 So we want to discuss one loop amplitude at least for us. So we've completed, we discussed this, the major issue for closed-stream is outside. Let me quickly tell you about the similar combination for open-stream. So suppose we want to do a one loop amplitude for open-stream. That's this little string here that runs around in the loop. Okay? So the, the kind of stuff that we have is, is a cylinder. This side is identified as, it's a little string, the string is going in the loop. Okay? Now, we can ask, how do we mobilize the cylinder? Okay? This is the same as the number of zero-modes of the B operator that respect the boundary conditions of the cylinder. So why are the boundary conditions in the cylinder? The boundary conditions of the cylinder have been discussed in this direction. And at the edges, the boundary conditions we discussed before named it as B is equal to B. Okay? Now, on a torus, we had two zero-modes for the, for B and B to the left. Namely constant B and constant B. Okay? But then at the same time as zero-modes, it's going to be thought of as part of a torus. With the lovely trick it becomes, it becomes a torus. Okay? So we can have the same zero-modes except that billion people are not equal to zero-modes. Because B is equal to B to the left, B. Have the boundaries. Okay? Do you have the boundaries? I see lower and lower. No. Well, I mean, same line over here, it doesn't matter. But about other, I'm taking you with this little string, going around this string. But these are the end points of it. Of course, that's symmetric, right? You can think of it. I mean, that doesn't matter. That's not the way I think. Okay? So I'm thinking this is a sigma system. Okay? Fine. So there's one modulus in this situation. And what's that one modulus? The one modulus is very clear. It's the height of this. So suppose we always choose this to be zero pi, this is my direction. This direction is, let's say, tau. What is this? You probably want to call it t thanks to pi out of the table. And the modulus is simply the value of t. So that's the modulus of the game. What's the other angle? You see, the angle is fixed by boundary. You see, the boundary conditions we're putting are at the edges, there's a sigma of x is equal to 0. If you change the angle, then you'll have to put different boundary conditions. Okay? So this is this with less sigma x equal to 0. Okay? You can't change the angle. And if you change the angle, you change the boundary conditions. That's not a symmetry. So the modulus here is t. Okay? Now, once again, we want to calculate the measure of the boundary line space. So it's the same angle we had. You know, as we've seen last time, del g z z, if you remember, del g z z divided by del tau pi, I think it was, was equal to, what's proportional to 1 by d tau. And del g z pi z pi by del tau, because it was proportional to 1 by d. This is what we saw last time for the talks. Okay? Now, what are we doing here? What we're doing here is basically the same as a special case for the angular force. Okay? So in this particular case, we do more or less the same calculation, except we focus it at tau equals i time, and we're actually part of tau equal to i d. Okay? And because the tau inverse is rectangular, you know, because our metric always takes the form d s squared is equal to d sigma 1 squared plus tau squared times d sigma 2 squared. You know what happens? Basically, all we're changing is g y 1. But we change t. We put the necessary changes in g y y. And g y y has equal components of g z z and g z bar z bar. Okay? It doesn't look that way. The question we're asking is, suppose we have the w, the z coordinate, and we have g z g z bar is equal to, so the metric is equal to d s squared d z d z bar where sigma is equal to, belongs to the range 0 to pi, and tau belongs to, as we already formed it, in 0 to 2 pi. Now we're going to go to a new way, suppose, now, suppose we take d 2, so this is the general set. Now we go to the variable w such that d s squared is equal to d d z d z bar. Okay? And these ranges were signalized the same above, but tau belongs to 0 to 2 pi t plus n. And now we take this metric and rewrite it in terms of the metric, in terms of the variable z, where the periodicity is of z. Okay, let's, I'm sorry. To be consistent with the relation we used last time, we call this the z thing, but, okay. Like, okay? Whether we rewrite it in terms of the variable w, where the periodicity is of w, was that of 0 to tau, 0 to 2 pi t, without any effect. So what's the change that we have to do? The change that we have to do is, real is said is equal to real part of w, but the imaginary part of z, okay, is equal to the imaginary part of w, where the imaginary part of w reshapes, right. The imaginary part of z, so z where it goes to t times the t, so t plus delta t. So we take this metric, these definitions of the