 Last class, we worked out the partition function for three scalar fields of the torus. And we found that z of tau is equal to, I think we've got 1 by 4 pi to the power of 36, alpha prime and in tau to the power of 13. And then we got everything else that we got from the oscillator, which was that this formula denotes. One over QQ bar to the power of 15 by 12 is to 1 minus Q to the power of 8 by 6. And you remember that we also worked out last time that the action to understand your string theory is d tau, d tau bar by 4 pi. Well, 4 pi we didn't keep track of. I wasn't keeping track of constants, but we just kept care for the track of that. I'm not going to go through the number with, you know, you can keep care for the track of this number through all the manipulation of the formula. All right. Times in tau and then times the insertion of a C, a C bar, a B and a B bar at some fixed positions, fixed positions don't matter. Okay. So the partition function of the top of the theory volume matter goes with these insertions. So we've already computed the partition function of the matter theory. All the insertions develop hosts and the insertions don't matter as far as the partition function of the matter theory is concerned. Now what remains is to compute the partition function, or more precisely the partition function with these insertions of the host function here. Okay. Now as a warmup, this is not the right calculation, but as a warmup, let's try to compute trace each of the bar minus B times H of the B C system. Okay. So this is bar minus B times H of the B C system. What is that? But each of B and C is a fermionic, as I said, a fermionic oscillator, whose frequency goes from 0 to 30. A fermionic oscillator of frequency n and energy, you know, energy n, you can only have occupation number 0. Okay. So we'll take it back on 0, my energy and so on in a bit. But ignoring that, it would simply be product n is equal to 0. N is equal to 0 to infinity. 1 plus Q to the power named square and then time. Okay. Because for each of the B and C systems, you have one analytic and one analytic oscillator. The analytic oscillator, everyone sees an analytic, and I think the one sees is an analytic oscillator. The analytic oscillator of B and C can be Q-charged and Q-bar-charged. Okay. And then they have energies from, they have frequencies from 0 to infinity. Each oscillator can be occupied from 0 to n for one time. And that's this function. We ignore 0 point energies, which I'll fix up in a moment, but first is the square. Okay. Now we have to include the contribution from 0 to energies. Okay. So the contribution from 0 point energy is two sources. The first, that energy is on the cylinder, differ from scaling dimensions on the plane by a factor of minus C divided by 26. And then that L0 differs by L0, the plane by minus C divided by 26, and L0 bar divided by 26, and the other scaling, and the empty scaling of the plane by C divided by 26. C divided by E equals C. Okay. So that gives you a factor of Q-Q-bar. Now, what was the exception charge of this theory? So I'm trying to C divided by 24. Because all of these theory formulas have that L0 is equal to C divided by 24. And what is the central charge of this theory? Okay. What was the central charge of the whole theory, matter plus cost? What was the central charge of matter of it? 20? 26. We have six different matter bits. So this theory happened. We don't have central charge minus 26. That's how it all worked out, right? This thing is minus 26. Okay. We get a plus sign. Now, the ground shift energy shifts by minus C. So this is plus 13 by 1. Now, these two signs now both will be the same. So there's a problem. They have to be opposite. It was the same shift. So, okay. Which did we mark up on? This would have negative energy. So this should be minus C. The energy at the ground stage of the free force of state. What is free force of energy? I'm sorry. Both of them? Oh. You're totally right. Wait. Just a minute. You're completely right. Thank you. Thank you. Thank you. Thank you. Thank you. As well as you. Thank you. Okay. Yes. Okay. So that was one more fact theory that we haven't done properly. It's something that we want to discuss. But if we compute in this thing, just try to use the power minus required, this is active. That's not a problem. There's one more thing. Zero point energy. What's that? What was the scaling dimension of the lowest energy operator of the BC system? Minus one. We started counting as a per vacuum at scale dimension zero. We actually had minus one. Okay. So we need to add an additional factor of q, q bar to the power minus one. Is that okay? So the final answer of a trace, so let me just write it again. The trace of e to the power minus beta H is equal to q, q bar, and I'll just keep these factors explicit. Eventually they cancel. Into q, q bar, into minus one, into one of the stuff. Into this bar. So the trace is not what we want to calculate. The trace is not what we want to calculate when we calculate the function function of strength. Now the reason for that is that the calculation that we want to do is the calculation of the part integral of the BC system on the docks. The question we should ask is with what boundary conditions do we do this part integral on the docks? And the answer is given from the logic of the subject of Gage-Shakes. B and C have to do with, if you want, morphism, fields and things to do with metric and the excess and so on, all of which would be periodic. Of course. So generally when we go through, when we review where all this came from, we