 spnice za kratnice zabilitega idejo na napravse, da Sorom načo obočje Fire, pa je to z napravkelju predstavila, da je to napratil, da je to za lango, da je to engajno zwrotno. Salo delanje, da je approvedvo, da je to z kalja, po to, da je to priden. ... stones for, because we have four complex dimensions, and the worth, oops, sorry, and the worth magnificent, stands for the belief that four complex dimensions is somehow the highest possible one where one can play this type of game. And my main goal during this talk is to present some conjecture for some supersymmetric partition function, Zelo še bejte, ki jim se volim, da sem imel dobro izgleda vzelo. Sestem izgleda je vsek kz, ki jim je zelo rejda, ki je, še je, nekaj, vzelo vzelo, in je vsek, ko je v 조čej dela delovnenj delovnih rejda, ki je vzelo v nekočej delovnih. So along RT times R8. With a constant B field, turn on. As I will be more precise in the rest of the talk. But OK. The statement is that this object, which I call zeta K for nD8, in the operator formalism, of course, I can write it as a written index, is a character valid written index. hk je vziv, spas, tudi, tudi je bila g. In sem tazimo, ki je superashelja, je nisem način, nezavodno v tembetarih beta. Po vkroju, ni ne zelo. Však je, zelo se zelo, kaj je motivacija za toz problem. One could be interested in studying the bound states of the brains, for example, the 0D4, the 0D6, the 0D8 are all interesting topics to look at. This is just a character valued version of those computations. Moreover, there is a connection. zelo, ki je začali noženih在a iz nas iz verbenih na vljana, katiljno mi, kaj čojnega začupaj n Primedesitega in z��ila predtitev dvoj delijo deličnih teori, na zelo po vseh izgledaj, da je ta ročnija odličenidela, da je ona začal začal začalam za njih deličnimi delimi deličnih deličnih deličnih, ker so pradi doličenih deljičnih počeri, zato mu je to, da, da ta ročnija bježi nožena iz boži državo. in analogovosti istantično občutij, prejste zelo olačiti, to bi je vsev spina veliko, tko vse vse izjelimo, ne bo vsev spina vsev. In vsev, in kratkivom matematijku, to je srednje do Donaldson-Thomas sprave, da je, zelo, tko kala-biao, torik kala-biao, nekaj sem in vsev počakal lači tko tko priplizirana tko torik kase. Pozorno je, Zeta, kaj sem stojnila, če sem pošljala vzelačenja atlokovana, o kjeva z 0 in finiti, pk, zeta k. Fitne, tkaj to je pletistične prizvače, zelo, da je pletistične prizvače, je tko drog treba, ker sem da bo vzelačenja. Zelo, da je pletistične prizvače. office finitely many numbers for early in so I call this function of n. Yes yes, so Jesus un group element which leaves in a global, a group of Rob In na svejmer. In ja. Tak, je. Tak, ja. Ja võlga. Nama, da vse da, da, z tem,än ta gaurja, pa, tako, da, da, ne, da, domaš več staro, zači, da gauri na odkod nekako, Vesu, ja. Trim sem že v pačjih spodobna, ali je bачiti ampak palecje, in taj palecje sem izbytno za glasboj zameranje. Vse sem je pogleda, nekaj let bo na vse glasboj, na salitku v drže in drže in v Hellenji. To bila zelo vautn Historia, na Fietiliopoloj ste. Je to pripečne, da je srednja dorenje. So. Yeah, and OK, not the way of looking at this is that this low moving approximation is the same thing as basically taking the alpha prime to zero limit. At the same time is killing the field as alpha prime. The remains constant and this n, 1, n, and one would like to study, what is the effective action for this system. Sve, ki sem postančen, porque sem ide, da kako se tudi tako zste kakv. Počakam, da je celo zelo, da je tako zelo, da je zelo, da je zelo, da je izelo. In ja, to je kvantomekanice, ker smo izgledali na teoriče na volume z zelo. To je počkaj. Prostavno, kaj smo počkali, that the 0-d8 system, in kako je skupnila na v10, je to supersymmetrične, kaj bi nekaj zelo, tako je zelo, da se počkala v bps-bondi stati v nekaj kazih. In ja zelo, da je vse zelo, da je vse zelo, da je vse zelo, Tako, zelo igram do vresnega, imam medanje in imam mene, izvečurjo r8 in ni zelo vse zelo. Neselimo, kaj je zelo, kot lahko je, kaj je vse, kot neče zdaj. Zelo igram vse zelo, izvečurjo r8. Vse vse zelo, da sem od 1 do 4 aj, imam 4 zelo, tako, imam vse zelo diagonal, na 4 zelo. Selo, da sem, taj si priberite teori tako zvoru premačenosti taj vzil. Tati, smo že me prirožjali je, anderiji labuSE,先 čas imelo okledoziv. Zato vergodit ja z okídem. Protoč aquí je odrečujela beta. Premačen je doditi supersimetri z vsovvenje kanici, da je čas in gažnja taga, čas je gaž,gaž na katom mody evac. Gažnji ampak je uk, kar gaž enak vzalj taj vzil. To vzaljana je jaz na