 notes at this address, or you can just look at my web page and there's a link to this, okay? So yeah, I think for now there's only the notes for lecture one, but I'll upload the ones for lecture two right after this lecture. Oh yeah, so I can see myself, so. Yeah, so yesterday we discussed how you can use, a Dirac operator to study interactions between geometry and topology of a manifold. Yeah, and we saw that our ideas, like the techniques I were discussing, fails for a very simple example, the example of the Taurus. Yeah, so today I guess we want to go a little further and study instead families of Dirac operators. Yeah, so we'll start by talking a little bit about a slightly more abstract way to think about what happens when you have families of Dirac operators and even more general families of Freiburg operators, okay? Yeah, so I guess let me recall, so recall, so we have H, you know, it does more general places in which you can look at this, but I'll think about separable Hilbert spaces, okay? So this is, they're all the same, but you know, if you want to think about a specific model, you know, you can look at L2 functions on your manifold. Okay, I think we always have a nice compact manifold that we look at and then recall definition, you have an operator T, which is bounded, is fredel, both the dimension of the kernel and the dimension of the co-kernel are finite. Okay, so this is some sense, you know, bounded operators on a Hilbert space are very complicated things, usually, but the class of Freiburg operator is somehow nicer. Then in some sense, they look a little bit like operators on finer dimensional spaces and I guess John talked a little bit about this in his class yesterday, too. Yeah, and we can think about the index of the operator. This is defined to be the dimension of the kernel minus the dimension. The co-kernel. Okay, and with the note by Freiburg H, this is the space of all Freiburg operators. Okay, and this is a subset, you know, by definition, these are bounded operators, so it's a subspace of the set of bounded operators. So yeah, so let me just tell you two basic properties of these operators. So properties, so the space of Freiburg operators inside the subspace, the space of bounded operators is, sorry, is open. Okay, so if you have a Freiburg operator and you changed a little bit in the norm topology, then you still get a Freiburg operator, okay? And then the other thing is that the index from the space of Freiburg operators to the integer is a locally constant. Okay, so not, no. So we know that if you perturb an operator, which is fragile by a little bit within bounded operator, it will still be fragile. And furthermore, we'll have the same index. So the index is a very robust object to study. And that's why it's somehow related to topology, because if you perturb a little bit, this index quantity stays the same. Okay, and this is very important because, you know, dimension of kernel and the co-kernel. Okay, so both these quantities, both the number jump when you move a Freiburg operator by a little bit, they stay finite. But yeah, their difference stays the same. So that's why Freiburg operators somehow are, in some sense, a very topological object. Okay, any question? Yeah, so this is a crazy space. It's a space of somehow operators on some infinite dimensional space. But yeah, it turns out to have a very interesting and describable topology that I'll now tell you a little bit about. Okay, so the basic tool to study, so you know, we'll look at families of Freiburg operators, so you can think of it somehow, study the topology of this space. Yeah, so the best tool to study the topology of this space is K theory. So let me just give a K theory. Okay, so K theory is some kind of algebraic topology invariant of a space. Yeah, so let's say we start with X. This is a finite CW complex. And for everything I'll say today, I'll think of it as connected. Okay, but you can define it also for non-connected, but it's a little more annoying. You have to pick base points and stuff like that. So let me just say that I'll look at connected. Okay, so you know, later on we'll take X to be, this X will parametrize the family of Freiburg operators. And in the case we look at is the torus associated to a manifold amp. So this is the torus of flat connections on the trivial bundle, so that's a torus. Yeah, and we look at this as our main example. Okay, so we look at, this is vectors, a vector bundle, this is isomorphic classes, vector bundles. Okay, so vector bundles, you have some operations on them. You can take direct sums and you can take tensor products. Okay, so this has operations, direct sum and tensor product. Okay, so this algebraic structure, it's very similar to, so this makes into, make this into a semi-ring. Okay, so what is a semi-ring? Well, it's the same as a ring. It's just, it don't have additive inverses. Okay, so, or multiplicative, well, you don't have additive inverses. So it's something like n. I guess the definition of semi-ring is something like n. Okay, so a ring you can take additive inverses, but in something like n, you cannot take additive