 Sem jaz za mene. In pa res sem pri redugačnji za drugo damagesenje, da je 2,2 supersimetri. V afterwards počutimo七o vitr-slasje intermenujega vzgočenosti, vzgočenosti za drugo misi narediti, in tudi imamo Studio-punje Habitino. Čestne prejena gravidino prejme tvoj form, in taj 2-dimensionalne redukcije o nekaj super gravidi. I taj dot, dot, dot, as I said yesterday, taj taj term, ko ima gravidino, če, če nekaj, in so, so je 0, when we set to 0 the gravity in itself. In we have these two complex parameters, they have opposite r-charge, so v mu is the gauge field that couples to the r-charge and then this h and h tilde, so that represents a complex scalar that we have in our graviton multipletism in auxiliary field. And so, so now we could take this equation and study what are the most general solutions. So we will not do that. This has been done by but I will just present two simple classes of solutions which are almost exhaustive. And, okay, let me also make one observation that since we are in two dimensions, rotations are, group of rotations is SO2, which is U1, so the spin connection is a standard abelian gauge field, so it's convenient to introduce this omega mu without further indices, which is just the contraction of the spin connection with anti-symmetric indices, the joint of SO2 with an epsilon tensor. And so in this sense, you see that spin in two dimensions is not different from any other abelian gauge charge. In fact, the covariant derivative of a field of spin s, if you take a certain component of spin s, just so s plays the role of an abelian charge and this is the standard abelian covariant derivative. So what solutions we can find of these? So one class of solutions is essentially what we do. So we set to zero this scalar, so we don't use it. And then, so in here we have the spin and here we have this abelian charge and as you see the spin is in another abelian charge so we can just arrange for this gauge field to exactly cancel the spin connection, such that this becomes a standard derivative and so we can just choose a constant for this spinor. So what do we choose? So we choose V mu to be equal to a half omega mu, a half because epsilon is spin one half and then epsilon we just choose a constant but of course in this derivative is a spinor of two components, plus one half and minus one half spin so this can only work for one of the two components, not both. So one of them. So again I'm using a base in which I diagonalize the carality. Here is the opposite story because the R charge is opposite and then this H is zero. So this is a simple class of solutions. This in fact works on any orientable manifold on any Riemann surface of any genus and in fact this is called a topological A-twist. So this is a solution that has been known for a long time and once again allows us to put any two-dimensional theory with this amount of supersymmetry on any orientable Riemann surface and correctly I mean this solution that was not found in the past in this machine in office to transfer this framework and it does. So if I write in this way using the spin connection just a U1 gauge field then a spinor has two components plus one half minus one half so in this formula I really have to put the Z component of the spin. So if you want in these two components for one I have to put plus one half and for the other one minus one half but on the other hand I can only cancel one component and the other one is still a section of a bundle and on a Riemann surface there are no non-benishing sections of these bundles because of course one of the conditions that this is nowhere vanishing otherwise at that point we don't have specific transformations. So once again this was constant and in this topological A-twist the simple algebra so now we should take this solution plug it back into the supergravity variations and if you wish in the supergravity action and for instance the algebra is very simple there are these two variations in relation with respect to epsilon in relation to respect to epsilon tilde the square to zero and the anticommutator is zero so it's a very simple algebra. And in fact for genus bigger than one this is essentially the only solution. Ok, this is not new however there is another solution which is also interesting and in this solution instead we turn on this H but we don't turn on this V so if you want this is an untwisted solution and so this is the solution so we set V mu to zero we turn on this H where R is the in fact in fact this solution works on S2 on the round S2 so R is the radius of the sphere and then if we plug in and we see what we get we get these equations one for epsilon and a similar one for epsilon tilde and in fact what these equations are these are Kean-Spinov equations on the round sphere and it turns out that in fact there are four so each of these solutions if these equations are two solutions so in total there are four solutions so I'm not going to write the solution but you can go ahead and solve it on the round sphere it's not difficult so there are four solutions and so in particular something that didn't say here so in this case there are two solutions so one solution is epsilon minus and one solution is this epsilon tilde plus so in the topological A twist we break half of the super c so we can preserve only half of the supercharges of the supersymmetries but in the case we find four solutions and so in fact so we started with four and we can preserve all of them