dosimetrij. ta sekne accim husband ne bo. Prejč covering Lyn Head can come from to type of Alle strings either this is strings stretching from 0, shuttle brain or string stretching from the SVD Nu destructive brains. The other question was about G. I denote this G as an element of the global symmetry group which are called G-flavors and I will explain what it is for this particular model. In... Ja. So let's see. Let's come actually to that question. What are the symmetries of this problem? In principle, so this g-flavor could be... So this g-flavor has various factors. There's an SU4, which is the SU4 inside spin8. So they are 8 as a spin8, but we have to take SU4 in order to preserve the two supercharges that we had before. And then there are two UN, the first UN, small N, I call UN capital N. This N is like a meant to remember, to remind the champaton bundle of the 8. So this is the symmetry group of ND brains, ND brains champaton bundle. And then there is another UN, so same small N, but I denoted with sub capital M. This guy, this UN is not manifest in the geometry. And so in the next few slides I will motivate how it comes about. And there is also a speculative interpretation about it may come. And is that one could introduce in this system another stack of N anti-D8 brains and that M will be the champaton bundle. So same dimension N, small N. And then, I mean, I don't know how to prove this, that's why I wrote speculative, is that then by... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... sestem, so let se, in also, there is another U1, which is a global U1, and which is anomalous. And the charges of the fields, I also list there to show you that there is an anomaly, and such that I will introduce something to compensate it. So it's not a gauge anomaly, so it's not, the system is not broken. It's just that the correlator otherwise will be zero. And so for these two guys, a plus minus five, say the complex combinations, the charge is plus minus two, and it's minus one for these fermions. Yes, yes? Yeah, so that the measure of the, you can write a part integral for the quantum mechanics, and the integrants in the measure will transform with this U1 charge. Sorry? You can think of it as like a ghost anomaly if you want. I mean, these charges, you can read as the charges of the fields in the transverse direction. So this is that interpretation. And then once you write down, I mean, the written index that I wrote before, you can write in the integral language, and the measure, many of these, all these fields will appear in the measure, and that measure will transform. So that's what I meant. That's what I meant. You can, actually, you're not forced to introduce this, because now you can evade it in another way, which is given, which you do anyway, which is given a weight of this type also to the omega background parameters. But one way to avoid it is to introduce an auxiliary system, which I will later call Upsilon. It's just a decouple fermi multiplet, and that does the job of... So at the level of inserting some observable in the partition function, you're inserting some chair classes weighted by those masses m that I talked about before. It's like you make an extra insertion to make the observables non-zero. That would be a way to... I don't know if this satisfies you. Yeah, okay, I don't have another explanation, it's just the measure transforms this way. And yeah, the other relevant multiplets are the current multiplets, and they have two origins. So one comes from the zero to zero sector. And again, I have... So from the zero to zero sector, the main players will be these matrices, B sub A, and this dagger. This index A will always believe the... Always live in the four of SU4, and again, they are K by K matrices. So their representation content is that joint for UK and the four of SU4. And another way of phrasing this is that each BA will live in endomorphism K, and their fermions. And then from the zero to the eight, there is a... So for NS sector, okay, let me say first, for a month sector, you have a bunch of fermions, which will always be there even without B field. But now, because of B field, you have a massless state in the Neveshvat sector, which I call I, and I dagger. And they will live in the B fundamental of UK times the first UN, N. So K times N bar. And again, another way of thinking in terms of vector spaces, they will each I... So I will live in on N, K, and I dagger on K, N. And for all these Keral guys, the bottoms will have charge zero, and the fermions will have charge plus one. And finally, there are a bunch