So let me just remind where we left off the last time. So we were considering symplectic dual pairs. So we had two symplectic resolutions. And we saw that symplectic duality was some list of relationships between these two pairs of symplectic resolutions. And the one that I sort of discussed the most was this relationship about the homology. So let me just remind you about that. So we had torus actions on these two symplectic resolutions. And we considered the attracting sets for these torus actions. And then we said that the homology. And in fact here, let me just say that I said the right thing last time. So this, so recall here that H of some variety of Z means the top or more homology. And the last time I said, and it was correct that if we take the attracting set of in Y then it's top real homology actually coincides with the total Braille more homology. So just for this Y plus we can either think top Braille more homology or total Braille more homology. And this top Braille more homology, this more total homology decomposes by the decomposition theorem. And after using the decomposition theorem and this hyperbolic stock function, we decompose it into the strata downstairs, the symplectic leaves downstairs. We see the top homology of the attracting set in each strata, closer in each strata, tensor the top homology of the fiber over a point in the strata. So recall here, F alpha is pi inverse of some point, little x alpha, little x alpha lies in some stratum, capital X alpha, this math is called pi. And under some lectic duality, we have a bijection between strata in the dual varieties, order reversing bijection. And also what's reversed is the roles of the fibers and the attracting sets. So we have some shrieks here. Shriek is just my notation for symplectic dual. And then we have a quality is going like this. And these equalities are sort of not just the qualities of vector spaces, but they're actually given by bijections between irreducible components of those two. There's a variety. Okay, I guess I can move this guy back a little. Okay, so that's just a quick recollection on that. By the way, I just wanted to point out, so I listed a number of things that are structures common or not common, but structures matching on both sides of the symplectic duality. And I wanna point out that there's some more structures that I didn't talk about. Let me just point out two more structures of continued numbering. So this will be four. This four I may come to later if I have time. So in the end of the lectures, that's called something called the hechidic conjecture. So maybe if we have time, we'll come to that. And number five, I'd like to point out is a matching of stable envelopes, particularly elliptic stable envelopes. And this will actually be the topic of Richard Romani's lecture next week. So you should tune in for that. I think it's toward the end of next week. And I should say that we have these different structures that match and you might ask, what's the relationship between the matching of structures as certain ones imply other ones? And usually you could say that there's, mostly I would say, no, we don't know that matching of certain structures implies other structures, but usually there's a kind of compatibilities between them, but not a perfect implication. I mean, I think the situation is very analogous, very similar to in the usual mirror symmetry, 2D mirror symmetry, where as you know, you have different structures like matching of Hodge diamonds or what are mirrors of Hodge diamonds or homological mirror symmetry. Anyway, many other things that I'm not an expert on, and I believe that it's not that one is like, implies all the rest of them, but that there is interrelations between these different matching. So it's a similar story here. And I would expect that there's probably more things that people just haven't thought about, or maybe they have thought about them and I just don't know about it. Okay, so today, what I'd like to do is talk about a particular example of some particular duality, some particular dual pairs, which turns out to in some sense be, I don't know, the main example, at least for me it's the main example. And it concerns core varieties and affine gross mining slices. And after doing that then, if we have time today, if not tomorrow, we'll get into this Barman-Finkelberg and Kagema construction. Okay, so to set the stage, I'm gonna start by reminding a little bit more about core varieties. So let me fix the algebra. So G is semi-simple, the algebra, and I'm gonna require it to be simply laced. So in other words, an ADE type. And let me associate some data to this semi-simple, the algebra, which would take two dominant weights. And we'll write, all these dominant weights will extract two vectors of numbers. So we write the first dominant weight as a linear combination of the fundamental weights. And we write the difference between these two dominant weights as a linear combination of the simple roots. And if it's, I would like to demand, so like my lambda and mu are not just arbitrarily chosen, but they're chosen so that these VI's are integers. So WI's and VI's, not just integers, but natural numbers. And sorry, here I, in both these sums, I ranges over capital I, which will be the vertices of the Dink and Diagram. The fact that VI's are natural numbers, that's saying, I mean, that's equivalent to the fact that lambda is greater than equal to mu in the usual partial order on the dominant weights. Then one more piece of data I'm gonna fix is I'm gonna write lambda as a sum of fundamental weights. So each of these are fundamental. So also those omegas that appear there are fundamental. So there's gonna be W1, omega one, W1, omega ones among this list, W2, omega twos among this list, but I'm fixing and ordering on them. So there's a tiny bit more. Choice fixed here. And the last thing I'm gonna require is that all these lambdas actually are not just fundamental, but so soon that they're actually minuscule. So in type A, this is not a requirement at all, but in types D, there's only three, there's three minuscule fundamental weights. So it's a very strong requirement. And in type E8, there's no minuscule fundamental weights. So it means I can't deal with E8. Now, this assumption, well, there's a few, this assumption is not strictly necessary, but we'll see later why I did it on both, both on the covariate side and on the other side. And if you're familiar with covariates, you might ask, why did I bother saying that mu was dominant? Well, we'll also see the answer to that question. So associated to this data, we're gonna consider representations. So we have these fundamental representations and then I'm gonna tensor them together. So this tensor product, the fundamental representations of our Lie algebra. And I'll write this as V lambda underline. So lambda underline denotes this list, V lambda underline denotes this tensor product. And then I'm really gonna be interested in setting the weight space in this tensor product, okay? What about this weight space? Well, we can take this tensor product and we can decompose it into ERAPs in the way we decompose any representation. So that these multiplicity spaces which give these columns and answer the ERAP over new dominant weights. And then in particular, we can look at the new weight space in there. So it's this, this is the kind of decomposition of a weight space of a tensor product into multiplicity spaces tensor weight spaces of ERAPs. So let's go to quiver varieties. So what do we do? We choose an orientation which won't really matter later in the day, but for now we'll choose an orientation of the Dinkin diagram. That oriented Dinkin diagram is gonna be called the quiver. So it's just a directed graph. And then we have dimension vectors coming from our V's and W's. Let's draw an example. So this is maybe for SL4. So the Dinkin diagram is a type A3. So I have three vertices, my arrows go to the right, and then I have three framings. And last time somebody pointed out I wasn't very consistent in which direction the arrows go to the framings that, anyway, maybe this time I'll be more consistent send them to the framings. So W determine the framings, V's determine the gauge vertices. So last time I explained that from this choice, we got a group G, which is the product of the GLVIs. And we got a representation of G, so which I'll call N, which is the direct sum of of CVI, CVJ, where I, J is an edge in the quiver and harms of CVICW. So then we form the neck and gym quiver variety and I'll use this new notation. So I'll call it M lambda mu definition.  So the first parameter refers to the level of the moment map we take and the second parameter refers to the GIT parameter for GIT quotient. And so Kai here is a character of our group G and we just take it to be the product of the determinants. So that's the definition of the neck and gym quiver variety. Well, that's the definition of the smooth one and this is the definition of the affine one. So just the same thing, but at the zero level. And now we can state theorems of Nakajima. So one is that M lambda mu is smooth and I will separate them up. M lambda mu is a resolution of M zero and M lambda mu and T, thanks. Let's find any many fixed points. Oops, I didn't describe T. Back up one second. So I have a Taurus acting on my quiver variety. Where does it come from? Well, I have these framing vertices in the squares. By the way, Alan Knitzen told me why they're called framing because it looks like a frame, or maybe why they're called, why they're going in squares, right? So we have those squares, those frames. And so the, of course, the general linear group of those vertices, of the vector space of those vertices acts. In particular, we consider the Taurus to be the product of the diagonals inside those vertices. That's where it seems. So that's our choice. And access finding the many fixed points. So by the way, right here, we use that mu is dominant. And right here, we use that lambda I are minus. So those assumptions are used in this term. Two. The syntactic leaves this affine quiver variety, affine quiver variety, the singular guy, are just given by regular low side and smaller sub-singular guys. Joel? Yeah? We have a question. Oh, great. Is