 Okay, so are there any questions from the previous lectures? Okay, so let me start just by reviewing what we're doing at the end of yesterday. So the idea is to derive the pure spinner and green Schwarz formism from gauge fixing, an action which is classical action which is purely bosonic. So it's just built out of the usual space time variables, X and P, together with this projective pure spinner and its conjugate momentum. So just to remind you, projective pure spinner intended to satisfy this constraint, and the conjugate momentum has this gauge invariance, and it's defined up to these scale transformations. Okay, so this is this twisted like constraint which will replace the mass shell constraint. L alpha is Lagrange multiplier. So this is a constraint whose on-shell which is imposed on-shell, but of course off-shell when you do BRST, use the BRST method, this will not be imposed anymore off-shell. Okay, so the BRST method is of course you look at the gauge transformations generated by the constraint, and the gauge transformations are of this type where theta is the gauge parameter, and in the BRST method the gauge parameter becomes a fermionic ghost. So the theta, which is normally thought of as a classical variable, is now thought of as a ghost variable, which makes sense if you look at its world-line spin, which is a world-line scalar. Okay, so this is the transformation of X, this is the transformation of the Lagrange multiplier, so it has the usual term del theta. There's an additional term you can add here because lambda gamma lambda is zero, and because of this term you have a gauge for gauge symmetry, which means that if you gauge theta in one direction using phi and gauge rho in the opposite, well okay, because of the minus sign, it's in the same direction with del phi, these two terms can cancel. In other words, the q squared is zero even though theta and lambda vary in this way. Sorry, theta and rho vary in this way. Okay, so this is part of the BRST transformations, but this phi is a ghost for ghost, because rho carries ghost number one, which means that phi should carry ghost number two. Okay, and w alpha, it picks up a piece because of the canonical conjugate to lambda here, but it picks up another piece depending on the gauge fixing fermion in order that q squared equals zero. So if you don't have this piece here, q squared will only be zero up to equations of motion. Okay, so we did the gauge fixing, which led to the pure spinor formalism, which was of this type. So p alpha and beta are anti-ghosts, and what we showed is that if you add q of chi to this classical action, and solve the equations of motion for the auxiliary field, you get just the usual pure spinor gauge fixed action, but this is in the gauge where you use the scale symmetry to fix phi equal one. So because you've fixed phi equal to one, that shifts the ghost number so that after doing this gauge fixing, theta carries ghost number zero instead of ghost number one, and lambda was now going to carry ghost number one instead of ghost number zero. Okay, so that's a review of what I did at the end of yesterday. So is there any questions? Okay, so now we're going to choose a different chi, which is supposed to give you the green Schwartz super particle. So the chi we're going to choose is going to have the property that q of w alpha is going to be zero. So we're going to choose, the reason is because we want to land on the green Schwartz super particle action, which of course has no w's or lambdas in it. So if we look at the transformation of w, we can see that it's possible if it's going to be equal to minus one-half L times phi inverse. So this has the property, obviously, if you take the derivative with respect to L and multiply by phi, it will cancel this term. Now we're still going to have this term here in chi. That of course doesn't depend on L, so it doesn't spoil the property that q w L equals zero. Okay, so now we're going to compute the action that we get by just hitting with q chi. So in this case we'll get, so it's trivial to use the BRSC transformations to compute what q of chi is. So from the transformation of L, this term, and from the transformation of theta, you get this term, and finally from the second term you get this from the transformation of rho, and from the transformation of beta, you get where n is just the Nakanishi-Lautro field, so q of beta is equal, q of n is equal to zero. Okay, now if you look at SC, you can see that the two terms here have the same form but with the same sign, so they don't cancel. So this is equal to detail. And now like before, we still have the scale symmetry, and we're going to gauge fix it in the same way. So we're going to gauge fix phi equal one, which again will imply that theta has ghost number zero. Now once you do that, that implies the equation of motion as before. Now if I write this, how the gauge field couples, the gauge field here couples to w alpha lambda alpha, then you have a coupling also to beta phi. So if you set phi equal to one and vary with respect to beta, you get of course that a equals zero. So this becomes an equation of motion and of course you also get that beta is equal to minus one half w alpha lambda. So plugging in these equations of motion, this action becomes now p m x dot m plus alpha. Now it's just an ordinary derivative. Okay, and the others are this zero and this becomes just an auxiliary equation of motion. I forgot a term. It's important. Theta alpha dot. So rho equals zero using this equation of motion. Sorry, this term should have been in here also. So this term here contributes just minus one half theta dot piece. Okay, any questions up to now? Okay, now because q of w alpha equals zero, one can ask if this term, so