of fermion multiplets, one fermion multiplets, many fermion multiplets coming from the zero to zero sector, which I call KAI AB. Again, they are K by K matrices. So they stay, they live in their joint of UK. And then in the six, which should remind you of the previous, the composition that I reminded you here. So the four are the fermions here, and the six is the here. So you could complain that KAI AB, the index AB run from one to four, but there is a constraint that KAI AB is equal, say, to epsilon ABCD KAI DC dagger. So this gives you the six of SU4. And another way of phrasing this is that they stay, they are living in endomorphism of K with pi. I mean, some projecting because they are fermions. And similarly, and so they have charge under DC1 minus one. And if you add the charges, you see that if you introduce now another fermion multiplet, which we call Upsilon and Upsilon dagger, which is with its auxiliary boson, this time valued in the bifundamental of UK and another UN, which is what we, I hope this read about, is this UN sub M, which charge minus one, then these charges are canceled. And so we just add this auxiliary system to our homological field theory. And again, one can say that it's valued in homomorphism of M to K. So this is more or less my field content. Can you say again? What is the U1? This one? You just invented it. Ah, no, these charges you can just read. I think you can read by observing that this U1 is the transverse one to the array plane. So this should be the correct charges of the field. And then this circle one you assign it. And yet another, okay, now I think about yet another argument is that, so this minus one, I mean minus one times K times N times two, because there is a Upsilon and Upsilon dagger. So this is minus 2KN. And this precise the virtual dimension of the module space that I will, of the instant module space that I will come in the, so that's another way of checking that this should have that dimension. I will come to that in a moment. So, yeah, instead of, okay, like I said, I mean, the action will have a zero to zero sector, which is this reduction, and then the zero to eight sector, which you can, for example, compute, which will have kinetic terms, and then a scalar potential. The interesting part is actually the scalar potential. So let's go on the X branch. Namely, X branch means that there will be a complexified connection, and we go to the branch where this 80 say plus phi is zero. And then we can read off this scalar potential, and it will be positive, and it will be the sum of squares. So I wrote here the first term is trace SAB, S dagger AB, less than B, from one to four, where this first term is just involves the d zero, d zero sector. Namely, it involves those matrices that I just introduced BA. It's like commutator of B, ABB, plus epsilon ABCD, BCBD dagger, and it's actually related to itself in this way. And you can see here already that the SU4 tensor appears, is epsilon ABCD, which means that this guy won't be invariant under a full SO8, but we have to pick an olonomy, and we pick SU4. So this is just, I mean, it's not we pick, it's like the system picks it for us. And the second term is trace some real moment map that comes from the d term in the vector multiplet, which I call mu r, and we can actually, actually V10 itself instructs us to do so that in the presence of V field, we are allowed to introduce a phyethyliopolis term, which I write, which I denote by zeta. So we take it positive for reason that will be clear also later. And this mu r, now is as cross terms, we did zero d8. So it's sum of the commutator and plus ii dagger, where i is rd 08 matrices that I introduced before. And now, the model space I was talking before is this m, which is just the zero set of the potential, of the scalar potential, modded out by UK gauge transformations, where the UK action is the obvious, I mean, is the one that I introduced before, so the B transforms in the joint and the I transforms with the factor. And on this space, okay, I will have a few comments about this, but let me first say that, first of all, it's quite clear that on the space of solutions to these equations, the B is commuter, which comes from the first piece. You can manipulate this SAB and SAB dagger using the epsilon and so on. And you will, this will imply that all the terms, I mean, that all these terms alone should vanish, so that it's a simple computation. So B is commuter, as is the case in the Zilber scheme of points stories. And the second one is that you can, I mean, mu itself has to be equal to zeta, so we