4/5 Symplectic Resolutions, Coulomb Branches, and 3d Mirror Symmetry
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Transcript: English(auto-generated)
00:15
So let's, as I left off last time, we're about to start explaining
00:21
this Robert and Franklberg Nakajima construction of the Coulomb branch associated to gauge 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.
00:45
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 GLN, and N is the HOM space indicated by that arrow. So it's HOM from CM to CN.
01:06
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 degree has a Higgs branch and a Coulomb branch.
01:23
And so the Higgs branch is easy to understand mathematically, it's just a cotangent bundle of N. Then we take the Hamiltonian symplectic production, Hamiltonian reduction by the action of G.
01:43
And so in our running example, this will give the cotangent bundle of the grass monument of M glands in CN. And now we're about to define the Coulomb branch,
02:01
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,
02:22
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 model of the grass monument is a resolution of the N minus M slow to a slice. I won't write that. Okay, so I explained last time
02:43
the rough definition of this Coulomb branch. So the rough definition is that it's speck of a certain algebra, that algebra is constructed as the realm or homology of a space. And that space is a moduli space of maps
03:00
from a ravioli curve into the stack quotient N over G. I remind you that this ravioli curve is a non-separated curve made by taking two formal disks and gluing them together along a puncture disk. So it looks like this.
03:21
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
03:40
how to think of this moduli 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 thing, this moduli space parameterizes pairs, P, S, where P is a principal G bundle
04:11
on this ravioli curve. S is a section of the associated N bundle.
04:23
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 G. Now this description is not so helpful. So we're gonna break it up into the two disks.
04:44
So 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 P one. Well, not just pairs, but two both. P one, P two, phi and S. So are these guys. So P I, they're both principal bundles on a disk.
05:14
Phi is a isomorphism. P I on the punctured disk.
05:21
So P one on the punctured disk, with P two on the punctured disk. And phi, sorry, 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,
05:41
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.
06:08
By the way, I meant to say this right at the beginning, I meant to do this example, this sneak it in. So in the example of G being GLM and N being this hom space, I just, maybe it's kind of obvious,
06:21
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.
06:42
And what's, sorry, on this bubble, on this, I really heard. 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.
07:04
So that's just what that works out to in the example. I'll come back to the 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.
07:22
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 five will simply be the data of a trivialization. So then it leads me to the following two spaces,
07:40
I'll call it TGN and RGN. So TGN is the following space. P phi S, so I've said P is a principal G bundle on a disk, phi is a trivialization on the puncture disk.
08:09
So this P zero means a trivial bundle. So here I have an isomorphism, two principal bundles. Now I've required one of them be trivial, so I just get a trivialization.
08:20
And S will just be a simply a section of this principal G bundle or other section of the associated bundle. So that's my space T. Now inside that space T, I have a smaller space called R,
08:43
which is P phi S 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
09:03
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,
09:21
oh, I meant to introduce the notation. They have actually 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. And some kind of, I'm being a little sloppier,
09:41
but somehow spec O is this disk and spec K is this puncture disk. Okay, so this space RGN and TGN have actions of this G of O which acts by changing the trivialization. So that's the action of this G of O.
10:03
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.
10:25
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.
10:40
This part is a little abstract, but maybe some people like it. It's a pretty thing. Okay, so what did I want to 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.
11:04
So we get a map from TGN to what you might call T G zero. So we just set N to be the zero vector bundles and the data goes away. Then we just have a pair of P and Phi, principal G bundle and the trivialization.
11:21
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
11:42
because it's an isomorphism P zero and P zero, the trivial bundle on the puncture disc. And then we mod out by the possible changes of that trivialization. So this TGN maps to T G zero, affine gross mining, and this map is actually a principal,
12:02
no, it's not a principal bundle. This map is a vector bundle, infinite rank vector bundle, namely it's fiber over some point is just this space of sections. If we write it in this quotient terms, we can realize it as follows.
12:23
So this thing can be realized as the following thing, pair G S, so G is in this affine gross mining of G and S is in G N of O.
12:42
So it's fiber over some point brackets G is just G N of O. So it's an infinite rank vector bundle. And inside of here, we have this smaller space RGN,
13:01
let's examine it in this affine gross mining terms. So then it's pairs G S, G is still in the affine gross mining and S now it's in G N of O, but we have this second condition over here.
