 So, today I am going to talk about heterotic and type 2 string theory and since the lecture is going to be a little more technical than what we had earlier. I will try to work out most of the things on the blackboard although once in a while I will try to go back to the slides, but in case you have difficulty seeing the blackboard you can just let me know and I will put on the appropriate slide. So, we will be focusing mostly on heterotic type 2 is essentially more of the same. So, heterotic wall sheet theory. So, this is a super conformal field theory. So, it has several components first we have matter super conformal field theory and here we are not going to make any assumption about what the background is ok. This analysis will be basically valid for a general background. So, which means general matter super conformal field theory, but of course, it has to have the right central chart. So, CLCR will be 26 and 15. If you sorry 10 and 15 no this is fine to see 26 and 15. So, left moving sector will have 20 central chart 26 right moving sector will have central chart 15 ok. So, for the for example, if it is flat space time this 15 comes 10 of these come from the space time scalars space time bosons and sorry wall sheet scalars and 5 comes from the wall sheet formulas. Then we have the ghost in fact, there are many ghosts. So, BC these have I did here. So, BC these have CLCR as let us say in this convention 0 minus 26. Then you have B bar C bar these are the usual ghosts that I have been hearing about. So, for these CLCR is minus 26 0. So, these are diffeomorphism ghosts. So, these are diffeo and then we have beta gamma these are associated with fixing wall sheet supersymmetry and these have CLCR as 0 and 11. So, as you can see in my notation the holomorphic fields these are holomorphic fields these are right movers and the anti holomorphic fields are left movers. And you can check that the total central chart always adds up to 0 both on the left and the right sector. So, these are bosonic. So, these are anti commuting and these are commuting ghosts ok. So, from the wall sheet perspective these are bosonic fields. Nevertheless, it turns out that to describe formulaic string perturbation theory you have to bosonize the beta gamma system. So, we bosonize beta gamma and the way one does it is that we identify we introduce 3 more fields psi eta phi. So, these are anti commuting fields on the wall sheet this is commuting ok. There is a scalar on the wall sheet and the relationship between the original fields beta gamma and the psi eta phi are as follows. So, psi is eta e to the phi. So, gamma is eta e to the phi beta is del psi e to the minus phi and then we have the deltas chronicle deltas delta sorry not chronicle delta d r delta delta gamma e to the minus phi and delta beta ok. So, this is the relationship between the beta gamma system and the psi eta phi system ok. And all the operators that will be writing down will be constructed out of psi eta and phi ok. Whenever you have operator statement of beta and gamma you can use these relations to convert them back in terms of psi eta and phi are there any questions. So, it turns out that there are 3 quantum numbers which are important these quantum numbers are the ghost number then a new quantum number which is called the picture number and the final one of course, is the GSO property whether a given field is GSO even or GSO odd because all physical operators have to be GSO even in this formalism ok. So, what I am going to do is to write down the these assignments for various fields that I have introduced here ok. Now, first of all matter sector fields have trivial ghost number and picture number ok. No ghost number and picture number then GSO ok the while sheet formulas for example, are GSO odd the while sheet scalars are GSO even and for spin fields there is a complicated transformation which I will not write down but let us look at the ghost sector. So, C C bar these have assignment 1 0 plus which means it has ghost number 1 picture number 0 GSO even B B bar these have minus 1 0 plus because B's are anti ghost C's are ghosts gamma is a ghost. So, it has 1 0 minus. So, notice that this is GSO odd similarly beta is minus 1 0 minus. So, notice that all the original fields original ghost fields that we had in the problem all carry picture number 0. So, the picture number in fact is something that appears after you do this motionization. So, I will now write down the similar assignments for xi eta and phi ok. So, xi has minus 1 1 plus eta is opposite. So, 1 eta is 1 minus 1 plus e to the q phi ok these are the operators that will be considering this has 0 q and minus 1 to the q and if we have something like del phi these have 0 0 plus which means this GSO even and it does not carry any quantum number ok. And you can easily check from this that if you know the quantum numbers of eta e to the phi which is listed from the in the last column over there you can easily work out what the corresponding quantum numbers for gamma and beta are ok. So, these are not no longer independent once you have specified these quantum numbers for xi eta e to the q phi ok. And in this convention one can write down an expression for the BRST charge which I will not do here. So, BRST charge for example, carries ghost number 1 picture number 0 because BRST charge is constructed purely in terms of the original ghost phase that does not really require any xi eta and phi. And finally, one more operator have to introduce the picture changing operator which you have been talking about, but