 On the lecture by Ashok Sain on the instantons, please. Thank you. So last time we saw that every d instantor amplitude has its overall normalization fixed by the exponential of the annulus. So today we are going to try to see how to calculate this. And there are two reasons why we want to start with this. First of all, as we discussed, this is what appears in every d instantor amplitude, right? So you have to calculate this to get the overall normalization. And also, this certainly looks like the simplest diagram, simplest worksheet diagram. Now, even though the methods that I will be describing are quite general, and they can be applied to any theory, we will focus on a specific theory, so that everything is very explicit. So we will focus on a specific theory, and this will be the two dimensional string theory, Bosonic string theory. So let me describe what this two dimensional Bosonic string theory is, because it is probably not something that everybody is familiar with. So 2 d Bosonic string. So the worksheet has one free scalar, x. This will basically give the time direction. And then another interacting scalar, which is also called the Liouville theory. Interacting scalar or Liouville. And this is the parameters are chosen, so that it has central charge 20, 25. So total central charge of the matter system adds up to 26, because a free scalar has central charge 1. These are central charge 25, so total central charge is 26. And then of course, it has the usual B.C. ghost system, B.C. ghost system, with central charge equal to minus 26. That is the standard Bosonic string theory, except that 25 of the scalars I have replaced by one interacting scalar, which we call the Liouville. And this theory is very simple, because if you look at the spectrum of closed string states, this describes a single massless scalar on a half line. So this half line is basically parameterized by the Liouville direction, because there is an interacting scalar, a chi L, which from the target space view point will describe a coordinate, a particular coordinate direction in target space. But because it is an interacting scalar, there is a potential. So there is a potential, which takes this form. And because of this potential, essentially, the space is halved. These closed strings cannot really go on the other side, at least not perturbatively. That is the sense in which we have a theory on a half line. And the x, of course, describe a time coordinate. So it's a 1 plus 1 dimensional theory, of which the space direction has this potential, which effectively makes it into a half line. And the important point is that this theory has a de-instanton. And since by definition, de-instantons have Dirichlet boundary condition along the compact and along the non-compact direction. It should have a boundary condition along x and chi L. This is also sometimes called the z-z instanton, because Zamolochka discovered this instanton. So in this theory, we can calculate this annulus diagram. And later on, I will explain in a little more detail what goes behind this calculation. And this has the form of an integral that we have already seen. If you recall, I gave this as one example of an integral that appeared in the lectures. And this is where it appears, the annulus diagram for de-instantons of this theory. And this one clearly is divergent. It diverges as t goes to infinity. This one diverges exponentially. This one diverges logarithmically. And our task is to try to extract finite result of this. Now, often we may think of these divergences as arising because we are dealing with Gosonik spring theory. And one might expect that perhaps when you go to super strings, these divergences will go away. So you can compare what happens in type 2B. This theory has a de-instanton. And in this theory, if you calculate this, you get an integral like this. So I have written it as 8 minus 8 because this 8 comes from the Nebuser sector. These are the Gosonik state contribution. And this comes from the Ramon sector. If you are not familiar with this language, it is roughly Gosonik sector and the Formonic sector of the open strings. And indeed, it does look like the situation is better because this is a 0. And you would think that, OK, this is 0. So the exponential of the annulus is just 1. But if you really take this as the answer and go ahead and calculate various amplitudes, you will get results that are in complete contradiction of whatever you know from S duality, for example. S duality does give some prediction for certain de-instanton amplitudes. And the result that you find by taking this equal to 1 does not agree at all with what S duality predicts. So this is not the right answer. So the lesson is that even sometimes in super string theory, you may think that, OK, divergences have cancelled and hence we can proceed. You really cannot cancel divergences of this kind. You have to separately interpret the divergence that you see in the Nebuser sector and the Ramon sector and then try to get the result. So I just gave this example because once in a while I'll refer back to this. Just to show the similarities between the two situations. In these 2D bosonic string theory, are there tachyons? 