 So, we had this discussion in the last class of our three-level advertiser string theory. The last class, again, we computed four-point functions. We got this next formula that took pressures of gamma functions. And we discussed how these four-point functions have singularities corresponding to extreme particles. We discussed how unity allowed us to determine the pre-factor that appeared behind every particular vertex operator in terms of its unknown determinant of the sphere. And we discussed how this pre-factor is the same as that, the re-interpretation of the pre-factor is the same as that you would get by the state operator. We did work that out in the last class of our three-level advertiser string theory. Yes, yes, yes. Good. So, we also talked about the high energy behavior of stream scattering after this. And one of the things that we found was that if you did high-energy fixed angle scattering, you know, they've got some four-point function, something here, and then they come out. So, you get k1, k2, k3, k4. k1, k2, k3, k4, all they can be large in the sense of high energy. So, high energy, high spectrum, they come in and go out of that some fixed angle. Now, when this happened, we found that the amplitude went like e to the power minus, alpha prime s log s plus the same expressions with t. Okay? The point to know this here is that this amplitude is exponentially smaller than high energy, even more than exponentially smaller than high energy. Okay. Now, this gets contrast with the behavior of three-level scattering of the gravator. We did this calculation for calculables, but those more work, you could do the same kind of calculation for the scattering of four-valet terms. Okay? And you get the same kind of result. Okay? So, the point is the following. If you did a four-point scattering to your calculation for gravitons in Einstein's gravity, the result instead of decaying at high energies would blow up at high energies. Why is that? It's just because the effective gravitational coupling constant has powers of the energy in the event. You see, the gravity theory is a two-dimensional theory. Every interaction, because of two-dimensional theory. Okay? So, every interaction per second, that takes you to powers of k. So, in this net time, I mean at four powers of k, and the whole thing will blow up. I mean, it will become very large at high energies. So, this result is very different from the result that you get in gravity. Now, this is connected to the finite and the substrate theory. For the following reason, you see, the value of the divergence is in doing the relative value of the divergence. The divergence, of course, comes from the fact that very high energy gravity and running in loops give you very large concentrations. However, the idea is that any perfect factor involving very large energies is very small. The kind of divergences you get from high energies running in loops is all of them, from finite energies. However, the extra pressures we get from amplitudes are explanation. Of course, you've got some process in which you've got some loop running around here, and you have very high energies in these loop momentum. Okay? Then, this is proportional to some apparent divergence, but you have a potential to indicate the vertex factors. Okay? And these affected vertex factors, as we've just seen, have become very small. So, that's the key point. It's just that the scattering amplitudes die off at very high energies. Therefore, if we think, you know, I mean, that's correct. In fact, if we think, you get high energies at stake. Okay? And they have an exponent. Do you think there's a statement that the average public constant would expect it to run in a certain way? Yeah. No, it doesn't change that. You see, the exponential comes apart. It would not be there if it just had radotons, radotons, scattered radotons. And that part still nightly grows. It's just a classical definition, so that's what grows according to the classical graph. Okay? What cancels it? What cancels it? It's everything else from these points. You see, it's not just the exchange you were grabbing on, you have to look at it. You have to look at the exchange of all string states. And that whole thing is exponential. Now, said this way, it sounds like a million. However, there's another way to say it, which makes it sound less dramatic. You see? And the other way to say it is that, well, look, what is the size of a string state? Okay? So any string state has size approximately, I would say. Okay? Now, what? When you, when you scatter, you when you, okay? So suppose you had an electron and was sent to scatter again some big fluffy cloud of positive charge. How does the scattering look? But the scattering amplitude is proportional in the ball approximation. Suppose you set it in the amplitude K by 1. Right? If the ball approximates what's scattering amplitude is proportional to the Fourier mode, to the Fourier, the Fourier transform of the source function and momentum K minus K prime. Okay? Now, if K minus K prime is very large, compared to the length scale in the charge distribution, the Fourier transform is very small. Okay? Well, what I think is the power of my inside KX times minus X squared by 2, let's say 8, is where? So that is the size. What does that equal to? So we get something that's proportional to the power minus K squared by 8 squared by 2. So K is very small. This is an exponential small angle. It's the scale. But K is very large. This is an exponentially small angle. If your momentum is very small, if the wavelength of your particle is very small, compared to the size of the object you're scattering against. This A, this is A divided by 2. There's a smallness in the length, sometimes sort of 4 in fact. So if you scatter at very high energies, very small wavelengths, you scatter off something that's very big. Some scattering always makes more. As you can see, it's probably a Fourier transform. But what's going on here is that let's go into a frame in which we can do a string scale together. Let's go into a frame in which one of them is a dress and the other one's coming hard. So now that's a dress. It's like this guy. In some object, a string scale size, in