 I'm updating the videos a bit. Now finished yet, but OK. So if you think my look, I am updating the little videos that move the journey money. Link is in the description of the material panel in these lectures. It will be more useful if you want to look it up. OK, so let's get to you. So today, the first thing we're going to do, so you remember last time we put this no-boss here, the no-boss here in the photo of the world of the street. And the statement of that theorem was that if we look just at the first one-times stream with the Maca staff and then VC posts, we had many states in that department, really. And these states had negative norm, positive norm. However, if we focus attention to the cobalt energy at the behind-sv operator, and we also focus attention on states that will be not on state equal to 0, then there was a cut down from the states of 28 oscillators for states of 24 oscillators. In particular, we showed that this cobalt energy was the one-to-one correspondence with the Hilbert space at the line-code one-times stream. And we also showed the other products on the cobalt energy, which we're trying to do because any state with 0 norm, any state with qx act as 0 in a product with every number of states. And we also showed that the inner products in the cobalt energy were only one-to-one correspondence with this cycle one-times stream. So I'm going to start today's lecture by just giving you the example of how this works. And again, last time's lecture by pointing that in order when we, in the argument of cobalt energy, we actually found representable elements of every cobalt energy. We actually constructed a vector and a state that was in one-to-one correspondence with the states of the line-code one-times stream. And we noticed that that state had no n minus b of c oscillators. In particular, no b of c oscillators, exactly. So using that observation, we basically argued that there was a way of choosing elements in cobalt energy such that they were in the bc ground state and only the matter stuff was excited. Now, in this restricted class of states, we still have an analog of cobalt energy problem. It's not quite a cobalt energy problem, it's something like that. Because this restricted class of states still has 26 oscillators. And we know the right number of oscillators is 24. It has already said so far. In my restriction, I'm not imposing n minus b of c oscillators. So we select 26 oscillators, the right number is 24. So this is a physical subset of the whole thing. Because this course stuff, in the end, is some jazz, some funny cobalt-gain tricks in Christmas. It's the other one, conformity theory of physics. So it's nice that we can deal with representative cobalt energy that have nothing funny going on. Because it's just everything interesting happening with the matter. But already, even now, we have some conditions. You remember that we showed that the states of QA are still closed. It implies that ln on the side equals 0 for all n greater than 0. Whereas l now is just the matter sector, l. Except for l0, which is l0 matter, minus 1. When this minus 1 came from the normal order constant of l in the ghost area. The fact that it closed back from 0 equals minus 1. So we show that this has to be true. So that's one set of conditions. Rosalind has one condition for each oscillator. The other is that cluster 1, 1 oscillators. So that takes 26 to 25. What takes 25 to 24? And the point is that even in this set of states, there are many states that accumulate that. Now, these states are Q exact, but they're not Q of some state in this set. The Q of some state outside. So it's because consistently this set of states are analyzed over one meter. However, what we can do is to use the property of Q exact states, which was the Q exact states for always now. And all states that are null by the normal order of theorem and that are also closed are Q exact. And let me realize that any state because of the form l minus n on the side was null. And in fact, it was also null for every physical state. So these are the states that were identified. These are the Q exact states. And so in this physical representative set of states, the analog of Q exact states are states that have descendants of some state. The Q closed condition is simply the condition that the states be primed. States that are both descendants and primaries are physical, but not certain. They are the OK states. OK, so now I want to very quickly lead you to a little computation. And the lowest and first level of closed string theory to try to examine Q-core modality, we first do this in the physical sector and then also do it including the v-sequest. So we first do it in the restricted set of states. OK, so this restricted set of states, now let us look at states at a level z. See now this restricted set of states, the only oscillators we need to care about are the matter states. Again, that's the same because b and c is a matter of vacuum. OK, so level 0, what we have, the unique states, it's a vacuum, more precisely, that's a unique state up to zero mode motion, zero mode stream. The state labeled by 25 momenta medium. Now the question is, what is the condition that L, that is physical, that L a, and hiding the state from L? Well, let's remember L n is a product of oscillators whose index numbers add up to n. But since the n is positive, not zero, that always means that there is at least one unpaid oscillator. For instance, L 1 will have, we go like alpha 1, alpha 0, plus alpha 2, alpha minus 1, plus 1. And alpha 1 will have hiding the vacuum. Alpha 2 will have hiding the vacuum. And so at this state, it's annihilated by all L n's by most. We get the famous motion of the front piece, square root of 4, plus 1, or minus 1, to zero, to zero. The minus 1 and 1, you're getting the difference, the zero mode, the energy of the closed line. And we obtain it by L minus n of some other state. The answer is no. Can family suggest now? What is the, suppose it was, what is the relation between the energy of L minus n on psi and the energy of psi? This k energy of psi. n plus the energy of psi. Exactly. L minus n raises the energy, raises the active value of L naught by n. If zero was L minus n on psi, then psi must have energy any of the slower than the energy of zero. There's no oscillator state. There's no oscillator state, which has this. That's just very impossible. At this zero state, the lowest sector, we found that there is one state that's physical. And that it also has never to be exact. It can never be written as L minus n or something else. OK. So this one state is less than 1 state. It's the cat. Of course, we would have had the one state also applied to the one state, so that's good. Now let's move on to the first excited state. So the first excited state. The first excited state, the various states are a mu alpha minus mu minus 1 of mu. At level 1, the