on the S2 so this solution preserves more supersymmetries and so this is interesting if you want this is new with respect to 30 years ago this was not this before and then once again we can go back and plug into supergravity and then we find in this case that supersymmetri algebra is deformed as I said there is also this trial and error method that you can use but I mean it was not this was not discussed at the time this topologically twist was discovered and discussed 30 or something years ago but at the time only this was observed and these type of solutions are more new if you want is a two dimensional version of what past constructed on the S4 ok so what about the supersymmetri algebra ok so this time supersymmetri algebra is deformed so the anti commutator of some epsilon and epsilon tilde is equal to the following so there is a lead derivative along a killing vector this killing vector ok I didn't write it in my note but essentially is a sandwich of this epsilon and epsilon tilde this came up to some factors that I don't have here is essentially some epsilon tilde or mu epsilon or maybe there is a transpose here I'm not sure but it's a biliner in these two in these two spinors so a lead derivative is the genetic of translations so this means that this is this symmetry here is a rotation on the of the S2 so it is a symmetry in fact and then there is another piece which instead involves the r symmetry which is this vector like a symmetry we are assuming here so it's the one that couples to V well the other anti commutator vanish so you see this is a deformation of the flat space supersymmetri algebra because in the flat space supersymmetri algebra we do have of course the anti commutator of super charge is momentum and this lead derivative is this general translation so this is momentum this is deformation because now the r charge enters into the algebra and correctly in fact this is suppressed by power of r so we take r very large and we go back to flat space this deformation goes away as it should now in fact what this what this algebra is well this is as we say this is generates rotations of the sphere so this is SO3 is SU2 slash 1 let me call it A because it uses the vector like r charge in the same way as the A twist does and so this is a super algebra in fact the bosonic part of this super algebra is SU2 times U1 so SU2 is the rotations of the sphere and U1 is this vector like r symmetry ok let me also notice something that we mentioned yesterday about this part of the solution you see here as I say I am using this notation in which tiled fields are fields that would be complex conjugate in Lorentzian signature but then in Euclidean become independent and you see they are not complex conjugate because there is no Hd is not minus i over r it is still a plus so this is not the weak rotation of a real background in Lorentzian signature and so as we said this breaks reflection positivity ok then it doesn't solve the equations so I can try to do H equal i over r Hd equals minus i over r it just doesn't solve the equations ok so ok so now that we have as I said this is not exhaustive list but almost there is something else that we can do on t2 this leads to the elliptic genus and on the round s2 well of course on the round s2 notice that we have now two solutions we have the twist and we have the untwist so on the very same manifold still we have two ways of preserving supersymmetry this is what at the beginning of my lecture I call the background so even just specifying the manifold does not fix in an ambiguous way the way to preserve supersymmetry here we have two ways and they lead to different observables if you compute partition functions the twisted theory there is a way of making it equivariant with respect to rotations but I will not discuss that I'm sorry I cannot hear you could you deform the s2 as well please yes for instance something that we can do that you can do is to squash this s2 so if you do then you are back to two supercharges it turns out that the partition function does not depend on this cauching so there is no point in doing that but at the level of supersymmetry it would be half no sorry so I would not so you take this background and you ask okay this background how many, it gives me some super algebra how many supercharges does it have these two supercharges it has four supercharges the very same that you started with in the UV so this background is not breaking any supercharges it's just deforming the algebra and it's different from the twist because the twist breaks half of the supersymmetries oh okay yes so you can either just say okay I compute the partition function and then you preserve all the supercharges oh you can say I want to insert observables and observables are gonna break some of the supercharges for instance you can insert Wilson lines Wilson loops and they break half of the supercharges you can also insert local operators at two antipodal points and they are compatible with Wilson loops if the Wilson loops is special it's on the equator or something like that and so all this configuration preserves two supercharges Paston's version I think that since he doesn't have an offshore yeah I'm not sure I'm not sure if without the observables you will have more maybe you do have more supercharges Does it matter whether the parameters are real and complex to do the key vectors because having force fields I would have expected force fields of all the possible