 OK, so I'll continue where I left off. There was one comment I wanted to make about the green short super particle before I forget, which is when we split theta dot a into a theta a, and we call this equal eta j bar over p plus, one way to think about this is that you have a constraint, a second class constraint. And of course, you have eight p's and eight theta's, but they're related by this constraint. So this choice here, what it does is it chooses four of the theta's to be interpreted as coordinates and four to be interpreted as momentum. So in some sense, it's solving this constraint. And instead of thinking of it as a second class constraint relating eight momentum and eight coordinates, it just allows you to express four of the coordinates in terms of momentum and vice versa. So you just think of these theta's and p's. This is like p. So this is one way to interpret what we did in the morning. So you're solving the second class constraint by just writing four of the variables as coordinates and four as moment. So any questions about what I said in the morning? So now we're going to discuss the pure spinner super particle and how to get it from this twister-like action. So I wrote down the super particle, the action. And I told you this lambda variable is constrained. And this is the definition of a d equals 10 pure spinner. And there was a question about how to define this in any dimension. So in any even dimension, as I said, it's just the definition that lambda alpha lambda beta is proportional to the phi form, or the self-dual form. So in dimension d, it's m1 to md over 2. So obviously, contract. So all the components, all the forms with fewer than d over 2 indices is set to 0. So in 10 dimensions, this is the one form, which is set to 0. OK, now there's a feature of this action, which we'll get to later, is there's no term of this type. We'll see later that you could put such a term in, but it turns out to be BRST trivial, so it doesn't have any effect. But in order to get physical states, you have to include a BRST operator. So the BRST operator is defined to be, and this is what was called d alpha. OK, so what we'll show in the first part of the lecture today is how this BRST operator, when it computes the cohomology of ghost number one, it just describes 10 dimensional on-shell super Yang-Mills. But before we do that, let me say a little bit more about this pure spinner. So it's convenient. Of course, this is a quadratic constraint, and one can solve it, but not in a covariant way. So the easiest way to solve it is to wick-rotate from SO 9-1 to SO 10. In fact, even more convenient is to wick-rotate to SO 5-5, but I'm not going to do that. If you wick-rotate to five dimensions with positive signature and five with negative signature, you find that this has a solution where lambda is real. We're not going to do that just because it will make life more complicated later. But we're just going to wick-rotate to SO 10. In that sense, oh, I should have said, there's something similar that happens in four dimensions. So people that use twisters in four dimensions like to work in signature SO 2 comma 2, because their two components, spinner, can be chosen to be real. So in 10 dimensions, if you work in signature SO 5 comma 5, you can describe a pure spinner as something which is real. OK, but in any case, wick-rotate to SO 10, and it will be convenient to decompose in terms of U 5 representations of SO 10. So in terms of U 5 representations of SO 10, a spinner splits up in this way and splits up into lambda plus lambda AB with A and B going from 1 to 5. So this is anti-symmetric and lambda A. So this is a 1. This is a 10 bar and a 5 of SU 5. So of course, there are 16 components here. This is just how a 10-dimensional spinner splits up into under U 5. And an easy way to understand this decomposition is using the notation where a spinner in any even dimension can be written with 5 plus or minus signs, which tell you which of the gamma matrices annihilate the spinner. So what you do is you split gamma M into complex gamma A and gamma A bar. So this would be like gamma 0 plus i gamma 1, gamma 2 plus i gamma 3, et cetera, up to gamma 8 plus i gamma 9. And these would be the complex conjugates. And the plus or minus tells you, well, this will be an eigenvector of gamma 0, gamma 1, gamma 2, gamma 3. So acting on this spinner with gamma 0, gamma 1, gamma 2, gamma 3, et cetera, will give you either plus or minus a half. So this sign here would be the eigenvalue of lambda when hit with gamma 0, 1. This sign here would be gamma 2, gamma 3, et cetera. And for example, if it has all plus signs here, it means it's annihilated by all of these gamma matrices, whereas these gamma matrices don't annihilate. Similarly, for example, if the first index was minus, then it would mean it's annihilated by this one and not by this one. So you can think of these as raising an index from minus to plus and just lowering it from plus to minus. OK, so this is convenient notation in any dimension. But now in 10 dimensions, obviously, you can have 5 pluses, 4 pluses, 3 pluses, 2 pluses, or 1 plus. Now it