2 dimensional string theory has? Are there tachyons? Are there tachyons? Yes. No, there are no tachyons. This massless scalar actually is what we call the tachyons in bosonic string theory. But in 2 dimensional, OK, this is a Liouville direction. So the Liouville direction has this linear deleton field. And that basically makes what would be the tachyon in bosonic string theory into a massless field. And so the tachyon is in this massless scalar field. So there is a tachyon in this. So this is a fully consistent string theory? At least part-arbitrally. Partarbitrally it's a fully consistent string theory. Any other question? Yes. It works. I didn't fully get where are the numbers 8 and 8 coming from? Is it the 8 and 8 directions you can move the? I'll explain this. I'll explain where this 8 and 8 come from. Because there is some subtlety here. Now, to make sense of this, we'll need to make use of string field theory. And as you heard yesterday, it's a bitter pill. So everybody wants to run away when string field theory is mentioned. So what I'm going to do is we'll try to give you the dose in small amounts so that not everybody runs out at the same time. So let me, at this stage, just give the bare minimum of what you'll need. At least for the next few steps, then I'll give you some more details when you need them. So for now, need to know the following facts. First is that string field theory is like a regular quantum field theory. It is a regular QFT, infinite number of fields. And basically, it's infinite number because there is one field for every more of the string. And since string theory typically contains infinite power of states, that's why you have infinite number of fields. Of UV divergences? Yes, it is free of UV divergences. Yes indeed. It's a free of UV divergences. But otherwise, it's like a normal quantum field theory. It's free of UV divergences because the interaction vertices have some kind of exponential suppression in momentum. And that's why it's UV finite. But otherwise, the Feynman rules, et cetera, are all like in normal quantum field theory. The second point is that it is designed at least formally reproduce the world sheet result for scattering amplitudes because it's not a new theory. It's just a formalism which is designed so that you get back the usual scattering amplitude. But of course, there is some difference. That's the sense in which it's formal. And this, hopefully, will become clear later. And the final point that is specific to D in sentence is that the string fields describing open string modes on the D in sentence live in zero space time dimension. And that's because the D in sentence are point-like objects. They are localized in space and time. And since open strings live on those D in sentence, they also really move in zero dimensions. Sometimes, there were the D in sentence may have compact norm and directions. But those compact directions, of course, you can decompose into appropriate harmonics. So as a field theory, it's a zero-dimensional field theory. And this means that a path integral reduces to ordinary integrals. Because this is what you should expect. That when you talk about path integral over open strings living on D in sentence, we should really see some ordinary integrals. Maybe over infinite number of variables. Maybe over finite number of variables in special cases. But it will be ordinary integrals. They are not really path integrals. Now let me write down the general formula. That one was written specifically for the case of two-dimensional bosonic string. But the general formula for this takes the following form. And the z of t is given by trace of minus 1 to the f e to the minus t times l0 is 0 c0. I'll explain this. So the trace is taken over all open string states on the D instanton. This f for us will be ghost number plus 1 of the state. So I have to define the ghost number. So ghost number is defined so that c has ghost number 1. I introduce a bc ghost. b has ghost number minus 1. And matter has ghost number 0. So x and chi l have ghost number 0. Ghost number 0. And once I have given the ghost numbers for the fundamental fields, you can find the ghost number of any state that we have. So that's what enters here. The important point is that the statistics is related to ghost number in an opposite way. So if the ghost number is odd, this is thought of as a bosonic state. If the ghost number is even, this is thought of as a harmonic state. L0 is the scaling operator. So this basically measures the conformal dimension of the state. This is the standard variable generator. And this b0, c0, these are the ghost 0 modes. bc on the annulus, both b and c have 0 modes. The reason that this is inserted in the trace is because this is essentially projects. This is a projection operator. This projects to states satisfying states that satisfy b0 acting on psi equal to 0. Now this is sometimes also described as that you have to insert the ghost to soak up the 0 modes. On the annulus, there are ghost 0 modes. And unless you soak up the ghost 0 modes, you get 0. A simple way to see that that will happen is that imagine that you have a state s that contributes to a trace. Imagine that you have a state s such that b0 acting on state equal to 0. Then you can build a whole tower of states by s and c0 on s. If s is there, c0 on s is also another state in the spectrum. b0, of course, you cannot apply any more because that annihilates it. So these two have exactly the same l0 eigenvalue because they are applying a0 mode of c. And they have opposite statistics because c has ghost number 1. So if s had even ghost number, this has odd ghost number and vice versa. So they have opposite statistics. Their contribution will cancel out in this trace. To get a non-zero answer, we have to insert this projection operator. So that this state is not there. We basically, this projection operator removes this state. Now this z of t is not identically 0 because this is sometimes what is called that you have to soak up the ghost 0 modes. But this also comes as part of the wall sheet