I have not told you what it is and now I am going to write down explicitly the form of this. So, this is the anticommutator of so let us say we insert it at g it is the anticommutator of q mu with xi and I can write down the explicit form of this. So, this is this is c del xi plus e to the phi tf minus one quarter del eta e to the 2 phi b minus one quarter eta e to the 2 phi b. Where tf is the super stress tensor of the matter fields super stress tensor matter fields matter SCFT ok. So, again I want to emphasize that we are not assuming what anything about what the matter SCFT is. So, this is valid for flat background as well as cloud background as long as the background is a solution to equations of motion. But as long as it is a super conformal field theory it will have a super stress tensor and that is what we have to use here ok. So, this is the picture changing operator. And because BRST operator is nilpotent this the q b commutator with this gives you 0 and you can easily work out the quantum numbers of this. And in the same notation as they are I will write the quantum numbers it is 0 1 plus ok that goes number 0 picture number 1 and it is yes we are there any questions. So, this will be one of the main building blocks in describing super swing part motion theory what is it that you have to compute to get the physical s matrix elements ok. So, yesterday we introduced the notion of the Hilbert space of states. So, there is a similar notion of what is the correct Hilbert space of states in this case. So, the Hilbert space of states are essentially states in matter plus ghost super conformal field theory, but we impose the same condition that this is annihilated by B 0 minus B 0 bar and L 0 minus L 0 bar ok. This is the same condition that we imposed for the Bosonic strength theory that you only work with a restricted class of states we satisfy this ok. So, this Hilbert space we can divide it into infinite set of sub sectors. So, we call h n the subspace of h subspace of h carrying picture number n n and GSO even oh I should have said that we impose this all the time that you will only consider states which are GSO even ok. So, even which means that even though you use operators which are not always GSO even these for example, are GSO odd operators the actual vertex operators which will correspond to physical which will correspond to off shell string states will be taken to be GSO even and satisfying this. And then if you want to impose on shell condition then you also have to demand that the vertex of operators we are a same variant, but that is not something that we are going to impose now because we will be working with off shell amplitudes ok. So, in this rotation now I can be more specific and introduce the notion of off shell states that we will be working with ok. So, off shell in a sector states by definition are states in h minus 1 ok. So, these are states which carry picture number minus 1 and off shell Ramon sector states belong to h minus half ok. These are states which carry picture number minus half. If we are dealing with super strings instead of heterotic string then the story will be repeated both on the left and the right sector ok. Here we have introduced the beta gamma system only on the right sector in super strings there are also a beta bar gamma bar ok and you will have a notion of NS NS which will be in h minus 1 minus 1 ok. NSR will be in h minus 1 minus half right. Ramon Naboo Swartz will be h minus half minus 1 and then finally, Ramon Ramon will be in h minus half minus half ok. So, basically the there will be four sectors instead of two sectors ok, but otherwise it is the same. So, I will not do that explicitly it is a good I have to just focus on the heterotic strings ok. Are there any questions? So, it turns out that when you compute correlation functions on a Riemann surface there are anomalous conservation laws for both the ghost number and the picture number. So, on genus G on genus G Riemann surface we need total ghost number and total picture number equal to 2G minus 2 ok. So, only those amplitudes are non-zero for which the total ghost number is minus 6G minus 6 and the total picture number is 2G minus 2 for bosonic strings ok. This condition is still there and that in fact is a reason why we can only work with the 6G minus 6 forms. If you have for example, n equal to 0 if you remember we had 6G minus 6 insertions of b ghosts right. I mean those are precisely what was needed to produce this total ghost number of minus 6G minus 6 ok. So, this is a statement you should keep in mind total ghost number has to be minus 6G minus 6 total picture number has to be 2G minus 2 ok. Now, as you will see this will be taken care of automatically ok when you define the form ok. But let us see what it means what this statement means total picture number has to be 2G minus 2 ok. So, let us suppose that we are computing a an amplitude with m ns sector states and n Ramon sector states ok. So, an amplitude with m ns and n r sector states ok. As long as you take even for G equal to ok G equal to 1 and G equal to 0 and G greater than 1 as long as you have for example, for G equal to 0 3 punctures and for G equal to 1 1 puncture this statement is true ok. It is if you do not have any puncture then there are additional 0 modes and you have to be a little more careful ok. So, if you take an amplitude with m ns and n Ramon sector states each ns sector states carries picture number minus 1 right because ns sector states belongs to h minus 1 ok. Each Ramon sector states carries picture number