combinations of the liners well I mean if you construct these by liners this tilde you construct T out of this you get 3 no I mean you can try you get 3 because this gives you the the rotations of the sphere which are SO3 so you don't get more these parameters here no they are complex yes yes maybe there is a dagger here I didn't write these in my notes but yes you can try to plug in your killing spinos and you get solution for those and you get the killing vectors of the sphere I can look afterwards okay so now I would like to let's see okay we will be quick so now I would like to do this to see how to perform the actual localization and so I would like to focus on a class of theories so again we are in 2D n equal 2 comma 2 so let's focus on gauge theories and in particular let's consider simple theories in which we only use vector multiplet and chiral multiplet there are more general theories there are more multiplet so there are more general theories that you can consider but let's just consider this simple example which is interesting enough and these multiplet are just a dimension reduction from 4 dimensions so I'm not gonna spell out what these multiplet are because you can work it out yourself let me just list the data that enters in these theories so of course there is a gauge group then the matter is in chiral multiplet so these chiral multiplet are in some representations of the gauge group of course the representation in general can be reducible then an important observation is that the supersymmetry as you said is deformed is this SU2 slash 1 and so in particular so here we have the symmetry group but we also have the r symmetry so this is different from flat space so if you wish is in some sense more similar to super conformal symmetry the r symmetry is part of the algebra and because of that the theory does depend on the r charges that we choose it's not true in flat space in flat space the r charges are some assignments but they do not affect the theory they do not affect the Lagrangian so we will have to choose some r charges they control some curvature couplings and the reason is that this r symmetry does enter into the algebra so we have to choose r charges for these chiral multiplets then there are interactions and we have superpotential interactions so this is just dimension reduction from four dimensions so it is the same story some holomorphic function of chiral multiplets there is also a new object in two dimensions which is a twisted superpotential and this twisted superpotential is an holomorphic function of twisted chiral multiplets that are another multiplet that exists in this dimension and so we are not putting other matter in twisted chiral multiplets but it turns out that out of the vector multiplet you can construct a twisted chiral multiplet unfortunately I do not have time to to spread these out in details but essentially you can rearrange the same fields in the vector of the vector multiplet into a twisted chiral multiplet if you wish this is similar to what you do in four dimensions you start with a vector multiplet and you can rearrange them in the w alpha which is a chiral multiplet the field strength so this is similar but it is a twisted chiral multiplet and out of this you can specify an holomorphic function this is called twisted superpotential and this is something that you can add to the Lagrangian and then there is another object which again I will not go into details these are called twisted masses and these are related to to flavor symmetries and so here the idea is that every time that you have a global symmetry you can couple well, first of all you have a current of course if this global symmetry is continuous then you can couple to background gauge field this is something that we always do however now you are in a supersymmetric theory and so we should actually couple to a full vector multiplet a background vector multiplet and since in two dimensions the vector multiplet contains a scalar now there is a scalar in this background multiplet to which we can give an expectation value and if you work out in the Lagrangian what it does it gives mass to the current multiplet under this symmetry ok so these are masses but they are interesting because they are related to global symmetries somehow they take values in the cartan of the global symmetry of the flavor symmetry you cannot do it for the r symmetry ok, any question on this? ok, so this is the data that we can play with so you see that even with these two types of multiplet we have already quite a rich structure and then we can go on and write actions and let me just write down the kinetic actions just to give you some example once again this action how do we obtain this action? either we take this background that we worked out and we substitute in the supergravity action or if you don't want to play with supergravity because we are lazy we can do this by hand method which is still correct and the final result is the following so the kinetic action for the vector multiplet looks like the following so all the fields are in the joint representation so of course there is the kinetic term for the gauge field and let me well, here I need to tell you what are the fields here in the vector multiplet there is a vector there is a geigino in fact there are it's a complex geigino and then there is a complex scalar that I will write as sigma 1 plus i sigma 2 and then there is the D term that you also have in four