rules. Wall sheet rules automatically insert this ghost 0 modes into the correlation function. So with this understanding, we can also write z as sum over b e to the minus hb t minus sum over f e to the minus hft. Where hb and hf are the l0 eigenvalues, bosonic and formulaic. Is this OK so far? So now you can see where the problems will be. If you look at that integral, typically from the lower end there is no divergence as long as you have a consistent string background. On the upper end, we can see that the problematic cases are when hb or hf is non-positive. So problematic cases are when hb or hf are less than or equal to 0 because that's where the large divergences come in because z is not sufficiently damped at large. So what we'll do. Sorry, what about the possible divergence at t equal to 0? Yeah, so what happens, you can see that there is no divergence from t equal to 0 and here. In this case, yeah. So typically, the t equal to 0 and is the closed string tadpole channel. And as long as we have cancelled closed string tadpole, which is necessary for consistency of the string background, the t equal to 0 divergences get cancelled. So it happens here, but this is generic. As long as you have a consistent closed string background, you don't get any divergence from the t equal to 0 end. So the strategy will be to start with cases which you understand that consider first hf greater than 0 cases and rewrite the expression in the string filter language. And then once you have done this, then we use the same language then use the same language for hb, hf less than record the 0 cases. So that's what we'll try to do. How is the strategy different from the standard analytic continuation on the metamorphic plane? The point is that here there is no parameter, you can do analytic continuation. Normally, there is some momentum that you use to do analytic continuation. So this cannot be interpreted as analytic continuation in hb or hf. That's right, yeah, exactly. In fact, we'll see that for hb negative, we can in some sense think of this as analytic continuation. But for hb and hf equal to 0 are the problematic cases. There you really have to use the full insight of string filter. But strictly speaking, even in hb, we cannot do analytic continuation because string theory gives us fixed value of hb. And there is a consistent theory with a slightly different value of hb. So now let's write down some identities. So these are identities for hb, hf equal to positive. First one, the integrals 0 to infinity dt over 2t square root of hf over hb. These are mathematical identity you can check. I'll not prove this. If you don't believe it, you can put it on a computer and check. Put it in values of hb and hf, it will come up to be right because everything is finite here. The second identity is hb to the minus half is can be written as integral d psi b over square root of 2 pi. But psi b is a Grasmann odd, even variable. They are standard Gaussian integral. And then similarly, there is an identity involving hf in the numerator, which is integral dEf dVf. Why these are Grasmann odd variables? So sum is over all states which are physical or unphysical. Yes, everything. So I'm just trying to understand what are these problematic states. I would have thought that physical states which are BRST, cohomology states should have positive weights with respect to some physical L0. We'll write down explicitly those states. But are these physical states which have this property or these are mostly BRST trivial states? No, they are not BRST trivial. They are not BRST trivial, but yeah. So certainly, there will be zero modes, for example, which are physical states, which have this property. Many of these, zero modes are typically physical states. OK, zero modes. But the negative ones? Negative ones are not physical, but they are still there. If you just throw them away, you don't get the right answer. OK. I mean, you really have to include everything. And then here, there are no positive ones. But often, they are also positive ones, which are again not BRST invariant, the unphysical states. But again, you have to take into account their contribution to the partition function. So because it has something to do with the zero modes, is it some way of implementing the collective coordinate quantization and field theory? Partly it is collective coordinate, but partly it is something else that we will see. Partly it is something else. OK, thanks. Now one technical point that you see, we have in this identity, you have square root of hf. But hf always comes in pairs. This is a general observation that, I mean, when you have this sum, same hf appears twice. So even though individually you have square root of hf, once you take the product, you always have integer parts. OK, that's how you can make use of this. So with this, we can write this exponential of this as integral factor over b, let me write this prime, just one minute. There are logs when you are there in exponent. Oh, yeah, yeah. In the log hf over log hb, yeah. Exponential of this, yeah. So there are logs in this integral, right. But you have to exponentiate it and that's what gives you this. Thank you. Any other question? So this prime basically means that for the formulas, we take the product over pairs. OK, that's why we, because each integral gives you hf and not square root of hf. So you sum over, you organize the sum over all the formulas into sum over pairs of formulas. And now the idea is that this integral, we are going to interpret path integral over open string fields. Later, we'll see, we'll write down the open string field for reaction and then