13:20
Then after transporting it using phi, it extends to a section. So, well, this S is actually really like phi S from before in some sense. So the condition is simply that this is an interval. And again, this will map to that from this money. So this map,
13:42
so this guy is not a vector bundle anymore, but over each G of O orbit in affine gross mining, it is a vector bundle and it's again, infinite rank, but it's finite co-dimension in the trivial bundle.
14:05
it is a infinite rank vector bundle of finite co-dimension in the trivial vector bundle.
14:30
Oops. Okay, so that's RGN. So maybe it's helpful to see it in the example.
14:42
So again, in the example, and G is equal to GLM and N is equal to HOM, C M, C N. Then we can think of the affine gross mining of G as being, instead of this quotient,
15:01
we can think of it in terms of lattices. So it's the same as O lattices in K M. And in these terms, the T G N is a pair consisting of a lattice
15:23
and A, what's this A? So A here is an M by N matrix over the field K and I require that A of L inside of O N. So that's T G N and R G N is almost the same thing,
15:42
but now this matrix A should lie in over O. So we can give relatively concrete descriptions of these funny spaces, R G N and T G N.
16:04
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, this generalized Springer fibers. So these generalized Springer fibers are precisely the fibers.
16:25
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 Jim, I won't talk more about this,
16:59
but since Eugene mentioned it.
17:02
Okay, so we have our space now, R G N. 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.
17:21
We rewrote the space of maps in these affine Gershmanian terms, like so as this quotient. So studying the homology of the space of maps would be the same as studying geoboic ovarian homology of R G N. Is there a question? Does the space of maps of the ravioli to N of G
17:43
emit a square root of the canonical? I'm not sure, sorry. Oh, and then there was another question.
18:01
What is G N of O? I think hopefully the under answer that and I think your answer looks good. So G is an element of G of, 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.
18:26
So after acting on it by G, you get some other subset of N of K. Please, please ask again if that was unclear. And maybe the like slight subtlety is that this G is like sitting in brackets because it's modular G of O,
18:41
but of course G of O acts on N of O and to preserve N of O. So this is invariant under the right multiple. Like if we change G to G H or H lies in G of O, you won't change G N of O. Okay, great. So now we come to the theorem of V of N.
19:12
So we form the following algebra, A G of N. So it's just by definition,
19:22
well, the G of O equivariant homology of R G N. So actually this definition requires some care because this space is very infinite dimensional and the group is also infinite dimensional. So it takes a little bit of care to define this space.
19:40
And essentially in their work, they work with cycles, which are finite dimensional along the affine gross mining and along the base.
20:02
So R G N is like a not quite a vector bundle over something a little like a vector bundle. They work with things which are fine dimensional along the affine gross mining and then finite co-dimensional in the fibers. So we'll soon see what example of what I mean by this.
20:29
So with some effort, they define this equivariant homology of R G N. And the theorem is that this carries a structure of a commutative algebra.
20:53
So where does the algebra structure come from? So sort of morally or intuitively,
21:04
here's one way to think of it. I mentioned the space T G N a few minutes ago. It's bigger than R G N, but actually if you, so T G N maps to N of K
21:23
because in T G N we don't require that S be in N of O. So it just comes to map to N of K. In this map from T G N to N of K, it's an analog of the map from the cotangent bundle of the flag variety to the nilpotent column.
21:42
So it's natural to form the analog of a Steinberg variety. I'll show you what's the analog of first, this Steinberg variety. So we would take T G N cross with itself over N of K.
22:06
We call that thing ZGN. So this is not, the reason why this is actually not what's formally done is because this leads to more infinite dimensional issues. So, but if you imagine doing this,
22:22
then you'll see that the stack portion of Z G N by the action of G of K is actually the quotient of R G N by G of O. So that's the thing we're interested in. So, and this guy, you could study his homology
22:42
or the G of K echovarian 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 convolutional algebras, you could apply it to Z G N. If you were not like worried about infinite dimension,
23:03
not that infinite dimensionalities. So if you avoid Z G N, which is what BFN do and work instead with R G N, then you do basically the same thing, but things become a little like more complicated or less symmetric looking because of this,
23:22
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 R G N. So we consider triples G one.
