 Okay Okay, now we can finally start I guess okay again Okay Shorter version geometry and topology very interesting understanding their interaction Yeah, so I guess the the the tool that I want to explore a little bit in this three lecture is the The the Iraq operators, okay, so and I guess the The way the thing I want to discuss a little bit today is like what the rock operators are and how you can somehow relate Them to these two fields so so we really see some nice interaction today between geometry and topology and Just as a as a warning so here. It's really the analysis of Of the rock operators that comes into play I'll be very sloppy with the analysis I want to focus more on the interactions, so I'll say maybe something which is imprecise, but you know nothing falls, okay? Yeah Okay, so let's start with what the rock operators are so Yeah, so I guess before starting to talk about the rock operators Let me just kind of remind you that the essentially the most fundamental operator is the Laplacian And this is the Laplacian, okay, and well, maybe you're used to seeing the one with the pluses but I Prefer the one with the minuses because this is the positive one. So yeah, I guess we'll you know This is better for geometries in some sense Yeah, in particular, you know, this is you know the the solutions To these equations are called harmonic functions Okay, and they're very important objects, for example, they They come up everywhere like in physics or even you know, for example the the potential in electromagnetism or stuff like that So, you know the study of harmonic function is a very important topic of Study in in geometry Okay. Yeah, so I guess the idea is that you know, this is Very nice. It's a second-order operator in both second derivatives, and I guess the rock 1928 asked asked himself I Guess what what he asked himself is that he wanted a first-order operator D this square is The Laplacian, okay, he was motivated by trying to study the the quantum mechanics of electrons and stuff like that But yeah, I guess from our perspective is just well, can you find some operator that squares to the Laplacian? Okay, and of course, you know, even from a very busy perspective is very nice to have a first-order equation because it's much easier than a second-order equation Okay. Yeah, so I guess, you know just to be very So, you know to be Love first guess so suppose Let's try, you know, he was guessing well how such an operator D might look like so let's guess You know, let's say D has the form a 1 d in the x 1 plus a 2 x 2 plus plus Okay, and a n let's say let's say a n are constant Okay, so let's say, you know for simplicity. Let's try the simplest thing you can try You know just write a general differential operator first order with some constant coefficients Well, and now let's let's set, you know, this is the equation we want to satisfy This one so let's set it up. So Yeah, so if you write the square of this Well, we're assuming that AI are constant. So so this square turns out to be, you know, the sum of AI squared plus the sum of AI a j the over Okay, so if you want this to be The sum of minus square the Xi square you get a system of equations Which is AI square equals minus one Well, the derivatives commute. So What you get is that, you know, the excited xj is the same as the xj the Xi so and the term does not turn We'd mix the derivatives here. So what you get here is that AI a j plus a j AI is 0 Okay, so this is a very easy simple idea Yeah, so let's see if you can solve these equations somehow. Okay, so Yeah So if n equals 1 Because n equals 1 Well, you know, I just want You know, it's it's very easy. So you just want something A1 squared equals minus 1 Okay, yeah, so, you know, you cannot do this in real number. I guess this we learn in high school But you can do it in complex numbers, right? So yeah, it's very nice because then you can set the number Okay, and then you can get your the the operator you want. So this is called the Dirac operator D So we're happy with this. Yeah, so this is and you can check easily that if you square this you get minus first There you at this good the second there the index one square Okay Yeah, and if n equals 2. Well, I think that's a little more interesting, right because Well, this is saying that H2 is minus h2 a1 Okay, so this you cannot solve, you know If if you work with complex numbers, you cannot solve this just because well complex numbers are commutative and we know What why these implies that they're non-zero? So it seems like you kind of you know, there's no solutions of these such Equations in complex numbers, but the rocket Dirac's idea is like Well, you can solve this if you you look at more complicated algebraic structures and in particular you can look at two by two matrices So you can set Yeah, you can set for example a1 Equals Zero minus one One zero two okay, and they satisfy the equations you want Okay, so then your Dirac operator turns out to be You know a1 b and x1 Yeah, and this is just minus d in the x1 plus i One plus And you can check that these squares to the Laplacian if you you know We so here at the beginning we're thinking of the Laplacian acting on functions But you know it turns out that the Laplacian has a square root if you think of it as having values in C2 valued functions Okay, so if you try to write a square root of the Laplacian on function you fail But it's nice because you know this square equals Laplacian acting Yeah, so I guess the question is like how do you go back here? Yeah, instead of thinking you as a like a function I just think of it as a as a column vector for example. Yeah, so it's a C2, you know more than Vector value function. Yeah, so this is exactly we got so I guess that the idea is that you know You have to move on from functions number value function if you move on with vector value function, then you can find a square root Yeah, exactly. Yeah, and you know just to keep on going. Let me tell you maybe I write them here and they'll be Yeah, just as a reminder, you know, just as point out this these things very look very closely to You know the anti-holomorphic and holomorphic derivative of a function, right? So Yeah, so the Dirac operator in n equals two dimension is very close to