variables, yeah. Okay? And finally we look in terms of w and w, is this correct? Okay, it's the same exercise we did last time. Okay, thanks. So if we want to work it out, we will get d r e w total squared, because that hasn't changed. Plus t plus 2 delta t, okay, so just 1 plus 2 delta t by t for the first order, d in that thing. Because I can't substitute it. So the change in the metric is what? So delta g is, we have delta g imaginary imaginary is equal to delta t divided by t. That's correct. But now we can translate them to g z z and g z bar z bar. But g z z, you know, some linear combination of the real and the imaginary part. There's no change in the real part. So the change in both g z z and the change in z z bar are proportional groups. This is the single modulus in the game. What we have to do is to find where integral d to z or not, it doesn't matter. v times delta times d g z bar z bar by d t plus v bar by delta g z z by d t. In this insertion, only the zemo matters. On the zemo, v is equal to v bar. And each of them, each of them, and the whole thing, and the whole constants, v is equal to v bar the whole constants. So the whole thing can be written as the volume of this thing times, okay, so it's d to z times v of 0 times 1 by 2. Is this good? But this d to z is pi into t. It just adds some pi into t. So this thing is proportional to v of 0 without any factors of t. It's the volume of this thing. So the v insertions are just v of 0 with no factor of t. They're not the same as in the donors' view, if you raise that. Okay, two copies. Should be okay. However, what about the c insertions? We've got the same thing with the c insertions that we did last time. We have just one unfixed formula here in vector. It's the same thing that we said for v's, because there's double c in a c to the left, both constants, but now c is equal to c to the left. It's the volume of this thing. The one unfixed formula here in vector is translations along this section. And we could fix it in that one unfixed formula here in vector. Fixing the one unfixed formula here in vector is like choosing one boundary operator to be fixed. But because the insertion point of that c doesn't matter since the constant, zero model c is the constant, we can separate where we fix, where we put that c, and where we put the integration of the operator. And then integrate it over the vertex operator up to a division by the volume. Okay, so the next result is that instead of having the fixed vertex operator, we have all vertex operators integrated, but with an insertion of c, zero, divided by volume, and the volume is c. So the mind of conclusion is that we have dt divided by t, except they have to be value of b, zero, c, zero, time whatever, or time to integrate the vertex operator. You see where this one over t came from. Where this one over t came from, it's actually very intuitive. You see there was one gauge invariance that was left unfixed. The gauge invariance was translations along the strip line. So because the volume of this translation along the strip line action is out of bounds, we could forget about fixing it. That's what we're saying there. We're treating all vertex operators as integrated. But in forgetting about fixing it, we are treating gauge equivalent configurations as discrete. Now each gauge equivalent configuration gives us the same answer, but we get the volume of the gauge of it. But if we just treated that everybody's operator as integrated, and forgot about fixing it, then we would be overcounting by a factor of the volume of the unfixed gauge. That volume was t. So we did it right. That's the intuitive reason for this 1 by t. The intuitive reason for the 1 by t. Again, you need a b0 and c0 to make any sense, and this is the unfixed gauge intent. Is this clear? So remember this is the answer for the open strings, and the answer for the closed strip is d2 tau by d2 tau, and b0 is the bar 0, c0 is the mass int predictor. Now the rest of this class is to actually evaluate the closed strip partition function and the open strip partition function for the simple theories we can study, and then interpolate these objects. The candidates will show us some of the basic properties of string theory, as you bother. Then I sort of qualitatively move on. So we've done the closed strip thing before. But just quickly to remind you of our event and to remind you of the interrogation. So what do you have to put in here? We have to put in the partition function on the BC system with the partition function of the x conformity theory, and that's it. So now let's write down the partition function of the x conformity theory. The partition function of the x conformity theory with what? It was 1 by d tau to the power 26 by 2. So let me try to hold to the power of 26. Then we had 1 over q to the power minus minus 24 minus 1 by 24 1 minus q to the power 8 for the rest