 So, let me begin. So, I will start by reminding you of certain facts and definitions. So, recall that we have defined this Hilbert space of H of states this is annihilated by B 0 minus and L 0 minus and I will introduce these notions, notations today. So, B 0 plus minus in my notation will be B 0 plus minus B 0 bar L 0 plus minus will be L 0 plus minus L 0 bar and we will also use C 0 plus minus which is half of C 0 plus minus C 0 bar ok. These will be the definitions we will be using. So, the Hilbert space of states that we had introduced to begin with are all annihilated by B 0 minus and L 0 minus ok. Now, in this Hilbert space I introduce a grading based on the picture number ok. So, H n are the states carrying picture number n and then I said that the Nebuschor sector states H n s will be identified with H minus 1 ok. These are states which carry picture number minus 1 and the Ramon sector states of general off shell Ramon sector states will be the ones which carry picture number minus half. Now, one of the important information which I am not going to prove, but it is easy to verify has to do with the grass morality of this vortex operators or vortex operators ok, which basically tells us whether a given vortex operator is a commuting vortex operator or anti-commuting ok. For example, if it has odd number of formulas it will be anti commuting, if it has odd number even number of formulas it will be commuting ok. And typically the result that one finds is that if we look at the n s sector then the grass morality is even for Gauss number even Gauss number or odd ok. And I should perhaps say that I will considering only GSO plus sector because all off shell states will also have GSO positive GSO priority ok. So, this is with GSO projection plus ok. In the Ramon sector it is opposite. So, this implies for example, for physical states ok the BRST invariant physical states which are always in a Gauss number 2 sector ok. We will have the Nebuchadre sector vortex operators will be commuting and the Ramon sector vortex operators will be anti commuting and which is what you want because Ramon sector should describe formulas and Nebuchadre sector should describe those ones ok. Now, I will start with the definition. Suppose, we have a set of n states let us say start with a 0 to a n which belong to h n s plus h r ok which means that it can either be an n s sector state with picture number minus 1 or a Ramon sector state with picture number minus half. And let us suppose that with there are m n s sector states m n s and n which is capital N minus m r states ok. We will define a bracket a 0 to n as follows. So, this sum is over genus ok this is a formal definition. So, to get any given order in order in order version of course, you always stop at a finite order in genus. So, sum over G this is the string coupling ok we put the weight factor of string coupling to the 2 G and then integral r G m n omega 6 G minus 6 plus 2 m plus n of all these states oh I should have if you are not trying this should be capital N plus 1 minus m ok because the number of states here is N plus 1 I have started it from a 0. Next I remind you is the 1 p i subspace 1 p i subspace had I replaced it by the full section this would have been the full off shell amplitude involving the states a 0 to n ok because that is what you mean by off shell amplitude you take the genus at genus G we integrate the corresponding omega then weight put a weight factor of G string to the power 2 G and then sum over all genus ok. So, this should have been a full off shell amplitude if this had been the complete section instead we are using the 1 p i subspace of the section and hence this will be 1 p i amplitudes. So, this describes off shell 1 p i amplitudes is this definition clear ok. Then we are going to define another quantity which is also going to be useful. So, given states a 1 to n in H we define define a 1 to n ok this is a state in H by the following relation the relation is simple its inner product with any state a 0 in H is given by a 0 for every a for every a 0 in H is this definition clear ok. So, you have to define we are trying to define a state in the Hilbert space and to define a state because the B P G inner product is non degenerate it is enough if we specify its B P G inner product with every state in the Hilbert space and this is a definition ok its B P G inner product with any state a 0 in the Hilbert space is given by this bracket this is just a number that I have already defined here ok. Now, I should also share this point that even though we are applying it in a somewhat different context of 1 p i amplitudes and for super strings and heterotic string the structure is essentially borrowed from old work in string field theory in particularly the work of Barton Zwaiba right. So, if you want to just learn about the structure his article is the best article to read which is the reference is given in the slides which are on the net ok are there any questions ok. So, there is turns out that these brackets have some very useful properties and since we are going to use those properties I will have I will put it put up a slide containing all those properties and explain what these properties mean or what they signify. So, the first property is simple it just tells us that if we exchange the a i is inside