So this is now where I stopped in the second lecture and I wanted to just make, you know, a comment about the deformation theory. So this was the question was about how to write explicitly, I didn't really say explicitly how to write this deformation theory for fixed domain. I'm not going to say much, but what I tried to say was that, and here I wrote it now carefully with better handwriting is that on the modular space of maps of a fixed curve, what you have the only structure you have, and I screwed it up here. Thanks. The only structure you have is this universal map over the modular space to X. And from this universal structure, the deformation theory is this object here this our pie lower star f ever start tx and in the dual. So we have to map it to the cotangent complex and how to construct this map. So, in some sense the question about what the deformation theory is specifically is how to construct this map. This map tells you everything. If you want to know why this map tells you everything then I said, then I recommend reading the paper by Fantechi and Baron and Fantechi. The question is how to construct this map, and you construct that complete total logically from from this geometry. And the only thing you have is differentials tangent bundles and differentials. And I was going to write out the whole thing but of course it doesn't fit in the margin here and but the fact of the matter is that I had written this thing out for a class that I taught in in the fall, and I gave Andre the link. So if you go to if you follow this link for my class notes in the fall, you will find two or three pages which just tell you how to go from this geometry to this map. And it's not original I mean that stuff was already explained in Ky Baron's paper in slightly Terser language and I think it goes back to illusi. So that's, that's a longer answer to that question. And another comment about this is that if you're, if you are new to the subject to deformation theory of maps, which is very similar deformation theory of sheaves but has a slightly different flavor of course, then a place which is things are written out very explicitly and nicely is in the first chapter of Janusz Koller's book on rational curves and varieties. So I recommend an explicit discussion. Of course it's not exactly in this, he's interested in Hilbert schemes but there's there's a pretty nice discussion very explicit discussion of deformation theory of. Well you can interpret them as maps because a map can be interpreted as a sub scheme. So I recommend you look at that. Alright, so that's the finishing the business of the previous questions, and I encourage people if you actually look at these slides to send me corrections. I can't say there's been a lot of that. Okay, so I last time I finished basically with a discussion of the formula for this very sorry conjecture and then some philosophical comments about whether it could be true or where it where it's true or why people think it might be false. But I wanted to make a couple more points. The first one is that you see that this matrix here in the definition is the action of the canonical class or the, sorry, the first turn class of the tangent model the dual of the first turn class. So you might think that the simplest case is when you have, if the first turn class is zero. And it is a very, it's a very nice thing if the first turn class is zero because it wipes out a lot of these terms, not all of the terms because you could have an eye to the zero. That's the matrix to the zero that's always identity matrix well raised or lowered. But still if you have this the first turn class is zero it wipes out a lot of this makes a formula much better. But as I tried to explain last time unfortunately for club you have three folds. This, this, the various our constraints you won't learn anything interesting, but you could think about other examples, who else has C one zero well of course a point does and then an exercise just to make sure that what I've been writing makes is you should, you could try to reduce the equations here to the equations that I wrote in the first lecture. I mean this is just basically an exercise and understanding the definitions. But then a case which is interesting, which is genuinely interesting and new with C one zero is of course the elliptic curve equals elliptic curve and that's extremely interesting case. It has another feature which it has the odd comology, but the very sorry takes a rather simple form there. Simple non trivial form and you can use it to solve the entire problem with the elliptic curve. And this is what is done in one of my papers with on break on golf but I, if you're interested in very sorrow. The case of the target being elliptic curve is a very good case to study because a lot of terms go away. So we've gotten through dimension zero one and three. Well, you could say what about K trivial varieties and dimension two, why am I skipping them there's for example the K three surface, or the abelian variety. And sadly, well it depends on what you're trying to do. Either sadly or happily the grammar with invariance they're mostly vanish, because they're holomorphic symplectic. And so these cases are essentially wiped out and there's there's nothing really much to say they have an interesting theory. That's this what's called the reduced from open theory, and one can ask about various our constraints in that reduced geometry and that's a different direction and it's not really fully understood. So that's the discussion there. Okay, and I wanted to give a couple examples of how one actually uses these constraints so in Gina in the case of a point. I want to explain this that they're extremely useful and in particular they can you can even code them, and they can very practically solve the problem for lots of gender I mean really pretty high genus. But I wanted to give an example where it's not a point. And of course the easiest examples the curve but I wanted to give a higher dimensional example so we're going to consider the example of P two, and that's a beautiful example it's a classical example. Next to lots of things in mathematical life. So the project of playing P two, and the basic grandma Whitman variance their accounts of very degrees. And this is the language of these brackets and this I'm putting a zero here, I'm only going to think about zero now for this example. So that's just point conditions. So we asked for these genus G degree D plane curves to pass through a certain number of points how many points well this many number of points because that's the virtual dimension. You get a number, this number is NGD, and this number is the classical. So it's a little theorem that one has to prove but that the grandma Whitman number here is actually a numerative. So it actually counts the number of these genius G degree D curves through these many general points and in fact they all have to be nodal curves. And these numbers in algebraic