23:41
So that's a point in G of K. G two, that's a point in the alpha ingress monion is by putting brackets. And S, and S lies in N of O. And also the S lies in G two N of O and also G one S lies in N of O.
24:02
And this space maps to three copies of R G N. 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 like unnatural
24:21
about the fact that we have like G one outside of brackets and G two inside of brackets. And there's unnaturalness because of what I just mentioned about the fact that we're going to R Z. Just for completeness, I don't know if it and it cares that much, but I'll write down the three maps that go with this.
24:41
If you open up the FN, you'll see that my notation is a little different than theirs, but it's equivalent. So those are the three maps.
25:01
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 GL one acting on C to CN,
25:23
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 reversing error. Let me just take it this way. It doesn't really matter.
25:42
And then what's our affine was money. And well, first of all, that fingers mining of this group C star is just Z, Z because we're just taking K star mod O star and the invertible series mod invertible power series.
26:02
So that's just given by the order of the last series. So we can write it just like T to the P. And then what is our R space looked like? It looks like the district union over P
26:20
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 going to consider? I'm going to take one of these chunks.
26:42
So one of these pieces of this district union, for example, I take one, which is the fun. So I take the fundamental class of where I put T one and it's my fingers mining element. And I put O N as my main element. I allow all possible choice of S.
27:01
So it's an infinite dimensional cycle, but it's finite dimensional and after I guess my direction, infinite dimensional in the fiber direction. And I call that the U and my other element will be the characters, will be the cycle of T inverse cross T O N.
27:26
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 HEN.
27:44
Not very consistent with the brackets. And W, my third element will be the generator of the C star covariant homology of a point. I didn't say this, but the unit in the ring
28:03
is just the class of the identity element in the affine grass mining and cross the fiber. So maybe I should write it somewhere, but because the reason why I'm mentioning it right now is because it's really W acting on the identity. Okay, I won't write it.
28:21
Understood? And then to compute U, V, the product of these two cycles, so 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.
28:46
And you end up with a class of P zero. You end up with P zero because you multiply this T one and T inverse as your affine grass mining.
29:03
And then you're doing some kind of intersection of these things. So you end up with T O. It's like, it looks very straightforward. It looks like it really just intersected, but well, anyway, it's basically. So you end up with that guy and that guy sits,
29:21
it's like in the, so again, that identity element in the ring is one. The identity element in the ring one is just a class of T zero cross O N. So this is almost like identity element,
29:41
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, but I'm hearing a little echo.
30:01
Then, so in fact, this guy is actually equal to W N because that's the Euler class, equivariant Euler class of O N over T O. So that's the origin of this thing. I mean, so therefore we conclude that U V
30:23
is equal to W N in this one. And this is pretty much the only computation one can do. I mean, there's a variance in this computation, but this is basically the only computation you can ever do inside this algebra.
30:42
And I'm just joking, but then maybe I should say it's 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 U V W modulo, U V minus W N, and therefore our Coulomb branch
31:02
is C two mod two mod N. So, let me say that in a more positive light, this computation can be generalized to the case of arbitrary G being a torus and N being any representation.
31:20
Generalized in a straightforward way to show that you've got a hypertoric variety. And so maybe I should just write that. So I guess we can say also due to B of N. So if T is a torus,
31:44
then during the same computation, we see that M C G N is an affine. It's not just N, but the affine hypertoric variety,
32:04
symplectic dual or GAL dual. So Michael explained about that in an exercise session yesterday to the Higgs branch.
32:21
I mean, if G is a torus. Okay, 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.
32:41
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 front of make use of this computation as well. I'm not gonna write that down here.
33:00
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 move back up to the definition. Definition was here.
33:21
Well, that was the theorem, but it's commutative algebra structure and the definition of this R space was sitting here. So it's like, I don't know, sort of complicated, but maybe once you work with it enough, it becomes like something person can handle and get their mind around.
33:43
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
34:01
to be the symplectic dual of the explanation. Okay, so the first one, maybe isn't quite of that form. Anyway, so we have a map. We have a map from, of course, from RGN to the F address mining of G. And this can be used to define a map, not in a completely straightforward way,
34:21
but you can use it to define a map from the algebra AGN to the algebra AG0. So AG0 means we have no representation. So we're simply stating the echo-rank homology of the F address mining. And therefore a map backwards from this
34:41
Coulomb branch associated to G0 to the Coulomb branch associated to G and N. So having written this, did I write this the wrong way?