it possible to realize tensor products of Verma module similarly? Yes, similarly, not exactly in the same way, but similarly. It's sort of, in some sense, this will actually come up. It's a good question, but it sort of will come up a little later when we talk about this affine-gross mining slices and cooling branches. So there's kind of maybe two approaches you could use. One is you could take these framings to go to infinity. Of course, then the representation gets bigger and bigger and it becomes more like a Verma. Another approach is to get rid of the framings altogether, but then do a slightly different, realize the representation in a little bit different way. So if you get rid of the framings altogether and do something slightly different, then people sometimes call that the loostic, no-ponent variety, and that can be sort of used to realize a Verma module. But it doesn't quite fit in the same framework that I'm talking about, so I won't talk about it. Okay, the symplectic leaves of these singular guys are given by the smooth loci or regular loci. I'm just finding a good representation in the other such proper varieties. That's a new, by the way. And here are new ranges of over-dominant weights that are trapped between lambda and part three. So there's an action of our Liegeber G on our friend from before, so that we take this smooth square variety, take its attracting set, take its more homologous and action of G on that thing, such that. So here we have our fundamental decomposition that I talked about last today in the beginning today. So we have the attracting sets of strata downstairs, tensor, homology of fibers. So here F new denotes the fiber over a point in this new strata inside the poor variety and lambda mu. And this matches the tensor product, so the decomposition that I mentioned just a few minutes ago. So this is isomorphic, this homology of the attracting set in the total space is isomorphic to the mu weight space in the tensor product, and this matches this decomposition. So let's be, make sure I careful about which way around it matches it, yeah, this way around. So this attracting, homology of attracting sets becomes home spaces into tensor products, tensor product multiple space spaces. Quality here, quality here, quality here, so on. We have another question in the Q and A. Go ahead. Oh, sorry, I mean, what's the map? Well, there's always a map from a projective GIT quotient to the corresponding affine GIT quotient. This is prod of some ring and this is spec of some ring and I have to think for a second why that it gives them that, but if you think of, I mean, it's, oh, this is I think the degree zero part of the ring that this is prod job. So that's why there's no, yeah, yeah. So there's like spec of some A zero in the prod, graded ring. So there's always a map like this. I didn't define the projective GIT quotient, but I mean, I can if you like, but it's too, take us a lot of time. Okay, let's look at an example. So this will be my running example in this section. It's a very simple example. G to be SL2 lambda to be N copies of the first fundamental weight, I mean, there's only one fundamental weight in SL2 and mu to be lambda minus alpha, the simple root. So the representation in question is, well, we have the fundamental representation is C2. So we're interested in C2 tensor N and then looking at the N minus two weight space of C2 tensor N, C2 tensor N, well decompose into many representations, but only, only two of those representations have an N minus two weight space. So this is, there's two factors in this direct something composition. We can, we have the N dimensional representation and it's N minus two weight space and it occurs once. I mean, let me consider with my home notation, the N into C2 tensor N, answer the N minus two weight space of the N plus of the N minus two into C2 tensor N, answer its weight space, answer its weight space N minus two weight space. And so everything in sight is one dimensional except for this factor here, which is N minus one dimension. So what's our quiver in this example? Well, we just have a one here because one, the coefficient of alpha is one. So W is equal to N and V is equal to one N. So we just have the cotangent bundle of P one as our quiver variety. Again, it's like P N minus one is our quiver variety, again resolving N by N matrices of rank less than equal to one N squared zero. And we saw this before, there's two strata. And we saw before that over, that this N minus one, so that the decomposition of the positive locus here looks like, well, the only really interesting part is here where you have the, where you have this, all these orbital varieties. So you have this topomology of this space of upper triangular matrices, we call this the X I suppose. So here X plus is gonna be those upper triangular square zero rank less than equal to one matrices. And that's responsible for this N minus one. Here we have fiber one, I guess I called it last name. And here I have X zero, which is plus, which is the point. And here I have fiber zero, which is just P N minus one. There's another point. And here I have N minus one components. So there we see the same decomposition. But by the way, I don't know if I really said this before, but any co-attention bundle of a partial flag writing type A can be realized as an academic variety, as well as any resolution of a sort of a slice in type A, as well as any resolution of an intersection of a sort of a slice with a no-phone orbit closure. So in type A, lots of, I mean, anything you can think of can be realized as an academic variety. Well, not really anything you really can think of because in a few minutes we'll think of some other things that you can't. Okay, great. So that's the core varieties. Now I switched to these ethnograph mining slices. So let's take G to be the Langland's dual group to G, which looks a little weird, but it's not such a problem because actually the G is isomorphic to its Langland's duality algebra. Since it's an AD type, the Langland duality you don't really notice, we can kind of ignore it. But I mentioned this for thinking about more general situations. And then we're gonna be interested in the affine-grasse mining and of G. So which means I take G over the Laurent series and I take the quotient over G over power series. Affine-grasse mining will play the essential role in their remaining talks in the series actually in two different roles. So this is gonna be the first way and later we'll see a different appearance of the affine-grasse mining. So because I took Langland's dual, the reason why I'm mentioning that is because this Lambda I'm gonna think of now as a co-weight of G. So a map from C star into my group into the maximal torus of this group G. And therefore we can define a point T Lambda in the affine-grasse mining of G coming from this Lambda. So for example, example would be in say the SLN and this Lambda would be some integers adding up to zero. And T Lambda would be the matrix T to Lambda one. Up to T to Lambda. A lot of as a point on this affine-grasse. And we're gonna be interested in orbits in the affine-grasse mining. So two kinds of orbits. The first they call a grilamda. So then I take the group I co-ordinated by this power series group, it's orbit through T Lambda. And this is something called spherical Schubert variety. It's an analog or spherical Schubert's cell in agency. It's an analog of Schubert's cells in a finite dimensional five variety. So it's affine-grasse mining. One way to think about it is think of as a kind of G mod P and there's the orbits of P on G mod P. And these guys are finite dimensional. That whole affine-grasse mining is infinite dimension but these orbits are finite dimension. In fact, the dimension is given by the pairing of Lambda with two row, but it's not very important for our purposes. Oh, minor the important for our purposes, I suppose. And then a second thing I'm gonna be interested in is called W mu, which are transverse to these first orbits. And so to find their orbit, I take a group transverse to my first group, which I did note like this and take its orbit and then we have to find this group. So I take G now of polynomials in T inverse. I have an evaluation map to G sending T inverse to zero. And I take the kernel of that map and that's G1 T inverse. So you can think of as matrices where the coefficients are polynomials in T inverse and a module of T inverse, it's identity matrix. So I take the orbit of that group through T mu and I got W mu. So these are transverse orbits. In particular, they're infinite dimension. So these are like some kind of Schubert cells and these are like the opposite Schubert cells. And then the main objective study will be denoted W lambda mu, which is the intersection of verland with W mu. And maybe even more importantly, will be W bar lambda mu, which is the intersection of verland of bar with W mu. So this guy, this guy here is an affine variety, finite dimensional, in fact, the dimension of this W lambda mu bar is two row, here it was lambda minus mu. Row is the half the sum of the positive roots or maybe positive cohorts, positive root cells. Okay, so one more construction. So this is gonna be, this W lambda mu bar, this is gonna be our affine plus unvaried. It's also possible. And now we're gonna construct, it seems like a resolution. So to do that, we need one more sort of construction. We form a gr lambda underline. So I'm using this list now. Another, this is just notation. So it's the notation is suggesting that it's a kind of product of these gr lambda eyes, but not exactly a sort of twisted product. Okay, so that's just notation. Here's the definition. So it's a sequence of points in the affine-gross money with the condition. So these G is the node elements of G of moron series and brackets G, you know, it's a corresponding point in the affine-gross money. And the condition is that G nine minus one inverse GI brackets is in gr lambda I for all I. Okay, so I admit this probably like a little confusing if you haven't seen this before. The way that I like to think about this, one way I like to at least like to explain it is that these, we're setting points in the affine-gross money in like this. So either my G one, G two, G three and so on. And they have distances between them described by these lambdas. This is distance lambda one, distance lambda two. So it's a variety of polylines in the affine-gross money. So sequence of points with prescribed distances between them. So this condition can be thought of as distance between GI minus one and GI. And this is a sequence of points minus one and GI.  