we'll see that this is indeed the open string field for reaction. But for now, we have already got this from our starting point. So let's just proceed with this. OK, we interpret this as a path integral and this as the action. Are there questions? So now, first let's look at the, now, OK. So now the idea is that we have got this for hb, hf positive. But now, we are going to use this as a definition of the whole thing. Even for when hb and hf are negative or 0, we'll take this as the fundamental description of what this, the exponential of the annulus partition function. OK, if we had started from string field theory, this is the kind of expression you would have gotten in the first place anyway. So first, let's look at the hb less than 0 states. hf less than 0 actually is not a problem at all, right? You can see here because there are last one integrals. So you just get back hf. hb less than 0, right? What is it saying? For hb less than 0, if you have an integral exponential of minus half hb psi b square, this exponent grows along the real psi b axis. But this means that the, it's a saddle point, right? This is not a minimum of the action. It's a saddle point, right? Because along the real axis, the action is growing. But whenever you have such situations in a path integral, normally the way we deal with this is to integrate on the steepest descent contour. So you don't integrate about the real axis, you will integrate over the steepest descent contour, which in this case is the imaginary axis. So integrate, so for hb less than 0, since that is the steepest descent contour. And the answer that you get by doing that is basically 1 over square root of hb, right? The same as this, 1 over square root of hb, except that because hb is negative, this is imaginary, okay? And the fact that it's imaginary, you can interpret by saying that because of the fact that they are integrating along the imaginary axis, the integration method has a factor of i, right? That's what is responsible for this being imaginary, okay? But the answer otherwise is just 1 over square root of hb. And this, in a sense, you could have said that you are just doing analytic continuation from this expression, right? To negative value of hb, right? So that sense, it is indeed equivalent analytic. Sorry, isn't there some sine ambiguity? Yes, there is sine ambiguity, right? And that sine ambiguity has to do with the fact that the steepest descent contour can be chosen either along the positive imaginary axis or along the negative imaginary axis, right? So this is the choice, ambiguity is the choice of contour. And then you have to use physics to fix the correct choice, right? You have to use input from physics to fix the correct choice, right? You have to start with the original integration contour and then see how you can deform the original integration contour so that you get part of the steepest descent contour, okay? This is the same as in quantum field theory. Is there any natural i epsilon prescription that one can use here to fix any ambiguity of? You can in the sense, I mean, we will see that the unitary for example, right? We will put constraints on what kind of sign you should get, right? I mean, in particular, the de-instantons, right? We will give imaginary part of the, into the closed string amplitude, right? And the way we interpret the imaginary part is that we saw that the closed string fields live on a half line, right? And the imaginary parts basically tell us that the particle can actually leak through that wall, okay? Go to the other side. But that means that the actual scattering amplitude should be less than one, right? s dagger s should be less than one, not larger than one, and that fixes the sign. But if you are, if you don't worry about the physics, right? It's up to you how you choose the contour, right? When you have defined an integer, I mean, that's true in all quantum field theories. If you are given the action, right? You have to also tell us how to, what contour you are doing the path integral over. Now, you notice that this kind of argument doesn't solve the problem when HB and HF are zero. Because when HB is zero, for example, right? You are just doing integration over psi B without any damping factor. When HF equal to zero, you are doing integration over the UF and VF without any dependence on UF and VF. So HB integral, sorry, psi B integral, ni, ni will diverge, and HB integral, UFB integral, ni will give you zero, right? And this is where the insight from quantum field theory comes into play, right? You have to understand the origin of these zero modes, and then use the insights from quantum field theory to deal with these zero modes. And this is what I'll try to describe now. So origin of bosonics, origin of zero modes. So bosonic zero modes, okay? These kinds of zero modes also arise when you are doing standard instant calculation in quantum field theories, right? And those typically describe, reflect the fact that the instanton solution breaks certain symmetries, and these are the goldstone bosons of that symmetry, right? For example, in four dimension, if you have an instanton solution, it breaks transition invariance because it's located in space and time, right? So it will have four zero modes that tells you the degree of freedom to translate the instanton along the space and time directions. So at least one class of bosonic zero modes will come from broken symmetry. Typically, I is from, so for example, for the instantons in 2D string theory, 2D bosonic string, we have, we break transition invariance along x, so we expect one zero mode. Along the kaiel direction, there is