want the BC system, we wanted this part integral of the BC, with the boundary conditions of the BC system to be periodic. Of course. Okay. However, there is this fact about part integral of anti-community fields, which is that if you want the part integral representation of trace into the part minus beta H, then what you should be doing is, I should say more clearly, this is trace of Q to the power L0 Q to the power minus. Okay. If you want the part integral representation for that, then that part integral representation is given by just the anti-biotic as you go up in the time circle. So the part integral that we're interested in for both of the string theory is not the part integral that computes this object. Firstly, probably. Secondly, we've not put in, we've not put in our exertions. Remember what we wanted was to stick with the, with the CSE bar, the energy part. Let's try to fix that up. Let's try to fix that up. First, let's try to fix up the boundary conditions. So let's try to find a Hilbert space, a Hilbert space, an operator expression for an object, for the object that's an independent object. And it's clear what that is. That's simply trace of minus 1 to the power x. That's Q to the L0 Q bar, by L0 bar. Where f is defined as an operator, where minus 1 to the x is defined as an operator, then anti-convius with both the B and the C fields. But thereby, the use of binary is an object. You know, because this object gives a minus, gives an additional minus sign, as you close out the part integral. When you go from T equals 0 to T equals beta, it changes the boundary conditions from plus to minus for the fields that are charged at this side, both B and C are charged under the sun, so it fixes up the boundary conditions. Okay. If you're interested in just the part integral with no insertions, that would be computing this object. That would be computing a Hamiltonian language in this object. Now, let's have a look at the value of this object. What you say is completely right. In fact, it's just this thing is equal to QQ bar to the power minus 1, QQ bar to the power 13 by 12. And then let's write this with a little more care, product of n is equal to 0 to infinity 1 minus Q to the power n. It's a minus sign, because these are the data that changes the sign. Yeah. 1 minus Q bar to the power n, and the whole thing. If you want that in mind, pay attention to the limits of this product. The limit of this product. Zero to zero. Why is that? We already knew that we got zero then, and we got zero anyway. We knew that because we know that we do the part integral with periodic object conditions. Around these stores, we have C0 more than P0 more. We integral over the C0 more than P0. Okay, integral over P0. So, zero is physical. It's physical. Okay? The part integral with the daughters without insertions is just zero. Okay? Just to say, here, here in this notation here, we also included the contribution of the zeroes. The zeroes will carry any energy. It just gives you a multidimensional. You see, in the BC system, in the BC system we had the upstate and the downstate, both the left movers and the right movers. So, total of four states have any energy. If we were just computing crazy to the power minus beta H, we would get just a degeneracy factor of four times what we get from our actual status. Okay? However, if we were computing crazy to the power minus minus one to the F times this, all these states, if under some convention, two of them are positive from your number, the other two have negative functions because they relate each other to the action of... So, for instance, if we say that down, down carries positive from your number, then up, down, up, close carries positive from your number, but up, down, up, and up, down, that would be negative. At every energy level, because as many states have positive from your number, that would be a negative energy number, they're about crazy to the power minus one, up from minus one to the F, but minus one to the F, it just finishes. Okay, so go, calculate. But that's not much use. The quantity we want. The quantity we want. Okay? The quantity we want involves insertions of these guys. Insertions of one factor, a C and a B, a factor of C bar and a B bar. Okay? Now, as you've seen before, it's only the insertions of... although these insertions have a specific position, it's only the zero ones that matter. And then you take that, write that field out and expand, and it's only the contribution of the zero that matters. It's anything that matters. All the time I was like, okay, what will manage anything? Okay? So what? So this thing is proportional to C zero, B zero, C bar zero, C zero, B bar zero, C bar. Now, let's look, let's figure out the right numbers. Let's just look at the left numbers. Okay? This quantity evaluated of us, you see, remember that all the states in the BC system come in two flavors. There's the down, which is annihilated by B zero, and then there's the up, which is annihilated by C zero. Okay? All up states don't contribute to this binary function. I mentioned, because we have the right-moving sector. Something with down-state right-movers. These insertions, the only role of the insertions is to take that, is to calculate the same object that we had here, except to keep the contribution from only one of those four states. So my prevention, sir, is downed out. So it's a particular task for me on number of charges. The only difference that comes about from adding these insertions is that this formula gets modified, too, and you would want it. The other answer is that ZBC, the