 We'll discuss, the material we'll discuss in this lecture is directly taken from Pudzinski chapter 4. It's the known course here. In the next lecture, we will complete our discussion of the material Pudzinski chapter 5, discussing values. We had this lovely formula for scattering amplitude, it was on a straight day, but there were many questions about that formula that we left out. You know, does it remain invariant if you choose different vertex operators to be fixed or is it BRST invariant? That was not a question because we could address when we put on the scattering format. I mean, we didn't know if it was BRST or entry cost, but now we formula it and we go to the question and answer. So, these next two lectures were planned to clean up a formal discussion of string amplitudes, but that will start actually trying to revisit string theory. We'll try to calculate the scattering amplitude. We have found. So, let's connect. So, let me remind you what we discussed last time. We had this general discussion about Fadi-Pokov gauge fixing. And now, whenever we Fadi-Pokov gauge fix a symmetry, there is, at least formally, at the classical level, if you know what the problem is, there is a phonionic symmetry naturally defined. There's a phonionic symmetry naturally defined associated with the gauge fix action. So, very briefly, to remind you of that discussion since it's been a month-long break, you remember the fields in which I suppose you get fields which are phi i, that this gauge symmetry is generated by delta alpha, delta alpha, so phi i is already a function of phi i. And this delta alpha, we have the symmetry delta alpha delta beta is equal to F alpha beta gamma. Gamma is the F alpha beta structure of the theory. And we choose to want to have this theory set up by fixing the gauge conditions, F a of phi equal to 0. Then the Fadi-Pokov procedure gave us an action that was s was equal to s, again, I'd call it, yeah, it was just a gauge variation action of phi i. Plus the term that was the delta function, then we get plus minuses. The delta function associated with gauge fixing, plus the term that was the Fadi-Pokov determinant, here, that's something else. Signs, right? So, the term associated with Fadi-Pokov determinant was what? There was the determinant in this issue of which gauge fixing condition and what variation of the gauge. So, there's a Bac alpha, these are the fermionic, and then delta alpha. And this is a very nice compact general form of the action for the gauge fix later on. Notice that this action has a symmetric of the following transformation. The delta phi i was a gauge transformation generated by the gauge parameter C alpha. C alpha, alpha phi i. Okay, and if we fix, if we stick to a collision state convention with a minus i here, there's also an overall anti-communic element. Then the change in V a was 0, the change in little v a was equal to epsilon times v a, sign right, yes, and the change in C alpha was equal to i by 2 F alpha beta gamma C beta C. And under this transformation, we saw many things. We firstly checked that this transformation was little different, so that the generator, this transformation is square to 0. We checked that by checking the action, by just doing the action twice in many of these fields, we checked how much physically and others would have to do with the next example. Okay, so the next example we checked was that this action here, to make some good form, it could be written as S phi i of phi i plus q active times V a of a, probably without a minus, with some conventions with a minus i. The important point about this action was that the details of the gauge fixing function entered the action in a Q-A's action. That is, the action changed by Q acting on something. We wrote this in operator language. We wrote that we can only keep quantizing the system and wrote down an operator that generates this Q symmetry where that Q acting on this would become the anti-communicator of Q with V a of a. Isn't that what we asked on BRSC fixing, on BRSC gauge fixing of the particle? We said, well, suppose we're dealing with a gauge theory. We want to know what the Hilbert space associated with that gauge theory was. So, I asked this question from first principles, you know, that we used to understand it. You take the particle integral of the gauge theory, slice it and see what Hilbert space, this gives you a trace. You can answer this question honestly. However, in doing the perturbation calculations, in the perturbation calculations, we introduce all this other machine and all the other causes and so on. We're going to take this action and face that. And just chronically quantize this action. You're going to consider the path integral, slice the path integral of this action, associate a Hilbert space with it. This is clearly sort of a larger Hilbert space than the Hilbert space of the original theorem. But it's not all face. It's a constant. So, the question was, what is the connection between these two Hilbert spaces? So, the question was, is this reality method merely a trick for computing path integral? Which it clearly is. You can ask how we introduced it. Or, I think we do better. Can we give a nicer physical interpretation of the reality method? Can we also give an Hilbert space interpretation of the path integral that we get using reality techniques? That's the question. There's no pressing logical reason that the answer had to be yes. But, in fact, it turns out to be yes. So, what we're going to do is to identify a special subspace of the Hilbert space of BRHT quantized theorem and to make a candidate identification of the subspace as the physical Hilbert space, the Hilbert space that we would have got, in fact, be honestly done in the canonical quantization of our original theorem. So, that's the question we're asking. We don't know a priori about the answer to this question, but we do know that the answer to this question must have a certain property. Okay? Matrix elements, a time evolution matrix elements of honest physical states, honest states of the Hilbert space must be independent of our choice of the matrix. So, if we dub this all correctly and we presume we have, this must be true. However, you know, can you see that the path integral of Hamiltonian of the theory appears to depend on the gauge fixing path, on the choice of gauge fixing