 I didn't know because it was just that I drank Pulchinsky wrong, so I didn't know how to guess. So I thought that he had a C.EV, his action. He actually has a C.EV, which is what we had. So there was no confusion between them. Okay, so our discussion for the point particle last class was accurate. We're looking at it just now. It just looked like it broke. Fine. Okay, so now, so today's class was going to go on. Now we're going to generalize our discussion of this point particle to the string. Okay? The discussion proceeds basically an exact analogy to the discussion we had for the particle. And I'm just going to give you the final answer for the BRSA transformation laws from which we will deduce a comment. And we look at the answer to C.EV. Okay, so the answer, so, you know, I'm going to give you a final answer by which I mean I'm not going to go through the intermediate step of introducing metric and the ground-multiply integration. Okay, so first we have to say what dx mu is. So remember, dx mu changes out of the gauge transformation equation. The gauge transformation equation can have different forces. C is the different forces in the field. So how does it change? Well, it has an i times epsilon, which is the anti-commuting parameter, plus c times del plus c power times del r. Okay. The change of the scale of field. Okay. Now what about the change of B? Now you remember that in our general discussion the change of B was proportional to the ground-multiply. Remember that the ground-multiply is what multiplies the metric. Okay? And we solved for B by the equation of motion from the metric. So when you differentiate with respect to the metric, you get two terms. You get B and you get whatever else you get. But whatever else you get is by the equation of the stress tensor under there. Okay? So the solution of B is to set B equal to the stress tensor. Okay? So this just gives us the stress tensor, which we divide out into two parts. For convenience, the part that comes from the matter piece, and the part that comes from the basic process. So dm is the contribution of the stress tensor in the matter piece, dg is the contribution of the stress tensor from the basic process. Okay? And what about c? What about change of c? But last time, remember, we had c times c power. There's an obvious extension of that. The obvious extension of that is this. There was no half, because there was a half in f, but then f had two terms. That's a half, and then you write the formula, and then you evaluate it as two terms. Okay, great. So this is the matter piece. Okay. We got this symmetry transformation, which you can check if you'd like to directly by, you know, this is the symmetry of the theory. But we do it in more sophisticated way. You see, we know that out here, this is the geometry of the theory. Okay? And I should, of course, say it by the way that there are analogous rules for db bar and dc bar. Right? You just replace this by d bar, dg bar, and we're going to say it by c bar. Right? I'm not writing it down. It's the obvious extension. Okay? So now, the question I'm going to ask is the following. Can I write down... So, let's suppose that I do the usual thing. So I should try to write down brst transformation in a charge as a modulation by ai times dz of jz minus dz bar, dz bar. Let's call it, let's call it jj tilde down, b for brst. So I'm writing the usual thing. I try to write down, but there's a charge and it's an integral over a current. And I'm going to try this out. I'm going to make the assumption to start with that we have an analytic current, you know, the jb is analytic, jb bar is analytic. And we'll see if that's correct with the answer. Okay? So the question I'm going to ask, the question I'm going to ask is, what is... Yeah, again we write down, look at these transformation properties, let's look at the transformation properties one by one. The first thing we want is that the change in x mu is that, it's just the derivative of x mu. So that's the analytic part first. We only want to generate this part. Okay? So the change in x mu is the derivative of x mu times the same thing. We know that... that we know that on general problems, that the change in a field is obtained by looking at the pole, the residue of the pole, which creates the generator question and the field. We derive this and use this many times. Okay? So we want some term in jb, such that the product of this term, and x mu has a pole, which is c times del x. Okay? But we know what that is. It's c times the stress tensor. Because we know that the stress tensor, active on any function of the matter of fields, has a pole part, which is derivative of that matter of field, that function of the matter of fields, back by pole. Then we know that it's true for x. For instance, for x we can change it explicitly. Quickly remember the stress tensor is proportional to del x del x. We have active on x, xx for the log, del xx gives you the pole, proportional to del x. The residue is proportional to del x. We would also want to see that. I'm just doing the analytic part. So that goes on through g1. Okay? So I'm just trying to find the analytic part. This is g1. Great. So we've got one piece in the realistic column. It's c times g1. Okay. Now what? Now let's move on here. Whatever