 Let us start we are happy to have Ashok Sen with his last lecture about the instant tones in string theory. So, let me remind you that in the last lecture we had introduced this open string propagator and also how to deal with this and the idea was that this was the expression that would have gotten from the wall sheet picture and this is the string field theory picture. And in the string field theory picture ok we can resolve the divergences for the negative L0 states and for L0 equal to 0 states we can I mean somehow use insights from quantum field theory to deal with this corresponding divergences and for positive L0 of course both are finite. So, you can use either the string field theory or the wall sheet picture. So, this is for the open strings now we also have closed strings. So, in the current problem we also have closed we have opened and closed strings and if you recall the discussion in the first lecture the external states are only closed strings, but the internal states and we open or closed. So, this means that we need closed string propagator also and it has a structure very similar to this it is proportional to L0 plus L0 bar inverse where these are the homomorphic and anti-holographic generators and this we can also express as integral 0 to infinity dt can the wall sheet picture gives you this and the string field theory picture directly gives you this and this is the connection between the two pictures. So, in principle we can deal with the divergences associated with closed strings also in the same way, but there is certain simplification for closed strings at least for the current problem that if you look at the L0 plus L0 for Eigen values they have the form k square plus m square where k is the momentum carried by the closed strings up to a proportionality constant. So, the difference is that the closed strings actually carry momentum which open strings did not and because closed strings carry momentum we can adjust k. So, we can analytically continue k k square plus m square positive at least for at low order this is always possible the order at which you will be working this is always possible at which means that you do not really need the power of string field theory to deal with closed string divergences you can just do it by analytic continuation because by analytic continuation you can make this integral convergent because once this is positive then of course, this integral is convergent and the wall sheet has no problem. So, you do not need string field theory please at low order and equivalent you have saying this is that we can integrate out can integrate out the closed strings. So, in the language of string field theory this will mean that any contribution to diagrams that has a closed string propagator we can redefine into the definition of the vertices interaction vertices ok. So, just closed strings they do not will not appear in the propagator. So, this means that effectively the we have external states as closed strings closed strings and internal states will be open strings. And we will use the notation where closed strings will be denoted by a wave line and open strings will be denoted by straight line ok. So, we will see that in all the diagrams that will be drawing these will be the external states and these will be the internal. Now, let us recall the discussion that we had on the first lecture. So, a leading order the only divergent contribution was the annulus partition function right that we dealt with. So, at first sub leading order so, we had several divergent contributions. So, first so, you always have e to the minus c over gs. So, let me write down the whole expression then exponential of annulus and we had a disk 2 point function of disk 1 point function ok. So, n minus 1 of these sorry n minus 2 of these and 1 of this ok. This is on sub leading because compared to the case where you have altogether n disks with one closed string partition of partition h this has ordered gs. Will there be some combinatorial factor like n minus 2? Because this you know yeah we will write down the combinatorial factor in a few minutes right. So, I will write down the expression for these diagrams I think then it will become clear. Thank you. Then you have e to the minus c over gs exponential of this and then sorry I should separate out the diagrams. So, there is a confusion first diagram the second one will be annulus 1 point function and then disk 1 point function ok. So, we will have n minus 1 of these and 1 of these and then there are 2 diagrams that I will draw together. Then you have the product of disk 1 point functions as n disk 1 point functions n of this and then you can have a disk with 2 holes ok or a torus with 1. These 2 I am drawing together we will just take the sum of this because both of these have Euler number of minus 1 ok. So, this is the leading contribution and these will give extra factors of gs because of this Euler number minus 1. So, these are all of the same order. So, to write a convenient expression for this we will introduce 2 or 3 quantities gs times f omega 1 omega 2 will be by definition the disk 2 point function divided by the product of disk 1 point function 2 disk 1 point functions and omega 1 omega 2 are just the energies ok. So, let me label. So, this is 1 and 2 here and this is 1 and this is 2. So, that is the definition of gs times f gs factors have taken out because this goes as 1 over gs whereas, this goes as 1 over gs