which these things could lie. But if you go to energies much higher than a string scale, so that the wavelength associated with the projectile is much smaller than a string scale, then we're exactly in this situation. The guy that's colliding, he's colliding with a big, big, fluffy cloud of size strings scale, but it's not much smaller than a length. And we get exponential suppression, basically. This one. That's the basic thing that's going on. What's going on? You see, the limit where gravity dominates over there, is that you can think of the string as a point. It's a smaller size state. On the other hand, at very high energies, this thing where you know, excited all the time just for string states equally. It's very misleading to think of the string as a point. In fact, it's a big, fluffy cloud. In which all the excitations up to the energy that's available to you are equally being excited. So, clearly, it would be a bad approximation to use just the gravity approximation. The answer, but we've got an exact formula. We know what it does. And this shouldn't be so much better description of what's going on. Then grab it. You would expect that gravity dominates over all of the literations. I mean, it's the energy that we see when we're understanding how it works. Yes. But here, what is happening? We understand that there are new interactions which have been added, which together need to stop. Exactly. Exactly. The new interactions are all the other excited string states. And there are also, like, three degree separate interactions that I would not add. The net sum of all of it. You know to separate the interactions. The net effect, the net sum of all of this is very highly suppressed. You see, what we've done is we've taken the full 4-point function. Full 4-point can be answered here. Now, separate it in the bad part that comes from gravity on exchange. The bad part that comes from the first state of the state on exchange. Each of those will be summed out. We sum it all up and then you get this. You get this again. More than that, more than that, it sounds like a miracle that this happens. But if you think of the object, this picture that makes it intuitive. You see, we've got a kind of failure in that way. You see, it's always true that when the gravity, this dominates over all other interactions, the stream needs to be smaller. I didn't understand the question. When gravity dominates under all other interactions, say we are working, if we want to work in an approximation where I just have that, I don't want to add any other interactions. Okay, good. Then it is necessary that when stringing and in that approximation we small compact all other things. That's fine. That's fine. You see, why is that true? It's basically if you want to ignore keep only gravity on it. You know everything else. You want to be in an energy regime where producing other particles in intermediate states is suppressed by the energy. That would be the case when your energy is smaller than the lightest other state that exists. The lightest other state that exists has a mass of a value of 1 on the square root of alpha. The longer your energy is that much smaller than the one on the square root of alpha, gravity and the other masses interactions are. And how do you know that? Okay. This just will tell you if you want gravity to be dominant you have to be an energy smaller than the other lines. Other questions, comments? Yes. It's not stupid, right? It's not extreme constant. We do tree levels actually. Basically, classical string theory. You know, anything that's tree level is essentially classical. There are no powers of age constant. I mean, there are many things about this approximation but the way I wrote it was not very interesting, sir. But the appropriate version of that nothing would apply to us. Once we started doing loops we don't do classical. Really? And so the loop collection is over in, of course, we were suffering like this way. It's a surprise by G string. It's a surprise by G string. But the question, the kind of question that makes that variation tricky to be there is that formal suppression by G string in artifact is the number that multiplies applies at infinity. And we get divergence in field rate. And that often causes the naive perturbation expansion to be available. Okay. So we'll see that's not the case in string theory. There is, you know, everything is completely well-behaved, completely mind-bending. This is our first indication of that. Had the sign been the other one it would basically have been impossible to get finite motivation. Never had to go beyond tree level string theory to get these, correct. So this kind of energy separation. Yes. So what is, I mean, in the sense that we would expect that the gravity would get suffered in some way. Okay. There is a distinction of what is classical and quantum in gravity and what is classical and quantum in strings. Right. How so? So how do you get one bit, I think, how come you know we just did tree level in that unit? Yeah. Yeah. All we're doing is really, we've not used loops at all. One of what we're finding is that the exchange of the granite is cancelled by the exchange of all of them. Look, let's see, let's see how that works in the final formulas. I'm going to ask somebody can we get the formula to give you the product for gamma? It was gamma of A, gamma of B, gamma of C divided by gamma of 1 minus A, 1 minus B, 1 minus C. Okay. So it's A, B, gamma of C over gamma of 1 minus A, gamma of 1 minus B, gamma of 1 minus C. Okay. Now, somebody had a deal with what the A and B see. A was minus gamma of 1, 1 minus S by 4. S by 4. And then, then G to B. Okay. A, B, and C are these. It's good. Now, how do we see that gravity scattering grows with the energy from this form? Okay. To do that, what we have to do is to work with the graviton pole. We did this a lot. We saw that when we when we sat with the graviton pole, we got contributions that were proportionally remember, K to the 4. We got contributions that were proportionally K to the 4, which was like T squared or something. T plus 1 is got T. It's a polynomial of T. T plus constant. Right. By graviton, you mean all the massless gas. All the massless