state doesn't exist. It's just less than 1 state. It's actually the same story at the time. Is it going to be times any ratio of the state? Oh, OK. Again, I'm just asking what division we can say on the left moving part of the state. There's immediately the conditions of the right moving part, and then the level of the state. The level of the action actually was right. Yeah, got it. That's basically automatic, because level of action is L0 equal to L0 half, but each of L0 and L0 bars will be sectoral in this part. So the level of action part of it will be really automatic. Let's get out of it. We'll get out of that. OK, so just let's look just at that. So if there are any mu alpha minus 1 mu 1, OK. Now, first question, what constraint can we get from the fact that the state doesn't exist? Firstly, there's a constraint on L0. But now, the L0 constraint just tells us that alpha minus p squared by 4 is equal to 0. So plus 1 from the fact that we are acting with oscillators will be 1. p squared, alpha by p squared plus 4, plus 1, minus 1. So alpha by p squared plus 4 is equal to that. OK. Now, we're going to get at least oscillator number 2 i, L2 with alpha 2, alpha 0, plus alpha 3, alpha 1, plus 1, plus 1. The only issue is 1. The only interest of N1 is the number of the form, alpha minus 1 alpha 1 mu alpha 1. See, the only thing that is not automatically acting on this state, and what do we get? See, alpha 1 heats up the alpha minus 1 mu, and we get something proportional to alpha mu, like alpha 0 mu, but alpha 0 mu is i equal to 4 to p. So alpha mu, p mu, 1, plus L1 on the side, L1 on the side, is proportional to 8. I can prove it out of this state. 1 on the side will be equal to 0. So that implies that alpha 0, right? Exactly. Now, what is the interpretation of this condition? The interpretation of this condition is simply that every on-shared state of a photon or a graviton is transverse polarization. That's transverse to the direction of moment. We have got, like, string theory, the only computing on-shared space matrix, and that's basically the interpretation of that. OK, good. So how do you understand the physical state condition? We've understood the physical state condition, the condition that L n highlights the state. But what about the condition that the state does not be at L exact? OK, so if the state were to be L n on some other state, tell me what L n must be and what that other state must be. Suppose we had a state at level 1 that was L minus L n on some other state. What would that state be and what must the n do? L minus 1 at the order of. L minus 1, and what was the state be? Ground state. Exactly. You see, because this state has energy 1. So it's the second lowest energy in the theory. So the only way it can be L minus n on some other state is if it was L minus 1 and L minus n on a state of lower energy. There's only one state with lower energy. It's the ground state. But when I say energy, I mean the oscillator part of the energy, just all that these L's can never get about. Right. So the only thing it could be is L minus n of the ground state. Now, it has to have energy 1. Ground state has energy 0. The state has energy 1. The only thing it could be is L 1, L minus 1 of the ground state. So let's work out what L minus 1 of the ground state is. So L minus 1, the only doubt that contributes is that there are no annihilation operators on the ground state. So the only doubt that that's at alpha minus 1 mu, that's alpha 0. The such a term acting on p mu, it gives you its proportion to p mu alpha minus 1. The such a state does exist. It's the state with a mu proportional to p. The state would be of the form L minus 1 of the 0. Now, is a state with a mu proportional to p mu is such a state of physical state? Well, let's see. The physical state must obey the condition a r p is equal to 0. But a mu proportional to p mu is a p r p is equal to 0. But that's true for the partial condition. p r p is equal to 0. So such a state is a physical state. However, it must be moded out. It's 0, I must be excited. What's the interpretation of such a state? Such a this statement? Well, in the angles here, you see, if we did this with the open stream, which I've not discussed in class, what we'd be getting at the massless level is the angles here, four times. We do it for the closed stream, we'll get it out of here. But I'll make the statement for angles here because it's more familiar. The statement for angles here is simply that a shift in a mu, proportion of the p mu, is generated by gauge concentration. a mu goes to then a mu plus del mu chi. It's a gauge concentration. If you take two different photon states, they should be the same physical answer for gauge transformations. That's what we're saying here. That gauge transformations in the external states, the difference between two states that is generated by gauge transformation, is a state that is q exact. And therefore, external states that are different than before only by gauge transformations. Our world shift states that before only by q exact, and therefore are the same scattering amplitudes. So we see this work not very nicely, and it's not as clear as we expected as it should work out physically. So at the first level, we lose one polarization because of physical segregation, and we lose the second polarization because of state equivalent. You see how nicely space-time concepts that we're familiar with tie in with these world shift concepts that we just did at it. Now, though we have an abstract proof that this works so hard on this, we're not going to bother to try and check it at higher and higher orders. It's worth it while you try to do so. I mean, I'm not going to do it in class. It's worth it while you try to do so at level two. You'll find that every level there are states that are killed by the physical state condition and states that are killed by the non-active condition. So the fact that we have an extra non-state at every level tells us, using each non-state's correspondence to gauge, the fact that we have these extra non-states at every level tells us that stream energy has much larger gauge in that it's the lower gauge in that field. The non-states in the mass of this level are the gauged variances we're familiar with in the entry-level field. But it's very much larger gauged variances. It's not that a very intriguing statement is certainly correct in some sense, but it's never been put to any concrete, perhaps except for open string theory, but it's never, I mean, it's not been as useful as statement to us. It's not, this must be a deep statement in some way. It's not the net concrete or error. Used to say something. Very interesting about string theory, to my knowledge, yet. There's probably something to it down here. There's some discussion. Okay, let's talk about this one just a bit. You've said any statement that if you want to have some