holomorphic geometry Okay, so this is more or less, you know depending on your convention. This is more or less okay, so That's some very interesting relation between this natural the Dirac operators and holomorphic geometry Okay, so maybe I'll leave it here. So I guess n equals two. I'll leave it here for later Two zero and then if you want n equals three Well, we go back to the we got the famous Pauli matrices so you get sigma one is I zero zero minus one sorry a one one zero Okay, so this is solution again in two by two matrices you can solve what we want It's eyes. Yeah Yeah, so we wanted to square to minus one So this does the job and then n equals four Can people see here? I guess you can see from there. Yeah, you You get some matrices. Yeah, you need four matrices and what we do is just you look at four by four matrices With blocks. So you get minus the identity Identity Okay, so this is one and the other are obtained by these AIs in some way So, okay, these are try this means transpose conjugate the star Okay, so once you get to four by four things You cannot find solutions in two by two matrices anymore But you know you can solve it using four by four matrices and the four by four matrices have this nice block structure Okay, and you know if you if you look at these maybe at some point You can guess how to get five six and so on you just keep on blocking Any question. Oh, I guess well Oh, sorry. So I guess the original question is why do we want this? Yeah, so I don't want to say silly things because we're not theoretical physics building. I'm not a theoretical physicist I think yeah, so to to to find the relativistic quantum mechanics. You need a first-order equation So he wanted something which was related to the Laplacian, but was a first-order equation So actually his paper original paper for net 28 is very readable. So I suggest it's very nice He's using the same notation we used today essentially, so it's like so it's one of the few a hundred year papers all Paper that still very modern. So the question is like, well, what's the dimension you need? Yes So actually every two every time you go up, you know You multiply by two and then you stay the same you multiply by two you stay the same So I think it's two to the floor of n divided by two or something like that I'll say more formally what the thing is But yeah, just draw the the first few low-dimensional cases just to show you you know the occupator Usually people Introduce them by oh spin bundle clifor algebra. It's something super concrete You just write down the matrices and you know you work with those Yeah, so I won't talk about spin bundles and clifor algebra So Yeah, you can think of yeah, so this is yeah in some sense, you know this this is I wrote down a solution to this problem But of course not the unique one, but this is in some sense the minimal one. So in a specific sense Yeah, yeah, so the question I guess the people in the chat can read that so I don't have to repeat that I have to repeat for them. Okay. Yeah, so the question is like, yeah any operator Which is this is positive because so you can't find Square root in the functional calculus sense But yeah, it's it's much nicer to work with differential operators because those are more close We'll see that that would be much more closely related to geometry. I guess that's the answer. Yeah, but yeah, of course Because it's an operator, you know, which which is positive that Appalachian has a square root as a pseudo differential operator But you know, that's hard to yeah any other question Yeah, let's move on then. Yeah, so let's keep this in mind Yeah, so maybe I'll Yeah, so we we have this the rock operator So, you know, I just wanted to say something super concrete and but this is in our end So let me tell you what what happens on a manifold Okay, so I'll just tell you the formal properties and you should think of it Oh, it's something that you know has these formal properties as and looks Something like this at each point. Okay, so this is we'll go to Yeah, so the rock operators Yeah, so the actual definition. I was mentioning I guess it's pretty complicated You know, I would have to spend probably the fall the three hours just to define the derock operators properly But yeah, let me just tell you like the formal properties of So this is a Riemannian manifold Yeah, and we let's see we choose a spin structure So choose a spin structure suppose Yeah, so I Won't even define what a spin structure is it's something about taking square roots of your bundle Which is you know, it makes sense because here we're somehow taking square roots of your most fundamental operator on your manifold It's a plushion But yeah, let me let me tell you what what's how to check if a manifold that meets a spin structure so This this is if and only if the second stiffer Whitney class of your All the tangent bundle, which is an element in Each two is two coefficients banishes. Okay, so it's a very easy thing to check if your Manifold is spin, you know, for example, if the tangent bundle is trivial, then you admit a spin structure Yeah, and it's equivalent so if And the dimension of your manifold is at least three this is equivalent to Yeah, I'm restricted to this to skeleton Yeah, so this is it so we'll always assume that there is a spin structure. It's not unique So the operator will write out will depend on which spin structure we we choose Okay, so let me just tell you what Formally the Dirac operator is then oh, yeah, by the way, I typed out notes for these lectures and I'll post them afterwards This afternoon, I guess or right after class on my website Yeah, so the the lecture notes are more or less a transcriptional way I read them with some more detail and references and Maybe point out some details that are important, but I don't have I'm not going to say in class Okay, so what what does the speed structure give us? Well, I guess The Dirac operator had this thing that you had the target space, you know We're looking not at functions, but Cm valued functions. So in general on a manifold that will correspond to some kind of bundle Okay, so this gives us the C to the M. Okay, so what is this? Well, this is a bundle. So this is a S over M. This is a hermitian bundle. So it gives you the