gas. Graviton, B, and U did it all. Just by sitting on the pole, we can't distinguish them. But if we want, yeah, we would have to do more K for the experiments. You know, keep track of polarization dependence. But, at this level, that's almost the same. Okay. Now, let's look at the structure of the poles more carefully at higher and higher excitement. Okay. Just to see how the amplitudes grow with energy and at different excitement. Fine. So let's go instead of the pole. Remember, we sat at A equals 0 with the graviton. Equal to 1. Minus 1. What's the gravity? Okay. So let's go instead of A equals minus N. A equals minus N is if we can understand what the residue of this pole is. Okay. So, remember, we had A plus B plus C is equal to 1. Okay. We're setting A is equal to minus N. Let's use B as our variable. So we get C is equal to 1 plus N minus 1. Okay. So we have A equals minus N. That means some sort of pole here with something vectorial which I will be talking about. Okay. Now, we have to put in all these other things. So, this thing is 1 plus N vectorial. I mean, 1 plus N gamma is N vectorial. This thing has a pole of quantum factors. Our vector is ignominal. These guys are different. They're different. So this is gamma of B times gamma of 1 plus N minus B over gamma of 1 minus B times gamma of 1 minus 1 minus B minus B minus B. Okay. So the rest of the pole is this ratio of gamma. So we will put this ratio of gamma. We do that because probably gamma functions that tells us gamma of B is B minus 1 times gamma of B minus 1 and so on go all the way down to B minus N. So that captures that with the product B minus 2 up to B minus N. So which is the larger generally this gets the larger one. So once again we get we get factors up to 1 minus B the factor we get is N minus B. All the same factors occurring with the minus the other way around. So how do we get it? Again B minus 1 squared B minus 2 squared up to B minus N. So that B is much larger than 1, 2, 3, n. This is just B to the power of N. Remember there is something like T. Okay? There are orders in the mass level in exchange. You get higher and higher powers by reaching the vertices. I think well we should set ourselves up for a huge blow. Gravity goes up look at the high energy and end up the exact opposite. It's an exponential all we are seeing if we go to very high energy is that if you take you know if you take something like E to the power minus X squared and you take the X factor you look at X squared X to the 4 X to the 6 and you say oh my this must mean that when anything is classical the exchange of power a higher mass level gives you higher power so it's not true that they have an exchange of power the other things are certainly important in fact more important in this popular way but they all sum up to be something like it was meant to be a throwaway intuitive statement that will be justified by new calculations since the vertex contribution is one inch of what the groups factor is the effect you see something rather than a low it's very large has to connect to the rest of the factor that connection factor will be explanation so it will lead to very small contributions of that in now I mean you know such arguments are many ways of going now so you should trust such arguments just give the worst most and visualizing something once they convert the formulas so this is not a sphere for doing new calculations I want just have somewhere thinking this is a very unusual thing if you think that's a field theory because you just put a lot of non-denominational intentions yes you're very honest that's why string theory is a special you know it's not to have string theory in some way you see as we will see we will see as we go that string theory is very tightly constructed you know for instance if you take a field theory without gravity you can often change it into you but not changing much okay for instance if you take a theory and you have one cutoff you change the cutoff to something else you take dimensional regularizations something else is it one of the things that we discussed when we were discussing the consistency of string theory amplitude you remember was that beyond as the invariance of string theory required that you know was true only up to total derivatives in modularized space do you remember that if you look at the expression of string amplitude there was an integral of modularized space and then there was you got some total there now as we will see the integral of a modularized space corresponds to the loop detector of a of a we'll see that in this let we will discuss the loop detector well you will see that okay so if changing the cutoff is like adding some factor to the loop detector if you change it properly every time something throws very large with K you can put a theta function there put an e to the power minus K squared okay so changing how you do that integral is the cutoff of field theory if you want to do anything other than not string theory asks you to do you will not decouple the RST that is if you would just take the modularized space and decouple it just cut it off so oh I want to integrate over I because you will get some other terms to this to this BRS so now it says won't decouple anymore theory won't be getting sense it won't be unitary okay so as you will see more and more as it goes on string theory is not any over the quantum field theory in particular then certainly nobody has understood how to take the string theory effected at action and change it a little bit as we get consistent theory it's not that kind of theory it's very tightly constrained from its internal structure so from a point of view it looks like America from that point of view it doesn't make but it happens because the field theory part of use an effective disruption of value in certain regions but but when you are high energies that's a very bad way to think it's so many difficult whatever I said in detail but you may rather be confused in three points rather than scattering and we got an answer that had various powers of K I just wanted to remark that it's only the lowest the lowest order in the powers of K that comes from the Einstein approach the powers of K come from derivative