of the n of 1 over q to the power minus 1 by 24 product over 8 1 minus q to the power 8 where q to the power 8 should be negative. Okay, now somebody help me understand this expression. And that's the whole thing is raised to 26. Somebody help me understand where every girl in this partition function is. This is just z z matter on torus of modulite of modulite is equal to this. We're not integrating over any modules. This is the partition function of the matrix conformity theory on our torus modulus channel. Firstly, give me a Hamiltonian interpretation of this partition function. What? Hamiltonian interpretation of partition function on the torus is that of a trace over the spectrum of the conformity. Is this clear? Does everyone know this? We discussed this last night and this is the basic one in theory you should know. But somebody explain this to me. How do you compute the trace over a trace into the power minus beta h? You know the Euclidean path integral of what matter? The space for the circle and the size of the circle is beta. Okay? Is this clear? This is familiar. Suppose we have like 10 kilo atoms. We have like 10 kilo atoms. Okay? Then what would that the path integral so this is 0 to 2 pi this is 0 to 2 pi tau. Okay? What would the interpretation the trace interpretation of this of this path integral be? Same to of 2 to h over beta. Let's call this 2 pi t. That's tau. 2 to pi t as beta and it is living on a circle. Right. So let's say that the system has to be living on a circle of size 2 pi. So it would be trace of e to the power minus 2 pi e as h. How do we make this the strongest skew? How do you have a suggestion for a for a for a Hamiltonian derivative? So let's call this 3 t. You see, instead of just translating it to 9 you also translate it to space. So how do you get the skewness? You put a generator of translations at the last step of the bottleneck. You do a trace up to a generator of translations. Okay? So what is the generator of the translations? And leave it up by 2 pi times the real path. Okay? So you add e to the power 2 pi times the real path. And then I that's what generates. I'll go into it. And then the generator of translations. But h is L0 okay? So this was L0 not L0 and this was L0 minus L0 minus 0. So look at the terms of course which are the L0. So this is e to the power i times 2 pi times tau L0. You see? Times e to the power i to the power tau bar L0. So the Hamiltonian interpretation of the publishing function of the tallest of non-linear pattern of the tau is this. Okay? This is what we've chosen about q to the power 0. This is what we've done about q to the power 5. In this graph this is the thing we have to calculate. Now you understand by using these terms. Okay? So let's do the easy thing. What is this 1 minus e to the power L to the power 0? Exactly. So the oscillators. What is the spectrum of us is the L0 spectrum is the spectrum of oscillators frequency 1, 2, 3, 4, 5. Each oscillators frequency m as published in the publishing function 1 over 1 minus e to the power n. Because the occupation number of oscillators. So we've got all these oscillators and we have to take the product. Is this correct? The q bar was the same thing. Now what is this q to the power minus 1 to the power 0? L0 equals minus 1 to the power 0 by 0. Very good. And what about this intelligent power let's make the different number into 0. No, no. That came from the nature of it. You know just under the partition function of the bosonic sense. In just the partition function I would suggest the data is excited. I meant to you guys like let's say end of this is a chemical right. Okay, last week of October we have an exam all of consciously volume okay. And I'm going to be very careful. At least I know it's right. Yeah. That's what is the input? We've accounted for what? We've accounted for the oscillators. What is there at the moment there? It's a 0. 0 more is what? 0 more behaves like a heart. Remember there was 1 over 4 we have 1 over pi as alpha prime times back of 2 pi 6. Okay. There is this part of the Hamiltonian of the problem that we've not yet dealt with. Now in dealing with this part it's easier to barely use this form of the expressions of Hamiltonians than this form because 0 more not to both times 0 minus 0. So we have to calculate trace of A. So what's the spectrum so what's the Hamiltonian? The Hamiltonian is what comes from here is p squared alpha prime by 2 p squared by 2 n. This is exactly the grand deal for a non-electric particular mass n. Then it happens to be 1 by r. Is this clear? I'll write it down here p squared by 2 n. So this is spectrum. So what we have to do times 2 pi eta this is what we are going to do. Now what is this? Even if you take an expression of the stress if you take a d this is a constant exact. So this is we take a dp e to the power alpha prime squared p squared by 2 