For example, if we take G to be equal to PGL two, then the affine-gross money of G is an infinite tree or more precisely the vertices of an infinite tree with P one branching. So that means it looks something like this. So have the tree but infinite tree and at every vertex there's many, many edges. In fact, a whole P one of them coming out. So that's, and then it continues. And affine-gross money is just the points in this tree. And this distance is just the distance in the tree measured as like aligned edges. Okay. So that's the definition of this gur lambda underline. And then finally, we define W till the lambda underline mu by definition of one piece of notation, one piece of notation. This space gur lambda underline comes with a map which I'll call M lambda underline to the affine-gross money and it takes this sequence to the last point in the sequence. So just remembering the last point of this line. And then I define W till the underline mu to the M lambda underline inverse of W. So I just want that last point to lie in this sort of slice W. Okay. So here's a theorem about these spaces. I guess this theorem basically do to myself with Ben Webster, Alex Weeks. So the first point is that W till the lambda mu is some section of resolution of W lambda. So in particular, it is a resolution and it has a symbolic structure. And the W lambda mu has a personal structure. And then there's Taurus, oops, okay. So Taurus here will just be the Taurus of the group which is acting just by left multiplication. So Tx on this W till the Mw with finally many fixed points. Oh, and I promise this, to tell you, this guy, this first fact, this uses that lambda I are miniscule. And here we see a manifestation of something I mentioned last time, which is that when this lambda I miniscule on the full variety side meant that was needed to ensure finally many fixed points. And on the dual side, this needed to ensure that we have a resolution. So that's interplay on both sides. And one structure existence of resolution matches the other structure finally many fixed points for the Taurus section. Okay, continuing this vein. So another result is that the syntactic leaves of this W lambda mu bar, Rw new mu bar, or maybe new mu delta bar, I suppose. And that's for news dominant ways trapped between lambda mu. And here we see a promised feature of syntactic duality bijection between the leaves. So both cases of leaves are indexed by those dominant ways trapped between them. And so part four, and this part four basically just uses the, it's not really our theorem, this is just part of the geometric syntactic correspondence of chemical etching belonging, and says there's an isomorphism between, well, let's move on as follows. So between the top homology, also full homology of this W tilde, lambda new R attracting set, and this weight space in the tensor product, compatible. So just an isomorphism with spaces but compatible with these decompositions. So recall from the representation theory we have this decomposition by the isotypic components. So no subject there. And tensor, the weight space is there. And geometrically, we're gonna have here sitting the homology of the fiber, which will be all right, M lambda mu inverse of this point. And sir, the homology of the attracting set in the leaf and we have qualities. This guy, I mean, is sort of a famous guy. This guy is called the, or the earliest components of this guy are called the American Ritual and Encyclicals. Joella. Yeah. A question? Yes. Is this situation the simple dual of the quiver I to picture you presented the first? Yeah, yeah, that's a whole point. Yeah, that's exactly what I'm, about to say. But before I say that, I'll just do a quick example. If we take lambda mu as above, so lambda is n times the first fundamental weight and mu is n times that fundamental weight minus alpha. Then just w lambda mu bar is just this C2 mod C mod n. And it's in this guy's, it's resolution. And okay, I already discussed previously how this decomposition of homologies works in this case. I won't bother saying it again, but in fact, well, you might be, you know, you might say, how do you see this so easily? Well, you can actually write down explicitly in matrices, this isomorphism. Another fact that's true in general fact is that, well, the dimension formula that I gave predicts so we have lambda minus mu, which is rho. So we just get two, the dimension of this covariate will just be given like this, which is just equal to two. So we definitely get something, not covariate, afringer's mining slices. We definitely get something two-dimensional here. And in fact, you always get sort of C2 mod C mod n. As for all two-dimensional afringer's mining slices are of the form C2 mod C mod n. Okay, so now I come to what Francesco said. So this claim is that these guys are subtracting dual. That's subtracting dual. And well, what does it mean to say this subtracting dual? I'm just saying I had some lists of things which are generally supposed to match. And I should check them off one at a time. Well, I did explain quite a few of them, but let me explain actually go back to almost the first one I said, which was a matching of the algebra of the tourist with the H2. So what tourist do we have? So recall that I have this C star to the sum of the WIs acting on the covariate. Try to make mechagemus covariate. Nothing, there we go. Acting on the covariate. If you're careful, you'll notice that this actually maybe is not effective, like some portion of this C star to the sum of WIs actually acts trivially, but ignore that for a second because actually it will match on the other side. And so the Lie Algebra, so that was our tourist here. So it's Lie Algebra, well, C to the sum of the WIs. Now, what that's supposed to match H2 on the other side. Well, this guy, this WM that told us mu is by definition a sub-variety of the affine-grasminium to the N, just because it's by definition, N points in affine-grasminium. So I end up with a map backwards from H2 of the affine-grasminium to the N, which of course just H2 of the affine-grasminium directs on itself N times. I mean, if I took the cohomology, I would get the cohomology, full cohomology, but so I get H2 directs on itself N times because I have the H0s tensor many, many H0s and then one H2. So I get this H2 directs on itself N times. So I map backwards like this. And this sum of the W is actually equal to N, almost beneficial. This N is the same as the sum of the Ws. Because remember, what was N? It was used to make this list. Well, first of all, this, let me just write this down and you'll see it's obvious. Okay, so the N here appears because it's the list of these fundamental weights adding to lambda and the W is how much of each fundamental weight occurs. So of course, N is the sum of the Ws. So in this way, we see that this torus is Lie algebra is C to the N and also this H2 here is also C to the N. I hesitate to write their isomorphic because they're not isomorphic, but somehow they actually have exactly the same sort of kernel, so they match. So, oh, I didn't really say this, but the H2 of the affine-grasminium is one dimension. And it's Picard group is just Z, it has a canonical line bubble. Okay, so that's the matching of the, remember to try it as an equation. So we get this matching of this, matching of this Lie algebra of the torus with this H2. And let's look in the opposite direction because it's also quite instructive. So in the opposite direction, let's start, I guess, with affine-grasminium, let's start with the quiver variety again. So how do we get line bundles on the quiver variety or how do we get comology of the quiver variety? There's this Kerwin map. So because it's a quotient, we get these tautological line bundles coming along. So we end up seeing that H2 is just C to the I. So I here is the vertices in the thinking diagram. So these, the line bundles correspond to determinants of the tautological vector bundles. Anybody who's worked with quiver varieties, this will be familiar. And if you haven't, this is part of the more general phenomenon called this Kerwin map. So that's the H2 of the quiver variety. And on the other hand, if we take our affine-grasminium slice, I have, this is a, this is living in the affine-grasminium of this group G. And so the torus of the group G, as I mentioned before, will act here on the resolution. And this torus, this group G has dicken diagram with vertices I. So this torus is just C star to the I. So we get H2 of this quiver variety, nice and warm to the C to the I, which is the the algebra of this C star. And so that's the matching there. And what's, you know, beautiful about this is if you could look at some examples, you might say, okay, well, sometimes maybe this tautological line bundle is trivial because of, you know, the nature of the quiver variety. Maybe it's not really, doesn't use that vertex. And so the tautological line bundle is trivial.  So everything always matches. Okay, so that's the matching of the tori. And then of course the co-logical matching is what I mentioned. So we saw, sorry to scroll a lot, but we saw these two isomorphisms. So this Mirkovic-Velonen isomorphism here, we're seeing this representation theoretically composition shown geometrically here and using this Mirkovic-Velonen result and then going further back up. The same thing, but now with flipped. Flipped the rolls. So I here I have tracking set fiber and down below I had fiber tracking set. So I flipped the rolls and that's exactly what we expect with some like duality. And I already mentioned that we have this bijection with leaves or reversing bijection with leaves. In fact, you can even go a step slightly further and you can produce a bijection bijection between the irreducible components of say one of these fibers. This is in the perverite side of the perverite with the irreducible components of this attracting set in the African-Gracian slice. I mentioned before that these guys are called Mirkovic-Velonen cycles. And one way I know to produce such bijection is using theory of Mirkovic-Velonen polytibes. So I wrote some many papers about this topic. But anyway, I'm not gonna bother explaining that. But there is exact bijections. I see a question. Where do the plus on structures and some like instructions on African-Gracian mining come from? Thanks, that's a great question. Of course, I've been like not really discussing these African-Gracian structures too much in general in this talk, but I apologize. So the answer to the question is we have a man in triple. I'll just write down these. Let's see. African-Gracian slices come from the man in triple. So there's just an aside. And this is explained in our paper, first K-W-W-Y paper. Okay, great. So, okay. So now I have a few minutes.  