 The plan is like this, there's four lectures. So lecture one today, what is a BPS state? I don't expect you know what a BPS state is, but by the end of today, hopefully you do. And then lecture two is BPS states in sort of the first interesting example that I know which is BPS states in two dimensional quantum field theory. And there we're gonna encounter this object called a spectral network and it's a relation to Stokes phenomena. So this is something in kind of classical mathematics. The spectral network is something a little newer. Then in the third lecture, we're gonna move on to BPS states in four dimensional quantum field theory. And there it's gonna be related to, there the kind of mathematical buzz phrase is Donaldson Thomas invariance. And then in the fourth lecture, we'll talk about some applications of all this stuff in hypercaler geometry. So very roughly, if I would make a graph of the amount of physics in the lectures, versus time, that's gonna be kind of like this. Well, I might go a little bit. And the amount of Higgs bundles in the lectures, none today, but already tomorrow we'll have Higgs bundles and that'll kind of go like this. Okay, so that's what you can look forward to. Okay, so let's get started. So today I'm supposed to tell you what is a BPS state. By the way, what I tell you is gonna be a very condensed version of what's in the notes. I am trying to, I am following these notes, but the notes have way more than I can possibly say. So if I leave a gap, hopefully you'll find it there. So what's a BPS state? So the short answer, the kind of one sentence answer, is what we're gonna do is we're gonna study, we're gonna study some vector spaces, they're gonna be representations of Lee's super algebras. So a Lee's super algebra has an odd part and an even part. So I'll use zero and one for the even part and the odd part. So the representation is gonna be also graded. So it'll be a vector space H0 plus H1. And the BPS states will be some subspace, I'll call it HBPS inside of H. What makes them BPS, what makes them special is just that they're annihilated by some of the odd generators. So in kernel of sum of A1. Okay, so we'll see various examples of this as we go on, but this is like the basic structure. Fundamentally, what I'm gonna tell you today is basically just a story about representation theory of some specific super algebras. Okay, so let's start out with quantum mechanics as you learn it maybe in like an undergraduate quantum mechanics class. So a quantum mechanical system has, I'm not gonna try to say what it is, but some things that it has, it has a Hilbert space. And right now I'm not yet in the supersymmetric world and so my Hilbert space is not yet gonna be Z2 graded. So at this moment it's just a Hilbert space. It has a Hilbert space and it has an unbounded, generally unbounded Hermitian operator. Unfortunately, I also call it H, I just can't avoid this. So it's script H for the Hilbert space and regular H for the Hermitian operator. So that's called the Hamiltonian, that's why it's noted by H. So what's the example that you do first? The example that you do first in your quantum mechanics class is called the particle on the line. So let's fix a function, capital V from R to R. Then I'm gonna give you a quantum mechanical system. Well, I'm gonna give you these two things. So what are they? So the Hilbert space will be just the space of L2 functions on R and the Hamiltonian will be the operator, Schrodinger operator, minus a half d squared by dx squared plus multiplication by V. So this is an operator that takes functions of the functions. Just take d squared by dx squared and then add V times the function. Okay, so we could think of this already as being, we could think about this already as, I'm giving a representation of an algebra. The algebra has one generated, the algebra is R, it is one generator, which is H. Until we can say this Hilbert space is a representation of that kind of stupid algebra just generated by H. Okay, well then you have a representation you ask, how does it decompose to think about that representation? I should think about how it breaks up into irreducible representations, right? Well, that's a question that physicists care a lot about. That's the question of finding the energy spectrum of the theory. So the decomposition of this representation, in other words, the spectrum of the Hamiltonian, so that means I look at, I can say it's, I look at solutions of the equation, the time-independent Schrodinger equation, H psi equals E psi, so E is called the energy. What does it look like? Well, what it looks like depends on what I pick for that function V over there. But for example, like the example that everyone does first is the simple harmonic oscillator, right? We take the function V equals a half x squared. And then the spectrum is just, if I got my normalization right. So here's the eigenvalue, I'm playing the eigenvalue. I think the lowest one is one half. So the ground state, you look for, you just look for the eigenstates, you find the lowest one has energy one half, then the next one has energy three halves, next one has energy five halves, and so on. Particularly, there's a discrete spectrum and if you just take the eigenvectors, they give you actually a basis for the Hilbert space. You know, more generally, if I just take some V of x, where V of x, let's say I make it smooth and I say it goes to infinity plus infinity as absolute value of x goes to infinity, then I'll have some picture which is quantitatively similar to this, although the details