1 by eta because the stress we do in this is a fact that completes yes okay and now we are going to see the integral from having a 1 over square n. Now all the other factors that Wynchowski has we have been going to keep track of whether we have done them correctly we are not bothering you. So that is this paper for the glossary let's simultaneously put them together let's we write them for the objects let's make sure those things get back okay now we are going to ghost partition now some of the things we can say about this partition function immediately some of the things that we can say immediately about this partition function is that it is modular container that is z for instance realized that Or more generally, it's invariant as a full module. We discussed this module in a group. This SL2 is the same module in a group for last semester. That's right. But why is that? That is because this is a conformity. So the published function is invariant at the scaling. The basis of modular parameter tau is equal to geometry of the daughter's modular parameter minus 1 by tau, up to an overall scaling. And the published function is just invariant at the scaling. That's what we discussed. Great length, last semester. So why is that? Obviously, it's algebraic. We get a very interesting property in this function. So this published function that's defined by this in tau and power out, the product of q that's product of q bus, is invariant under tau goes to minus 1. This is something we use when analyzing the open string spectrum in a little bit. So I just want to remind you of that. OK. Fine. Now let's return to the, let's return to the cost. OK. So what is the published function of the cost? Now I'm going to tell you the winner of the question, write down the answer, and I'll ask you to help me figure it out. This q to the minus 1, q bar to the power minus 1, q to the power minus, what is it? 26 by 24. And then product of n is equal to 1 infinity, 1 minus q to the power n squared, what is q bar? This published function was a b, b bar, c, c, y insertion, the cost part. The q to the power n is 0. So the spectrum of n is 0, so what? Where do we get this from? What are the modes of the c and b and c? Part of the links in here, and I'll get it straight as I said. This should be an enumerator. The b and c system has oscillators just at the x thing, starting from 0 to the n. But b and c keys were anti-communal. So the occupation number of each oscillator is either 0 or 1. I mean, because they're both. There's an extra minus sign because what are we computing? We're computing the partition function with periodic boundary conditions for billions. Okay? Why are we computing with periodic boundary conditions? Because they were of course for the homophisms, which have periodic boundary conditions. Well, the models have periodic boundary conditions. But the Hamiltonian interpretation of the boundary network with periodic boundary conditions for a field that is anti-communal, is traced with an extra factor of minus 1 to b. What we're computing is not tracing to the power minus beta h. We're computing trace of e to the power minus 1 to the x times e to the power minus beta h. Okay? So this is the vacuum. No oscillators. One oscillator excited. Since it has an opposite formula from this guy, it has one extra. One extra anti-community field excited. It comes into the minus sign. It's an enumerator that says that there are only two possible states. Occupation number 0, and occupation number 1. Is this correct? It's square into your z. Whatever we have to see here, I'll say, in the 0 mode sector, had we not had these extra insertions of v bar, cc bar, we would have got 0. Because the 0 mode sector has the slew state system with opposite formula number and the same energy. So trace of minus 1 to the x times e to the power minus beta h. But these insertions had specific altities. There's 1, 2 state system for left-overs, 1, 2 state system for right-overs. So there's a 4-fold degeneracy in the 0 mode sector. These insertions have become exactly one of those 1 states. Okay? And we don't get 0. There's no more acceleration. It's the state rate, which is obviously 0 on the side. It's 0. It's 0 state. We discussed this last year. It involves going the right scale up rather than the same. I won't get it wrong. Okay. So, fine. So we get this. Now what about the x bar? What about this q to the power minus 26 beta? Plus 26. I'm not a medium. I'm sorry people. I'm getting it wrong. So, I should have crown speed up q to the power, energy. Yeah. It's q to the power of energy. And energy. I should have had q to the power of energy. And so this was the new rate. I mean, sorry. This q was with minus 1 by 24 was the numerator. But down the stairs it was. I would have plus 1 by 24. Because the crown speed, then, was minus 24. Minus 1 by 24. In the latter sector. And we