else we have, shouldn't have any non-singular opinion with matter of fields. Because we've got to get all the transformation of the matter of fields from this part. So the simplest assumption remains that all the remaining parts function with the coarse fields. Okay? So we want the coarse fields to have this problem. Now to get this, to get this straight, let's remind ourselves what the coarsest stress there is on this. You remember we worked this out in generality when we started the BC system. And we worked out that in general it was tg was equal to del b times c minus lambda that is del of bc. In our situation as we've discussed, lambda is equal to 2. It's up to you to remind me why lambda is equal to 2. It's simply the way of saying lambda is equal to 2. So lambda equals 2 gives way to 2 to b. It's minus minus c. Why is that the correct answer? I mean, sir, if you look at the vertex operators, they have to be one month in a row, or I mean, you have to have one month away from that. The c insertion gives you way to minus 1. The simple thing to say is that in our generation of what b was, remember b was something with two ANDs that came out of the water, and therefore we had to do it. Whereas c was something with one AND that came out of the top, and that one came out of the top. You remember c was negative, b was a 2 times. That's the basic sense. Can't be seen in that question that c is negative and b, this was the vertex operator, which was the derivative integral of b squared. Yes. So it had to be consistent, this had to happen. But you see, that's the consistency of the final answer. Where in logic did it happen? It came from this index function. You remember we wrote this action in terms of b with lower indices and saw that the action was explicitly wider than that. So the rate under conformity transformations which in general is the rate under derivative operators plus wide transformations was purely the rate under derivative operators which now is the decision that can't be made. And that was our variation. Okay fine. I should have said by the way that this room was needed in 313, so it's going to be a brief class. We should have it. There's some chocolate, something. Okay. So the course system in the concentration had this stress answer. So let's see. So what we want for b, for the change in b is that del b is equal to let's drop the i epsilon. So what we want for the pole. So we want jb times b to have a pole on the right end of the left pole piece. We want that to be so now that's simplified. So this is del bc with a minus 1 minus 2 b del c and we call this z and this 0. We want this from here we want that from here we want that jb of z and c of w times c of w of w goes like c del c of z. So now let's try to see if we can come up with a guess for a comment that will do this. So let's write down the answer and check if it works. So consider bc del c also this thing had to have of course there have to be a plus k because there is this part Now the plus dm is trivial because we already have from the part that we added already to the post column we've got to term that c times b. So c times b is 1 over z minus w so the dm is automatic. So this part is automatic for what we already have. The new part is to give us this. So let's see first let's see roughly we want this new term to have an op we want this new term to have an op with we want the new term to have an op with b that has one b and one c. So clearly the new term has two c's and one b's because it's c and b that are not trivial contract. So we also want one derivative between f and b so it looks like a reasonable thing to start with. So now let's check. So what we want to do is to look at bc del c at z acting on bs of c. So what is the op? We can have either this guy contracting or that guy contracting. This guy contracting that has no minus sign but it has one over z minus w the whole thing squared with a minus sign because that differentiating produces a minus sign times dc at z. Now what we are interested in is the pole which goes like z minus w and we want the operator to take rest at the point w. So we need to cancel one zero so this gives us the contribution of this contract first contract. Minus del of bc by z. First we have minus sign because the c goes through the del c. But then there's nothing else. b del c times one over z minus z and the minus sign because the c is constant del c since the sum of these two so we add them up and what we get minus b del c with a factor of two minus del bc and this term is del of b times c. I added a factor of this and a factor of this so I would think there's much. So if we didn't try just the pole part we've got exactly one. Is this clear? Okay, let's also check that this thing that's what I think we'll see. Is that jb times c is equal to c del c divided by z. This is the z, this is the z. Okay, so this one's very easy because we take b at the end that goes through two and you can see that it picks up no sign then b with c is one over z so what we get is jb with c is c del c over z minus w. Exactly. We've got the moment of the current because we've got a current that has light poles with everything so that's a fun fact. So a provisionally final answer is that jb is equal to c times dm plus bc then c. There's