 Explain the general program and motivation for why you want to study off shell amplitudes and in particular one by amplitudes ok. And our eventual goal of course, will be to study super strength theory and super strength and heterotic strength, but today I will do a warm up exercise ok using Boson strength theory because some of the things are simpler of course, in Boson strength theory. But as you know it is not really fully consistent because of the presence of the tachyon. So, let me begin by reminding you of the tasks that we have to do perform in order to carry out this program and this is the slide that I showed you at the end of yesterday's lecture. So, the first goal will be to find a parameterization of the space P g n and its tangent space. So, P g n let me remind you is the total space of the modular space of a genus G Riemann surface with n punctures together with the choice of local coordinates around each of these n punctures that is the space P g n. So, P g n contents information about where you are in the modular space as well as what kind of local coordinate system we have chosen. The next task was that for any n given external states not necessarily on shell or BRSA invariant states and these n external states we collectively denoted as some phi. We want to construct a P form omega P on this space P g n with the desired properties and the desired property is this one namely if we replace the argument of omega P by Q B acting on phi where Q B is the sum of the BRST charges acting on all the states then it satisfies an identity that relates it to a P minus 1 form omega P minus 1 acted on by the exterior derivative and the final task was to construct blue incompatible sections and one P I subspaces of P g n ok. So, this blue incompatible sections are what we call S g n and given S g n we also introduce the notion of one P I subspaces which are called R g n ok and hopefully some of these things will become clearer as we carry out explicit examples today. So, are there any questions on this program true for any, any phi yes yes. So, phi is some any general state which is in the product of n Hilbert spaces right there are n external states. So, that takes value in the product of n Hilbert spaces and it is true for any phi yes. So, that is all we will introduce the Hilbert spaces first and then we will take phi to be an arbitrary state in that Hilbert space no no because on on shell states of course, this will be 0 right. So, it is for a general state we are not assuming phi to be on shell pardon P P is any arbitrary integer right. So, we will construct this omega P is for every value of P ok even though we will actually actually use it only for few values of P, but we will give this construction for an arbitrary P. So, omega P this is an argument of omega P right it is it depends on the state in this Hilbert product of Hilbert space, but omega P is a P form valued in P g n. Yes sir. Yeah. P for on that. On that space. On that space. Yes exactly yeah. So, you get a real number given phi or complex number given phi yes. Ok. So, let me begin by giving a description of P g n in particular how you parameterize P g n ok. So, P g n let me remind you again is so any given point in P g n labels where in the modular space you are in as well as what kind of local coordinate system we have chosen around the punctures right there are some meaning of P g n having a point in P g n ok. So, let us suppose that you have a Riemann surface ok. So, here I have definite this is a Riemann surface with two holes ok and two punctures ok, but given any Riemann surface of genus G and n punctures it is easy to see that we can consider this as a union of n discs ok. So, each of these n discs are cut out around these n punctures. So, around each of these punctures you cut out a disc ok. So, it is a union of n discs which are denoted by D A and 2 g minus 2 plus n spheres which are denoted by S i each of them we have three holes ok. Now, this number counting it takes a little thought to see that these are the correct numbers but this figure illustrates what we are doing. So, around each of these punctures this is a puncture here and this is a puncture here we cut out a disc which I had called D 1 and D 2 and then the rest of the Riemann surface I just decompose into spheres with three holes by drawing circles ok. We just cut the Riemann surface along arbitrary circles of this kind ok. So, that it splits up into collection of spheres and discs ok and you can count that the number of spheres that will have is 2 g minus 2 plus n. So, for example, here in this example we have g equal to 2 it is a genus 2 Riemann surface we have n equal to 2. So, you expect four spheres and indeed we have four spheres S 1, S 2, S 3, S 4 and you would see that each of these spheres have three holes ok and you can also count that there will be 3 g minus 3 plus 2 n circles which form boundaries between these different regions ok. So, there are some circles which form the boundary between discs and spheres and some other circles which form the boundary between two spheres ok and this counting again is a simple exercise that you can carry out ok. In this particular example this number I think is 7 right g is 2 so 3 times 2 is 6 minus 3 that is 3 plus 2 n that is 4 right. So, this number is 7 and you can see precisely there are 7 circles over here. So, is this clear? So, this is just a geometric fact that Riemann surface you can think of as a collection of discs and spheres ok glued at their boundaries ok. So, now, given this kind of decomposition we introduce WS which I call the complex coordinates on DA ok and we will choose a convention that see each of these DA's surround the puncture right eighth puncture. So, D 1 surrounds the puncture number 1, D 2 here surrounds the puncture number 2 ok. So, WS we will call the complex coordinates on DA and we will choose the convention that the eighth puncture is situated at the origin of the W coordinate system ok. This is the convention we have been following yesterday and we will follow it today as well and on each of the spheres we can introduce complex coordinates which I will call Zi ok. So, we have complex coordinates on each sphere complex coordinates on each disc and it is clear now that if Si and DA share a common boundary then on that boundary we have some functional relationship that relates Zi with WA ok because these are holomorphic coordinates. So, this is a