corrections to the granular granularity direction as we will see as we go along these higher order corrections are very different it's a remark I don't think I made okay any other questions about classical tree-level stream scattering before we get terminated a little I want to study the simplest non-tree-level kind of I'm going to use the big M string here and these will be amplitudes on the tops and it's in Gs1 okay so the first thing we need to do is to study the torus just as a geometrical manifold to the extent we need it okay so first what is a torus you know in coordinates in which the metric has been made is why equivalence to something that's flat okay so one thing of the torus is a metric as well an electric Gs1 www okay okay with the identifications w is equal to w plus 1 that's the same to the y w is equal to w plus 2 pi and w is equal to w plus 2 pi that okay so think of the complex w plane so complex w plane and all this complex w plane so the first one is in the real axis direction and the second one is in some complex direction and this factor is the complex number in the final in the fundamental regions each of which is related to parallelogram by this identity okay easy to cure anything anywhere with complex fate if you put it inside parallelogram by using these identifications it protects okay so uh this is what the problem is if this the region of the complex lets qualify the identifications this side and this side to identify and this side and this side okay so if you want to think of it pictorially while this is in general an arbitrary complex number this is in general an arbitrary complex number let's think first of all a angular torus okay so identify this side and this side because this becomes a circle thanks a lot and then identify this end and this end identify this end end of the cylinder that's the closest one okay so we've got the torus that's your region it is here now is suppose we've got uh suppose we've got a chloride with different values of this this identification parameter okay are these different chloride in fact while and if you're over some in equivalent objects or equivalent here they conform with it they conform with it okay get to know okay the answer to that in general know but in order to see that let's try to do a calculation this calculation will be useful for us taking on this one okay so suppose we've got two different tori okay one which is this object another one which is w' w' less than 5 and w' with the w' less than 5 and another way to do is to make a coordinate change in w' uh it was these two tori are different in this the metric of ds squared is equal to d w' dw' wow so now what we do is to make a coordinate change in w' in order that in the new coordinates in the new coordinates I have the same identifications as I did for w okay and then we check when two metrics in the coordinate vector is this correct so what do I do so I'm going to define w' is equal to aw plus b uh w bar and these things choose and then choose a and b such that the identifications of w and w are the prime points okay maybe I should and maybe I should stop and get another there there just not to confuse these okay and I've got a demand that when w' goes to w' plus one implies z goes to z plus one and w' goes to w' plus two five seconds plus one is always two five okay and w' goes to w' plus two five tau implies z goes to z plus two five uh tau prime so z goes to z plus two five two five tau so we're not trying to change the identification model I have this the stores but it won't change okay so let's see what I get so these numbers actually that's a bar of yes okay so this gives us the equation that a plus b is equal to because if I say one and one here I should get one okay so none so for the other guy we have the equation that's tau we also have a plus b equals one so let's write this as we find that a is equal to one over tau minus tau bar in two minus I imposed these conditions so that I changed the periodicities appropriately okay and b is equal to one minus a is equal to tau minus tau bar tau minus tau prime okay let's check if we chose tau equals tau prime we should get equals that works and we should get a equals one can be written as minus tau bar times z plus minus tau prime times z bar over now let's take the metric dw prime dw prime bar and rewrite it in terms of z in terms of z1 what do we get well we get z bar but we also get terms that are dz squared and dz bar squared let's write down what the dz squared and dz bar squared is so the term that's dz squared comes from taking z here and z in the other one that comes from there okay so so that so that would give us tau prime minus tau bar over tau bar minus tau z squared this could be dz bar squared metric does not change when we change the identifications when we change the identifications so as so as to change tau bar right once we change identifications the metric in this new coordinate is not conformally related to this guy these two spaces now in terms of z z and w we have the same range of coordinates but different metrics are not just the metrics are not just different they're not it's possible to change the identification range by working transformation that change changes the metric that change changes the metric by by a factor that's not in general a wide transformation so these two different objects are different metaphors as far as up to wider anti-film offices are concerned is this clear so we've got this one complex parameter set of tau right which are distinct complex different conformally equivalent which are conformally equivalent metaphors and though I'm not going to try to demonstrate it to you so this is important because that tells us that for the for the case of the tau is unlike the case of the sphere we now have more uniformity there's not one sphere up to conformant transformations but there's one complex parameter set of tau right I'm not going to try to demonstrate to you but okay you can probably demonstrate there's nothing else where you can before we start with a little bit thank you and we ask what are the modularity now we know that the equations of motion that the modularity is literate actually the number of modularity