geometry have some history and they're called the severity degrees, because they were studied by And one can ask to what extent these equations we know, constrain or determine the severity degrees. And the topic I mean if one wants to give a lecture about this the very first place to start is of course Gina zero. And there's a fundamental equation here from quantum comology. That's very well known it goes back. Well it's an application the WDV equations and goes back to can save it. And you get this very nice equation that calculates the China zero invariance. And here's the first two of them. And there's many, many good in good expositions of how to find this equation from the properties of grimova theory. And the answer how to find this equation is that you have to look at the splitting axiom and grimova theory. And what that means is we take this we take the modular space of curves the modular space of maps maps the modular space of curves, but the modular space of curves have some divisors as we've discussed before. For example, a boundary divisor or I split the genus in the two parts and I connect with one node, and I can take some kind of fiber product. And we would like to, this would be the, we would like to have some rule for how to compute the grimova invariance. After intersection with this divisor in terms of the grimova invariance of two pieces. And that's what's called the splitting axiom. And it says that, well it says that if you want to know the grimova invariance of the full curve you can sum over the grimova, the virtual class restricted as divisor. That's this pullback, you can sum over the distribution of points and degrees on these two sides. This is a splitting axiom and grimova theory. And it's kind of interesting it says that in particular it says that if you know things about mgn bar then it gives you some constraints on the modular space of stable maps and grimova invariance. And this is this is explained in terms of splitting axiom. I'm sorry I'm going through this a little bit quickly because there's been, there's lots and lots of places where you can read a very nice discussions about this. And the WDV equations are obtained from the splitting axioms by taking the simplest possible relation m04 bar. That's the relation where this boundary point is is equal to that point in H2. And so I leave this basically as an exercise derive this recursion from from this relation and this behavior of the splitting axioms. And honestly this requires some geometric ideas you'll have to think about it, but this wasn't somehow my point to do the genus zero again. My point was to apply the virus are constraints in P2. And the interesting application of that as happens in genus one. Before they do that I'll make this comment that you see here the comology of the modular space of curves constraints grimova theory that if I using the splitting axiom. If I know things about for the relations about the modular space in come out of this curves, I can get some actual constraints on numbers of grimova invariance. But in fact the opposite is also true that's what. In some sense, in the last decade there's been a lot of work on that the opposite is true that the geometry the modular space of stable maps constraints the comology of mg and bar. And examples of this relate to the Faber-Zagier in Pixon's relations that's not the direction of that I'm going in these lectures. But in fact the relationship is kind of mutual that the modular space of curves constraints maps and maps constrain the modular space of curves. You can learn about you can learn both things, you can learn things, going both ways there. Okay, I'm sorry that was a bit fast in the genus zero. But nevertheless we go so that what I really wanted to explain from very sorry is genus one and this is a much more subtle thing. It's extremely beautiful equation it's for the. It's an equation for CP to the protective plane, and it's an equation in genus one, and it says that there's an equation of the same flavor. So the flavor here, this associativity of the quantum product equation that somehow it's a recursion and has a quadratic flavor with some polynomial and binomial coefficients. And this is the equation in genus one. As I said this equation is much more subtle, and it's less well known it was it was written in one of the papers of a Gucci Hori shong, because they themselves saw it as an apple made their derived it as an application of the various our constraints, which I'll explain about but one of the reasons it's kind of interesting as you might think okay if this is true. I should, I should try to prove it the same way that the genus zero equations proven by taking some maybe some equivalence in the boundary geometry of m one n bar. I'm taking taking some relation and h two of m one n bar and pulling it back. But this doesn't work actually. Part of the reasons is there's no interesting really relations, there's no interesting boundary relations in m one n bar. So you can't even really start this idea. And then in m zero and bar that was this cross racer relation I explained but an m one n bar there just there isn't even one to start with. So that then it leads to a little bit of a puzzle why could this relation looks like it's a shadow of something like that. But there's a, there's no so to speak object to be shadowed. Nevertheless you can look at it as I said that it appears in this paper of which you Hori shong as an application the various our constraints, when maybe when they wrote it maybe wasn't so clear the various our constraints were true. But they did it and then they calculate some numbers and you see that it gives the correct numbers. This is maybe the first interesting one. And so I want to explain how you get this as a consequence of, of the various our constraints. You can also do it from gets those relations later found by one for but I don't want to explain that. That's more complicated relation and the derivation there is much more complicated the best derivation of this equation is from the various various our constraints for CP two. And, and since those are proven this is actually proven equation. So how to prove this. So I give you some steps. So the first thing is what which for the various our constraints am I supposed to write down what I'm telling you write down L one. And then what do I do, while I take various our constraints say L one of this partition function is zero. So I'm going to study this L one partition function divided by the partition function. This is a standard thing to do. And I put some slides in the first lecture to explain why does the standard thing but it's easy to explain so explain again here is that the Z is kind of X of f we like f f is the connected invariance and they're a little bit nicer. So think about, and when you take some differentiation of Z. This is the same thing and it will not surprise you and I write this. It's the rule for differentiating the exponential