dimensions while this scalar comes from reducing the vector along the two direction that you reduce from 4 to 2 so these are the fields that should appear in this action and here so for instance let me write it and then I will comment and maybe I will not write the fermions ok, so this is the bosonic part of the action so you see if you send r to infinity this is just a flat space the standard Lagrangian so there is the young mill's kinetic term there are the standard kinetic terms for these scalars well, there is this commutator that if you wish comes from the fact that the group is known a billion in four dimensions so there is a commutator in the young mill's and then becomes a commutator of the scalars and there is the square of the D term but now because we are on the sphere there are these relevant deformations of the action and they appear as this deformation term here and this deformation term here ok and what concern the matter Lagrangian so this is the kinetic term for the Karel multiplet and it looks like the following so once again let me write it and then I will comment and then we have the fermions and so about the same comment applies if we send this r to infinity we are back to the flat space Lagrangian that just follows from taking the kinetic term for a Karel multiplet in four dimensions and reducing and here so we can look at this now these are the relevant deformations we obtain on the sphere this small r is the r-charge of the Karel multiplet so the fields in the Karel multiplet are this complex scalar phi and this is the complex auxiliary field f that you have in four dimensions so this is the r-charge and as I promised the r-charge does enter into the Lagrangian this is not so in flat space but this is so in this curved action in fact it controls you see some curvature cappings as I was saying so these are some masses or some cubic term here and these are masses for the scalar they do enter into the Lagrangian because the r symmetry is part of the supersymmetry algebra okay without the r yes not with the r without the r is just that yes but not with the r can this be then the reduction of the Pestum Lagrangian already deform in four dimensions but not at Pestum you could try to do as 2 times T2 but this would work for the twist but it doesn't work for the untwist so unless if I remember correctly I'm not 100% sure but I think that you could try to put the untwist here and here it's just flat I don't think that this supersymmetric solution is but 95% sure okay so now so these are the kinetic actions then there are the interaction I'm not gonna write it as a similar story there are various deformations so now we should choose a contour and we can choose the integral contour by this what I mean is that simply fields are real in Lorentian now it gets complexified in Euclidean and the contour is just that they are real and complex field in Euclidean in Lorentian they become two independent fields in Euclidean we just choose the contour where one is the complex conjugate of the other so it's the most basic thing that we could try so these are a contour that makes the patenteral convergent because it makes these two actions positive definite and so and now we choose so we have many supercharges we choose a supercharge to do the localization we don't have to use all of them so it's convenient to use some supercharges we call delta q which is done in following way you choose one epsilon and you use one epsilon tilde you choose one and you just localize with respect to that and if you do that you discover the following that in fact the super Young Mills action is q exact delta q of something that you can work out and the matter kinetic action is also q exact so this is quite interesting for two reasons so first of all this is telling us that I forgot the gauge coupling so this is telling us that in fact if you compute a partition function this theory is not going to depend on the gauge coupling and this is important useful because in two dimensions the gauge coupling is dimensionful so it sets a scale if you want above the scale the theory will be coupled and below the theory is strongly coupled but since there is no dependence on the gauge coupling it means that there is no dependence on the RG scale and so what we are going to compute is independent of the independent of the RG flow yes phi tilde equal to sorry yes the complex conjugate let's see I mean it doesn't have to be reflection positive so we know that there are the formation breaks reflection positivity so the claim is that this is just the trivial so we go for Lorentzian to Euclidean and this is a contour that is the rotation of that but it does break I mean the background does break reflection positivity sigma 1 and sigma 2 are real sigma 1 and sigma 2 are real in this contour yes ok so as you said there is no dependence on the gauge coupling there is also no dependence on the wave function normalization of phi this follows from here and then the second observation is that in fact we can directly use these two guys to do the localization we don't have to find a v that does this well essentially we can use this so whatever this is that I'm not going to write we can choose this as our v ok because these are q exact and well we also know that they are also q closed because these are super symmetric but we can also check it ok so the interaction Lagrangian what about the interaction Lagrangian well we mentioned by this types