holomorphic function that relates them ok. It is of course, holomorphic only around that circle it can have singularized elsewhere ok. And similarly if Si and Sj share a common boundary then on this boundary we have a functional relationship that relates Zi and Zj ok those I will call Fij ok. So, given a Riemann surface you can always find these functions from this ok that giving this functions is a meaning of telling you what Riemann surface is what Riemann surface you are considering then any questions on this ok. So, these functions in fact, contain complete information on the Riemann surface and the choice of local coordinates ok. So, once you are given these functions ok that is equivalent to giving the Riemann surface ok in a MGM what particular point in MGM you are at and also what local coordinate system you have chosen right because these contents information about the W's ok and W's are the local coordinates around the functions so a given point in PGN. So, giving a point in PGN is equivalent to giving all these transition functions and the number of such transition functions is the number of circles which we have counted as 3 G minus 3 plus 2 N ok. So, 3 G minus 2 N 3 plus 2 N functions will specify the Riemann surface as well as the choice of local coordinates. So, this gives a characterization of PGN. So, we can choose as coordinates of PGN these functions ok of course, it is an infinite dimensional space which we already expect so it is a space of these functions which describes PGN ok, but it is not the full set of functions that describe PGN because there is a lot of redundancy in this description ok and the redundancy comes from the fact that if we make any parametration of the z i's ok. Suppose you change the coordinate system z i to something else that is not going to change the Riemann surface ok it is just a change of coordinate system now that is it going to change the choice of local coordinates on the disks because those are the W coordinates ok. So, that means that given these functions there is a set of equivalence relation that is induced by re-parameterization of the z i's ok. Of course, these re-parameterization should have to map those spheres into spheres right if it singular inside the sphere then it is not a valid re-parameterization ok, but there is a lot of freedom here ok you can re-parameterize your z i and you get back the same Riemann surface is this point here. Now, in general if we re-parameterize the W's they will give different points in PGN right. In fact, re-parameterizing W's will keep you at the same point in the modular space but it will take you to a different point on the Riemann on PGN because changing W corresponds to changing the local coordinate system ok. So, re-parameterization of W is not a re-equivalence relation ok. However, we have to declare that if we just change W by a phase ok. Suppose you consider re-parameterization where W a goes to some phase times W a then these are defined to give the same point in PGN which basically means that what I told you earlier ok was not completely correct I said that PGN labels a point on the Riemann on the modular space as well as the choice of local coordinates ok. What PGN specifies in fact is a point on the modular space and the choice of local coordinates up to phases ok phases do not count. If two choices of local coordinates differ by phase we declare them as being same point in PGN ok and this is necessary for this program to work because yesterday there was a question of whether it is possible to choose a section ok we needed a section for carrying out the integral of omega ok and one can show that the sections do not exist ok if you include in the description of PGN information about the phase ok. It is impossible to choose a section that also keeps track of the phase ok. So, you have to forget about the phases ok and only under that condition that we can define we can find a section in PGN ok. So, this is something that we impose explicitly ok and you have to there is a price to pay which I will come back later. Yes, so it is enough that if you choose if you just say that a change of constant phase ok is identified that you identify different points in PGN I mean apparently different points related by this constant phases as the same point ok that is enough for us to find a section ok you do not need any further identification ok. You can try to do further identification, but that will restrict yourself to smaller Hilbert space ok as you will see that this already will restrict your Hilbert space to be not the full Hilbert space of the conformal field theory, but a subspace ok and I will describe what subspace you have to consider right. The more condition you put on this W right the restriction comes from the fact that then the states that you consider on the Hilbert space must be invariant on the that reprimandization right. So, you do not want to impose too many constraints because you want a general off shell state ok are there any other questions? Yes. Yeah, so the fact that you are integrating theta from 0 to 2 pi right basically means that the phase does not matter right whatever phase you have chosen for this W right if you have chosen another phase ok you will get the same result after you carry out the integral over theta from 0 to 2 pi yes. So, this says that theta is ambiguous, but as long as you are integrating the over the whole range of theta ok this ambiguity is irrelevant right you pick any coordinates any local coordinate W right fix some phase right if you have chosen a different phase that will correspond to a different theta, but after you have integrated over theta the result is in unchanged. Yes. Yes. Yes. Well no number you cannot vary the choice of spheres and circles you could vary right. So, you pick some particular choice of spheres and circles well the what I said is that you this is just a convention that you choose 3 punctured spheres right spheres with 3 holes you can. So, as long as you choose spheres with 3 holes right then the number is fixed yes you could right then you will use a different parameterization of the more of P g n right this is just one particular to parameterize P g n, but there are other parameterizations it is not I am not saying that this is only you have to parameterize P g n is this clear ok. So, this is just a way that you parameterize your space P g n because yesterday I think a lot of people asked about the space P