no transition invariance anyway, right? So you don't get zero modes from that direction, because there is a potential. The instanton is fixed along that kaiel direction. In 2B, we break-break 10 transition invariance because you have a point-like object in 10 space-time dimension, right? You can move it anywhere along those 10 directions. And so you expect that there will be 10 bosonic zero modes. So now let me describe how we deal with these zero modes, okay? And let me again focus on this case, 2D string theory. So the fact that we expect that there should be a transition zero mode, right? Basically, another way of saying this as the D zero mode variable, let me call this psi B zero, right? Because general psi B have used for all the modes. So let's say psi B zero is a particular zero mode. This is related the D instanton position along x, y along x. So you expect a relation like psi B zero to be equal to some constant, k1, y. And there may be higher order connections to this relation that you will not worry about right now. k1 is some constant. Later we will see how to calculate this constant. So once you have understood this, right? Then integral psi B zero or D psi B zero is k1 times dy. And now we can see that the integration over y has the interpretation of the, as the integration over the D instanton position. So the fact that the D instanton has to be integrated, the position has to be integrated, is already encoded in the word sheet expression in the form of a zero mode, okay? There is a divergent quantity here, right? And that's what tells us that we have to integrate over the D instanton position, right? Because one of the zero modes has the interpretation of this transition mode. But once you have identified this, then you can use the insights from quantum field theory. When you have an incident in quantum field theory, here also you have this collective mode, the position of the instanton. But the rule is that you calculate the, you do the integration over the position of the instanton at the very end of the calculation, okay? You have to put everything together and then you integrate over the position of the instanton. Which in this language means that once you have separated out the integral this way, then we don't try to calculate this directly, okay? We calculate the full diagram first. This with all these, for example, right? The leading contribution as we saw has annulus times all the disks. We first, so we first separate out the zero mode integral. Then the integral, we calculate the product of everything, okay? Not just the annulus, but annulus and product of everything. From the annulus we have separate out the zero mode. These have Y dependence because the results for these depend on the, if the instanton is placed at position Y, right? This diagram knows about Y because the boundary condition is dependent on where the instanton is put. So what one can show is that the dependence of this on Y has the form of e to the i e times Y, where e is the total energy of all the closed strings. So you keep the Y integral undone in the intermediate stage. At the end, when you do the Y integral, we, this Y integral gives integral dy e to the i e Y, which is the energy conserving delta function. So in other words, even though individual disk wall sheets didn't have the energy conserving delta function, right? Because I told you that, because the boundary condition breaks energy conservation, but the energy conservation is restored at the end after you do the Y integral. And in a very precise way, right? You get this 2 pi delta i and you have this factor of k1 that tells you how to relate psi b0 to Y, okay? This is something you have to calculate. In the case of type 2b de-instanton, you basically get 10 factors of this k1, right? It may not be the same k1, maybe some other constant, but you get 10 of these factors because there are 10 transverse interactions. You determine k1? I didn't get it. How do you? Yeah, how do you determine k1? We'll do it. For that, you need the second dose of string field theory. So that's how I'm postponing it, right? Some constant that you have to calculate. Could I imagine that maybe I would cheat and I just put a small potential along the x direction, compute the integral and get rid of the potential afterwards? You could try to do that. There are various ways. I don't know how that you can do it with small potential, but basically the way you, okay, the philosophy will be that we know that Y dependence of the amplitude is of this form, right? It will be I, U, I. Now you look at how psi b couples to the amplitude. Psi b is an open string state, right? You have to correctly normalize it, find this vertex operator, and basically calculate the wall sheet diagram to see how this couples to the rest of the closed strings. And by comparing these two, you can find the relation between, we'll find the relation between psi and y, right? But normalizing the psi b requires a little more detail about string field theory, which I'll do it, do eventually. Thanks. Now, again, although super strings are not what I am going to discuss here, let me just say that for type 2b super string, instanton n 2b super string, we have 16 fermions 0 modes. This is because the instanton breaks 16 super strings, okay? So there will be 16 fermions 0 modes. And so you expect a minus something there, that's the contribution from Raman sector. So this is responsible for the minus 8. Now, why it's minus 8 and not minus 16 is a technical point. Basically, what happens is that we wrote down the expression as this, but for the Raman sector, actually this is square root of