calculated q to the power of energy. Okay. So this q was the new rate. Sorry. Okay. Now here, people have. Here I can read that directly. But let me write it down. I'm sorry. Let me write this whole thing down. This would be q to the power of minus 1 by q to the power of plus 26 by 24. And product over n, 1 minus q to the power of n squared, 1 minus q to the power of q. Sorry. So this is the same cost. Now then, what does this factor? It's the same as this factor. Just explain the statement. It's a cost that is no other. There is no zero point in the problem. Right. Right. Every conformal theory has cascaded energy. You see. The vacuum, the identity operator, has dimensions here. So that 90 translates into the vacuum having energies. That statement is 90 because it doesn't adopt a conformal object. That means cascaded energy. And the correct version of that statement is that the energy of the state corresponding to the identity operator is minus c by 24. In this case, c was 1. So that's minus 1. L0 was 1. Minus 1 by 24. L0 by 24. Minus c by 24. L0 by 24. Minus c by 24. Minus c by 24. Everything is to the power of n, 6. Right. What is the conformal? What is the centrifuge of the bc system? It is minus 26. You know that because that's how we got the critical language, right? The full conformal theory of the centrifuge of 0, matter at charge at charge 26, bc system at charge minus 26. So this is the energy of the state corresponding to the identity operator minus c by 24, c is minus 26, minus n is plus 26. Is this clear? Finally, what is this? The word? It's 0. Okay. Yeah. So it's 0, but it's not. You know, this is not a miracle. It doesn't have the hugeness out of 0. When is the difference in this theory between the vacuum and the state corresponding to the identity operator? Do you remember the state corresponding to the identity operator was done the same as the vacuum of the bc? The vacuum of the theory, in fact, was c, was the state corresponding to the operator c. And since c has dimension minus 1, the vacuum has energy minus 1 compared to the state corresponding to the identity. That's this. All of the oscillators accounted about the vacuum, which would be right in the other energy of the vacuum. And we'd have that value. Minus 1. Minus c divided. That's the energy of the vacuum. It's much slower than I thought. Okay. All right. So now let's put all this together. We have to modernize this guy and this guy. Okay. So these oscillator factors of 26, 2 of them both. Okay. What about these other factors? You see, this and this can. But this remains. And this is exactly enough to be a q to the power plus 1 by 24 in the denominator for each of these factors. Okay. So let me write down what it means. What it means is 1 by 24 times 1 by q to the power of 1 by 24 product over n, 1 minus q to the power of n, product of q to the power, say, product of, okay, 1 minus q to the power of n times square root of q to the power of 24. Exactly. This is a partition function of this. This very powerful. In the life of the worker, there's a zero motion. You know what 26 is. Apart from in, down factors. This is what we were talking about. Okay. Yes. But if you actually do this extra in, down, it's exactly. And then we combine this together with the measure of a modernized space he has in the course. That was d to the power over n product. So that becomes d to the power over n product. And then square times wherever that's in the bracket. This is our final expression for the partition function over the darkness. And we were very happy with this expression, if you remember, because we wanted this expression to be modulated variant. And we checked that d to the power of n times square root of modulated variant. And, of course, this is modulated. That would just happen. Because whether it's the power of 26 or 24, it's modulated. The same thing that we have there, but the power of 24. Yeah. It's the partition function of 24. Bohr's answer to the problem is there, isn't it? Yeah. So we've got a very nice, we have this very nice body there in town, sir, for the partition function of your strength and the darkness. Okay. Now. Okay. This is very nice. Now, the thing that we want to do is basically get this. Okay. So, what is the partition function in strength theory? Well, in field theory, suppose you calculate the partition function. Okay. Suppose you calculate the partition function with no insertions of vertex operators. What are you getting? This is a correction to vacuum. Don't think that, well, in this problem, we know what the correction to vacuum energy might be. It must be. If the moles of the shear strength are done by field theory. That's the answer to what we would get by summing over the vacuum energy graphs of all the particles of free stream. Okay. So that's the exercise we're having. Okay. So, suppose we have a particle of mass m, a scalar