appropriate Z, we want to calculate, it's pretty good. This product, N is equal to one infinity, one minus Q to the power N, one minus Q bar to the power N, all things square, times QQ bar to the power of N by 12 times QQ bar to the power of N. It's just oscillators like anything else. Except that they're adding no energy. So all of the zero-modes they did was to generate the four states, and the projector projects to one of those four states. No, no, we see they're working on these zero-modes. You see, quantization, you're working on these zero-modes, two states, that are four, to this function, because there are four. There's multiplicity of four, for example. But since you're projecting to one of those four, it's none of them. It's just one. It's just a degeneracy. It was clear. Okay, so now let's put it all together. Let's compute z-matter, that is 70. We get, you see, that all of these infinite products, we have 20 cents in the denominator and two in the numerator. So what we have for this four, 24. We have all of these factors of QQ bar to the power of 13 by 12, the numerator to the denominator and the denominator to the power of cancel. We still have to say, is this a factor of QQ bar to the power of minus one? Yes, we still have to have this is this a factor of QQ bar to the power of minus one? And so we can rewrite this thing and then we have where we got the zero-modes, the magazine. So we can rewrite this thing as one over 2 pi to the power of 26, alpha prime to the power of 13, in down to the power of 13, times 1 by QQ bar times product n is equal to 1 to infinity, 1 minus Q to the power of n, 1 minus Q bar to the power of n, the whole thing. We're going to see it in many other contexts, and we need to use the notation. This notation is like eta. So definition, 2 to the power of 1 by 24. Q to the power of 1 by 24, product n is equal to 1 to infinity, 1 over 1 minus Q to the power n, is defined as eta. Partition function becomes, with this definition of partition function becomes 1 over 2 pi to the power of 26, alpha prime to the power of 13, in down, it's convenient to write this as its name. 1 over in down, 1 over 2 pi to the power of 26, alpha prime to the power of 13, in down to the power of 12, and eta out. Now, a small, let's undertake a small diversion. A small diversion about, this eta function was defined long ago, long before string theory, by mathematician, and it has Ginzpa said in one of his lecture notes, one of the lecture notes in which I learned all of the stuff that's given me. You might think that it was a big coincidence that Jacobian friends defined exactly the function that he will need for string theory, including the Q to the power of minus 1 by 24. But why did they do that? They did that because they were interested in functions with good modular properties. So that's the thing that I'm going to tell you about for the next 10 minutes, but modular properties of functions. You say, when you write the function of down, you've got a function of a variable, find functions of variables, but a function of renouncances. That is, you're interested in a function that evaluates to the same thing, or two dory that are actually the same renouncances. We've already seen in the last lecture that different values of tau can correspond to the same renouncances. So different values of tau correspond to the same renouncances if they're related by sl to z propagation. So they're interested in defining functions on renouncances. Functions of the space of dory that have good modular transformation properties. That is, transform it in some nice way under modular transformations. That is, under the variable change, tau goes to something sl to z. Is this yet? If at the moment, as we will see in string theory, we will be interested in modular invariant objects. Because once again, in string theory, we are going to integrate over the partition function over dory. We should get the same answer for the partition function. If the same tau is parametrized in some different way. So the full partition function that appears in our, you know, the full integral that we do has to be modular invariant. Otherwise, there's a cut down that we talked about last time from integrating over the space of all tau to integrating only over the fundamental domain. Makes no sense. If it was not modular invariant, it would have been better in which fundamental domain. So both of the mathematicians and string theorists are interested in functions with good modular transformation. Okay? Now, what can we say about modular transformation properties of eta? Jacobian friends and, you know, all of those characters used interesting and extrude mathematics work from my work here, extrude mathematics to prove the following formula. To prove that even tau to the half times eta of q times eta of eta of q bar is modular invariant. You know, this only works when you define eta with this factor of q to the power of 1 by 24. That's why I did that. That was the reason for the edification. Now, I want to give you a physicist's proof of this statement. We've already seen the proof, but I just want to, I just want to simply the following. Remember that we have argued that the path integral of the free-mode after common field theory. The path integral of the free-mode after common field theory, let's take a single free-mode. A single free-mode path integral of the free-mode after common field theory, all of those modular mathematics at tau is simply the inverse of this object. We had the zero point energy shift that is q to the power minus 1 by 24. So that's q