turns out if it's vial, then it has an odd number of plus signs. And if it's anti-vial, it has an even number of plus signs. So this one will be denoted the one with all 5 plus signs. So that's this component here. This component here will have 3 plus signs and 2 minus signs. And obviously, there are 10 different ways to choose it. That's why it's 10 bar. And this one will have 1 plus sign in the other's mind. OK, so this is different ways of thinking of a spinner in 10 dimensions. So are there any questions about this notation? And now what's a pure spinner? So a pure spinner satisfies this constraint here. Now obviously, the constraint is Lorentz covariant. But we're going to try to find a solution of the constraint, and we can't do that in a manifestly Lorentz covariant way. But we can say, suppose this component is non-zero. So this is what's called on the patch where lambda plus is non-zero. So suppose we're on this patch, one can ask, what are the constraints on the other components of lambda? So of course, it has 16 components, and this is going to constrain them. And what you find is that it doesn't put any constraints on lambda AB, but it implies that lambda A, why is it lambda A, is equal to 1 eighth. So obviously, this only makes sense when lambda plus is non-zero, because this is in the denominator. And you see that these components are not independent. They depend on these. Now this is on this patch, but of course I can do a Lorentz transformation to any other patch. So we see that it has 11 independent components. Instead of having 16, it only has this and these are independent. Now one can say this in a better way, which is that the space lambda alpha parameterizes the space SO10 over U5. This is a 10 dimensional complex space. So this has 45 real components, it has 25, so the difference is 20 real or 10 complex. And if you multiply by a complex scale factor, those are the 11 components. So you can think of these lambda ABs, there are 10 of these, these parameterizes space SO10 over U5. And then finally you have the scale factor, which you could think of as being the value of lambda plus. OK, in general, in any even dimension, it will always be SOD over UD over 2. Now this complex scale factor, we're going to require that it's non-zero, which means we're not going to allow all of the components of lambda alpha to vanish. That's why I put the star here. So it's the complex plane minus the origin. And this space here is what describes a projective pure spinner. So a projective pure spinner is lambda alpha defined up to a scale factor. So a projective pure spinner has 10 complex components instead of 11. OK, the only question? No. So if you add two pure spinners, you do not in general get a pure spinner. Because you have this quadratic constraint, which is obviously not satisfied. If you just, you get a cross term precisely like you said. It turns out in fewer dimensions, so if you go down to four or six dimensions, it turns out you can solve this constraint algebraically. So in four dimensions, the solution to this constraint is just a vial spinner in four dimensions. So it just has two components. Because lambda gamma and lambda in four dimensions is always of this type, lambda a lambda a dot. And in four dimensions, you can just set lambda a dot to zero, and you've defined up pure spinner in four dimensions. But once you go above six dimensions, in eight dimensions, for example, you can't algebraically solve it. So it's really a quadratic constraint. And 10 also. Any other questions? Yes? Yes? 16 bar of SO10. So you have SO10 as a vial. And antiviral representation. And that's right. It's the antiviral. Yes? OK, so let's look at gamma 0 plus i gamma 1. So we know that, first of all, gamma 0 plus i gamma 1, the square of it is 0. So what it's going to do is it's going to take, it's easier to say how it acts on it. So acting on lambda, which has a plus sign here and anything here, it gives me 0. And acting on something with a minus sign here, it gives me lambda plus, where these signs are the same, of course, is here. So you see that eight of the components are zero, and the other eight will take this to this. So I don't know, it's not one. But I think you understand what I'm going to write. And here, of course, it's zero and zero. So it doesn't, is the notation clear? So I think this notation is probably better. OK, and similarly for the other gamma matrices. There are other questions? OK, so if you ask how, so lambda here is a vial spinner because it has an alpha index up. If you have an alpha index down, and you ask how it splits up into under u5, it turns out to split in this way. So actually w alpha is something within an x down. So let's do that one. So that gives you w plus, wab, and wa. So this turns out to be like a 5 bar. This is a 10. And this is a 1, if you like, you can call it a 1 bar. And this object here is something with all, you wanted to have, sorry, you wanted to have an even number of plus signs, is that what you asked? OK, sorry. So this one would have zero plus signs. It would be minus, minus, minus, minus, minus. This one would have two, so the