the bracket you pick up a sign ok that is just dust manality ok if you exchange 2 odd objects inside the bracket you pick up a minus sign. Second one in fact is the most important identity ok I am not going to prove it it can be proven using conformal field theory methods ok, but I will tell you the physical origin of this identity right what this identity represents and what it represents is the following. So, if you remember that we drew this picture. So, this is our whole e g n this is m g n and the 1 p i subspace was something like this. So, let me put m n 1 p i subspace was some subspace like this r g m n. Now, what was the important property of this 1 p i subspace ok it is that that at the boundaries of the subspaces ok it smoothly joins over to some 1 p i contribution built out of lower Riemann surfaces ok Riemann surfaces have a lower genus or lower number of punctures ok that is the way in fact we define this 1 p i regions that we first built all the 1 p i regions and then whatever is left over we put 1 p i regions ok we just interpolated by 1 p i region. So, the fact that the boundaries of Riemann have to be matched by some 1 p i contribution that information is fully encoded in the second identity ok. So, the second identity is the algebraic description of this geometric fact that the boundaries of Riemann are connected to 1 p i contributions are there any questions ok. Intuitively you can see that the left hand side which has a single bracket ok that is like inserting q b gives a total derivative on Riemann. So, it peaks a boundary contributions the right hand side has 2 brackets right you can see a square bracket inside and then it is inside a big curly bracket ok that basically says that it joins up with something ok which is a combination of 2 1 p i regions. I introduce this notion G ok which is going to be very useful ok. So, G is an operator which is identity when you act on the NS sector states in the Hilbert space and which is the zero mode of this picture changing operator that I introduced. So, what was I know this was just integral d w over w for the picture changing operator. So, on the Riemann sector states it acts as that I know and this G appears in this various identities simply because the way we describe the gluing operation ok. So, the definition of gluing compatibility was by inserting this chi zero mode of the picture changing operator. Third identity also is simply simple to prove basically its proven by writing it. So, we have a 1 to a k and then G on a 1 tilde l tilde. So, what you do is that you replace this by a complete set of states and then you can write this as a 1 to a k phi r and then phi r c and then by suitable manipulation you can show that it is actually symmetric under the exchange of a 1 to a k and a 1 tilde to l tilde with some additional sign factors which I have written there. So, the point I want to emphasize is that these identities can be proven for any matter super conformal field theory ok. It does not require assuming that we are working in a flat background in particular if you have a Calabi out ok. It is exactly the same identity which hold there ok even though the correlation functions themselves of course have to be calculated the definition of omega which involves a correlation function that of course will change because you have a different conformal field theory in that case ok. Are there any questions? So, next I will introduce the concept of string field ok because if you are going to write down an action ok we have to write the action as a function of some fields ok. So, how do we introduce string field? So, here is a definition in a sector string field ok. The definition will begin looking very abstract but then I will explain how it is let the usual fields that you are familiar with. So, NS sector string field is a state which I will call psi NS belonging to H minus 1 of Gauss number ok and GSO plus ok. GSO plus will always be assumed and an R sector string field is a state psi R of Gauss number 2. Now normally you are used to thinking of fields as being functions of space time coordinates. So, what do you mean by a string field being represented by a state ok. So, what I will try to explain is how are these abstract definition of string fields related to what normal you think of as fields. So, this is done by choosing a basis of states in H NS. So, suppose phi R is a basis of Gauss number 2 states 2 states in H NS ok. Then since psi NS is an arbitrary state of Gauss number 2 we can expand it out as a linear combination of this type ok. This is a fixed basis of states and you have the coefficients of linear combination which are calling AR. So, when I say that a string field is represented by a state basically what I mean is that the dynamical variables in the NS sector are in one-to-one correspondence with this variables AR ok. These coefficients of expansion of the string field are the dynamical variables of a theory. Now you still do not see a field ok, but that you can see by noting that in this sum over R ok, one of the thing that you have to sum over or more precisely integrate over is the momentum carried by the state right. This is a complete basis of states ok. So, in particular it is characterized by the momentum along the non-compact dimension right because the state will be characterized