parameter data, but they appear to depend on it only through this Hamiltonian function. So, if I change the gauge fixing parameter, then I change the Hamiltonian, but the anticombinator of q will be a delta fq. Choose our physical states of our, we choose our, we choose to restrict attention to those states that are anti-Hedges by q. You don't know, but we also discuss the qs as a wishing operator. So, acting, it acts both left and right. Okay? So, if we choose to restrict attention to those states, chi related by q is under a change in the gauge fixing path. Because the change in such a matrix element is simply, the change in this is simply chi A, more than this matrix element, we might be interested in operating that in sessions, yeah? But once again, if we choose, they operate as commute, or respectively commuter, anti-commuter with q, depending on where they come in, it goes on like that. Then, the second conclusion goes through. This thing just goes through all the operators, and that highlights the physical state either left or the right, in the sense of combination, line of combination. Again, if you view these states, these matrix elements of these states, and matrix elements of these operators are invariant under change in gauge fixing condition. Our physical labor space, the first space it could be, the space of states that are annihilated by q, and the largest space of physical operators that could be, is the set of operators that are commuter to q. There could be further restrictions that don't come out of this simple, the simple condition, but this way would make sense. Okay? So, that was the first thing. We restricted attention to states that were annihilated by q. The q on the side was equal to 0. That was our first conclusion. The next thing we noticed, was that suppose you had any state, kari, that was equal to q of some zeta, some other state zeta. Then, the inner product of kari with any physical state, inner product of kari with psi, which is the same thing as zeta of q psi, was equal to 0, as long as psi was just, because q annihilated psi, it's commision. This q square is 0. Any state of this form, any state kari of this form is physical, in the sense that psi is annihilated by q. However, it's all physical states including itself. The matrix elements between any such state and any other physical state is 0. And that continues to be true for matrix elements including operator insertions as long as the operators commute with guilt. Therefore, any two states that differ by a multiple of such a state have all matrix elements equal. So, all probabilities of doing starting with something in one state, ending with something in another state are identical to all of these states. Clearly, these states should be thought of as described in the scheme of physical system. That takes a clean solution. So, our final prescription was as follows. The true inverse space of the theory is the equivalent is made up of equivalence classes. First, we define this physical inverse space, the space of states that was annihilated by q. And let me look at equivalence classes where states would consider an equivalent if they differed by a multiple of q actually not a half state. And we identified our, as a trial, we identified our Hilbert space with this equivalent, the set of equivalence classes, which has a name, which is likely to call this q-core homology. So, this was our trial identification of the physical Hilbert space of our system. And in the quantization of the free particle and inherited stream theory, we argued that we needed one further condition, that in addition to states with VRSK invariant, we also required these states to be annihilated by the zero more than v0. You remember when we did the free particle, we got twice the right answer and we killed the wrong part by asking for the states to be annihilated by the zero more than v0. Great. So, this is why I wanted to remind you of the general discussion of VRSK invariants and any questions or comments about this. Great. Now, then the next thing we needed was to discuss VRSK invariant of the VLSK of the stream. So, we managed to produce we sort of went ahead and implemented the general procedure on the VLSK of the stream and we managed to produce this nice VRSK output which was equal to C times T, T-magnum, plus B. It was half C times T cost but the more useful way of writing it was B-C density. And then plus it was a better term that we added to know how to make current tensors. 3 by 2 density. We had this nice VRSK current which generated VRSK transformations in the sense that the OP between this current and any and all the fields had poles in the rest of the poles which gives the transformation of that field and the symmetry operation generated by Q was the correct thing. And we computed many things about this VRSK current. We talked about how this VRSK current in OP with itself had zero residue of pole only when D was in it, 26. But the OP of this VRSK current in itself is related to the Q-Q anti-component. So, the fact that Q squared to zero by classically always though was quantum mechanically true over D equals 26 of the critical dimension. So, from the VRSK point of view this was the special thing of the critical dimension. You know, that Q squared to zero by the way Q is always a symmetry of the theory of anti-dimension. Always generated a symmetry of the theory of anti-dimension. You can see that just by the fact that the VRSK current was analytic. And therefore, it was conserved. Okay, and J z was analytic, J z bar was analytic, therefore J z bar J z plus L z of J z bar was zero. Now, which is so romantic. However, the special of the critical dimension was that Q squared to zero. Okay, we also worked out the OP's between the VRSK current and all of the other fields in the theory. The model that will be of relevance from what I have to say today, I'll rewrite. Z with B of zero at many terms but the pole piece the pole piece here was dm plus tg on the side where tg was the the energy point of the course. And this is particularly inside that where dm with B m was equal to lm where lm was the the Fourier mode and in particular the thing that we would really use today is that Q would be zero as you said. Where zero is the full zero-thoratorium mode of the ghost. Okay, so this we're going to use heavily in a little