square. So, the net ratio is of order gs then gs times g omega will be annulus 1 point function divided by disk 1 point function and then I will define gs times c as just the sum of these 2 and the point function is that all of these are divergent. So, f g c are all divergent. So, let me write the expressions for divergent part. So, f we will write as f finite plus f divergent and g also will write as g finite plus g divergent and c will not speak c is also divergent. This is perhaps a slightly tangential question, but is there an unambiguous answer for f finite or g finite or is it somewhat scheme dependent as in quantum field theory? No, we will first I mean once you define what if divergent is then this is finite this is unambiguous. So, you will just subtract of the divergent integral and then declare the rest as finite. I see, but that what we define as a divergent quantity is typically scheme dependent. Well, that is after we do the integral we will separate out. So, the point is this will be an integral f is an integral with a diver with an integral which diverges at certain ends right. We will take a particular quantity and subtract it. So, that the integral is finite then that becomes unambiguous right. We have to define it that way. Then all the ambiguity comes in defining the divergent part and we will see. So, the whole purpose of today's lecture is to see how to extract the divergent part right. We will try to make the divergent part into unambiguous finite pieces. So, that will be the main goal. Thank you. What is the difference between the two diagrams? Between the two diagrams that you drew. Oh this this is a disc with two holes ok. This is a torus with one hole right. So, this is this is not a hole this is a doughnut think of a doughnut and cut out a hole in that right. But as this is like having a disc with two holes, but both have on a number minus one that is why I have put them together ok. But this is how nobody has calculated this so far ok. So, we will not this will not be of too much importance today at least ok. So, the divergent parts I write them as integrals of this integral. In fact, we had seen before I had just given it as an example ok. So, basically what happens is that f is an integral and once you subtract of this part from the integral then the rest of the integral is finite. So, f finite that is unambiguously defined ok because you have just subtracted of the divergent pieces ok. And let me write this as a f plus b f ok where a f and b f are both divergent integrals ok and our goal will be to figure out what a f and b f are. Similarly, g the divergent part of g this is double integral again this is something we have seen before. Because these are this is divergent integral from v equal to 0 and x equal to 0 and this is divergent from the v equal to 0 and. So, again we will write this as a g plus b g ok where these again we do not know these are all both divergent integrals. Let me write the amplitude in terms of these quantities. Minutrile factor also become clear. So, let me call it a 1 a 1 because this is the first sublating order one stands for its first sublating order. So, if we look at these definitions ok this. So, you are trying to sum up sum these and if you look at these definitions you can easily convince yourself that this has the following structure is g s times the leading order contribution. The result is basically product of disks with this exponential of minus c by g s and exponential of the annulus partition function times. So, this comes from these diagrams and the point is that these two could be any or any pair out of the right that is why we have those sums and similarly here this could be any of those ends right that is why you have that sum. And now if we use this fact that if we write as a finite plus f divergent and f divergent and g divergent has this kind of structure. So, we can write this quantity as finite contribution plus minus 1 by 2 is 5 right. The finite includes a finite and g finite contribution which I am not worried about. There are these many f's because of this each pair gives an f n edges as you can see right c is only once then we have this energy dependent contribution. So, bg times sum over omega square and af times sum over omega g omega. But this quantity after I use the energy conservation the energy conservation will come at the end right we have seen the integral over the y will give energy conservation. So, this leading order result has is delta function. So, once you use this these two can be combined one term which is bg minus half right I just use the sum over omega j equal to 0 to convert this into sum over omega j square. In this af sum g can be less of g less than k. So, the point is you include the j equal to k term right in that case you can put a half and they just do free sum over j and k that for that term vanishes right. And the extra term that they are including is the j equal to k term that you have to subtract off that is how you get this. Sorry this is bf af has already gone there ok. So, now let me disturb the strategy. So, our goal will be then calculate these numbers right af and gg and bf and bg ok. As I said c will not calculate, but fortunately because we have n dependence here. So, the way this comparison goes with the matrix model is that matrix model result gives a result for all n. So, once you know the result for all n you can just take appropriate linear combinations to eliminate c. So, by