lamp, yeah. Yes, OK, so this isn't zero plus signs. This is two plus signs. This is four. Zero plus signs, two plus signs, four plus signs. Is that OK? Other question? OK, so why does this describe Super Yang-Mills? So first of all, q squared equals zero, as I showed you at the end of last lecture, because lambda gamma lambda equals zero. So you can define the cosmology. And I'll say much more about this tomorrow. So maybe I'll just try to convince you that this is Super Yang-Mills, but I'm not going to go into details, because I'd like to get to this Twister-like description today. OK, so v, if it has ghost number one, so the claim is that the cosmology of q goes number one. So at the moment, this will have ghost number one. This has ghost number minus one. And all others have ghost number zero. So of course, that's natural from the BRST operator. So it has ghost number one. So if it has ghost number one, it has to be, let's think of the lamb as being coordinates, and the w's as momentum. So it's only a function of lambda x and theta. It's linear in lambda, because it has to have ghost number one. So this is essentially the most general thing you can write down, which has ghost number one. And no one wants to ask, what does it mean to be in the cosmology of q? So here's q. So q of v is easily seen to be lambda alpha, lambda beta, d beta v alpha, where d beta is equal. So p alpha just acts like dd theta. And p, of course, is like ddx. Now this object here is just a 10 dimensional supersymmetric derivative. So one can easily show that this anti-commute with space-time supersymmetry, exactly like it does in Fortiment. So this is a constraint on this super field. Furthermore, the super field is defined in the cohomology. So it should have a gauge transformation of this type. Now omega is a super field of ghost number zero, because of course q of it should be a super field of ghost number one. So this is just equal to lambda alpha, d alpha. So we see what does cohomology mean. So this has to be zero, of course, which implies that d beta, a alpha, when contracted with lambda alpha, lambda beta is equal to zero. But the definition of a pure spinner is that this combination is only proportional to the five form. So this implies that gamma m1 to m5, alpha beta, d alpha, a beta equals zero. So this is the equation of motion on the super field. And of course, the gauge transformation from here tells you that delta a alpha is equal d alpha of omega. It's clear that this leaves this equation of motion invariant, because the anticormutator d alpha and d beta is proportional to the gamma matrix with one vector index. OK, now these two equations together are enough to describe super angles. So what you do is you expand this in powers of theta. The lowest component, of course, only depends on x. You can write terms of this type. This lowest component, it turns out you can remove a version called lambda. You can remove by choosing omega appropriately. So this turns out to be pure gauge. So if you choose omega to be equal theta alpha, let's write it like this. It's equal to f plus theta alpha g alpha. Then it's easy to see that d alpha has the term d d theta. So you can just choose g alpha properly to kill this. So this could be gauged away using g alpha. Now f is just a gauge transformation that generates the usual gauge transformation of AM. And it turns out that if you work this out, you find that this only satisfies this equation if AM satisfies the usual super Maxwell equation. So this here implies dm. So that's the usual Maxwell's equations written in terms of the gauge field. And of course, the gluino should satisfy the usual on-shell Dirac equation. So this just describes the on-shell, d equals 10 super angles. But of course, it does it in a covariant way. Now this is in superspace, but you should be aware that it is just equations of motion. So you might have heard the story that you can't do 10-dimensional super angles in a manifestly supersymmetric way. It's true. We don't know how to write actions in a manifestly space in a supersymmetric way. But we do know how to describe equations of motion. So these are just equations of motion for d equals 10 super angles written in superspace. Any questions? OK, so this is the answer. But what is the starting point? So I already mentioned, it's a little bit mysterious that this action, although it's gauge fixed, there's no EP squared term. And furthermore, there's no BC ghost, as you might have expected if it came from gauge fixing some world line reparameterization in variant action. So similarly in the BRST operator, here there's no CP squared term. So it doesn't look like the usual thing you get from gauge fixing a world line reparameterization variant action. So here comes the new feature, which is the claim is that it comes from gauge fixing a purely bosonic action. So the idea is the theta alpha variable, as I already mentioned here, is a world sheet scale, or a world line scale in this case. And it's going to come from gauge fixing a constraint, which is spinorial in spacetime, which is constructed from p slash and lambda. So the constraint, which