bit. Questions or comments before this questions or comments before this are going to be presented. Do you agree? Okay, great. Now, I just want to remind you about the structure of l0. So what is l0? So l0 included the zero mode part so that was the part where we get coefficient straighter to it but schematically at the moment of some the part that was alpha prime p2 and it was 4. Alpha prime p2 plus the number operator the number operator associated with x oscillators so let's call that nx. So this thing was sum over m sum over m alpha this m alpha mu m m is equal to one infinity and remember that these alphas had their commutation relation such that they commuted to m. So each of these modes, the m is alpha mode has energy m. And plus there were number operators for the c-modes and number operators for the b-modes. So analogous definitions with the c-modes being a b plus m with a c minus m and the b-modes being a b minus m with a c plus m because c plus m destroys a b minus m. So there was an analogous piece like this but there was also a normal logic. So for the macro parts of L0 was zero. We worked this out in detail. But if you remember we were just discussing the bc-cos system. We also discussed what the normal altering constant was for the stress tensor the bc-cos system. And remember we got the answer lambda into one minus lambda. In this system lambda is equal to 2. So this thing is 2 into 2. So that's equal to minus 1. So when you rewrote L0 in terms of oscillators you took these oscillators and you oscillate a normal order. Remember it's different from conforming an order. That was the origin of this normal order constant. It didn't make a difference for the macro part. Except for zero modes. First it made a difference. For the course it made a difference. And if you take L0 and rewrite it in terms of oscillators to normal order and so on. In terms of these number orders there's an additional minus one which comes from the zero mode of oscillators. You remember in two ways of deriving this look at you remember this normal order constant? In two ways of deriving this. First one was just looking at the difference between conforming a normal order when we defined L0 operator and operated a normal order. Using that to be directly computed so it was easier or less conceptual to use the fact that L0 is the commutator L1 and L-minus one. I just wanted no normal order in the utility. So those were completely well matched. If you use that to define L0 you've got the right answer with the normal order. Same answer. This course has been a little staccato so you may have forgotten this but I'll actually go back to your notes in a moment. Since the minus one is very poor. So this is because you see you remember that we started more or less this lecture by saying that we made an imported condition B0 and Q inside L0. Q would be 0 to L0. So it follows that if both these conditions are imposed it follows that we also want an imported condition L0 and L0. This would imply L0. But remember this is a full L0 including the normal order. So this of course is the this condition L0 inside L0 is the mass shift. Because this is the mass squared in units of 4 divided by alpha prime. So this condition tells you that the mass squared is equal to 4 divided by alpha prime and then there's a minus one plus also make an alpha prime. It's such a condition from both the left and right and the left and right. So this is what we try to do in more details to understand the Q-core homology on the version of the string. But we've got attention to those states that await the Q-core homology that await B0 on side equal to 0 and L0 on side equal to 0. And this is the consistent subset of Q-core homology because of the following combination. Because Q would be 0 and Q would be L0 is equal to 0. It gives a conserved charge commute for the Hamilton. Because these two relations are true the restriction to states that are annihilated by B0 and L0 is self-consistence. You start with the states that are annihilated by B0 and L0 you add to the Q on it and it will continue to be annihilated by B0. Because these two combinations are on the states. So we don't need to simplify a problem by instead of looking at all the states looking at those states that are annihilated by B0 and satisfied the martial action. In the subset term we wish to examine we wish to examine Q-core homology. A very formal argument of this no-host term and then we look at examples to try to understand. So the final thing that we are reaching to prove what we are going to show is that this Q-core homology with the suggestion of B0 and L0 is the right answer. Now what is the right answer? The right answer is what we got to in the second and third lecture of this course. Namely the spectrum that came out of the right-core modernization of the boson extreme. The right answer is a bunch of particles one associated with every oscillation of the 24 transverse dimensions. So I don't know how much more in it. It has 28 different kinds of oscillators. B oscillators, C oscillators and the 26 X oscillators. So we have to get rid of 4 oscillators somehow or the other. And the claim is that Q-core homology does this. The answer to Q-core homology is precisely one of our thoughts once with the correct excitations correct excitations, these transverse excitations are straight. So let's see how do we go about it. Is the goal clear? The argument I'm going to start with is formula and you may not see exactly where we're going to start with but please hang on and have patience and you'll see where we're going to be where we reach my feet. So the first thing I do and this is totally removed to start with we want to get correspondence with the transverse degrees of freedom. Out of these 26 degrees of freedom time is special of course but in addition we choose one other special direction like we need for likelihood like a quantization and isolate it. So let's define alpha plus to be alpha 0 plus alpha 1 by squared make this definition for oscillators so there's an m-branche m-branche for minus 20. Minus but this is alpha 0 plus alpha and this is the one special special direction. You might want to call it alpha 25 but let's call it 1 just to save it. And let's define alpha minus m is equal to alpha 0 m minus alpha 1 m. Now let's figure out about the combination of these alpha plus