And then we'll really get into it next time. So what's the idea? Well, we have this notion of simple African-Gracian pairs. We've seen that many examples now, but now you might ask, how is there some systematic way of constructing the dual of if you have one simple African resolution, is there some systematic way of producing the dual simple African resolution? And well, in general still, there's no really good answer for that, but here's a sort of partial answer for that. So the answer, let's look at our list of some of these synthetic resolutions. We saw, we had a hyper-torque varieties. We had river varieties. We had something like T-star G-mini-pese, and we had these African-Gracian slices. So for these classes of examples, there of the following form, you take the contingent bundle of some representation of G and then take a Hamiltonian reduction by G for some group G, right? They're always of this form. And whenever our simple African resolution is of this form, that's when we can use the BFN construction. So it's this class of examples. So to use a slightly physically language, or well, not yet, so let's start with the following data, G a reductive group. So usually this group G will be product of some GLNs or a torus, which is just a product of C-stars, GLNs. So G will be a reductive group and N will be a representation. And for the physicists, they would say that this G and N define, well, one of these theories, a 3D gauge theory, which is one of these N equals four supersymmetric theories. So they define it as 3D gauge theory. And I mentioned before that from any, these N equals four supersymmetric field theories, the physicists associate two spaces, one called Higgs branch and one called the Coulomb branch. And this Higgs branch is just, for mathematicians, it's just this Hamiltonian reduction. So just take the cotangent bundle of N and take the Hamiltonian reduction by G. So that's the Higgs branch. And the Coulomb branch, I guess was more mysterious both of the physicists and mathematicians until, well, work of some physicists and I don't know the physics literature very well, and then the work of Barbara and Finkelberg-Makajima.  We're many five minutes. Okay. So for this rough description, let's introduce the following weird scheme. So sometimes called the ravioli curve, sometimes called a bubble, I'll call it B, raviolo curve, maybe more precisely. So you take two copies of the punctured disc and glue them, sorry, two copies of the formal disc and glue them together along the punctured disc. So it's a non-reduced, it's a non-separated curve. The usual way I like to think about this thing is you probably know that if you wanna construct P1, you should take two copies of A1 and glue them along C star. It's a usual description of how you build P1. And usually when you do that, you don't glue them like directly, but you glue them using the inverse map. So somewhere in this gluing thing, there's a T goes to T inverse. If you forget to do T goes to T inverse, well, you can still glue them, but then you get a non-reduced, a non-separated curve, sorry, something like this, right? You end up with A1 with double origin. So it's a kind of bad version of P1. So this bubble curve is like the bad version of P1 and then you just look in a formal neighborhood of this doubled origin. So that's the bubble curve. So it's pretty close to P1. If you don't like it too much, you can just think about P1. In fact, in Nakajima's first paper about the school loan branches, he explained basically that the natural thing to use would be P1, but for the purpose of, well, for some purpose, we'll see soon. It actually only works if you use this bubble curve, but you can think of it as just P1, okay? So what are we going to do with this funny bubble curve? We're going to consider the following modular stack of maps from this non-separated curve into the stack of N mod G. So N here is a representation of G, we'd like to group G and we consider the stack quotient N by G and consider maps. So I emphasize that you have to use the stack quotient here, whereas in constructing the Higgs branch, we've always used the GIT quotient. And when we, in doing so, we'd like thrown away the unstable locus, but here we need to use the full locus, full everything. So we take this full stack quotient here. We take such space of maps, then we take the homology of this mapping space. And this homology carries an algebra structure, the convolutional algebra structure. So where does this algebra structure come from? We'll see, like, I'm going to redefine this thing and it's like different language next time. So in that, and then I'll explain the algebra structure in a different way, but intuitively, it comes with some kind of gluing of these bubble things. So it comes from considering like a disk with tripled origin. And so considering such a modulus, maps from such a modular space into the second, the modular space and maps from such a tripled disk or disk with tripled origin into the stack, that's used to define this convolutional algebra structure. And that somehow is like related to why we don't use T1 because here we have this possibility of this triple origin. And this algebra structure, so it's similar to the, if you've studied the Steinberg variety inside of the product of the quotient, the square of the quotient of bundle of the five variety, you know that this Steinberg variety, its homology has a convolutional algebra structure, like in the book of Chris Ginsburg. What's a little different is that in this case example, the convolutional algebra structure is commutative. So this is a commutative algebra. And since it's commutative, we can then take sprang and we can get a steam and that's what we'll do. So that's the definition of the Coulomb branch, at least the singular. So this is going to be the dual of this Hamiltonian reduction. So these guys are going to be selecting two and this guy is going to be the Coulomb branch. Too much space, MC. Okay, at least, so this is how to produce the singular guy and then you can produce the smooth guys one. Okay, so next time we'll define all this stuff in here a bit more precisely or this is pretty precise but I'll define it more in a way that makes it easier to work with. So, and we'll see some examples and hopefully. Okay, we'll stop there. Any questions? You have a question and you have two questions now in the Q and A. Okay. Mentioned that stable envelopes play a role in something like this. What is known about stable envelopes for slices and affine gross mine in? So this is subject, I guess, of Yvonne Danylenko's thesis, which hasn't yet been published. So it has been studied by Yvonne Danylenko, but I mean, sorry, not being published. I mean, not even appeared on the archive. So it does sort of exist the study but not in public yet. And another question from anonymous person. Can we define WLMU for non integral London U? No. No. I don't know how to define it if they're not integral. One thing we'll see soon though, is how to define it when MU is not dominant. So in the definition, at some point in the beginning today, I said, we assume MU is dominant. And if you're paying close attention, you would see that actually on the curve variety side, much of what I said goes through, even if MU is not dominant. It's still a smooth, it's a resolution, but not exactly of this singular guy, not of M0 London U, but it's a resolution of something. So there's still a smooth curve variety, even if MU is not dominant. And going on the outfingers money inside, well, actually it looks like much of what I said also goes through if MU is not dominant, but we'll see soon how to really good definition of W London MU when MU is not dominant. But for not integral, MU, I don't know. What is G of the Laurent polynomials? Okay, let's back up then. Andrei answered that question.  Yes. So don't worry. He answered it? Yeah, Andrei. Interesting question about the patient. Oh, great, great. I took the liberty. Any other question? Thank you. So I have a question, maybe. So before you say that queer varieties for finite dinking diagrams is understand where corresponds in particular to slices. So if I have any Nakajima variety, this corresponds to some slice and, I mean, what it would be the sympathy dual. If you have a curve variety outside of finite type, is that the question? Yes. Yeah, so, well, we'll get to that later. Yes. That's that. Then we're in the realm of generalized affine grass money and slices. And those are defined in this coolant branch way. So we'll, we'll get to that later in the talk. But in, maybe I should say though, in affine type A, which might be the case that you're most interested in. In affine type A, then the symphactic dual of Nakajima curve variety is again a Nakajima curve variety. Well, sorry, when at least when new is dominant. The reason, the reason for that is, and it also happens in finite type A. And the reason for that is there's this funny rank level duality, which happens in finite and affine type A, which, which, which ends up manifesting itself as an isomorphism between affine grass money and slices and quiver varieties. So something's called Mirkovic VWarnov isomorphism. Okay, but I'll, maybe I'll make sure to address that question a little more precisely a little later. Any other question? You're Mark, let's see. No. So I guess we can thank Joe again.