particle of mass m, and we want to calculate its contribution to the, to, we want to calculate its contribution to this vacuum energy business. You know, we want to calculate its contribution to the vacuum energy. We want to calculate its contribution to the vacuum energy of this area. Okay. So, what do we get? What we want is to integrate BGP at a log of K squared per cent. Right. Because the easiest way to say this is this graph. And that graph represents just that, that logarithmic development. But a more physical way to say it would be, okay, I'm just going to, can we move forward from here? This, if you were to accept this graph, this graph is just a lot more. We have factors of all, depending on the stream of the, A will be the number of degrees of freedom. Okay. Suppose you were doing just a scalar particle, it would be exactly this. And if we had additional, you know, let's say vector particle would be multiplied with effective number of degrees of freedom. Okay. That's always how it works in a 1, 2. Yeah. So, we're going to get this ninth here. That's a true statement. So, there may be this ninth here just to sum over all the effective degrees of freedom. Okay. So, this calculation is correct just for scalar particles. But it's true actually, but over every kind of particle, if you multiply by effective number of degrees of freedom. The state would have been. This, sir, this effective vacuum energy graph. Okay. Can be treated in a sort of pretty okay, that helps us to make contact with what we have in the stream of freedom. Okay. And that way is to rewrite this in the following way. To rewrite this as dT by T times exponential of minus T and for fun, you put it in 2 years at the level at which we're working in learning factors at this time. DT by T exponential of minus T by T You see that suppose you didn't have this T here. Then this also could be proportional to 1 over square root of T by T times 1 over T by T okay. Now, integrate both sides with respect to T by T okay. And that will be 1 over T here okay. And there's another way of saying it actually the state is a bit a bit careless because this integral is not really very defined. You have to define it with respect to a subtractor. People have cut off here. Lowest part of epsilon. Another way of saying it is the formulae. Another way of saying it is that if you look at the formally this object is independent of k squared by x k squared plus x squared by T depends on the speed times k squared plus x squared this measure is invariant formally this object is not dependent on k squared okay. But the reason it actually does is that there's a cut off here. There's a divergence at smaller things. You have to cut that cut off. That cut off state is very small. The better than the cut off state we don't have to think. And when you make this change your value you change that cut off by this factor of k squared that's k squared and that's the option at the small term. So that's the second way of saying this thing. You know you get another way of repeating that. Without this T the fact that this is one of the k squared plus x squared is my initial analysis. And the same my initial analysis that makes it a log limit as well as the equivalent of this. Okay. Now this is a very useful reference. It's a little careless that something you have to do is cut off and so on and so forth. Let's just forget about all those things which we can read. Okay. So now two representation and it has an interesting interpretation. Suppose we're interested in looking at the theory in first one language. Okay. Then this is sum over all particles parts. T is the modulus of that surface the length of that part. DT by T is a sort of measure that you would have calculated by the other of all the processes that we've been dealing with for the strict theory. And this is simply the fact that you're computing the trace of the Hamiltonian weighted with the effective temperature of this quantum mechanical system So from the point of your first quantum language this is very natural. Okay. And that makes it clear that this is the right expression to use in comparing with the strict theory with a strict theory. Because again, it's clear that it's the right expression to use because because because it's strictly working effectively first. So now let's see. So what would we expect from this whole view the vacuum energy of our string theory to give us what we expect is that we would have a minus 3 by 2 into k squared plus m i squared with some i over all the particles state. We should sum up the particles weighted by effective degrees for you. So bear in the same we should sum up over all the states of the physical states of the deformity. Each of those gives you some polarisation of the item. Okay. Then it deals with the orientation. Okay. Now how do you relate m i squared to something important? Well, remember