 This is an abstract argument which just goes through unchanged. And therefore, all of s plus u go homology, all of the kernel of s plus u is equal to q go homology. So the arguments saying that s plus u go homology, the kernel of s plus u is q go homology, and the same is the arguments that we used for the kernel of s being q1 go homology. Those arguments completely abstract in general. That was the advantage of being the abstract. The new argument that we got is that we got exactly compute q1 go homology, meaning the kernel of s, because that was the trivial. And finally, one was the map between the kernel of s and the kernel of s plus u. What is the top-list of the actual interest of the argument being the kernel of s plus u? Well, we just done it, so it will be short. Do you know that these states are 0? That was identity by s. When you follow the state side, that's identity by s plus u. The reverse constructions are also true. But the same reason is you can write s is s plus q minus u. And the same thing goes for s. Same. Basically, this map that we have between kernels is very general property of lower triangular matrices. This is the lower triangular matrix. This is lower triangular in the sense of under eigenvalue of nsc, because u is a negative eigenvalue of nsc. So the kernel of a lower triangular matrix is isomorphic to the kernel of its time limit. Let's just see that to cross the matrix. Yes, suppose we know the policy. We know that because I get that is 0 here. Actually, it's going to seem trivial if you promise to cross the thing, but let me just say it. Suppose we have 0 again. So what's the kernel of this matrix? It's a one-dimensional kernel, and the kernel is all in the elements of the form half star z. Now, if you have 0, 8, b, and a is not equal to 0, then this is lower triangular. And the matrix diagonal thing was invertible apart from in this 0 rock. This is the structure we had, remember? x was invertible outside the 0 eigenvalue. Then it's clear that this element's no longer the kernel, but you can find more kernel elements. Because in order for, let's look at alpha beta. So let's see. So 0, 0, the first addition gives you nothing, the second addition gives you b alpha plus m beta is equal to 0. Now, it's not true anymore that you can choose beta equal to 0, but beta is, you choose somehow, then beta is determined by b alpha by the minus. So it's a is not equal to 0. So if you've got a lower triangular matrix, such that in this region, the space where you have no 0 is here, though. All elements in the lower triangular are like this. Suppose you've got a matrix of this structure, but these are zeros. This is really zeros of any any diagonal matrix, zeros, zeros, a and b. Then there's one to one map between the kernel of this matrix and the kernel of the operator a. And that's exactly what we've done. You see, this operator a is 0 is s. a is 0, b, blah, blah, blah is s. It's a larger idea in a sense. We're using an eigenbasis of n and c now, where this b comes down here. It changes the, it changes the eigenvalues of n and c. Okay, so you can just think of this in terms of matrices. In that sense, it's totally obvious. So this abstract construction in sets of matrices is just an entirely obvious statement. Is this clear? Then there's an isomorphism between the kernels. Okay, there's one more thing that I want to remind you of and that is that, okay. So we have therefore concluded that the, we've concluded two things. First, we've concluded that the cohomology of q is equal to the kernel of q plus s, of s plus u. We've concluded that the kernel of s plus u is isomorphic to the kernel of s, which was equal to the correctional space. Like, good quantization. There's one more thing we have to worry about. That's enough. But let's worry about that. What I want to clarify is, now, suppose we build this map, I have a cleaner product between psi and let's say we've got psi 1 and psi 2. Let's call this, let's call this guy. The cleaner product between psi 1 and psi 2, okay. It's impossible to have non-zero, you know, a contribution cleaner product from, if you have two states, both of which have nlc equals 0. And some operator in the middle, it just has a negative value of nlc. Just because you won't be able to, you can try it. You won't be able to eat up all the oscillators. Some of the oscillators left over, which are annihilated one at a time, one at a time. Okay? So, when you try to compute the inner product between psi 1 and psi 2, all terms are going to be used. The only term that survives is equal to, we not only have produced a map, that's, you know, an isomorphism between states, we've actually, in fact, produced a map that respects the inner product. I understand that it could be proportional. No, no, but with this one, it's exactly equal. Because everything else is just 0. Of course, we could have made a proportion by changing this constant to something else, but multiplying itself. With this choice of 1, with this one, it's equal to, now the loco is filled with the