like this. So it's nice to consider terms like that because this is equal to just the derivative of f. Okay. So it's kind of standard when looking at the various our constraints. So I say you take this L one, you look at this expression and extract a particular coefficient why lambda zero because I want genus genus one and the exponent there is to G minus two. Why this well this these are 3D minus one point conditions. So it's one fewer point condition but this L one is an operator and it'll put one more point condition. So if you take this particular exact expression and extract a particular coefficient you'll find an equation. And this equation will have leading term in some in you know in some aesthetic sense but anyway one of the terms will be exactly what you're searching for. So this is the term we're searching for this is our goal, you know have some coefficient and when you do this you'll be happy to find a nine. That's good news because this has a nine. This formula has a nine and there's the nine. And this nine is produced from the these combinatorial coefficients here. And various are maybe I'm being too specific about this but you know it's produced by these, these terms. Actually it's produced by this one, exactly at the Dillon Dillon shift. So that'll make you happy you'll find the nine. That's not like half of the proof already finding that nine. And then there'll be a lot of other terms, and you have to go. You have to go exactly write every single other term that occurs because after you finish writing it then you get to say that the sum is zero and and all of those other terms will be inductive terms they will fill out. These terms. And when you do that you'll find that you'll need you'll need to be able to remove things like this. A tau one with a hyperplane class or a tau two with a identity class, and these insertions are hard to remove in general, but we're in genus one and then there's another idea about how to remove them. I mean the old ideas were stringing divisor equation, but there's a new idea called the topological recursion, and I've written it for you, and it comes from a certain boundary equation and m one one that the cotangent line is is related to the boundary like this with a 112 I guess. So in the end you do use a little boundary geometry them one one. I've written that equation for you, and this is a really good exercise to do. If you really want to know what's going on. And in some sense I've told you every step. If you just apply it you can't go wrong but it takes some time to get all the constants and everything correct. And it makes you believe the various our constraints more if you do it.  At the end of these notes. I've appended the full derivation written by long thing so he just does it all it's not that long but in order to get it right you have to be pretty careful and get all the coefficients right. All right so that's the first in some sense, one could say the first application of very sorrow past the point. And it produces something that's really geometrically non trivial in fact, I only know to two proofs of this one from the very sorrow one from gets there's relation in code of mention, an algebraic code of mention to and that gets there's the derivation using gets those equations is pretty complicated. This is much better. But for example, there's many other approaches to counting curves in the plane. So I get this equation from any of them so maybe if someone knows you could tell me like for example the tropical curve counting or the Caporoso harvest recursion and things like that if you know tell me, but it's a it's a little bit of an interesting equation. Okay, so that was. That is more or less the end of what I want to say about. Yeah, you could ask for genius to. Sorry. We discussed genus zero and genus one. It's natural to ask for higher genus are there any higher genus recursions. And the answer is yes but they're much more complicated they're not of the simple form that I just explained for genus zero and genus one or at least I don't know any. They are more complicated forms and they involve add additional recursions for descendants. And it is the case that you can solve everything from normal written theory in many different ways but in particular using very sorrow and some generalizations of the TRR and got one has a nice article about that. There's other ways to do the severity degrees. And these other ways tend to have to do with degeneration of p two, and you can do that within classical algebraic geometry or using the degeneration formulas, and you get. Well you get some degenerations which are effective and solve the problem also, but these generations also introduce many more new problems you have to solve recursively. You never find these very clean things that we found in genus zero and genus one. All right, so that's the end of the discussion there. And now I think I finished a lecture from yesterday. Yeah, so I, yeah, I, I encourage you to either do this yourself or read long things that derivation here carefully. Okay, so I have to get a new slide, and I think that for this, let's see if I can do it here. Maybe it's this one. All right, so we finished two lectures now maybe a little bit late but we did finish them, and they were about modular curves modular maps with a lot about descendants. Maybe I'd say a focus on descendants. I tried not to be too distracted and going in other directions. And the reason for that, and of course about Vera sorrow. And the reason for that is now we're going to switch to sheaves. And again the focus eventually is going to be on descendants and also Vera sorrow and some somehow that's the, the part of this here, which is relatively recent. It is is about transferring the virus our constraints to sheaf theory, the descendant theory of sheaves, and you get some, you get some very nice equations there and I want to explain that what happens there but So to do that a first I have to explain modulate curves modular maps and girls are there and this is complete. And now I have to tell you about the numeration of sheaves so we start from the beginning. So a fundamental property of government theory which one takes for granted one one studies government theory is that it's uniform for all x. And it turns out this is very this is a really great thing about government theory that if I take any non singular projective variety of any dimension. So there is a modular space of maps. It has the same definition. There is a virtual class, it has basically the same definition, and the theory of integration against the virtual class exists, and that definition is uniform for everybody. And as I said that one takes that for granted, but it becomes clear that that special one one starts counting sheaves. The sheaf problem is a little bit different so in an idea of a map is that you map your your counting maps from curves to x. If the sheaf counting problem is more delicate and the standard series of sheaves