35:10
Sorry, I think I wrote it backwards. I mean, of course the map goes this way, but I think the map on algebras goes the opposite direction. Yeah, sorry, I think everything goes backwards here.
35:39
Well, reverse arrow here.
35:43
So 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.
36:03
And then we sort of have this Gizin 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 when I'm about to say it. But I think the map goes in the opposite direction.
36:22
Anyway, so this guy has been extensively studied before the echo-rank homology of the F address mining. And it was proven by Benjamin Kavnikov, Finkelberg, and Mirkovich. So the FM, and the B is a different B.
36:43
That this is isomorphic to the universal centralizer space in the Langlands dual group. So the following space. So pair is consisting of G, X. So here, these lie in the Langlands dual group across the Langlands dual algebra, regular locus,
37:06
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 this is called universal centralizer space.
37:22
And this maps to the just remembering G check, which is the same as the T dual W. And the generic fibers of this map are tori.
37:46
This map is a tori. So this is a very much in the framework of the geometric satake correspondence. And we see this actually,
38:01
so to see how this is working. So this T star W is the same as the G of O echo-rank homology of a point. Which is the same as the G of O echo-rank homology of a point. And this G of O-rank homology of a point embeds
38:21
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 on the F of this one. So I expect it's, and so I get a map like this. And so we, this thing, actually we have a diagram like so, and also the same maps here.
38:41
And okay, this M C G N is not the same as M C G zero, but in this way, it behaves quite similarly. So the generic fibers of this map from M C G N to T star W is also T check.
39:06
So in particular, the dimension of M C G N is twice the dimension of, this T as usual is the maximal torus. T check is a maximal torus of G check. So it was equal to twice the dimension of T check
39:21
or also known as the rank of G. Okay, so this is the first property of this M C G N that it's pretty close to this universal centralizer space, and in particular, it has this dimension.
39:48
This map here is a, yeah, I didn't say this, but this map here is a birational map. Okay, so that's like the first basic property.
40:03
At least we have its dimension. So the second thing, we can define a non-commutative version of the BFN algebra, non-commutative Coulomb branch algebra, something it's called. So we simply tweak our construction. We take a covariant homology, not just respect to G of O,
40:24
but respect to G of O semi-direct product C star. So we call this thing A H bar G N where H bar is the C star covariant generator, C star covariant homology. And how does this work? Well, this is C star acts on RGN
40:43
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 variable T.
41:03
Or if you like, it comes from an action of C star on the formal disk. So this is well studied action in the affine gross money and gives an action here.
41:21
We take a tech variant 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 UV thing would be the class of this T O N. And we'd end up with sort of,
41:41
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
42:03
mod O N. And then that would be, that has action of this bigger, the C star acts non trivially there. So we'd get something like W plus H bar today. So that's where the non-commutative function. Yeah, a question.
42:22
But Andrei, you had a question. So we're going to remark about the ring for more general G. Do you get something like an variance under the value group inside the ring for T? So the equivalent branch would be something like a quotient of a hypertrich variety.
42:45
Yes, but it's a bit more complicated. So you, there is for every, so the person is asking about the map from A.
43:07
So we can embed A G N inside of A T N. Hopefully I get this right on, yes.
43:20
And therefore spec, and therefore embed, and therefore to find a map from M G N to M T N. And there is a relation between these spaces and there's a map, but it's not just like invariance
43:41
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. I didn't prepare this part, but yes, this is one important tool for studying these spaces to this abelianization.
44:04
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 is just acting on by scaling on O, but then if we add this loop rotation C star,
44:21
that acts differently because they're sitting in different positions because of this T business. And so this H bar G N is a non-commutative deformation of A G N and therefore, so this is like the verse, usually we start with an algebra with a Poisson structure
44:42
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 carries a Poisson structure.
45:02
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 UV and VU, which if I did it right in my head, would just be N W to the N minus one, I think.
45:27
Okay, just expand these out. No, yeah. Okay, so A G N carries a Poisson structure.
45:42
So it's kind of interesting that from the very beginning of the whole lecture series I was talking about, like the resolutions, Poisson priorities, and then I quantization, but here we get the quantization and the Poisson structure after. Okay, so that's the second like property,
46:01
the non-commutative, non-commutative deformation slash the Poisson structure. Now it's probably a good time to, somebody asked me, how can we make a symplectic resolution? And we can, and we'll get there, but somehow it's not this next property because the next part, we have the one step at a time.