strongest form. That is, states in Kyoko homology are one-to-one map with the states of the Lycone oscillator, the states of the Lycone quantization, transverse oscillations of the spring, and all states in Kyoko homology have positive derivative. Would you know what happens if there's some operation of psi 1 and psi 2? Yeah. Okay, now, you see, that would require us to specify what an operator does have. Yes. That's a different level of, at the moment we're just looking at the inward space. We do have to worry about that. Let me look at strings, cap and amplitudes. Okay, but let's not be so ambitious. The result shows you the power of abstract mathematicalism. You know, I mean, if you were actually to enumerate this Kyoko homology level by level, let's say, it's actually, it's quite beautiful. But after mathematical reasoning, it can sometimes give you very powerful results. You know, physicists, like me, tend often to dismiss this kind of abstract thinking as often pointless. Well, one should take that too far. You know, it can be very useful sometimes. It's a very important thing. Okay, so, you know, the problem, of course, is that it keeps getting interrupted. Next Friday, there's a department meeting at 2 o'clock. You guys have an alternate time you could find. Maybe it's, okay. We might try to come up with an alternate time offline, otherwise we won't have time. So can I just take 15 minutes more now? Okay, we just keep getting interrupted, so it looks less than we've ever been given. And of course, those who want to leave, please go ahead. Okay, let me just take 15 minutes more just to talk about the remaining things I wanted to talk about today. It could be the next time we meet the next Wednesday. Don't get too far. Okay, now, the next thing I want to do that I want to tell you about, the next thing I want to tell you about is about an explicit construction of states in cubo. You might say this is pretty explicit, and it is explicit, but it involves this inversion of, you know, this action of one minus X inverse U S inverse. Well, this is really explicit, because it involves acting with software that is used by a lot of people. But it's explicit. It won't be totally museum. It's a more explicit characterization of states in cubo quality. It's a way that covariant quantization makes contact with what's called the old canonical quantization. Just one of the first ways in which covariant quantization of the state was done in analogy with the thumb law of quantization of electrical analytics. So it's a useful thing to keep in mind in that I don't know how I did it. Before I... So in order to get to that, in order to get to that, I want to tell you about one property. I'm going to tell you about one property of the states that were constructed by this construction. And the property is that these states all have zero eigenvalues under the operator 2n minus plus n is C plus n. So take... Consider the operator that counts the number of... the number of... twice the number of minus oscillators plus the number of C oscillators plus the number of B oscillators. Okay? The state can be constructed by this construction always zero eigenvalue under this operator. This is the k. This is the k that I have. Let's first count... Let's first count the eigenvalues of... Let's first compute the eigenvalue of r. Let's remember r. It's equal to k plus 1 over k plus sum over m Bm alpha plus n. With this n index here being negative, that creates one B oscillator but it destroys one minus oscillator because we have alpha with a plus of n. Alpha plus with a plus of n. It destroys a minus oscillator. So, such term is the value of minus. Is that clear? Because it has minus equals minus 1 and then B equals 1. On the other hand, we also have terms where this index goes negative. In that case, it creates a plus oscillator so it doesn't carry any n minus. But if this index is negative, this index is positive but destroys the C oscillator and therefore nC goes negative. It is minus 1. In both cases, we have this quantity in performance. So again, r is the eigenvalue that's the statement. And it's easier to see what you get. See how much it will add to the eigenvalue of the state if you add it a bit. That's how we can do the computation. We can do it by doing the computing. So r is eigenvalue minus 1 under that. Now, what about what about now I'm interested in q0 and q1? In the same way, in the same way q1 has eigenvalue easily checked at the time because of this directly check that s has eigenvalue 0 under this eigenvalue. Okay? Well, what about q0 and q1? q0 and q minus 1 have terms of all kinds of values of r or all kinds of values of this operator but there is no there is no term in either q0 or q minus 1 that has eigenvalue greater than plus 1 under this operator. This is my case. Now, the easiest way to check this is to try to produce a function. You know, it's going to be important for what I say that this is true. That there is no double eigenvalue 2 or 3 years. So the easiest way to check this is to try to try to