Hello and welcome to another session on sequence and series and in this session. We are going to discuss Selection of terms in an AP. What does that mean? There are a few observations that You know few given terms are always in AP. So, you know, what are these terms we are going to discuss? So let's say when can we say that three terms are always in AP or four terms are always in AP or Five terms are always in AP. So is that is that some kind of or is there some kind of? structure of or let's say some kind of terms and In you know inherently they are in AP. So are there such terms? So that's what we are going to discuss today So let's take up Three first first the case of three terms. So let's study this three terms are always in AP three terms are always In AP, okay, let those three terms are a Or instead of A and B, let me use terms like P Q and R Okay, so PQ are are in AP. So do we have some, you know, generalized PQ are so that it will they will always be in AP, let's try and understand So when is P and Q and R in AP clearly when Q minus P the common difference must be equal to R minus R minus Q or Two Q must be equal to P plus R Okay, so do we have some selection of P and R and Q so that they are they are always in AP. So What we can do is let us say this is equal to or you know, so if you select P as let's say a minus D and Q as A and R as A Plus D where A and D are any arbitrary real numbers where where A and D are real numbers Are real numbers. So if you see A minus D A and A plus D will always be in AP, isn't it? Why because their common difference is always D Common difference is equal to D Let's take an example. Let's say A is 2 A is 2 and D is 3 Okay, A is 2 and D is 3. So first term is 2 minus 3 Second term is 2 itself and 2 plus 3. Clearly these three terms are in AP, isn't it? Minus 1, 2 and 5. What is the common difference? Common difference is 3 So we see that if you have A Minus D and A and A plus D three terms are there in this form They will always be in AP. So is it restricted only to two terms? No, let's go for what if there are four terms in AP Four terms always in AP Always in AP what all they can be.  The common difference is 2D right in these in these four terms common difference Common difference is equal to 2D Okay, so they are very much in AP with common difference 2D. So you can treat this as first term and Common difference to be 2D and then you can write any AP in this form.  So hence Let's say there is a problem where you have to consider four terms in AP So you can always consider these like them instead of taking them as PQRS or A, B, C, D or A plus A, A plus D, A plus 2D, A plus 3D instead of that you can take A minus 3D, A minus D, A plus D and A plus 3D Okay, so example, let's say A is equal to 5 and Let's say B or D is equal to 1 then what are the terms 5 minus 3 5 minus 1 5 plus 1 and 5 plus 3. Is it it? If you check this is 2 This is 4. This is 6 and 8. They are in AP, isn't it? So hence Whatever be the values of A and D, they will always be in AP Again, if you have let's say you have to pick up You know 5 terms in AP, 5 terms 5 terms in AP. Okay, if you have to take up 5 terms in AP then clearly you see A minus 2D, A minus D, A, A plus D and A plus 2D with a CD of or with a common difference of D. Correct? Again, you can write 5 terms in AP like that. You can check on it, you know It's a common difference between every two consecutive term is D correct Okay, so What if we have a selection of 6 terms So if there is a selection of 6 terms then again A minus 5D A minus 3D A minus D, A plus D, A plus 2D, sorry 3D and A plus 5D So whenever there are even number of terms so we have You know A minus some odd multiple of D like that, right? So this is this is odd multiple of D, A minus odd multiple of D or A plus odd multiple of D These are 6 terms in AP all the time. So 6 terms in so these are 6 terms in AP always in AP.  There is a reason why we are writing in this manner. Why because let's say when we you know Take up problems where some of the terms of AP is given Then you can eliminate one variable that is D by adding all of them if you see if you add all of them Adding all of them just eliminates D and hence we have You know represented them in this format, right? adding all of them Eliminates D and if you know some of three terms or four terms or five terms or six terms or seven terms or for that matter any end term You can take terms in such a way that you know you eliminate D by adding So if you add all of them D is eliminated you get A directly and hence the reason why we are expressing them in this Fashion it will be clear when you solve problems on this subsequently


What's up everybody Jason here for jazzy's case breaks calm 2021 flawless collegiate football one box break your team number five cells Out with this random number block for the Chargers Dolphins Packers Cardinals colds Ram Seahawks Bills Raiders Lions Titans in Washington Football team and again you get a piece of all of these 12 teams for a fraction of the price and 10 spots in this format random number block from 0 39 everybody gets a random number And of course we go by the left side of the first number the left side of the serial number dash That's your examples right there any one-to-ones go to spot one any redemptions go to spot zero and a non-numbered cars event If there is any we'll go to Somebody in this break will randomize it So here's a dice roll there's a customer names from gale down to mark and zero through nine there And there's the teams just to represent all the teams in the break Dice roll it and we got ourselves a six and a three nine times. Good luck one two three four five six seven eight and Nine. Oh, yeah, Jim, but I arch manning is gonna be the best manning and that's that's it's hard to say For me because I love paint manning. I think he's obviously the best manning, but if arch can run like his grandpa and be as Talented with his arm as like paint manning Watch out But of course his last name is gonna get him a lot of publicity, but I think it's motivation, you know He's still working hard, right? You're not gonna get it easy six and three nine times nine times My thing is that are they gonna treat him like they treat a pay manning if he doesn't go to Ole Miss Like are they gonna like really just want to send death threats as They did for Peyton when he went to Tennessee I don't know if you guys ever seen the mannings like documentary on ESPN. It was really good five six seven eight Nine, but again, there was some sanctions coming his brother not playing there factored into it You know, it's a lot of people just didn't really take all that into account They just thought arch get your boy here, man. Get your boy arch That's okay. They got Eli a few years later All right, so Tommy you have nine niggas with three mark with eight and seven niggas any one of ones would be yours Donald with four Matthew and his redemptions be yours at zero Paul with six Matthew with five and gale with two So there you go guys. That was nine times coming up next is the break itself. Good luck everybody