g n this is a concrete description of what you mean by P g n it is a collection it is an information that contains information about all these functions up to these 3 parameterizations ok. Now, it turns out that we also need to parameterize tangent vectors of P g n ok it is not just enough to specify a coordinate system on P g n, but it is also useful to understand tangent vectors of P g n because you have to define forms on P g n right and for those for defining forms it will be useful to keep track of tangent vectors. So, what are tangent vectors in any space right a tangent vector corresponds to infinitesimal deformation of coordinates. So, you have to understand what you mean by infinitesimal deformation of the coordinate of P g n ok you go from 1 point in P g n to a neighboring point how do you do that you do that by changing these functions by infinitesimal amount. So, take for example, these functions f i a right we can take a given f i a and we change it by an infinitesimal amount which we call delta f i a ok. So, this means that in the original relationship was z i equal to some function f i a of w a then the new relationship will be z i as a f i a of w a minus delta f i a of w a and these changes ok. So, this delta f i a is naturally a function of w a, but I can re express it in terms of z by simply relating w i w a in terms of z right it is a inverse relation is w a is f i a inverse z i. So, that is I have written delta f i a of f i a inverse z i ok. So, this is an infinitesimal vector field on the Riemann surface right because this is telling you a small change in z i ok not everywhere, but on that particular circle right because this transition function is well defined only on a particular circle that forms a boundary between d a and s i ok. So, let me draw the figure here ok. So, here you have a circle ok on the one side of the circle you have d a on the other side you have the s i let us say this is the d a is the interior s i is the exterior ok. And f i a tells you how to match the coordinates inside with the coordinates outside change in f i a ok is a small change in z i for a fixed w right. So, a change in f i a will correspond to a small change in the real sense between z i and w a and that you can think of as a vector field defined around the circle ok. So, the vector field that you have defining this way is non singular around this circle, but if you try to analytically continue it either outside or inside you may run into poles singularities we do not care all we need to know is that vector field will define around this circle ok. So, those changes those kinds of vector fields parameterized infinitesimal deformations of coordinates of p and hence they level the tangent vectors is this point here ok. So, tangent vectors of p g n are infinitesimal deformations of the coordinates of p g n and this is the way you characterize infinitesimal deformations ok. So, let me summarize the lesson again. So, the lesson is that the infinitesimal deformations of p g n are characterized by tangent by or the tangent vectors of p g n are characterized by infinitesimal vector fields defined along these boundary circles ok along these various c i's that I have introduced. Is this point here? Yes you can well it is a change in F i a right now you can think of this as a change in z i for fixed w or change in w for fixed z right. So, the because the functions the which level p g n are these functions F i and F i j right. So, what you change are these functions ok and now you have to reinterpret this ok what that function means what changing that function means right. It is a change in the relationship between w and z you can either think of this as a change in z for a fixed w or change in w for a fixed z right. It gives the same vector fields as a vector field there is no difference is this point here. Now, again this is a redundant description because there are some changes ok which are induced just by reparameterizing z i's ok. I told you earlier that reparameterizing z i's do not move in p g n right they leave you at the same point in p g n ok because that is just a change of coordinate system ok. Nevertheless, if you keep the w's fixed and you reparameterize z's then these functions will change. Then in general imagine that we have let us draw this sphere with 3 holes here is our z i and there are various 3 transition functions here. If we reparameterize z i that will change the transition function here the transition functions here and the transition functions here ok. So, that is some general linear combination of the vector fields one vector field here one vector field here and one vector field here ok. So, that should be declared a zero tangent vector right because reparameterizing z does not change the coordinate system the coordinate of p g n right this is a gauge transformation in some sense ok. So, those kind of the changes that are induced the changes in these functions that are induced by these re-parameterizations should be declared as zero tangent vector ok. So, not every vector field on c i represents independent tangent vectors of p g n there is a relationship between these tangent vectors which are which correspond to re-parameterizing the z i's ok. And the same is true of course in changing the phase of w ok tangent vectors which are induced by changing the phase of w should be declared as zero tangent vector is this point here. So, this is what I have said here these must be declared as zero tangent vectors of p g n ok. So, this is basically all we need to know about how to parametrize p g n ok and the tangent space of p g n ok. In fact, what is going to be important is the parametration of the tangent space. So, now we move to the next task namely construction of this p form on p g n. So, for this let me remind you again that if you have a bosonic string theory it is based on a matter conformal field theory of central charge 26 and a ghost conformal field theory of c equal to minus 26 ok. And in particular this ghost conformal field theory contains 4 sets of ghosts b c and b bar c bar. So, b and c are holomorphic and b bar and c bar are anti holomorphic ghost fields. Now, these ghost fields as in any conformal field theory ok these are holomorphic and anti holomorphic fields. So, you can decompose this into modes ok we can for example, we can decompose b b of z as sum over n b n z to the minus n minus 2 ok. These are have conformal over 2 and c of z these