hf, right? So there is another factor of square root because the Raman sector involves what is called the super conformal generators, right? So that's the reason why even though you have 16 0 modes, you see it as a minus 8 in the partition function. But anyway, this is not what we are going to discuss today, but the general philosophy is that whenever there is 0 mode, okay, you have to take into account in the same way that you do in quantum field theory. You have to integrate over the 0 modes explicitly. And again, you do it at the end of the calculation. Are there questions? Now, if you just go back to the expressions that you had written down, you will find that the numbers don't quite match. Because we said that we found 1 0 mode, right? For in 2-D Boson X string, we found 1 0 mode, 1 Boson X 0 mode. But if you look at the expression for z of t minus 1, e to the t basically means that there is an hb equal to minus 1 state, right? That's the reflection that this is a saddle point, right? There is an open string tachyon on the D instant on. But you would have expected a plus 1, right? Because we found 1 Boson X 0 mode. Right? So, you should have gotten plus 1. Instead, you have got minus 1. If we look at 2B in 10D, okay, there you have written down the expression for z t as 8 minus 8, okay? This we understood. This is Raman sector. This we understood from the 0 modes. Where you see, you would have expected 10 0 modes. Because there are 10 directions in which you can translate the D instant on. Right? So, these should have been 10, right? Instead, you are getting 8. Okay? So, somehow we have not counted all the 0 modes, right? Something is missing. And what is missing is the same value in both cases. We are missing a factor of minus 2, right? Because you would have gotten plus 1. Because the natural expectation, we are getting minus 1. Here, you should have gotten 8. We are getting minus, we are getting, sorry, you should have gotten 10. We are getting 8. So, what I will do now is to write down the other 0 modes, right? Which are missing. And that is the point where this is the case where things differ a little from usual instantons in quantum field theory. The additional 0 modes. This is I will write down the states in the CFP or the vacuum. And C1, C minus 1 vacuum, okay? This is the SL2R infinite vacuum. So, B0 annihilates this. And B0 also annihilates this, right? There is a C0 to protect this. So, these satisfy the correct projection condition. These are L0 equal to 0, right? This clearly has L0 equal to 0. This is the vacuum, right? And this has L0 equal to 0, sorry, C minus 1 C1. These are L0 equal to 0 because total L0 of these two add up to 0. So, these 0 modes are present in the open string spectrum. And in the language of string field theory, they are going to give two Grassmann integrals, right? And you have to understand then how to deal with these two Grassmann integrals, right? So, these 0 modes are responsible. These are responsible getting some integral like du, dv without any weight factor. Goal will be to find the interpretation of these. Now, it turns out that these arise from wrong gauge fixing. Because these arise from wrong gauge fixing of string field theory. But what I will describe do first is to explain this by using just the gauge theory description. Later on, we will see explicitly how this happens in string field theory. But we will first understand what is happening just by using gauge theory description. Can we explain what you mean by the word wrong here? We will see. That is what we will explain, right? That is what I am going to explain now. What is wrong with the gauge fixing, right? So, for this, we have to go back a little. And instead of talking about D instantons, let us consider just general D p-brain. Then p equal to minus 1 will correspond to D instanton. So, take a D p-brain, D p-brain. This is any theory, p plus 1 dimension will work all over. D p-brains always have a U1 gauge field. This is probably one of the first thing that was learned about D-brains that D-brains carry gauge fields. In particular, if you have a single D-brain, it carries a single U1 gauge field. And one would expect just from gauge invariance that it will be, the effective action will be given by Maxwell action. So, string field theory action for these fields is integral D p plus 1 x minus 1 quarter f mu nu. This is where you would stop for a Maxwell action. But string field theory gives you a little more. This phi here is an auxiliary scalar field. Question? Yes. Typically, on the D-brain world volume, we have a DBI action. So, why don't we use something like that? You could have used it, but you don't, I mean for this purpose, we are just calculating the angular diagram, right? So, this quadratic terming action is enough. For perturbation. Yeah, because this diagram, right, we are just calculating one loop vacuum diagram, right? For that, you only need a quadratic piece, right? Exponential of minus I B square or whatever, right? Yeah. So, DBI action just becomes this. In the low energy limit. I think we are using Euclidean, so let me put everything plus. Now, this is an auxiliary field. In fact, you can just integrate out phi, right? And eliminate this and you get back the usual Maxwell action. But string field theory gives you this. So, let me keep it as it is. And if you are wondering where this, what this phi field is, right? Let me just say, I will also maybe give the Mu field, what states in string theory correspond to, right? This is the, these fields probably you have already seen, if you have gone to some of the voice sheet calculations. So, Mu