come from number of zero modes the number of zero modes of the field B in this quantum system number of the zero modes of field B has a periodicity functions so we're going to list all the analytic functions that respect the periodicity of the tau there's only one analytic function that respects periodicity of the tau namely the constant so the two complexes of volume of the constant are the two modularity these are the two zero modes of the operator this is the two zero modes of the operator so this is another way of speaking of the modularity if you think of it as zero mode of field we're just inurating the zero mode of field B okay fine just so that we have in our hands for when we need it in a moment let's I'm going to need to compute what dz squared is in this situation but dz squared is at first order at first order in the variation of this in the variation of this complex okay so at first order we get delta of factor of i times imaginary part of the tau okay if you compute the the change in the dz squared of the metric when you do this what the change the change in the metric is you see that that's that this is the place where you get the first order of variation the nature where that's that tau tau minus tau minus tau bar is twice i times actually okay that's all I want to say about those I don't think I'm going to say about the tau I'm going to say about the moduli space oh no sorry let me say one more thing he should say about the moduli space oh he might think that well good so now that I know I understand the moduli space of the tau right is written by one complex i to tau however that is wrong that is not a complete description of the moduli space of the tau okay how do you see why is it wrong let me say I've got powers which have identification on here I'm going to suggest another i another set of identifications that will give me actually the same physical space I'm trying to say that there are discrete identifications in the space of tau so that the discrete equivalence is in the space of tau see you identify w goes to w plus 2 pi and w goes to w plus 2 pi tau that is the same thing as I did doing some other identifications any suggestions for what that's good okay good so the first thing is that tau goes to minus tau it's a selection that's good but anything else I guess we have to have a space well suppose w is equal to w plus okay so pi is equal to w what say say so if this is tau then it's very good this guy you see because this guy at the same point as this guy same point as this guy that implies the same point as this guy okay you can get to this guy by shifting here so I think suppose that w is equal to w plus 2 pi w is equal to w plus 2 pi you can add any type of this equation plus this equation and you'll find w is equal to w plus 2 pi and w is equal to w plus 2 pi equal to w plus n these two are completely further than each other because to get back to this system you can take this equation and subtract the entire this is the step so this is a statement that we've identified think modular now let's just start with one integer worth of a meter okay tau goes down plus n that's not defined in this space okay what will become the basic for instance if you now go to tau plus 1 think of this is the basic cycle of that tolerance these two are the basic cycles okay and this is the complicated composite of this guy and this guy okay so what your basic cycles are the choice that's your choice of description the space in there has to change simply the same space okay but it's not obvious from this expression of everything okay from this expression of the metric no it's not okay so that's the definition you could you can take all of these different types and write them in a space where the identifications are to pi this way and to pi this way okay and then you could ask well and now suppose I when you do that you can complicated metrics as you see okay now you can ask how I did two different spaces with one with tau and the other with tau plus n all these two things good and you can show it's a coordinate change that relates these two each other without changing the without in this case without any life factor and without changing the identifications what are the changes then what are the changes of the form so suppose you take sigma 1 sigma 2 it's a coordinate change of the form A B C D where A B C D are all integers and the determinant of A B C D is good in this particular case it's a special example of this the special example being 1 1 1 okay this particular coordinate change I am not going to work it out yeah but this particular coordinate change will change the net its manifestancies are all integers but it's not changing the identifications sigma 1 to sigma plus 2 pi times n we just we're finding with integer coefficients it doesn't change the identifications but the special thing what it will do is take a take the metric with one particular value of tau and re-value there's another particular value of tau and this is in the example that we talked about here 1 n 0 1 all coefficients determinant equals 1 this will be tau goes stateless we just saw that we just argued that it was not related to that metric by varying you know like but this is an example that you know like there will be some coordinate changes that we did say no this is not even while in the example we talked about here there's no while factor yes that is what I'm saying is that you know like what does it do for the previous example what was the previous example that you know like that the metrics are not yes good good you see what will go on there will be that it's possible to find a coordinate change change you see you see this was the coordinate change that we tried to do was the bottleneck it tried to keep the fundamental cycles of the two tosses missing you see we wanted to work in coordinates when we took z goes to z plus tau we had w goes to w plus tau in those coordinates it doesn't work however you can try to find z goes to z plus tau it's not w it goes to w plus tau prime but w goes to w plus tau prime to z coordinate change which changes which identifications you see in this analysis firstly because we fixed what coordinate change but these other