this term is invariant under the BRST transformation, someone can ask if it can be written as something BRST trivial, in fact it can. It can be written as q of w alpha theta dot alpha. So there's also a phi here, but since we've gauged phi to one we can just ignore them. So if we add this term to chi, so actually in chi it would be delta theta alpha, then we can cancel this term. So now this action here is very similar to the Green-Schwarz action. So remember the Green-Schwarz action had this term here, but instead of this term here, the Green-Schwarz action had the term e p m p m. So I guess I wrote e over 2 in my notes. So you don't quite get the Green-Schwarz action, but you get the Green-Schwarz action with this term here replacing the usual term implying the mass shell constraint. Now of course one still has the BRST transformations. So let's see how the BRST transformations act on these fields. You can of course read them off from here. So you get q of theta alpha is equal to lambda alpha, q of x m is equal to one half lambda gamma m theta, q of l alpha is equal to theta dot alpha. Now this term here has the property that when you vary l you get p slash lambda equals zero. So equation of motion p slash lambda equals zero. So in this case unlike in the pure spinner formalism where the gauge fixing removed this term here because you've gauged l to zero, in this gauge fixing you don't gauge l to zero. So the equation of motion is p slash lambda equals zero. Now this implies obviously that p squared equals zero because if you hit it with p slash you get p squared times lambda alpha equals zero. And it also implies that lambda alpha equals p slash kappa for some kappa, for some kappa alpha. So that just follows from the fact that assuming that at least one component of p m is non-zero p slash lambda implies that lambda can always be written as p slash. Now if you compute the BRST transformation, the BRST charge, the nerder charge associated with this symmetry. So this is of course a global symmetry of the action here and you can compute its nerder charge. What you find is that the nerder charge associated with this is zero. So the nerder charge which we could call q is actually zero. What that means is that this is actually a local symmetry. Although you thought it was a symmetry just for if you put epsilon here being a global parameter it's actually a symmetry if this epsilon is a local parameter. So what that means is you haven't completely gauge fixed the symmetries. Now that's of course related to the fact that this is a symmetry for any kappa. So there's a local fermionic symmetry which has survived this gauge fixing. And it's easy to see this is just kappa symmetry by just replacing lambda equals p slash kappa here. So this becomes, well I can write this as one half delta theta gamma m theta. And this is precisely the usual kappa transformations of the Green-Schwarz action where this term here implies that this term here transforms like this is equal to minus theta dot p slash lambda plugging in the solution for lambda here this is just equal to minus theta dot alpha kappa alpha times p squared which is precisely the kappa variation of e. So this replaces kappa transformation. Okay so we see that this action here which comes from gauge fixing the Twister like action using this gauge fixing fermion reproduces the usual Green-Schwarz super particle action but with this term here replacing the usual world line variable times the mass shell constraint. Okay, are there any questions? Thank you. Which line? Yes, minus one half delta theta minus rho lambda alpha. What's the problem with the indices? This is p slash. It's okay. This is a problem. Oh here also. Which m are we talking about? Oh sorry, this is kx dot m. Sorry, this p slash should not be here. That was more serious. Are there any other questions? Okay, so that's the Green-Schwarz. Now of course this was the particle. Now we want to do the string. So of course both the pure spinner formalism and the Green-Schwarz formalism can be constructed as a string theory and in fact there are also pure spinner and Green-Schwarz formalism for other kinds of brains like the super membrane also has a Green-Schwarz and pure spinner formula. So in fact this method works not just for the super particle but it works for the super string and for the super membrane. I don't think I'll have time to discuss the super membrane but now I'd like at least to sketch how it works for the string just in the pure spinner formalism because that's the one I'm going to be using to compute tree amplitudes and multi-loop amplitudes. Okay, so the idea is of course to generalize this to the string. So I'll just discuss the type 2, A and type 2B but for heterotic it's trivial to generalize. Treat the right moving variables the same like you do in the bosonic string. Okay, so it's simpler to generalize using first-order form but I will sketch how to write the action in terms of the usual second-order form with just X's later. So you have lambda alpha and W alpha. These are projective pure spinners as before but now you also have another set of lambas and W's which I'll call lambda hat and W. Alpha and alpha hat go from 1 to 16. They have the same chirality if you're doing type 2B type 2B an opposite chirality. Now as before we're going to introduce a constraint of this type. Of course these are independent pure spinners so lambda gamma m, lambda equals 0 and lambda hat, gamma m, lambda hat. They have scale symmetry similar to this. You're going to have scale symmetries independent on both sides so you'll have scale symmetry under omega of lambda and let's call it omega hat. So you can scale them independently and instead of having the constraint p slash lambda equals 