inverses. So in fact, yeah, there is a natural map, this from vector bundles, there's a natural semi-ring homomorphism given by the dimension. So in some sense, we have two nice semi-rings and there's a natural map between them. But they're somehow different, let me just, so I guess this is a warning. So n has this very nice property that's, well, that's not, has a nice property that's not true for the vectors on x. So on vector of x is not conciliative in general. I, the basic thing that we do with natural numbers is that if a plus c is the same as b plus c, well then a is equal to b, but that's not true for vector bundles, okay? It does not imply a, okay, so you have to be, that's the only thing that's, from a point of view of algebraic structure, this is one big difference between vector bundles and natural numbers, I guess. So that doesn't have, you have to make things, you have to be a little careful about that. Okay, so you can think about an example. For example, yeah, the tangent bundle of S2. Oh, sorry, well, I guess, yeah. I, today, I guess, I mostly think about complex vector bundles, so I guess that's not an example. Okay, okay, so, yeah, so what's, yeah, so it's, same rings are not as nice, you know, we don't study them in our basic algebra class because they're not as nice as rings. So if you have a same ring, it's, it would be, you know, we have a same ring, but it would be nicer to have a ring. And there's a natural way to get one. So yeah, we have, you know, in the same way you get N to Z. Okay, so if you remember, you know, maybe it's from your real analysis class or something, that how do you define this? Well, Z is defined as some kind of equivalence classes of formal differences of natural numbers. So in particular, in general, if you have a same ring S, we can perform the same construction and denote the same ring. And we can denote, get the same, an analog thing called the Grotten-Dick ring. Okay, and this is the ring consisting of formal differences. A minus B, formal differences. Okay, and you mod out by the relation that A minus B is equivalent to A prime minus B prime if A plus B prime plus E equals A prime plus B plus E for some E. Okay, so I guess if you do this with N, you don't need this extra E. But in general, if you want to get a good definition for non-consellative monoids, sorry, semi-rings, you need to add this extra E in your definition to make things well-defined. Okay, so this is a very nice ring. Okay, so I guess my definition, so we have our finite connected CW complex and by definition, the k theory of X, this is by definition, this Grotten-Dick ring. Okay, so this is a ring associated to your topological space. Any question for some E? Yeah, okay, sorry, the question is like, do you need some assumptions on E and the answer is no. There is some E that makes it work. Okay, let me just tell you some properties. Yeah, so an element in k theory is has a shape, it's an equivalence class. You can think of them as some kind of formal differences of vector bundles. Okay, up to some equivalence. Okay, we had a semi-ring homomorphism given by the rank and this construction is very natural, very functorial. So in fact, you get a homomorphism from k of X to integers given by the rank. Okay, so this is a ring homomorphism. Okay, and finally, this gives us a functor. This is a functor from CW complexes to rings, I guess. Okay, I guess it's a contravariant functor. In particular, if you have a map between CW complexes, there's a map on vector bundles given by pullback and homotopic map give you the same map on vector bundles. I'll give you isomorphic vector bundle maps. Yeah, so you get a nice, you know, something that looks like homology, okay? Something functor to rings that's, you know, induce map the panel only up to homotopy on your map of the homotopy. Okay, yeah, so these are just the very basic properties of k theory and let me tell you, you know, a little bit about, I guess this is in the spirit of John's discussion of representability and k theory is a representable functor. Yes, oh, yeah, I guess, sorry. The question is I find I CW connected, I guess. Sorry, that was, this is my working, yeah, thanks. Yeah, that's a, I might forget to say it at some point today, but that's always the working assumption. Yes, yeah, so representability, okay? So rank k vector bundles, k vector bundles that are represented by the grass manian. So this is the k plane grass manian, okay? So this is how you represent the k dimensional vector bundles, yeah. Oh, the question is like what happened if you drop, you know, some things still sit out true. You can go out more general, but let's stick with the simplest case for today. Yeah, I guess, you know, really we only use it for the toro. So, you know, that's not, that's not, yeah. I guess, you know, it can go with para, compact, I don't know, something like that. Compact, compact house dwarf or something like that, should be. Yeah, so actually, you know, a basic fact, again, in this setup we have, if you have a vector bundle E over X, then there exists another vector bundle such that the direct sum, okay? So if you think