coordinate changes are discrete so infinitely what we've done here is completely correct but it misses the possibility some big discrete identifications by more imaginative coordinate change you understand what I mean in reality what's happening we demanded we demanded more equivalence we demanded equivalence plus that the two cycles of the taurine match to each other that's too strong a demand what you should demand is a sum of coordinate system in which in which the two things match how about that is this correct if you clear that there's how much actually what will it affect oh yes because infinitely what we did was completely affected the only other ambiguity we have is it suppose we make smaller change then you can't change this cycle this cycle then it's that good if you make a big change menu so infinitely if two taurine are infinitely related then the two cycles that must match to each other on what I know this argument is certainly correct infinitely but it leaves open the possibilities that there are discrete identifications is this correct so we certainly got two dimensional modulations that's for the earlier argument show this okay already this this two dimensional modulations is not the whole problem let's play just because we didn't try hard enough you know with discrete of the UD strategy but we're doing that it's more systematic we could have approached a little bit more is this clear still unhappy but can we do this this kind of identification of 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x 2x Now, if there are some some way are you that it is that there is no other integration, except that you know it is not very that these two are not this is problem this is right this is this is I accept that this is these two are the same species, what I am not convinced about is that the previous argument that they are not very difficult, because I thought that you know like previously that you know this was a canonical coordinate transmission and we see that they are not very important so they are not very important, but here what is happening is that you know there will be some of the coordinate applications which may do in this argument and it is not I mean firstly I want to convince you that there are two the two different towers are related infinitesimally are very near each other, then the only possible way in which it could work is the kind of coordinate transformation you tried, because what are the possibilities? many of you know how would we get back to this, but many of you know how you can systematically approach this, one way of systematically approaching this is taking this different dog and switching around in the same space, so space of identification to pie to pie, that means you have mentioned that at that time, now your identification is going on with that, now you have said to food around, now you ask what kind of coordinate changes you make, then respect these ideas and that basically is the identification of this sort, like this do that identification, remember these are in teachers, so if you can't make something that infinitesimally different even in that stratosphere, you use a full class of things that you thought of with different proportions, in a systematic way, is this clear? it's sort of clear that from that point of view that this thing would be parametrized by these matrices to be used. and to the possibility that there would be some other variations also that is not yet exhausted? we've not yet exhausted, not from what I've said, from what I've just told you I don't think that would exhaust everything when you did the exercise, but getting to the full exhaustion, that would be more or less proved, certainly if this exercise is done all of this you spend 10 minutes to do it, there's no deep end to it ok, now I'm going to ask you, up to why transformation seems to be the thing of any other okay, let's talk to the rectangular course, this is 2 pi and this is 2 pi i times some a, so tau is i times a okay, now I'm claiming that if you took tau to minus 1 by tau suppose you have another rectangular course in which the tau was equal to i times a now let's look at a bonus in which tau is equal to minus i divided by a minus of 1 by i a, which is equal to i by a the minus sign is the magnetic finish so I'm claiming that these two tau are i and y in this space here, let me write this in sigma 1 sigma 2 so the metric here is d sigma 1 squared plus d sigma 2 squared this is an example but the identification radius of sigma 1 is equal to sigma 1 plus 2 pi and sigma 2 is equal to sigma 2 plus 2 pi i 1 by a squared, so let me write this as sigma 2 by the whole x-square by sigma 1 by 8 whole x-square transmission over all factors of matter d sigma 1 tilde squared plus d sigma 2 tilde squared where sigma 1 tilde is equal to sigma 1 by 8 and so as identification, sigma 1 tilde plus 2 pi so sigma 1 tilde is sigma 1 tilde is sigma 1 by 8 sigma 1 by 8 sigma sigma 2 so we have sigma 1 tilde is equal to sigma 1 tilde plus 2 pi a and sigma 2 tilde is equal to sigma 2 tilde plus 2 pi I switch my labels I switch what I call 1 and 2 and I got exactly this tau because this tau has had sigma 1 sigma 1 plus 2 pi and sigma 2 sigma 2 plus 2 pi a you see that the map tau is equal to tau goes to tau minus 1 by tau so here is very simple, I am going to send you a message suppose you had a long tau like this and then you rescale such that this long edge becomes 2 pi this edge then becomes very small that's the a goes to 1 by 8 this is the x-y clip you change what you regard as the x-axis and the y-axis and then rescale everything so the x-axis has identification 1 so those are our connections so if the x-axis was originally the y-axis was originally very long once you flip and contract the new y-axis will be short like this a goes to 1 so this operation of tau goes to tau goes to 1 I am going to leave you to check that more generally this will be more generally than 4 times the tau for any time the tau goes to minus 1 by tau is always important so this identification here these two identifications are related this one is sometimes called t this tau goes to tau plus 1 sometimes in the name t well this tau goes to minus 1 by tau sometimes