gradient values, you should have so much. Yeah, I mean, you should have so much of it. But I don't mean. That's as I go. There's probably been people who consider and discuss this question, okay? I'm not aware of the discuss, but I'm not familiar with this. For example, can you just, you know, like, though it is not possible to do anything, can you just stick to something that's been studied, that's what they do, whether they can get something in history? Yeah, you could try that. Well, you see, if you do that, as a non-interactive method, that would then be consistent, because all these different states are physical, they don't talk to each other. But once you put internal interactions, nothing good will happen to that theory. So, I mean, I don't think there's some general statement that if you don't ask the good properties in the end, that you need another state to answer that. That you need to answer this. But somehow, if you want the theory to be good in some sense that you have to make precise, you know, that's unitary, that's where the fire, blah, blah, blah, all of that. Then this may be required, but I shouldn't say much more, I don't know much. I mean, I'm sure there's been an interesting discussion on this question. I'm not really aware of it. But I don't think I can tell the concrete stuff. So all I wanted to say about the implementation, so this is how the no-post procedure works out about, you know, level by level of strength theory. Okay, the exercise is trying to get analyzed into at least one eye order, which I stopped here earlier. But what we're going to do now is to try and now move on and see how we can do the same thing, how we can do the same thing, just for, I mean, for the full reverse phase of the worksheet theory, including the BNC course. Let's look at the full input space, including the BNC course. So what do we have to do? What we want to do is, now, because the states are Q closed and states are not QF, first, start at level C. Now, according to this BRST operator, okay, the current corresponding to BRST operator was C times D plus, okay, half straight, C j's. Let's already start. Let's already start at the, let's already start at the, at the Loster. Two-quadrucent as Cn L minus n plus, blah, blah, blah, contributions coming from the previous state, from the previous state, right? Okay? But if you see that, if we can't, if we can't overall it, if we can't both matter as well as, as well as a closed level, Q does not have a Q as a zero. There's a Cn in these n minus, at Q on a state, it doesn't change the level of that state. So this discussion of Q-cohomology is in some sense, even easier than the discussion of the L-cohomology. Because if you maybe have a state that is Q exact at some level, it must be Q exact, it must be Q acting on a state of the same. Because Q doesn't change the level of the state. All right, so let's, let's try to solve this one. Now first let's work with level zero. Level zero as a form is only one state, it's a mu, okay? Now it's just zero, because either, either C is an, it is an annihilation operator, or L, the only place where we have neither C zero or L zero, but L zero on the state is here. That pops a little bit. This part is L zero matter, but when you, when you act on this part, you also get C zero times L zero cost, okay? And it's L zero matter minus one, which is just zero. Okay, when you work that out, definitely. This part, which is Q acting on P, is simply zero. Now the HGC test has the answer to go home on the state of the state. So what's the answer? We don't need to do a second calculation. What's the answer? I will give you a second one. So what's the answer? Somebody? This is a physical state, how do we know it's not Q excited? Yes, yes, yes, yes, yes, yes. Make that precise. Of some other state, what would the level of another state be? So it should be the same. The same. But we found that every state at level zero, it's only one. But okay, every state at level zero, is a very little non-trivial state of the form Q of something at level zero. This state is physical, it's also not pure of each. So say it's a state state. So they can be A times C minus one plus B times alpha times C minus one plus beta times B minus one plus a mu alpha mu minus one on P. P one percent. So let's go to the side. Now let's see. The kinds of states that we will, they're in doing the calculation. There are two kinds of states that one could keep track of. There are states which had a C zero, have an explicit factor of C zero. And there are states that don't have the explicit factor of C zero. Now we should do the explicit C zero must end up canceling each other once we impose the physical state definition, once we impose the physical energy definition. Can somebody tell me why? I take this answer that the, that if we impose L zero on states equal to zero, which of course we follow from given variance, as we can see. The entire contribution of C zero, all states corresponding to C zero will just vanish once we impose L zero on states equal to zero. This is something we saw last class, let's have a reminder about that. So remember the state that has an explicit factor of C zero in it. It's not annihilated, but it's, okay? Why do you remember that we saw the set of states which obeyed L zero equals zero, L zero minus one then? Normalized L zero equal to one. And obeyed B zero equals zero. What closed up the queue? Human class that L zero with Q was equal to zero. But L zero with D, now I'm having to sign it up. Sorry, I don't remember which way it was. It was equal to L zero. We saw that from the Q, B, O, B, remember? This is the whole L zero, this L zero matter, customers. The problem of Q homology can be done sector by sector, you know, you can consistently restrict to states that are annihilated by L zero by plus G and V zero. In particular, if you add to a state which is annihilated by B zero, with Q, state obeys L zero, the state that you get, sides if B zero on side equals zero and C zero on side and L zero on side, then B zero on Q side in this calculation, we get any contribution proportional to C zero. That contribution proportional to C zero must be a multiplying factor of L zero. So the only role of things proportional to C zero will be to impose the L zero equals zero constraint. Let's do it by calculation and say this because it saves us information. The state's proportion to C zero are a little tricky to do. I mean, they're not tricky, they're a little bit of a pain to keep track of. Other states are totally straight. Computing Q one side is equal to whatever. I'm going to ignore all states that we get proportional to C zero. Just knowing that the thing proportional to C zero must be proportional to L zero. L zero matter is L zero most. Now keeping this in mind, keeping this in mind, let's do the calculation. At Q one of this state. So what terms in Q are