comparison the matrix for comparison the matrix model we can actually figure out what af, ag and this combination is. And the important point that will become the significance will become clear later is that this combination is what appears in the expression for the S matrix element this bg minus half pf. We want to express the sum over Feynman diagrams. Once you do that then L0 less than 0 contributions automatically become finite right because you will get them as 1 over L0 instead of the integral. But there will be divergent contributions from the L0 equal to 0 states ok because 1 over L0 is infinity. But we have seen how to deal with this L0 equal to 0 states. There are the origin of the L0 equal to 0 states first there is a Bosonic 0 mode which is the translation mode. That mode we saw that you are instructed to integrate at the very end the integral over y right. So, the path integral should not include that right. The path integral which leaves the Feynman rules that should not include the y integral because y integral has to be done separately at the very end ok. So, we remove the L0 equal to 0 contribution from the propagator. So, the L0 equal to 0 contributions will come from the translation modes and also the two gross modes which are there which are there in the signal gauge right and the string perturbation theory gives in the result in the signal gauge. But you saw that those modes also should be removed because those are not physical right. We are treating the integrals differently by look integrating over the gauge invariant variables and dividing by the volume of the gauge group. So, all the L0 equal to 0 modes will have to be subtracted from the propagator. So, that gets rid of all the divergences because the divergences came from L0 less than 0 and L0 equal to 0 modes. But this is not the end of it. Phi if what you call was out of signal gauge mode right the C0 equal on the vacuum mode ok. This is not included in the wall sheet expressions because the wall sheet gives the result in the signal gauge. So, this has to be explicitly added into the propagator. And finally, this is something that I will try to explain later, but let me write this anyway. We have to account change in the Jacobian or corrections to the Jacobian psi b not to y theta to alpha change of variables. Namely we had seen for example, that we had psi b0 equal to k1 y right and we have calculated k1. But this is going to get corrected and because of these corrections the Jacobian of change of variable from psi b to y will have additional terms ok. And those additional terms will also contribute to the amplitudes and they have to be taken in back up because they are not accounted for in the wall sheet formulation. And the same thing is for theta to alpha k2 times alpha. These all also get corrected. Describe the analysis of the f ok. And then I will just write the result for g. Because once you do the analysis for f it will become at least part of the procedure will become clear. So, f this was the disc 2-point function. It is it will be more convenient to represent this as a 2-point function on the upper half plane ok. We have a map from the disc to upper half plane ok. For example, u equal to 1 plus i z over 1 minus i z. So, if this is the disc coordinate unit disc this is the upper half plane. And it is very convenient yourself that this as z takes value on the upper half plane u takes value on the disc unit disc ok. So, there is a map from disc to upper half plane. So, we will work in the upper half plane. So, in the upper half plane there is an SL2R invariance. So, we can fix three quantities. So, we will fix one of the closed string vortix operator at the point i ok. i you can see from this that that corresponds to u equal to 0 right. That is like placing the vortix operator one of the vortix operator at the center. And the other one I will place at i times y ok. And then let y vary from 0 to 1. If y goes out of this range then you can do a z to minus 1 over z transformation to bring it back. So, the reason that I have parameterized this by y is because this is a variable y that appears in the expression for f ok. So, this integral of course, came from the y sheet. And the y that you see here is exactly that y ok that I have introduced. And so, you see the divergence as y goes to 0 here corresponds to the divergence when this goes to the boundary ok. That is the that is the source of the divergence and that is the that is what you have to deal with. So, let me draw the Feynman diagrams. And again I have integrated out the closed string. So, there will be no closed string internal propagator. So, two Feynman diagrams that you have to deal with this I will call diagram A. And the other one is I will call diagram B. So, this is like a contactor no internal propagator this is with one open string internal propagator ok. And we will see that this is the one which is going to be responsible for that for divergence because this has the 1 over n 0 right where the divergence will come from. So, this 2 point vertex roughly is a 2 point function of one closed string and one open string ok. So, in the upper half plane we can think of this as a 2 point function of one closed string and one open string ok. Because this is one closed one open. And by S L 2 R invariance we can always choose the closed string at I and open string at 0 ok. Because now there are only few things to fix we can fix all of them. Unlike in the case of 