I'm Terry Neill. I work for the Commonwealth of Learning. We work across 49 developing countries in the Commonwealth and our focus is increasing access to education and training with the use of open and distance learning and technologies. My role there is unresponsible for the technical and vocational skills development work that we do. In terms of why I'm here, at one level because the Commonwealth of Learning was invited to send a representative so I'm doing that because of my skills focus. I'm here to learn and contribute and also we would really like to explore how we might work with the sailor in some way as well. Okay, well I think at one level everything that we do is focused on addressing the skills gap. We have a number of initiatives so I talked about three this morning, Lifelong Learning for Farmers and Girls Inspire and Skills in Demand which are two of which are very focused at grassroots level people and informal skills development. Then the skills in demand is looking at formal qualifications but again targeting marginalised people and I think that's across all of our work. We're really looking at marginalised people and how to give them skills rather than people that would be fine anyway. But we also have an open schooling initiative, a virtual universities initiative, a teacher training initiative and we have a virtual university which is across the small states of the Commonwealth so we have many strands of work but the common theme would be that open and distance learning and technology are solving the challenges facing these developing countries in skills development. Absolutely and they range from Nauru in the Pacific which is one island in 10,000 people a whole country through to India, the most populous country in the world and everything in between and because of those differences they face very different challenges and the potential for open and distance learning and technology to solve those challenges is quite different. Some it's about scale, for others it's about dealing with tiny numbers and particularly in the Pacific and the Caribbean distributed populations with big chunks of water in between so yes, we can't have one model for all of them so we are looking at models and then with those kind of guiding principles we then look at replicating and scaling those once we've proven them in one place so like with the lifelong learning for farmers that was developed in India and then has been spread to Africa and is now being piloted in the Pacific but the specifics for that would be quite different. But I think what we also have is now 30 years of history where we have not just tested and tried to get these things happening but have an approach of evaluation and gathering evidence about the effectiveness of those things so we can share those stories with the rest of the world too. So one of the things that I've been reflecting on over the time so far at the summit is that there's quite an emphasis on higher education and university and degrees and my personal passion around closing the global skills gap is recognising the value of TVET I think it's often seen as the poor relation and if you think about the kind of the triangle of the world's population higher education is this bit and this bit is where the TVET sector sits and so the more people, the deeper down you can go and get into the really large numbers of people and you start to really transform more people's lives and transform economies and particularly many of the countries that we work and that's what can take them from a low performing economy to a middle level and eventually hopefully to a high level economy so it's both the numbers of people that you can change the lives of but also the technical skills that you can contribute to these economies in this country because we're not just trying to give people skills for skills sake we're actually trying to give people skills to improve their lives and to improve their countries


Let's talk about dietary fats or the fats that are in the foods that we eat. This has been such a huge topic of debate over the past 30 years that I'm sure I'm going to offend somebody in the next four or five minutes, but I'm just going to tell you what I know and hopefully give you some good common-sense advice to help you navigate these waters. Okay, first I have to tell you that Dietary fats are really important for the healthy growth and development of our children. They're especially important in brain development and development of the nervous system. They also help to protect and cushion vital internal organs and the kind of fat under the skin called subcutaneous fat is both an important storage form of energy and it also serves to insulate our children's bodies. In addition to that, when our children consume a meal that has a certain amount of fat in it, they're less likely to feel hungry soon after the meal and it actually even helps regulate the blood sugar levels in our children because foods that have a certain amount of fats in them will slow the speed at which glucose or sugar is released into the blood. Over the past three decades, we've read a lot about the role that dietary fats might play in the development of heart disease and specifically, there's been a lot of discussion about the role they might play in the buildup of plaques in our arteries and this led to a whole sort of fat-free, low fat craze. But what we're seeing now is firstly that if we replace all fats in our diets with, let's say, more sugar or more refined carbohydrate, that may also be contributing to arterial plaques and that certain healthy fats in the diet may actually protect us from this kind of disease. So let's look at the different kinds of dietary fats and see which ones might end up being better for us and for our children. So we can divide dietary fats into two major categories and I bet you've heard these terms. We usually divide them into saturated fats and unsaturated fats. Saturated fats get their name from the fact that the fatty acids in this kind of fat are very straight and flat so they can pack together quite densely and these kinds of fats usually end up being solids at room temperature. So the first thing that comes to mind, of course, is our good old friend butter and the other thing that people often think of when they're talking about saturated fat is red meat. Unsaturated fats can also be divided into two types. So we have, I'm going to call them natural kinds or naturally occurring unsaturated fats and then we have what I'll call man-made unsaturated fats. So they exist the way they do because of the fact that we've altered them somehow. Naturally occurring unsaturated fats are found in oils, vegetable oils like olive oil, for example.  The oils in nuts are also usually quite rich in unsaturated fats. The man-made fats that are unsaturated are found in things like margarine and even in, for example, oils that are reused and reheated as we see in French fry machines. These kinds of unsaturated fats are definitely less healthy for us and they're the ones that we've altered and I'll show you how they've been altered in just a minute. So the fatty acids found in naturally unsaturated fats are kinked in a very specific way and that means they can't stack together as tightly as the unsaturated fatty acids. And so these kinds of fats tend to be liquids at room temperature. When we chemically alter the fatty acids in unsaturated fats like we do when we make margarine or when we fry French fries over and over again in the same oil, the kinks in these fatty acids end up looking really different and we end up with something called trans-fat. Now these trans fats do two things in our bodies, neither of which is very good for us. The first thing that they do is they increase the amount of what we'll call bad cholesterol or it's called LDL cholesterol, the kind that actually promotes those arterial plaques that we're so concerned about. And the other thing that trans fats do is that they reduce the good cholesterol in our bodies, the kind of cholesterol that is believed to be protective against these arterial plaques. And while saturated fats have also been shown to raise the amount of bad cholesterol in our bodies, they haven't been shown as much to lower the good cholesterol. Also, if we replace all of the saturated fat in our diets with foods that are high in refined or simple carbohydrates, things like sodas and white bread and other processed foods, then this may also be contributing to the buildup of arterial plaques. So the best advice probably is, again, going back to moderation, eat small amounts of naturally occurring foods of all kinds with a focus on unsaturated naturally occurring oils like those found in olive oil and avocados and nuts and small amounts of saturated fat in our diet are probably okay as well.