important? Well, we have the C times T. So we'll have terms like C minus one L zero one. And if we were to keep track of states when C zero equals that, terms like C zero and C zero. But we're not keeping track of the state, so just forget it. You already see that. L zero is a variable, that's easy. This is a matter of part, the most part that we'll be able to go through. Okay, so please understand. This is the only thing that will be of interest to us in the part of Q that comes from here. The part of Q that comes from here. Again, since we're not interested in C zero, we have to answer the equation. Because levels must add up to zero. We should never get more than one. Okay, the levels must add up to zero. Should never get more than one, so what do we get? Either one of the C's is a zero, or one of the B's is a zero. Because the levels must be zero, one by two, okay? But B zero, and that is the state. Not keeping track of the states that are C zero. So none of those states add up, none of those terms add up. Is this clear? So the only thing we have to worry about is this, this is the only part of the operator Q that will be of interest to us. Apart from getting L zero equals zero condition, which we know is way to go. So, we've already argued. We have Q on this, we only have to worry about this. Now, what do we get when we worry about that? So first we, there's also here the, I forgot, of course there's the, in the matter part there's a C one and L one. This is also a four, right? We have to think all of the things where nothing is more than one. Because it's one with the C one. With the C one, okay? So now let's, suppose we add this on this. This term is zero. But this term gives something out of zero. It gives us, it gives us, oh, how? This C one acts with a B minus one, okay? So, on this we just get C one, okay? So all of us, if there's a mistake, with this B minus, we get a factor, we get some state that's proportional to, and not if you have a sign in the factor of two, this wouldn't matter at all, okay? So we get a state proportional to B times L minus one, on B, we will get a state when this L one hits this alpha, this one. The only term in L one that's of importance as we saw previous times, is alpha zero, alpha one. Now we'll make it, the alpha zero will contract with a mu, but alpha zero is P. So there will be some number times this plus some other number times P dot alpha into C minus. And if you work it out carefully, including also the C zero terms, if you work it out carefully, including also the C zero terms, you would get some contribution which has C zero and then something else, something else, some factor, basically. So this is the action of Q on states at level one. Now we want the state to be physical, that can be annihilated, okay? Therefore, we first demand that L zero is on state zero, that's martial condition, fine. In our general discussion last time, we said we would only do the states in a way that are martial condition. So let's omit that problem, okay? So there's a martial condition. But there is also, now, the fact that B times P zero, you see these states can cancel each other. So those are C minus one and the first L minus one, which is matter, L minus one, is totally different. So beta must be zero and beta are all of us. Physical condition, signal is exactly what we saw before. But in the initial, we have this beta. It's the same. In fact, that's where it came from, right? C minus one, sorry, C minus one was multiplied by L one, L one acted on that state, okay? So that's the sense in which, you know, on those other states, C i is, can just be followed by spectrometers on the first set of states. So since the position of every C i must vanish, that's where we got Ln variation. So this must be a subset of the positions, right? As we see. Yes, this problem we've just been taking our 28 hours and they present and cut it down to 26, okay? So we must also have some other states. So now let's compute which physical states are none. Okay? So are there any physical states? So states of this form, but we can do better. We actually go. Any physical states which are Q exact? Well, we've already done the calculations. Tell me which state, which physical states are Q exact? So what condition are alpha? What? Alpha equal to theta, proportional to P. Say again? Alpha equal to P equal to P. Alpha proportional to P. No, sorry. So the states which are operating by acting with this way of Q are so these two are. No, what was that? L minus one matter of P is equal to alpha minus one alpha zero of P, which is P dot alpha minus one. P mean alpha minus one mu, one mu acting of P. Okay? So now can we, what we said was completely right. Let's say again, what are they? So there are, there are, there are, there's two dimensions like this states of Q exact states. Tell me, describe these two dimensions like this states. There is one K-sharp solution which is there. Which is this one? Which is this one. Which means in terms of A beta and alpha, what is that? So path of second beta is zero. Alpha equal to zero, beta equal to zero, and A mu is in proportional to mu. That's one of the vectors in this space. Okay, what's the second vector in this space? And what is that? In terms of the stuff? K of A beta is zero. Alpha is, you know, there's two P of A. I'm convinced you like, sometimes writing alpha for P. Yeah. I think that's it. So yeah, alpha is P of A, and P of A. Now for what's P of A? What does that mean? But why? You can get any value of any real number by doting and arbitrarily in with an arbitrary P. But with an even P. It's P of H. It's Q exact. Okay? So any value of alpha in a combination of these two. So Q exact states are alpha C minus one plus P mu alpha minus one here. Or, say, in one. So the Q exact states are these. The Q exact states should be of how much they are equal to equal to equal to one. Can you, can you, the only option that is between level one and level one, can something argue for me that these states are indeed of how much they are equal to equal to one? C minus one, we're equal to zero. Yeah. So this is Q exact. The physical state for the equal to, let's call it alpha. Little plus C minus one, plus let's call it B mu alpha minus one mu on P mu, where alpha of P is equal to one. Okay? So now that it's obvious that it's what you said. To our knowledge, no B, but there is no B, but we're going to be equal to zero. Exactly. Sir, because that's what the absence of B much for every physical state, this guy has nothing to contract with, so zero. And similarly here, the only thing, this one contracts with this, but it was a P dot B in the contraction, and P dot B is equal to zero at the physical state. So states that are Q exact are common to every physical state. Equally, by the term of the states, this is the set of Q exact states, we're just computing the matrix of enough products on this 