2 closed strings where there are 4 things 4 variables. And so, we could fix 3 of them, but one of them had to be integrated. But to calculate this Feynman diagram we need more ok. And we need more because we see that you need off shell there are off shell states right this internal propagator is off shell ok. So, just by saying that we have vertex operator insert sense at I and 0 does not specify the amplitude fully ok. We have to specify little more and I will now explain what extra ingredient that you have to we need for calculating this Feynman diagram. So, for calculating off shell amplitude off shell interaction vertex ok. So, interaction vertex is external state on shell off shell ok. So, local coordinate basically some arbitrary complex coordinate system you can it is up to you to fix ok. But you have to say what coordinate system we are going to use around sorry around around this point and what coordinate system we are going to use around this point ok. Because in you have to insert the vertex off part in that coordinate system ok. And that makes a reference if the vertex operators are off shell ok. Because if the dimension 0 primer is then it does not matter ok. So, for on shell vertex operators which are dimension 0 primer is it does not matter what coordinate system you use right because they are controlling invariant ok. But for operators which are not dimension 0 primaries right which is typically the case for generic off shell states we have to specify which coordinate system you are using you are inserting the vertex operator. So, that is the meaning of local coordinate z is the global coordinate z the UHB coordinate typically around the function z will be a function of w some function that you have to specify. And in general this function can be different around different vertex operators ok. Around the one vertex operator you choose 1 w another vertex operator you choose another w. Once you have chosen this local coordinate ok and you have found the functional relationship between z and w then the vertex off part is inserted where if this notation is basically the conformal transform of v. So, in other words you do not insert the vertex off part directly in the z coordinate system you specify some local coordinate inside the vertex off part in the local coordinate system. But your correlation function final correlation function is being calculated in the z coordinate system right z is the global coordinate on the upper half plane. So, you have to convert from the w to z and this f dot v basically does that that v is inserted in a w coordinate system, but this tells you what it becomes in the z coordinate system. So, I will write down some examples of what we mean by f dot v this is the ghost field c we know has dimension minus 1 right it is a primary, but it has dimension minus 1. So, this is f prime of w inverse on c we can write down explicitly what this is and the f of w of course is z right. So, if you are trying to calculate the correlation function of a c if one of the off shell states was c then this is the vertex off part this is what we insert in the correlation function of the derivative of c right this is not a primary. So, you have to do more work, but this is the w derivative of f prime w inverse times c okay and you can work it out by chain because in principle this is not simple, but you can always work it out what this f dot a given operator is. So, the idea is that if we take for example this vertex right what suppose we are trying to calculate this 2.4 x this is a this 2.4 function. So, this will be given the upper half length 2.4 function this will be given by some f c f o or w prime f o dot okay we have to choose this functions f c and f o it is completely up to us to choose and then the 2.4 function that this off shell 2.4 function will be given by this. So, now you can ask that I mean if there is a so much choice in this choice in the how we choose f c and f o right that looks like the off shell amplitudes are completely up to us to define right it is not a unique object and indeed it is not a unique object. So, different choices of these functions give rise to different apparently different string field theories. But what one can show is that these different choices that gives rise to apparently different string field theories are all related by field definition that you take string field theory constructed with one particular set of choices of this and string field theory constructed with another set of choices of these functions. There is a change of fields a redefinition of fields that takes you from the first string field theory to the second string field theory. So, all physical results at the end should be independent of how you choose this even though the intermediate steps will depend on how you choose this. Maybe if I do some explicit calculation it will become a little clearer. Can you just choose w equal to z as a gauge and about this? Yeah, you can. But there is a better gauge than that otherwise. Is there a better gauge because otherwise we would not have discussed this no you just say w equal to z and. Well, we will choose almost w equal to z right. But I mean final result should not depend on whether you choose w equal to z or not. No, no, okay. Yeah, but you can certainly choose w equal to z. We are interested in so our goal is to compute this diagram right. This in this diagram the external states are closed strings these are all on shell right