 So that's, as I left off last time, we're about to start explaining this Reverend Frenkelberg and Akechima construction of the Coulomb branch created to a gate theory. So let me remind you of the setup. So we had G, a reductive group, usually a product of GLNs and N, a representation of G. And for this talk, the main example I'm going to consider is associated to this quiver like this, which I'll remind you means that G is the group G-I-M, and N is the HOM space indicated by that arrow. So it's HOM from C-M to C-N. That's going to be our running example for today. And I mentioned last time associated to this GNN. There's a physical theory, and this theory has a Higgs branch and a Coulomb branch. And so the Higgs branch is easy to understand mathematically. It's just a cotangent bundle of N. Then we take the Hamiltonian symbiotic production, Hamiltonian reduction by the action of G. And so in our running example, this will give the cotangent bundle of the gross money of M, G, and N. And now we're about to define the Coulomb branch, which I'll denote M, C, G, N. And I'll remind you that these two guys are supposed to be symplectic dual, which means they have a whole host of structures, different structures on each side which match. So we'll see a bit of that today, although in general it's kind of conjectural some of those properties. By the way, I never said this so far, but the dual of the cotangent bundle of the gross mining is a resolution of the M, N minus M slow to a slice. I won't write that down. Okay, so I explained last time the rough definition of this Coulomb branch. So the rough definition is that it's spec of a certain algebra that algebra is constructed as the Braille more homology of a space. And that space is a modular space of maps from a ravioli curve into the stack quotient N over G. And I remind you that this ravioli curve is a non-separated curve made by taking two function, two formal disks and gluing them together along a punctured disk. So it looks like this. It's like an A1 with doubled origin, but just locally. So this description is correct, but we're gonna unpack this thing now. Okay, so in order to unpack it, I'm gonna first tell you how to think of this modular space of maps. So it's a stack quotient by this G action. So a map to a stack quotient by G means you have a principal G bundle on the source. So this modular space parameterizes pairs P, S, where P is a principal G bundle on this ravioli curve, S is a section of the associated N bundle. All right, that N sub P. And by definition, that's the associated bundle. So I take P cross N and divide out diagonal action and G. Now, this description is not so helpful. So we're gonna break it up into the two disks. So a principal bundle on this union is the same as a principal bundle on each disk along with the gluing data. So this is the same as pairs P1, well, it's not just pairs, but two both. P1, P2, Phi and S. So are these guys? So PI, they're both principal bundles on a disk. Phi is a isomorphism. PI on the punctured disk. So P1 on the punctured disk, with P2 on the punctured disk. And Phi, that's right, S is a section. So it's a section, let's say it's just a section on one side on one of the disks, such that after transporting it using Phi to a section on the other disk, it extends. So now it's just a section on the punctured disk, but now it extends to a section on the other disk. By the way, I meant to say this right at the beginning or I meant to do this example, let's sneak it in. So in the example of GLM and N being this HOM space, I'll just, maybe it's kind of obvious, but just to make things more concrete, this is a pair V, A, where V is a vector bundle of rank M, and vector bundle on this disk, and what's, sorry, on this bubble, on this ravioli curve. And A, it's a section of this associated bundle. So it's just a map from V to the trivial rank N vector bundle. So that's just what that works out to in the example. I'll come back to an example in a few minutes. Okay, so back to the general case. So I have principal bundles on these two disks and I'm gonna trivialize one of the principal bundles. So ignore one of the disks for a second and just think about the first disk. And then, well, sorry, if I trivialize one vector bundle, then this phi will simply be the data of a trivialization. So that leads me to the following two spaces. We call it TGN and RGN. So TGN is the following space, P5S. So as I said, P is a principal G bundle on a disk. Phi is a trivialization on the punctured disk. So this P0 means a trivial bundle. So here I had an isomorphism, two principal bundles. Now I require one of them to be trivial, so I just get a trivialization. And S will just be a simply a section of this principal G bundle or rather a section of the associated bundle. So that's my space T. Now inside that space T, I have a smaller space called R, which is P5S as before, but with the additional requirement that this phi, when I apply it to S, I now get a section of the trivial N bundle and I want it to extend over the disk. So what's