will be essentially the starting point for this description, is p slash lambda equals 0. This is called twist you like, because if you work in four dimensions, so pen rows, of course, only describe twistures in four dimensions, but the idea is that in four dimensions, if you have a null vector, you can write it in the form lambda a dot sigma m a dot. And this, of course, identically satisfies p slash lambda equals 0. So if you try to solve this constraint, you find that p can be written in terms of lambda. Now, in 10 dimensions, things are a little bit more complicated. So this lambda has now 11 independent degrees of freedom. So we're not going to try to solve the constraint, but one thing you can see immediately is that this constraint implies p squared equals 0, just as it did in four dimensions. And that's trivial, just if you hit it with another p slash. So literally what it implies is that p squared times lambda alpha is equals 0. But we already decided that lambda alpha cannot be vanishing because of this c star. So this implies p squared. There's one more thing that I should say is that, of course, this constraint does not require the scale part of lambda. So this lambda is going to be a projective pure spin. OK, so that's the starting point. And now we're just going to write down an action built on that starting point. So the action is very simple. There won't be any thetas in the classical action, but you're going to have this constraint times the Lagrange multiplier for the constraint. This is the Lagrange multiplier. And the claim is that quantizing this action in different gauges will give either this pure spin or description or the green Schwartz description. OK, now the surprise, of course, is that this action does not have any fermionic variables in it. So the claim is that the super particle is going to emerge, which I put in a minus 1 half, yeah, I know. It's going to emerge from this only after the quantization. OK, now there's one thing that I didn't say, which sorry, there's a mistake here. I want the action to depend on a projective pure spin. So the action should be invariant under scale transformations of lambda. So scale symmetry will be imposed by introducing a covariant derivative here, where the covariant derivative is defined to be. So A is just going to be a world line gauge field. And the scale transformation will be lambda alpha goes to omega lambda alpha. W alpha goes to omega inverse W alpha. The Lagrange multiplier will transform with omega inverse so that this is scale invariant. And finally, the gauge transformation will transform in this way to leave so that this scale is covariant. Sorry, there's no alpha on the A. OK, so that's the action. There's a scale symmetry of this type. And there's also a gauge symmetry, because here you have a first class constraint. So in addition to the scale symmetry, we're going to have a first class constraint, which generates the following gauge symmetry. So we want to describe the particle. And we want to end up in this gauge fixed description, or green Schwartz. All I can say, OK, why did I introduce it? Because it works. I don't know. There's not many things you can do which are simple. And this is certainly simple. Of course, we already had the idea that a pure spinner is in the game. And it turns out that in the 80s, some people, not Penrose, some people were trying to generalize twisters to higher dimensions. And what they found is that pure spinners are the generalization of twisters to higher dimensions. OK, I don't have too much time to go into this, but papers by people like Lane Houston and also Lionel Mason worked on this in the 80s. So of course, pure spinners in this language, I should have mentioned also, it didn't come from that. It came from actually some work of Paul Howe, who showed that integrability along. So Witton showed in the 80s that integrability along light-like lines was related to caposimetry in the green Schwartz super particle action. And Paul Howe showed that in the 90s that integrability along pure spinner lines were related to super Yang-Mills also. OK, so in some sense, it was natural to think of pure spinners as twisted variables from those two points of view, either trying to generalize what Penrose did in four dimensions or to understand what Witton and Howe did in the sense of integrability along these spaces. OK, so of course, there's been a lot of recent work on twisters that I guess Freddie will talk about. I don't know if somebody talked about already. I guess Simon probably mentioned twisters also in four dimensions. And usually, when people describe N equals 4 super Yang-Mills in four dimensions, I would say 90% of the features of that theory are also present in 10-dimensional super Yang-Mills. So of course, dimensional reduction reduces 10-dimensional super Yangs to four-dimensional super Yang-Mills. You lose four-dimensional conformal invariance when you work in 10-dimensions, but you gain lots of new symmetries. So it wouldn't be surprising, I think, if twisters play an important role in 10-dimensions just as they played in four dimensions. OK, so by the