closed strings are not internal propagator. So, external closed strings are on shell. So, you do not really need a choice of this. So, because this is on shell whatever you take this to be right it is a invariant. Okay, so the result does not depend on what choice of fc you have made. So, we really only want this we have to make a choice of f4 and we will choose open string vortex operator equal to z by lambda sorry lambda z. Well, lambda is an arbitrary constant and if what I have told you is true then the final results better be independent of lambda right because this is a choice that you are making. And so it is always a good idea to use some parameters so that at the end you can check that the result is independent of that parameter to make sure that you have not done something wrong. Okay, so this lambda will solve that purpose for us. So, with this then f dot v f dot some v open of w if this is primary of dimension okay you do not need primary because this is just simple scaling. Okay, so this will be let me just make sure that I have the lambda to the minus. So, now with this we can now try to calculate this diagram right that diagram sorry. So, there is one straightforward we are doing it which is that you take closed string vortex operators are fixed. So, you take every possible open string state calculate this two-point function using this formula right because the closed string is inserted at i and the open string is inserted at 0 but with this scaling right so that you will become factor of lambda to the minus h for every open string state. So, you calculate this you calculate this take the product multiply by 1 over l 0 and sum right that is what we are doing it. However, the water identity is a conformal field theory gives you another way of doing it right and I will tell you what the result of that water identity analysis is. So, for this let us write this 1 over l 0 as integral 0 to infinity dt e to the minus t times l 0 and what I am going to give you is to say what happens for every t right you have to replace the propagator by this. So, what we insert here is not 1 over l 0 by e to the minus t l 0 right the t integral will be done at the end okay and I will give you an algorithm to calculate this. So, CFT water identities tell us the following. So, we take two copies of the upper half plane why two copies because why we have one upper half plane coming from here one upper half plane coming from here okay. So, two copies of upper half plane let us call this the z and z prime as the upper half plane and so correspondingly we have a local coordinates w. So, w is will be equal to lambda z and here you have w prime equal to lambda z prime okay these are the two diagrams corresponding to these two vertices. So, each so this has a closed string vortex of water one inserted here and just two inserted here okay and the open string vortex of water inserted here and they are being and you have this matrix element of e to the minus t l 0. So, what the CFT water identity tell us is after you sum over all intermediate states that corresponds to identify blue and w prime w w prime equal to minus q where q is equal t to the minus t. So, CFT water identity tell us that this diagram after you sum over all intermediate states but the propagator replaced by to the minus t l 0 okay is equivalent to calculating a correlation function on a remote surface where you identify w and w prime by this. Now, since w and z have this simple relation this can also be translated as lambda square z z prime as minus q or z prime equal to minus q over lambda square z. So, this is now a single upper half plane parameterized by the coordinate z we can also use z prime it does not matter take coordinate z and z prime is given in terms of z. So, z prime is not a new coordinate anymore. So, now look at the vortex of water of upper positions. So, one is at z equal to i and if you are using the z coordinate system it remains as z equal to i. Two was at z prime equal to i but that corresponds to let me write as z as q square over lambda. So, this z prime is i that corresponds to z equal to i q over lambda square. So, this means in the upper half plane we have one vortex of water at i and the vortex of water at i times q by lambda square. And if we recall this is what you had called y. So, we conclude that y has to be identified as q over lambda square which is equal to minus t over lambda square. So, we were free to pick lambda as we like. So, how do you guarantee that q over lambda squared is necessarily between 0 and 1 if we are. Well, you cannot ok I mean you are free to choose lambda as you like, but you should choose it so that things are singular. Right. Right, but ok you can choose it in any way you like let me complete this and it will become clear right what the rest of the integration region is right. But it is that that you have two singularities which have to subtract it out right. So, it is better to choose lambda sufficiently large so that that does not happen yeah. So, that is otherwise it will be like a singular field definition. The theories are still equivalent, but under a singular field definition. So, this means that as t goes from 0 to infinity y goes from 0 to. So, this basically says that this diagram A that you see this covers a region I had told you that the way string field theory gives rise to the usual once it amplitudes is that each diagram covers part of the integration region. And by this