of interaction Lagrangians so it turns out that super potential is q exact and so in particular there is no dependence on the coefficients in the super potential yet super potential is not completely trivial because the super potential fixes the R charges and the Lagrangian does depend on the R charges so still there is a small remnant of the super potential it's not completely independent of the super potential but it does not depend on the coefficients that you put there twisted super potential this is not q exact and so in fact this twisted super potential does give us parameters and the partition function will depend on these parameters in particular a simple super potential if you want the simplest super potential is a linear one just because a constant drops out doesn't do anything the linear one is a phi etiopolis term and so the partition function will depend on phi etiopolis terms and so what else did we mention and the other one was twisted masses and there is a dependence on twisted masses so upon this is also not q exact and this is because when they deform algebra that they wrote they didn't put the central charges but essentially the twisted masses are central charges so they deform the algebra and because of that they do affect the result sigma 1 plus sigma 2 this is what is the definite because commutator it's sigma 1, sigma 2 dagger this is sigma 2, dagger, sigma 1 which is if I take them real so I get a minus sign from the dagger so it's imaginary if you want sigma 1, sigma 2 commutator is imaginary yes ok ok so so we can try to use these two these two actions as localizing term now here because of lack of time I'm not going to do full details so using these super yamils is perfectly ok because along the contour this is positive definite now this one is not strictly speaking it's not positive definite so for instance if you fix this object here it's positive definite in the fields of the current multiple but of course here there is a cubic term so in fact if you want something that is really positive definite you do have for the current multiple to construct the qv so if you want the v which is given by the sum over fermions of q psi double dagger psi and you can construct this object so it's really positive definite I'm not going to write it because it will take me some time I also don't have my notes ok so so what do we do so first step now is to or whatever step we are is to figure out what are the bps configurations because the particle localizes to the bps configurations and so let's see so this will be zeros of the localizing action and turns out that this is very simple now because you see these are sum of squares and so the bps locus is just given by we have to set to zero all the terms and so so the first term gives us this equation the other term get this equation and then we have so we have this so see this is very simple because this is telling us that sigma 1 and sigma 2 so first of all they commute so we can diagonalize them and so this means that also f and d should be diagonal since sigma is in their joint now if everything is diagonal these are standard derivatives so sigma 1 and sigma 2 are just constant and then also d and f are constant so we find a very simple from the gates sector so this equation in which so these guys so all the fields are constant and diagonal now sigma 1 now becomes so can take any real value for each of the carton directions in the gauge group but of course from here we get a constraint because sigma is a real field but this is the field strength and there is a direct quantization condition and as a result of this sigma 2 should be quantized and so in particular if we want we can parameterize d which is equal to sigma 1 over r we can call it a over r squared now a is dimensionless because these fields were dimensionful and f12 which is sigma 2 over r we can call it some m over 2r squared and this m is a magnetic flux because essentially is the integral of f12 so this is the m is precisely the result of integrating the field strength so I hope so this f is the field strength so I hope there is no confusion maybe you want to call it some calligraphic f this is the auxiliary field but so this is the magnetic flux and this has to be quantized this should satisfy gno quantization condition in other words if you want this is an element of the algebra now if you exponentiate it this is an element of the group and the condition is that this in fact is the identity in the group this is the gno quantization condition for you one is obvious just tells us that this flux is integer but if you have a generic group this is the condition ok and then ok now since in the gauge sector this is the solution if you want we can plug this into here and we can see what is the condition on the matter fields if you want to be more precise you should use the version on the nose but then something very simple it's just that phi and phi tilde are zero and the auxiliary field is zero ok so we get something very simple and this is nice because as we said in yesterday lecture or maybe we didn't say it but if you through the localization of the integral to a simple problem because we only integrate on a subspace of fields but first of all if you want to get some mileage this reduced space should be finite dimensional if it is still infinite dimensional it is a simple field theory so first of all we want that this subspace is finite dimensional which is not always the case