the relationship between these spaces? Well, these two spaces, TGN and RGN, they have action of the group. Oh, I meant to introduce some notation. They have action of the group G of O. So for today, I'm introducing the following notation. O will be the ring of power series, K will be the field of Laurent series. Some kind of, I'm being a little sloppier, but some kind of spec O is this disk and spec K is this puncture disk. Okay, so the space RGN or TGN have actions of this G of O, which acts by changing the trivialization. So that's the action of this G of O. And if we allow ourselves to change the trivialization, that's the same as like untrivializing the second bundle. So we find that RGN divided by this action of G of O is actually the same as the mapping space that we're interested in. So from now on, we're gonna kind of ignore this mapping space and just think about RGN with its action of G of O. Any questions? I promise you things will become sort of more concrete shortly, this part is a little abstract, but maybe some people like it. Well, it's a pretty thing. Okay, so whatever I wanna say next. Okay, so let's just examine the structure of TGN and RGN for a second. So we can forget the information of this section. So we get a map from TGN to what you might call TG0. So we just set N to be the zero of RGN bundles and the data goes away. Then we just have a pair of P and Phi, principle G bundle in a trivialization and maybe people know that this is the same thing as the affine gross mining of G defined to be the quotient G of K by G of O. So the reason for that is because we can trivialize P, if we trivialize P then the data of the trivialization is the same as an element of G of K because it's an isomorphism of P0 and P0, the trivial bundle on the puncture disc. And then we mod out by the changes, possible changes of that trivialization. So this TGN maps to TG0, that fingers money and this map is actually a principle. No, it's not a principle bundle. This map is a vector bundle, infinite rank vector bundle. Namely it's fiber over some points is just this space of sections. If we write it in this quotient terms, we can realize it as follows. So this thing can be realized as the fine thing pair G, S. So G is in this affine gross mining of G and S is in G and O. So it gets fiber over some point, brackets G is just G and O. So it's an infinite rank vector bundle. And inside of here, we have this smaller space RGN, let's examine it in this affine gross mining in terms. So then it's pairs G, S. Well, G is still in the affine gross mining and S now it's in G and O. But we have this second condition over here that after transporting it using five, it extends through a section. So, well, this S is actually really like five S from before in some sense. So the condition is simply that this is an interval. And again, this will map to the affine gross mining. So this map, so this guy is not a vector bundle anymore. But over each G of O orbit in the affine gross mining it is a vector bundle and it's again infinite rank, but it's finite co-dimension in the trivial bundle. So it is an infinite rank vector bundle of finite co-dimension in the trivial vector bundle. Okay, so that's RGN. So maybe it's helpful to see it in the example. So again, in the example when G is equal to GLM and N is equal to HOM, CM, CN. Then we can think of the affine gross mining of G as being, instead of this quotient, we can think of it in terms of lattices. So it's the same as O lattices in KM. And in these terms, TGN is a pair consisting of a lattice and A, what's this A? So A here is at M by N matrix over the field K and it required that A of L inside of ON. So that's TGN and the RGN is almost the same thing, but now this matrix A should lie in over O and have the same pitch. Okay, so we can give relatively concrete descriptions of these funny spaces, RGN and TGN. In fact, well, I didn't go to all of Eugene's talk but I fortunately went to the right part where he was mentioned a little bit about this BFN, well, generalized Springer fibers. So these generalized Springer fibers are precisely the fibers. So Eugene mentioned briefly these generalized affine Springer fibers, or I like to call them BFN Springer fibers and they're the fibers of this map. I mean, sorry, in greater general, I mean, not just for the fibers. I won't talk more about this, but since Eugene mentioned it, it's not a huge. Okay, so we have our space now, RGN. So let's just rewind a second. So where were we? We wanted to find the Coulomb ranch. We wanted to define it as the homology of some space of maps. We examined the space of maps. We rewrote the space of maps in these affine Grismanian terms, like so is this quotient. So studying the homology of the space of maps would be the same as studying Geo-Vo-Echeverian homology of RGN. Is there a question? Does the space of maps of the ravioli to N of G emit a square root of the canonical? I don't know, I'm not sure. Sorry. Oh, and then there was another question. What is GN of O? I think hopefully they're under answered that