in the name s tau plus n and you have tau goes to minus 1 generators, elements of a group generators of a group and the full of group is whatever you get when you close all operations that you can make like combining any arbitrary set of these so let's see what you get so it goes to e tau plus b over c tau plus b clear let's just see that to convince you of this well first let's see that the two operations that we looked at were tau goes to tau plus 1 is what it's the matrix 1, 0, 0, 1 that goes to tau plus 1 so a should be 1, d should be 1 b should be 1 c should be 0 so this is the matrix for determinant 1 okay goes to minus 1 by tau tau goes to minus 1 by tau a0, d0 and let's say b equals minus 1 c equals 1 this is got to be a group of entries and integers okay but whose entries are integers and whose determinant is 1 now the next thing you can check is the what suppose you've compounded two of these transactions you have tau goes to a, a tau goes to b plus c tau goes to d and then you have another this tau prime was equal to let's say tau double prime was equal to this with tau prime and then tau prime itself was equal to a tau plus b over c tau plus d then you can ask what is the net transformation that's generated by this by this operation and you'll find it's of the same form so I'll leave this video to check you'll find it's of the same form with the new matrices of a, b, c and b being the product and matrix sense of the other two matrices is that clear so you've got the first transformation generated by one matrix the second transformation generated by second matrix if you compound the transformations that is matrix multiplications for these matrices group of the set of two cross two matrices okay now what did we say about the software it's clear that if you take something of this form and multiply it with something of this form so you always start with integers it's something of this form okay so the entries of the group will always be integers more of a sense of determinant the product of determinant matrices the product of the product matrices the product of determinant matrices the fact that both these generated elements have a determinant equal to one that means that what you'll have to do is determinant matrices okay so clearly every integer entries and determinant equal to one and you can also differentiate that every such element every such matrix where this integer entries and determinant equal to one can be generated by by taking the sufficient by taking powers of these these two matrices okay this is the formula and it has a name it's called s n 2 c s n means the terms equals one z means integer entries two cross two matrices where determinant equals one so the modularized space of dory will be h c it's the space of all tau's modular s n 2 z transformations of tau again modular there's the space of all tau's and there's an s n 2 z identification that acts of tau's according to the formula and the space of inequality dory are given by sets of tau's by an x n 2 z I can understand in some sort of geometrical let's try to draw a picture of the set of equations okay so let's draw the tau clearly now we know that tau goes to tau plus one is an identification so clear any tau and put it in the line between minus half all of them you'll think of as initial observation we don't have to worry about this bottom side because whether you identify like that or perhaps it's the same thing okay so clearly everything just by using tau goes to minus tau and a negativity set of limit we're only at just the top half we can do better because you see that suppose something was tau goes to one by tau plus tau goes to minus one by tau when it takes tau goes to minus one by tau in particular what you do to the modulus of tau is invert it so let's draw the circle of modulus of tau the circle which is mod tau that's something here if you activate the tau goes to minus one by tau you'll get something with modulus of tau greater now either this would be here or it would be here now if it was outside you can once again use you can once again use tau plus one to bring it inside now this when you do this operation when either end up here or it's a hollow point here but basically you can show though I've never tried to do it myself no I'm sure it's not here you can show us this process I can always be activated so that in the end you end up for some point outside here so clearly the fundamental domain the fundamental domain on the space of tau is the space between minus half and and exterior that any tau in the space can be brought to the tau of that form by an eccentric transformation of the range of the space for instance what you do is take this fundamental domain and see what it maps to and the tau goes to minus one by tau it maps to something like this that's another fundamental domain of this of our space and that's the need for the number of them okay but you know when you do extreme theory we have to do an integral over our moduloid so if we do an integral over all equal to the tau right we pick some fundamental domain the domain we would usually concentrate on is this guy about the moduloid of tau right any questions about this understood moduloid of tau right that stands as a conformity so I'll ask you a question about conformity vectors question of conformity vectors is question of how many zero bonds of the of that p operator acting on c on there now once again in this coordinate system that means analytic that operators simply in the z bar so we're looking for analytic vector fields that respect the identifications of the dors so these vector fields are given by z at the base of these vector fields n times z but n has to be equal to zero respect the identifications okay so now we're looking forward to getting back to the conjugate of 2 that's z z z z bar 1 analytic, 1 anti analytic it's the same as the number of moduloids ckb's let's check that this matches everything that let's check that this matches everything we know about the surface of dengue consideration you remember there's a Riemann-Roch theorem that we had this theorem that came from counting zero that came from the anomaly of the bc system it told us well in this case it's actually