minus half ok. So, you get a picture number. So, the vortex operators vortex operators give picture number minus 1 minus m minus n by 2 ok. But we need total picture number 2G minus 2 to get a non-zero answer ok which means that you need to insert an operator which carries picture number 2G minus 2 plus m plus n by 2. So, what are those operators? Those are the picture changing operators that I wrote down here ok. So, we need we need ok. So, these picture changing operators in fact, arise naturally when you if we start with integration of a supermodelized space and then try to carry out the formulaic integrals. But we will not follow that route we will simply use the fact that just from picture number conservation it looks like you need something which has to supply this many picture number the this much picture number. And picture changing operator is a natural object because this is BRST invariant. So, when you insert this into a correlation function none of the BRST properties ok will get spoiled ok. So, we will try to compensate for the missing picture numbers by inserting these many pictures in the operators ok. Are there any questions? e to the phi carries g of ghost number yes yeah. So, ghost numbers are all being carried by psi and eta ok. So, now, let me remind you what we have to do now ok. So, this is exactly same task that we had for Masonic string theory. The first task is to find the parametration of the space p g m n ok. I am now calling it p g m n because it is genus g m n s punctures and n Raman punctures ok. We have to distinguish between n s and Raman punctures ok. So, let us start with that task. So, parametration now there are two kinds of coordinates of p g m n. The first kind of coordinates are what we introduced yesterday ok. These are identical to what we had for Masonic strings. So, the module i and choice of local coordinates choice of local coordinates these are identical you can be done in exactly the same way as in the Masonic theory ok. Because it is the same parametration right you just take the Masonic version p g m plus n ok. So, let us write this. So, you take p b now stands for Masonic ok. So, it is the same coordinates same way you introduce this coordinates as on p g m plus n ok. m plus n total number of punctures genus g and you introduce coordinates in that way ok. So, you basically divide the Raman surface into disks and spheres look at the transition function and those transition functions are what level this ok. But now we have a new set of coordinates which are the locations or the PCOs this is an extra data ok. So, these are denoted by y k a complex coordinate y k for k equal to 1 going up to 2 g minus 2 plus m plus n by 2 because so, there are really two coordinates for each k because it is a complex ok. Now, tangent space ok. So, what are the possible tangent vectors of p g m n. So, tangent vectors are essentially infinitesimal moves in p g m n. So, again there are two kinds ok. The first kind are the same as the tangent vectors of the Masonic ok. You just change the transition functions by infinitesimal amount and those as we have seen can be labeled by vector fields on the Raman surface ok exactly as we had yesterday. But then there are few more ok those are the tangent vectors corresponding to deforming y's ok and those we can denote these are normal ordinary coordinates right. So, so, why each y k I should have set takes value on the Raman surface ok. It is a complex coordinate, but it is on the Raman surface. So, these are del del y k and del del y bar 3 ok. There is a basis of tangent vectors on associated with the y deformation is this clear ok. So, this basically finishes the discussion of how to label how to co-ordinate is p g m n and how to parameter is its pet its tangent space ok. So, that basically finishes the first task. Now, you go to the second that you have to construct a omega p ok and this is the identity that will that guides us in constructing omega p ok. Again what I will do is that I will write down that result for omega p ok and the proof that it satisfies those identities again is a straight forward conformal field theory manipulation ok. You have to basically use various conformal field theory what identity is and you can prove those ok. But let me now describe how to construct omega p. So, omega 0 we start with omega 0 phi ok and phi let me remind you belongs to n fold let me call it h n s m. So, basically m copies of n s sector Hilbert's space and n copies of the Raman sector Hilbert's space right because there are n Raman sector states and m Nebuser sector states. So, phi is some state in this product Hilbert's space. So, omega 0 in fact, is simple it is more or less the same as in the case of Bosonic strength theory. So, 2 pi i 2 pi i times. So, the correlation function on the Raman surface phi s, but now we have to insert a product of picture changing operators. So, i equal to 1 to 2 g minus 2 plus m plus n by 2 chi of y j y i ok. So, you see if we did not insert this the result will be 0 right because phi together carries picture number minus m minus n by 2 right that does not give a non-zero amplitude on the Raman surface. So, in order to get something non-zero we had to insert something and what we insert is a product of picture changing operators ok. And you see that the result depends on why because where we insert the picture changing operators of course, affects what the correlation function is. Is this definition clear? So, this is essentially generalizing what we did for Bosonic strength just inserting some extra picture