a fancy way of writing because of c times the cost current. It's not very useful. This is the same thing as c times the cost stress tensor divided by 2. That's because the cost stress tensor has two terms. One which is c del b, the other is del you know b del c. The term that is c del b is when multiplied by c is zero. So all that remains is the term that is b del c and then so that it's getting proportional to the cost stress tensor the fact that it happens to be half. That's not very useful to me. So let me just write it down as tm plus tg ctg by 2 that's not very useful to me. Now as far as transformation properties go as far as generating the right transformation properties go there's nothing more to say. However, it turns out to be that you have to add a total derivative to this definition. Because it's a total derivative it doesn't affect the q, b, r actually. Again, because it's a total derivative precisely because it never affects the polar germs in the in the OP. Let me write it down first. All terms are any singular OP between c and anything is already a total derivative of it becomes 1 over z minus w cubed. So this is the polar way of saying that it doesn't affect the total charge. So it does but as a total derivative if you want a current that generates a particular cement you can always add total derivatives to that current because it doesn't affect the charge therefore it generates the same cement. However, we chose this so that in the critical dimension the costant will become a primary operator. It turns out that without this term that would happen. It's a convenient choice of additional total derivative that affects nothing physical if it requires an activation. I'll give you the answer. I won't give you the answer. Okay, great. So at the moment it's just some choice of compute that we fix in a particular way that we find. Okay, great. This is the current and now we can ask is the current observed? Is the current observed but the answer is obviously yes. Why is it always that it's a non-conservant current? Is that right? So what's the current conservation equation? And is it picked out for the start probability? Okay. So why is it obviously concerned? I don't know if it's a hydrophobic field but it's a hydrophobic current. When you work out the same algebra for G bar that's obviously an hydrophobic current. So without saying anything else it's a concerned charge. That's very nice. That's a nice operator. But there are a few more things we want to we want to ask. The most important of these is the question of whether Q is a question. Okay. Now how do you address given the current, how do you address if you know the contents of a charge how do you compute but Q is an anti-commuting object so the square of the operator is the anti-commutator of the operator itself. So now what kind of thing you do to compute an anti-commutator from charges? Remember what we had and what about the OP tells you the answer? You look at the you look at the operating session and there's a thing there and then you remember what the other what terms I think matter which is the four, the rest of you also. Okay. So the question of whether Q is quest to zero or not is a question of whether Q comes in or addresses you to the JBG okay. So so there's a bunch of OP that we should calculate for the service current. Now we've got a lot of OP calculations in class all of this is pure algebra I'm not going to do it I'm just going to present a lot of answers for you and leave it for you to check for these things. So let me read out a set of some time through this course it can't be that part of it we've compute the whole parts but at some point we have all the singular terms. So let's look at JB times B you remember we've already found that the whole part is one by Z so this is at Z and let's say zero okay. On the other hand the whole part is one by M plus DG one by Z because there were more singular terms than there was G the cost current by Z squared okay. What's the dimension of the left-hand side? What's the dimension of JB? What's the dimension of the cost current? What's the dimension of any kind of space? It would have been two had energy but this is What's the dimension of charge? Zero expression for charge in terms of current? Right. So the dimension of JB must be one comma zero because integral dZ JB gives you zero. DZ is minus one comma zero. So it has to be dimension one comma zero. So we've got the sanity that has dimension one comma zero let's confirm this explicitly count from this term del C is one what's the dimension of B? give me all that B is one two C is one del C is one at the left one what about C times TN? C is one okay good one by Z two is dimension what about the cost current? What's the cost current? two minus one one one one by Z squared is dimension two plus one is three dimension one by Z is dimension one stress tensor is dimension two and dimension three this is the kind of thing you should become routine through you see it on TV first thing you do is check it you make basic sense that okay let's do this one what's the dimension of the right contact? zero zero because del C is zero C is minus one one by Z is one and so on okay so these are actually very easy to derive they take commotion of work now we've done most