counting sheaves on x are for dimensions less than equal to three recently, there's been worked by Richard Thomas and now very interesting work on clubby our four folds. And so you can say why is this why is it for grandma wouldn't you can we can take any target but for she's counting we have to take a lot of national target. Well one way to say that is that well grandma wouldn't theory is really getting around the problem because we're in government theory we're only contained we're only counting maps of curves and when you count sheaves. You know sheaves could be also any dimension that could be supported on any dimensional sub varieties. This is part of the answer. It's not the full answer, because even the sheaf counting problem which is related to one dimensional sub varieties does not work in all dimensions. That's that's only part of the answer, but when we get to the kind of the the the real nuts and bolts of the answer it's about this deformation obstruction theory. So what's the reason for the difference. So, as I explained before, when I look at a map from a curve to a target x, the deformation theory is controlled by H naught. That's the deformation the tend to the, the deformations are given by tangent fields, and the obstruction space is given by each one have the pullback the tangent sheaf. And this is always two term. And say what about the infinitesimal automorphisms and this is those are killed by maps stability maps stability kills those so we have exactly this two term obstruction theory and that's a uniform for all targets. So what goes wrong or what happens for sheaves. So for sheaves we have some x and let's just say it's x is nonsingular projective variety, and we want to count some count sheaves. So that means we have we should have some modular space of sheaves and have some kind of counting theory. And so let's look at what the deformation theory for that is well first of all there's infinitesimal automorphisms. And sheaves kind of always have some, and that's given by harm of the sheaf to itself, and she's always have some harm, unless you I mean just if you have just a naked sheaf will always have some harm, because you can, you can always have the scalars the complex numbers acting by scalars. If you have sheaves with sections and other things you maybe can kill the harms. But if I just have this bear sheaf it always has in a harm, and we try to kill these harms and this is mostly killed by sheaf stability. If you if you have a stable sheaf then the only harm can be the complex numbers of scalars that almost kills it but still it leaves a little bit more we have to fix. And this leads to the idea of traceless x and things like and some, you know, some, some additional complexities. And the definite the deformations of the sheaves or get x one, the obstructions and x two so this is I think the subject and Richard spent some time on this at least that's what I've heard. And then there's some higher obstructions. So you have to at least consider these x. The traditional way to confront this problem as you kill the infinitesimal in the infinitesimal obstructions, you kill this basically by stability in the traceless condition, and you kill these guys, the higher x by dimension constraints on x so you somehow run out of room there. And then, in good situations you have only these two, and you got a deformation obstruction theory this two term, and you obtain a virtual class and have a good counting theory. So you see that this is now much more delicate than what happened here. So I wanted to do some examples. And these are fun examples so this, this is what we're aiming for is dimension three and we'll get there. But there's a lot of nice things along the way I want to point out. So the first example is dimension one that's when x is dimension one has dimension one. So what is dimension one this. A question here. Yeah. So these these higher obstructions. Yeah, I think we discussed last week that like, I think we said something like all the obstructions live in the x two space. But this is for like extending from like some order k definition to k plus one definition. So maybe this is mean something else. Yeah, I mean it's yeah I think that you have to consider if you want this. Well, if you want the this x theory to give a perfect obstruction theory in the language of Baron and fanta key, then you need that you need something like this high you need these higher obstructions to vanish. But it is true that somehow they're not all necessary. It's not the case that they all vanish. I mean, if it's not the case I can just forget this and just say that that I have a obstruction here a perfect of two term perfect obstruction theory for sheaves for x one and x two for any say so that's just false. And one, one symptom of that being false is that one thing about a two term obstruction theory is that you always have virtual dimensions are actual dimensions are always higher than virtual dimensions. So when you have these x higher you can have some problems with that. But I think it is true that you don't really need to consider all of them. But to the extent exactly which ones that you have to consider maybe a little bit delicate. My view is that I'd like them all to be zero. Okay, so dimension one is a is a, the, in some sense the easiest case, and then x is a non singular project of curve of genius G. Sheaves on a curve of genius G and this is a really old subject. Well, 50 years old, maybe more by now, 60 years old. And, and the, what, what you get there, the first, if when you apply ideas of stability you get the modular space of stable bundles and the simplest case is when the rank and degree of co prime. And this is already non singular with expected dimension, since this higher x all vanish because we're on curves. So from this point of view, everything's perfect for the modular space of stable bundles, the virtual counting is the actual counting the from the from the theoretical problem point of view there's no problems. And of course there's many variations of this like Higgs bundles and things like this. And other problem in, in dimension one, not as well studied as the modular space of stable bundles of the quote scheme. The quote scheme is you have some non singular, or you have this curve and I take some vector space CN, you can make more general ones this is the basic one. So you have some rank and trivial sheaf and you consider the quote scheme of quotients of that. And then you have to fix some numerical invariance and for a curve you just have to pick the rank and the degree of F. And this obstruction theory was considered by Marion and opera. And the idea there is the deformations. So now we're not doing deformations of sheaves exactly it's a quote scheme, the quote scheme is a deformation of this quotient sequence. And but that has a deformation theory that's given by X of the sub to the quotient and the deformation spaces this home space and obstructions