Hello, I'm Stacy, I'm a GIS analyst with a natural capital project, and I'm glad that you've joined me to learn some techniques for working with the geospatial data that's used by the Invest Ecosystem Services Modeling Toolkit. This episode provides an introduction to working with vector layers. To get the most out of this tutorial, I highly recommend following along in your own GIS session. In this video, we will demonstrate techniques in QGIS, and we'll be working with some sample data. The webpage for this video provides a link to the sample data that we'll be using. So if you haven't already, now is a good time to pause this video, download the sample data, unzip it, and bring up a QGIS session before continuing. Most of the data that's used by Invest Models are rasters, but most models also have vector data as either required or optional input. A vector layer may contain either points, lines, or polygons. The coastal vulnerability model provides a good example of all of these. Wind and wave data is given as a point vector, the continental shelf is defined as a line vector, and habitats are defined as polygons. Here are some examples of vector layers that are used as inputs to different Invest models. The ones with an asterisk next to them, like geomorphology or watersheds, also require additional parameters for each feature, which we must define in the vectors attribute table. As usual, I recommend consulting the Invest User Guide to learn the specific requirements for whichever model you're working with. Something else to know about vectors is that there are two primary formats that they come in, Shapefile and GeoPackage. Shapefiles were developed by Esri many years ago, and they've historically been the most common way of defining vector data. Now I use Shapefiles all the time, and you'll find Shapefiles in many of our models sample datasets. However, Shapefiles have a lot of limitations, just a few of which I've listed here. First, Shapefiles are defined by at least four different separate files with different file types, and all of these files are required for the Shapefile to work properly. Shapefiles also have a file size limit of two gigabytes, and their attribute table field names are limited to ten characters. And if you work extensively with vector attribute table data, you know how frustrating that limitation can be. GeoPackages are a modern, open, platform-independent data format that is becoming widely used. They consist of only a single file, have no limit on their file size, and you can use much longer more descriptive field names, which is really nice. There are also other differences between these two formats, which you can learn about with a web search. NatCap has been increasingly supporting GeoPackages as both inputs and outputs of our models, and we encourage you to try using GeoPackages if you haven't already. Now let's look at an example of both of these formats in our GIS. Open a File Explorer window to the sample data that you downloaded for this tutorial. In the folder called Vector Basics Data, you'll see two different versions of the same data. One data set is called BuildingsOSMGeoPackage.gpkg, and this is vector data of building footprints from the OpenStreetMap project in GeoPackage format. The other four files are called BuildingsOSMShapeFile, and they are the same OpenStreetMap building data but in ShapeFile format. Now all of these files together make up a single vector layer. Let's bring both of these layers into the GIS. Drag in BuildingsOSMGeoPackage. For ShapeFiles, we can bring them into the GIS by using the file that ends in .shp. So let's drag in BuildingsOSMShapeFile.shp into the GIS. We can see that these layers contain the same polygons if we turn the layers on and off. If we right-click and open up the attribute table for each one, we can also see that they have the same field names. We can work with these fields in a similar way. Because many of us are still using ShapeFiles, we will work with ShapeFiles for the rest of this tutorial, but most of these same methods will also apply to GeoPackages. This vector layer contains polygons of building footprints that can be used as input to the Urban Flood Mitigation Model for calculating infrastructure damage to flooding. I downloaded these data from OpenStreetMap, clipped it to our study area, and projected it to the coordinate system used by some of the other sample data provided for this model. To learn more about coordinate systems, please check out our previous tutorial that covers that topic. Now that we have our basic data set to work with, let's look at the Invest User Guide to see what the requirements are for this layer. The webpage for this tutorial includes a link to the Urban Flood Risk Mitigation Model Chapter, which you can open now. Under the Data Needs section of this chapter, there is an entry for Built Infrastructure Vector. In reading this, we learned that the Vector's attribute table must contain a column called Type, with integer values referencing different infrastructure types. For example, a value of 1 might correspond to residential buildings, a value of 2 might correspond to office buildings, etc. It also says that this column is related to the Damage Loss table, where we provide values of potential damage loss for each type of infrastructure. So going back to the GIS, let's look at the Shape Files attribute table. By default, it does not contain a column called Type, so we need to create this column. To do this, first click on the pencil icon to turn on editing mode. Then click on the icon that says Open Field Calculator. If you haven't used Field Calculator before, I do recommend learning about it, since it is a very useful tool for assigning values to vector fields. And it has a lot more capability than we will use in this exercise. By default, there is a check mark next to Create a New Field. And in this case, this is what we want, since we are creating a new field called Type. In the box where it says Output Field Name, we will enter the word Type, just like it says in the user guide. And note that exact spelling is required by the model. Our building type values will be unique integers, so we can keep the default field type of whole number or integer. Now we can use the expression tab to assign a value to the type field. For a real-world project, you might need to assign different type values to different polygons, and there are a variety of ways to do this, which often begins with selecting a subset of polygons before opening the Field Calculator tool. At the top of the window, we can see an option that is grayed out that says Only Update Zero Selected Features. If we had a subset of features selected, we could check this box to only assign values to that selection. But for this simple exercise, we will assign just one type value to all of our buildings. In the expression tab, there's a white box where your calculations are typed. In this case, it's pretty simple. All we want to do is assign a value of one for all polygons. To do this, we just need to type a one in the box. This is telling the tool to create a new field called Type and assign all features a value of one for this field. Now we can click OK, and then let's look back at the table, and we can see that the column called type has been created, and all of the entries now have a value of one in that field. To save our edits, click on the Pencil tool or the Editing Mode button again. It will ask if we want to save our changes, and yes we do, so we'll say save. Now that we have the vector layer prepared, let's create the Damage Loss Table that goes along with these building polygons. Open a spreadsheet program. I'll use Excel, but you can use something else if you prefer. And first, let's go back to the User Guide to read what it has to say about the Damage Loss Table. It says that the table must contain two fields, Type and Damage. The Type field has values that correspond with the ones that we just added to the building layer, and the Damage field contains values of potential damage loss in currency per square meter. So let's add the two required columns to our table. First Type, then Damage. The Type field has values that correspond with the ones that we just added to the building layer, and the Damage field contains values of potential damage loss in currency per square meter. In our vector layer, we only assigned a single value of one to all of our building types. So under the Type column, add the value one. For the Damage Loss per square meter, let's add an arbitrary value of 100. This is telling the model that any residential buildings that get flooded will incur a cost of $100 or whatever currency you're using per square meter. These values will be used to calculate the value of avoided flooding due to natural infrastructure. And of course, in a real world project, you'd be researching this Damage value much more carefully. Let's save this table by clicking on the Save button. Where it says Save as Type, we want to select CSV, delimited. This is the format that is required by the Invest models. And then for file name, I'm going to call this DamageLoss.csv. And then click Save. The next step of this process would be to test out your building layer and this Damage table by running them through Invest. We won't do that now. You can do that for homework. The sample data that was provided for this tutorial includes a folder called UFM model data, which contains the rest of the data required by the Urban Flood mitigation model. So please try out your new data by running it on your own. Next, we'll look at some of the Invest errors related to vector layers. The building's layer that we just looked at should work fine in Invest. But what if it doesn't? What if we get an error? Listed here are several of the errors that you might encounter when running a model if Invest has problems processing the polygons in your vector layer. Now, as you can see, there are a variety of these errors and they are not particularly user friendly or self-explanatory. So when we're trying to decipher these errors, key words to look for are topology, geometry, polygon, and ring, all of which you see highlighted here. Errors like this usually indicate that there's something incorrect about the geometry of one or more of your features, usually polygons like building footprints or watersheds. If you get one of these errors or ones that are similar, you'll need to go back to the GIS and use the check validity and fix geometries tools. So let's try that now. First, let's find out which features are causing problems by using check validity. Under the vector menu, go to geometry tools and then choose the tool, check validity. For the input layer, make sure that it says buildings OSM shape file. And then we can leave the other options as default. And this will produce several temporary files with the results. So we can just click run. In the tool output window, the log already shows us that it found one error and one invalid geometry. So let's now close this window and look at the spatial results. The check validity tool produces one layer with valid features, a separate layer for invalid features, and a third output for errors. Right now, let's just look at the invalid output. Right click on the invalid output and go to zoom to layer. It zooms to one particular polygon. So now we know which one is problematic. And if we open the attribute table for invalid output and scroll to the right, we see a column called errors. And in this case, the problem is that the polygon intersects itself, which we can easily see by looking at the polygon. Now that we know the problem, we need to fix it with the fixed geometries tool. So under processing, let's bring up the toolbox. And then up here in the search window, type the word fix. And you should find a tool under vector geometry called fix geometries. Let's double click to launch this tool. For the input layer, again, we're going to select buildings OSM shape file. And then for fixed geometries, let's navigate to the folder where we want our outputs to be saved. So we'll say save to file. And I'm going to call this layer buildings OSM fixed geom.shp. You can call it something else if you prefer. And we're going to save this because late for homework, you're going to run this through in the invest model. So click save. And then click run. After you've used the fixed geometries tool, try running the model again with this repaired layer. Usually that gets rid of the invest errors. So give that a try for homework. One other error you might get related to vector layers happens when you enter the shape file or geo package into the user interface. Red X's will appear next to the layer. We can see this next to building footprints. And you'll have a red X at the run button. And when you click the X, a message will appear telling you what's wrong. And in this case, the error says the fields are missing from the layer. And the missing field is named type. If you get this error, you need to go back to the GIS and check the attribute table, adding the type field and populating it with values, like we described earlier in this tutorial. Okay, that's enough for this session. Remember to do your homework and run your new and fixed buildings layer and damage table through the urban flood mitigation tool. If you have any questions or comments about this episode, we'd love to hear from you on our community forum. There's a link to the forum in this video's webpage, where you can search for previous posts and create a new post under the category of training. I and other techies at NACAP will see your post and respond as soon as we can. Thanks for following along.


