the index of F, and this works again in a greater generality, but we can assume this is zero. And now the proposition is that for any W1, W2 has the pre-image, so this is a finite number of points, so I can consider the number of points in the pre-image, mode two, this is the pre-image, so it's number of points in FW2 minus one of Y, and they are equal. So, this is the pre-image of FW2 minus one of Y, and they are equal. So, maybe the last part of the theorem is a proposition. In fact, this equals degree mode two of straight F. Now here are two proofs of the theorem, one is a little sort of complicated, not quite, one is very easy and one is sort of a little bit more complicated, but since we are doing, so the motto or the slogan of the today's lecture is let's complicate things, so let us do the complicated proof. So, here is a proof. I consider the space of passers connecting W1 and W2, so let it be gamma in CK say one, two, two, W, such that gamma one is W1 and gamma two is W2. Now, this is a non-MG Banach manifold, right, so let's, and so I can pick a point here, gamma, and I can construct a map F hat from X times gamma times the interval into Y, so what I do is say so F hat of X gamma T is just F of X gamma T. And now it's easy to check that actually Y is just Y. There's a regular value for F hat, that is, well, maybe let's do that again, so we take F hat minus one of Y, this is now a sub-manifold in X cross gamma cross the interval. We have here natural projection onto space gamma, and this map is again a friend-hole, so this means that for generic gamma in gamma, we will have, so let it be again pi, pi minus one of gamma, this is simply a non-MG Banach manifold, pi minus one of gamma, this is simply what? This is the space of source XTs, such that F of X gamma T equals zero is a sub-manifold, and this is now a parametrized version of the space that we had so far, and in fact, this is a manifold of dimension one. What we have pictorially is a simple picture, so we have something like this, so this is, so we have a bunch of points, say, like here, some bunch of points in here, and this manifold, let it be M gamma, is just a one manifold which connects the boundary points, it may be something like this, right? It's not just an abstract manifold, so this manifold is boundary, and the boundary of M gamma is just F inverse of Y, W one, if you wish, cross one, and union with FW two, Y cross two. Okay, but now, you know from this picture that any smooth one manifold has at least two ends, either has two ends or no ends at all, which means that if I count points here, mod two, this will be the number of points, mod two on this side of the boundary, right? So this immediately tells us, so this implies the statement. Okay, so now, what is the, let me give you an easy proof of that, so here is proof two, since we know that FW one is homotopic to FW two, W two, because we can connect W one and W two by a pass, this is again homotopic to F, then the degree of FW one, so mod two equals the degree of FW two, mod two, and this is the degree of F mod two. And that's the whole proof, right? So of course, you may ask why did I consider this proof rather than the simpler one? The point is that the construction of the proof here is more general, right? The, what we have done here is that we allowed Y to be a fixed point, so we didn't need to vary Y, and this will be valuable for the covariance setups that I will consider later today. Okay, are there any questions to that? Yes? Okay, I see, I see, I see, you're all right, yes. Any other question? Okay, now, as you already know very well from smooth topology, the degree can be defined as an integer, not just as a number, mod two, at least in the case when we have a map between oriented manifolds, and so we may ask, is it possible to define the degree also as a number, as a Z value degree? And so this requires certain tool which is called the determinant line bundle. Okay, so the setting is this. Assume we have a family of red hole maps, so linear red hole maps, so this is where P is just a parameter, lives in some parameter space, so as I've said, each DP is linear red hole, and P is just in general just a topological space. Now, if you have this, at any point P, I can take the determinant of TP, which means the following, so this is top space, and if P is fixed, I will take that TP to be, there is no enough space, so let me continue here. So this is the top exterior power of the kernel of TP. So we know this is finite dimensional, so we can take this top exterior power, answered with the top exterior power of the core kernel of TP, and let us dualize the second space. In any case, what we have is, at any point P, we have a real line, and the important fact is that this family of lines is in fact, is there any question? Zero is always, yes, why is zero? Okay, what I'm trying to say is that we can take that TP, so that TP, now this is a family of lines parameterized by points of P, and this is in fact a locally trivial real line bundle. The local reality here is not sort of quite obvious, because what can happen is if you vary P, the dimension of kernel and of the core kernel may vary, and it's not quite clear that you can choose a local trivialization, but in fact, this is always the case, and the proof is again, not really hard, but I don't want to talk about the proof of that, so you can look it up, either in the literature or in the notes, but let us just take this as granted, and with this, let me assume that we have, again, a family as before, fx times w to y, and the properties are very similar, so fw is for the whole, for any w, say, index of fw is d, I will assume that y is again a regular value of f, what else? I will assume that fw, so the pre-image of y, is compact, the last assumption is that the determinant line of d, x, fw is trivial, well, let me say trivial over x times w. So this is a little bit more than I really need, but let me say what I'm saying is, whenever you take a point xw in this space, I can take the derivative of this function and I take the determinant, and this is a trivial line bundle, yes, and now, actually even by saying that the bundle is trivial, I actually mean that the bundle is trivialized, so there is a preferred trivialization of this bundle. Now, the theorem is, and you already know the proof, more or less, if a holds, then for a dense subset of w's, mw, so fw is, w inverse image of y as a compact, oriented submanifold of dimension d. And again, for any choices, w1 and w2 as above, mw1 and mw1, w2 oriented, so what I mean is that this manifold is oriented, this manifold is oriented, and there is a d plus one dimensional submanifold whose boundary is mw1 plus mw2. Okay, so here is a proof, actually you could yield a proof on your own, the only thing that we need to discuss is actually the orientation, so the determinant of dxfw, right, by the definition that this lambda top of the kernel, you know, tansored with lambda top of the core kernel and dualized. But now, since we have assumed that y is a regular value, there is no core kernel, this is.