moreover it might still be a complicated space maybe some modular space of some complicated partial differential equations we don't have handle on but in this case it's very simple you see we can parameterize in a very simple way and so this is going to lead us to a very concrete formula for the partition function and of course we are not guaranteed that this is the case there are examples in which the problem remains complicated so ok this is our BPS locus then next step is to compute the so we should take this configuration plug them into the action now we are not going to get anything from the kinetic terms because we say those are q-exact and so we just get zero so the only place where we get something non-zero that I raise is the twisted superpotential is the only term which is not q-exact ok yes so one way ok I just I had this expression if you want since we found the solution for the gauge so take the solution plug it back if you keep fix this that was positive definite in the matter fields if you keep fix this and then you see that I mean it was quadratic and so the only solution is that they are zero but if you want to be more precise you can construct one which is really positive definite ok so this was the BPS configurations that contribute to the path integral then the next step is the classical action so it's very simple you take this superpotential the action that you get from twisted superpotential you plug in this and I'm just going to write the result because I didn't write anything before so this comes from twisted superpotential so if we specialize to only a linear twisted superpotential which is just a phyethiopolis term then what we get is something is something like the following so this psi is the phyethiopolis term since this is a linear term this is a coefficient in front of it and then the rest depends on the so there is a sigma 1 and there is well the twisted superpotential the phyethiopolis term in two dimension is complexified so really this phyethiopolis term contains a real part and an imaginary part which is a theta angle in two dimensions you can turn on a theta angle and so this is so this depends on the parameters in the action and depends on where you are on the bps locus so if I draw again this picture where we have this space of the configurations and there is a special locus which is the bps locus this is the classical action that we get a certain point on this bps locus and we will have to integrate over this and this is a function of this space ok so almost to the end so now we need to compute one loop determinant and so these one loop determinants are computed from the localizing term because the term dominates in our deformation so they are not computing with the original action they are computed with the qv however we chose the kinetic term as our qv so in this particular case the kinetic terms that I wrote now as I said one can either try to compute exactly the spectrum in general this is very actually impossible or I can use more sophisticated tricks or methods, mathematical methods they have to do with index theorems and they use the fact that in these super determinants there are a lot of cancellations and they usually solve simpler equations so I will not describe this term these methods maybe Samir will say something but in this particular case we have the round as 2 one can actually compute exactly the spectrum because there is a lot of symmetry so for instance if you look at the current multiplet just to give you an idea so this one loop determinant as I said this is nothing else that the determinant of the fermionic operator or the determinant of the bosonic operator and these two operators are the quadratic expansion of the in this case the kinetic term for spinors and the quadratic expansion of the kinetic term for scalars at a given point this also will be a function of where we are on this manifold so we expand around that point that quadratic order we get this operator and we should compute all the by the spectrum of this and as I said in this case we have a particularly simple situation so we can actually use spherical harmonics and we expand these fields in harmonics and we compute the spectrum so one uses the so called monopole spherical harmonics that depends on some angular momentum on the sphere so j will be the angular momentum j3 is the z component but it also depends on a spin and in fact these objects also take into account a billion charges of these fields because as we said in two dimensions there is no distinction between spin and other a billion electric charges so all these fields each components of this field is some line bundle on the sphere and the spherical harmonics are simply the spherical harmonics but not for a function but rather for sections of line bundle on the sphere so they take into account both the actual spin and electric charge under the magnetic field and yes in particular these are the egg and functions of the operator d mu squared where in this d mu as we said there is both the spin connection and the connection under the the gauge connection so these are just the egg and functions and so just to give you a flavor because we don't have time to go into the details of this computation well maybe this is useful to keep so just to give you a flavor for instance we take the scalar operator we expand the quadratic order and we get some laplacian the contribution of this d term this object is here and so this is our