correspondence we see that this diagram A corresponds to the integration region 0 less than y less than 1 over lambda square and hence the rest of the region has to be covered by the diagram B. So, this must cover 1 over lambda square less than y less than. See this will be the definition of the of the of these partings. Because this is the sense in which I said that a string field theory is designed so that it reproduces a once it amplitudes. So, the general procedure is that you calculate all Feynman diagrams and you see how much of the modular space you have gotten and whatever you have not got you declare as a new contact diagram there and you do it at every stage. And here you see that if you have taken this lambda in a wrong way right. So, that y extends beyond 1 you basically have to subtract of that. So, this will have a negative contribution right which have to subtract off. So, that should be the definition of these partings easier to take lambda to be sufficiently large so that that does not happen. So, now we are in a position to do the analysis. So, let us take the original integral right let me I will explain how this is done. So, you have this divergent part half integral d y y to the minus 2 1 plus 2 omega 1 omega 2 y. This I am not going to split into 2 parts integral 0 to 1 over lambda square d y y is to the minus 2 plus. So, second integral is of course, is e this will be half. So, this gives you half times 1 over lambda square minus 1 plus 2 omega 1 omega 2 log ok this is an ordinary integral. So, no problem with this right this is the divergent part is here you can see the y go y equal to 0 region is here. So, here you have to change variable from y to t. So, here you change variable from y to t and that relation was y y is to the minus t over lambda square. So, if you make this change of variables you find that this is half integral 0 to infinity dt lambda square e to the t plus 2 omega 1 omega 2. Now, you can see that this is e to the power h with h equal to minus 1. So, this corresponds to L 0 equal to minus 1 state ok because it is like e to the t minus t L 0 that is what we are comparing right. So, this corresponds to L 0 equal to minus 1 state ok. So, you should replace this by half lambda square times 1 over minus 1 1 over L 0 right that is this contribution. This one is L 0 equal to 0 state, but L 0 equal to 0 states you are supposed to remove this contribution means that this is a contribution L 0 equal to 0 state in the propagator, but you are supposed to remove them. So, just drop this. So, you can add this to this right the 1 over 2 lambda square just cancels between this minus half lambda square and here. So, total minus half plus the result that you get from this Feynman diagram are there questions? Now, we are not done yet we have to do one more thing which is that in this diagram now you have to take the contribution of the phi propagator right because you say we have to remove the 0 modes right, but you have to add back the contribution from the phi propagator right which is not there in the original worksheet expression. So, you have to calculate a diagram like this. Now, if you recall the action had phi square plus phi is a real field. So, from this you follows that the phi propagator ok if it has half phi square you have been 1 right that is the for real field that is the standard convention right because it is phi square it is phi propagator is half and then you have to calculate these 2 point functions right, but you know how to calculate this right the 2 point function. So, this is what x. So, we basically have this half of this times. So, this is 1 times. So, this is half of a upper up plane 2 point function of the close to in vertex operator i and f dot phi if you recall the phi was the state it was multiplying c 0 on the vacuum which is the phi vertex operator is the derivative of c this is the first vertex operator the second vertex operator. But since this has conformal at 0 this is the same as and you can show that this correlation function actually is 0 ok that is by some symmetry of the close correlation function. So, because in this particular example the phi exchange just vanishes ok, but this is not generally true I mean generally we will get contribution from phi exchange. In fact, even if you had taken a different look coordinate system in instead of this if you had taken w equal to lambda z over 1 plus a z for example, right then this will no longer vanish. This will no longer vanish, but this will also be different this analysis is also be different right at the end this dependence on a just cancels out. Anyway so, because it is 0 this is the total f we get f divergent omega 1 omega 2 is minus half and now if you use the fact we have we have defined this as a f plus b f omega 1 omega 2 ok we see that f is minus half and b f is log lambda square are there questions yes. I may be losing my mind, but it looks like the first integral over here has a lambda squared and the second one has a 1 over lambda squared and I am I do not see why they cancel and you only get this log lambda squared term. Oh I think I made a mistake somewhere this is actually lambda squared. Okay cool yeah yeah thank you because it is a dy over y square right so it is a 1 over y and the lower limit is 1 over lambda square ok. Now I will not have time to describe the full calculation of g ok, but let me just say so g divergent gets contribution first the Feynman diagrams ok that