dimension we're discussing so to the guys if I equals one it's a Like a one-dimensional. So here it's C. So here the target space was C. I guess the same color is C2 C2 and C4 Yes Yeah, and this also comes with a Connection so this is a hermitian Okay, so once you have a bundle, you know, we're not in our end anymore. So you need a way to Talk about how the fibers relate to each other. That's a connection Yeah, and that that's determined by the metric and then you have the Dirac operator And the Dirac operator an operator but D from sections of S to sections of S Yeah, so, you know, this is a little more abstract, but you know really If you get lost you should always think oh, this is the same thing that we saw in our end So we have our operation or operator between section of this bundle in in in our any is just a trivial bundle So it's CM valued functions and then, you know, there's your standard derivative and you know, the rank is what it is Yeah, and you know if you if you're careful in some sense in local coordinates this operator in low-dimension looks exactly like This these operators that I wrote. Yeah So the question is like where is this bundle coming from and the question is like locally it's Essentially exactly what I wrote here The problem is that you to do it globally you need to make it make sense of it globally You need to be very careful and that's where the spin structure comes into play Unfortunately, I won't have time to tell you where the bundle comes from but yeah There's very detailed references that in the notes. So if you're interested you can look at that Yeah, so I guess the bundle is like, you know, if you try to glue this this Yeah, the bundle. So here. Yeah, I always think of M and oh, sorry What does the but the question is like what does the bundle depend on? Yeah, it depends on the manifold And I guess the actually it depends on the metric to and the choice of the spin structure Yeah, so this is an object associated to a spin Riemannian manifold. Thanks Um, okay, let me tell you two key properties of the Dirac operator so two key properties first of all it is a Formally Self-adjoint. Okay. So what does this mean? It means it's essentially you should think of it as a symmetric matrix But if you pick two sections of your bundle, which are compactly supported, for example, if the manifold is compact then So if you do d phi Psi the L2 so, you know, you take the inner product and integrate it over the whole manifold So maybe I'm here. So, you know, S is a Hermitian bundle. So you can take inner product of things So with the node Okay, this is you know, it's a symmetric matrix. So you can bring the D to the other side Okay, it's a fun exercise in integration by parts to show that this is true for the operators I drew Down there and this is also integration by parts then a more important. Well, I guess another very important property is that It is elliptic. Okay, so this is a very important analytical property of the equation I want to find okay. I want to find Okay, but we will use a lot the consequences of this ellipticity So for example, a very important consequence key consequence. For example, is that, you know if d phi equals zero Okay, so if you have a solution to the to the to the Dirac equation. So this is a question here. So This is like phi is the harmonic And the same sense in the same way that if you have function who's Laplacian is equal to a harmonic function This is all I didn't I didn't give the name to this S. This is called a spinner bundle Okay, so the sections are what we call spinners and this is a harmonic spinner. Okay, so, you know D is a first-order operator. So for this equation to make sense You need to be able to take one derivative and that's enough Okay, so this makes sense for example for For a C1 function, okay So, you know if you if you have differentiable maybe with continuous derivative then this equation makes sense The very cool thing is that if this is true, then you're you're automatically a smooth So this is one of the key proper, you know It means that it's a very nice equation You know in general, this is not true You know if you write a random differential operator, it's not true that a C1 solution is necessary smooth But is this true and you know maybe because of what I said this is Shouldn't be too surprising for you. You should compare this Well, this is true for example for the bar equation This is what we learned in complex analysis that holomorphic functions are necessarily smooth Okay, even if you know the definition you really use very few properties and also for the application So harmonic function also have this property and spinners are kind of closely related. So so this is not to surprise Yeah, yeah, yeah. Yeah, I guess if you square it, it's yeah. Oh, sorry. Yeah, so I guess the question is Yeah, so Well, it wasn't the correct one more like a calm like Yeah, is that the the fact that this is true for the Laplacian implies this for the Iraq essentially Yeah, so that's true in our end in on a manifold. We'll see that this square is not exactly the Laplacian It's very close. So yeah from what I say later. That's that's That's true. Yeah Yeah, any other question so now we're you know, this is our Gadgets these are out. This is the tool that will allow us to somehow connect geometry and topology on a manifold Okay, so let me just let me start by telling you how this can be related to Okay, so what another key property of ellipticity? Is that also implies? Another key property that if on a compact manifold KERD. Okay, so this is The file in the space of harmonic spinors is finite dimension Okay, this is not true for the general operator So, you know, if I give you a random differential operator the space of solution You know, there's no reason for it to be finite dimensional But this is true for general elliptic operators on a compact manifold So yes in particular you get some numbers out of your operator Which is some dimension and it turns out that that number which is obtained in this analytic way So it's trying to solve some differential equations Is very closely related to the topology of the manifold. Okay, so Yeah, so we need a little bit more to state what the relation actually is I need to tell you a