What is going on guys welcome back to channel critical overload here since I'm talking about Jeepers Creepers 4 again Jeepers Creepers We're born so just to go back a couple steps with one of the earlier theories I was talking about as relates to who stew might be in the film Peter Brooks character Initially and what I know a lot of you have in mind is that he is Jack Jr he is Taggart's other son his oldest son and He somehow I guess from what we are expecting if this is true He would have made it out of the barn attack somehow Taggart senior unfortunately died maybe passed away before the creeper even awakened and Now Jack Jr. Is out of his going out of his way hunting the creeper down Get revenge for his for his brother get revenge for his father not because the creeper killed his father But because the creeper was why his father just had never been the same since Billy's death 23 years ago so if that's who Peter Brooks character of stew is assuming stew again is not his actual name his name But actually of course end up being Jack Jr. If that's the case even if it's not I know I've talked about who he could be in plenty of other videos But the premise of what I want to talk about really regarding everything else for Jeepers Creepers 4 is that what if it actually takes a route where all of the characters that we've we're going to meet Michael Chase Lane Ocean of Arrows character who still doesn't have a name as far as I know and I think there's another character named Sam that I brought up who was coming to Sam is gonna be played by Gabriel Freilich I know he's the one of the friends that comes along with chase and lane to the festival at one point in the movie So what I was thinking was that what if all of these characters have some type of familiar tide? to old creeper victims for any of you who are Fans of scream if you also simultaneously watch my Jeepers Creepers videos or you also watch my Jeepers Creepers videos You know the formula that they're kind of going with when it relates to the killer's formula for screen 5 where He's they're going after people connected and tied to the original killings and anything related to the original killers So with this what if it's like something similar? Where all the characters that we have they are tied to someone that we've met in the past Whether they'd be from the second film or the first film in this instance It would probably be a lot of people from the first not the first but the second film Maybe some characters from the third film like for instance what if chase is Actually someone who is related to someone from that bus He could be related to double D because I know a lot of people have wanted to see that character return I've made a video talking about how double D would not be able to escape the creeper and How he would ultimately be like an opening kill if I were to come up with some type of concepts For how a Jeepers Creepers 4 should open Set 23 years after the events of the last two films What if chase is a relative of double D could be a relative of I think there's another guy Kimble Could be relatives of someone else that was on the bus but let's say he's just related double D and what Chase's gripe would end up being is that he grew up in a household where His father in this case. Let's say that's his father Double D was obsessed with this thing. Just is just his target was Double D of course being upset obsessed with this was stemmed from Minxie telling him You know this thing isn't gonna stop coming after us and of course since the thing is in hibernation That that leaves the question mark for the next 20 years. It never stops coming after you So does that mean when it awakens? It's just gonna come right back after double D Maybe double D would just spent the last 23 years obsessing over it Despite him having his son and having a family. He just couldn't take it one day. He ended up shooting himself and His his trauma and everything just got the best of him and he ended up taking himself out leaving behind his son Chase who I've made videos talking about this like a video talking about this who his son chase Learning from his father what this thing might be supposedly that's where the legend is originating from him He could have heard stories about the creeper and then after his father commit suicide He goes online finds little forms and things like again Who knows how wide the the legend of this thing has spread online the the bad out of hell in po-hole County on East 9 in the old Taggart barn and basically he's trying to research this thing that his father was just so obsessed over and Again, he comes out to the horror-hound festival with his girlfriend with the hopes of going out there to see if the thing that kept his father up At night and kept him just in constant fear for the last 23 years. It was actually true And then of course he could come to find out that it was indeed true Etc. The other thing that would have to happen is with Lane What if I've seen someone make a comment about this and I don't know if I touched on this in another video I don't think I have what if she is somehow a relative of ming see what if she's a relative of ming see and Maybe those two don't even know that they're connected Maybe those two since they're dating chasing Lane. Maybe they just don't know that they both come from people who were Almost victims of the creeper in jeepers creepers, too Maybe she's ming see's daughter or some other relative a part of ming see's lineage Which would explain why she has visions if that's the case where ming see just comes from a line of people who have insight as well like Giselle and Then as far as like Sam Sam could be let's say a relative of Scotty Let's say he's a relative of Scotty and maybe that just doesn't make him an instant fan favorite Because given how Scotty was portrayed in the film the type of character he was The other thing I would like to just add is that this wouldn't be like things that they kind of just are I don't want I wouldn't want them to use these as crutches But I just feel that we're gonna get a film that's not very much focused on character development is gonna be very much more so focused on just having fun with itself having a premise that is Very familiar to those of us who are already fans of this franchise But it's not gonna be anything that is topping what we got in the original with some solid character development I Just don't think I think if they were to do this where they had all the people trapped in that new house of pain connected to Old victims that they we wouldn't get to learn too much about them. They would kind of just use that as crutches I would hope they don't do that But I just don't I something in my gut doesn't tells me that this movie isn't gonna be very heavy on Giving you a reason to care about what happens to these characters just gonna be Characters that are one-dimensional at best And again, I hope it's not because I would love to just learn things about these characters first And then when they're in the house of pain together They can tell their own personal stories and about how they're all How they all have family members that they've lost to this thing Etc. And then we can all grow attached to them based off of not only learning some likable traits about them Now we also know that you guys are related to some of the people that we've seen deal with this thing in the past from the bus film Maybe from the original it may be the third one somehow if they bring in relatives of Kenny Brandon and darla clewate But let me know what you guys think about this down in the comment section below Would you like to see relatives tied in somehow if you haven't already of course make sure you subscribe turn on post application In this video in the description. I've links my social media accounts my face But Twitter and Instagram you can message me there, of course Any movies news or views you'd like me to cover in the future and with all that in my guys, I will see you in the next video