by let us say what are the possible basis states right. You can have psi mu ok let us consider the Bosonic case ok which may be more familiar with alpha mu minus 1 alpha tilde mu mu minus 1 k ok. So, you will one of the terms will be something like this h mu nu of k integral d 26 k ok. So, h mu nu of k in that case will be the dynamical variables this integral over k as well as sum over mu and nu are the summing of R ok. And you are now recognized that this is nothing, but the Fourier transform of the graviton field ok. So, when I say that this dynamical variables the ARs are the dynamical variables these are fields in this sense that once you decompose this index R into a momentum integral and a sum over discrete set ok. This momentum integral represents the usual Fourier integral ok and you interpret the corresponding coefficient as a Fourier transform of the field in the position space ok. And you can do this for every level ok. So, there is a field for every level for every possible index structure ok you have a field yes. Yes, except when the first quantization you will not include the complete basis here right you will restrict to some specific choice of basis which will eventually represent physical states right. But if you expand it out on a complete basis right without worrying about whether it is physical state pure gate state or anything then that is what gives you the strength fields at the complete set of strength fields ok. But this notation is useful because I am going to write down everything in terms of psi ns if you want to work in terms of explicit component fields you have to just substitute for psi ns this expansion and carry out the expansion ok. But all the formulae will be written in terms of this psi ns ok. Similarly, we can expand psi r in a basis Br let us call it phi tilde r ok. This is the basis of states basis in H r or goes number 2. No, we are all working in a small Hilbert space. So, C 0 is not there I introduce this before, but I will restrict to states which are in the small Hilbert space I should have probably said it that whenever I am defining H ns and H r they are all in the small Hilbert space. And I introduce before yesterday I used C right, but it always came in as a difference right. So, it is always in the small Hilbert space. So, none of the correlation functions ever go out of the small Hilbert space ok. This is a technical point that all the correlation functions that will be ever dealing with all involves only differences in the field C between two points ok. So, you will always get expressions that involve C in the combination like this. And similarly all the states that I will be including here in this expansion ok. We will only involve derivatives of C right. The vortex of potential never involve a C by itself ok. It will only involve derivatives of C there will be arbitrary number of derivatives. And this is what is normally referred to as a small Hilbert space of the CFT. Yes, it is like like that exactly right. So, because you are we are writing you know one pi action right I will write a classical action. And they are the classical action you have to just do three level analysis to get the S matrix elements right. But in a string field theory you would first write down a classical action then you will quantize it. You will think of a a and a's and b's are dynamical variables and then you have to quantize it ok. Now, the a's are commuting variables. These are anticommuting variables because we know that the bosonic fields or the fields which carry integer spin are supposed to be commuting ok. And fields which carry half integral spin which are the Raman sector states these are supposed to be anticommuting. So, let me write that down here. So, AR these are commuting or Grasman even and these are anticommuting here Grasman order which basically means that both psi n s and psi r are Grasman even ok. Because if you remember I told you that the vortex operator here is Grasman even in the n s sector these are carrying host number 2 ok. And you are multiplying it by a Grasman even quantity. So, this product is Grasman even here the vortex operator is Grasman odd ok. But you are multiplying it by coefficient which is also Grasman odd. So, the product is again Grasman even and this is going to be important. So, the string field is always Grasman even even though it is representing the formic fields or the anticommuting fields and bosonic fields which are commuting fields ok. Are there any questions? Yes, because we will not flat we have chosen some conformal field theory that describes a matter sector right. And you are basically doing string field theory around that background. The background is fixed by what super conformal field theory you have chosen right. So, that provides the background. And so, I mean the string fields equal to 0 will correspond to that background ok. So, with this I am now ready to write down the 1 pi effective action in the n s sector ok. I will include the Raman sector at later 1 pi effective action ok. This is a 1 pi effective action in the sense that these generate the 1 pi amplitudes. And if we calculate the 3 level amplitudes from this 1 pi effective action you get the full amplitudes in string theory. So, because you are considering n s sector it is only a function of psi and here is