little more about a little bit of structure Yeah, so now let's assume and even okay, so for the first Sure, I will mostly talk about the and even dimensional case and then we'll switch to our favorite dimension Which is three and then of course we have to go to an odd In the last lecture, but yeah, let's look at an even what happens Yeah, so it's very interesting. So, you know an odd, you know, there's this diagonal term So, but you know if you like that and even The diagonal terms are always zero. So here is also always zero Okay, and it's a bit so the dark operator somehow splits in two parts and you know just and only has off diagonal terms Okay, so you also here if you write down explicitly what it is There also be some how do the composition in two pieces and the operators exchanges the two pieces Okay, so when n is even Your bundle splits There's some natural splitting. So that's the case of our end and actually it holds also on a general manifold that your bundle splits Okay, and your Dirac operator You can decompose it as a block in two blocks zero minus D plus and zero Okay, so you should keep in mind You know the case n equals two that I wrote out explicitly Okay, so the Dirac operator you can write it as these blocks with zeros and two of the other terms and this Yeah, so you have You know sections of this bundle plus and section of this bundle minus and this is plus and this is Okay, and these are called the the chiral. Yeah, so now Because these adjoint Sorry this self adjoint Okay, it's because these are symmetric matrix It means that D plus and D minus are adjoint to each other. Okay, so in some sense, you know D plus five psi Okay, so well, they're not self adjoint because you know, they don't even go to from the same place to the other But somehow they're one of the transpose of the other Okay Yeah, so if you are So and now oh Yes, sorry, this is also yeah, thanks. Yeah, the question is like should we integrate and your answer is yes Yeah, so in particular, you know if you remember from linear algebra So the if you have a matrix the image of the matrix the orthogonal This is the kernel of the transpose Okay, so this in our situation. No this still holds in this situation and we get that the dimension of the co-kernel of D Plus so the everything modular the image of D plus This is the Dimension of the okay, so this is and this is finite Okay, because well the kernel of D minus is containing the kernel D plus so and these are all finite So we can form this quantity I guess definition The index of D plus is index of the plus this is the Dimension of the kernel of D plus Minus the dimension of the kernel Sorry, well, I guess Okay, and this is a natural number Okay, so here we use ellipticity in a very very strong way in general neither of these two numbers is finite But in this case because of ellipticity and you know on a compact manifold So for now on all it was work with compact manifold. This turns out to be an integer. Okay, so what's the what's the magic here? Yeah, so the point here is that These two things These are not invariance spin structure Okay, so if you you know manifold is what it is, but if you change the metric and the spin structure, so maybe Yeah, sorry under changes of G So if you change the metric is not a topological invariance on the dimension of the kernel of the operator and the dimension of the Cockernel the operator. They are not invariance of the many. They're not topological invariance So, you know our goal is to somehow Understand interaction between topology geometry. So these quantities separately. They're not good Objects to study just because if you change the metric you get something different Okay, but the miracle is that this index it is it is an invariant. Okay, and there's a formula for it in terms of characteristic classes okay, so Yeah, so let me write this formula. So this is as very special case of the The idea singer index theorem To the apply to the case of guess this is Theorem due to what here and singer okay, so then under assumption we have so I'm a compact spin manifold Then the index So the index of the plus which is a integer Okay, and it's an analytic quantity because you know to compute it in principle You know you need to solve the Dirac equation find the space of solutions and compute dimensions So this is an analytic quantity. This is analysis can be computed in terms of the Pontiagin numbers Yes, so the question is like what happens with spin instead of spin see what's been see depends on an extra line bundle Yeah, there's a more complicated formula. So I'll just stick with the simplest case. Yeah Yeah, I guess the idea is that for any elliptic operator does a formula. So yeah Sorry, there's someone in the back first. Yeah, it turns out it's the same I read the formula and then you'll show that it doesn't depend on the spin structure Yeah, the question is like does the index depend on the choice of spin structure and the answer is no But there's no a priori reason. It's just the formula doesn't involve the spin structure Yeah, in the case of spin see the question. Yeah, but in the case of spin see structure depends on which spin see structure Yeah, and it's very cool this theorem is very cool because it relates analysis on one side and Here you have topology, you know Pontiagin classes are topological invariant of your of your manifold So somehow this is the way that the thing that will allows us the first step going from topology to analysis And let me just tell you a little bit of formulas. So n equals 2 Well, then the index Is always zero so any equals 2 mod 4 guys or 2 6 10 so somehow nothing interesting happens So if n equals 4 Well, the the index is Minus the first Pontiagin class Of TM so this lives in H4 so we can evaluate it on the top class and Then divided by 24 Okay, and you know for a format if all the Pontiagin class is related to the classical signature So you can also get you know, but here's the book you get to minus signature of M Divided by 8 guys. So the index of the dirac operator Is related to the signature and you know that there's crazy formulas going on So I'll just write the next one if any cost 8 the index is Minus 4 p2 plus 7 p1 square. So this is a class in H operate so you can evaluate it