corresponds to this state C1 alpha Mu minus 1 on the vacuum. So, this is the vacuum that carries momentum k and these are the oscillators of. So, in the same language, the phi field corresponds to C0. No, this is in the gauge invariant form. This is gauge invariant. I am going to write down the gauge transformation laws. Then you just get this because this is quadratic. So, you can just shift phi called declare this as phi prime and just integrate out phi, right? That just gives you some constant factor in the problem. So, this is completely gauge invariant form of the action. The only thing wrong about this phi is that this does not satisfy the B0 on phi equal to 0 condition, okay? But that from the perspective of string field theory that B0 on state equal to 0 is a gauge condition, okay? Here you are writing the gauge invariant form. So, you do not impose that, okay? So, that is why this phi is there and also what are all the couplings, you find that the action has this structure. Yeah, so that is what is normally called the Siegel gauge, right? So, the voice sheet expression that I gave you, right? Is something that comes out of string field theory in Siegel gauge, right? So, the particular voice sheet action you had this exponential form, right? That you will get as a string field theory action in the Siegel gauge, okay? But right now let us not worry about those. All I am saying is that the string field theory gives you an action of this form, okay? With an extra scalar. And it in fact does a clever job I will explain why, right? In fact, it is probably a little more clever than just Maxwell's form of the action. So, the gauge transformation laws are delta Mu, okay? I will just normalize the function square root of 2. This is a standard gauge transformation of gauge fields. Theta is the gauge transformation parameter. This is clearly invariant and you can easily convince yourself that this is also invariant under gauge transformation, right? Because this goes as box theta and this goes as box theta. This transducer box theta. So, it is completely gauge invariant form, right? Now the way string field theory gauge fixes, okay? Which is what is called the Siegel gauge and that is what gave the annulus partition function. Gauge fixing in string field theory just sets phi equal to 0, okay? Phi transforms under gauge transformation. So, you can set phi equal to 0 and that is the gauge choice. So, once you set phi equal to 0, the action simplifies. The action just becomes integral. You can do some integration by parts, minus half Mu box Mu. You recognize this is a Lorentz gauge action. This is a gauge fixed action. And you have Hadyapoko ghost, right? And the ghost kinetic term is just the determinant, as I just, it is just computing determinant of this box operator, right? Because you have set phi equal to 0 gauge. So, the ghost, it is easy to read out for the ghost kinetic term is. So, the ghost kinetic term is box. So, A's ghost will be integral. So, if you are doing on a DP brain, U and V, these would have carried momentum, okay? And string field theory actually precisely give you this U box V kind of term, okay? So, string field theory automatically knows how to gauge fix, okay? How to introduce Fourier power of ghost, etc. And you would have gotten, everything would have been consistent, yes. But this does not fully fix the gauge, because you are still free to perform gauge transformation with box, box data equal to 0. Yes, but those are residual gauge invariance, right? You do not have to fix them. Because for doing Feynman diagrams, it is enough that you fix the gauge transformation law, transformation that has parametrized by a function of four variables. Okay? So, the residual gauge symmetries are always there, right? Even in normal electrodynamics, you have this. But for perturbation theory, you do not need to fix them. So, the, if you are done the same annulus calculation, not on a D instanton, but on a DP brain, okay? You would have gotten a similar expression that is what he got. And that corresponding action would have included these terms, right? Everything is, I mean these, because these will be the L0 eigenvalues of the various states. But now, you see what will happen if you try to apply the same procedure on a D instanton. See, if we apply this on D instanton, first of all on the D instanton, there are no gauge fields. Because mu runs from 0 to P plus 1, but P plus 1 is 0 for D instanton, okay? So, there are no gauge fields. And there is no space time, because these are 0 dimensional fields. So, box, u box v is 0 and no gauge fields. So, if you took the same formalism that works for DP plus DP brain and just naively applied it on the D instanton, you would have gotten an integral like integral du dv with no action, right? If you just take a formalism that the string field theory gives for the gauge fields and just apply it naively on the D instanton, this is what you would have got. And this is the source of the problem, right? This is the divergence that this is the 0 more integral that you are finding. But now that you have understood where these u and v fields are coming from. We can go back and check what is wrong, right? What we did wrong there, right? Why is it not working for D instanton? And the problem goes back here. You see this, because we have delta phi equal to box theta, right? We fixed phi equal to 0 gauge. And this works for all DP brains, right? But on D instanton, delta phi is 0. So, this means that you should not fix phi equal to 0 gauge, right? It is not possible to choose phi equal to 0 gauge because phi is gauge invariant. So, this is the