changing operators. Now, we come to omega p omega p here again we have to specify omega p by saying how to contract it with p tangent vectors p arbitrary tangent vectors. And as you have seen the tangent vectors are of two kind the first kind or the second kind. Now, first let us suppose that all the p tangent vectors are of the first kind then the answer is exactly what we did what we gave last time for Bosonic strength because the first kind of tangent vectors are what is inherited from what we had for the Bosonic strength. So, for the first kind so, these are all of the first kind the result is ok we insert these products of these ok. If let me remind if you have forgotten how d v is b was the contour integral. So, we had seen that this first kind of tangent vectors for every tangent vector of this first kind we can associate a vector field v i of z on s around some circle around some c i and then you had defined b of v i as I presume it was plus let me go I think 1 over 2 pi I did not write in fact, it is understood when you do this contour integrals. So, let me use the same convention ok. So, we just insert this b of v i's. Now, the interesting question is what do we do if some of these tangent vectors are of the second kind ok either del del y k or del del y bar k what is the rule for constant contraction of omega p with such tangent vectors ok. So, first of all we for each tangent vector of the first kind we do exactly the same thing so, what will happen is that some of the b's will be removed right because there will be less number of tangent vectors of the first kind. So, instead of having this product this is j equal to 1 2 p. So, instead of having the product over j equal to 1 2 p it will be less than p because it will be the number determined by the number of tangent vectors of the first kind ok. So, those b's will be removed ok, but instead what we have to do something to show that it is being contracted with one of these tangent vectors ok. So, what are the rules? So, let me now write down the rules. So, if we contract del del y k what effect does it have the first we have to remove chi of y k ok. So, in this product that we had one of them would have been chi of y k the particular del del y k is being contracted you remove that from this product and then we insert ok. So, where you remove b at that place you insert this factor of minus del psi y k. So, this is the rule for contracting with del del y k much simpler than what we had for this tangent vector of the first kind. Contraction with del del y bar k is even simpler it is 0 ok. So, there is no contraction with del del y bar k is this clear ok. So, omega basically has no component along the del del y bar k. Now, this is of course, definition ok. This definition is concrete ok there is nothing no ambiguity in this. However, it requires some work to prove that with this definition omega p satisfies the required identities ok. All the required desired identities that it is supposed to satisfy it does in particular that identity which is crucial for proof of gain variance and various other things yes. So, m is just the number of ns punctures and n is the number of Raman punctures oh sorry this is m plus n I am sorry this should be m minus m yeah. Everywhere you are you see this this is as a total number of punctures yeah. So, I just borrowed it from the Bosonic this it is always the total number of punctures p well not apriori at this stage no ok. But the ghost number conservation of course, will give a non-zero answer only for a specific choice of p ok and that choice is the dimension of the modular space over which you integrate right with the 6g minus 6 plus m plus n. So, that will come only when you fix the ghost number of phi right at this stage I have not said anything about the ghost number of phi I have only said fix the picture number right. But once you fix the ghost number of phi only one particular p can contribute are there any other questions yes the picture changing opera. So, I mean if you remember the relationship between x and x was basically q b with xi right ok. So, q b carries zero picture number ok. So, x and xi both carry picture number 1 ok. So, there is no confusion. So, indeed yes total picture number has to be always the same right because otherwise you will get zero answer ok and when you replace x by minus del xi you are not changing the picture number of the operator ok are there any other questions ok. So, if you are interested in on shell amplitudes ok the convention on shell amplitudes then this is the end of the story ok you can take this definition of omega that I have given and you can compute on shell amplitudes. So, you just calculate the same this omega 6g minus 6 plus 2 m plus 2 n ok with taking this state phi to be our strain variant ok and then we can prove by using exactly the same method that I used earlier that the result of the integral of omega over a section is independent of the choice of the section. So, you choose any section and integrate ok and you will always get the same result and that is the on shell amplitude ok. So, if you have not seen how to compute the on shell amplitudes in super strain perturbation theory or heterotic strain perturbation theory this is a complete description of how you can compute on shell amplitudes in heterotic strain theory ok. And for super strain as I said you basically have to repeat this also for the left moving sectors ok there will be additional beta gamma both system beta bar gamma bar correspondingly additional picture changing operators ok. But