and I think your answer looks good. So G is an element of Geo. Yeah, so I guess the question is about this notation right there. So G is an element of G of K. It acts on N of O and N of O sits inside of N of K. So after acting on it by G, you get some other subset of N of K. Please ask again if that was unclear. And maybe the slight subtlety is that this G is like sitting in brackets because it's modular G of O, but of course G of O acts on N of O to preserve N of O. So this is invariant under the right multiplicity. If we change G to GH or H lies in G of O, you won't change G N of O. Okay, great. Okay, so now we come to the theorem of the FN. So we've formed the following algebra, A G of N. So it's just by definition, well, the G of O equivariate homology of RGN. So actually this definition requires some care because this space is very infinidimensional and the group is also infinidimensional. So it takes a little bit of care to define this space and essentially in their work, they work with cycles which are finite dimensional along the aphangarous monion along the base. So RGN is like a, not quite a vector bundle over an aphangarous monion, something a little like a vector bundle. They work with things which are finite dimensional along the aphangarous monion and then finite co-dimensional in the fibers. So we'll soon see an example of what I mean by this. So with some effort, they define this equivariate homology of RGN. And the theorem is that this carries a structure of a cumulative algebra. So where does the algebra structure come from? So sort of morally or intuitively, here's one way to think of it. I mentioned the space TGN a few minutes ago. It's bigger than RGN, but actually a few, so TGN maps to NFK because in TGN, we don't require that S be an N of O. So it just comes to the map to NFK. In this map from TGN to NFK, it's an analog of the map from the co-tension bundle of the fly variety to the nilpongon. So it's natural to form the analog of a Steinberg variety. I'll try what's the analog of first, this Steinberg variety. So we would take TGN cross with itself over NFK. We call that thing ZN. So this is not, the reason why this is actually not what's formerly done is because this leads to more infinite dimensional issues. But if you imagine doing this, then you'll see that the stack quotient of ZGN by the action of GFK is actually the quotient of RGN by G of O. So that's the thing we're interested in. So, and this guy, you could study his homology or the G of K equilibrium homology of Z. And that thing would become an algebra structure just like the same way it works for the usual Steinberg variety. So just like the construction from Chris Ginsburg of convolution algebras, you could apply it to ZGN. If you were not worried about infinite dimensionalities. So if you avoid ZGN, which is what BFN do and work instead of with RGN, then you do basically the same thing, but things become a little like more complicated or less symmetric looking because of this, because you're trying to avoid this infinite dimensionality problems. So one way to implement this algebra structure is to consider the following thing, which maps to three copies of RGN. So we consider triples G1. So that's a point in G of K, G2. That's a point in the affine-gross monion, that's why I put it in brackets, and S and S lies in N of O and also the S lies in G2 N of O and also G1 S lies in N of O. And this space maps to three copies of RGN. And then you can use this space to define the multiplication basically by pulling back your cycles from two copies to this space, intersecting them and then pushing forward. And again, there's something unnatural about the fact that we have G1 outside of brackets and G2 inside of brackets and there's unnaturalnesses because of what I just mentioned about the fact of argument R and Z. Just for completeness, I don't care as that much, but I'll write down the three maps to go with this. If you open up BFN, you'll see that my notation's a little different than theirs, but it's a code. So those are the three maps. So let's see an example now. So all this was a little abstract. Let's see an example. So it's a sub-example of the example I was running. So I just take the M to be one. So my group G is just C star, GL1, connecting on C to CN, which is just CN and the C star because it's acting from HOM, C to CN is actually naturally acting by the inverse scaling. I could get attacked by scaling by a worsening error. I just think of it this way. Doesn't really matter. And then what's our affine-gross money? And well, first of all, the affine-gross money of this group C star is just Z, Z, because we're just taking K star, mod O star, and so we're taking invertible Lorentz series, mod invertible power series. So that's just given by the order of the Lorentz series. So we can write it just like T to the P. And then what does our R space look like? It looks like the disjoint union over P of such a point T to P. And then the choice of S is given by the intersection of T to the minus P O intersect O and at N of them. So