and alpha minus. So first tell me why is alpha m plus alpha m plus something that where we get 0 exactly. So two ways of thinking about this directly the second way is to just be right the covariate form of the combination relations. You know the covariate form of the combination relations had g mu nu on the right hand side but there's no g plus plus m. Or directly we get something where m is equal to minus m. We get something from the 0 term but we get the opposite thing. We get something from 0 term with the opposite thing from the one term. Because the combination relations involve alpha m 0 alpha n 0 is equal to delta m minus n times the factor of m times g 0 0. And g 0 0 and g 1 1 are negative. The thing is equal to alpha minus m alpha n minus is equal to 0. However alpha plus m and alpha minus n that is the fact that we have a g plus minus on the right hand side. And let's compute it from let's write this as m delta m minus m and from the 0 terms we would get a minus sign but one term we also get a minus sign. So we have this 2 square root of 2 is in the denominator cancelling. So we got this nice combination relation. We got this nice combination relation for these alpha pluses and alpha minuses. Ok, now we're going to define whether you define an operator we call m l c the number of light cone oscillators. And this operator is defined as 1 over m 1 over m 1 over m alpha plus minus m alpha plus minus m alpha minus m and m is sum m is equal to minus infinity infinity but not equal to infinity. It's summed over all integer non-zero integer values and let's give a physical interpretation to this m l c operator. I want to clarify this m l c operator acting on a state counts the number of minus excitations minus the number of plus excitations in that state. Let's see this at work in a simple example suppose psi was equal to alpha minus of minus k that came off the number on zero. This state would be activated. The term of this operator that would be activated was when alpha plus, alpha plus came with a plus k. That term would have alpha plus we want to do this to compute n and c on alpha minus of minus k on c. Now I'm claiming that the only term you see this is a sum of terms with one index plus and one index minus. So the plus index would annihilate the reaction unless it is killed by the commutator going through again. So the only term that can contribute is when something that can click with this alpha plus of k alpha minus of minus k and then we want to divide this by minus k. Because we're supposed to divide it by the negative of the index of the plus. This thing acting on alpha minus minus minus k according to this formula gives you a minus k because the commutator has a minus. So we take this through here that makes a difference. We take this through here and analyze the factor that you up to the factor of the commutator the commutator has a minus k. We can minus k divided by minus k then the alpha minus of minus k on 0. Then you want a minus k. This is the key factor of alpha minus k alpha minus k2. We get out a factor of 1 for each of these guys. Now if instead you had alpha plus k you would get a factor of minus 1. The signs work out like that. Is that possible or would you be like too uncomfortable and you want me to check it? Great. So let us just count the number of minus oscillators minus the number of plus oscillators. Now very good. The commutator is almost you know is almost a symmetry of the theory. It's all we won't use this and we won't try to justify it in any detail. It's almost the generator of Lorentz transformations in that plus minus boost to that equation. It's not quite that because through the contribution of 0. But what we find here is something that's almost a symmetry generator of the theory except that it's not quite because we're not including contribution of 0. So what do I mean by that? So let me look at a particular operator, the operator that you are interested in is q. So let me look at the operator q and let me look at what the value of n and c is on the various terms of q. So q of course is part of the dispute of course because there's no plus minus aspects. So the interesting part is the thought that came from the integral of c del x del x is c del x stress mass. Now what is c del x del x in oscillating language? In oscillating language that is c del x minus m minus k alpha k sum overall alpha mu alpha mu sum overall and k. The factors are correct here because the expansion of c del x is simply expanded in terms of order of sum. It has the utility. The fact that this formula is really just correct. Okay, so now let's look at this formula. Of course any term where there is mu means that n and c equals zero. We don't care about those. So what we care about instead are the terms in which this is plus minus. Yes, we have sum over all of them. Now the terms, so let's there are various cases. One case is that m is positive and k is positive. These are modes that destroy a plus oscillator and destroy a minus oscillator. So for such terms n and c is zero. Which m is positive or n and c is just zero. Let's do the two cases again. Suppose both of them value on alpha minus is a destruction operator for alpha plus for a plus oscillator. See, we've got the physical interpretation of this thing. It actually counts the number of oscillators with no factors because these factors nicely cancel. It's just that the operator that counts the number of oscillators. Number of minus oscillators minus number of plus oscillators. Suppose both are positive then this is a destruction act on some state. Either if it's zero or it destroys a positive oscillator. This is a destruction for a negative oscillator. So it reduces the number of positive oscillators by one reduces the number of negative oscillators by one so that n and c equals zero. That's clear. You might think, oh, whatever if described as negative it creates a minus oscillator. But this time positive destroys a minus oscillator. n and c is zero. So the statement that the n and c come in is... Hang on, hang on. So that is eigenvalue of the commutation of this operator. There is this operator where everything is either positive or negative is zero. However, let's say if we find out n and c to exclude zero. So there are terms here in which one of these is zero. Suppose we look at the term where this guy is zero