our goal of course is to construct off shell amplitudes and for that we have to also complete the third step construct the gluing compatible sections ok. And again here it goes in the same way as in the case of bosonic strings with some a little addition of the subtlety which I am going to describe. So, basically we gluing compatibility means that whenever we have a 1 PR Riemann surface where we have glued 2 Riemann surfaces ok. Then the choice of local coordinates around each puncture have to be what is induced from the lower genus surfaces ok that was the bosonic strain constant and that constant still has to be satisfied for super strings ok. The choice of local coordinates have to follow what we had used for the bosonic for the lower genus surfaces. But now there is an additional constant ok and additional constant comes from the fact that the choice of section involves not only specifying how we have chosen the local coordinate system, but also specifying how we have chosen the picture changing operator locations ok. So, these crossed circles are picture changing operators ok in this notation. So, there is a certain set of picture changing operators here depending on the number of vertices and gluing compatibility now requires that on the 1 PR surfaces the choice of picture changing operator locations should also be the ones that are induced from the choice on the component surfaces ok. So, if these are 1 PI ok on these you have chosen some section which means you have also specified how to locate this picture changing operators. Then on the new Riemann surface that you get by gluing by plumbing fixture of these 2 the choice of picture changing operators would follow exactly the same rules as you induce from there. Is this statement clear? Well it turns out that even though the statement may be clear it is wrong ok it is not possible to do this ok and I will explain why it is not possible to do this ok and it is a simple counting exercise. So, let us do some counting. So, let us call this surface sigma 1 and let us imagine that it has genus G 1 M 1 N S punctures and N 1 Ramon punctures ok that is the convention we have been following and this half as a sigma 2 and suppose it has genus G 2 M 2 N S punctures and N 2 Ramon punctures then the number of PCOs on sigma 1 is 2 G 1 minus 2 plus M 1 plus N 1 by 2 number of PCOs on sigma 2 is 2 G 2 minus 2 plus M 2 plus N 2 by 2. So, total which is the sum of the 2 is 2 G 1 plus G 2 plus M 1 plus M 2 plus N 1 plus N 2 by 2 minus 4. Now, suppose sigma is the new human surface which is obtained by this plumbing fixture and you have to count the required number of PCOs on sigma. So, you have to consider 2 cases separately the first case is that the plumbing fixture is N S puncture ok. So, then on sigma we have number of N S punctures equal to M 1 plus M 2 minus 2 2 have gone away because we are using the N S 2 of the N S punctures for plumbing fixture ok. Number of Ramon punctures is N 1 plus N 2 nothing has happened to Ramon punctures and genus of course is G 1 plus G 2 ok G is G 1 plus G 2. So, the required number required number of PCOs sigma is 2 G 1 plus G 2 minus 2 plus M 1 plus M 2 minus 2 plus N 1 plus N 2 by 2 ok and you can easily check that this number is the same as that ok. So, there is no problem here. So, this is fine, but it is easy to check that if you glue at Ramon puncture then the required number on sigma ok I will just write down one can just look at this answer and let it done. So, it is G 1 plus G 2 minus 2 ok. Now, you see the total number of N S punctures is still M 1 plus M 2 and the total number of Ramon punctures is N 1 plus N 2 minus 2 because 2 of the Ramon punctures have been used up. This one is one more than this number that is minus 3 this is minus 4. So, which means that we are missing if you just follow this procedure then we will be missing one piece picture changing operator ok. You do not have the required number of picture changing operators on sigma ok. So, somehow we have to find a rule to insert a new picture changing operator on sigma ok just stating this rule is not good enough. Nevertheless, we have to make sure that you do not destroy the nice properties that we described earlier ok which will generalize the Bosonic string results and it turns out that there is a natural choice that one can make ok. I will just write down the result for this. So, if the plumbing fixture is using this relation and what to do is that you insert the picture changing operator not at a point, but take the average of picture changing insertion of chi of w 1 which is the same as integral d w 2 over w 2 chi of w 2 ok because you can always go from w 1 to w 2 coordinate system and this relation is keep this integral unchanged ok. So, you insert a picture changing operators in the in the vicinity if you have these two Riemann surfaces join do you insert the picture changing operator in this vicinity it does not matter where you insert it ok and you average take the average of those insertions over the whole circle ok in this fashion and this one can show satisfies all the desired identities. So, that you do not violate any of the nice properties that you wanted for factorization. So, this is what one means by doing compatible choice of sections for type 2 for heterotic strings ok. For type 2 there is a similar unjust you have to do insert both integral of chi x as well as integral of x part ok because there is also Riemann sector from the left. Are there any questions? Yes. Well, I do not know how unique this is I know this works and a generic