or this X one. So there's no higher X. And so we get a two term obstruction theory there and it's not exactly this one, because we've, it's not the same problem it's a quotient problem. But anyway it's a it's a it gives us a virtual class and this is an interesting virtual class, unlike here. The virtual class just gave the ordinary fundamental class, which is interesting its own right but it's not new and interesting. Here, the quote scheme is in general singular of mixed dimension. But because of this deformation theory this the deformation theory analysis it carries a virtual fundamental class which is a very interesting thing. Even though the geometry might think is simple it just, it's just sheaves on curves which is a, you know, very under well understood, I would say by now geometry. But this virtual class is a new class there. And the first exercise, if you want to, if you want to have any idea of what I'm saying or maybe you already do but you want to learn it compute the virtual dimension. So it's not a hard exercise just have to apply everyone rock to the right thing. And you find the virtual dimension is given by this formula. On an open set. This quote scheme is just a modular space of bundles with sections that. And Alina Marion and drag or pray I use this idea to transfer integrals from this modular space of stable bundles to. An open set is a modular space of stable bundles with sections. That's why we're open set. They use this idea to transfer integrals on the modular space of stable bundles to this quote scheme that's a really interesting idea. Because the quote scheme you know just you just look at it from far away it's just like bundles with some sections. It's it's a tricky thing to transfer the integrals because you have to make sure whatever inter whatever intersection problem you're doing. If you want to transfer it you have to make sure well maybe there's a dimension shifts you have to shift the dimension the problem. And also you have to make sure that all the action is happening in the locus where it's stable bundles with sections so there's some analysis of you have to throw out the pathologies but they get it to work exactly and and then they can actually then you can use this path to transfer integrals to this against this to this virtual class. And then and then they can actually compute integrals this gives a proof of the verlander formula there's many proofs the verlander formula but this one uses the virtual class this code scheme. And why is this useful because the point is once you get to this code scheme, then there's a C star action. So you can look at this tourist localization and the virtual, the localization of the virtual class I haven't discussed that here but there are techniques. So this localization as a basic technique by a T a bot bot and a T a bot that tells you how to calculate intersections and or integrals on varieties with tourist actions in terms of the fixed point data. There's a version of that theory with respect to the virtual class that I developed with Tom Graber. And why that's useful is because while these faces are kind of complicated, the fixed points with respect to the C star action C star action scales the different different coordinates of this complex vector space. The fixed point actions are very simple they're just products of they're just symmetric products of the curve. And so this idea is used to transfer complicated integrals to maybe complicated integrals and then localization the trans transfers them to integrals and products of the curve. And there they can be sold so that's the that's something in dimension one. And I wanted to just explain one more thing in dimension one because it's kind of fun is that. So, why you get this metric product of the curve is that if I take the quote scheme with just one copy of the of the trivial sheaf. That's a symmetric product. But you can do something kind of a little bit stranger which is you can do quotients of end copies of the trivial sheaf. That means you look at here. This middle fellow is still CN and I want to look at some quotients, but I asked for this rank to be zero so it's like some kind of generalization the symmetric product in some strange way. I look at the rank zero quotients of the n dimensional sheaf for curves. So this is some kind of punctual quote scheme I don't know exactly the name for this thing. But it's in some ways generalization this metric product. And exercise here the deformation theory exercise is that this object this punctual punctual quote scheme. If someone has a better name you can tell me what that is this punctual quote scheme is a non singular of this dimension and this virtual class is the usual usual fundamental class. So the virtual counting here agrees with the actual what's physically happening. And then there's a lot of stuff that happens on the space of example there's total logical bundles. And if you're familiar with the theory of the Hilbert scheme. None of this will surprise you if you take a vector bundle of rank e on x, I can move that to be a certain total logical bundle on this punctual quote scheme. And it's, and it's demand the rank of the total logical bundle increases by a factory of D. It's the bundle whose fiber over particular point of this quote scheme is a H naught of F tensor e. This is a kind of standard move in the study of Hilbert schemes of points. So this has not been so well studied I would say and I'll give you an interesting property that came up in a paper with Dragoosh. It's some kind of exchange property it says that if I want to calculate the integral over this punctual quote scheme for this total logical line bond for the total logical bundle associated the line bundle, and I take the segue class. That's the same thing up to some sign as doing it in rank one over the symmetric product, but taking n powers of the same segue class. The segue classes that you should segue of a bundle is equal to one over the total turn class. So there's some kind of interesting symmetry here, and we proved it by doing some calculations but I always wondered whether there should be some kind of proof of this without doing any calculation so that's a challenge. Understand what I've said and find a conceptual proof with no calculations. Because you're kind of swapping this end. This is some kind of higher higher rank punctual quote scheme and this property swaps it down to rank one but of course a little bit. There's a price to pay you to put the end here. All right, so that was dimension one dimension to this. There's a lot of interesting stuff in dimension to and I think a lot of Richard's lectures were about that. I will take go to this dimension to with a slightly different focus. So let X be announcing the project to surface. I think the simplest theory for this dimension to is again the code scheme. And precisely this quote scheme. So X is the surface now, and I again look at quotient the trivial sheaf. The question