if this guy is positive it destroys a minus oscillator. If this guy is negative it creates a plus oscillator. In either case it has n and c equals minus one. On the other hand this guy uses it. This is x for zero we will get n and c equals plus one. Because we have creation operators for minus oscillators or destruction operators for plus oscillators. Then we can write q is equal to q minus one plus q zero plus q plus one. The index reflects the charm of this operator and the commutation within x and c. Now q zero is most if we operate. If it is all the path that is pure ghost if it is all the path that has transverse oscillators then we are in oscillators of plus and minus. However q one and q minus one contains that small part of q that has a zero one contribution to minus. And we just decided that q one written as sum over c minus m alpha plus zero alpha minus minus infinity infinity q one We look at we are going to be acting this q on states so we choose our states to be in a basis such that they are in eigen states of the zero mode moving to operators. So that then we functions that are in the bi. In that case alpha zero plus acting on a state is simply k plus the plus component of the momentum. Because remember alpha zero is the momentum operator up to an overall normalization. So we won't worry about the overall normalization because it has a square root of two alpha prime and place it over so that is not quite enough. So q one up to an overall normalization is k plus sum over c minus m alpha minus alpha minus m acting on a state which has momentum plus one plus q zero plus q one the whole thing square is equal to zero. We decompose this equation into those operators that have all possible allowed values of nlc. When there is zero it's projection on to any given ideal basis zero zero. So the parts of the operator that have what are the allowed values? The allowed values in q squared are nlc equals minus two, nlc equals zero and nlc equals plus two. The part that's minus two is q minus one the part that's plus two is q one the part that's zero is something more complicated. But each of these must independently be zero. The operator is zero it's like you have a state zero and some basis all these coefficients will be zero. So each of these is independently zero and in particular q one is squared. We could consider instead of studying the complicated problem of studying vrxc and kohmah we could first do a warm up that is the problem of computing q one and kohmah. What we want to do is we're going to construct q one to kohmah and then we're going to show that q kohmah is actually isomorphic to q one. And thereby construct q kohmah. The problem is it's a huge simplification of q kohmah because q one is very simple operator. Let's go right. Maybe I should try just using the oscillator definition just directly construct the kohmah construct states and kohmah. But we're going to do this to give an abstract argument but q one is simple enough that you can just directly sit down and construct a kohmah and kohmah. But I'm going to work a little abstract abstracting is useful because it will help us in generalizing the q one. So how do we proceed? The first thing I do is to define an operator on each I just put out a hat of q one over k plus and then replace alpha minuses by alpha pluses and c minuses by b minuses. So sum over b minus m alpha minus n one. So this is my definition of this new operator and you're thinking my god this is going crazy definitions upon definitions. We're almost near the end of the definition. We're almost at the end. Compute the anti-commutator q one and half. So let's do this with the computation. First thing is k plus which cancels to one over k plus so that doesn't enter. So what we want to do is to compute the anti-commutator of c minus m alpha minus m and anti-commutator with b minus k let's call it alpha minus all everything sum over 0 minus of m and k. Say again alpha plus k. Yes. If this was alpha, I want it to be tm. So now what are we going to do? The only thing to be careful about is sums. I need to remember some abstract rules in doing anti-commutators or you just work it out by pulling things. So let's work it out by pulling things. So we get q one r that is equal to sum over m sum over k c minus m alpha minus m b minus k alpha plus k. Now let's pull this alpha minus m through this back. You pull this through free and pulling this through you get you get a factor of you just force m to be minus k. So this is equal, this thing is equal to a term when you pull it through which is sum over c minus m b minus k alpha plus k alpha minus m plus the term that you get from the commutator which is sum over sum over okay, this term has two parts. It depends on whether m is positive or not because we have alpha you will always be there you will always be there because of the factor of one over m. In fact for m, yes. Yeah, yeah, yeah, yeah. We'll just keep the factor there. So let's do that. So this thing is plus m times c minus m b minus k times delta m is equal to minus k. So let's set this sky to be equal to m. What we are going to do is to pull this at this sky through this sky also there. In doing that we get this is equal to r times q1, that's what we get once we pull this guy through. Plus this that's q1, that's what we wanted. Thank you. So commutator r is equal to this guy and commutator r is equal to this guy plus what we get pulling this through of this c minus m and b minus k. Okay, now that guy simply gives you that guy simply gives you what. Now what are the formulation relations of c and b? So right. So we have that when this is positive it basically depends on when this is whether the guy is positive or not. Right? The creation is equal to 1. So we are going to go over the simplification of c and b word. c, m this is from the audience that just is just c, m, b, m will simply get out m. Okay. So now that we do that we get an additional factor of this object. We get an additional factor of this object. When k and m are equal so let's write down the final answer. This is sum over m m c minus m, b, m plus sum over m, alpha plus m, alpha minus minus everywhere m is not equal to c. Let's let's interpret the result. This guy is the number of plus oscillators weighted by each level plus the number of minus oscillators weighted by each level. The number of