26-dimensional state. This place, we would have found two non-states, everything else looks like. And then we've got the same onset of four. Because although physical states have a C minus one, you see C minus one on the sphere range. So you can variably go from one to the same to zero. And then physical states have transverse momentum, but moving the amplitude by fact at the bottom of the momentum gives you zero. Okay? Fine. So that's all we're going to say about this, our post-scherm and quantum energy itself. And now we're going to have a discussion of the scattering answer to you. I'll let the arm roll into knowledge. But any questions or comments? Okay, in this exercise, what's doing it when I know the sign that you get the same onset of C? All right, that's the scattering answer to you. This discussion is largely of whether it's not that deep, but we should go over it just so that, so that you know the scattering amplitude has on the consistency of these elements. First, let's remind us of the formula of scattering. So we have scattering amplitude of various states and an integral of a monuniline, okay? Then had an integral for the position of all the unfixed vertex operators. For the i not belonging to fixed vertex operators, okay? Okay, then we had a product of, and let's put a factor square root g, the matrix. So, now for each of us, they put the right measure. Then we had a product over i belonging to the fixed vertex operators, okay? And we had a factor of c and c tilde. C and c tilde for each of these operators at, and then we had a factor of the vertex operator. Then we had a factor of the vertex operator, these factors of b del k g. These factors of b del k g for every modules. And then we have the exponential of minus x, and find the formula for the scattering amplitude. There's been some time since we wrote this down before. So it's every term here familiar. So there's any term here that is, that you can't remember what it means. Please tell me, we'll elaborate on it. The k g was? The k g was the derivative of the matrix with respect to the variation of monuniline, of the kth modulus. So this k, okay, so good question, sorry. i is equal to 1 to m, that's the number of monuniline. And k, there's a product here, the product over k. k is equal to 1. That run bracket meant the inner product, was the inner product of the space of the product. That we defined, right? This, the difumofically very inner product between b, which has two indices down, and g, which has two indices down. So the inner product was b alpha beta g alpha a g beta b, that it gave g, kb, where g is the integral of the space. And you remember that if we chose a matrix that was locally widely proven to holomorphic matrix, then we were stressed this, which meant that it's only now zero components for bzz, and bz-paz-z-pa. Also remember that b and c, but b was taken to have lower indices, and c was taken to have upper indices, who wanted to find it. Remember, I'll also remind you that that's the cost proportional to alpha beta, but where would g be alpha beta? It's operating at p, elevators minus p. p on c was del mu c mu minus del mu c mu minus g mu del. So p is an operator that takes an object with one index to a traceless object with two indices. Okay? p and c, you get an object with two indices, it's traceless, and then you just take the inner product. Let me write it. So let me write this. p and c is the two index object, you take usually the inner product. So when you get a flat space, a flat space, what does this, or any more, any wider space? You remember how this is inner product go? Well, in that space, we got del, second form, or c, z, okay, that's what it's like. So let's actually work out how this works, when let's say we have a flat space, and we've got complex coordinates. So there's no square root g. So there's a b, z, z, but this has a g, z. Now the only couple of points are z bar, g, z, z bar, and then there is del, z bar, but that's a flat space which is del, z bar, of c, z bar, d, z, z bar, right. So this was b, z, z, del, z bar, c, z bar, which is the same thing as b, c, z. Okay, and this b, z, z, then we call it b, c, z, which we call c, okay? And we also check that we got all of this out, that when we took the variability like this, this actually was actually what it was like. You remember that if we took the metric to b, each of the bar, five, 10, 10, 10, 10, 10, but you got the same answer for this one, which is b to the bar, five, six, seven, eight. Okay, there you have it. So, okay, just every term of this on the backboard here, v is the number of the formally-filling vectors. It's the number of agents that are not yet fixed, m is the number of moduli, it's the number of directions that, extra directions that you have to integrate over the extra, the space of metrics that couldn't be reached from a particular choice of metric by the formal transformations. My wife. My wife. Okay? Questions, comments? Well, there's a single c from let-and-go-vers and a single c from right-and-go-vers. The, the, there was, let's see, so if now you want, basically, you see there were two diffeuomorphisms we had to, there were two additional diffeuomorphisms we had to fix. Right? You remember where the C insertions came from. They were far, far the pope of posts for gauge fixing, putting operators at particular points. Now, we're putting operators at two, at a point of, points of a two-dimensional effort. Okay? So there were two, there were two, there were two dashed functions, two, two conditions from that. That's, that's what, that's what the, the two, the two things. There's one, there's one c for every unique, every component. Good. Other question, comments? Now, what I'm going to do is take some various properties of this object. Okay? So, hey, in order to do that, I'm first going to tell you about, so let me keep the formula at the moment. Okay, so in order to do that, let me first tell you about another representation of the, of the course part of this formula. Okay, so first here, also now, what are v's? You see, we have a much better understanding of what these v's are, than we did when we, when we, when we threw this calculation last time. Okay? You remember where we got all this from? Where we got all this from was by scattering physical states. But there's one physical state for every element of, of the RSD code. Okay? In the simplified way of thinking of it, there's one physical state in the C-Ghost vacuum, in the C-Ghost vacuum, and corresponding to every primary matter operator. Corresponding to every primary matter operator that is not itself, like I said, so the v's here should be thought of as primary matter operators that were not descendants of something else. These were descendants of something else, that would be fine, it wouldn't change anything as we would say. So, so more precisely, we use any primary matter