has the quotient F and has the kernel G, and unlike the case for curves which I didn't have to make any assumption I just take any quote scheme for curves. The obstruction theory is two term and I get a richer class now for surfaces because the dimension is increased I have to, I have to pay for that some way in the sheaf theory. In the sheaf theory you always pay for increasing dimension. And how I pay for that as I asked for this quotient to be rank zero that means generic rank is zero and what that means in practice for a surface is it supported on curves. So I don't ask for every possible quotient I just want for course I don't consider quotients that are supported on curves. So such a quotient has a first term class, and it has the other characteristic. So the rank because the rank is zero. Well, and in papers with Mario and a prayer. We investigate the obstruction theory for these quote schemes. And this is motivated by several things but in particular their work in dimension one, and the deformations are as before. It's given by the harm. And then there's the obstruction so that what we have to worry about is killing these higher obstructions. And so this x to here is shared dual to a certain home. And it's precisely to kill this home that we have this rank zero, because then F is a torsion sheaf. But the right hand side is a sub sheaf of some torsion free sheaf. So this home is killed since F is a torsion free since F is a torsion sheaf. And this generality I said here this x is killed in this.  You could say can we ever remove this torsion hypothesis and the answer is yes. If x is fauna I think we can remove it or maybe to some other small assumptions but, but the idea, but this direction of removing this torsion hypothesis and paying for it with some additional geometry on x. And that is an interesting direction that's mostly unexplored I think we did a few calculations there. And you can calculate the virtual dimension of this quote scheme and there's a formula and it's essentially everyone rock calculation which I leave to you. And the interesting thing is what are the integrals here and this is going to be some kind of foreshadowing of what we do for three folds and descendants. And there's different ways to study this theory you can study it and numerical theory k theory, but the way to get some total logical objects on this quote scheme is by this the construction and sometimes we already saw. We started with a k theory class on x that's what we did for a curve, the k theory class and x with a line bundle a vector bundle, you pull it back to to x cross the quote scheme and x cross the quote scheme has a universal quotient that's a kind of drawn it is curly. And then I can take our pile or star and get a K theoretic class on this quote. And that's the way of taking a K theory class on x and getting a K theory class on the quote scheme. And that's just that this is rather familiar operation, and we're going to interpret this eventually as descendants. And if you want to write an integral, the most general kind of descendant integral you can write is I take this quote scheme. I have to sum over something so I'm freezing the curve class, and I sum over the degree of I sum over the order characteristic that's this degree some. And then I put in the churn characters of this K theory is total logical K theory classes I've constructed. And as I said that we will later interpret these kind of things as descendants that are parallel to our descendants and come up with in theory, and that parallel will be made rather precise. But here it's a little, it's a little here it's in some sense more just words. So my view these is a descendant insertions. And since I have a growing, I have a growing virtual dimension. I need to put some other class in so the standard thing to put here is a total churn class of the virtual tangent bundle. And the motivation for this comes from different ideas. But one of them is if I just get rid of this like this thing gives me the, if I don't have any classes here this gives me the virtual or the characteristic. That's the, that's the, you know, full theory for quote schemes of surfaces and the generality that you can do for any surface. And you could say, can you solve this theory. And the two basic ideas that come in this in the, when one tries to solve it, there's two basic ideas. The first one is rationality. And that says that this generating series in queue, the one we've just defined here. And then you thought that this integrans were inserted haphazardly. But there's something about that which says this is definitely not the case because the first conjecture is that this is always the Laurent expansion of a rational function in queue. So this is it's born as some kind of infinite series it can have some negative parts, only finally many negative parts. And in fact, the, we conjecture that this is always a Laurent expansion of rational function queue. And in fact, this this conjecture is proven in many cases I didn't list all the cases we can look in these various papers. The last one with with Noah. So there's there's there's four papers here, who none limb and Johnson and the last paper with no harvest fell that's in a key theory, but it proves this kind of rational things in many, many cases but not exactly all cases yet. You have to go look and see if you want to find out where we know we don't know it's a rational function, but most cases it is this thing's actually true for the most part. So that that already shows that there's something magical going on with this kind of insertions. And then you could say can you actually count calculate it and that's one thing nice for these quotes scheme theories which I indicated before, and in fact was what Alina and dragosh used is that they're very computable, because there's this tourist action that reduces to rank one and sometimes that's the, that's the kind of nice thing about these quotes scheme theories and in some sense that they're higher rank because they allow for. These quotients to be higher rank objects, in this case higher rank on the curve. But on the other hand, they have this tourist action which reduces them to some kind of rank one object. And so you can actually do this calculation and I gave one example we've done many calculations in different configurations but the calculation the most general calculation that shows the theory of solvable is this exact solution the case when x is a simply connected minimal surface of general type with non singular rational non singular canonical curve. So you can forget all of this you can say it's a minimal, if you have a surface of general type that's a lot of surfaces, and to write this formula nicely, and you don't need to have all these hypotheses I just wanted to write it nicely. It's a minimal surface with a non singular canonical curve and the non singular canonical curve and of course as a genius so I'm going to write the formula in terms of