any case, I'm doing it because I can and because it's simple. And it will generalize to the string, of course. Otherwise, I wouldn't waste your time. OK, so the gauge symmetry generated by this, you have the usual transformation of the Lagrange multiplier. With the hindsight, I will realize the gauge parameter here is going to be interpreted as a ghost, which I'll call theta, so I'll already call it theta here. The transformation of x is just, this is the gauge parameter, remember, and p slash lambda just generates this type of gauge transformation for delta x. And of course, delta w, because w and lambda are conjugates, it will generate a transformation of this type. Now there's another gauge symmetry which is not generated by p slash lambda. So this just comes by using the canonical commutation relation of p slash lambda with x and with w. And l is the usual way that the Lagrange multiplier transforms so that the action is invariant. But there's an additional symmetry which is of this type where rho is a new gauge parameter and that just comes from the fact that lambda gamma lambda equals zero. OK, so that will also be relevant. So these are the gauge symmetries and also there's an extra gauge symmetry coming from the lambda gamma lambda equals zero constraint. OK, there any questions? There's one more thing I can say which is suppose lambda is constant. If lambda's constant, of course, this term goes away. And this term, if lambda's constant, just looking at how lambda acts, for example, suppose that the only non-zero component is in this direction here. So lambda's constant, the... So let's say it's just in this component, one, zero, zero. This implies that the constraint has only five independent components. So if you split Pm into Pa and Pa bar, it implies that Pa equals zero for A equals one to five. OK, we'll have split Pm into... So just into five complex components. Now, for people who work with a topological string, this is essentially the constraint for a topological string. You impose this becomes dxA equals zero. So what lambda does, because it parameterizes SO10 over U5, it parameterizes how the complex structure is chosen. So the usual topological string, what you do is you fix the complex structure. And then you say, for example, that dxA vanishes where A is the holomorphic exit. What I'm saying here is that if you start with a topological string and you integrate over the choices of complex structure, you generate the super string. So it turns a topological string, on this case topological particle, into the super string. So that's another nice idea that follows from this construction. OK, any questions? OK, so let's see how to quantize this. OK, so in addition to these gauge symmetries, there's a further gauge for gauge symmetry. The gauge for gauge symmetry comes from the fact that if you look at this transformation of L alpha, if I shift rho and I shift theta in opposite directions, I can get a compensating gauge for gauge symmetry between them. So the transformation is what's called delta prime. If I shift theta in the direction proportional to lambda, and I also shift rho to be equal to del phi, then these two terms almost cancel. So first of all, this term doesn't change because theta lambda gamma lambda equals 0. This term will shift, but there's a gauge symmetry associated with lambda gamma lambda that I should have said from the beginning. This constraint here implies that W alpha should be identified with W alpha plus. So lambda gamma lambda implies that only 11 of the 16 components of lambda are independent, which also implies that only 11 components of w should be relevant. And those are the gauge invariant Ws under this gauge transformation. So this transformation here should be thought of as a gauge transformation of W. And you can see that shifting theta by lambda is of that type. But this doesn't quite cancel because you can get a derivative here acting on the phi coordinate. So this is a gauge for gauge parameter. And it turns out that this is not quite invariant. You find that delta prime of delta L is equal to phi del lambda L. But this can be canceled by doing a shift of W because, of course, the equation of motion from W is del M equals 0. So finally, there's a shift of W. OK, so what is chi? Chi is the gauge fixing fermion. So what we're going to do is when we find gauge fixing in the usual BRST fashion, by choosing chi as we like and saying that the gauge fixed action is equal to the classical action plus q times chi. So for people that aren't familiar with this, OK, this is standard in any BRST or any textbook that discusses BRST. And now if chi depends on L alpha, this thing here implies that q acting on q chi, or if you like, delta prime of q chi is going to be equal to delta chi delta alpha times phi del L. So that's why we need this shift of W in order to cancel this shift of q chi. So in other words, the BRST operator's nil put it up to equations of motion. Equation of motion is del lambda. And to get rid of that equation of motion, I have to shift W. OK, so putting this all together, we're going to find the BRST transformations as