this is where the S stands. And what kind of cycles am I gonna consider? I'm gonna take one of these chunks. So one of these pieces of this disjoint union. For example, I take one, which is the fund, so I take the fundamental class of where I put T one and it's my affine-gross money in element, and I put O and I allow all possible choices of S. So it's an infinite dimensional cycle, but it's finite dimensional in the affine-gross money direction, infinite dimensional in the fiber direction. And I call that the U and my other element will be the cycle of T inverse cross T O. And my last element in this ring will be W. So these U, V, and W all live in the homology of RGN. So that's my algebra, AGN, if I'm not very consistent with the brackets. And W, my third element, will be the generator of the C-staric or very chronological point. I didn't say this, but the unit in the ring is just the class of the identity element in the affine-gross money and cross the fiber. So maybe I should write it somewhere, but because the reason why I'm mentioning it right now is because really W acting on the, I guess. Okay, I won't write it, but okay, understood. And then to compute U, V, the product of these two cycles, I use this correspondence diagram that I wrote above. And if you do it carefully, which I just did a few minutes ago, you'll find that you end up with the class of, let's say use correspondence diagram, and you end up with the class of P zero, you end up with multiply this T one, T inverse as your affine-gross money. And then you're doing some kind of intersection of these things. So you end up with T O, it looks very straight forward. It looks like a really disintersected, but well, anyway, it's basically straight. So you end up with that guy, and that guy sits, like in the, so, what should I say? Again, the identity element in the ring is one. The identity element in the ring one is just a class of T zero across O n. So this is almost like identity element, but it's not because it's sitting as a sub variety. And so it differs from that by vector space. Sorry, I'm hearing a bit of, I don't know if you guys are hearing that too so much, but I'm hearing a little echo. Then, so in fact, this guy is actually equal to W n because that's the Euler class, Euler class of O n over T O n. Okay, so that's the origin of this thing. I mean, so therefore, we conclude that UV is equal to W n in this one. And this is pretty much the only computation one can do. I mean, there's some variance in this computation, but this is basically the only computation you can ever do inside this algebra. And I'm just joking, but then maybe I should say this a little more in a positive light. The variance on this, let me just write the conclusion. The conclusion is actually that this is the only relation in the ring and that in this case, our algebra is UV W modulo UV minus W n and therefore our Coulomb branch is C2 mod 2 mod n. So let me say that in a more positive light that this computation can be generalized to the case of arbitrary G being a torus and N being any representation, this generalized in a straightforward way to show that you've got a hyper-toric variety. And so maybe I should just write that. So I guess we can stay with this here. Also do to be a thin. So if T is a torus, then during the same computation we see that MCGN is an f-line, not just N, but B-f-line hyper-toric variety, symplectic dual or Gale dual. So Michael explained about that in exercise session yesterday to the Higgs branch, I mean, if G is a torus. And the proof of this theorem is just exactly the same computation just done more generally. Oh, actually they have a completely different proof of this theorem too. But anyway, one way to prove this theorem is to do the same computation if I do it. And another thing I wanted to say is that if G is arbitrary, you can embed it in, you can embed this Coulomb branch algebra in the course planning one for the torus in kind of make use of this fact, computation as well. I'm not gonna really write that down there. Okay, any questions about this definition? So that's our definition of the Coulomb branch and an example of a computation. Any question about it? So let me move back up to the definition. Definition was here. Well, that was the theorem, but it's commutative algebra structure and the definition of this R space was sitting here. So it's, I don't know, sort of complicated, but maybe once you work with it enough, it becomes like something a person can handle and get their mind around. Okay, great. So let me, but feel free to stop me at any time with questions about it. Now I'm gonna describe some properties of this Coulomb branch that follow from this definition. And in doing so, we'll see why this thing looks like a candidate to be the symplectic dual of the X branch. So the first one, maybe isn't quite of that formula. Anyway, so we have a map. We have a map from, of course, from RGN to the f-ingress mining of G. And this can be used to define a map, not in a completely straightforward way, but you can use it