46:25
So let's consider now a Taurus action on, let's see, on its little branch. Where is this Taurus action gonna come from? When it come as follows. So the connected components of the affine gross mining
46:42
of G, because it's sort of like loops into G, its connected components are 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 really sneeze up some thinking diagram, for example, then the fundamental group will be Z to the I,
47:02
the connection of the components in the affine gross mining will be Z to the I. And let's just assume that this fundamental group is free to be the episodes, and let's let T check be the Taurus, whose weight lattice is pi one of G.
47:27
And so then because the space, so this pi zero of RGN, so that's the space whose equivariant homology we're taking is actually, well, it's the same as pi zero,
47:40
the affine gross mining, 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 AGN, this homology algebra, this is a pi one of G graded algebra.
48:03
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 AGN, since almost by fear because it's weight lattice is pi one of G, so it acts there.
48:21
And therefore the Taurus acts on the quilin bridge. So this funny Taurus made using the pi one of G. So let's think about this from the viewpoint of semantic duality. Oscar asked a good question. What the algebra also has a grading
48:41
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,
49:01
but I'll mention it before the end of the session. So let's see why this is the right Taurus from symplectic duality. So recall from symplectic duality, the Taurus which is supposed to act on the one side is coming from the H two or coming from the Picard of the other side.
49:21
And notice that this, we have this current map. So if we take a character of G, then it will give us this current map or topological line bundle on the Hamiltonian reduction. So in the case of quiver varieties,
49:44
this is just the determine line bundle of the topological vector from this quiver variety. So this is on the pig's branch side, we have such a map and up to like finite possible finite little groups.
50:00
This hom is precisely the co-character lattice, co-8 lattice of our Taurus T. So we actually do get the Lie algebra of this T
50:21
is this, well, at least it comes equipped with a map, which is often an isomorphism to H two. So that's as, so this Taurus is the right one from the viewpoint of symplectic value. And Oscar asked a good question.
50:42
So Oscar asked, what about the grading on the homology by homological grading? So also carries a homological grading
51:02
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 in the very beginning, I talked about conical synthetic resolutions. So I expect, we expect an additional C star action,
51:24
this conical action. So this is the conical C star. To be a bit more precise, this action is not always conical.
51:45
Like maybe it's actually fairly conical, but in fact, in physics, they have a funny, they have a funny name for when this is conical, which is they call it, they distinguish between, they call it bad and ugly theories.
52:01
And conical, this conical condition corresponds to either good or ugly. So to be even more precise or slightly more precise, you can tweak this. I mean, you have a two, two torus, you have the C star and you now have a T check acting on your Coulomb branch.
52:22
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, as you should say.
52:44
And you can actually try to tweak them and find a sort of C star, which is 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 radians
53:00
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 combinatorics 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.
53:21
Okay, so this was the discussion about the second property. And now, or I'm sorry, I guess the third property. So now let's discuss a fourth property, which is a deformation or resolution.
53:42
So this comes from a different source, which is called the flavor torus. So let's extend our G to a G tilde. Assume that we can embed G here with G tilde, which also acts on N, and such that G tilde mod G is called F is a torus.
54:06
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.
54:22
So for example, in the basic example, like this, we would take G tilde to be GLM cross C star to the N in this part to be the F. This was G, okay?
54:40
In this case, it's split doesn't have to be split. And then the natural thing to study would be, I'll call it the tilde GN, which is the G of O equivariant. So G tilde of O will now actually act on our,
55:03
this bigger group G tilde of O will act on our space, our GN, then we can study equivariant homology with respect to the bigger group. And this will give us our commutative deformation of our equivalent branch, algebra and therefore deformation of our equivalent branch.
55:22
Then we would take spec, this bigger guy who would map the spec of the equivariant homology of respect to this torus to identify with the algebra, and we'd get a deformation of this.
55:46
And this is like the scripty X that I mentioned at the beginning of the lectures. So remember this scripty X was this deformation of the symplectic conical, symplectic singularity,
56:02
and it would map to H two of the Y. Well, if you were a little bit more precise, you would map to H two of this Y mod W and 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.