our Marshall formula. Our Marshall formula was that alpha prime a squared by 4 was equal to L0 minus 1. But that was also equal to L0 bar minus 1. Right? This is alpha prime X squared by 2 is equal to L0 minus 1 plus L0 bar minus 1 by L0 now it's pure opposite we take it out of 0. Okay. Remember that twice alpha prime X squared by 4 is equal to this. You see every oscillator state of L0 that corresponds to particle in the state there. The Marshall is given by this formula. But that corresponds to particle only but the arbitrary oscillators of the left and of the right but some gets in an error machine. L0 is strictly weighted by the house that you have with you. So the first thing they do is to replace P by a tau variable. So P is all alpha prime tau. So that's not the problem. Okay. So we get integral of these tau by tau that hasn't changed then exponential of minus tau alpha prime T squared by 2 that will be out. Now exponential minus tau alpha prime means squared by 2. Okay. Further down to some for independent organization. But that some now will be replaced by trace over the World Sheet Hamiltonian and it will replace alpha prime squared by 2 by exponential of minus tau minus 1 plus L0 power 1. Okay. Now we I'll call this T again. It's useful for me to have a T. It's tau but we've got x things. So this is getting DT by T. Okay. L0 and L0 power will now be implemented later than actually. So now what's important should say L0 is equal to L0. The ideal is that L0 and L0 power are integers. For example, representation of the delta function with integers over all space integers. We multiply this by d theta integral over theta to the left of 0 to the right and multiply exponential of i theta into L0 power 1. That's the next function that says L0 to L0 power and enforces that. Now it's imagine your variable tau is equal to theta plus i times t. That this gives us q to the power L0 q bar power L0 bar. Okay. We have q, q bar bar to the power minus 1. That's the minus 1 stat. And we also have I mean we can try to say L0 minus 1 and I say L0 bar minus 1. Okay. And then we do the integral of a k and get our in tau it was t, t was in tau to the power 1 and 13 by d to tau. It's d real tau that's the actual one. Now this was imaginary part of tau so we recover exactly what we want to do. Exactly what we want to do. Exactly what we want to do is what we want to change. We have d to tau by in tau now this in tau this is in about 13. So this is square and then we put 12 in there. So 1 by square in tau times q to the 1 by 24 q bar to the 1 by 24 and this is proper. 24 because we're doing this is like like like like physical spectrum and then happy post we have exactly the same expression that we had before except for one change we never tell you about that one changes the changes are the original integral integration in string theory you integrate tau over all equivalent tau and you remember from last semester that was tau from minus half in the real axis times the exterior of the human circle so the region of which we integrate is this the fundamental domain of the torus the set of all equivalent torus set of torus are not equal to each other up to a modulator in the string theory combination here in the string theory combination here what you get is the integral over the whole upper upstream because theta 0 to 2 pi if you rescale theta by 2 pi 0 to 1 minus half that's the same method and T remember remember T that's the source difference so apart from this change in integration domain apart from this change in integration domain we got in the same this change in integration domain is actually very important it's actually very important to the following case you remember from our theory calculation the part of the calculation that was sensitive to UV divergences was large k now large k transits to small t why is that because we have this e to the power minus t that's k squared so the only way the large k is very small otherwise large k is exponential suppressed so the part of the diagram that corresponds to UV divergences to field theory was small t as present in the integration that field theory seems to give us but it's absent in the integration according to the rule of the string you are thinking this is a bit of a thing because I happen to choose this particular modular modular domain I have chosen this in that case we would have had a small damage and a potential divergences but you see all those things are the same so let's try some what they say what is this point under the modular problem the modular transformation that takes a familiar years tells us about better so this point part is infinity again so what you are saying problem what you got there so what about the problem and then at large at large t the only values of k that contribute are very small values of k so this is one of the great things about string theory you