operator and we just have to take all these harmonics and we act on them so we know what is the egg and value under this because these are precisely the egg and functions and all the other ones are just well they just multiply by this value because there are no derivatives and so one gets some expression for this determinant which would be some product of egg and values I mean the expression is not important because we are not going into the details and here there would be some expressions so this determinant would just be some product of all the egg and values that we get from all the egg and functions so we do the same thing for the fermionic operator and we get a similar expansion once again we use the very same harmonics they work for any spin integer or half integer by the way, you can find these monopolar harmonics on a paper if I correct it, of V and Young but you can also find them on Wikipedia and then from there you can find references to papers ok, so once you have these two determinants then the determinant is the ratio and as I promised there are lots of cancellations because of supersymmetry so many of the egg and values that you find here you also find here and so in this ratio it's much simpler so let me actually write this expression so this is the expression that one obtains there is still an infinite product on this N and then here there are the R charges that do appear in the quadratic operators and there is this parameter A which parameterizes these continuous directions that we have to integrate over and then there is this parameter M which is the magnetic flux which instead is quantized and so it's more like a sum so if you want this modular space looks like this for each value of M there is some continuous space which is parameterized by A so you get an expression like this now this steel does not make sense strictly speaking because this product is negligent so one needs to regularize this and for instance one way to regularize it is to use zeta function regularization and so how this is done one takes in this particular case the whole width zeta function is well for generic values of the R charges there are no zeros for generic values of the parameters there are no zeros but it's not prime nothing has to be excluded and then by analytic continuation this might have poles of course but in general there are no zero modes to remove so this is the whole width zeta function this series converges if the real part of z is large enough then you can continue by analytic continuation on the whole complex plane and then essentially you use that if you compute a derivative d of this object what you get is and you evaluated z equal to zero this you cannot really do I mean you can compute this but you cannot use this expression because this expression is not valid at z equal to zero but if you try to formally do it so let me put this expression in quotation mark just because this is wrong but it's formally what you get which up to a log is precisely what we want to compute n plus something but in fact the correct expression for this now I'm not using quotation marks is a gamma function and so this expression which I should write in quotation mark this is not really mathematically well defined is regularized by the gamma function in the end of the day the one loop determinant for the matter of fields is given by an expression so of course here we'll be a bit with quick so of course the matter fields are in the gauge representation so for each weight of the gauge representation we get one of these one component in this equation and so really we get a product over all the weights in the gauge representation of some expression that involves the gamma functions now the specific expression is not important in this example but what is important to note is that first of all it's a very explicit function so we do get some function it's also really simple it's just a ratio of gamma functions this function does depend on the R charges and it does depend on the points where we are in this BPS manifold so it depends on A and M and then the end of the day we have to integrate over A all the weights in the gauge representation any question? ok, then something similar has to do to be done with the gauge sector so we're not going to the details but if I just can I take my five minutes since my last lecture so so now we have to do the similar thing for the vector multiplets so the only detail that I want to add in the gauge sector is that one has to do gauge fixing of course and so one has to add a gauge fixing action and and essentially so what you use the gauge fixing in a background field because also with the gauge field we have to separate between if you want a classical part which is the classical configuration that gives you the BPS plus oscillations these oscillations are again awaited by 1 over T to the one half because we want to keep canonical normalization and so this gauge fixing action that involves the ghosts these ghosts I remind you these are scalar but they are anticommuting and they are adjoint so this is the standard of gauge fixing terms so I'm not going to explain this but the important point is that in these covariant derivatives only the background part enters and the oscillatory part is basically so in all these covariant derivatives only the classical part enters and so in this sense this action is actually quadratic in the oscillations so as I said this is the standard gauge