analysis basically is the same as what I described except that it is slightly more complicated because there are more diagrams and more complicated diagrams and then the phi exchange ok. Once you have identified the Feynman diagrams then every open string propagator you can possibly replace by phi right so that you have to calculate separately, but the calculation is exactly this that you have to calculate the phi propagator is half and then you have to calculate the appropriate vertices and put them together, but there are two more contributions that come from this Jacobian and that is the last thing I am going to say. So recall that we had this relation psi b not equal to q1 y and this was based on the study of how the psi b couples to a bunch of closed strings right y of course always couples as it to the i omega y so that there is a confusion, but psi b not how psi b not come couples to a bunch of closed strings and the point is that once you have some additional closed string background that changes how psi b not couples changes and as a result the relationship between psi b not and y also changes gets modified at higher order ok. So typically at the next order it takes the following form. So this f is some known function and the c of c omega is the massless closed string field the Fourier component of the massless closed string field. So basically this says that when you write the make a change of variable from psi to y you get a Jacobian that gets an additional factor of 1 plus gs integral d omega which you can exponentiate and write an exponential of gs to this order. So it is as if you have gotten an additional term in the action which is linear in c, c is the closed string field and if you have an additional term that is linear in c then you will get an additional one point function additional contribution to one point function right. So besides the annulus diagram that we studied this is an additional this will give an additional contribution to the one point function that is not captured by the wall sheet Feynman diagrams right because wall sheet does not know about the fact that you are making change of variable from psi b to y to carry out the integration over y and a similar effect exists and again there are connections of a similar kind. So these Jacobians have to be taken into account ok and at the end of it so you have to sum over all of this. So the Feynman diagram phi exchange then the one point function of c that you will get from this Jacobian and the one point function of c that you will get from this Jacobian. So after you add all that the Jacobians so you get g divergence omega as omega square log lambda square by 4 but this is supposed to be equal to a g plus b g omega square ok. So by comparing this we see that a g is 0 and b g is half omega square very half log lambda square by 4. So at the end of all this you have a f b f a g and b g c as I said has not been calculated ok because that is a more complicated calculation. Now you see that f and a g are independent of lambda ok which is what it should be right they appear in the S matrix formula for the S matrix b g and b f are not independent of lambda but if you recall the S matrix involved not b g and b f separately but only in a particular combination that means that was we had a b g minus b f by 2 right this was what multiplying omega square sum over omega j square and you can easily convince yourself that if you take this combination right this half log lambda square gets cancelled and you get this combination as minus half log 4 which is lambda independent. So this lambda independence of the final answer is a consistency check that you are doing things correctly ok and in fact you can introduce many more parameters because every time you have a new cortex right you have a lot of freedom of choosing the local coordinates. So you can try to introduce as many parameters as you like and make sure that at the end everything cancels right because that is a check. So there is a balance that if the more parameters you introduce the more complicated the calculation is right but the advantage is that then you can at the end you can check if the things all cancel the extra dependence ok. And these results actually agree with the matrix model results the matrix model results are evaluated numerically ok. So the comparison is numerical but it is quite a good accuracy this minus half log 4 0 and the half or minus half right they agree with what you get from the matrix model calculation which is the dual calculation ok. So I think I will stop here I think there is a time. Is there also this third contribution which you mentioned it has not been which contribution I mean this torus yeah yeah so that has not been calculated and matrix model gives an answer I mean it is in fact says that this should be 0 the sum of the 2 but that has not been calculated. But is there some essential difficulty or is it just like. So it is like a genus 2 calculation right I mean it is a square root of a genus 2 calculation so that that is the difficulty but I do not think that is any essential difficulty in that. So we can leave it as a homework for students here. Yeah and the other thing I should have said that I listed those 7 cases where the annulus partition function has been checked right this has been checked only in 2 and half examples. So there is a lot to do here. Okay these are normal questions let us take a shock for the great set of lectures.