the result. So, see that I have told you the definition of this. So, I should have said that what this means is that this is basically n copies of psi n s inside ok. And for this to make sense it is important that these are all Grasman even ok. If they are Grasman not then of course, it will vanish. So, what does this tell us ok? What this allows us to do is that it given a set of ARs which means given a field configuration it tells you how to get a number right. Because I have told you this is just a normal BPG inner product and this I have told you how to compute right in terms of those integrals. And that is what an action is supposed to do right that given a field configuration it is supposed to produce a number and that is what it does ok. And you can see that this is indeed the generating function for the 1 pi n point amplitude because this integral this curly bracket after all was defined to will have erase the definition this curly bracket was defined. So, that it gives the 1 pi amplitude are there any questions how is oh because the definition of the curly bracket involves omega right and that omega involves the correlation function. So, that is how the background see up to enters. Now, one can show in fact, this is a an exercise which you can do and all that you need to do to prove this is to use those identities. You never need to know in fact, what these omegas are or what the definition of this curly bracket is just that those identities and the relationship between the curly bracket and the square bracket which defines square bracket in the first place. You can use to show that delta s is 0 for delta psi this is infinitesimal the n s again u where lambda n s belongs to h minus 1, but it has it carries host number 1. It has to carry host number 1 because otherwise this does not make sense right because this has host number 2. So, it better carry host number 1. So, this you can prove you can just take that action that I have written down there make an infinitesimal variation thinking of lambdas as infinitesimal parameters and you can show that delta is vanishes. And this is the infinite dimensional gauge transformation that I referred to earlier why this is a gauge transformation again to interpret this you have to decompose lambda n s in a basis states. You choose some basis state phi hat r of h minus 1 of host number 1. So, this c r's are the gauge transformation parameters. These are the variables which implement our gauge transformation and again you can see that because the sum of r includes this integral of our momentum these are really functions of k or equivalent functions of x by Fourier transform. So, this is gauge this is local symmetry. And this include in particular if you look at the linearized transformation where you just ignore these terms it includes linearized general coordinate transformation and also various higher spin gauge transformations. But what these terms tell us is the full version of those transformations including quantum corrections because this in defining this bracket we also have taken into account the quantum corrections because the 1 pi action involves already contribution from loops. So, you can think of this as a full quantum corrected version of the gauge transformations that we see in strain theory. Yeah roughly except that the second term you can see interaction term normally you think of as starting at cubic and higher order right here the expansion starts at n equal to 1 and that is basically because on the torus on a higher genus surfaces even the 1 point function is on 0. But you can always choose to call it interaction term because these are of order g s or a g string. String tension again it is in encoded in the correlation functions right the CFT correlation functions of course knows about the string tension and everything else. But at this abstract level you will not see any other parameter. So, next we can ask for equation of motion that you can derive from this 1 pi action and that basically means that you vary your psi n s infinitely and demand that the first order term variation will always give you 0. So, again this is an exercise you can check that the equations of motion take the following form q b on psi n s plus you notice that here the expansion begins at n equal to 1 ok that is a simpler reflection of the fact that here the sum begins at n equal to 1 ok. And this happens because the linear there are linear terms in psi n s because on Riemann surfaces general Riemann surface even the 1 point function of a generic field of this kind is nonzero ok. This is typically always true ok you can always find operators whose 1 point function on Riemann surface is nonzero ok. So, only on the sphere at tree level that the 1 point functions are all vanish. So, which means that there is an n equal to 1 term here which is a constant piece right which is just this no psi n s inside. And that means that psi n s equal to 0 is not a solution to equation of motion. It is a solution at tree level where at tree level the expansion starts at order psi n s cube, but not otherwise. So, even to find the vacuum the normal transition invariant vacuum you have to do some work you cannot simply take this action expand it around psi n s equal to 0 and start calculating tree level s matrix ok. You have to first find the vacuum