can trace it. And so we discovered that the IA dagger must be non-zero and must be equal to K times zeta, where zeta is the FI. And again on this space, which is the space whose virtual dimension is, say, real, is took again, I was talking about before, there's a stability condition type that holds and is that we can take, for every vector V in a capital K that we can take, capital K is that K dimensional space I just introduced. You can, there exist a polynomial, F in four variables, Z1, Z4, and the vector V in N, N is the other relevant vector space, such that this equation is satisfied. So F of B1, B2, B3, B4, applied to I, applied to V is equal, can generate all these. You can generate all these in this way. So now, okay, this curly line is meant the following. I mean, this type of condition is very reminiscent if you are, for example, familiar with the 0D4 case, namely the four dimensional gauge theories, this will be the Hilbert scheme of points on a, say for N equal one, so N capital N one dimensional, this will be the Hilbert scheme of points of C2. And this type of, so for example, if we denote the map phi as this map here, then there is a corresponds by kernel phi and the ideals and vice versa, I mean, vice versa into this space of equations. So this, I mean, this is not a proof, but this is somehow meant to suggest that in this case, the space, I mean, this can be proved, it will be the Hilbert scheme of points of C4 and in higher N will be generalization thereof. And as in 2D, the, now the torus fixed points of this space was the ideas parameterized by the partitions. In this case, like I say again, this is not a proof, that's why you're at this curly arrow and I will come back to it. But this, in this case, the relevant fix points of my actions will be the solid partitions. So for dimensional partitions and moreover will be colored because in general the rank N, small N as rank N, so I take color solid partition. So that just mean to suggest that we should think about this solid partition and that's meant to motivate my next slide, which is an introduction, a brief reminder of this story. So we can think, let me see. And also, yeah, it's very important, I forgot to mention that the silver, I mean, in general this space and already the case N equal one, the Hilbert scheme of C4 is very different from C2. I mean, it's not smooth, it's not type of caler, it's not caler, it's just maybe symplatic quotient. So it may be much more singular, but still we can, this type of localization goes through and the fix point one can extract and are these solid partitions. So there are many analogies with four dimensions, but up to a certain point. For example, in the 0D4, this data somehow is playing the role of desingularizing that space. But here, Hilbert of C4 for i enough k is already singular, is not irreducible. So there are some differences, just to point out. So now the solid partition, let's, yes, yes, yes, you mean, ah, you want the microphone? This one, that's fine. Yeah, I'm just saying that this i and i plus in the middle of the thing, in i dagger, yes. Yeah, i and i dagger. If you really quantize the 0D8 string, because you have co-dimension one, there's an ambiguity of the GSO projection. That's correct. And on one side this exists and the other side is just all massive, if I remember correctly. That is correct, but this word that you are talking about is in the B field space. So I'm talking in the B field parameters. Even when B is zero. No, when B is zero, there is no such thing. Yeah, that's my point. Yeah, okay. When B is zero, this guy's, in this sector is massive, so there is no i. When I turn on B instead, I have it. I see, so you're always working with B non-zero. Correct, yes, yes. I always assume this B is turned on. That's fine. Right, right. Otherwise there is no, and that you may ask, you may ask about wall crossing, for example, like you did. Yeah. And so just quickly, how much time do I have? 15. 