see if you have problem at large t it means that then must have been in field theory language a problem at very small a value of k or restated if there was no problem at very small values of k as long as this is big day there cannot be a problem at large t so now what I am saying is that string theory links good i of a here to good u v u v v let's see this in multi-day oh we have to run this thing yeah let me just one minute let's see this in multi-day you see let's look at this partition function let's look at this partition function at very large values of t okay at very large values of t only in the ground state problem the ground state in this problem particularly in a problem has negative energy okay so what I am saying is that you take this partition function expand it and the power series expansion I mean the power series is equal to power k and as we get e to the power 2 t that's the ground state plus 1 that's the 0 the the the contribution of muscle particles plus extra energy dispersion and we have to do the integral d to so let's create a theta action just the t action so d t over t squared and if you remember we have t to the power 30 okay just in the square of 20 so d t over t now this guy here he's just wrong he blowed up extra energy at last so we you know in the basically if you got a state of negative energy effectively okay that's going to give you the energy now why it's because it's because you're trying to do a path integral of path integral of e to the power plus m squared i squared minus k squared i squared so k is smaller than m these moles these suppression these moles the amplitude for the moles k is smaller than m want to grow rather than you know if the path integral is maximized by these amplitude both will be very large all will be very small and that gives you the divergence that's the divergence basically we see the fact that if you take any negative path particle and try to do perturbation there okay you'll get a divergent contribution there it's from k is smaller than it's an infrared divergence so we are closely telling does have this infrared divergence because it has a vacuum that we knew however in the absence of the tachyon suppose we managed to make a stream there where there was no day in the past state in the probabilities no vacuum then this onwards is final this is a stream theory which has no pattern so we want to have to make this apology about the infrared divergence the ultraviolet range can always view the path in no way so that the ultraviolet region is just not there all the possible problems are infrared and if we know that the theory is a stream theory stream theory for this particular question of energy that's something very smart it cuts off effectively high-moving technologies you see because one last thing I said and that says you might think that well that's sort of easy we just put some cutoff in momentum space somehow for you and we get some sense I want to emphasize that's not true that's not true because one thing I want to emphasize is that and the tau goes to 1 by tau you see these points are invariant and the tau goes to 1 by that boundary of what's linked by tau and a point here is back to the point here the boundary is but actually this edge this thing is the same as this thing so it's a cylinder and this part is the same as this part so you see it's like a cylinder whose edges it's back okay and infinity every point and infinity should be thought of as the same point you know where this can be so what this is is like a sphere the topology is like atmosphere and this from the very beginning here it's a compact matter that's really important because you may remember from last semester that all our our our no-ghost sphere was the fact that you know BRST equivalent amplitudes BRST equivalent operators gave us the same amplitudes and relying on there being no total derivatives and module amplitudes you know BRST equivalent operators are amplitudes up to total derivatives and module amplitudes if there are bounds of these two materialized places it's total amplitudes but we can't do it if BRST equivalent objects give you different amplitudes you have made non-states interfering and your theory breaks you down okay and this is the problem that is played every previous attempt to quantize you try to do a cut-off by hand it's very hard to do that way that's not great if you're bothered and therefore it introduces negative non-states okay strict theory has a very clever way of putting this cup off in a way that preserves you in an area if you violate it in this nanny way if you try to integrate it over some other part of the fundamental domain you will introduce negative non-states in your theory and the theory who makes this it's a very particular way of cutting off the theory that manages to make sense this is a very important okay and this starts the open string also