to the power 1 by 24 on the denominator. And then we got this from the zero point, from zero point, one over square root. Imaginary power of tau. Again, every function of the torus. Reparameter is the torus of different way. You'll get the same answer for the path. Good. Diplomophically invariant theory. So that integral of this theory out of certain space makes sense. This answer must be modulated. We come back to this human being with fermions. It's one term we automatically through with fermions we try to understand why. Diplomophism is an anomaly. But we come back to that when you step forward. But for the world's understanding, you understand what I mean? You've got some space. The path integral of this theory is well defined on the space and as a function of what space you do the path integral. It's not a function of how you parameterize space. Provided that the theory is Diplomophism. Actually, Diplomophically invariant. Which the free-mode of theory is at least. Is this good? How do we get this result? From the path integral of space to find as a function of the Riemann surface? No, that's a parameter. It must be the same. How do we parameterize the Riemann surface? So this thing, all different values of tau that we do the same Riemann surface. Okay? Must be the same amount. There's one thing I've been in here in this group demonstration. It is that modular transformations don't actually do the same metric. It always gives you the same metric up to a conformal factor. So I've used not just the three-pointer problem of the field theories. Diplomophically invented. But also there is an invented conformal transformation. True statement. We understand this we understand it's a conformal field theory The non-zero central chart does make no difference. It makes no difference on the same. You're just taking two different tau that are related by whose volumes are related whose metrics are related by constant factor. So all that you require is scale and variance. Which is natural. If you have to do some one minute and a change, it might be. But for the statement, although you might hide it hard if you want to go directly using mathematics, which I'll show you in a minute. There's a huge branch of mathematics that deals with certain things. And this is the way the example of these things are. I've recently come to more, you know, forms of genius to do among circles. Some people eat them for breakfast. They're quite a lot of things. But, you know, it's a way the example of that. But, you know, it's true and anyway you might find it hard to prove it's not that hard. This might take root. I don't know. But from part of your physics, you know that from the argument we just get it. Okay, so now let's look at this. So our final answer we've written in this form infinitely I've kept in part 12 here. So let's make this thing explicitly modulated. The final part of a particle of strength here. The final particle of strength here was the tau bar and form of pi in tau. Let's come back to the value of tau of this in tau. So we get 4 pi e to the power of x squared. That's all the substance. It's a simple algebraic exercise. I'm going to leave it for you to check that this integration measure is also modulated. So what do you have to do to check that? You just do the change of variables. tau prime is equal to a tau plus p over c tau plus b. You do it now, you do it now, you do it now, and you check that it's true. It's a two-figure exercise that we work on. The original measure that we got from public form of H-fixing which had one factor in tau in the denominator was not modulated. Extra factor of in tau in the denominator, from the fact that while the VC system cancelled the oscillator part of the XCF it did cancel the zero load part of the XCF. So the XCFT, there were 26 directions. Two of the oscillators of those 26 were cancelled by the VC system. But each pair of dimensions added a factor of 1 over the denominator. So there was one overall hanging in tau of the denominator that wasn't cancelled by anything. So the in-taus that went with 24 oscillators that survived combined with those oscillators it was not even complete. Because one over in tau was hanging around that combined with the in-taus of the measure to now be very explicitly modulated. So that was very good. So we have finally now fixed our formula the formula that we are interested in that we're going to study for the partition function of the bosonic string of the tautos. Any questions or comments? To try to explore a very important property of loop amplitudes in string theory. Here's this formula to try to explore a very important property of loop amplitudes in string theory. Namely, the unified likeness of these components. Again, so in order to do that in order to do that we're going to try to really interpret this formula interpret this formula in terms of a vacuum remember that we discussed the things that we computed by computing this formula as of now we're not going to as of now the thing that we computed by computing this formula is a vacuum energy string. We're doing this at one loop but at one loop in built theory it's simply given by free theory it's this sound over a very particle that can run in that. What we're doing is the vacuum energy calculation in string theory which we should now you might think the following you might think well I should be able to do the following I should be able to but this calculation should be closely done to the following advantage I know the answer I know the answer for how to do one loop calculations for vacuum energy for free