on the fundamental class of M and then you divide by 5,760 Okay, so, you know, it's already very non-trivial that this number here is divisible by 5,760 so you can you can get a lot of interesting divisibility for Pontiagin classes and a lot of cool theorems Just by using the fact that this turns out to be an integer Question that yeah, so the question is like is there a homotope invariance So I always forget the statement that the Pontiagin classes rational contract in classes are homotope invariance, right? That's not because something so yeah, they're ration their homotope. I guess Paul is Okay, yeah, they're homotope invariant The integral ones are not right, but the rational ones are yeah. Yeah, any other question? Oh The question is that why the index defined for D plus well, you can define it for D minus they're adjoined to each other So the index of D minus is minus index of D plus. So yeah, so one of them is the you just need one of them Yeah, thanks. Okay. Yeah, so I won't have this is one of the you know That yes in your index theorem is one of the most important theorems of the last century and I'm not yeah You know, I don't have time to discuss anything here other than the formula that you get so but yeah I'll there's references in the in the notes if you're interested in learning more about it Okay, so this was the link, you know our goal was going from topology to a geometry and somehow we managed using the Index here to get to analysis. So now let me tell you how to get from harmonic spinners Analysis to geometry. Okay, so yeah, so now this brings us back to the beginning of the lecture. So Geometry, okay, so recall the Dirac's motivation Was we want this square equals we want something that squares to the Laplacian, right? and In our end, okay So Dirac the place we started with is that Dirac wanted an operator that squares to Laplacian In our end and yeah, so somehow well we did a lot of steps Right, so we we wrote it down in our end and then I told you oh you can do the same thing on a manifold But not clear now that you know what what this even means and if it's true on a manifold Okay, so in fact, it's almost true, but we I need to tell you in which sense is true So on a manifold. Okay, so on a manifold. What's the Laplacian first of all? Well, the Laplacian You can think of it, you know Laplacian on our end is Minus the divergence composed with the gradient. Okay, so to make sense of these I need to tell you what the divergence and the gradient are on a manifold. Well, so the gradient, you know so this is You know, we have our connection on our bundle. So that's the analog of the gradient That's a connection picks sections of your bundle and spits out one forms with values in your bundle and this is You can think of the connection as the analog of the bundle analog of the gradient Okay, so this is Yeah, I guess again The very nice thing is that the divergence is the adjoint of the gradient in the integration by part sense so You can take You can define simply by thinking this is the adjoint formal adjoint is the analog of minus Okay, so when you integrate by parts you remember there's always a minus sign that pops out Which is now which is annoying and that's why we we Include them in our formulas. So you don't have to worry. So yeah, very very concretely what this means is that So I keep on looking for erasers, but Yes, so so what this means, you know again L2 adjoint means that you know L2 Okay, so it's a you define it by integration by parts and this is True in our end if you replace this by gradient and this by diverges Okay, and in particular, you know this expression we can define the Laplacian So, you know if you do NABLA to go here and NABLA star you go back to the original place you start with so Is our the analog of our Laplacian? Okay, so this is the analog of the Laplacian on your bundle. So this makes out, you know, this here I didn't use didn't use anything about the bundle. It's you know, the spinner bundle This is this you can make this operator for any Hermitian bundle with a Hermitian connection or like, yeah So then this is usually called the the bachner Laplacian. Yeah, so this is the natural Laplacian on any bundle with a connection Compatible with a metric. Yeah, so the key result that connects The analysis of the rock operator with Geometry is the following version of this for a manifold So this is a theorem due to the rockets Yeah, by the way, I didn't I didn't give you the time frame. This is all stuff that happened in the 60s more or less Yeah, so yeah, so this formula is almost true But not quite so there's a you know when you take squares There's always second derivatives and on a manifold, you know unlike a random second derivative I see don't commute on a manifold so you can expect that when you do things There's some curvature term popping out and it turns out that the actual formula is very nice So this is this so you can think of you know, this is exactly the formula But then you have some correction which is given by the scalar curvature of your manifold. Oh, so the question is like is there a nice proof of this theorem I think Yeah, so I guess yeah that you either you just write out I think we talk about with this Otis a lot I think by pure thought maybe one can try to figure out some just symmetries of tensors and stuff one might Figure out that it's related to the scalar curvature somehow and that's the only thing I haven't tried it Yeah, but I guess I don't know how to guess the over four though from that So, you know, am I able to somehow figure out they only depends on the scalar curvature, but then yeah, I guess that's Yeah, yeah, I don't know. Yeah, I haven't tried though. So there might be a way Yeah, so okay, I won't do the computation for you because I don't need to define properly the dirac operators But yeah, so what why is this interesting? I think you know, this is a very You know, it seems like a very innocent formula But I think what's the magic behind this formula So the magic So, you know, we were looking for a square root of the Laplacian So it's not surprising that, you know, so this square nabla star nabla our second-order operators With the sphere with the same Second-order part by design. Okay, so we were looking