oscillators weighted by level. This guy here so let's see, suppose m was positive then we will have b this m was positive counts the number of c oscillators weighted by level. You see that? Because if you have a c oscillator, b will destroy it with factor 1 replaced with this with a factor m. On the other hand if m was negative we get a minus from here we get a minus from the factor we have to flip this in order to first destroy. So m negative counts the number of c oscillators weighted by level. This quantity is q 1 to the power r is simply equal to number of c oscillators because number of b oscillators weighted by level number weighted by level. Plus number of plus oscillators weighted by level plus number of minus oscillators. The minus of n minus w is minus alpha minus okay, how is n minus w? n minus w is defined as n minus w is equal to sum over m greater than c also n is equal to 1 infinity of alpha plus m alpha minus minus. It's the thing that counts the number of minus oscillators weighted by that. Okay? And plus w is equal to sum over m is equal to 1 to infinity alpha plus minus alpha minus okay, n wc is equal to sum over m m is equal to 1 infinity then c minus m dm and n wb is equal to m is equal to 1 to infinity m b minus m c is the square and technically if you turn the broken up into the path where m is positive the path where m is negative alpha minus seems to have gone on the other side. Which side? Which side? These things you are worrying about the normal order in business. Yeah, okay, good. Yeah, okay. Good. So one should work out the normal order in independent. Sorry, you are completely right. This calculation was was modular order. Now we have to independently work out, I am claiming the right answers actually when you write them this way. How do you know that? You know that because both q and r and I need to like you see what was this thing in order the easiest way to fix normal ordering constants. Okay, when something is defined by a common data is if you know individually what the elements of the what each of the two guys do to a particular state. Okay, so now let's let's, so that's a good question. Let's remind ourselves about the definitions of q minus r. So what we are doing is there will be normal ordering constant which is iris term the exoscillator, the plus minus exoscillator which is exactly the exoscillator. It will be exactly can it's sometimes hard to keep track of these normal ordering because they are infinite. Yeah, by the way, term by term if you allow yourself to think that way that's what will happen but that's dangerous. That's dangerous because infinity minus infinity could be anything. Okay, so it's much safer to work, you know, in some consistent way. So the way we would actually do it really, you know, one way to do it is by looking at the current corresponding to these operators appearing on please and so on. There's another way that we could work which is also safe, it's completely correct and that way is basically the operator question is being defined the operator question is being defined by the anti-computation between two operators. Okay, the only ability we have in question is about we have to go to operator up to a constant. So let's just say that we know what we got up to an unknown constant if I look at, if I see what the action of the original operators was on a particular state. Can make that both the original operators Q1 and R annihilate the vacuum. Therefore it must be true that the anti-computator of Q1 and R annihilates the vacuum. This form of the operator does annihilate the vacuum. It's a normal moment. So all I have to do to convince you that this is the correct answer is to demonstrate that both Q1 and R annihilate the vacuum. Let's remember about Q1 and R Q1 was equal to K plus times sum over M alpha minus of MC minus So in IKEA, basically it's one positive mode of one way. One of the two states of the way is that it's the back. Since the sum of R runs over M down equal to zero the sum runs over M down equal to zero you don't get any any possible contribution from zero. Similarly with R R was equal to one over K plus sum over M down equal to zero alpha plus MC minus M sum runs over M down equal to zero in this case you can add one over M equal to zero V zero into annihilating state. We don't have to go over that. Sum runs over M down equal to zero is a state annihilates the vacuum. This is always one destruction. Therefore anti-computator of Q1 and R must also annihilate the vacuum. We can fix the normal ordering constant by normal ordering of items. This is unambiguously determined. Determinously. That's a good question. Thank you. Now let's focus on this actually. The first thing that we see is that the right-hand side here is positive. The entire Hilbert space of the string oscillation structure can be decomposed into the right-hand side that there's a zero item value. That's the state of no cost no minus, no B, no C oscillators. And then there are positive. Q1 Q1 with Q1 commutes with this thing up in that given name with power x. So now I tell you Q1 commutes with x. Or more precisely, anti-comput. Why is this the case? Well, Q1 has terms with either one position, operator of minus and one annihilation operator of B or vice versa. No matter the moment in which one state has been created, the other state has been annihilated, this thing democratically counts the number of all states. This is once we created the other state that has created is equal to level to the state that's annihilated. It doesn't change the quantum number of x. It doesn't change the problem of Q1 go-on g. We can discuss the problem of Q1 go-on g separately in every eigenvalue of s. Then Q1 on s of Q1 on side is equal to a is equal to a times Q1. See, we've got a state that has eigenvalue a under the operator s acting with Q1 Q1 and it maintains that. So the whole problem go-on g splits up block by block. 