choice will not work ok it will violate the factorization property I mean it is possible that somebody else will find some other prescription at some point right and then one has to show that this that is equivalent to this ok. So, now, I come to the final topic and this has to do with spurious singularities whether the choice of section exists or not ok and issues of this kind. So, spurious. So, the picture that have given you so far ok is that we have this p g m n ok with a base which is m g m plus n ok with a fiber and we have some section ok. So, you can either choose to look at the full axial amplitude or 1 pi amplitude in which you will get a part of the section and you have to integrate omega p over this section ok that is the general procedure for off-shell amplitude and this is the same procedure for on-shell amplitudes except that for on-shell amplitudes phi is bear a state variant and the issues that I am discussing now the spurious poles they exist both for on and off-shell amplitudes ok. So, it is not that these are special to off-shell amplitudes even if you are trying interest only in computing on-shell amplitudes in the conventional sense you will run into this problem ok. So, let me explain what the problem is the problem is that omega p is not well defined everywhere in p g m n ok it has spurious poles and those spurious poles occur on a co-dimension two subspace in fact, on many co-dimension two subspaces ok some are easy to understand some for example, you have these insertions for the picture training on pointers right the x of y j's if two y j's come together ok that is the co-dimension two subspace of this ok because one of some of the coordinates of these are the locations of the y j's. So, if two of the y j's come together then there is a singularity due to operative singularity the standard operative singularity. So, those co-dimension two subspaces have to be avoided, but it turns out that there are other kinds of singularities which are not directly related to coincident point singularities even when there is no coincident point anywhere the correlation function suddenly may diverge ok and that also happens on co-dimension two subspaces. So, these are what are called spurious poles ok, but in our description we will also include in spurious pole the ones that come from collision of picture training operators ok. So, the absurd is that on p g m n we have to avoid certain co-dimension two subspaces ok in order to make this integral well defined, but in general that is not possible ok. It is not guaranteed that it is possible to choose this section to avoid those co-dimension two subspaces on which the integrant becomes singular and this has its reflection in the problem of integration of also called modernized space that came up in the last lecture ok. So, all of these problems are related to each other ok in this picture changing operator formalism the problem appears in this form that in general it is not possible to choose sections which avoid this bad co-dimension two subspaces. So, now the question is how do we define these integrals if there are these bad co-dimension two subspaces on which the integrant diverges ok. So, this is what I will discuss last and then I will stop. So, it turns out that it is possible to do this using a notion of what I will call vertical integration and I will give only a very rudimentary description ok the full description is much more involved. So, let us suppose that the base is two-dimensional ok I cannot draw multi-dimensional space here. So, suppose the base is two-dimensional m g n and the fiber is one-dimensional ok which is labeled by one of the y i's ok I am just ignoring the other directions here. So, in this case so, this is a three-dimensional p g n and a co-dimension two subspace is a line ok. So, let us suppose that you have a line going like this which is a bad line ok you have to somehow avoid it, but how can I avoid it because the section has to run like this ok. So, if a section has to run like this there is no way you can avoid this co-dimension or subspace right has to hit it. So, what you can do is the following ok you choose you draw some circle here enclosing this co-dimension two line which in this case is a line does not matter how you orient it you just draw a circle like this. Now, you integrate you choose a section where you integrate outside this surface without hitting the line that you can certainly do right you just cut out a hole ok. If you are trying to extend the section through this hole you would have hit the singularity ok, but you just consider the integral outside this circle then there is no confusion. Then you go vertically up which means that you draw a cylinder based on the circle and you go sufficiently high up so, that the spurious pole goes out ok. So, this is let us call it S 1 this is we will call S 2 and then you fill the hole here. So, now your section that you choose it is not a section anymore because it has this vertical segment, but it is a continuous manifold linear. So, let me explain what a continuous manifold is. So, this is a three dimensional space you come like this in two dimension the spurious pole location is going like this you stop here then you go vertically up then you fill the cylinder there ok. So, the spurious pole crosses the vertical segment right that is the important point ok and we will see why this helps. But is this construction clear ok in this if you think of the three dimension you come the spurious pole is going like this you enclose it by a circle do the integral