will surely be different. So I take some confining potential, some potential that goes to infinity, I diagonalize the Hamiltonian operator there, then I'll see some discrete spectrum, lowest state, the ground state, next state, next state, next state, and so on, up to infinity. Now it doesn't always happen like that. So another thing I could do is I could take the potential V of x to be zero, suppose I do that. Then this Hamiltonian doesn't have any bound states, it doesn't have any sort of, if by bound state you mean like an honest L2 normalizable eigenfunction, it doesn't have any. So maybe I should just make this picture blank. On the other hand, it does have what you call continuous spectrum. So if I look at h of the function e to the ipx, it's not normalizable, but never mind for a second. So that's p squared over two e to the ipx. And so if I allow these guys, then I would say I have every possible eigenvalue. Every real number occurs as an eigenvalue. So that's called continuous spectrum, so I'll draw it like this. For us in these lectures, mostly this is gonna be something that's bad for us. We don't want this situation. We do want this situation where you have an honest decomposition, orthogonal decomposition into discrete summands. But at some point we'll have to wrestle with this just a little bit, which is why I mentioned it now. Okay. Now let's talk about one more example before I move to the supersymmetric world. Let's see. This one's wet, so if I use this, I'll be in trouble, right? Also wet. So next example is the particle on a Riemannian manifold. So let's fix am a Riemannian manifold. And then I'll take my Hilbert space to be, again, L2 function. So now it's on the manifold instead of on R. And I'll take the Hamiltonian to be a half the Laplace operator on the Riemannian manifold. Here, let's get rid of the potential. So you could have added a potential, but let the potential be zero. And then, so now the good situation is for M to be compact. And if M is compact, then I'll have one ground state exactly at energy zero. That's the function psi equals constant. That's the ground state. And then there'll be a discrete spectrum above it. Okay, and this discrete spectrum, this spectrum here knows every little detail, well, not every little detail, knows lots of the details of the Riemannian metric. So if you change the metric, typically, these guys will move around. Oh, and this one won't, the constant function is kind of frozen here. Okay, so so far these are just some kind of number one and number two examples of quantum mechanical systems that you might write down. Now let's talk about the supersymmetric version of this. So let's do the super particle. So again, it's gonna be on a Riemannian manifold, but now it'll be the super particle on a Riemannian manifold. Okay, so like before, fix M on Riemannian manifold. And now the little change that I make is gonna be, the Hilbert space before I took just L2 functions, but now let me take L2 differential forms. So the whole complex of differential forms. So how do I write it? Differential forms L2. And the Hamiltonian will be the Laplacian, the differential forms Laplacian. Let me remind you what that is. So you can write it like this. It's the bracket of the exterior, the usual exterior derivative and its adjoint. So this is just the L2 adjoint, the formal adjoint of D. Now by this bracket, I consider these as kind of odd objects. So by the bracket, I mean the anti-commutator, I mean DD star plus D star D. So that's the usual definition of the Laplace operator on acting on differential forms on a Riemannian manifold. Okay, so now in this example, now my Hilbert space really is Z mod two graded. I mean it's actually Z graded, but I'm gonna ignore that Z grading and just use the Z mod two grading. So I write, it's the direct sum of two pieces I could call H0 and H1. H0 is the even degree differential forms. And H1 is the odd degree differential forms. Okay, so I've got now a Z mod two graded thing. And it's acted on now by an algebra which is very slightly more interesting. So before we just had the algebra of H. Now we have a very slightly more interesting algebra acting on this, I guess I'll rate it here. So we have a few more kind of canonical operators that we can write down in this setup. So H is a representation of a least super algebra. Least super algebra. I'll give this super algebra a name. It's N equals two supersymmetry in dimension D equals one. But for a minute, don't worry about why it's called that. It's just gonna be some explicit super algebra. So, sorry, it's gonna have an even part and an odd part. So the even part is just generated by the operator H, the Hamiltonian. The odd part has two generators which I'll call Q and Q bar. So the even part is one dimensional, the odd part is two dimensional. And the brackets, so let me write the brackets of this least super algebra. So the brackets are, so H commutes with Q, H commutes with Q bar and Q with Q bar, the bracket of Q with Q bar. So these are kind of two elements in the odd part so their bracket is gonna be in the even part is two H. Okay, so I, sorry, can you still hear me? So I told you the, so this is, now I've given you a least super algebra. You can check it satisfies the super Jacobi identity. So there's a least super algebra. And now I claim that it acts on H, how does it act on H? What acts just by, yeah, exactly, exactly. So, yeah, so, right. So A acts indeed on H by Q is D, Q is D, Q bar is D star and H is the Laplacian. So all I'm saying, you know, the complex of differential