genius at that. And it says that this this series, and now I'm not putting any of the descendants and this is just the virtual this is a series in some sense for the virtual Euler class. The basic series is given by some sign, which is an interesting number related to the cyber written invariant, some q shift, and then some functions and this should be a rational function how are we going to get a rational function. It turns out you have to look at the roots in w of a certain polynomial it's this polynomial equation. This is a polynomial equation for w in terms of q and has n distinct roots, and to get the rational function you have to sum over all choices of these many of the distinct roots. So what function do you sum it's well it's interesting function it's this function that's a symmetric function and you can this this so this shows that the answer is a rational function, because it's a symmetric function of these roots. And it gives a complete solution this problem is for these virtual other characteristics, in all cases, in all of these. Sorry, sorry, sorry, sorry. Simply connected general types of services. This is the kind of showing the series really computable. And in the papers there's lots of other computations, but what what are we after in these computations well so the first thing that comes in this is that that you see in this is that there's this factor minus one to the order characteristic of the holomorphic characteristic of Oh, and this is the turns out this number is very simple number is the grommov is the grommov written invariant. And this is an old relation that goes back to this grommov people cyber written theorems of tabs, but for such a surface surface of general type. There's one basic environment which is counting the genius G curves in the class in the canonical class, and that's given by a sign. And actually that's the whole action in this, this whole at the whole action this formula is that the rest is something universal. So what this says is that actually that the grommov written theory, and this very I mean that this is saying somehow that this enumerative geometry of this quote scheme is some kind of complicated manifestation of the curve counting in grommov written theory. So that that already shows that there's some kind of connection between these curve and she's counting. And what we're going to do for threefold is somehow a much richer exploration of that. Now, I wanted to make some comment about some of the topics that Richard talked about. That's the she's counting on surfaces, a more sophisticated theory, significantly more sophisticated was proposed by vaughan witton and defined specifically by Tanaka Thomas, and it takes a lot more delicate work to define the virtual class there. And one of the things that you get out of it which is nicer is that it's more closely related to the modular bundles on on the surface. The idea for this vaughan witton construction is that you should move to dimension three, you somehow approach this she's counting on this surface by looking at sheaves rather not on the surface but the total space the canonical bundle. So it's harder to define, and it's harder to calculate. And one of the, one of the things that comes out of it is the rational functions that are found in the code scheme theory are replaced by model in some sense replaced by modular forms in this vaughan witton theory. And I refer you to various results and conjectures of good and cool. But one of the hopes I've always had is that there should be some way to transfer. Not the virtual or the characteristic but transfer integrals from the modular bundles to this code scheme theory as what happened in dimension one, and the advantage for that would be that these are very computable. All right, so I leave you with some ideas there you can explore. So this is to show that this sheaf counting is already quite interesting in dimensions one and two, but we are primarily going to be interested in dimension three. This is the most interesting dimension for counting it's like the perfect dimension for counting. And there are many directions of study, and I'm not going to cover these because there's too many. You can think about this in terms of mirror symmetry or DT wall crossing stability conditions refined and variance. There's a long list and it's an incredibly interesting subject that's going developing many different ways, but I'm going to start at the simplest place. I count sheaves in dimension three, and the simplest modular space of sheaves in dimension three which more or less is familiar in some ways to most algebraic geometries is the Hilbert scheme of curves. The Hilbert scheme of curves so I take X to be announcing in the projective three fold, and I use this notation here. I eyes for ideal sheaves because I'm going to view this Hilbert scheme as a modular space of ideal sheaves. There's two discrete invariance and beta. So and is the holomorphic or the characteristic of the quotient curve and beta is the fundamental class in in homology of the curve. So an element of the Hilbert scheme is a quotient where I take. Well, the quotient has a quotient and an ideal sheaf and it reads and I it's an ideal sheaf is because it's, it's a one it's it's not the quote scheme it's the Hilbert one copy of oh, we can consider the Hilbert scheme as a modular space of ideal sheaves this is kind of important conceptual point so I would say that normally an algebraic geometry. When one looks at the Hilbert scheme, one looks at the Hilbert scheme as some parameter space of quotients. We can also look at it as a parameter space of ideal sheaves, and not always but in this case it turns out that's a very profitable thing to do so if you help him as a modular space of ideal sheaves. And you could say is it so this is this is this statement has some kind of theorem in it, which is to say that if I take this ideal sheaf. It's intrinsic modular, I try to deform it. This this theorem is saying that actually in this case of curves on three folds that modular space of ideal sheaves is this actually the Hilbert scheme which is to say that when I deform this ideal sheaf. When I finished deforming it actually it stays an ideal sheaf in a canonical way and gives me a quotient. So there's something to prove there and of course you might object you could say that I could change the first turn class of ideal sheaf. And of course it won't and that's true that's why we have to say actually trace free deformations. So in actually in the modular sheaves is, there's always the issue about whether you're controlling the deformation of the determinant or not. Okay, but so this is a kind of technical point but it is the case it's true here and it's a little bit remarkable that the modular space of ideal sheaves of this modular space of this sheaf coincides with the quote scheme. And just to make sure that we're on the same page we really consider the entire Hilbert scheme. So if you take X and