then expand the action around the new vacuum and then from there you can calculate the tree level s matrix elements which will give the full string amplitudes. Is this clear? Although 1 pi action because you given a 1 pi action right in a like in a quantum field theory you are supposed to do tree level analysis, but you have to expand around this extremum right. Otherwise the tree level amplitudes do not make sense right you will get tadpoles if you do not expand around the extremum ok. So, the first task if you want to do anything with it ok. The first task is that you have to find a vacuum the vacuum solution and this you can do try to do iteratively iterative solution vacuum means 0 momentum right. So, basically only look at states which carry 0 momentum ok. You just solve this equation in the 0 momentum sector because you want transition invariance ok. So, here you can prove the following that suppose psi k solves equation of motion to order g s to the k. Then you can show that we can find a solution to order g s to the k plus 1 ok to when you say I say it solves equation of motion to order g s to the k that means the equation of motion possibly has a correction as error terms of order g s to the k plus 1 ok order g s to the k term in the equation of motion vanishes. Then we can find a solution to order g s to the k plus 1 of the following form ok I should introduce some notation. So, p will be the projection operator to L 0 plus equal to 0 states ok L 0 plus let me remind you is L 0 plus L 0 bar all states are have L 0 minus equal to 0 and I am introducing an operator p which projects into L 0 plus equal to 0 states ok it will become clear why. So, the solution takes the following form now I should sir even though I am saying here n equal to 1 to infinity in actual practice you only have to stop at a certain order because this psi is already of order g s right and you are only try to compute the solution to a given order in g s. So, this upper limit is not really infinity ok, but I am writing it as infinity anyway you just pick whatever to whatever order you have to go for a given solution plus ok. So, what is phi k plus 1 phi k plus 1 satisfies p phi k plus 1 is equal to phi p k plus 1 ok. That means, it says L 0 plus equal to 0 state and it satisfies the following equation q b on phi k plus 1. So, q b plus 1 ok. So, what you can shows that if psi k satisfies the equation to order g s to the k then if you choose psi k plus 1 in this way ok then psi k plus 1 satisfies the equation of motion to next order ok. So, this gives an algorithm to solve these equations iteratively. Now, you can see the importance of having this projection of operator p ok. In fact, the conventional string perturbation theory if you translate it into the in this language what it does is that it does not introduce this p and it does not have this separate factor ok. The danger there is that if this if you did not have this p oh it is not p sorry here this is 1 minus p ok and I think this is n minus 1 because I put a 1 or n minus 1 factor So, if you did not have this 1 minus p then the contribution from the L 0 plus equal to 0 sector we have a divergence right. So, you should not expect a non-singular solution it you may be lucky and then there may not be any L 0 plus equal to 0 component of this, but if there is then you run into a divergence because there is a 1 over L 0 plus setting here. So, the role of p is that we separate out the part which has L 0 plus equal to 0 and treat this separately that is this 5 k plus 1 which has to be solved separately. So, in the L 0 plus equal to 0 sector you have to separate the solve the equation that is this equation and then you add it back to psi is this point clear ok. So, in other words if we just proceed nively ok and that is what conventional string partition theory amounts to that will correspond to solving this equation iteratively without this and without this p and the divergence is the tadpole divergence is that I mentioned in my first lecture that essentially is the divergence that will get because this may contain L 0 plus equal to 0 state which diverges yeah that is the 1 over k square equal to 0 that is all. Now, of course we cannot hope to create magic right if it here you had a tadpole you cannot get rid of get rid of it by just changing the formalism ok. So, how do we see possible existence of tadpoles so possible existence of tadpoles must come from some kind of obstruction to solving this equation. Now, this is of course this part is straight forward right once you have removed the L 0 plus equal to 0 states this is completely unambiguous ok. So, there is no confusion in what this object is the possible obstruction may come from the inability to find the solution to this equation right this is an equation which tells a q b on phi k plus 1 equal to something how do you know that there is a phi k plus 1 which is which satisfies this. So, the first consistency check that we have to perform is that because q b is is nilpotent ok we know that q b acting on this better give you 0 right. If q b acting on this did not give you 0 then there is no hope that you can ever solve this equation because left hand side on a q b acts on the left hand side it is 0 ok. But this you can prove in general that as long as phi k satisfies the same