are conformal over 1 minus 1. So, here the decomposition is c n z to the minus n plus 1 ok. So, b n and z n c n are just operators in the Hilbert space of this conformal field theory. So, what is going to be important for us are b 0 ok and maybe I should also write the expansion of t the energy momentum tensor this has weight 2. So, this expansion is l n z to the minus n minus 2 ok. And similarly, there is a expansion for the b bar c bar and t bar where all of these are replaced by their bar counter point. So, we will consider h we will define h as the Hilbert space of this combined CFT ok, but we will not allow for a general state in this Hilbert space ok. We work with a restricted class of states ok and that will be included in the definition of h ok. Those are the states which are annihilated by this combination b 0 minus b 0 bar and l 0 minus l 0 bar. And these conditions are necessary ok because in the definition of p g n ok we allowed we identified local coordinate systems up to phases right. We say that if w and e to the i 3 i alpha times w ok these two local choice of local coordinate systems are declared as the same point in p g n ok. And this is guaranteed ok that this is in order that we get a sensible result we have to impose these conditions ok. For example, l 0 minus l 0 bar right what does it do l 0 minus l 0 bar essentially is a rotation acting on the state and because this annihilates the state it shows that whether you choose a local coordinate system w or e to the i alpha times w it makes no difference right. The state does not know about it because the two choices are related by l 0 minus l 0 bar. So, the fact that we have i we have defined p g n by saying that w and e to the i alpha times w correspond to the same point in p g n requires us to impose this condition ok. And we will see that there is a similarity reason why we have to impose also annihilation by this are there any questions on this. So, given this Hilbert space h we will denote by h to the n the n fold tensor product of h this is needed because if you want to compute the n point amplitude right then there you need n external states which of course, naturally take value in this n fold tensor product ok. And I will also denote by this the BPZ inner product between CFT states ok if you have not seen it before it is basically a 2 point function on the sphere. So, if you take a sphere with the vortex operators a and b inserted ok. So, a b is basically this ok you put 1 at 0 1 at infinity and that is what you define as this inner product ok. So, now, we will start defining our omegas. So, let us consider some point in p g n ok again giving a point in p g n means that you have specified all the transition functions ok. The moduli space is you have picked a point in the moduli space and also the choice of local coordinates. So, this corresponds to a Riemann surface a sub genus g and n punctures and it also contains information about the choice of local coordinates at each function right that is what I mean by saying you pick a point in p g n ok. And again when I say a choice of local coordinates it is understood that we are I am referring to choice of local coordinates up to phases ok we forget the phase information. And now suppose you are given a state phi in this product Hilbert space which means a collection of n states ok each of which satisfies these conditions that b 0 minus b 0 bar annihilates it and l 0 minus l 0 bar annihilates it. Then we define by this symbol phi s just as a correlation function of this n vortex operators of phi on the Riemann surface ok and the vortex operators are inserted with the chosen local coordinate system ok. The standard conformal field theory measures tells you how to calculate these kinds of correlation functions on arbitrary Riemann surface ok. This is a standard technology in two dimensional CFT that if you are given a state of vortex operators if you are given a Riemann surface ok. And if you are told how to insert this vortex operators on Riemann surface ok the vortex operators are not conformal invariant ok the result will depend on how you insert them in what coordinate system or in what equivalently what metric you put near the punctures. Once all that information is given the standard conformal field theory technology tells you how to compute the 1 point function or the not n point function on this Riemann surface and that is what it will denote by phi s. And now we define our omega 0 of phi the 0 form ok. Simply as this quantity phi s ok the n point correlation function on this Riemann surface is but with this normalization factor ok this normalization factor is necessary to get this various identities right. No off shell I am not assuming phi to be on set right because as I said phi is an arbitrary state in the Hilbert space ok with only satisfying these conditions right. These are not on shell conditions right on shell condition will correspond to q b acting on phi equal to 0 ok. And a particular representative will require that L 0 and L 0 bar ok and I highlight the state and also all the l n's and I highlight the state right if you want to have a primary state representation. So, none of those conditions are imposed. So, phi is the general off shell state. Are there any other questions? Yes it is yes in this given theory yeah. So, you have to give the conformal field theory right take for example, suppose you have bosonic strain theory in 26 flat space time dimension right. Then the conformal field theory is a free field theory right and in that theory you know how to compute it. If you are given a vasominoid in theory right then the conformal field theory is interacting, but again there are standard techniques which tells you how to compute this correlation function on Riemann surface right. No no just conformal invariance is not going to fix it yeah. Certainly even on the sphere right and the conformal invariance does not fix anything other than the form of 3-point function right for from for from 4-point function onwards there is a non-trivial information in the conformal field theory. Even the spectrum the conformal weights allowed conformal weights differ from one conformal field theory to another conformal field theory. So, certainly conformal field theory a generic conformal field theory will not tell you what the answer is right, but once you are given a specific conformal field theory there are standard techniques which tells you how to