produce a contradiction. So you see, the place where you might most easily be able to produce a contradiction is to raise an n minus by a large one. But that's as you get because of this factor too. So maybe we should be looking at the path in c. The path that is c minus m alpha plus minus alpha minus minus n alpha plus m plus n is as written as positive. Now that would raise n minus by 1. This is a contribution to q. Yeah, this is a contribution to q. This is not q0 by this one. But depending on what the values of m and n is, it's q0. Because it has to be your selector. It could be a company training. I'm going to say that in all of q, there is no term with the value of this operator being larger than 1. Okay? In particular, we know that it's 1 for q1 but there's nothing that's larger. Okay, so see, suppose n goes positive. Then this piece would raise this quantity by 2. That must mean that this operator is our creation. Because the sum of the indices adds to 0. Okay? Suppose this guy was a destruction operator. Then it would be destroying a minus of it. So it leads out to benefit of having this creation operator. It would be a creation operator but it's just an ideal value 1. The pieces in which you get this one would be ideal value 1 from this kind of creation, this kind of destruction then this cancels and you get 1 for k. All you would have had this kind of creation, this kind of creation, this kind of destruction. This kind of creation, this kind of creation, this kind of destruction. This creation just creates a plus operator that doesn't enter this operator. This creates a minus 1. This is ideal value plus 2 but this, because it's destruction gives you an extra minus 1. Example all the terms. You can convince yourself that there is no problem with ideal value larger than 1 of this operator. Any value which you have to compute are ideal value minus 1 and the queue who has ideal value ignores equal to 1. So u has ideal value less than equal to 0 under this operator. It's number of numbers plus number plus number. At size 0, the states of the kernel of s have zero ideal value because there is zero ideal value under n plus n minus n c d m b. Therefore, the states that we get by acting with the series x divided by zero ideal value under this operator. Therefore, the states that we get by acting with 1 plus s u inverse plus s u inverse, all of that is zero ideal value under this operator because the ideal value is less than or equal to the ideal value of the state that we start with and there is no ideal value less than zero. But the construction that we have here is the state in the kernel of this operator. It's the state. Taking that u has ideal value less than equal to 0, does it mean that u has ideal value 0 because of the series? You can have destruction operators. An operator can have under this operator. A state cannot. Let me repeat the argument. u has negative ideal value under this operator. So, the states i zero and zero ideal value under the operator s has zero ideal value. Negative or zero. Consider the state we get by this series operation of 1 plus s s inverse u plus s inverse u s inverse u. The ideal value of the state must be less than or equal to zero. But the operator on that state is about to change u. So, the paths which have negative u, much like it covers a lot of destruction of it. What is the inverse u which has negative ideal value? Which has either zero or pieces that are zero and pieces that are zero. There is no singularity of the s and u zero ideal. The state we are looking at is well defined. That will be already out. What about the state being well defined? It is just a question of what the ideal value of the state is under this operator. The ideal value is less than zero. The state with ideal value is less than zero. So, the state with ideal value is equal to zero. We conclude that all our states, all the states that we have got by this construction obeying the condition n minus on size equal to zero and z on size equal to zero and b on size equal to zero. What we focus on is these two things. Not only does that exist, not only have we characterized the co-mology, we have also shown that there is a representative of the co-mology that is made up entirely out of the meta-oscillators of the state. Because if this representative is an operator, next 10 minutes, next 5, 10 minutes, we are going to study states we are going to study states of this form. We are going to study the states of the form, states that are in the c and b vacuum. So, we are going to map the states but c and b vacuum. Can you say what the condition of the co-mology is less than zero? Size and exact, create an exact only with transverse oscillators and plus oscillators. And the point is how many plus of what? Given the transverse oscillator structure, the plus oscillator structure is different. It is not independent. You know, that is because the state will only transverse oscillators and the rest of the plus oscillators will put it in the co-mology of q. It is possible to do that. So, we are going