15, okay. Okay, so quickly, just a reminder, in four dimensions, the relevant things are, so four dimension for the gauge theory, so d0, d4, and two dimension for the young diagrams. I mean, the relevant objects are the young diagrams, which is just, I call it y, is a collection of non-negative integers, and it's decreasing, and it will be relevant to have its size, which we denote size y, it's just the sum of the integer, and there is an inclusion relation, which is just the objuzer. So this is the picture. Then we can go up one dimension in 3D. I mean, there are many ways of thinking about, I just presented the recursive one, which I think is the most natural, for example, for computation. So the next step is 3D, namely plane partitions. These are relevant for six dimensional gauge theories, namely d0, d4, d6 gauge theories, and you can think now of being in three dimension, stuck in cubes under the, say, going the first octant and put some gravity and stuck in cubes under that force, and they will assemble as a 3D partition. So a 3D partition raw is now a collection of young diagrams, such that it's again decreasing, in the objuzer sense, and we can define its size, which is the sum of the size of the young diagrams. And again there is an inclusion between, say, row 1 and row 2, if all the young diagrams of 1, each young diagrams of 1 is contained in the corresponding young diagrams of 2. And then the ones that are relevant for us are the four-dimensional or solid partitions, which are relevant for these eight-dimensional gauge theories, or d0, d8. Okay, now I'm stuck in hypercube, so I do not know how to draw it, but this is meant to be the analog in four-dimension of the previous picture. So you want to think, yeah, presumably, there's many ways of thinking, yeah, yeah. Yeah, yeah. I mean, this way of thinking is like taking a slice. You cut it by a plane and you read the slice of plane partition. Zoom up, zoom up, yeah, yeah. So for these, I can again define the size, which is the sum of the sizes of the plane partition composing it. And I will actually be interested, like I said, in colored solid partition, say of rank n. And this is just a vector of solid partitions, pi 1, pi n, where each pi i is a solid partition. And for, say, such an object, I will be interested in its character, which again I call capital K, which is the same letter I use for the vector space, so I'm identifying both the vector space and its character. So now the character associated to this pi will depends on the q's and on the n's. And it will be a sum over alpha of the n alpha times the character of each pi alpha. So sum over a point, and this is, again, another way of thinking about, is a two-pulse, four-dimensional two-pulse of non-negative integers, a, b, c, d in pi alpha, and then I take q1 to the a minus one and q2, q2, q3, q4, and so on. And, okay, just, it's not an observation, it's more like a curiosity, is that for 2d and 3d that represented before, the generative function is known by MacMau, I think. And, but in 4d instead it's not known, and people have only worked up to some size, say, 672, or that order. So it's not known how to do it in general. And now I come to the actual computation. So let's go back to the index we want to compute, which I called in the beginning zeta k. And this, in the integral language, as, I mean, let's write formally, is an integration on variables phi, and there will be an action, which I didn't write, which is, will be a functional of these phi's. And, I mean, I also didn't write the supersymmetry transformations, but are the more or less the standard ones. But anyway, the way that one can compute is that, in principle, this would not have periodic boundary condition, because it was twisted, but if we weakly gauge this g, then we can, there's a standard argument that the integral can be written as a periodic boundary condition. And this omega is meant to remind you that now the action itself depends on the parameter. So the same as in the four-dimensional omega background. So I call it s omega. And now, in this step, we can apply, I would say, the standard localization argument. For example, I mentioned before, there is a, I mean, there are various papers studying this. For example, there is a nice paper by Ori, that summarizes this construction for all equal to two system. There is also the previous work by Muren Krasov-Stratashvili, that deals with this kind of... So depending on your taste, you can... Anyhow, the procedure, the first thing that you have to, that one has to do is to, I mean, compute the wall-loop determinants of the field, and one gets a rational function. So just by just inspecting the field content I presented in the table before, one can write down this rational function, kaj k, which is, so, this product of a very different than j, this comes from the vector, this comes from the four, so the chiral multiplets, this comes from the Fermi multiplets, where here A goes to one to four, and here only to three, and this QAB is just a notation for QA times QB, and then there is... So these are the i's, another chiral multiplet, which give you product over i and product over alpha and alpha minus xi, and these are the Fermi