fields for free fields here and I also know that since at this level you don't see any interactions I also know that at least in some way free string theory is simply a sum of free field theories one free theory for each each of the particles of string so let me take this answer that I got and compare it to the vacuum energy that I would have got from from each of the particles of string theory running in this loop so I'll be exercising what we're going to go through we're going to try to do the combination for a particle of mass m a scalar particle of mass m running in this loop we can compute the vacuum energy and then I'll take the answer and sum it over take the answer and sum it over all of the spectrum I'll take the answer and sum it over the spectrum of particles in string theory and compare it with me now we don't, in fact these two will differ let's get to hope that questions will be getting to you will be getting to you that is not fun but all of the all of the states are the same yes yes yes you'll see I'll go ahead and take it just do the calculation so what are we going to try to do so firstly let's try to do the calculation of the vacuum energy let's try to do the calculation of the vacuum energy for now I'm going to present the formula present the formula written in a way that will be convenient for us and for this calculation and modulate it in two separate ways well maybe I should just give you so firstly if you were just doing this in field theory what would you do you see the the calculation is good suppose if you look at d5 exponential of minus del pi squared by 2 minus m squared by 2 let's put a minus outside so now what we want to do is to do the positive angle of course we are only looking at connected diagrams so we want to we want to make the logarithm of the variance then so if we do the positive angle what we get mod by mod into momentum for each mod we get a factor of 1 over k squared per cent what we actually want to do is calculate the log so we get a product of such modes so we get the sum of logs okay so keeping track of with what with what factor 2 pi is on but sort of clearly what we are going to get by 2 pi is the power k times integral of the log of k squared per cent per cent that would be more convenient for our our form of writing the sum of stuff is the form that you would get if you thought of field theory in first quantization there is a sum over our particle parts okay so first let me give you the answer let me tell you this you think will be mathematically equivalent but then we will also see so the form is 1 over 2L exponential of minus half k squared plus 2 squared this is basically the three so how do we know this is correct but so what if we didn't have any so suppose we take this expression suppose we take before we do the integral over it suppose we just look at this look at this expression and we differentiate it with respect to it we take this expression here and yeah sorry suppose we look at this this e to the power minus this is L is integrated from this is integral from 0 okay so e to the power minus Lx integrated from 0 to 30 is is this how okay so now if I am not here instead of 1 over x if I am interested in the log the integral gives me 1 over L inside the integral of the left hand side gives me a log with the right hand side that's 1 over 2 goes with that so this is an integral representation and I will be writing of this of this volume so the stress that this is that has a physical interpretation just so that you see you see what's going on here is what you are doing is suppose we were doing field theory in first quantize format that actually is the first couple of classes that we discussed that we had in this course when we were discussing how we quantize a relativistic particle in first quantize language you remember the action that we were using you remember the action that we were using had this m plus this integral d tau there was a factor of m plus x dot squared by 2 I think there was an m here so there was an action somewhat like this on the website of the free particle you remember we started with we started with the square root g with just the square root g action and then we meant to pull that off and the pull that off of the action had an additional constant here and then this thing with this is the way this is where we gauge things so that the matrix on the on the on the word line is 1 so we got something somewhat like this I can't remember the factors of 2 in my insights okay now so if you were doing a one loop calculation with this format what you are supposed to be doing is to choose every possible length see the only modulus now we are supposed to take parts of the particle to the close up in the circle the modulus in this game now is the length of how to do the calculation of this quantum mechanical system of length L this part here is the partition function of this quantum mechanical system of a circle of length L e to the power minus beta times Hamilton of quantum mechanics as far as quantum mechanics is concerned this is just the normal elitistic quantum mechanical system right beta length L is like the temperature of the system that's why I am not defining by N is this clear and then you are supposed to solve all the particles of all of the modules you will have to take a little bit of what the measure is but you know the answer to field theory is the measure of the N by N the N by 2 you would have to go through the kind of exercises we went through for string theory in order to determine the correct measure of the modulus of these parts but we know