very easy to remember the result because you remember that the number of b insertions minus the number of c insertions was fixed to be a certain number because of the anomaly but you remember the anomaly was proportional to the scalar curvature on the one sheet of the string and on the top of the scalar curvature the meter was going to be zero so the integral of the scalar curvature is zero so in this case there's no anomaly there's no inhibition in the anomaly it's a number of b minus the number of c modes not to get non-zero amplitudes in this case must be zero so the number of c zero modes must be equal to the number of b's and what's beautiful is because we found that the number of moduli was 201 complex and the number of confunding techniques was also two so in the considerations of the torus now we're going to try to set up strings here in the torus but any questions upon it the second not string amplitudes was to work out a way to get to the closed insertions again and when we were working through the sphere we had to worry about the closed insertions because of the b insertions because of the anomaly now we have two moduli so we have to figure these out let's remember what we did with the moduli the insertion connected to the moduli was d to sigma and square root g g alpha beta g beta phi b alpha beta and there where this is the moduli and g beta phi and then there's another factor there's another factor associated where there's another factor associated with down bar bar was at the modulus and we changed the down bar here isn't it? now this is probably the reason we're not changing coordinate ranges we're changing metric the last one we're doing was to relax the two pictures it's one picture in which you keep the metric and you can't change the coordinate ranges the second picture in which you don't change the coordinate range but change the metric rate okay now suppose we had this guy with identification tau suppose we had this guy with identification tau and we changed tau a little bit okay we had this guy with identification tau and the metric was dz dz bar and we changed tau now the metric was dz dz bar so every un-reformed metric here has only dz dz bar to both so this guy is one and this tells you that this is bzz which we call it b times del tau of g z bar d bar that would be bar times del tau bar and del tau first insertion by this point it was this guy so we had to keep a d tau there so this is d tau of g z bar z bar times b plus b bar times d tau of g z z and this insertion here was integral of b times del tau bar of g z bar z bar plus integral of b bar times del tau d tau bar of g z bar is this correct these were the two insertion factors that we discussed in our general discussion which I just worked out in this case using the fact that I enjoyed Constance in this discussion using the fact we know what the lecture is like now at the beginning of the class we worked out that delta of g z of g z z delta g z z delta tau bar over in okay so delta g z bar z bar delta tau in tau from here we get this guy clicking this guy clicks and then g z z this guy clicks each of which gives you a factor of 1 by what we include that the net insertion the product of these two insertions is a product of one by imaginary part of the power that's it the insertion of integral of b of z times the integral of b bar of bs you see we have to integrate over the metric of modularism we have to integrate of modularism tau b tau bar integral but the insertion of bs tells us that that's not the only tau dependence this thing is in fact divided by imaginary part of tau numbers then if you don't write it every year you have to keep track of all the numbers I'll give you the final answer but when it gets up really old as far as I can see what we've done so fast today but I've omitted my point of view I'll fill this up this is integral here I said in terms of the zero just an insertion of one point that integral all you want to do is probably have an extra factor in numerator which therefore gives you one over in tau instead of one over in tau but let me not take that yeah I think that's probably what's going on I think that's probably what's going on but let me just confirm that that makes sense okay I think as it's written here it's correct there's one more because if you can't make these b insertions only refer to the zero over the zero over is a constant as we've seen so we can actually do this integral with the zero over the net result has just one factor of in tau in the denominator but let me get back to a little bit more okay good but I don't want to become the integral to become this so that's one set of that's one set of insertions that we've dealt with now let's move on to the next set of insertions okay so the next set of insertions are you know we need to put one vertex operator with a c another vertex operator with a c we need the c tilde and have this fixed okay however once again here you see once again here once again here the the only way to get the c insertions is the zero and we've seen that the c zero over is constant it's a conformal killing vector to us so the issue of whether the c insertion is fixed firstly we don't care about where we fix it we don't care about where we fix it but when we fix it we multiply by the zero of the conformal killing vector at that point the zero of the conformal killing vector at that point is a constant so it's perfect you remember what the general formula was the insertion the insertion is from c giving you the determinant of the insertion is from c is giving you the determinant of of of c i sigma g right about c i is over the two over the two zero once again and sigma k was a point in this mission but yeah so what so let's so let's go to complex one the two the two conformal killing vectors c one z equals one c one z by zero and c two z equals zero c two z by equals y something in a purely ordered direction or something in a purely accurate okay now we should do that doing these things in general they would have been functions of where we were but in this case this is just a constant it's independent of these coordinates yeah okay so we just get this determinant of one zero zero one so the point is is it