0 the constraints we're going to be using are p slash plus d sigma. So this will be the constraint on lambda and the constraint on lambda hat will be with the opposite sign here. This constraint here of course implies the usual left and right moving via resort constraints but it turns out that unlike in the particle where p slash lambda commutes with p slash lambda because p and x do not commute it turns out this constraint does not close the algebra of this constraint doesn't close by itself. So you need to add also the constraint d sigma of lambda alpha equals 0 and d sigma or more precisely del sigma lambda hat alpha where del is going to be defined in the same similar way to what we did for the particle we're going to have del i of lambda alpha is equal to d i of lambda alpha plus ai of lambda alpha where i is either equal to tau or sigma and similarly for lambda hat but now we need an a hat because we have independent scale symmetry for both lambda and lambda hat so we have two world sheet gauge fields now so it's easy to show that this constraint together with this constraint that now closes to an algebra okay so the classical action now is going to be very similar to what we had before so you'll have in Hamiltonian form you just have so this is in the tau direction but now you have these constraints so you have a Lagrange multiplier for each of these constraints now times p slash lambda you have l hat times p slash minus d sigma x slash lambda hat and you also need a constraint for okay so this is the starting point so if you like the constraints imply that lambda alpha has no sigma dependence but it's not a constant okay so if it were a constant direction if lambda alpha were a constant direction it would be like the topological string where okay lambda and lambda hat in principle you could choose two different complex structures but for the string now you're going to integrate over those choices of complex structures okay so the claim is the gauge fixing this is now going to give the super string okay so any question sigma or tau so i can either be sigma or tau so the constraint only involves the sigma direction so this action here when you write things in Hamiltonian form the Lorentz invariance on the worldsheet does not manifest in a second I will write this in a way in which the worldsheet invariance is the reprimandization invariance would manifest any other question okay so to make the worldsheet reprimandization invariance manifest what you have to do is you have to integrate out the PM okay and go to the second order form if you try to write this even bosonic string in first order form you can't do it in a manifest Lorentz invariant way now a term that is that you would like to have here which is not here is the term this here is the stress tensor so this is equal to ET and this is equal to ET bar so within a Hamiltonian form if you're doing bosonic string theory in Hamiltonian form the left moving stress tensor just p plus dx squared d sigma x squared and the right moving is p minus d sigma x squared because we have these extra degrees of freedom they also contribute to the left and right moving stress tensor but it's clear the way we defined L and L hat the way we looked at the constraints that these are actually these terms here can be removed just by a suitable shift of L and K so that's just because this constraint here implies this constraint here and this implies this so adding this term to the action actually does nothing it adds a trivial gauge symmetry which you can remove it you can remove this term just by shifting L and K but it will be convenient to add these terms in order to integrate out p and then we'll be able to write the action in a more contrastly Lorenz invariant way okay so later let me tell you exactly what the shift is because that will play a role tomorrow when we discuss how to do loop amplitude so the shift we have to do is of this type so where lambda bar you can think of at the moment as just being a constant this should be lambda bar hat sorry for all the notation lambda bar alpha is a constant spinner of the opposite chirality of lambda so you can choose it to be however you like because when you contract this with p slash lambda or p slash plus d sin x slash lambda you'll just get a factor of lambda lambda bar over lambda lambda bar so for any choice of lambda bar alpha where lambda bar has a down index this term here will produce this term here I guess as a factor of a half maybe okay so this is the kind of shift we're going to have to do we're going to have to shift L by something proportional to this and L hat proportional to something like this and of course also we're going to have to shift K okay so this shift here it will be relevant in the next lecture when we discuss the b-ghost but I don't want to do that yet but in any case adding this term doesn't change the action but now you see p appears quadratically in the action and it becomes an auxiliary field it can be integrated out now it's a homework exercise to do the integration and you'll find that as C it can be written in this form okay so this is manifest the world-sheet re-preemptorization invariant the relation between this E and this E bar this E is of course the world-line verbine which is a 2 by 2 matrix which you can choose to be written in this form so the action is of course invariant under world-sheet conformal transformation so you can get rid of the scale factor it's also invariant under the U1 transformations local U1 transformations of the world-line verbine which allows you to make these two components equal and the remaining two components are just this E and E bar so all of these derivatives here are defined in the usual way using the world the world-sheet verbine