operator, but two different operators that differ by a descendant, you may be of the same kind of descendant. Doesn't matter which descendant you use. Okay? Before we go on studying this formula in more detail, let's look at certain, certain aspects of it. You see, what's going on here, we did this in self-generality, but now let's look at the answer. What's going on here is that those operators that were fixed to points, those operators that were fixed to points, also have a factor of C from the left to C to the left of the right, and I'll tell you about it. You see, that's so satisfying, this factor of C and C till they're satisfied from another point of view. And that point we use the following. If you remember when we worked on the BC host system, we worked out the state operator. And one of the things that we found was the operator dueled to the vacuum of the theory was what? It was C and the left, and C till the left, and the right. So if we did it here, it's precisely inserting the operator, so the fixed operator. You're precisely inserting the operators that are dueled to the special states that we consider in today's discussion at the end of the last class. Maybe the operators that are on the coast vacuum, the operators that are on the coast vacuum, and the operators that are on the coast vacuum, and for an arbitrary natural state that is simple natural state. Our procedure is to do that, but we should do that for some of the operators, for the remaining operators. We don't need this factor of C. We don't need this factor of C that they will be integrated over the word sheet. This may worry you, you may worry that this operator will now not be VRSD embedded. Because we know that something of the form that is C tends V, where V is a matter of primary, is VRSD. Check that. There are these Vs without the Cs and so on. All these things are, this is the kind of thing that today, that this lecture is going to be about. We'll clear up all these ones. So let's go systematically. The first question we're going to ask is, well, I'm involved in the integrity of our modernized space. But nobody has told us no God-given measure. There's no God-given measure on modernized space. Or perhaps there is, but it hasn't entered our discussion. So with what measure are you going to integrate over modernized space? This looks like a serious question. What does this formula even mean? If you don't know how to measure how you're going to integrate over modernized space. The great thing about this formula, as we've seen, is that it's invariant under shaker form that some body of space is. Okay, so that's the first thing we're going to, we're going to try to understand. All right, so invariant under understand that, we're going to take the coarsest part of this path and take it and process it in a way independent, right? But a white product would be this understanding. Let's see. Okay, what we do is ask the following question. Suppose we look at this coarsed action, okay? Of taking the B and C fields and expanding them in an eigen basis of something. You know, so what I want to do is the equivalent of, the equivalent of a Fourier transformation. The equivalent of Fourier transformation for a free field theory on space, that's best. What if the ends are a Fourier expansion? B ends and C ends are a Fourier expansion. That's exactly right. You see, now I'm trying to be completely general. We do this on an arbitrary manner for me. Okay, so this, so exactly, so what we're going to do is, you see the B ends and C ends have property that they have good transformation properties under the quadratic operator, under the operator that you get, well, hang on to five, see what we're going to do and you'll see the value. That's, it will reduce to that in flat space, but it's okay. Okay, see, this is operator P. It's an operator that pays a one in the X object to a two in the X. But this is it. This is an operator that maps one space to another. So it's not some sort of square matrix. And it's not mapping one space to the second. However, we have neither product defined in the space of object in this case. Okay, consider the operator P-tiger. P-tiger is defined by the, by the, by the form of delusion in one. Okay, and it's easy to find an explicit formula for P-tiger. It just takes a derivative, an appropriately lower derivative, and contracts away one of the indices of P. So, we define this, this, this, this conjugate operator P-tiger. Now, let us consider the operator P-tiger P. The operator P-tiger P maps objects with one index to objects with one index. Because P-tiger is the object with two indices back to objects with one. So, it's a square operator. And it can be diagonalized. But it makes sense to consider the ideal functions of, of, of this operator. Let's suppose that we have some, this operator has some ideal function, let's call it CN, with ideal value of it. It's the, it's that square, that space. Yes, exactly. But, you see, we want to make this, this discussion as general as possible, because we want to apply all of the assumptions. So, I don't want to specialize in that space. This will be a left-wing of that kind of thing. If we look exactly what this is, you see it's, this is a left-class CN that we, right, it will be essentially left-class CN, but acting on vector, or acting on vector fields. Because CN is, so it's not a scalar. Okay, we've got this, we've got this equation. So suppose CN is equal to P-tiger, P-tiger P is CN is equal to Gaian CN. Now, by and large, similarly, the discussion that we had for P-tiger P, we can have a discussion for P-tiger P. So let's let the N run over all eigenstates and all eigenvalues of the operator P-tiger P and the operator P-tiger P. Okay, now I want to show that there is a one-one correspondence between eigenstates of the operator P-tiger, and eigenstates of the operator P-tiger P, provided that the eigenvalues of these operators are not even the same. So P-tiger is always acting on the vector space of the operator P-tiger. P-tiger acts on the space of the operator P-tiger. Okay, so what? So these are the P-tiger problem, there's a square matrix on the space of human life object. Essentially, we're not asking this, okay? There's a square matrix on the space of one matrix. There is a solution to this equation. But the P-tiger is not an eigenvalue. The P-tiger, I think it essentially is. P-tiger, P-tiger, one of the matrices was one character. Yeah, you're right, you're right, you're right. But let's see, how about it? It would be sort of, P-tiger would be there's a mu of, mu of. You find that we get mu, mu, but we get mu, mu, mu, mu, mu, mu, mu. Yeah, you're right. It's not, you're right. It's whatever you get by doing P-tiger. Sorry, I will not say. Wait, is the other one also like P-tiger? Yeah, but that, no, no. And then this other one we don't contract. Right, you don't contract. And when