about what this means from the point of view of the definition using the growth index ring, this tells me that any A in k of X has the, has, is represented, is of the form. So, you know, in general, you need two bundles which are non-trivial, I guess. The definition in principle you need two non-trivial bundles, but it happens that you can always pick the negative one to be trivial. Okay, so this, you know, if you think about what this means, you know, it takes a little work, but you can show that this implies that k of X, you know, you have only one bundle here, really, and you know, the dimension here is always well defined. So, this turns out to be the same as homotopy classes of maps from X to BU. So BU, here is the union of all the gross money. Okay, if you have a k plane, you know, there's a k plus one plane just by adding a new direction, so you can define this interesting space. Okay, this is the, the, so yeah, this functor kX is representable by this kind of complicated space. Any question? Yeah, okay, so now let me tell you, so this is one classifying space. So, it turns out that, so yeah, why did I introduce k theory, you know, so far this has been a kind of an aside, so we started with freedom operators, right? And so now we, and I told you this nice algebraic tool, which is, you know, turns out to be a homology theory if you extend it properly. So this is somehow algebraic topology. We're studying some algebraic topology here. So yeah, so the key, the key point is the following, so freedom. The key point is that, so you know, here I gave you a classifying space for k theory in terms of gross monials, but yeah, it turns out that the space of freedom operator is also a classifying space for k theory. So yeah, so this is the following theorem. Okay, so under our assumption, so there is a natural bijection between, so k theory of x, maps from x to the space of freedom operators, okay? So, well, this tells you somehow, you know, the freedom operators has the same topology as BU time, this complicated, super complicated, gross monia. Okay, so, well, the space of, you know, coming from functionality somehow is very closely related to the algebraic topology of your space, okay? So I guess in some sense, this is a new, one of these correspondences, I think, you know, on this side you have analysis, and on this side you have algebraic topology, okay? Okay, so in particular, you know, one of these is like, so in particular, families of freedom operators, x, parameterized by x, these are in bijection, k theory of x. Okay, so our goal later will be to study family of dirac operators parameterized by some space, the torus, and that gives out naturally an element in the k theory of the torus, okay? And it's a one by two one correspondence. So our family of dirac operator will give us an element in the k theory of the torus. Yes, there's some question. Oh, h, sorry, the question is which h? So h, at the beginning I was talking about the separable Hilbert space, they're all the same, so they're all isomorphic as Hilbert spaces, so just pick your favorites, yeah. Like L2 sequences or something like that. Mm-hmm, yeah. Yes, the question is like, well, I have two classifying spaces, BU times Z and Fredlum. Yes, the question is like, is there a relation? Yeah, well, they are homotopic equivalent, yeah. Yes, so the question is like, is there content? Well, it's a little more, you have to be a little careful because, you know, you need to show that this space, you know, it's not a CW complex a priori, so you need to do a little work to show, you know, this tells you essentially, you know, the proof will show you that they're weekly homotopic equivalents, and then you need to show that something's a CW complex to show that they're actually homotopic equivalent. Well, sorry, can you repeat? Mm-hmm. Yeah, so the question is like, how do you see now? I'll tell you what this bijection is. That will need also in our application, so I'll tell you what this bijection. Well, it will be essentially in the, you'll see in the construction, sorry, the question, yeah, so, yeah, I'll talk about that now, yeah. Okay, let, are there other questions, while we stare at this beautiful theorem? Maybe we'll just take two seconds to contemplate more? No, sorry, sorry, what's? Oh, the question is like, what's, is this special, this fact? It's also true for real vector bundles, quaternionic vector bundles. It's key that the space is compact, though. If you look at non-finance CW complexes, then this fails, and in general, K-theory is very complicated for non-compact spaces, so that's why we will just look at finite CW complexes. Yeah, other questions? Yeah, yeah, sorry, the question is, is H complex? Yeah, we're always working complex spaces, yeah. Okay, yeah, let's move. Yeah, so let me just tell you briefly how, how to, how this bijection works. I'll just tell you what the map is. Yeah, so suppose we have a family of operators. Okay, so this is a family of operators that goes from H to H. Okay, so this T is not surjective, but it's almost surjective, you know, the kernel is finite dimensional, and because X is