so this is the manifestly world-sheet re-preemptorization invariant way of writing this action so here with the particle this first order form is convenient so in general when you work with constraints the Hamiltonian language is convenient okay are there any questions? okay now the gauge fixing procedure yes so of course you can start with this action so you can fix the gauge in which E and E bar is equal to 0 so that would be conformal gauge light cone gauge but you would fix something like X plus is equal to 1 but you still have this Lagrange multiplier so you still have additional gauge symmetries the world-sheet re-preemptorization symmetries are not the whole story because you have other constraints related to this Lagrange multiplier not, I don't know any way to fix them without getting ghosts it might be possible but I don't know how to do it so of course if you could do it you would have no fermions in the game except for the usual BC ghosts but okay the only way I know how to gauge fix it is the ways I'm describing here okay any other questions? okay so I'm not going to go to what I did yesterday for the particle but it's essentially the same method again you're going to use the scale symmetry to fix phi equal 1 in this case of course you'll have phi and phi hat for independent scale symmetries once you do that of course the theta's become ghost number 0 and you get to just in a flat background you just get to a quadratic action using the pure spinner formula of course you can also do the green shorts gauge fixing and then you land on the green shorts super string action so this is all discussed in detail in the reference I gave so for people who want more details it's disgusting so I just in order to save time now I just like to go to the answer and show how to use the answer to compute scattering amplitude okay sorry yeah good point so in this in the sorry I erased the green shorts so the usual green shorts action of course you can also write it in first order form the green shorts action and now of course instead of being just PM squared equal 0 you had a term you have a term of this type PM plus d sigma xm and there's also some theta dependence here so you have this type of term and again when you write it in terms of the L what you'll get is something of this type so but now you do the same trick as before you you use this equation of motion for lambda to imply that lambda alpha can be written as this object here times gamma m kappa and then this is just the usual kappa transformation so you have to do the same the same procedure okay any other question so this is all discussed in that paper okay so now I'd like to just give you what the final gauge fixed action is so again all of this is going to be in the notes that I'm writing up so if I'm fixing too fast you don't have to worry because it will be on the web page so tomorrow it will be on the web page for sure okay so what I'm going to talk about in the rest of the lecture today is things that I said 10 years ago but obviously it's important if you want to compute multi-loop amplitude so this is not related to the gauge fixed pure spinner string so after doing this gauge fixing and you also have the left the right moving sector so obviously the theta and theta come from the ghost for L and L hat and the P and P hat come from the anti-ghost now just to reduce the amount of writing I'm going to leave off the added variables okay let me just concentrate on the unhatted variables so the right moving stress tensor is minus one half dxm dxm plus wl for d lambda alpha minus pl for d theta and the first thing you can ask is is there a conformal anomaly so if you take the OPE of t with t it's just a free field theory because it's just quadratic in the variables of course lambda is constrained and I will tell you in a second how to work with that constraint but the central charge is just obtained by just looking at the conformal weight there are 10 scalars these all have conformal weight 0 and 1 so there are 11 of these that are independent so this contributes plus 22 and there are 16 of these each of which contribute with minus 2 because they're fermionic so the central charge is 0 as desired so there's no conformal anomaly now there's another test you can do of the theory which is you can look at the Lorenz generator it's more precisely the Lorenz generator for the spin so in RNS of course it's just psi m psi n in pure spinners this gets a contribution from the W's lambdas and also from the p-thetas so it's trivial to write down the Lorenz generator and now we can compute the OPE of m with m now it turns out these of course are holomorphic on the worldsheet but there's a double pole here which in general has a level so there's a term which is proportional to eta mq eta pn which is a double pole if this is at y and this is at z it goes like y minus c squared and then there's another contribution which is of course just the usual the usual algebra of SO91 which is proportional to m which is just a single pole now the level for RNS is k equal 1 that's trivial just you just take the OPEs of psi psi with itself you'll get just a single the double pole just has a numerator of 1 if you do it here of course you get a contribution of this with itself and this with itself now the contribution of this with itself if you have no constraints you get plus 4 from this part but computing with the pure spinors is a little bit trickier because of the constraint one way to compute is just to solve the constraint in a non-covariant way the way I discussed yesterday you solve it in terms of lambda plus, lambda a b and lambda a and then you compute the OPEs in this non-covariant way but what you find is of course the OPE of this with itself is fully