for an operator that's worth a lot plus in our end So, you know, it's not surprising that the second-order parts are the same But if you take the difference of two operator, which are the same second-order parts, you know You will expect that to be a first-order operator Okay In principle, okay, but the magic is that the first-order term is zero Okay, so this is a zero-thorder term is multiplication by the scalar curvature Okay, now you'll see that, you know in the argument I'll give you that if you had a first-order part and things wouldn't work out nicely at all Okay, so this is really the magic. Okay, so the magic part is some magic consolation of first-order terms Yeah, so s is just the actual scalar curve. It's a function. Yeah Okay, that's it What was the second? Oh, the second question. Well, can you stand over this board? Okay Okay, yeah, yeah, I can Yeah, any other question while I stand in front of this board, I guess I need to erase this board anyway So I guess, well, it's a good place because I forgot to the general statement. So it's a good place to write it Yeah, so I guess We said that the index The plus is zero if and is 2 mod 4 and In general, it's a combina, you know I wrote the formula for n equals 4 and n equals 8 but in general there is this crazy expression in the Pontchagin class It's called the a had genus This is called. This is a hat. This is called a had genus. This is an expression Pi of tm that you can write out if you want Okay, so let me put these two things together I guess it's a I guess it's nice because I'll just erase the middle part and So yeah, I guess yeah, you can think of this somehow relate differential operators so analysis to geometry so you can think of this as analysis Okay, so if we put these two things together, we get a relation between geometry and topology as follows So I guess this is theorem so suppose Mg a compact spin manifold and suppose that the scalar curvature is positive everywhere So they will say positive scalar curvature. Okay Then the index of the plus is zero. Okay, so this is some particular for example a G if It's a four-dimensional then, you know the index of the plus we know that it while I raised it by was the signature Than the signature is okay So for example, you know, this is in general, you know, there's a very a lot of metrics you can study on a on a Riemannian manifold and you know that it is a it's relatively easy to give constraints on the geometry of positive sectional curvature and positive itchy curvature, but scalar curvature is very hard to Study in particular. I think this is was the one of the first thing that the first intense in which you can give topological Substructions to the existence of this positive scalar curvature metric. So for example, you know There's this famous for dimensional life, which is called the K3 surface This is spin and has signature equals minus 16 Then So no PSC metric on the K3 surface. Yes Yes, that was the main What about Cp2? Yeah, that's my next example. So example Cp2 Has signature signature equals one and has has PSC metric Okay, the Fubini study metric. Yeah, so why is it that why this is not a counter example to the theorem? Yeah, let's not spin Cp2 is not a spin manifold. Okay, so these are very sharp theorem. You really need to use spin structures to spin Yes Yes, in particular, you know, if you write let me rewrite this theorem Using the previous one. So this implies that this tells you that they had genus of TM the top part Okay, so this is a topological constraint and this is a geometric constraint Okay, so somehow we managed to relate Geometry to topology using Dirac operators Okay, any question about the statement. Yes, we have a question there Well, the question is like can you have a hat non zero? I guess The question is like well, the the way this is the top class So by Pongare, you know the duo the top class has to be zero I guess because it's a perfect pairing It's like a one-dimensional vector space the top dimensional. Yeah Yeah, I guess Yeah, the lower dimensional the hey had genus really it's it's a lot of classes They're not just and those I this doesn't tell you anything about those Yeah, so this is only tells about the the the combination put drag class in the top So it's a linear constraint One linear constraint on your contrary classes. Well, I guess it's not linear in the contract. Okay. Yeah, so Yeah, let's prove this it's It's it's integration by parts Yes, so maybe I'll keep this here and I'll erase this one I guess let me call this the proof of theorem a Okay, so we show You know we want to show that the index of d plus is zero the index is the dimension of the kernel and the dimension of The co-kernel so, you know the easiest ways to show this is just to show that both those numbers are zero, okay? So we'll show that Care D is Trivial, okay, and then you know this contains care D plus and D minus Okay, and the recall index is the dimension of care D plus minus Okay, so if you show this then this implies that the index of D plus is zero Okay, so suppose it's not zero there exists Phi which is not identically zero such that D Phi, okay, then we can use this formula here. So apply it. Let's apply this formula here. So so d square Phi equals nabla star nabla Phi plus scalar curvature for Phi Okay, but of course, you know, this is D of D Phi. So this is zero. Okay, so now now we'll do our you know, let's take We take I Don't think I need this anymore. So maybe I can move so we have this formula and let's take The inner product now to with Phi. Okay, so we have zero equals nabla star Phi to But now, you know Phi is not identically zero. So it's positive. It's on zero at some point and The scalar curvature is positive everywhere by assumption. This is where we use our assumption So this term is strictly positive. It's an integral and at some point. It's positive. So this is strictly bigger then nabla star, okay, so here is where we use You know, I can't change but it's positive everywhere and now we can use, you know by definition nabla star You can bring it as a nabla On the other side. This is the definition of the fact that nabla star is a joint of nabla So you can bring it to the other side, but without the star Okay, and now this is an inner