the answer because we know the answer it's the N by N for the vacuum energy from field theory in terms of parts okay it's compute the partition function for this quantum mechanical system compute the partition function for the quantum mechanical system on a circle of fixed atoms that is a very easy calculation because that's a hand-drawn interpretation of that formula it's simply trace e to the power minus sine of circle integrated over the spectrum that's what we've done we've got the spectrum integrated multi-platform okay so that's the partition function from this one to get the point of view and then it's integrated over all the module and that gives you the field theory answer for vacuum energy of the system it's quite beautiful I'm just independently varying it's quite a beautiful fact there's like a final way of thinking of field theory there's a free, say Mars okay of a free particle of Mars and looks like now suppose we had a whole bunch of free particles of Mars Mars suppose we have a whole bunch of free particles of Mars and what would we do well what we would do is to sum these disconnected components of vacuum energy so some of these vacuum energies over all Mars okay so you see we can interchange the summation and the integration and perform the summation first for the engine of minus half m squared hey we'll put that on that's all outside the summation so you don't want that outside the summation so that's the thing that makes it just okay this L has some energy with the modularity of of stream theory that's the MZL number as the modulus of stream theory was a complex it's like the LF dimensions LF dimensions of 1 over L squared okay there's the modulus of stream theory periodic to pi so no dimensions okay so the first difference we'll see how that's management in a moment but let's take the second difference immediately by performing a change of tables on this model because now we should have change of tables correct and the one that we use will so that is let's let's write tau 2 equals L times what do we get? we get tau of minus m squared alpha prime pi 2 into tau same integral to these tables and let's rewrite everything here in terms of tau 2 so we have tau 2 alpha prime yeah and here we will get well this cancels so this is just to be tau 2d this cancels the measurement let's remember what we can say about the mass square of states of stream theory you remember that the mass square of states of stream theory was given by the following formula the mass square was equal to was equal to 4l of alpha prime into L0 of alpha prime and the condition of left movers is right so as we do this we have to make sure that the left moving and right moving and lots of the same because I'll make these two formulas at the same time but we can write this formula in a more symmetric fashion symmetric between L0 and L0 of alpha by saying that actually m squared into L0 plus L0 of alpha sum over all particles L0 of alpha prime minus 2 you know now you might be tempted to say the problem might be tempted to say well between the oscillator states since that's a one to one thought response between the oscillator states of 24 of free boson system and the particle of the stream theory okay that you might be tempted to say well all you have to do is to replace the sum over particles by a trace over the input space of the 24 free bosons of the problem in theory of free boson every state of the system of the oscillator of the 24 free bosons those states that way L0 of alpha you choose any L0 you must use the same L0 of alpha once you do that you get a particle of stream theory that was one of the constraints we solved all of the constraints except this one just by writing this formula so we have to include that constraint before just doing the blinding the trace over the free boson theory that's equal to L0 of alpha now remember that L0 and L0 of alpha are integers so the input area difference will be L0 and L0 of alpha so we want a delta function something that we want to insert a delta function of L0 of alpha into this formula and you might think that the way to do it is to integrate e to the power i theta L0 times L0 of alpha integral of theta to minus infinity the only possible value is L0 and L0 by integers and what we want is not a delta function that becomes infinity when L0 is equal to L0 of alpha which is what that integral representation would have given you but a delta function that becomes 1 when L0 is equal to L0 of alpha when only allowed values are integer because there's a very convenient representation for such a delta function which is 1 by integral from 0 to 2 by e to the power i theta L0 times L0 I'll call this tau tau 1 that's called as minus pi see when L0 times L0 by integers you don't need to do the integral over because it copies what you get so that's redundant to do the integral over all the space and in fact the redundancy allows you to normalize things to 1 it's clear that when L0 is equal to L0 alpha 1 over to pi is correct so this quantity is equal to delta L0 let's post this into then do the trace over the free bosons so we've concluded that the part of the formula that was summing over the masses of particles is given by trace of trace of 1 by 2 pi 1 by 2 pi and delta minus pi by d tau 1 e to the power i tau 1 L0 minus L0 by into and then we add e to the power minus tau 2 into L0 plus L0 by now let's group together terms of L0 so terms of L0 are multiplied by so we get e to the power L0 into i times e to the power minus L0 into tau bar 2 to the power i q bar to the power i minus 1 q bar to the power i