source of the problem, right? This is the reason why you are getting these goals with no kinetic term. But once you have identified the problem, we also know the solution is to do not gauge fix, right? Because I mean you are not supposed to gauge fix because phi is gauge invariant. So, this integral du dv, you have to replace an integral d phi exponential minus s. Let us look at what minus s is, right? This is phi square, right? The action is phi square, right? We write to minus s. But s is just phi square. So, you have to do the phi integral and you have to divide by the volume of the gauge group. This is the volume of the group. So, when you add this term, I guess the number in front of s is very important, right? It is minus 1 times phi square. Yes, yes, indeed. But is it obvious that this 1 is fixed because you write this action but I could multiply this action by another number or not? Yeah. So, you could multiply by make it into 2 phi, right? But if it is 2 phi, then this will become 2 box theta, right? You want to bring it back to box theta. So, theta also has to be multiplied by 2, right? So, in this integral, then you will get 2 here and 2 here. So, this ratio will be independent of that normalization, right? It is true that I mean phi you could rescale, right? But then you have to make sure that theta is also rescaled appropriately, okay? So, this is an elementary integral, right? So, this is root pi over integral d theta, okay? And now you have to figure out how to calculate this, the volume of the u1 gauge group, okay? But the volume measured in this parameter theta. So, the strategy for calculating this is that we have to compare gauge transformation generated by theta to the standard gauge transformation law. Here psi corresponds to any open string state whose one end lies on the d instant, right? Those are the ones which are going to transform under the u1 transformation, okay? So, this is a standard gauge transformation law, right? In terms of alpha, the period is 2 pi, right? Because alpha, alpha plus 2 pi is generated at the same gauge transformation law. So, if we can find out the relation between theta and alpha, we are done. So, let me state this here. So, if this k2 times alpha, this is some constant again. And then again, this has higher order corrections, okay? But which will not be relevant for this particular calculation. We have integral d theta as k2 times integral d alpha which is 2 pi k2 because alpha period is 2 pi, right? Alpha and alpha plus 2 pi give the same gauge transformation. So, then the final result, well it is not final yet because we have not done all the calculations. So, final result is the exponential integral. This is to be replaced. First there is a factor of i, right? That is the hb equal to minus 1 mode. Then there is a factor of q1. That came from the psi b to y, remember psi b0 is equal to q1y. Then you have this square root of pi from the phi integral divided by 2 pi k2. That is the theta, that is the volume of the gauge group. And then you have finally, integral dy. But this we know will eventually generate the 2 pi delta y term, delta y term at the end of the calculation. So, if we can calculate k1 and k2, we are done, right? That gives the overall normalization. Now, to do this calculation to calculate k1 and k2, okay? We need a little more details on strength field theory, okay? Basically, the point is that we have to normalize, we have to find a normalization of this 0 modes properly, right? And for that, you have to understand. I mean, we wrote, we said that psi b0 is a 0 mode. And what do you know about psi b0? Okay? That it appeared in the action that the general way we normalize psi b, okay? Whether it should appear in the action that it will be minus half hb psi b square, right? That is the general normalization that we use for psi b in writing the annulus partition function as they say. So, this in principle fixes normalization of psi b, okay? Once you compare this form, so we compare this the open strength field theory action. This will fix the normalization of psi b. The normalization of all the psi b that once, okay? In particular, the psi b that particular mode will have a associated state or a vortex operator, right? Which will be fixed including its normalization, okay? Once you fix it, okay? Then compare of psi b to closed strings data y to closed strings. Y coupling we already know. That is e to the i e y, right? That is the way the y couples to closed strings. So, you by making this comparison, we will be able to fix this constant k1, right? That appears. So, this gives psi b, psi b0 equal to k1y and k1 will be fixed. And a similar strategy will be used for comparing theta with y. Right? Theta as it stands is a string field theory gauge transformation parameter. But you have to understand. So, string field theory gauge transformations are of course known. So, you have to see how a string field theory gauge transformation parameter theta as normalized as you have given, okay? Actually transforms in open string whose one end lies on the dn center. Once you find that, then you compare with this transformation, right? Say for infinitesimal alpha. And by doing this comparison, we will find the relationship between theta and alpha. And that will give us the k2. Okay? And once we calculate k1 and k2 and then that is it. We have calculated the normalization. Okay? Are there questions? I think I have 7 minutes. So, let me begin. Actually, it was until 1215. Oh, really? Yes. Oh, sorry. I was a little bit... Okay, sorry. Is this a good time to stop or...? Yeah, yeah, sure. Yeah, yeah. Okay. So, let's thank Ashok now.