equation at one lower order q b acting on this is identically 0 ok this is an identity which you can prove. So, at least that is not a problem. But even when q b acting on this is 0 does not mean that you can solve this equation because this equation may have components along non-trivial elements of the BRST homology ok. It can have components along states ok it can contain states which are annihilated by q b but which cannot be written as q b on some on another state. And if those states ever appear in this right hand side then you cannot solve this equation. So, that is the way tadpoles appear in this formalism ok you will not see a divergence because the way it is set up you will never encounter a divergence here or anywhere else in the perturbation theory ok. But what you may fail to do is to find a vacuum solution ok. And if that fails then that means that the theory is inconsistent you give up at that stage but not at a stage where you suddenly find a divergence and then you declare the theory inconsistent. Now this in fact is also suitable for discussing the vacuum shift problem. So, let me briefly discuss this. So, if you remember in the first lecture I described an example where there is a potential for a scalar field of the following form phi square minus let us call it some chi square minus k gs square whole square ok. I wrote a complex field but let us suppose it is a real field. So, from the low energy effective field theory view point this has 3 solutions chi equal to 0 and plus minus gs root k. One can ask how will that show up there in this analysis and the way it will show up there is that when you are trying to solve this equation you will find that it does not have a unique solution ok. To order gs you will find that it has 3 solutions instead of 1. So, there will be 3 solutions for phi 1 and this you can check explicitly by working out this particular example for phi 1 ok at this order. But then of course you have to keep going up to construct your phi 2 phi 3 etcetera and then you discover that at some point the solution that starts with this ok one of the 3 solutions that solution develops a tadpole which means that that solution runs into an obstruction at higher order because this has non-zero energy which essentially induces a delay on tadpole. But the delay on tadpole of course you do not see at this order right you will see it only at order gs to the 4 ok. So, once you try to solve the equation at order gs to the 4 you find that you cannot proceed anymore ok that you run into an obstruction of the kind where the right hand side contains a physical state. So, it contains a state which is bearstain variant but not bearstain trivial ok and that is the way you see that only one of these 3 solutions can really be or only 2 of these 3 solutions these 2 can be continued to higher orders, but not the one corresponding to this ok. So, now let me go to Ramon Sector. So, it turns out that for technical reasons it is not possible to write down a an action involving Ramon Sector and Amouchevoir Sector states ok. However, this would have been fatal if you are trying to do a field theory genuine field theory which you have to quantize because having an equation of motion is not enough to quantize a quantum field theory. However, we are constructing in one p i action from which you are only supposed to do tree level amplitude ok. You are supposed to calculate tree level amplitude and that is supposed to describe the full set quantum corrected amplitude and strength theory. So, for calculating tree level amplitudes it is enough to write down equations of motion ok because after all tree level amplitudes are nothing but solving equations of motion ok. So, one can write down the equation of motion for N S and R Sector states together ok and it has a very similar structure to what we wrote. So, let us say we take psi as psi N S plus psi R ok. Just formally sum the 2 states Ramon Sector state describing this Ramon string fields and the N S string fields and then the equation of motion takes a form Q B psi plus ok. Where G let me remind you again it is one acting on N S Sector states and this 0 mode of the picture changing operator when it acts on Ramon Sector states ok. So, this is a simple generalization of what we had for N S Sector there of course, you did not have a G because when you are acting on N S Sector states you do not have a G, G is identity. So, by decomposing this into the N S direction and R direction you can get the equation of motion for the N S Sector string fields and Ramon Sector string fields. Gauge invariance is also simple to state. So, gauge transformation that leaves these equations of motion invariant is this ok. Where lambda now belongs to H N S plus H R ok and it has goes number 1 both in N S and R sector ok. So, this now includes for example, local supersymmetric transformations ok. Earlier lambda was in the N S Sector lambda I said includes general local general coordinate transformation. Now, it includes local supersymmetric transformations as well ok and you can see manifestly that these equations of motion are invariant under these transformations. Are there any questions? Well the point is you are shifting background by string loop correction. So, it is not a necessarily described by conformal filtering anymore right. See the