compute this and that is what you do want to do right in string when you want to calculate amplitudes in string theory right you first fix a background and you calculate s matrix in that given background and any other questions. Okay now we turn to a more complicated story we have to specify omega p right. I only gave you omega 0 which are simple just one with the n-point function on the Riemann surface what is omega p okay. So, to before I say what is omega p okay let me try to describe what are what doesn't mean by specifying a p-form right a p-form is not a number right. So, how do you specify a p-form right what is the meaning of giving omega p and one way of given a p-form on a generic manifold is the following. So, let us suppose that we have some arbitrary space okay level by coordinates x and if you have a code chosen a coordinate system then a p-form can be specified in this way I want to Ip dx I1 wage dx Ip okay this is what one means by p-form p and if we give the components of omega okay then that corresponds to specifying a p-form but this requires choosing a specific coordinate system. Now there is an equivalent but a slightly different way of specifying a p-form which is a following. So, suppose we consider a vector field okay vector field in a on a on this space we will take the form vi del del xi right this is a vector field okay on the same manifold. Suppose we have so this I will call v okay v is defined to be this. So, suppose we have p vector fields okay. So, for example, v1 is v1 i del del xi v2 is v2 i del del xi and so on. In this case you can define something called the contraction of the p-form with this vector fields omega v1 to vp is defined essentially as omega I1 to Ip vi1 to vip okay this is the definition of contraction of omega with this p vector fields and if I tell you how to compute this for every any arbitrary choice of p vector fields okay that is equivalent to giving omega right if I tell you how to compute this for every any arbitrary choice of vp vector fields okay then that is equivalent to specifying this components omega I1 to Ip okay even though we have now we are choosing a specific coordinate system to level omega is this point clear. So, specifying a p-form means we have to specify how to contract it with p vector fields p arbitrarily chosen vector fields okay and this is easier to specify because this is an arm body okay. So, what I will tell you now is on this pgn okay when I say I will I will define what omega p is okay what I will tell you is how to construct the contraction of omega p with any arbitrary set of p vector fields on pgn okay and this is the definition no okay so yeah yeah so the phi omega all of these p-forms are functions of these product of n Hilbert spaces right that is always there right so so I can write it this way yes exactly so this is a different thing all together right it is always a farm you have to first specify your phi and that gives you omega p right but now on top of that this omega p okay will be contracted with this v vector fields well it is a linear map right I do not know how they are once called no omega does not have any additional index omega is a complex number this is a complex number that given phi you get a number let us go back to omega 0 okay where this confusion does not arise okay this says that given phi it is just giving a number right that's that's what it means right this is omega 0 right this is not this is not a vector right this is this are no other index right this is just a number complex number yes omega is linear in phi yes two complex numbers yes exactly yeah but I mean the point is there is no hidden index anywhere here right yes if you remove the phi then there is hidden index yes okay but once you put the phi then there is no hidden index yeah yeah right okay any other questions okay so what I have said I have to explain here is that giving omega okay giving a p form okay basically means that we have to specify how to contract it with p vector fields now let us recall that the tangent vectors of p okay they are labeled by vector fields vi on the Riemann surface around these circles so given a tangent vector vi of p gn we pick the corresponding vector fields vi on the Riemann surface okay because each tangent vector is level by one or more of these vi's around these separating circles CSI okay and we define this operator okay which is just a contour integral of these ghosts b ghosts and anti b ghosts okay b bar ghosts weighted by vi and vi bar okay and these are naturally integrated because b of z okay is has dimension 2 okay vi being a vector field has dimension minus 1 so this is a dimension 1 object which can be integrated along a contour okay this is independent of how you parametrize the coordinates end so this is the definition of b of vi okay and if there are p such vector fields for each such vector field you can introduce this operator b of vi okay and now I give you the omega p of phi contracted with this sorry this should have been p okay it is not v and it is vp okay omega p of phi contracted with this p vector fields okay any vector any set of p vector fields that you can choose because every vector field has a representation like this is given by this correlation function okay so here it is not simple correlation function of the external vortex operators you also need to insert a set of b ghosts in this form so this means that given any p tangent vectors of pgn okay I have told you how to get a number okay how to come compute the contraction of omega with those tangent vectors and that is equivalent to giving omega how do you say the anti-symmetry well it is easy because these are all anti commute right so if you exchange v1 and v2 right v of v1 v of v2 will become v of v2 v of v1 right that just that just picks up a minus side right so anti commutation just follows from the anti commutation relation of the b's any other question okay and if you think about it a little bit okay this will also explain why we needed this condition on phi that it has to be annihilated by b0 minus b0 bar okay because these tangent vectors which correspond to rotating w okay you can show that those essentially amount to acting b0 minus b0 bar on the phi's okay but those should be declared as zero tangent vector right because changing w changing a phase of w doesn't move you in pgn right so those should be declared as zero tangent vector and they better remove back the