follows. OK, now remember theta is a ghost. Rho is a ghost. So these are fermionic. These are fermionic ghosts. But phi comes from the gauge for gauge symmetry. So phi is actually a ghost for ghost, which means it's a bosonic and not fermionic. So every time you have a ghost, you change statistics. So if you have a ghost for ghost, it becomes bosonic. OK, so it looks a little bit complicated, but it's pretty straightforward. So the idea is you start with this action. You describe the symmetries. It has a scale symmetry, including this world line gauge field. It has this gauge symmetry generated by the first class constraint. But then it also has gauge for gauge symmetries. And this is the gauge for gauge parameter. And then putting all these symmetries together, you find the BRST transformations. OK, so that's essentially the game. There's one more thing I should mention, which is under the scale symmetries, of course, this is how the matter fields transform. But of course, theta also has to transform. So under the scale symmetry, theta goes to omega inverse theta alpha. And phi goes to omega minus 2 phi. So that just follows by requiring that all of these transformations are scale invariant. OK, so that's the story. And now we're going to gauge, fix, and get either pure spin or green schroids. But are there any questions? So I think what you're asking is there are different ways to gauge fixes, gauge for gauge symmetry. One way to gauge fix it is just as you said, for example, we can remove rho. So if you look at how rho transforms, you can of course choose a phi so that rho will be gauge to 0. But you have to be worried that that will have some ghosts. Because rho transforms with the derivative, that will introduce ghosts. So yes, the gauge for gauge symmetries are essentially there in order to remove some of the gauge symmetries. But the way do we move them, it's important to keep track if you get derivatives so that you might get ghosts coming from it. As you like. So I will tell you the ones that give green schroids and pure spinners. OK, any other questions? OK, so let's first do the pure spinner. If we have time, we'll also do the, OK, I don't need this. Yes, please? Yes? Yes, I should have written it if I forgot to. So this action here. So I wrote down this PRS-T operator, but I didn't tell you where it came from. OK, so what's the gauge fixing fermion we're going to choose? So for pure spinner, we're going to choose chi to be, to gauge fix the Lagrange multiplier to 0. So what that means is that you write chi to be L alpha times an anti-ghost, which multiplies that Lagrange multiplier. So in general, if you want to gauge fix something to 0, you just take the object you're going to gauge to 0 and multiply it times the anti-ghost. So we'll see exactly why that does it. And we're also going to gauge rho to 0. So this rho here, we're going to gauge that to 0. So we're going to have an anti-ghost for that. Now, of course, this is bosonic. This is fermionic. But this is fermionic, so this is bosonic. So this is really an anti-ghost for ghost, whereas this is an anti-ghost. OK, so q chi, we have all the PRS-T transformations. The only thing I haven't told you is what is q acting on the anti-ghost. So the standard thing to do is to introduce something called the Nakanishi-Lautru field of this type. So this transforms into an auxiliary field. This transforms into an auxiliary field. And then the auxiliary field of PRS-T invariant. So these are called Nakanishi-Lautru field and L field. So this is all completely straightforward PRS-T language. OK, so there are two contributions. One when the q acts here, one when it acts here. So obviously when it acts on p-alpha, you get m-alpha l-alpha. When it acts on l, OK, it's fermionic. So you get a minus p-alpha times. So that just, I'm reading off from here what q of l is. I guess a head minus. Then you have q of beta, that's n. So you get n-row. And then finally you get beta del phi. So you see the q of chi multiplies the Nakanishi-Lautru field by the l-alpha. So obviously the equation of motion for m is going to be l equals 0, which is the gauge we wanted to impose. OK, so the full action is equal to sc plus q chi. This should be del. I don't know, sorry. OK, so these contributions, of course, were from the classical action. This comes from this term here. This comes from this term here. And I've explicitly written the gauge field that appears in these connections. And of course I've used the equation of motion that rho equals 0 and that l equals 0. Now we still have the scale symmetry. And now we have a choice how to gauge the scale symmetry. But a Lorentz covariant choice would be to use a scale symmetry acting on phi to gauge phi equal 1. Now this is only possible if we assume that phi of t tau is non-zero for every t, for every tau. So we have to assume that there's non-vanishing on the world line. If it is, then we can just use the scale symmetry to gauge it to 1. Now this is a bit bizarre because phi carries ghost number. So phi carries ghost number 2, in fact, because it's a ghost for ghosts. So what the scale transformation does is