of the derivation see this part this part we've already derived this part we've derived the rest or any any tensor made out of matrix it would take a volunteer to derive this that's why it's a primary operator a made out of matrix of weight in which C times Tm times Lm is what? Tm times Tm times minus minus two so this is why I wanted an h which is okay so let's put this as Z let's put this as W and now this is what? okay first an h go on Z squared good what do they do? wait what do I have to do? where are they located? they're under here is Cw or? oh C is it now you extend it and get the Z minus good we'll do that C is Z and then Lw is W and then perfect and now as you said we can expand this and we get a del C which has a pole which is h times of okay so it's all that simple we've done this a hundred times we must not get useless okay I could ask you okay we're wasting time okay good let's just stretch more there's one more P that will be of the two more algebraic factors what you did with the first term what? this also gives the first term yeah exactly I want to do one more P it's for practice there are two more that will be algebraically useful in the last okay so first let me write down the last two oh the last two will be Jb and Jb has Z Jb, Jw but let's see every operator has a zero minus now I'm not going to try any more to do any more deriving with your PAs okay let's just take these answers as given and exact the only important thing to look at in this OP is the coefficient of the pole the coefficient of the pole the term Cm the matter central charge minus 26 okay so that tells you that by classically you can check that the transformation rules that we we specify classically square to zero but classically this is Q something that squares to zero in any dimension quantity taken Q squares to zero only when the matter central charge is 26 so from the BRSC point of view this is how you this is another way of getting the critical dimensions frequently the BRSC can't square to zero in terms of the co-energy problem only with the critical you'll see as we go on everything was only one thing creates one supposed to be like that okay correct there's another related statement but every statement that will be important for us and that's this you remember we started the ghost we found that it was not quite it was not quite a conserved current it wasn't that but it wasn't quite in curved space because and that was related to the fact that the t j OP didn't have only a one by z squared but also had one by z cubed okay we had an exercise in which he was supposed to relate this one by z cubed piece to the failure of current conservation in an archery curved space it's quite a general statement that the only way that current can be conserved in an archery curved space is that it's a primary operator okay so this only here at the C L minus 26 over 2 z to the 4 okay which would make this a knockoff primary operator was one zero but in the critical dimension there were many fathers that had the OP with a primary operator so in the critical dimension j is a primary operator of dimension one and therefore it's a conserved current on any matter this would be important for us because we will be when we do string theory on higher genus remand surfaces that we have a conserved cost current not just in transfer so we've seen two interesting things from the OP's cost current firstly q squared is equal to zero but this works on critical dimension secondly the cost current is the primary operator of dimension one point zero this also works on critical dimension that's all I can say about I think that's all I want to say about the cost current oh no there's one more thing we have to add the the binding yes yes so the last thing I wanted to say was just an algebraic fact but maybe I should say two more things well first let's look at this algebraic fact we've seen that jb with b has an all part which is dm plus dg by z the anti-computator we've seen two fields related to the recipe of the code so this statement directly translates into qprsd anti-computator b b0 which is the chart b0 is equal to m0 plus lg and by appropriately multiplying both sides z to the appropriate power it also implies the slack also implies qb times bm would somebody okay I'm not need to derive this statement for me carefully something I wouldn't be tempted to do if I was sure you guys were practising opus at all okay but I'll leave it to you please you know go when we make statements like this without proof in the class please go and check if you have any problem with the writing please calm down I'll just move on assuming that you guys are doing well please do it okay very good it's the statement that we will find out immediately in our discussion okay and finally if anything else we would just say so now the next thing that I wanted to okay questions, comments questions, comments about everything now the next thing I wanted to do was consideration you see, it's all brsd quantization business and an essential element in it is the recovery of the physical spectrum by the construction of brsd go home, okay so now in the remainder of this lecture what I want to do is to prove for you that the brsd go homeology on the world sheet of the string is isomorphic to the spectrum