covariant so if I call this object n what you find is that nn it has the following OPEs it's minus 3 over y minus c squared a to a plus of course a to n so the level is minus 3 so it has desired these two match and that turns out to be important when we want to show that the amplitudes agree in the computations okay so this is the evidence that the theory is doing what you want so I'm just doing the left movers so for the moment I'm just ignoring the right movers but it's completely equivalent to the right moving computation any other question? okay there's no bc-ghost here yes but I chose the gauge so you could have asked this already for the particle so I chose the gauge remember I chose the gauge L alpha equals 0 if I had written it in this way and chosen the gauge E equals 0 I would have got bc-ghost and in fact as we'll see tomorrow the fact that these two are related will in fact relate the b-ghost to things in our theory so although the b-ghost is not the fundamental field the relation between L and E will imply that it can be constructed out of these variables so we'll see that tomorrow so that's to convince you that the theory at least the world-sheet action makes sense as a full quantum theory the next thing of course is to define physical states so physical states require of course the BRST operator so this is gauge fixed and the BRST operator following the procedure that comes from this twisted like action has the following form it's lambda alpha d alpha as it was for the particle but d alpha is a little bit more complicated now it's p alpha plus one half dxm gamma m theta plus minus one eighth and of course q bar the right moving q is of course constructed out of lambda hat and d hat so I'm not going to write that version now this d alpha although it looks complicated but the difference is that we have a simple opi which are trivial to compute using the free field action so d alpha of y with d beta of z has a single pole just proportional to gamma m alpha beta times pi m where pi m is defined to be the super symmetric version of dxm so it's dxm minus one half d theta gamma okay so d alpha with pi etc so in computing these opi of course I'm using that xm xn of z goes like minus log of y minus z p alpha with theta beta goes like delta alpha beta of one over y minus z now w alpha is not well defined because w alpha is gauge invariant remember that lambda gamma lambda equals zero and w has a gauge invariant so this type so if you try to compute for example opi of w with lambda you will get an inconsistency because it won't be invariant under the opi with lambda gamma m lambda so you have to compute opi of gauge invariant objects this object nmn this is gauge invariant so it's easy to see that if you shift w by this amount it leaves this object invariant another thing which you can construct gauge invariant is the ghost number operator now I just computed for you n with n so it gives you something like minus three over y minus z squared you can also compute n with lambda for example this goes like one half gamma mn lambda alpha times one over y okay so using these opi's you can compute the opi of any object involving a gauge invariant combination of w with lambdas and of course any product of x's and p's so now one can ask what are the physical states so we have the BRST operator we can ask what are the states of homology so for the open string the states so the vertex operators e satisfying of ghost number one satisfying of course qv equals zero and v is identified with v plus q of omega now we already know the answer to this because we did it for the super particle for the massless states it's just described by super Yang-Mills for massive states you're going to have dependence on derivatives of lambda and derivatives of x and derivatives of theta so for the massless case this is the only solution but if you want to do massive states you have to allow dependence on dx and d theta and of course also on p's on all the worldsheet variables so it's been shown that the spectrum reproduces the usual super string spectrum it's not trivial to show it but it's been proven now if you want to do scattering amplitudes you need more than just the open string unintegrated vertex operators you also need integrated vertex operators so you need something of the type dz of u if you're doing open string now normally u and v are related just by this usually has a c ghost and this you just pull off the c ghost in this form there's no b in c ghost but it turns out this is brst invariant q of u is equal dv obviously because if this is a total derivative then you can ignore it up to surface terms the total derivative now if you know what v is you can ask is there a u which satisfies this relation and in fact for the super Yang-Mills vertex it's straight forward to construct u so what you find is that u is given by d theta so these are super fields here but they're all related to each other so am is equal to gemm alpha beta d alpha v beta so this was a super field we saw yesterday this is obtained by hitting with one derivative w alpha is similarly obtained by hitting am with the derivative so more precisely it's and f is obtained by hitting w with the derivative and it's also can be related to am so this super field starts with the gluon this one starts with the gluino and this one of course starts with the derivative of the gluon so these are super fields that obviously you can construct in 10 dimensional super Yang-Mills and they appear in the vertex operator in order that you can convert them to this v so again this is a straight forward computation that I don't have time to discuss in detail but it's in for example the ICTP lecture notes that I gave out to the begin I mean that I gave the reference to okay