to define a map from the algebra AGN to the algebra AG0. So AG0s means we have no representation. So we're simply taking the equivalent homology of the f-ingress mining. And therefore a map backwards from this Coulomb branch associated to G0 to the Coulomb branch associated to G and N. Sorry, I haven't written this today, right? Just the wrong way. Sorry, I think I wrote it backwards. I mean, of course the map goes this way, but I think the map on algebra goes the opposite direction. Yeah, sorry, I think everything goes backwards here. Well, I guess I can reverse the arrow here. Sorry, the reason why it goes backwards is because this thing doesn't quite come from this map. It comes from the fact that this embeds inside of this trivial vector bundle. And then we sort of have a schism map going in the opposite direction on homology. Okay, I'll make sure to check that for next time. But it doesn't make very much difference from what I'm about to say, but I think this map goes in the opposite direction. Anyway, so this guy, this guy has been extensively studied before the equivariate homology of that famous monument. And it was proven by by Bezokovnikov, Finkelberg, and Mirkovich. So B, F, M, and the B is a different B. M, M, M, M. That this is isomorphic to the universal centralizer space in the Langley's dual group. So the following space. So pair is consisting of G, X. So here, these lie in the Langley's dual group across the Langley's dual, the algebra, regular locus, such that this element of G centralizes X. This element G centralizes X. And then we quotient this by the joint action of G. So it's called universal centralizer space. And this maps to the just remembering G check, which is the same as the T dual, not W. And the generic fibers in this map are tori. So this map is a, this is Taurus. So this is a theorem, which very much in the framework of the geometric's attack a correspondence. And we see this actually, so to see how this is working. So this T star, my W is the same as the G of O equivariate homology of a point, which is the same as the geographic homology of a point. And this geographic homology of a point embeds in these algebras, as I explained above, by acting on the identity element. So by acting on the class of the point T is zero and the F fingers money. So, so I expect this. So I get a map like this. And so we, this thing actually we have a diagram like so, and also this thing maps here. Okay, this MCGN is not the same as MCG zero, but in this way it's based quite similarly. So the generic fibers of this map from MCGN to T star, my W is also teacher. So in particular, the dimension of MCGN is twice the dimension of this T as usual is the maximal Taurus. Well, T check is a maximal Taurus of G check. So it was equal to twice the dimension of T check or also known as the rank of G. Okay, so this is the first property of this MCGN that it's pretty close to this universal centralizer space. And in particular it has this dimension. This map here is a, yeah, and it says, but this map here is a birational map. Okay, so that's like the first basic property at least we have its dimension. So the second thing, we can define a non-commutative version of the BFN algebra, of this non-commutative pool of branch algebras and it's called. So we simply tweak our construction. We take a covariant homology, not just respect to GFO, but respect to GFO, semi-direct product C star. So we call this thing a H bar GN where H bar is the generator of C star covariant homology at the point. And how does this work? Well, this is C star acts an RGN by what's called loop rotation. What this means is it comes from an action of C star ultimately on the ring O by scaling the variability. Or if you like comes from an action of C star on the formal disk. So this is well studied action in the affine-gross monion and gives an action here. We take its equilibrium homology and this gives us a non-commutative algebra now. And then we, for example, if we ran the computation I did above, we would find this new v thing would be the class of this T O N and we'd end up with sort of as before we'd end up with O N mod T O N and we'd get, well, this W to the N. But if we ran the computation in the other way around VU we would actually get something like T inverse O N mod O N and then that would be, that has action of this bigger the C star acts non-trivial there so we get something like W plus H bar today. So that's where the non-commutative function. Yeah, a question. Yeah, a question. But Andrei, you had a question. We're going to remark about the ring for more general G do you get something like variance under the value of inside the ring for T so the Coulomb branch would be something like a quotient of a hyper-triarch writing. So, yes, but it's a bit more complicated. So, there is for every, so the person is asking about the map from A. So we can embed A G N inside of A T N. T N, hope I get this the right way around. Yes. And therefore spec and therefore embed and therefore to find a map from M G N to M G N and there is a relation