one pair the construction of the one pair action of course, is based on conformal field theory. But once you have gotten the one pair action you never have to go back to conformal field theory right because you have already constructed a vertices. Now, you can look for vacuum classical solutions which are not necessarily described by conformal field theory ok. Like these two I described these are certainly not conformal field theories in the conventional sense ok. So, now let us look at how to calculate S matrix elements ok or renormalized mass ok. So, suppose psi v is the vacuum solution. So, what will you do? You will define a shifted field chi as psi minus psi v and write down the action in terms of chi right. So, instead of working with psi you will work with psi psi v plus chi ok and by construction because psi v is a solution to equations of motion you are guaranteed that this action when you expand in powers of chi will have no linear term ok which was the problem because having a linear term means you have a tadpole here by construction you have eliminated the linear term. Now, since the equations are a little involved let me just show the slides. So, basically it turns out that it is simplest to introduce a new set of brackets in this form ok. It is constructed out of old brackets ok. If psi v was 0 then only the old bracket term will survive, but now you add a whole series of terms involving the vacuum solution. Again I have written everything from 0 to infinity, but both in the analysis of psi v as well as in the sum you always truncate right. You never go up to infinity because you are only working to a given order in string coupling ok. So, this defines a new prime bracket curly bracket this defines a new square bracket and this defines what we will call a new BRST charge ok. Qb hat is a new operator act it is still a linear operator acting on a state, but that involves all this psi v's. In terms of these one can show very easily that the shifted equations take this form ok. So, it takes exactly the same form as the original equations except that all the brackets are replaced by prime brackets and Qb is replaced by Qb hat. In fact, there is something even more interesting which is that these prime brackets and Qb hat satisfies exactly the same identities that we are being satisfied by the unprimed brackets and the Qb ok. So, the essentially the whole structure remains unchanged and whatever identities we are proving earlier in using the Qb and the unprimed brackets also hold with Qb hat and the prime brackets. So, but let me come back to linearized equation. So, linearized equation is Qb hat chi equal to 0 ok. Why are you interested in linearized equations? Because the solutions to the linearized equations are going to tell as a mass spectrum right you compute the mass spectrum of the theory by looking at the solution to the linearized equations ok. So, this in fact turns out to have two kinds of solutions ok. The first kind there are solutions which exist at all k all momentum. These are of the form Qb hat lambda these are pure gauge ok pure gauge, but not in the conventional sense normally you call declared Qb lambda as pure gauge ok. Now, this is pure gauge, but this is quantum corrected gauge transformation because you are using Qb hat. The second kind of class of solutions are solutions which exist at spatial momentum square. So, for these minus k square gives you the physical enormalized mass these are the physical steps. So, this tells us how to compute the physical enormalized masses once you have the 1 pi action. In practice of course, you do not really follow this route ok. What this tells us which I will not have time to discuss here what this 1 pi formalism gives us is a very precise way precise operation involving the full correlation functions on the Riemann surface which allows you to extract for the physical masses. So, you never really have to construct the 1 pi action explicitly ok, but using this 1 pi description you can arrive at a precise algorithm of how to compute the physical enormalized masses from the off shell amplitudes. Any questions? So, I will not have time to discuss this in any detail, but one of the questions that we had raised earlier about is about the dependence on the section that if you choose a different section how does the action change? How do the enormalized masses change? And I told you that under a different choice of section the action gets redefined ok, that action is related to the action changes by field definition. And so, I will maybe end by showing you the specific field definition ok and here it is that imagine that you have two different sections ok, you have two different 1 pi sections. One is called r g m n the other one is called r prime g m n and I will take this to be infinitesimally close to each other ok. Then you can define this vector field u hat which connects r to r r prime and you can explicitly write down a field definition that relates the action that is based on the section r g m n and the action that is based on the section r prime g m n ok. So, in a sense you can explicitly prove that the choice of section amounts to just a field definition and hence the physical enormalized masses or this matrix elements do not change under this field definition. My time is up so, I will stop here.