to study states of this form. So, we are going to study states of this form. We are going to study the state that is in the b c cos vacuum. And we act on it with q. What does it mean for the state to be q closed? Well, you see because pieces that look like c m l minus m. Choose m to be, let's call it minus m. If we choose m to be negative, we have a c-destruction which analyzes states of this form. So, if we concentrate on states of this form, states that are in the b and c cos vacuum, then we only get contributions from anything positive. And so, what we get is c minus m on psi, which is something, times l m on psi. So, c minus m and c. So, c minus m just acts on the cos vacuum. This must be c0 if the state is to be in q-co homology. Therefore, the state must be annihilated by the matter part of the state. But for positive n, for positive n, but actually including c0 because c0 on psi is not c. Okay? So, we conclude that if we are interested in states of q-co homology of the form, then any state in the matter but for the field theory, times bc cos vacuum, then l m on psi must be equal to 0 for all n greater than equals. We have to stop taking all the states of the vacuum. We still have to model the area 0. There is 0 now. Okay? I know. So, we've constructed the state of q-co homology. You see, this is necessary and sufficient for the state to be in q-co homology. I'm sorry, to be q-closed. The states of this form might be q-exact. The states that are q-co homology of this form are states that are annihilated by all l m which is m positive. They are q-exact states. They have 0 in the product. So, we show that any state that is physical, that is annihilated by q and has 0 in the product with all the other states in particular 0 norm must be q-exact. Because we show that within q-co homology we've got a positive definite in the product that it knows where it comes from. The states that are q-closed that are q-exact among these. Physical states, states that are q-closed are states that are various. But not all of these states are in q-co homology because we select and remove the q-exact states. You have to remove the q-exact states, but basically it states of to the total d. The matter. There's also the first part that doesn't continue that kills that kills that power. Because it's the vacuum in the cost apart from a normal order in cost actually because you minus 1 in the vacuum which makes it this l-z of 1. So apart from the normal ordering thing. This one systematically, but an easy way to try to identify all states that are q-exact is to just identify all states that have this form that has 0 norm. What states are q-exact states? Let's try to use the same kind of reasoning that we used for q-co homology. You remember that states that have 0 norm were simply q or something else. Of course we had states that were l-n on some kind. Suppose we have got a psi. Let's compute the inner product of psi with any sizes. Let's call it apis. Let's compute the same. So this is equal to an annihilator physical state. So if you have a state that is a descendant that is l-n of some other state the state can sometimes also be primal. Okay, so let me introduce a problem. States are annihilated by all l-n for ins positive about primal states that are obtained from some other state that are actually of l-n. And positive are called descendant states. Of primal, a descendant. A state that is both primal and a descendant it has 0 norm. And therefore is a q-co homology. It's not q-co homology. It's q-exact. You're basically sure that I won't try to do it. That this is the set of q-exact states of this form. If we're looking just at the matter center then the states that are q-exact are states that are aware of the physical state conditions but are also descended from some other state. Now there's a very pretty structure of representation theory that I've done. Some of you are very familiar with from my study of superconductors. You see when a state goes where in a representation of a form it is both a descendant it's at the head of its own power of representation of the algebra but it's also descended from some other states that are 0 norm states. They should be thought of as pure gage. As we will see in the various examples. Now you have the matter that is varying. The equation of q-co homology is that all states that are primary that ends with n positive. Modular states that are descendants. In q-co homology the descent is down to this little sub-sector. Now I think we've gone on for far too long where I had also planned to do was to work out both q-co homology and this matter-co homology in at lower levels for the state at well to be frank just at the master's level. Can I leave it to you as an exercise and actually hope somebody might do it? I am going to suggest it to you as an exercise and we will very quickly rush through it in the next class and whenever that may be my class may just arrive at the next Wednesday and after that start now turn to a serious study of the properties of the formula we learned today.