compact, implies that we can find a subspace of H, a finite dimensional, such that the image of TX plus W, you know, is everything. So we can find a subspace which is transverse to all these TXs, and that, you know, you can find it locally, and then, you know, just by compactness, then, you know, it's not hard to find the one that works for all Xs. Okay, so this implies that if you look at the inverse images of these Xs, this W, well, this is a, turns out to be a vector bundle over X. So this is, you know, at each space you have a subspace, and, you know, it turns out that they move continuously. So this is a vector bundle. Yeah, so, yeah, so then, you know, the map that, to a family, associates the element in K theory, sends TX, X in X, to get sent to V minus the trivial bundle. Okay, so this is a difference of vector bundles. One is given by these crazy pre-images, and then the other is these fixed vector spaces, you know, the trivial bundle. Sorry, dmw. Okay, and this is an element in K theory. So I told you how to build a map here, and it takes quite some work to show that it's well-defined and it's a bijection, okay? But, yeah, this is, the map is pretty simple. Okay, just look at pre-images, yeah. Yeah, so why are these all the same dimension that follows from Fredon-less and the index is locally constant? So the dimension of the pre-images will also be, you know, because we're looking at pre-image over, and they're transverse, the dimension of the pre-images will always be constant. So just a diagram chasing, if you look at it, yeah. Well, the question, the way the question is like, why is the kernel, why we cannot take the kernel minus the kernel of the T-axis? So the dimension of the kernel, the kernel jump, so you kind of just take the, they don't form a vector bundle, so this makes things into a vector bundle. Yeah, but morally, that's what you want to do, like, so just to make things proper, you need to do this more complicated construction, yeah. So this is, yeah, morally, this is just kernel minus co-kernel of this family. Let me just remark, this will be useful later. So remark, T of X is a family of isomorphisms. Okay, so if you have a family of isomorphisms, you know, you can take, you know, that means the image of T X is everything, so you can pick W to be zero, and then V, you know, is the inverse image of zero, but that's the kernel, and that's, this is also zero. So, you know, both of these things are zero. Okay, so if you have a family of isomorphisms, you know, T X gets sent to zero, K of X. Okay, these will be the key observation for our applications. Okay, so under this bijection, family of isomorphisms gets sent to zero. Oh, yeah, so the question is like the ring structure, how do you see them? Yeah, so, yeah, you can compose freedom operators. That gives you a composition, but then the sum, you know, the other composition is not as clear, I think, yeah, so, yeah, that's, yeah, too. Yeah, the composition of freedom operators is still freedom, but yeah. Okay, any other question? Yes, yeah, so, yes, okay, theory, it's very complicated, you know, it's defined, you know, you need to understand vector bundles and, you know, do this grotendic ring construction. So it's a pretty complicated space. Oh, sorry, before saying that, let me say. Yeah, so I guess another remark that I'll put here. Okay, so yesterday we saw that, you know, if you have a manifold M, you get this torus, M, this is given by H1, M, R, mod pi Z, okay? This is the torus of three, of flat connections on the trivial U1 bundle. Yeah, and we get a family of Dirac operators, parametrized by B, so this is the chiral Dirac operator, parametrized by these torus, and these operators go from, okay, and this is a family of freedom operators. So you got, this is a family of freedom operators, okay? So this is all the twisted Dirac operators. Yeah, so this construction gives us an element, you know, I guess I didn't say it, but you know, this map here, the bijection that goes from here to here, this is called the index map, okay? So yeah, so I guess the most confusing thing about this part of math is that everything is called index, and the theorem is that all these indices are the same, okay? So I called everything index, and they turn out all to be the same thing, and it's always confusing, oh, this is the, which index is that, which one are you talking about? But yeah, I guess so. Anyways, you get the index of db plus in an element in the k theory of the torus, okay? So this is, you know, this is, this essentially captures the whole information, all the information about the topology of this family of Dirac operators by the T.I. Yannick theorem, okay? Yeah, so I guess now the point we're going to say is that, you know, k theory, well, just introduced it, but it's a very hard object. It's very hard to compute in general. It's not like, you know, standard homology, you know, if you have a cellular decomposition, no, it's very, you know, you can compute it, k theory is much more complicated to compute. But yeah, the very nice thing is that there's a natural way to go from k theory to co-homology if you're allowed to