it shifts the ghost number of other fields. So let me explain how that works. So suppose we have an operator of scale weight s and ghost number g. So if you want to write something which is scale invariant and which gauges to this in the gauge phi equal 1, then the object to construct is phi to the minus s over 2, O sg, because this obviously has scale weight minus s, because phi has scale weight 2. So this is a scale invariant operator which in the gauge phi equals 1 reduces to this. But this object here has ghost number not g, because this has ghost number s. Because phi has ghost number 2, phi has scale weight minus 2. So this should have been a plus s over 2, sorry. So phi transforms like omega to the minus 2. So in order to be scale invariant, this should be a plus s over 2. So this has ghost number s or g plus s, because this has ghost number. So what we've found is that an operator which has scale weight s and ghost number g, after gauge fixing, it's actually the relevant state has ghost number g plus s. So giving a non-zero expectation value to a field with ghost number is going to shift the ghost numbers of the other fields. So what does this mean? So for example, theta alpha started out having ghost number 1, because it was a Fadea pop of ghost. It has ghost number 1. Just if you look at how it transforms. But it has scale weight minus 1. So that means the ghost number after doing this scale shift is going to have ghost number plus 1 minus 1 is equal to 0. Whereas lambda alpha started out having ghost number equals 0 because it was a classical field. But it has scale weight plus 1 because of this scale weight here. So after doing the shift, it's 0 plus 1. So it has ghost number plus 1 as we want. So you see that using the scale symmetry to gauge phi equals 1 because phi carries ghost number plus 2, it's going to shift the operators, shift the ghost number of the operators, into precisely what we wanted in the pure spinophonism, where this now has ghost number equals 0. OK, so the last thing we need to do is see how this affects the action in the BRST transformation. Let me do that here. So the action after gauge fixing phi equals 1. First of all, the equation of motion in the gauge phi equals 1 is that a equals 0. So if you just, the equation of motion for beta is just going to be a equals 0. And it's going to implies that beta is equal to minus 1 half, I guess, plus 1 half, w alpha lambda alpha plus p alpha theta alpha. So this just comes from varying a in the action. And of course, this term drops out if phi is 1. So the action is just, OK, it should be exactly the same as here. And we can read off the BRST transformations from that line there. So we find that q of theta is equal to lambda alpha. q of xm is equal to 1 half lambda gamma m theta. q of w alpha and q of p alpha. So these just come from that line there, plugging phi equal 1. OK, so these are precisely the BRST transformations that are generated by this. OK, so we've landed on the pure spinophonism. OK, the only question? OK, so if you had world line supersymmetry, I would say the answer should be s. We don't have it, so I can't guarantee. So if we were doing the stream, then you would have an equal number of bosons in fermions. Because the critical, it will turn out all the world sheet fields will have either conform weight 0 or 1. And so in order to get the central charge to vanish, you need equal number of bosons in fermions. For the particle, that's not true. So we can think about green Schwartz. So green Schwartz, OK, I don't know. We can look at it here. I'm sorry. q is BRST symmetry. So it takes bosons into fermions, but it takes it with derivatives and things like that. So it doesn't just take a field into another field. It takes sometimes with derivatives, et cetera. q squared is equal to 0, yes. That's right. But it's nilpotent. So it's not like spacetime supersymmetry, which is q with q is equal p. So I don't think q squared equals 0 implies that. Well, look at the bosonic string. q squared equals 0 for the bosonic string, but there aren't equal numbers of bosons in fermions. OK, so I don't think you have an equal number of bosons in fermions for the particle, for the string, you will. OK, are there other questions? OK, so I'm not going to have time to do green Schwartz today. I'll do that first thing next time. But let me just say what's going to happen. So what's going to happen for the green Schwartz is that you're going to choose a different chi. And obviously, this W alpha has no role in green Schwartz. So what we're going to do is we're going to choose a gauge in which W alpha is BRST invariant. What will happen is that's going to produce another charge, which is actually 0. So it will produce BRST transformations which don't completely gauge fix. So obviously, when you have a local symmetry, the other charge is 0. So the fact that the resulting BRST charge will be 0 will imply that actually you haven't completely gauge fix as a local symmetry that survives. And that local symmetry will just be kappa symmetry. OK, so we'll do that next time. I won't have time to do that. OK, so let's start.