we obtained from my book on this this prove, plus certain extensions that we will also explain when we study in multi-day the S main things okay has fancy names called the no host theorem so in multi-day I want to prove for you that if you look at the brsd go homeology on the world sheet of the string you have some sort of correspondence with the states that we obtained through quantum quantization not just in number, but also in in your product the last state was very important because we have demonstrated through this process that there are no negative norm states on the world sheet of the string you remember that when we just did like old quantization, there were negative norm states before we fix like old gauge because there were oscillators of x0 however when we went to the like or the only oscillators were in the 24 clock transverse n inches there were no negative norm states so what I'm going to prove is that we have the same spectrum that we get from like on quantization and the spectrum that we get from brsd quantization is cost free the sense that all the states that we have obtained are states with positive norm states and it's certainly not true that we didn't restrict the states that were of the form that were in the artistic homology that we would get cost free states we'll see negative norm states off of the place as we as we proceed with our analysis okay so my plan is first to give you the form of our result and then to demonstrate how it works at the first end of the string okay so let's start let's start by looking at the form of our result and let's see how far this goes my discussion of brsd quantization has wholly been mirrored in the Polynchian scheme anything you don't understand what a real Polynchian scheme is it's already a real discussion I'm not seeing that anywhere else it's very clear let's start so the first thing when we use brsd homology of string theory I mean brsd homology plus the additional extra input that is the analog of the additional extra input that we had for the quantization of the Polynchian maybe it will also require b0 one of the states you see because in particular if we just restrict the string theory to the zero mode we really recover the quantization of the Polynchian so just to get every oscillator mode gives us a new part but the zero mode part is isomorphic to the quantization of the Polynchian so just to get the Polynchians to get not to get doubling of the spectrum we must have eliminate half the modes so what is the state this is the state no this is just okay the level of quantization we've been doing for the spectrum so far so what will we since it works with one particle we want to do this it's clear we must put the same condition here b0 plays the role of the operator that we had before remember that in string theory also we've got the c0 and b0 that generate the two state system in a c0 once we've got this two state system which commutes with the energy operator so we would have had two states at a level we impose this addition so the first thing I do calling Polynchians is to define a restricted input space I divide a restricted input space in the worksheet for string as we the state of all of see we want the side to be 0 on size but we also on size so it must be qp on size but therefore it must be qp then the commutator be on size but we just compute the word qp and the commutator qp with b0 is it's the stress it's the well the l0 in matter and coast in matter and coast therefore it must be that where by a 0 I mean l0 matter plus l0 coast on size equal to 0 let me do the find a restricted input space we had to be consist of those states such that b0 on size is equal to 0 but cuts down to the original input space by half in the two states of the space and also such that l0 on size restricted input space has good action of the BRST operator because the BRST operator maps this input space itself now check that we know that qv that l0 is equal to what somebody BRST operator has a conserved charge and therefore it walks to the amateur to me and qb times b0 as we see is equal to 0 so this commutes when of light to the sector of states that I am okay so the if you act qbrst on a state belonging to HAC you recover a state belonging to HAC so qbrst is a good restriction to the subspace HAC okay so in all the discussion from now on we will be restricting our attention to the subspace HAC right from the discussion the discussion that we are going to perform now is a little abstract and it is very clear but because it is clever the motivation for what we are doing would be clear until we reach the argument so you have to have a you know until we reach the conclusion okay so the first thing we do is about let us go to the space oscillators and let us choose the time oscillator alpha 0 and the particular space direction alpha 1 and let us do the this set of oscillators now we will make linear combinations of these oscillators so we define alpha 0 m plus alpha 1 and then probably part of it and part of it as we are talking about minus the goal is to demonstrate is to define the co-opology of the qbrst operator acting on the the subspace HAC and I am sure that you will see this as light co-optization now because we are going to show this same light co-optization remember in