forget some information, okay? So I guess this is k theory, co-homology, okay? Well, so first of all, you know, we have a map k, from the k theory of space given by the rank to the, to z. So, and really this z, you should think of it as the zero of co-homology of your space, okay? And this, you know, here I'm using that the space is connected, otherwise that's false. This is h zero of your space, okay? So x connected. Okay, but in general, you know, there's a much, you know, this actually generalizes to a homomorphism called the Schoen character from k theory of x to the co-homology in even dimensions. So this is the direct sum of all the even dimensional homologies of x, but with rational coefficients. Oh, sorry, the question is what are s plus and minus. So if you remember yesterday, we had the spinor bundle s and it's split in two when the dimension is even. So s plus and s minus, yes. These are the two chiral bundles, yeah. Yeah, so this Schoen character is a ring homomorphism and it's defined using, you know, as the name suggests, is in churn classes. You know, vector bundle have natural homology classes associated to them, which are these churn classes. So let me just write the first few. So churn character of just an element of a vector bundle is defined by the dimension of e. So here you see we recover this, the zero dimensional part and then there's higher terms. So then there's the c1 of e. This lives in h2, this lives in h0. And then the next term is c1 square minus 2c2 over 2 and this lives in h4. And then you get a series, which is finite because we have a finite cw complex. Okay, so there's a natural way to go from k theory to homology. Yeah, so what's, yeah, in terms of that, you don't lose that much information. So I guess the theorem is that if you tensor, if you rationalize, so k theory is an abelian group. So you can tensor with q. Okay, so k theory, we define using this super complicated, you know, much more complicated way using all these vector bundles, the growth index rings. But it turns out that if you just look rationally, you don't really get much more information than homology. Okay, but as a remark, if you look at the integral version have very different torsions. Okay, so if you want to start, if you, there's actually contains different information and homology, if you look at integral coefficient. Okay, for example, the order of the torsion might be arbitrarily large in difference, yeah. Difference in the order of, okay, any question? Yeah, sorry, well, sorry, I missed the question. There were two, was it more like two questions, right? Yes, so that, yeah. Secretly, you know, the key, yeah, so the question is like, is there some problems when you look at families and try to identify things because, you know, their identification might not give you, be well-defined and stuff. Yeah, the point is that, and it's a key point behind the proof here that I swept deep under the rug, is Cuyper's theorem that tells you that the unitary group of the Hilbert space, infinite dimensional Hilbert space is a contractible. So there's no interest in topology if you look at Hilbert bubbles in that sense, yeah. Yeah, thanks. Is the torsion in K theory well-understood? Well, the question, not by me, by someone else maybe, yeah, I guess. Yeah, I don't know what, in which sense. I guess it's torsion in homology well-understood. That's also, I don't know what, yeah. Anyways, okay, yeah, so, okay, so this object is, you know, so, yeah, you know, homology is a much more friendly object, so we have this element in K theory, the index of this family. Yeah, so what's the Athea Singer index theorem? What was the Athea Singer index theorem for families? Okay, yesterday I gave you the Athea Singer index theorem for one element, for one operator, and that's just a number, just the dimensional kernel minus the co-kernel, which in some sense, you know, correspond to this part here. In general, you have the whole information coming from the whole homology. Okay, the Athea Singer index theorem for families gives a formula for the churn character of the index of db plus. Okay, and this is an element in the homology of the tors, even homology of these tors, of flat connections. Okay, so then homology of torus, we're happy with that, so it's just an exterior algebra, right, so it's a nice one, it's a nice thing. Okay, so before that, let me, you know, the formula, yesterday, you know, the formula for a single operator involved is a had genus, then let me remind you, recall, a had genus of tm, well, this is an expression in the Pontiagin classes, yesterday I didn't write it out, but, you know, it has the form one minus p one over 24, p one square minus four p two over 500, 700, 60, and plus, you know, you get all terms in, this is in zero, this is in h four, this is in h eight, and so on, get all the terms. Yeah, so we need a little bit more of information to write down the actual formula. So, yeah, let me, so now, you know, the torus, these torus flat connections is identified with m, r, mod h one of two pi z. Okay, so there's a natural identification between the, you know, this is, you can