we now act with the other one, with this we will say it's something like there will be, there will be a mu of minus trace. Yeah, you're completely right. It's very, very different from the plastic. In the sense that the indices, one index is contracted with the other index is free, rather than the other index is contracted with. Now that's what P, subtracting with a trace, this is what P-tiger is. It's this dimension. This is a P-tiger P as on CN is equal to gamma n-z. Suppose there's an eigenfunction of P-tiger P that obeys this equation. Now let's take this equation. Okay, I'm rubbing this out a little. So let's take this equation and act the both sides of it. Operators associated, we get P-tiger on PCN is equal to gamma n and PCN. So provided that PCN is not equal to zero. Provided that PCN is not equal to zero, every eigenfunction of P-tiger P much of the non-zero eigenvalue of P-tiger P leads to an eigenfunction of P-tiger P. This is here. The other reverse is also true. And between eigenfunctions with non-zero eigenvalues of P-tiger and P-tiger P, P of CN was equal to zero. Suppose P of CN was equal to zero, then this thing would be meaningless. Because this is just an eigenvalue equation which is obeyed by zero eigenvalue, which doesn't mean anything. Okay, so there's no statement about correspondence between the zero eigenvalues of the operator's P-tiger P-tiger. So if we look at the operator P-tiger, the operator P-tiger P, its spectrum of eigenvalues coincides with non-zero. But these two operators may have different numbers of zero eigenvalues. Okay, so it just seems like... But we can do it the other way round. The eigenvector of P-tiger P, this is proved with a gamma and an angle of zero. This is proved that there is a corresponding eigenvector of P-tiger with the same value of company. But now, this is how you perceive the selection. We can repeat it by starting with P-tiger, of eigenstates with non-zero eigenvalues of these two operators. In fact, let's be more specific. So let's say that the eigenvectors are C-zero, R-th, and R is equal to one to V. These are the zero eigenvectors of the operator. And in addition, there are C-n, okay, so the eigenvalues are right. And let us choose these C-n's to be normalized under the norm, this bracket norm that we've been talking about. The bracket norm is just contraction, multiplied by square root G to be equal to V. So, these things are normalized eigenvectors with eigenvalues combined for the operator P-tiger P. Let us see if we can now try to formulate using this result. Oh, by the way, something I shouldn't say, because this operator is, you know, this is a commission operator, actually. So you take the back of the center again, you take back the center. And also, it's a positive operator. The important point is that P of C-n is equal to zero is equivalent to P-tiger P on C-n's. Is it from that point? P-tiger on the side. So I want to show that all eigenvalues are positive. Okay? Yeah, so I just take it up wrong, you know, negative positive then. Let's see. Yeah. Okay, so what I want to say is that if the only thing that P-tiger P-tiger P can be zero, it's a P on the side of C. This is the usual value, right? Because P-tiger P on the side equals zero. So let's contract this side and use the positivity of the inner product. So this is the inner product of the state P-tiger itself. The only way this can be zero is the P-sizing. So there's two ways. When I say P is an eigenvalue of the operator P, P-tiger P vanishes. That's the same thing, the same P gives the operator. Okay, good. So these are the zero modes of the operator's P. And then these are the other eigenvalues, the non-zero eigenvalues of the operator P-tiger P. If the non-zero eigenvalue states, we use this instruction. We have that P-tiger P, P-tiger on P-c-n, as you see this gamma n-c-n, gamma n-p-c-n. So let me call, let's say that V-n is equal to some alpha n, some normalization that I don't know, times P-c-n. Let me determine some normalization. So let's compute the norm of P. So the norm of V-n is the norm of P-c-n-c-n, which is the same as C-n, P-tiger P-c-n, which, what, do you think we want this to be the same as this? Isn't it that... Because these are the products you do with C-n-c-n. Right, right, right. You know, P is equal to the dagger of P, and P is the dagger of A. This is the definition of C-n, gamma n-c-n. We chose gamma, c-n to be unit normalized. So we choose alpha n to be one time gamma n. And when we choose alpha n to be one time gamma n, this V-n is also equal to P. So V-n is a basis in the space of this other guy, which is of Pp-dagger, which is unit normalized. And let's also write down the cost, the cost action. Well, this is leading to a point, by the way. Okay, so let's also write down the cost action. So the cost action is V-p-c, okay? Now we can take C and expand it as... We need some nice variable, gamma n-c-n. But there's already a number, too. Yes, long-distance c-n sum over n. This is the Fourier expansion. And we can take B and expand it. Is there a beta n number? Sum over n beta n Bn, with the sum over beta n p-c-n, okay? Which is equal to... Okay, and I'll separate out the contributions from the zero words. Okay, lambda, zero, i, c, zero. And once again, I'll separate out the contributions from the zero words. Beta, zero, j, B, zero, j, plus sum over... The remaining guys are beta n, p-c-n, by the way. Plug this expansion of c and b into this action. What do we see with the zero words? Firstly, what happens to these guys? Well, they're annihilated. They're annihilated, exactly, exactly. Okay, they're annihilated, okay, just for... These guys are annihilated because p-c-st. Obviously, these guys are annihilated because we write this as p-diagon P and p-diagon gives these guys. So the zero words don't enter this action at all. The only contributions we get are from the non-zero words. The non-zero words. You see, we have that. It's the eigenvectors of a mission of rate. So we substitute this in. So what we have is sum over beta n p-c-n, by the way, n. And then lambda m-c-n. Because of this, this is also lambda. This is the same thing as lambda n p-c-n. Thank you. So this is the same thing as beta n lambda n, by gamma n p-c-n p-c-n, which is equal to... Wait, there was an extra factor of... There was something like this, right? No, you can read the P. Yeah, this P was this P. Thank you. Yes. So now we have to... Okay. I am going to get a lambda. Yeah, gamma n is the answer. Yeah, but I have this as well. It proves the final answer. Sorry, so that may stop somewhere. So let me... Yeah, this is square root, right? Yeah, sorry. Okay, so let me do beta. Since this is a positive matrix, I'll define this with a square. Okay, then this can be just be accurate. Normalization can be just be accurate. A 1 by gamma n can be just be accurate, yeah? But when we do this stuff, we