multiplets that they call the Upsilon, which gives you N-alpha at the numerator, N-alpha minus xi. And then the procedure, the first step is that the party integral, I mean, there are many subtleties related, for example, to having, I mean, one applies localization and somehow wants to, the result turns out to be a residue, that's because you wanna apply the somehow stock's theorem and what is about boundary terms, the same boundary terms that give you wall crossing, and then by stock's theorem you can write it as a residue, so since we are in higher dimensions the appropriate tool is this object called Jeffrey Kirwan residue, so the result is one over k factorial because of just vile symmetry, and then some over some poles, so I call them U star, just because for listing the poles is better to go to these variables, and then some chi and residue respect to the U, and the interesting thing is that, I mean, the important thing is that the poles will be leaving some m singular, some space of singular hyperplanes, which are determined by chi, I mean, by the field content, and so by chi, and so where at least k linear independent hyperplanes meet, you define a point U star, and then you can compute these objects, which, so for each hyperplane arrangement defined by U star, you compute these objects, and these objects depends on the, so these are another hints of wall crossing, depends on the field heliopolis zeta, not punctually on zeta, just on the chamber, it just depends on the chamber, so this is like standard argument, and then you can manipulate again this expression to write it as a sum, now over the real, the true fix points, which are these colored solid partitions that I just mentioned, of sides k, of course, of a standard residue, the standard iterated residue of the same function chi, now computed at the flag, now, I mean, the iterated residue, the appropriate way of doing it, one appropriate way of doing it, define a flag, this flag will be given precisely by the content of pi, I mean, there will be a correspondence by the content of pi and the flag, once we pick an ordering for the monomials in pi, and also, okay, I use value invariance to get rid of this k, this chi k is value invariance, so we can get rid of it, I mean, and also a curiosity is that this residue itself was also studied by other people, by Parshin, by other mathematician, and again, they come up, I mean, I say this because they, both this jk and this iterated residue, of course, have a representation in term of cycles, so one is really integrating on some cycles, this is well defined, and yeah, that's, that's, okay, let me stop at this, indeed this is like, yeah, yes? Yes, yes, yes, but remember that I had a condition product over q is equal to one, yeah. No, actually, are you asking about value invariance? Of permutation of the x i's, or permutation of? So you're asking about the value symmetry of su4, yes, in that case, you need to enforce the condition product of the q is equal to one, and then it will be, which I always assume it, from this point, oh sorry, from this point on, I assume it, otherwise it's not even supersymmetric, yeah. Let me see if I want to add something. Okay, yeah, now this is, the result, now I come to, back to the, the conjecture that I introduced in the beginning, and that, and the conjecture is that the zeta, the partition function, has a simple free-field representation, so let's define the platistic exponent of some function f of x1, xr, I mean, it's the obvious way, so it's pA, and it's defined as exp, sum over m to one to infinity, one over m, f of the product of the variables, x1 to dm, xr to dm, and then, if we do so, the conjecture states that, if we take the generating function, the one that I wrote at the very beginning, now the dependence on g is made precise in sense that g is parameterized by the qA's, the nA's and the m alpha, and the conjecture is that this guy is equal to the platistic exponent of a simple function fn of q1, q2, q3, q4, minus p, dependence on p, and then the dependence on the Coulomb branch parameters and the masses just through an overall product, n over m alpha, where f more explicitly, if we define bracket of t as t to the one alpha minus t to the minus one alpha, then f is the form bracket q1, q2, q3, q2, q3, which I remind you q1, q2 is q1, q2, divided bracket q1, q2, q3, q4, q4, then the argument I call s effective, so bracket of s effective divided by bracket of p times square root of s effective times bracket of p over square root of s effective, so this is like a simple conjecture that we tested up to