forms on a remandian manifold naturally comes to you as a representation of this, you know, trivial looking little least super algebra. Half the Laplacian, thanks. Yeah. Okay, but let's now explore a little bit what we can learn about differential form, you know, just this complex of differential forms by thinking of it as a representation of this least super algebra. Well, the first thing you might say, well, maybe A thing you might say is the following. Here's a little observation. Suppose I have a state psi and H. Let's compute the inner product, the L2 inner product of psi with H psi. What is it? Well, H is the bracket we said of Q and Q star, right? About Q and Q bar. So what's that? That's psi with Q, Q bar, psi plus psi with Q bar, Q psi. Oh, sorry, I screwed up the half. Yeah, I mean, don't worry about the half. Oh, no, I don't have a half in this formula, right? Oh, I do, because the commutator's 2H. Yeah, yeah, so, sorry, you're right. So half, half, half. Okay, so coming over here, psi with H psi is, so using the fact that, yeah, this is important, Q and Q bar are acting by operators that are adjoint to each other. So I'll say it's a unitary representation of this least super algebra. If Q and Q bar act by operators that are adjoint. So that's true here. So because of that, the formula I have there, you can rewrite like this. It's a half the norm of Q bar psi squared plus a half the norm of Q psi squared. And so in particular, it's positive. The first thing you say is it's positive, or at least not negative. So what you learn is number one, all of the eigenvalues of H are not negative. Because if it was an eigenvector, then what you get here is the eigenvalue times the norm squared of psi. So saying that's positive means the eigenvalue's positive. And the second thing you learn is that if H psi is zero, that happens just if and only if both of these things are zero. So Q bar psi is zero, and also Q psi is zero. Okay, so in this supersymmetric situation, first you kind of automatically get the positivity of energy. Second, you get a little more, the ground states, the states with energy zero are annihilated by this Q and Q bar, what I call the supercharges, supersymmetry generators. Okay, so now let's think a little bit about what the whole Hilbert space looks like. Thought as a representation of this A. I mean, I'm kind of recapitulating a standard story of Hodge theory on a Riemannian manifold, as many of you probably know. But the point is to cast it in this sort of supersymmetric language, and you'll see that all the later stuff about BPS states is just a jazzed up version of Hodge theory on a Riemannian manifold. That's kind of the point of doing it this way. But anyway, let's do it. So let's think of what the Hilbert space is like. As a representation of A. So what does it look like? Let me make the same kind of picture I was making before. So I'm gonna put here the eigenvalue of H. I'll still diagonalize the operator H. And now, let's see, so now H is broken in two parts, H zero and H one. So I guess I'll put those two parts separately. And then if you think about it a little bit, you realize that the structure is like this. First of all, we have the ground states. The ground states form just one-dimensional representations. They're killed by everything, by H by Q by Q bar. So they make little one-dimensional irreducible representations. And so they're down here at the bottom. Let me make an example where there's three of them in H zero and one of them in H one, say. Yeah, suppose it's like that. So let's call those the BPS states. So we said the ones that are annihilated by some Qs are gonna be called the BPS states. So here the BPS states are sitting at the bottom. And they just make little one-dimensional representations. And then you have the states of non-zero energy. And those states we just said are not annihilated by both Q and Q bar. They could be annihilated by one, but not by both. And if you try to make irreducible representations that those can sit in, what you realize is that they're sitting in little multiplets of two states. That's what I got worried about. I suddenly realized that you probably can't have that because of quadratuality. Someone tell me some numbers that are allowed. I can't do this thing on my feet. No, I don't want to, I want the number, I want the difference to be non-zero. Oh, just put two here? Oh yeah, good, that's a good one. Yes, thank you. Okay, great. So the representations that appear at non-zero energy look like this. There's two states, and for example, it could be that you have one state here and then when we act with Q on it, you get to this one, then Q squared of it is zero. Going this way, act with Q bar, take me back to here, or it could be the other way. It could be that I use Q bar and go this way and Q to go this way. Anyway, so what we have is these two different kinds of representations. We have the one-dimensional representations that sit at the bottom and the two-dimensional representations that sit higher up. So let me give those representations names. So at an energy greater than zero, we have this representation that has dimension two, two states. I'll call that a long representation and I'll call it VE zero or VE one. The one or zero just has to do with whether it's Q here or Q bar here. So this is a little classification of the unitary irreducible representations of this algebra. So at an energy greater than zero, you have this two-dimensional representation and then at energy equal to zero, at energy zero, you have just this one-dimensional