between these spaces and there's a map. But it's not just like invariance for the action of W. So it's a bit, there is a map but it's a bit more complicated to study. If I had more thought in my head, I would explain a bit more, but maybe I'll move on and prepare this part. But yes, this is one important tool for studying these spaces to this realization. Okay, so here we see the origin of the non-commutativity by basically because these two things are the same as representations of the C star which you're just acting on by scaling on O. But then if we add this loop rotation C star, that acts differently because they're sort of sitting in different positions because of this T business. And so this A H bar G N is a non-commutative deformation of A G N. And therefore, so this is like the reverse. Usually we start with an algebra with a Poisson structure and then find its quantization. So here we find it's, we start with the quantization and we use the quantization to endow A G N with a Poisson structure. A G N, here is a Poisson structure. So again, it's the first order non-commutativity of this algebra A G N. So for example, I think we find from this like that the Poisson bracket of U and V would be the first order difference between U, V and V U which I did it right in my head with just the N, W to the N minus one, I think. Okay, just expand these out. No, okay. Okay, so A G N carries a Poisson structure. So A G N carries a Poisson structure. So it's kind of interesting that from the very beginning of the whole lecture series, I was talking about symplectic resolutions, Poisson prides and then a quantization but here we get the quantization in the Poisson structure actually. Okay, so that's the second property, the non-commutative deformation slash the Poisson structure. Now it's probably a good time for somebody to ask me how can we make a symplectic resolution? And we can and what we'll get there but somehow it's not this next property because the next part we have the one step at a time. So let's consider now a Taurus action on MC, on the 4th of March. Where is this Taurus action gonna come from? When it comes as follows. So the connected components of the affine-gross mining of G because it's sort of like loops into G it's connected components of the same as the fundamental group of G. For example, if G is equal to some product of GLVIs then the fundamental like over some index at I which like we already see some thinking diagram for example, then the fundamental group will be Z to the I the connected components in sapphire-gross mine will be Z to the I. And let's just assume that this fundamental group is free to be the end of this. And let's let T check be the Taurus whose weight lattice is pi one of G. And so then because of this space so this pi zero of RGN so that's the space is equivalent homology we're taking is actually well it's the same as pi zero of the affine-gross mine in so it's pi one of G and so it's a disconnected space and so it's homology is a direct sum structure. So this actually you can see this gives it this algebra A-GN this homology algebra this is a pi one of G graded algebra. Guess what's not totally obvious is that it's the algebra structure compatible with the grading but it is and therefore this Taurus acts on A-GN since almost by Fiat because it's weight lattice is pi one of G so it acts there and therefore the Taurus acts on the filling branch. So this funny Taurus made using the pi one of G so let's think about this from the viewpoint of some magnetic duality. Oscar asked a good question what the algebra also has a grading by homological degree? What does this homological grading correspond to? Let me, well I guess I can mention it right away. Well it's, let me mention it in like about two minutes but I'll mention it before the end of the section. So let's see why this is the right Taurus from symphlectic duality. So recall from symphlectic duality the Taurus which is supposed to act on the one side is coming from the H2 or coming from the Picard of the other side. And notice that this, we have this Kerwin map. So if we take a character of G then it will give us this Kerwin map or tautological line bundle on the Hamiltonian reduction. So recall in the case of Quiver varieties this is just the tautology the determined line bundle of the tautological vector from the Quiver variety. So this is on the Higgs branch side, we have such a map. And up to like finite possible, finite little groups this hum is precisely the co-character lattice, co-atlattice of our Taurus T. So we actually do get that the the algebra of this T is this, well, at least it comes equipped with a map which is often an isomorphism to H2, so that's as, so this Taurus is the right one from the viewpoint of symphlectic duality. And Oscar asked a good question. So Oscar asked, what about the grading on the homology by homological grading? So also carries a homological grading and therefore an additional action of C star. We get on this Coulomb branch. And well, that's to be expected before as well because at