some order and so on. On the right, n, you are asking the rank, let's say again, yes? Yes, yes, I understand. Well, it's just, I mean, the form is the same, it will be just in the variable s effective, but formally it's... Yes, so it's in the form of this guy, but you are right, formally I could have dropped the 10, because it's in here if you want. That's true, that's true. And okay, I have a few minutes, right? Or... Okay, good, good, good. So I mean, it's just a variable that I write here to, I mean, just a dummy variable that I use. So the conjecture is that the guy is equal to PA with this argument, so that depends on what is in here. Just I just write a dummy variable. So I can affect it because it's a effective U1, it's a product of all the guys divided that way. So in the last few minutes, I just described some mathematical properties of this construction, and it goes as follows, we can introduce a tangent space to this instant-tomoduli space. Yes, yes, thank you. We can introduce a tangent space at a fixed point. So the fixed point will be fixed by some toru section, and the toru section was the maximal torus of g flavor. And the fixed point we know is given by a color, a solid partition, and the tangent I call t, and will be equal to n minus m star times k minus the product from one to three, one minus qa k k star, which we think of as a virtual character, namely, I get before the spaces n and m and identify them with their characters, which are just some of the n's and some of the m's for m. And again, k was the character of the solid partitions that I defined a few slides back. And star just means to replace each of these variables, qa and alpha and so on, by their inverse. And okay, you could guess that this is already, I mean, you could guess that this is like the square root of what you would expect to be the tangent to this moduli space. And in fact, I say this because there will be issues with orientations, namely, one will, since we are taking a square root, one will have to be careful in picking correctly the signs. Otherwise, this is just the trivial, I mean, this is also what happens in the 0D6, the 0D4. There, you don't have this issue. So that's why I'm pointing out. So the first factor is that if you, if we pick a generic parameters m, alpha and n, alpha, generic means that they are not functions of the, of the q's, which, by the way, was also an assumption in the localization computation. Otherwise, somehow you have to resolve it and go back to degenerate cases. Then the tangent at some solid partition is movable, where movable means that, okay, let me explain a bit better. So the tangent t, once we write all the characters, is a sum of monomials in the q's, q inverse, n's, and so on. Sum of monomials with plus or minus signs. And actually, you can, if you look at it, you can convince yourself that there is the same number of monomials with the plus sign and the same number of monomials with the minus signs. And there may be, and there may be, there will be in general cancellations among them. And there will be also, there may also be plus minus one factors. The statement that this guy is movable is that there are no, after you simplify, of course, there are no plus minus one factors in this sum. And this is a fact that you can prove rigorously. And the second factor is points that the, I mean, in words is telling you what you expect, so that this tangent space picture is equivalent to the residue picture. Modulo, the sign that I was talking about. So concretely, the statement is that the iterated residue of chi k, that was at the end of this slide, is equal at some color, the solid partitions pi, is equal to some function of the tangent at again pi. Up to a certain well-defined sign, which I call minus one to the h of pi. And this sign has to do with orientations, namely, I mean, it's a concrete function for the partitions that has the interpretation of orientation of the formation and obstruction bundle to the modulus space. And then finally, the last thing they said, this map, a hat that I wrote there, is just a map that acts on monomial. So if you have monomial r and s in the tangent, then a hat of r is just one over bracket r, and it converts sums to products. And notice that this thing is well-defined because t is movable, otherwise there will be plus, either zero or infinity factors. But since it's movable, this is well-defined. And this is just a map that converts the tangent to product of weights in the equivalent k theory. So, yeah, I think I stop here. So thank you for your attention. Questions for Nicole?