representation, which I'll call short and that's called V zero zero or V one zero. Again, it comes in two flavors because that it could be, the one state could be either here or here. Now, okay, so starting with our Riemannian manifold, we've produced a representation of this super algebra. The guy in the even part could be annihilated by Q or by Q bar. So here, act with Q to get to this one but it could also be that act with Q bar to get to this one and act with Q to get to that one. There is a, I mean, it's true of course that in this theory there's a U1R symmetry. We could talk about them as representations also of U1R symmetry which keeps track of the degree but I think even without keeping track of that, I never really thought about this before but it seems to me there's two different things. Anyway, it won't be important in the rest, so if it's wrong, correct me and I'll correct it in the notes. What's really important for our purpose right now is that there's two kinds of short representation depending on whether the state is in the even or the odd part. So, it's okay, so we've produced, starting from Riemannian manifold, a representation of this kind of stupid looking Lee's super algebra. Now let's ask the following question. Suppose that I now make a deformation of this system. So, the most obvious way to get a deformation would be to just deform the Riemannian metric. And let's ask what can happen? Well these representations, these representations that are sitting at kind of positive energies, they have no particular rigidity property. I can easily deform them just by moving the energy up and down and of course if you actually change the Riemannian manifold that typically will happen. All the non-zero eigenstates of the Laplacian will change their eigenvalues. On the other hand, the representations that are sitting down here, it's not so easy for them to deform. You can see that from this point of view. If I wanna move, imagine I wanna take this one state and move it up. Well I can't because as soon as I move it up it has to be sitting in a two dimensional representation, not a one dimensional representation. This representation just considered by itself, this representation is rigid. It doesn't have any deformations. So under deformations of the Hilbert space as a representation of this algebra, the representations v i e with e bigger than zero can move. e can move to e plus delta e. But v zero i is rigid. So we say it's protected. It's protected by the fact that it's sitting in this small representation of supersymmetry. The small representations are more rigid than the big representations. On the other hand, it's not completely protected. So under just a totally generic deformation of this representation, while this individual representation is rigid, the direct sum of two of them is not rigid. So what could happen is you start with this picture with two states here. So now I'm just drawing the ground states. You would start with this picture with two here and one here and deform it. What can happen is that two of them move off. So one of them stays down here and these two join up into a long representation like that. So that's allowed. That's a perfectly legit deformation of this. Just thought of as an abstract representation. Now you may say this won't happen for an actual deformation of a Riemannian manifold and that's true. But if we just think about representation theory of this algebra for a minute, this deformation is totally legit. So yeah, I don't understand. You mean it could happen that we have sort of, it could happen that we have two guys sort of accidentally sitting at the same energy. Yeah, so then, because you first diagonalize the Hamiltonian. First thing you do is diagonalize the Hamiltonian. And then you have the bracket of Q with Q bar gives you the Hamiltonian. Oh, I see, you're saying, okay, we have to think about this a little bit. So I think it's true that the only unitary reducible representations are the ones that I said. I understand what you're worried about. You're wondering if there could be some slightly more complicated structure at fixed energy. I do not think so, but I would not be able to work it out in real time. I think I may be set this as one of the exercises so maybe we can talk about it. I think that's, I can't do it in real time, but I'm sure you're right. Okay, anyway, the point of all this is to say that the individual numbers of states that appear here and here are not invariant, are not necessarily invariant because of this process. This process where two short representations join up to make a long representation, but we can define something that is invariant. So what we do is we define a number, the index, which is just the number of copies of V zero zero in H minus the number of copies of V zero one in H. So like here, in this example, this index would be two minus one equals one. Of course, when I deform, it changes, but it changes by one minus one, which is zero. So these individual numbers, A priori don't have to be invariant, but the index does. Okay, so what's the main point? You have a Hilbert space, which considered as a representation, the whole Hilbert space depends on every little detail, every little detail of the system. And so it's therefore somehow difficult to explicitly get your hands on. It's hard to get your hands on, all the eigenvalues of the little plus operator. But this index, which counts only BTS states is much better. In fact, this is the complete opposite. Of depending on every little detail. It's actually deformation invariant.