the very beginning I talked about conical symphlectic resolutions. So I expect we expect an additional C star action this conical action. So this is the conical to be a bit more precise that this action is not always conical. Like maybe it's actually clearly conical. But in fact, in physics they have a funny name for when this is conical, which is they call it, they did they distinguish between the call they call it bad and ugly theories and conical this condition corresponds to either good or other. So to be even more precise or slightly more precise, you can tweak this. I mean, you have a two tours, you have the C star and you now have a T check acting on your Coulomb branch. So this is coming from the homology and this is coming from the connected components from this pi zero G. I mean, grading on homology. Homological grading decisions. And you can actually try to tweak them and find a sort of C star, we're just sort of sitting using both factors and try to use that to make your conical action. So I think maybe this actually a whole lot of already doesn't usually give you a conical action but often you can tweak it to make an actual conical action. So there's an interesting sort of commentatorics about the choices of how you might do this to try to make a conical action. But roughly speaking, this homological grading corresponds to the conical C star action. Okay, so this was the discussion about the second property. And now, I'm sorry, I guess the third property. So now let's discuss the fourth property which is deformation or resolution. So this comes from a different source which is called the flavor tours. So let's extend our G to a G tilde. So assume that we can embed G to a G tilde which also acts on N and such that G tilde mod G is called F is a torus. So in our quiver example, we would simply take F to be C star to these framing vertices, some of the framing dimensions and then G tilde to be G cross F. So for example, in the basic example like this, we would take G tilde to be GLM cross C star to the N and this part to be the F. This was G. Okay, in this case, it's split doesn't have to be split with us. And then the natural thing to study would be, I'll call it the tilde GN which is the G of O equivalent. So G tilde will now actually act on our, this bigger group G tilde O will act on our space, our GN and we can study echoline homology with respect to the bigger group. And this will give us our commutative deformation of our Coulomb branch algebra and therefore deformation of our Coulomb branch. So then we would take spec this bigger guy who would map the spec of the echoline homology of the respect to this torus, which we identify with the algebra and we get a deformation of this. And this is like the scripty X that I mentioned at the beginning of the lecture. So remember this scripty X was this deformation of the symplectic conical, symplectic singularity and it would map to H2 of the Y. Well, if you were a little bit more precise you would map to H2 of this Y mod W or somehow got ignoring this mod W. So it's really maybe some base change of this X. And again, I'm gonna tell you this is what's expected from symplectic duality because this F will act on the Higgs branch and therefore we see a deformation over the Lie algebra of the, like this H2 will be this F. So this is like what's expected by symplectic duality. So the deformation space of the Coulomb branch is the Lie algebra of the torus acting on the Higgs branch. So again, expected by symplectic duality. So this is like the first thing we can do. The second thing we can do is we can use this to make a non-commutative version. So we just do the same thing. And this gives us our universal deformation, non-commutative deformation, universal quantization. I'll come back to this next lecture in more detail, these quantizations. And then the third thing is about this resolution. So let's choose lambda, the co-eight of F. And why do we need this choice? Well, I explained a couple of times that in symplectic duality, resolutions on one side are parameterized by C stars acting inside of this Hamiltonian torus. So choices of this C star and the Hamiltonian torus which were used to define the subtracting set and that's this C star right here. So that choice is gonna be, actually, I guess I called it row elastic. Let me call it row. So that choice is gonna be needed to define the resolution on the other side. So there are many resolutions of one symplectic singularity and they correspond under symplectic duality to possible choices of attracting sets on the symplectic dual. So what do we do with this row? Well, we consider this our space, but now for G tilde. So instead of G, now we change to G tilde and this will map to the affine-gross money note F. The affine-gross money note F is just the co-eight lattice of F. And so we can consider the pre-image of this over multiples of lambda. And then we form the lambda. We form the union over such multiples by natural numbers of these pieces. So in particular, R zero G tilde N is simply the original RGN. Then