light co-optization we chose the time direction and space one space direction to be special okay so we have to make a special let us make the same special choice so this is time, this is space and we are looking at the oscillators in those special directions which we want to make so what we want to show is that the bbrst co-opology is one to one correspondence with states obtained by acting only with oscillators in the other directions and no bnc oscillators okay but now it is a predatory work for this so hang on for a moment we are going to do this from idea of combination relations of these oscillators alpha 1m with alpha 1 minus m was equal to m times delta m plus n 0 which and the sign here was the statement that the positive ends were annihilation operators whereas the negative ends were creation operators and annihilation of creation is possible no in very good minds which is dead back then but remember the normalization of oscillators was such that m is equal to m times delta m now but remember that when we did canonical quantization the combination relations actually appeared with a g mu when you want to write it it is a predatory work because we want time in this nation so we hang on the same relations here for alpha 0 again and alpha 0 again because the same thing is minus because canonical quantization and additional minus from the fact that the time is equal to the metric that is okay so now let's take the reason and add them out to see what the combination relations of alpha plus minus plus with alpha minus suppose we add an m such that things click alpha plus m alpha plus minus m we get a plus m from the alpha ones but a minus m from the alpha 0 so you can see similarly with alpha minus and alpha minus so the thing is that if you have known 0s alpha plus alpha minus the combination relations are from these guys are alpha plus m alpha plus m is equal to alpha minus m alpha minus m is equal to 0 but alpha now let me read it out and check that it is correct alpha plus m alpha minus m is minus m let's check that out immediately alpha mu alpha mu mu mu that's right exactly that's exactly another bit of this and this minus sign is silent this minus is the main trick okay great so now let's consider the following let's consider the operator sum over m not equal to 0 okay alpha plus minus m this operator has a name it's called hs okay I want to understand what this operator does okay so let's see what this operator does we are acting on alpha minus k minus k of let's say minus of m okay yes it's a number operator that measures the number of negative photons in the minus direction minus the number of photons in the plus area let's take this okay you see suppose we act we look at a state with a minus photon so k is the positive now if we act on this operator we get something normal 0 only if we have alpha plus with a positive so m is negative in that case we get the stop mutator which gives us negative number the numbers cancel and then we replace it by the same state you see we got an m here the answer is the m here the two negatives we got are positive okay so hlc on this so let's go to the side hlc on the side is equal to alpha minus k over plus okay we can say the same thing but the opposite side so hlc on the side is equal to minus okay and in general what this operator does is measure the number of minus photons minus the number of plus see this operating is very close to being the boost generator the generator of boosts in in on the worksheet of the stream there's no role in what we're doing so I want to analyze the detail but I recommend you do the following exercise compute the no-thread charge corresponding to the generator of boosts in the one zero direction in comparison to this operator okay if I go up and over our constant it's exactly this for one fact you see here we demanded that m was not equal to zero okay the generator the no-thread charge that generates boosts in the one zero plane includes contributions from the zero so it doesn't have this m not equal to zero it doesn't have this m not equal to zero so apart from the contribution of the zero mode this is essentially the Lorentz generator of boosts in the one zero okay now so let's move on we don't have much time okay that's for one now the next thing that we want to do is a problem we want to take a beyond sd operator take qb so why I introduce this hlc at the moment totally empty it's going to be a formal object that will be used the player used to know so now let's take qb and decompose it it's hard to beat okay anyway there's chocolate in ten minutes okay let me just give you we may need many weeks from now so I probably have to repeat this but let me just tell you where okay you can take qb of brsd operator and decompose it into operators of different hlc network if you want to try this yourself check that the three hlc numbers that appear are zero minus one and one take the brsd charge a brsd charge does not have a definite quantum number on the hlc but you can split it up in the sum of three pieces each of which does have a definite quantum number and you can see clearly that the quantum numbers appear as zero minus one and one but the argumentation is that the brsd charges the range is very okay we'll continue next time okay so let's stop now