56:29
And therefore we see a deformation over the Lie algebra of the, like this H two will be this F. So this is like what's expected by symplectic duality.
56:41
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.
57:03
So we just did the same thing. And this gives us our universal deformation, non-commutative definition, universal quantization. I'll come back to this next lecture in more detail, these quantizations.
57:22
And then the third thing is about this resolution. So let's choose Lambda, the Co8 of F. And why do we need this choice? Well, I explained a couple of times that in symplectic duality resolutions on one side
57:43
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 this attracting set, that's this C star right here. So that choice is gonna be, actually, I guess I called it row last time.
58:00
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?
58:21
Well, we consider this R space, but now for G tilde. So this instead of G, now we change to G tilde and this will map to the affine gross mining of F. The affine gross mining of F is just the Co8 lattice of F.
58:46
And so we can consider the pre-image of this over multiples of Lambda.
59:01
Then we formed the, we formed the union over such multiples by natural numbers of these pieces. So in particular, R zero G tilde N is simply the original RGN.
59:24
Then we define, my notation getting a little hairy here, but maybe I won't introduce the notation for this thing. Then we consider this graded ring, which will be the direct sum over N of the G of O equivariant homology
59:44
of the spaces R and G tilde N. So this is a graded ring whose zeroth degree part is the original BFN algebra that I worked with before.
01:00:00
and the 0th degree part is the algebra I started with. 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 it's project.
01:00:27
And this guy will be the candidate to be a resolution. So it'll map to my spec of this AGN, which is my equivalent branch. So this is the candidate.
01:00:47
I mean, I should say that of course, I don't know, maybe it's not of course, but not every cool resolution. We'll see some examples maybe next time, but this is the candidate to make a resolution.
01:01:01
And it, again, depends on choice of some data, namely basically this choice of this C star, which is acting on the Higgs branch, which is the right data we want. So choosing this C star is what allows us to even define this graded ring in the Higgs branches.
01:01:24
And again, I should say there's no general result about when this thing will actually be smooth, right? Nor is there any, yeah. Okay, so I'll stop there. And next time we'll focus on the specific case of these generalized F-ngrass mining slices
01:01:41
and their any questions. Any question, or not? Yeah, you have something to. How does this space depend on N
01:02:01
with respect to sums and tensor products? I have no idea about tensor products. Tensor products of N's is somehow a slightly unnatural thing to do in this context. Let me make two remarks. The first remark is that if you take
01:02:21
an external direct sum, like take two groups, so this is maybe not what you're asking about exactly, but it's maybe, has a nicer answer. So I'll answer it first. I have two groups on two different representations. Oh, wait a second, yeah, that's right.
01:02:48
If I have two groups acting on two different representations like this, then the Coulomb branch will be the product.
01:03:03
If I have a single group, I've got myself in a muddle. Do I mean tensor products here?
01:03:21
Okay, I didn't see anyone. If I have a single group acting on a direct sum, there's actually a map, it'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.
01:03:43
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 for the, if you just took N2.
01:04:02
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
01:04:21
only depends on, for N and 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
01:04:40
and use the fact, use the delineation to do it for general G. Any more question? Yeah. Is there any geometric property that this bi-rational map satisfies beyond bi-rationality?
01:05:08
Like it's not proper, it's not a blow up of something or an app end blow up or? No, I think it's usually like, well, let me just do an example.
01:05:22
This, like you could take again, this one. I mean, here's a natural example. I'll just take this and some bigger one and plus K or something, right? These two, so here you get C star mod and then the good thing is we'll verify
01:05:42
if I wrote the map in the right direction. And then the map is like this U of V W goes to U W to the K V W.
01:06:03
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.
01:06:21
Ah, the map goes the other way. Sorry. This one better be more quotient to do this. Does that look right now?
01:06:40
Yeah, that looks right. Wait, what, I'm sorry. Here.
01:07:01
Yeah, sorry, this one goes here, okay. Okay, ah, there we go, okay, now it's okay. Okay, anyway, so to answer your question, I don't know, this map is not a blow up
01:07:24
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.
01:07:41
Yeah, thank you. We have a question with respect to G. So 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
01:08:01
on the, will definitely give you a map on that from Gershmanians.
01:08:21
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 Quillen 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.
01:08:42
So I won't write anything down. Any further question?