think of this as vector space, modular lattice, so the homology of this space is naturally the lattice, okay, so h one of em with z coefficients, this is identified with h one with m two pi z, okay, two pi z, okay, and, you know, just for my mental sanity, I'll drop these two pi just by rescale, okay, yes, so now pick a basis, so x one, say, you know, let's say our torus has dimension z, okay, and then this gives me a dual basis, the dual space, okay, and because, you know, the duals, these are identified canonically, this is the dual space of this, and the dual space of h lower one of the torus is just h upper one of the torus, okay, so we have, you know, once you pick a, once you choose a basis for h one of the manifold, you also get a natural basis for h one of its torus, the dual basis, okay, this is a general, general, okay, and then, yeah, we can define the elements to be the sum of xi tensor yi, and this, you know, this is an element, you know, the first one lives in h one of m, is it coefficients, tensor h one of tm, and this lives by Kuhn's theorem, this lives in h two, okay, yes, so really the index formula will be given in terms of this element here, and this element here, and I guess let me just comment, you know, I chose a basis here, and gave me a dual basis, really, and these elementers are not to depend on the choice of basis, okay, secretly I'm writing the identity in a basis, but yeah. Okay, we're finally ready for the statement of the index theorem, so theorem, so this again is a singer, and this is from 71, so we are moving a little forward in time, so you have an even dimensional spin manifold, then the churn character of the index of the family, DB plus, this is, okay, let me write out something here, so e to the omega times a hat tm, okay, so what is e to the omega? Well, omega is this thing here, e to the omega is the one thing defined by the power series, so here, so e to the omega plus omega plus omega square over two factorial plus omega three over three factorial and so on, okay, so these are, this lives in the even comology, so this e to the omega also lives in the even comology, so these are even comology class, and this is also an even comology class, yeah, this lives in the comology of this space here, yeah, and then you evaluate it on M, okay, so this lives in the even comology with rational coefficients, and then, you know, once you have a class like this, you can evaluate it using Kudai theorem, for example, on M, and you get a class in tm, okay, it's called a SLAM product, if you want to be more precise, and you know, this is an element, so this is in, so if you do this evaluation, you get an element in, okay, so at least they live in the right, this thing lives in the right object, but yeah, we get this, you know, if we didn't have a family, well then, you know, if you look at the zero dimensional term, which is just the index, you know, you just pick up one, so you just get the a hat genus, so this generalizes the formula that we had yesterday, yeah, yeah, I guess, yeah, you can multiply a class like here, sorry, the question is like that, the dot means cup product, and yes, more or less, you know, this is a class in homology of m, and this is a class in the homology of a product, but you can multiply them, yes. Any question? Oh, sorry? Yeah, the question is like, yeah, what is a class of m, yeah, it's a fundamental class of m. Yes, yeah, the question is like, does it live in a specific grading? Yes, it lives in a specific grading, yeah. Yeah, there will be, sorry, the question is like, yes, where does, where will this live? Yeah, there will be some mixed classes in every dimension, yeah, so it will be something, yeah. Yeah, I guess, yeah, it will be in a specific grade, but it will have some kind of complicated things in it, yeah. I guess in the example we'll see, it will be much simpler, yeah. Yeah, the question is why are there coefficients in Q? Well, here we have to take the exponential, for example, yeah, and the term character in general is only defined with values in Q, yeah. And the A had genus also has Q coefficients, so I think Q is really needed here. Well, sorry, it doesn't depend on. Oh, so the question is like, what doesn't it depend on B? Well, here I'm taking all of them. Yeah, I'm taking the whole family. Yeah, if you pick a single B, then, yeah, just because the index doesn't change, then, yeah, of course, it doesn't change with B. But yeah, yeah, okay, any other question? Yes, so okay, it's a very complicated formula. I guess we don't even, we can't even figure out the grading of things, like just by looking at it, I guess. But yeah, let's look at, you know, there's a very nice application to the problem I was mentioning yesterday, to metrics of positive scalar curvature on the torus. So let me just open Lawson, okay, Tn does not admit a metric scalar curvature positive, okay? Yeah, so I guess as a remark, I can assume even, okay, if n is odd, otherwise, just take the product with S1, if this has a metric with positive scalar curvature, this also has a metric with positive scalar curvature, so, and the still torus, and I would