product of an element with itself. So it's non negative Okay, so we get our contradiction because there's a strict inequality here and have two zeros at both ends Yeah, so I guess this is like one of the first intents like probably the first instance in which People realize that you can't you know if you want to study Geo the relation between geometry and topology or many for you can look at the Dirac operators and you know the the rest of this class will look at some More complicated examples Yes But yeah, let me tell you, you know, these are very nice theorem, but you know, it's well you need assumption that the manifold spin But yeah, of course, you know, it's it's too nice for you to expect to be for it to be sharp in some sense So let me just tell you a very simple example in which this theorem fit. Yeah, so I guess the question I guess the question. So I guess Let's take that and torus just Rn modulo alaris so z to the n and because you know the the matrix on our end the sense So this has a has a flat metric PC Scalar curvature identically is you metric? Okay, you know You keep our together scalar curvature from the sectional currency. Well, you know, you take all these traces things So, you know, you you you got zero Yeah, so I guess the basic question is does e to the n Admit scalar curvature always positive. Okay, so can you somehow bump the curvature up at each point a little bit? Okay, so let me just compare these so compare. This is Myers theorem Okay, so this is a classical theorem in a Riemann geometry that doesn't use much technology Like all the comparison But yeah, the theorem is that if you know if you have a If M has metric With positive richie curvature, then the pi one of M Is finite? Okay, so in particular, you know what what you show is that the universal cover is compact So in particular the the pi one is fine. Yeah, and it's very interesting that, you know, richie curvature. So Remember scalar curvature is the trace of richie curvature But somehow this is much more, you know, there's much stronger or some constraints on this on having positive richie curvature and People took people on time to figure out constraints on the existence of positive scalar curvature metrics Yeah, and of course the torus, you know, pi one of torus Infinite, so no no matrix with positive richie curvature. Okay, and you know our our Now so let me right here. So here the tangent bundle of Tn is trivial Okay, you just pick like the basis the standard vector fields on on a ran and they descend to a trivialization downstairs Yes, in particular this tells me that in particular Tn is spin and Tn is spin and It's been and Well, all the pronged shagging classes are zero Okay, so in particular also So they had genus. Okay, so this constraint Doesn't apply to Tn. So the the theorem we just proved doesn't allow us to conclude that The torus doesn't emit metrics with positive scalar curvature Okay, so I guess the idea is that you know, we've been looking at us So I guess this is I don't know as a spoiler next time will prove that the answer to this question is yes And the idea is that this theorem will look at only a single dirac operator But really there's a lot of dirac operators floating around on a manifold. Okay, especially when b1 is positive So let me tell you a little bit It's this obstruction, you know, let me raise it because it's not useful for this more complicated problem Um, so so theorem a Does not answer Okay, so to answer the question, you know a theorem a we look at a single dirac operator But really, you know as the title of the mini course suggests, you know, the title is families of the rock operators So it's much better in general not to just look at the single dirac operator But you can look at some families of twisted versions of that's a question. Oh because Well, the question is like how do you conclude that kernel of D is zero Well, if you have an element in the kernel of D plus If you have an element in the kernel of D plus, that's also in the kernel of D because it's a block matrix D minus D plus zero and You know an element if an element in the kernel of D plus has this shape So then also the you know the yeah, but I guess by the definition of the domain it has to look like this Yeah, thanks. Um, yes. Yes. So the idea to study Families twisted. Okay, so what do you twist the dirac operator with so I guess I didn't really define properly the rock operators So I won't even define properly what twisting a rock operator means. I tell you I just tell you what you can twist the dirac operator with Okay, so So on M. So Consider, you know, we have the trigger bundle on M The trivial bundle and we can pick a Flat connection. Okay, so you have a trivial line bundle and we look at the connections on this line bundle in general There's a lot of them At least if you have some b1. So, you know, these these are in bijection up to up to equivalence With The torus of flat connect. This is a each one Okay, and the correspondence is given by a whole on me around. Okay, so if you have a flat connection, you know You can hold on me part of transport things around loops and you know things will come back a little rotated. So that gives you Whole on me and you know a flat connection on a bundle is determined up to equivalence by its whole on me so we have a lot of Interesting flat connection on this trivial bundle. Okay provided. Well, this is this is b1 of M Yes Yeah, and once you have something like this you can create the twisted the rock operator This is by this connection. This is the twisted the rock operator and this goes from the same bundle Okay, so you we twisted our rock operator by this flat connection be on a line Okay, so we have a big family of these Operators on the torus for example. So the idea is like try to give up take obstruction out of this big family of the rock operators And this satisfy All things we said before you are the standard the rock operator. It's just a little twisted version Yeah, so all the statements. I made it for about D Also holds for DB Okay, here you have to be careful I'm twisting with a flat connection if you twist with a non flat connection and things change But you know if you just a flat connection of things remain. Yes, you will find out next time