get gamma n squared. Okay, so this is equal to a gamma n beta n lambda n. Okay, this is now an angle that I've seen in particular, that the lambda n is equal to 0. The lambda n is equal to 0. That's the advantage. And the lambda n is equal to... One of the coefficients of this is also on the basis. Okay, there's a p-c system that has introduced something very nice. Okay, you just have something between beta n and lambda n. Okay, and there's this factor of the eigenvalue behind it. So the p-c action is something very nice. The key point that I wanted to make in order to do this is that the zero holds do not appear in this action. That's important because when your path integral is c, now what does the mean to your path integral over a field? One way to give meaning to the measure of a path integral over a field is to take that field, expand it in an orthonormal basis and do the integral over the coefficients of this expansion. This is everywhere for the devil, right? It's like the expansion in path integral momentum. For an ordinary field. Means now that we've done things carefully, what does this mean? This means d product over n this, d, have I called it lambda n. Product over n d beta n. That is very nice. These are the coefficients of the expansion. And these are the functions. These are the coefficients of the expansion. But I also have the products over the zero ones. Product i is equal to 1 to v and product j is equal to 1 to m. d, what did I call that? Lambda zero, right? And d, I call that beta zero j. A integral over the cost means, okay? And this part of the path integral is no trouble at all. It has this nice Gaussian action with which to play. However, the part of the path integral over the zero is kind of problematic because the action that we're doing the path integral over is independent of the zero. So why is that a problem? That's a problem because as you know, the Grasman integral of the constant is zero. If you do an integral over a particular Grasman, with a particular Grasmanian coordinate over a function, it's independent of that coordinate at zero. Because the Grasmanian integral saves the Grasmanian derivative. So it looks now like we're in danger of getting zero. By doing the path integral that we said I have to do in order to complete the string, the string, the capital M, because we've got these integral over all the zeroes. So what gives? The thing that gives is that in addition to doing the integral over e to the power minus action, we also have all the insertions here. Now, how many insertions do we have? Remember, the number of insertions of c was the number of unfixed gaiting variances. Unfixed gaiting variances were those gaiting variances that left the metric unchanged. Up to, if you want, up to, up to while in the transformant transcription of a diffeomorphism on the main. The change in the metric of a diffeomorphism is essentially given by the p-operate acting on this diffeomorphism vector. So, the number of c insertions, which is the number of unfixed gaiting variances, is the same as the number of zeroes of the operator p. The number of variations of the metric that is of all the metric, is the gauge transformation. It's the kind of change in metric that you get out of GU by one of these gauge transformations. So, it's a metric variation such that it's orthogonal metric variation in a product change in the metric due to gauge transformation is zero. So, metric variation in a product p with any c is equal to zero, which means p back on the metric transformation is zero. The modularity of those changes in the metric, which could not be met by performing a gauge transformation, a diffeomorphism plus wild transformation on a standard metric. Now that we have an inner product, you can think of it as that set of metric variations that is orthogonal to every variation that you can achieve by gauge transformation. Actually, that you can achieve by gauge transformation of simply p or v for any vector p. This is how metric changed. Therefore, if delta g is going to be orthogonal to p and v for every v, this is zero for all v, zero for all v, which means p back on the metric is zero. It's exactly this. It's those metric, those changes in the metric that have no over that with changes that you can achieve by gauge transformation. Basically, you're saying that any change in modularity should be a change in the density of the vector or the state of the vector. It should be achieved with some parameter, some physical parameter. Now the p dagger delta g is basically a vector. p dagger delta g is a vector. That's correct. You're saying that the vector won't change. What do you mean by vector mode? Because the p dagger delta g is a vector. Yes. What you're saying is that that change in the metric is not a change. Actually, it's not a change in the physical degrees. The physical degrees of freedom are actually the tensor mode. No, this kind of way of thinking I'm not sure exactly what you mean by tensor mode because we're thinking we'll leave you to say that there's an infinite number of solutions to this thing. The p dagger delta g is equal to zero. If I understand you correctly by what you mean by it. Just tell me what you mean by vector mode tensor mode. There's a small finite number of solutions to this, to these equations. This is a model. Yes. There are no solutions to this on the square. So this is not some very general decomposition in some things. Just again, what is this? This represents all metric variations that you cannot reach. Suppose you've done a particular metric. Let's say some standard metric on your space. You look at all neighboring metrics on that space and you ask which of these are orthogonal to all the changes in the metric that are brought about by while a different mode is in transformations of your standard metric. That's what the zero mode should be like to give you the answer to that question. That's the question of what does it look like? Those metric variations have no overlap with metric variations that you get by performing symmetric, diffeomorphism and biotransformations on your standard metric. That's why you need to integrate over the model. Because you should be summing over all of the metrics most of which are while diffeomorphism and biotransformations related to your standard metric. But not all of them. That's why you need to do the integral of them. That's why the model you have to work on corresponds with the number of unfixed gauge indexes. The zero of the p-tigra and one to one corresponds with the number of monodilag. But in the formula for scattering amplitude we had exactly as many c-insertions as there were unfixed gauge indexes and exactly as many b-insertions as there were monodilag. Because we had b, del k, g in unfixed gauge index we had a c-density. So that's what makes it all work. You see, the integral of the zero mode of the b and c fields