1 for each eigenvalue of the eigenvalue. What? Actually Q1 should commute with s. Commutate it with Q1 and s. s is a Poissonic object. s is a Poissonic object. s is a Poissonic object. Now the next important clarity is that Q1 go-on g is an empty if a is not empty. In all sectors in space with non-zero eigenvalue Q1 go-on g is empty. Now this is an argument that appears again and again in our kind of physics. It was the argument Witton used in the study of the Witton index. It appears again and again in the study super symmetric because we have to look at this argument. It's a very simple argument. I want to claim that in some sector in which eigenvalue a Q1 go-on g is empty. By the way, in order to prove that I will prove that if Q on a state size equal to 0 then it is also Q inside. There is state size such that s on side is equal to a on side and a is not used. The fact that s on size if a on side is the fact that Q times r Q1 times r plus Q1 on side is equal to a on side. Q and i is the state by assumption. Therefore we get Q1 on r is equal to a on side. We have got an equation on the side. If r is equal to 1 over a Q1 on upside if size is not equal to 0. Of course there is a possibility that size is 0 in case you have r and how you think of size. If size is not equal to 0 then it is equally satisfiable. So what you have is that this state is Q1 on some non-zero state. There is a little argument it arises when discussing the representation theory of super symmetry. Because when you look in super symmetry in flat space, you have QQ dagger is equal to h. That Qh is 0. It is the same kind of argument that says that states of non-zero h are states. That kind of thing appears again and again in many subtexts. So I am focusing just in the study at hand. What are we showing? We show that if you have a state with non-zero value of this eigenvalue it is negated by Q that it is also Qx. Fantastic. That means that there is no state in the poemology there is no state in the poemology of Q1 with non-zero eigenvalue of s. The states in the app n and the poemology of Q1 have 0 eigenvalue of s. But what is the eigenvalue of s? The eigenvalue of s is the number of oscillators of plus, the number of oscillators of minus, the number of oscillators of b, the number of oscillators of c. Each of these terms is individually positive definite. The only way to get 0 is from so we have demonstrated that Q1 poemology lies entirely within. Q1 poemology lies entirely within entirely within the poemology of the line to poemology we were discussing. Namely the poemology of states created only by transverse oscillators of the state. We still want to demonstrate that Q1 poemology is equal to this. We have to show that every state with s equals 0 lies in Q1 poemology. So that's what we're going to do next. But we've already achieved a large part of our job, though only for Q1, not for Q1. We have to show that Q1 lies above the state. Now, it's to show that Q1 annihilates all states with s equals 0. Q1 annihilates all states with s equals 0. This demonstrates that all of s equals 0 is in Q1 poemology. Because if a state were to be Q1 exact, at s equals 0, that commutes with s, it must be Q acting on some state with s equals 0. Is that clear? You see? Well, let me say that existed an exact state with eigenvalue of s equals 0. Then this state is Q1 acting on some state. But since the eigenvalue of the final state is 0 and Q1 commutes with the operator s, the eigenvalue of the state that we were acting on to produce this exact state must also have been 0. For that Q1 annihilates all states with s equals 0, we would have in one shot shown two things. A, that all states with s equals 0 are Q1 closed. And B, that there is no exact state with s equals 0. Because having an exact state with s equals 0 would have required Q1 to have not annihilating some state with s equals 0. Is that clear? Two, let me demonstrate in the Q1 poemology is to try to know. I'm not stressing because of the generalization. But first let's actually look at the definition. By looking at the definition it's obvious. The definition remember was k plus c minus m alpha minus n. But states with s equals 0 have no c oscillators, no b oscillators, no alpha plus oscillators, no alpha minus oscillators. Since each of these is always an intuitive instruction operator, it commutes to all the other oscillators present with s equals 0. So it's clear that Q1 and all states with s equals 0 just make a instruction. But now there will be a slightly more abstract argument slightly more abstract argument slightly more abstract argument for this for this phenomenon. Now you can start by noting that s commutated at Q1 is equal to z. So this means that s Q1 is equal to Q1 s. Now suppose there was a state that was annihilated. Suppose there was a state that was that at s equals 0 was not annihilated by Q1 and had x equals 0. So let's say that that state is psi. Q1 s on psi is equal to 0 because s is equal to 0 at state. But by flipping the order we get s on Q1 psi is equal to 0. So if the possible equation the two ways in which it could work it could work that Q1 on psi is 0 or it could work that Q1 on psi is not 0 but s on Q1 psi is 0. We're assuming we're assuming that it's working the same way. Q1 on psi is not equal to 0 and checking to see if there's a contradiction. The point is that Q1 is an operator with ghost number 1. Remember ghost number was ghost number was number of c's minus number of b's. s equals 0. Therefore at number of b oscillators and c oscillators both equal 0. Since Q1 has ghost number 1 number of b oscillators and c oscillators equal to 0. What the number of b oscillators is and what the number of c oscillators is we can't see. So the difference between them is 1. Q1 on psi is some state which has s equal some non-zero value. It's a decomposition of states with non-zero values of s. But the kernel of s with non-zero values of s is 0. There's no state of non-zero value of s of course for which s on that state is equal. Therefore this equation must be true. The only way it could be true is that Q1 on psi is equal to 0. If we assume that Q1 on psi is not equal to 0 we would find a 0 mode of s but we have an independent argument that the eigenvalue of s is not still. This is the contradiction. If we have an argument by a head examination of the operator it's an obvious argument but it's a little difficult to state. We just said it. Just abstractly from the ghost number of Q1 that it's impossible that Q1 on psi can be anything but 0. If psi has eigenvalue psi has eigenvalue 0 on s. Now I'm giving you this more abstract argument each time because while this explicit operator Q1 we can just do everything by hand. Now as we do the generalization now to Q we can't do anything by hand anymore yet