result zero for this contraction okay and that happens automatically once you impose that condition on phi that it should be annihilated by b0 minus b0 bar okay that's the origin of the second condition on phi on the state in the Hilbert space are there any other questions so this defines omega okay but of course there is a reason why you define omega this way and the reason is that once you have defined omega this way one can use the standard manipulation involving conformal field theory on Riemann surface and prove that this omega p has all the desired properties okay so what are the desired properties okay so here is the first one that suppose any of the tangent vectors is zero tangent vector induced by reparameter of one or more z i okay that we can imagine that we choose the corresponding vector fields right we said that for every vi there is vector field right but there are some causes of vector fields which are actually zero right because those are induced by just reparameters in the z i's and I told you that those should represent zero tangent vector because you are not really moving in pgn okay and one can show that if you should pick a tangent vector which is of that kind okay then omega p automatically finishes okay this quantity that I defined namely this quantity okay automatically vanishes okay and you basically show it by deforming the contour integrals of the brst of the of the of b okay and showing that the result cancels okay so it's basically standard conformal field theory technique that allows you to do this that whenever the vector field is induced by reparameterization of one of the z i's okay the result vanishes just by identically second condition is the most important one okay this one right this identity again can be proven in general just based on that expression for omega that I gave okay and identities in conformal field theory okay I will not go into the details but let me just point out the reason for this okay the key key result that allows you to do this okay which is that when you take the anticommutator of qb with b you get back t okay so essentially what you show is that if we have an expression like this okay this involves contour integral of the brst current okay you have to deform the contour integral okay through the b's and manipulate it as you deform the contour integral through the b's you pick up factors of t okay and what is t after all okay t essentially generates a reparameterization emotion in p g n okay so that's the way we see that omega p of qbi phi is actually an exterior derivative acting on omega p minus 1 okay why omega p minus 1 because once you pick up a commutator and you have lost one insertion of b right so that's like a p minus 1 forming contracted okay but you replace it by t and insertion of t essentially gives you the d exterior derivative operation because this is what moves in the motorized space okay or in the in p g n okay so this has to be done carefully but this identity follows from standard confold filter analysis any questions and finally that if we do pick phi to be a tensor product of brst invariant states suppose number two okay which are the physical states then one can show that this integral with the definition of omega that I have given earlier this integral over any arbitrary section gives you back the usual on shell amplitudes in string theory okay in fact if you have never seen an expression for the on shell s matrix element of string theory okay the usual on shell s matrix elements okay this is it okay you construct the omegas by this procedure by inserting vortex of part of the external states and this b ghost okay and you integrate it over the section and what you get is what he normally called the string it is string s matrix okay that you compute from this CFT correlation function which is of course good because you are trying to define off shell amplitude but they better have the property that when the external states are on shell in the standard sense they give you back the on shell amplitude right otherwise it is not an off shell continuation of the usual on shell amplitude okay are there any questions they explicit form of omega s it will depend on the choice of local coordinates right but that is basically in the choice of section right in here that even depend apparent dependence is on the choice of s because omegas are defined everywhere on pgn right because in that way omega being a form on pgn means that given a choice point in pgn right which implies that you have already chosen your local coordinate system okay you define omega so the dependence of the find the apparent dependence of this integral is in the choice of hgn right over what subspace of pgn you are integrating over okay but as we discussed last time as long as the external states are bearish the invariant right it does not matter what subspace will integrate over you get the same result omega is unique yes this omega is I mean this omega is unique right this is this is a complete definition of omega it tells you given the tangent vectors how to construct this p form so there is no ambiguity in omega so omega is unique what is non-unique is in the choice of the section right the section can be chosen in different ways okay this is something this one should keep in mind omega there is no ambiguity in defining omega okay it is a p form on this big space okay it of course depends on the choice of local coordinates okay but that is just a statement that whether you define the p omega here or here okay you get a defined result right omega is a p form on this space is this point here okay so let us then go to the last task which is choosing doing compatible sections on pgn okay because this is what is needed for defining off shell 1pi amplitudes okay so far whatever I have done I have defined off shell amplitude okay but we did not require any additional constraint like having this 1pi formalism okay so let me remind you that here the procedure is that we okay so we begin with three punctured sphere and one punctured torus okay because they are the simplest systems okay why are they simplest okay because if we look at the the three punctured sphere which is a sphere with three punctures and a torus with one puncture you can easily convince yourself okay that these cannot be obtained by plumbing fixture of anything smaller okay so in a sense these are the basic building blocks okay these are all one particle irreducible okay