we define my notation getting a little hairy here, but maybe I won't introduce notation to this thing. Then we consider this graded ring, which will be the direct sum over N of the G of O equivalent homology of the spaces R and G tilde N. So this is a graded ring whose zero-th degree part is the original BFN algebra that I worked with before. And the zero-th degree part is the algebra I started with in G. Out of time, but I'll just take one more minute here. And so then I have this integrated ring, so then I can form its project. And this guy will be the candidate to be a resolution. So it'll map to my spec of this AGM, which is my Coulomb branch. So this is the candidate to be a resolution. I mean, I should say that of course, I don't know, maybe it's not of course, but not every Coulomb branch will get a resolution. We'll see some examples maybe next time, but this is the candidate to make a resolution. And again, it depends on choice of some data, namely basically this choice of this C star, which is acting on the big branch, which is the right data that we want. So using this C star is what allows us to even define this graded ring and take its part. And again, I should say there's no general result about when this thing will actually be smooth or anything. Nor is there any, yeah. Okay, so I'll stop there. And next time we'll focus on the specific case of these generalized affine-gross mining and slices and their coins. Any questions? Any questions? Remark? Yeah, you have something good. How does this space depend on N with respect to sums and tensor products? I have no idea about tensor products. Sensor products of Ns is somehow slightly unnatural thing to do in this context. Let me make two remarks. The first remark is that if you take that external direct sum, like take two groups, so this isn't maybe not what you're asking about exactly, but it's maybe has a nicer answer, so I'll answer it first. Now, if I have two groups on two different representations, oh, wait a second. Yeah, that's right. If I have two groups acting on two different representations like this, then the cool and branch will be the product. If I have a single group, not myself in a model, do I mean tensor products here? Most likely I didn't say any more. If I have a single group acting on a direct sum, there's actually a map that's related to this map. I wrote in the wrong direction last time, so I better write in the right direction this time. So there'll be a map like so. This algebra embeds into this algebra. Yeah, there'll be a map like so. And this map will be bi-rational. So I mean, I don't know if that answers your question. It's not, and there's also one further if you just took N2, but. And by the way, something you didn't ask me, but I'll say is that if you take the dual of the representation, then it's isomorphic. And this is maybe an important point because as you can see the Higgs branch only depends on for N and N dual are the same. In fact, the Higgs branch only depends on T star N. So it's natural to expect this property. And this you can construct sort of an isomorphism by hand in the Taurus case and then extend. So I think the only way to prove this is to do it by hand in the Taurus case and use the fact, use the guillenization to do it for general gene. Any more question? Yeah. Hi, Joel. Is there any geometric property that this bi-rational map satisfies beyond bi-rationality? Like it's not proper. It's not a blob of something or an affine blob or no, I think it's usually like, well, let me just do an example. So this like that you could take again, this one. I mean, here's a natural example. I'll just take this and some bigger one N plus K or something, right? These two, so here you would get C star mod. And then the good thing is we'll verify if I wrote the map in the right direction. And then the map is like this U V W goes to U W to the K V W. So that's an example. Like here U V is equal to W to the N plus K. So now maybe you can see I wrote the map the wrong way. And here U W to the K. Ah, the map goes the other way. Sorry. This one better be more quotient to these. Is that like right now? Wait, but, sorry. Yeah, sorry, this one goes here. Okay, okay. Oh, there we go. Okay, no, it's okay. Okay, anyway, so to answer your question, I don't know, this map is not a blow up and it's not an open embedding. So neither of those properties are satisfied by this map. So I don't, I hope that answers your question. Yeah, thank you. We have a question with respect to G. It's a good question. I never thought about, maybe I didn't think about it. Yeah, I think a map of groups will give you a map on the, well, definitely give you a map on the affine-gross monions. I hesitate to say something because I don't want to say something that's wrong, but I think there should be a map, yes, between the Coulomb branches. I guess going in the same direction as the map on groups. But again, I hesitate to say something because I'm not completely sure, I'll say something. So I won't write anything down. Any further question? Let's...