outside go vertically up and then close it ok. So, clearly there is no problem integrating in this part and in this part right. The only issue is how to integrate about this vertical section because this is the place where the spurious pole hits it. Yes, the result will be independent of the radius because that of course, is this continuous deformation right. The only issue is about the spurious pole right as long as it is not spurious pole it is like a continuous deformation of the integration cycle. And because omega is closed right for B R S T invariant states omega is closed otherwise the result is not independent right, but it satisfies the desired identities which will help us get through ok. So, now let us see how to do integration of what this vertical subspace. So, you see the integration of this cylinder can be done by first integrating in the vertical direction ok. You keep a point on the base fix just go vertically up you do that integral first and then you integrate over the base ok. So, let us try to see what happens when you integrate in the vertical direction ok that is the d y i integral right. So, imagine that you are going from u to v d y i ok u is this bottom most point. So, let us take one of these lines. So, u is here v is here. So, when you are trying to do integrate omega ok in this case omega is two dimensional right because we are taking a two dimensional section when you are trying to integrate omega to along this. Because here the tangent vector is along the vertical direction we have to contract omega 2 with a del del y i right, because we have to contract omega 2 with the tangent vector of the cylinder. So, omega 2 contract to a del del y i what does it do? It removes chi y i and it replaces the minus del xi y i ok. So, now you see that you can do the y i integral explicitly because nothing else depends on y i ok. No other operator in the correlation function has any knowledge about what y i is. So, we just do integral u to v d y i minus del xi y i and that is xi of v minus xi of u minus xi of v ok. The difference in xi between the bottom point and the top point. And now you see what has happened right there is a singularity somewhere in between because this line is hitting the singular the cylinder. But the result is expressed in terms of something which has no singularity because it only depends on insertion of xi on this line where there is no singularity and insertion of xi on this slide where there is no singularity. So, this answer is finite unambiguous and hence we can carry out this integral is this point here. So, this is the way you can avoid this previous pole problem ok. This is of course a very simple situation ok, but this illustrates the basic idea ok. How you can avoid this how you can define integration through this previous singularities ok. More formally what it corresponds to is that you choose some section on part of MGM some section some other section here ok. And then the fact that this is discontinuous you have jumped from here to here you compensate by adding a boundary term ok because after you have carried out the integration over Yi the result has to be still integrated over the in this direction ok. So, that is an integral over the boundary between these two parts of MGM ok. So, this is only the tip of the ice part ok turns out that the general situation is as follows that you have to divide the whole modular space into triangles ok. You triangulate the whole modular space ok and you can take the triangles to be as small as you like. On each triangle you choose a section ok that there is no problem in choosing a section because each triangle you can take to be as small as you like you take choose a section to be in whatever you like ok. Then at the boundary is between two triangles there is the section is discontinuous right just like here ok. Here we had one section inside we have another section. So, at the boundary between two triangles there is a discontinuous jump and you have to add a correction term to take into account of this continuity ok. That is the analog of what you have computed here, but then you can have a situation what three such triangles meet ok even in two dimension we can have a situation of three such triangles meet ok. So, here you have made a correction term here you have applied a correction term here you have applied a correction term, but it turns out that that is not always enough you have to also apply a correction term here ok. That is co-dimension two intersection of the triangles and then there will be co-dimension three intersection of the triangles and so on ok. So, you have to basically go all the way up to co-dimension k co-dimension k where k is the total number of PCOs which is 2g minus 2 the same plus n by 2 ok. And you can basically stop there and you have a complete one can get a complete prescription of what this correction terms are ok. So, this is I should have said this is some work with Edwerton which we are still writing up ok, but basically we now know that this procedure is complete and you can compensate we do not have to choose a continuous section ok. So, this also avoids this question of whether a continuous section exists or not right you do not really need to choose a continuous section ok. We just choose local sections on each small triangle and then add correction terms at the boundaries ok and that gives an amplitude which is which satisfies all the desired identities that we need ok. So, I think I will stop here. So, next time tomorrow I will discuss the 1 pi effective field theory.