fixing action that you can find on peskin for instance and so then you do the same thing so you take the Young-Mid's action around each point you expand the quadratic order and so in the set of fields you will also have the ghosts and so and so what you compute is something very similar I didn't say this xi is not the phi etilopoulos term is the parameter that you use in arc C gauge so at the end of the day there should not be any dependence on the xi you can either fix it to whatever you want or keep it and check that you're doing correctly and so and so here now you will have you will have three pieces here there will be the term a piece from the gauge field since it's real you have a square root you have a piece from the gauge genome and you have a piece from the from the ghosts but otherwise you do the same thing use spherical harmonics and you do exactly the same computation and it turns out that you get something very simple so once again the details of the expression are not important but what is important is there is a very simple expression so first of all here we have now a product over the roots of the gauge group which are if on the weights of the joint representation so once again we have the magnetic flux m and this parameter a this continuous parameter this time this function is very simple it's just a quadratic function but then because of the of the zero modes it turns out that there are some zero modes there is also this product over these factors in which you have to take a product over the roots that annihilate the particular magnetic flux that you are considering so at the end of the day there will be a sum over this I'm not going to explain where you get this but just comes from doing this analysis with spherical harmonics ok so once we have our loop determinants what do we have to do well it turns out that there are in fact some bosonic zero modes here and essentially these bosonic zero modes they span the carton subalgebra which is not broken by this magnetic flux so if there is no magnetic flux this will be the whole carton subalgebra but the magnetic flux lifts these zero modes so there is also some integral over the zero modes once again since I'm not going to the details here we don't really need expressions ok there is really no point in writing this but ok we have so this will be our one loop determinant from the gauge sector and then the final thing that we have to do is to put everything together and we get an expression like the following so this is the dimension of the vile group and if you want this is a discrete residual discrete gauge that remains that is the one that premiots the carton components after we have diagonalized but the important point is that we have our sum or integral over the VPS configurations so in particular there will be a sum over magnetic fluxes and there will be an integral so this is this parameter m there will be an integral let me call k the rank of the gauge group there will be an integral of these parameters a and then we will have the classical action that I wrote before and then we will have the one loop determinants that there is really no point in writing them again so one from the gauge sector and one from the chiral from the chiral multiplets and these functions that we wrote these are functions of a and m as well as of some parameters like the R charge or twisted masses that have not discussed here which are not integrated over so this is just a concrete realization of the abstract formula that I wrote yesterday ok, so let me just conclude by making some comments on this so one comment is that it is easy to generalize these in the case where we include operators and so in particular Diego was talking about this formula for computing the expectation value of Winston loops, also here we could include Winston loops the difference would just be that here we get some extra factor which is just the evaluation of the Wilson loop on the background so it would be on the same footing as the classical action, very simple another comment is that this formula is incredibly simple just an integral and some sum, very simple functions this is a ratio of gamma functions and so on but in fact this formula is exact stress once again is the exact result of the pat integral and so in particular it should contain perdubative corrections so is it true that it contains perdubative corrections and in fact one can see explicitly that it is because with some trick so this integral if you want this integral along the real line so suppose we are in rank 1 this integral along the real line now one way to evaluate this integral is to close the integral at infinity of course you have to check that this integral goes to 0 at infinity you can check it and then use the Cauchy trick they reduce this to a sum over poles and it turns out that these gamma functions they have many poles in fact there is a wedge if you want a two parameter lattice of poles so you can write this as a sum over the rest of all these poles and it turns out that each pole is an instant on contribution you can exactly identify it as the contribution from a vortex in two dimensions instantons are vortices and in fact you could do the localization computation in a different way and reduce it to a sum over all these nonperturbative contributions and you can explicitly see with this simple Cauchy trick that all of them are contained in this formula so this is reassuring so I think I will stop here