even damage. Okay, so, yeah, so last time, let me say, so yesterday, so I guess like proof, what we really showed is that if S is positive, then db plus has trivial kernel and co-kernel for every b, okay? We showed that the index is zero, but the way we showed that the index is zero is to show that there's trivial kernel and co-kernel. So this means that db plus is isomorphism, is an isomorphism, okay? And by the remark that I left here, if you have a family of isomorphism, well, the index is zero, okay, the index of this family. Okay, so in particular also, the churn character of this family is zero. Now, to conclude, well, so zero is the churn character. Okay, but we have a formula now for the churn character of this family. So yeah, first of all, a hat genus of t of the torus, sorry, this would be very complicated. This is the torus of the torus, okay? So even more next level of complication. But yeah, the hat genus, you know, the hat genus is one plus a combination of put dragging classes. Those are all zero for the torus because, you know, the tangent bundle is trivial. So this is just one, okay? So this tells me by index theorem, plus that zero is this expression here. But now a hat genus is just one, so I can forget it. So this is just omega, okay? And, you know, this is, if you need to pick the right dimension, and this turns out to be, you know, omega if n is two m, this is a two-dimensional class. So the right thing to evaluate, sorry, nothing. Omega to the n divided by n factorial evaluated on m, okay? Yeah, and this is x1 up to signs. You can compute, and this is plus or minus x1 xm tensor, y1, ym evaluated on the fundamental class. Okay, now x1 were generators, were generators of the h1 of m, which is the torus. So this is the dual, you know, in the torus, the product of generators of a basis gives you the dual of the top class. So this turns out to be, so this is the generators for these other torus. And again, this is a non-zero class, you know, if you pick a basis of a torus in h1, their product is the top class in h upper n, so this is not zero. Okay, so really the only thing we use really is that the torus has a trivial tangent bundle and the product of the class, the generators of b1 is a generator of the top class. So this concludes the proof. Ha ha ha. The question is how do you reduce it to t2? Okay, yeah, I haven't seen that proof before, yeah. Anyways, we'll see other applications of the index theorem that don't reduce to t2, I guess. Yes. Oh yeah, so the question is like, what's the trick? So if you have a matrix, if you have a non-torus and you have a metric of positive scalar curvature, then you take the product with a standard metric here and that still has positive scalar curvature. Yeah, I guess, yeah, I guess for every, for n, well, I guess n equals one does no curvature, I guess, so there's no metric, yes. So if the circle had a metric of positive scalar curvature, then the two torus will also have a metric of positive scalar curvature, I guess. Oh yeah, I think he was confused by how the logic, if the logic breaks down when n equals one, and I think it doesn't because it's an empty first statement, I guess. Yeah, I guess. I guess the question is, is there a geometric meaning to this index element? I don't know, I think the, it's somehow measuring how far you can, you know, I guess the, I think the most understandable meaning to me is that if the, you know, it's somehow measured how far you can homotope things from being isomorphisms at all points. It gives you obstruction to your family to being a family of isomorphisms. That's the way I think of it. So the question is the homology of TM. So that's always a torus. So it's a, TM is a vector space modular lattice, so it's a torus, always. So the homology is just, yeah, the exterior algebra on the first homology, yeah. Yeah, so the question is like the, yeah. So sorry, this is, I mentioned yesterday I would lie a lot about the analysis. Yeah, so the theorems are expressed in terms of bounded operators between Hilbert spaces, which are Fredon. Yeah, the Dirac operator is not bounded on L2. It's not even defined on L2. But yeah, so things work, yeah. So I want to open the, yeah. But yeah, for families of elliptic operators, there's a trick to make into associated canonical bounded families, yeah. There are really interesting manifolds for which this, yeah, I think it's pretty general. You know, you just have any manifold in which you can, you know, some combination. You can get obstruction in terms of the homology ring of a manifold and it's point-reacting classes and how they interact. So this is the simplest case in which, you know, the aeginus is the point-reacting class of L3, and there's this nice product. But you know, in general, you get terms here, which involves cup products of one-dimensional classes together with point-reacting classes. So if some of those are, you have some interesting combination of those which turn on zero, then that still tells you that you can conclude, yeah. Yeah, so if you have P1 and a cup of one-dimensional classes, which is interesting that this argument might still run.