But the answer. Yeah, I guess the answer is is no Sorry, the question is like is the answer. Yes or no Well, the answer is No, oh, yeah, we kind of proved the existence of metrics using this technique We can prove that things don't exist. So yeah, we'll prove next time at the end that things don't exist Such a metric does not exist Okay, let me just tell you know, this is a very Abstract thing or you know have this operator that I didn't define for you You know, but going back to the original thing and then we twist it, but you know, it's good because Everything can be done in a very concrete way going back to the beginning of the class, you know It's already been an hour and a half. If you remember I started writing down matrices and matrices, you know Yeah, you know this twisted the operator are not that much more complicated Listen simple examples. So let me just tell you the simplest example of this twisted Iraq operator Yeah, so basic example, right, let's say M equals s1 Yes, I guess sorry the question is like up to gauge. Yes, sorry So yeah, I guess I said it Verbally, but maybe let me record it in written form Up to gauge up to isomorphism up to bundle isomorph is there Causified by holomony. Yes There's a question there Yes, the question is that do they all square to the same thing? Yes, because it's a busy flat connection in general if it's not a flat connection You got a curvature of B term too. Yeah, but I guess in this in this mini course We'll always work with twisted by flat connections. So the the the Lickner of its formula wrote also holds for that And you used you why you need to use instead of nabla nabla B. I guess that's the yes So I'm because it's one if you remember the derac operator was very simple. It's I the let's say, you know This is Corian X. This was ID in the X from, you know complex valued function to itself Okay, and this twisted derac operators are not much more complicated. So, you know M. I think of it as R mod 2 pi Z Okay, and then I pick next here Okay, the twisted Well, all connections on on S1 are flat. So we were you know up to equivalence. They have the form the I D in the X plus C Okay for C From C infinity Complex number itself and you need to pick C Real number. Okay, so any of these turns out to be a flat connection. Oh, sorry a twisted derac operator and You know really there will be an S1 family and here we have a Real parameter, but some point they become the same so To to C prime, you know in general they look like different parators But well if you think of matrices there are like you can conjugate one to them So their equivalent does a change of basis that brings one to the other our equivalent C minus C prime is in Z. So this is an exercise in the at the end of the notes if you want there are some exercises that You know should help you grasp What's been going on in class? This is one of the exercises So, you know, we have this family of operators parent tries better real line And they are the same when the parameters the pan, you know, they're different by integer So we have an S1 family of of twisted derac operators Um, yeah, so I will and here maybe I guess I have two minutes. So I'll just Mention a Definition that I forgot to say I guess this fact of having I guess I'll say it a lot of times next time. So I guess it's good to for me to say it so we look at this deraco operator and It has finite dimensional kernel and co-kernel and those kind of parators have a name so and Equivalent like there's a nice change of basis between of them. So if you if you do a Yeah, I guess, you know at the end of the day This they become sorry Yeah, sorry, the the question is like in which sense are these operator equivalents Well, you know that equivalent meaning in the sense of matrices. So there's a unitary transformation That carries one to the other Okay, and at the end of the day, this is the say, you know You can think of this somehow as a some kind of this gives you a family of flat connection on circle And they are equivalent exactly when this is So you have this S1 family of connections. Yeah, but I think yeah the exercise Ask you to think from the point of view of being equivalent as operators Yes, there's another question up there Yeah, I might be no, I think it's pie Z right. Sorry the question is like is it pie Z to pie Z or Z? and I think The question the answer is I don't know, but I think it's Z. I thought about it before writing it, but yeah Okay. Yeah, that's yeah, that's too pie here. So the the Fourier modes are e to the And Z. Yeah, so there was a question there. Oh so the question is that is it obvious that the dark operator is Fredon and No, but it's in general true from ellipticity. So elliptic operators on a compact manifold Are Fredon between the right spaces? Yes Okay, I'm one minute over time. Can I take one minute to write what Fredon means? Okay, so definition But this so next time we've got deep into more No algebraic topology and functional analysis how they relate so this is a Word that I say instead of saying, you know, it's just a word to say find a dimensional kernel a pokernel I'm in a single wall. So so suppose you have H Hilbert space. Okay, so we'll always assume separable for us Okay, and then T from H to H a bounded operator dimension of kernel T and dimension of Cochernel T are both finite. Okay, and we can define its index in general index of T is defined to be so so here it's where I start to lie because the Dirac operator is not bounded and That there's a lot of lies, but you know, it's but yes follow has a question Yeah, so that I guess the question in general if you have the questions I do you need the image to be closed? That's automatic for Hilbert spaces or Banach spaces in general Yes follows from the open mapping Theorem, but yeah in general if you if you look at from operators between random topological vector spaces You need to add the condition that the image is closed. Yes Yes, and also yeah, I guess to be at some point. I lied to you, you know the fact that the the image of the The orthogonal the image of these the Cochernel also used that the image is closed and that I didn't even mention That is true, but also follows from ellipticity. It's a footnote in my notes. Okay. I think I'll stop here