remember this trace is down only over the oscillators the trace is down only over the oscillators states so that now we just spent the last lecture about this lecture computing this trace we know what it is it's in fact 1 over 2 pi 1 over eta q to the power 24 e to the power i in fact this minus 1 is precisely what's needed to make this exactly e to the power because there's a q to the power 1 by 24 times 24 times let's go back to the formula we started 2 by I'm sorry because we still have tau 1 then we get rid of a detail 2 over the integral now we've got extra 4 pi so we're going to make it 4 pi into tau 2 and then we have integral d tau 1 we have integral d now in our case this is 26k d26k over 2 pi to the power 26 and then there was exponential of we have minus l by 2 but then we redefined l with the factor of alpha minus k squared alpha prime tau 2 by 2 and then the remaining part is what we've done the trace so over eta of q to the power 24 eta of q power formula for sum over all the part it's a straight thing but actually we have to sum it over alpha means all that remains is to all that remains is to do the tau c integral of k and I'm sure I've made the factors of 2 wrong we didn't here we're dealing with tau whose periodicity is 2 pi I think in the last class we've been dealing with tau whose periodicity was 1 oh I'm sorry for that there are 2 natural organizations in the space of atoms you know where the other part of tau is periodicity periodicity 2 pi and in the central part of tau periodicity periodicity 1 and I use one of them in one discussion in the other one in the other I'm sorry we should redo this calculation with extra factors of 2 pi give me a minute to see if I can trace the one of 2 pi also if you remember last time we defined q to be into the power i and i to pi tau that's the same 2 pi sorry sorry give me 2 minutes so the main thing that will change is that we get a 2 pi here basically we have to take out this tau and we get 2 pi times the old tau that's the redefinition okay this tau we get 2 pi times the old tau okay then I think that everything we've done is correct if we mess up the 2 pi in the end that would return to this I think that everything is correct but I'm not saying that now let's make a change of variables tau is equal to 2 pi times tau goes to 2 pi times let's just make a change of variables and see what happens so we get d tau 2 d tau 1 by 2 tau 2 so we get d 26k by 2 pi by 2 pi and exponential of minus k squared alpha by eta eta 2 to 24 eta 2 pi okay now we have the Gaussian integral so once again the factors are pi because we get pi to the power 26 by 2 from the formula for the Gaussian integral we get 1 over pi to the power 96 by 2 because this is pi here we get 1 over alpha prime to the power 13 and so we've got d tau 2 d tau 1 by 2 tau 2 alpha prime to the power 13 2 pi to the power 26 2 pi to the power 26 and then we get where is that where we got the tau 2 so that's tau 2 tau 2 to the power 13 then eta from q to the power 24 eta 2 pi 1 2 by 2 this is exactly this is 3 2 tau and next up is 2 pi any 1 extra 2 pi and d 2 tau tau is the same thing as imaginary part okay so this is 2 tau 2 square times alpha prime to the power 13 2 pi to the power 26 and then times 1 over tau 2 to the power 12 tau to the power 24 eta 2 by eta 1 eta 2 minus 6 okay after 2 pi I've lost somewhere I've gone that far and I'm going to request you to try to take us through we just do it directly, we've lost 2 pi somewhere okay but after that 2 pi I won't speculate listen I didn't okay after that 2 pi after that 2 pi is even more exactly the same for me part with 1 key difference and this is the main point of this lecture the 1 key difference is the range of integration of tau this is the range of integration of tau in this model remember tau 2 was tau 2 was what was added because integration was 0 without new or minus half to half in that because the PDRS is one so it's a main one so let's look at these two integration ranges in string theory we're doing the integration over this switch integration between the integration over this switch so it's almost the key difference difference in integration though let's make the difference in integration okay now we're going to have to stop this lecture so let me just tell you the words we'll explain this before we can ask later but what the region tau 2 going to 0 was in this particle language the limit of very small particle parts it is the region of the integration space over one day life that gives rise to the ultraviolet divergences of quantum field whether it's the same calculation here except that the region that was ultraviolet divergences just does not give the big fake to say that in the PDRS as we will explain exactly interpreting the answer in which the UV divergences just does not so we're bringing it so you see you've hit the key part suppose we were to go to this guy then we would be approaching here so you can ask what is the map region of this domain there and since this is related by tau goes to 1 by tau you see it's there so all of the potential UV divergences of this picture have been pushed away to the higher distances in this region so we see a fundamental property of string activities that we will see again and again divergences from one point of view are ultraviolet in string theory and then we're related to something infrared in the infrared as we will see in the next class that's great because infrared properties are physical so any issue about a divergences potential divergences in string theory is related to some infrared phenomenon which are