there any question yes okay so this is conformally invariant because all of these objects have ghost numbers 0 have conformal weight 0 if you look at the tachyon the tachyon lets do bosonic string first so the ground state is not at k squared equals 0 which is why you need to introduce these ghosts which has negative conform weight in order to cancel it in RNS it's similar but that's because the ground state is not massless I'm not following your question but maybe we can leave it for the discussion session it sounds like a more detailed question so what we want to do now is compute scattering amplitudes so of course for the closed string it's just the left right product of the open string so scattering amplitudes of course are defined as in bosonic string let's do tree amplitudes for now you have 3 unintegrated and n minus 3 integrated and this Brian can't mean that I have to of course do a functional integral over all the worldsheet variables with the action in the exponential now naively this Brian Kett is going to involve an integral over all the lambdas and all the thetas of course all the x's too so there are two steps to this the first step is getting rid of all the non-zero modes of the worldsheet fields so v and u v only depends on the zero modes but u depends on the non-zero modes d theta and pi so you have to use these OPEs in order to eliminate all the non-zero modes of the worldsheet fields and that's going to leave an expression which only depends on the zero modes so the zero modes are of course going to depend on the x zero modes, the theta zero modes and the lambda zero modes to get rid of the zero modes now it's easy to see this as ghost number three it's cubic in lambda so this is going to have this type of form and now one wants to evaluate this so this f of course will depend on all the polarizations and all the momenta so all the information on the scattering amplitude is encoded in f so it just comes by doing all the these OPEs to eliminate the dependence on the of the non-zero modes after doing all those OPEs so f of course will also depend on the locations of these depend on the ZRs etc where you stick these vertex up okay but now we have the zero mode integral to do and we can see that there's an issue here because lambda now is some non-compact pure spin so lambda gamma lambda is zero but we no longer have the scale symmetry so we have the non-compact modes of lambda which will naively give you something so this d10x of course just gives you delta 10 of the momentum so that's painless but this is going to give you something like infinity to the 11 because you have 11 non-compact zero modes to integrate this will give you something depending on how many thetas you have so if this expression has m thetas this integral here will go like zero to the 16 minus m now it turns out that this infinity times zero can be regularized and we'll see how to do that tomorrow we won't do it today it turns out that the result is that the answer should involve an integration an expression here which only contains five tatas so of course that will cancel this zeros and infinities if m here is equal to five so the expression you get which we'll see in more detail tomorrow is of this type so after doing this integration over the lambdas and thetas you get the expression of this type d theta to the fifth of f alpha beta gamma where of course this object here contains indices and the indices here we precise the alpha beta gamma so the indices are going to be defined as fall so remember d d theta has a down index so this has a down index no that's right this has an up index so this is what is meant by this notation here so of course the expression is Lorentz invariant and using this definition it's easy to show that lambda gamma m theta lambda gamma m theta lambda gamma p theta theta gamma m p theta theta gamma m p theta this is just equal to one so this is the measure factor for tree amplitudes which plays the same role as for bosonic string theory is obtained by c dc d squared c so it's something of ghost number three this is in the comology I discussed this yesterday in the discussion session this is actually the unique state in the comology of ghost number three like c dc d squared c is for the bosonic string and we'll see tomorrow exactly how to derive this from this functional integral so let me just finish by just showing how this gives the three-point amplitude for super Yang-Mills so the three-point amplitude is of course you just have v1 v2 v3 which in this case is just going to be equal to lambda alpha a alpha 1 lambda beta a beta 2 lambda gamma a gamma 3 and of course a alpha contains both the gluon at order theta and it contains the gluino at order theta squared then of course there are high order terms so if you just pull out the term which is fifth order in theta what you find here is that this is equal to just the usual three-point vertex which is a coupling of two gluinos with the gluon and then of course you also get the coupling of three gluons through one of them through the field strength plus of course permutations so you can check trivially that the three-point amplitude reproduces super Yang-Mills you can also of course show that the higher points reproduce the three amplitudes of super Yang-Mills the most direct way to prove this is comparing with the rns formula so in the rns formula of course it's much easier to confuse in never short states where the states are just the gluons and not the gluinos in that case you use the fact that mmn that these two have similar OPEs and that makes it trivial to show that the end-point amplitudes agree with the end-point gluon amplitudes in rns and the gluinos here just fall from space-time super symmetry okay so I'll stop here