so that is why we start with this okay so we begin with three punctured sphere and one punctured torus okay so the three punctured sphere has zero-dimensional modular space and the one punctured torus has two-dimensional modular space right these two dimension is the modular parameter of the torus the tau okay which takes value in the fundamental region. Now here because these are 1pi okay these objects these are both 1pi we can choose local coordinates at the punctures arbitrarily but not totally arbitrarily it you have to be consistent with symmetries for example exchange of punctures on the three punctured sphere okay should preserve this choice of local coordinates and if you take for example this three punctured sphere if you make an sl2c transformation that exchanges the punctures okay you should make sure that the local coordinates also get transformed accordingly okay so that is what one means by exchange of punctures on the three punctured sphere and similarly on the torus okay we have to be consistent with modular transformation right if you take a point on the upper half plane and you go to another point in the upper half plane which is related to it by modular transformation you should ensure that your coordinate system also gets transformed accordingly okay but other than that the choice is completely arbitrary. So let me remind you again that choosing specific local coordinate system corresponds to choosing specific sections of p03 and p11 okay in fact p03 okay this is a little misleading for p03 because m03 is actually a point right m03 is a point right so p03 is in fact only vertical okay it is it is an infinite dimensional space okay but there is a base direction right on this infinite dimensional fiber direction you are picking a point okay that is what I mean by choice choosing a section of p03 for p11 this is more representative because mgn is actually two dimensional right so you have a section okay we have chosen a specific local coordinate system for every value of tau which means choosing a section okay so we declared this to be 1pi sub spaces of p03 and p11 okay these choice of sections and this means because we have declared all of them to be 1pi Riemann surfaces in this case there is no difference between R and S okay so R03 is the same as S03 R11 is the same as S11 okay R11 if you remember we defined R as the 1pi sub spaces S as the full section okay in this case everything is 1pi okay and hence S and R are the same okay so let me remind you again what we said last time that we said that if we have say in pgn this is your pgn okay this is mgn okay we have a smaller part which I call Rgn and the full section of course is bigger but the idea is that this dotted part this should be obtained by gluing the Riemann surfaces okay so because these are 1pi these are 1pi okay so there is a basic object which we called Rgn okay basic part of the set of Riemann surfaces and the rest are obtained by plumbing fixture okay since three punctured sphere and one punctured torus okay cannot be obtained by plumbing fixture of anything else we have to declare the whole thing as R is this point here but now let us consider two three punctured spheres and glue them using plumbing fixture okay so take one puncture from each of the three spheres and glue them okay let me remind you that if W1 and W2 are the local coordinates around these punctures then the plumbing fixture is done by setting W1 W2 equal to Q where Q is e to the minus s plus i theta and you take s to range from 0 to infinity and theta to range from 0 to 2 pi so all four punctured spheres that you obtain this way you declare this to be one p r four punctured spheres okay and you have to choose the local coordinates to be those which are induced from the three punctured spheres okay on this subset of four punctured spheres and you have to repeat this for any equivalent permutations of the four punctures for example if you mark the punctures as 1, 2, 3, 4 okay if this is 1, 2, 3, 4 okay you also have another contribution where is 1, 3, 2, 4 and 1, 4, 2, 3 okay so those are the s, t and u channel contributions okay or diametrically you can diagrammatically you can think of this s, t and u channel diagrams okay so these the one pi the four punctured spheres that we get by this plumbing fixture you declare as one p r the rest you have to declare as one pi okay so I will now write down some explicit expressions which we will perhaps make it a little clearer so suppose we take this three each of the two three punctured spheres to be described by a coordinate system z1 and z2 okay just as a plane and let us suppose that the punctures are chosen to be at 0, 1 and infinity okay we can pick them anywhere but this is the standard notation that on a plane you pick the punctures at 0, 1 and infinity okay the three punctures on the sphere and suppose that we have chosen the local coordinates around zi equal to 0 on each of these as wi equal to some function of z1, w1 as function of z1 and w2 as some function of z2 okay this function I am not specified it is some function that specifies local coordinate system it is easy to see from this that the plumbing fixture relation which was this w1, w2 equal to q gets translated to the following relation between z1 and z2 okay because after all z2 equal to 0 corresponds to w2 equal to 0 right w2 sorry z2 equal to 0 corresponds to w2 equal to yeah if z2 equal to 0 of course you map to z1 equal to 0 okay that that is not the point so let us see so suppose we take arbitrary point in z2 plane okay then w2 is given by f of z2 okay w1 will be given by q over f of z2 okay and z1 is f for f inverse of w1 so z1 is related to z2 in this form okay and now the question is given this relation okay where are the punctures in the z1 plane you could have also done it in the z2 plane but let us look at it in the z1 plane okay so you already had punctures at 1 and infinity which are originally there but now you have a puncture at z2 equal to 1 which corresponds to in the z1 plane a puncture here okay and another puncture at z2 equal to infinity which corresponds to in the z1 plane as a puncture here okay so this gives a four puncture sphere okay where the punctures are at 1 infinity f inverse of q over f1 and f inverse of q over f infinity is this clear okay so you get a four puncture sphere this way okay so this is again the repetition I have just written the coordinates of the four punctures that