these abstract elements can't do anything. So we completed our discussion of Q1 cohomology. The theory that Q1 cohomology is identical to the to the 90 code. By the way, identical in a very explicit sense our construction of the cohomology our construction of the cohomology was really excited only the X-ray only in the transverse oscillators. So in particular we have shown that the inverse phase is identical to the light cone and the enough products on this Hilbert space are identical to the enough products on that light cone space. So in particular we show that within the Q1 cohomology there are no negative 0 norm states. The matrix of enough products is positive definite within Q1. This is far from obvious because for the inverse phase with 28 oscillators there is a very zero norm and negative norm states as we will see as we consider examples. But if we restrict just the Q1 cohomology the Q1 cohomology at the moment we are sure of this. Now we have to check what happens with Q1 cohomology. But before that any questions or comments about Q1? For what? This one? They see what is 0 states are not the this argument of the earlier thing was why is s equal to 0 states why they can't be the exact why so that even goes in two steps. First we have shown that every s equals 0 state is annihilated by Q1. Therefore in particular every s equals 0 state is Q1 closed. That's it. Now suppose there was an s equals 0 state that was Q1 on the time. The identity of chi must be under s. What must it be? If psi is Q1 on chi, psi has s equals 0 what must chi have? Q1 has 0 charge of s 0 charge of s therefore chi has s equals 0. So then they must if psi is Q1 on chi it must be Q1 on chi where chi has s equals 0. We have shown that there is no state chi with s equals 0 that is not annihilated by Q1. So it's impossible that psi is equal to Q1. To understand Q1 homology we want to understand Q4 homology not Q1 homology. How do we do that? We need to use the operator r but to replace the operator s by what you get by Q let's consider the operator Q Q minus 1 plus Q0 plus Q1 now let's see let's remind us r was under nlc remember the original nlc operator that we looked at we forgot to know by the last 30 minutes the number of minus oscillators minus the number of plus oscillators remember that Q1 has charge 1 under nlc but r has charge minus 1 minus 1 under nlc because we replaced minuses with pluses r has charge minus 1 under nlc Q1 has charge 1 by the subscript in indication because r minus 1 Q1 the subscript gives you the charge of that nlc so Q0 that r has charge minus has charge minus 1 and Q minus 1 the r has charge minus 10 is going to be equal to s a good trend plus Q we're not going to say what Q is we're not going to work on Q in detail the only thing about Q that people keep track of is its charge under nlc and Q1 has negative charge it's size 0 consider any state size 0 consider any state size 0 and it's annihilated by s so s on size 0 is equal to 0 that is when you demonstrate it consider any state size 0 that isn't in Q1 now I want to give you a construction of a state that is in s plus u that is annihilated by s plus u see essentially what we're using is the fact that that the kernel of a lower triangle of nx is equal to the kernel of the diagonal elements of the matrix exactly so what I want to do is to show that there is one to one map by explicit construction of the construct one to one map that takes states in the kernel of s and produces states in the kernel of s so how do we do this well you see let me give you the construction let me try the opposite the construction is y which is equal to 1 minus s inverse plus s inverse u s inverse u now whenever you see a construction involving s inverse you should scream you should scream and say what do you mean s inverse because s has a space with 0 i in the map however remember that u has mnc which is a 2 on size has 1 0 i give value of s because if nnc is negative it needs the number of minus oscillators minus the number of plus oscillators is negative which means in particular the number of plus oscillators and number of minus oscillators come busy okay so i make u suppose we add u on size remember that psi 0 has s equals c let's look at what s can be for u on size okay psi 0 has s equals 0 therefore it is nnc so nnc of u on psi is some negative number therefore u on size some state with number of minus oscillators minus number of plus oscillators being some negative number in particular it is impossible to the number of minus oscillators number of plus oscillators is 1 0 therefore it is impossible to this state as i give value 0 under s so s inverse while it doesn't make sense when you put it back to the u in front now that this is annihilated by s plus u is obvious it is obvious for instance where you can write out the series if you want or it is obviously because we need to write this as so what is this this is 1 over 1 plus s inverse okay so that is 1 plus s inverse u so that you can write as right right but because the inverse goes to the net so it is it is so this is this is 1 plus s inverse u inverse which is the same thing as s plus u inverse and then when you add this to the next plus u it is that and you get s on size 0 what i am saying here is that this thing is equal to s inverse that is s plus u inverse the whole thing inverse and then add this to the net that is an algebraic equation otherwise you just get termed by that you should give us that is why i need to take way of understanding why we get this theorem that the action is well defined is clear from this formula but now this algebra will ensure that you will get this theorem okay and if you do not trust this manipulation you can just not apply you get cancellation term by term you can see this the first time u multiplies this s multiplies this u minus u then you get u squared so it is just term by term that there is a 1 to 1 math that is a 1 to 1 math between well for any state in s for a quantity we produce the state in now for what? any state of the kernel of s we produce the state in the kernel of s plus u okay and now abstract arguments all the abstract arguments that we used before continue to draw it remember that q was that q was that q commutator of was equal to s okay so by the same argument that we used before that any state with non-zero value eigenvalue of s plus u