we find that way okay and now from this we can calculate the cross ratio which is really the SL2C invariant combination okay which labels a modular space of four puncture sphere okay and basically once you have chosen the function f okay we see that as you vary as you let q vary over the region where mod q is less than or equal to 1 which corresponds to letting s go from 0 to infinity okay we cover certain range of sub certain subspace of the x plane similarly the key and usual in diagrams will cover other subspaces of the x plane okay which which will be related to these by exchange of x2 and x3 and x2 and x4 okay so this is what I had said that for the four puncture sphere okay the modular space so let us suppose we have here m04 so the s channel diagram covers part of the modular space the t channel diagram covers part of the modular space and u channel diagram covers some other part of the modular space okay so these you declare as 1pr four puncture spheres whatever is missing you have to supply back these you declare as a 1pi four puncture spheres okay and in this explicit example okay you can calculate once you are given a specific choice of f you can calculate explicitly what part of x space is covered by s channel diagram what part of x space is covered by p channel diagram and what part of x space is covered by the u channel diagrams okay so you can explicitly identify these regions okay so once you have done this then on the 1pr four puncture spheres you choose the local coordinates to be what is induced from the three puncture spheres and on the rest of the four puncture sphere we can choose the local coordinates arbitrarily but again you have to be consistent with the symmetries and continuity and we can we declare them to be 1pi four puncture spheres okay continuity means that we basically if we have ended here right we do not want to start from here right we end from we begin from here and go here so local coordinates in the neighborhood of this should be close to what it was here okay that is what on means by choosing it maintaining continuity similarly we can glue these three puncture sphere with one puncture torus okay you can easily see that this produces a set of two puncture tori okay a total so this has a topology of a torus with two punctures we declare these to be 1pr and we choose the local coordinates on these to be induced from the constituents so this will cover part of the modular space of two puncture tori and we then declare the rest of the two puncture tori to be 1pi and choose local coordinates on these arbitrarily maintaining symmetries and continuity okay so this is basically the general procedure you start from the simplest object okay and then we build the more complicated objects by plumbing fixture okay whatever is not covered you declare as 1pi and then you use those to produce even more complicated objects okay and go this way okay and this way you can cover the all the remand surfaces and you can declare some of them as 1pi the rest are 1pi so this is the half what I have said here okay that we by proceeding this way we can choose gluing compatible sections we identify part of the sections as 1pi space rgn okay and given this the off shell at 1pi amplitude is given by integrating omega over this 1pi sub spaces rgn and the full amplitude of course is the integral omega over the whole section is gn okay so that is the basic story are defining off shell amplitudes in bosonic string theory and off shell 1pi amplitudes in bosonic string theory okay so what we have to do tomorrow is to see how to generalize this to formalic strings okay two super strings and heteronic strings but the basic procedure will be very similar okay some of the technical details will be slightly different but the basic procedure will be similar are there any questions yes yes no there is no okay the point is that the on the on the sphere the number of punctures always has to be 3 right there is no okay remand surface with less than 3 punctures that is why 3 puncture spheres are elementary right and 1 puncture torus are elementary right there is nothing that you can do to build them from lower surfaces so those are all 1pi and then you build from there okay so the only way to get a 4 puncture sphere from the plumbing fixture is by gluing 2 3 puncture spheres right there is nothing else that we can do yes yes all possible channels yes so given a particular gluing you have to basically sum over all in equivalent permutations of the external legs right because external legs are not are marked right because you are inserting different states on different external states in general external legs right so if you have something like this if you have a remand surface like these are the punctures right these are the places where you are inserting vortex operators and those vortex operators are not identical states right you insert vortex operator one here two here three here and four here okay so this is clearly distinct from this contribution where you insert one and three here and two and four here so you have to sum over all possible in equivalent permutations of the external legs exchanging these two is not in equivalent right because you have chosen the vortex to be symmetric under exchanges so one two here are two one here is the same okay but one two here or one three here is clearly distinct so you have to sum over all possible contributions okay so all of these will contribute separate one pair gives separate one pair contributions okay and then whatever is left you declare as one yes yes no it doesn't give anything nice I mean if we just take the one pi amplitude and focus on the external option on shell states right yes so the point is the result will depend on the choice of RGN right see for SGN we argued that it doesn't matter what SGN you choose on shell amplitudes are all the same okay but on shell one pair amplitudes are not all the same right it's exactly as what has happens in gauge theories right suppose you have done a particular gauge fixing okay and now if you calculate the on shell amplitude okay and instead of considering all possible Feynman diagrams we just consider the one pi Feynman diagrams the result will be gauge dependent okay so on shell one pi amplitudes are not particularly nice okay so one pi amplitudes are useful but not directly in this sense maybe we should postpone our further question through the discussion session so let's thank Kashuk