you look at the Hilbert scheme of curves. A lot of fuzzy stuff can happen you can have some kind of smooth curve that looks pretty good but then you could have some embedded points. Or you could just have some non reduced points running around by themselves, or you could have some monster where you have some very singular curve, and there's a nilpotent structure generically and then additional nilpotent structure at the singularity. We're considering it all. And this was a topic developed in Richard Thomas PhD thesis, which was to show that I mean the first part to show that this object has a two term obstruction theory. And there is some subtlety in that. So first of all, as I said that we're supposed to first thing. This is, you know, if you go look in Richard's thesis you won't find all these statements I'm saying but the first thing is to consider this as this modular space of ideal sheaves. He did that at the beginning yeah I don't think he was he didn't write about this as the Hilbert scheme, the modular space of ideal sheaves. And then as I said we have a problem with these scalars. There's x zero there's x one x two we're happy with that. There's x three which we should kill. And how do you kill x three. Well, that. So it turns out both x zero and exterior killed with the same idea, which is that that you have to look at not the deformation theory of ideal sheaves. The theory of traceless deformations. And this is explained very well in Richard's PhD thesis I encourage you to look at it. So there's the traceless deformation theories deformation that preserves the determinant, and you want to express that into x just like normal deformation theory if sheaves is just allows determinant to wander how it wants and that's given by this if I want to fix the terminus given by traceless x and this traceless x kills the sea so it's going to kill this guy. And by ser duality it turns out also kills that. So by going in the traceless deformation theory it's two term perfect obstruction theory, and the traceless axis defined using x and the kernel of the trace map and as I said I recommend looking at what Richard wrote. That's developed very nicely there. So, the summary is that this modularized this this Hilbert scheme of curves in x which is kind of classical object. I would say something familiar to algebraic geometries. I do the first move is look at it as a modular space by deal sheaves, and then do the second move which is what Richard doesn't see says is look at the traceless deformation theory of the modular space by deal sheaves. Then this Hilbert scheme carries a virtual fundamental class obstruction theory in a virtual model class. So it's kind of a little different way of looking at it. Then what's traditional and algebraic geometry. It's the same object but you look at it slowly differently. And that's a beautiful thing. And I mean this virtual this deformation theory in a virtual class. And the first exercise is calculate the virtual dimension. And if you do you'll find something kind of remarkable the virtual dimension. This thing has two discreet invariance. This or the characteristic in the curve class. And if you calculate the virtual dimension that you'll find that it depends only on the on the beta. It doesn't depend on and it's independent of that. So this is kind of a little bit strange. You might think that as you're giving, you know this end is the other characteristic. And you could say you're somehow twisting your sheaves by some point somewhere, but the virtual dimension doesn't care. It doesn't give you any more virtual dimension for doing that. Even though the modular spaces might be growing a huge amount, but virtually they don't. So if I, if I want to integrate against this I'm allowed now as a virtual class I can integrate against this modular space, this Hobbit schema ideal sheaves and that's this integration is called Donaldson Thomas theory. And what about grimovin theory it turns out grimovin theory also has an independence property with virtual dimension. The model just face a stable maps to x rex is a three fold. It's also independent of one of the parameters now it's independent of the genius, which is also a strange thing. Because normally you think when you're mapping curves to varieties, the genius is going to matter. Because it controls how many line bundles and there's no other theory but virtually it does not matter. So these dimension formulas are the same, the dimension formula for the Hilbert scheme and the dimension formula for the modular space of curves. They're the same they're the same in the sense of the right hand side is the same. The left hand side is different and it's not even clear what I mean by they're the same because there's a G here and there's an N there. But I would say that they're kind of strikingly similar. So the simplest place to do this now is for club your three folds but maybe I should stop we can do that tomorrow. So it's clear from this character here we made no assumption clubby honest of x, but it's clear from if you the club you condition is going to help us, because it's going to make not only the virtual dimensions, independent of n it's just going to make them all zero independent of beta to. This is some kind of sign about the clubby a three fold. So first of all, if you're doing an innovative geometry dimension three is the best dimension. And if you're if you're counting in dimension three clubby ours is the perfect place. And roughly speaking the reason for that in one line is that if you're almost all the, I mean, okay I'll try to say it in the exaggerated way that all counting problems have dimension zero and clubby a three fold this the statement is kind of the first order that's true and that's why it's the perfect place for counting. So we'll start on that next time. So I'll stop now. Equations are one of the today. So the question is, how do you sort of constraints depend on stability stability of, you mean the stability of the curve. The question I did not say. So I think that maybe you could say, okay, maybe then I just tried to interpret it. So I was talking about a certain, I mean, like for curves is the Deline Mumford stability. You could try to change stability conditions for curves that you have choices there. Actually, I'm hearing some kind of echo. Should I, should I just ignore it or what. Anyway, so you can change stability conditions like you with the, if you have a mgn bar with points you can put some weights on those points and there's assets stability. And there has been work on how the cotangent lines change the integrals the cotangent lines change with this different assets stability. And I think even some work about the constraints, my memory is some paper of white P Lee and some others, but I'd have to look. But I think the answer somehow is that when you change the stability conditions in the way I was saying about mgn bar, the cotangent line integrals change in a more or less controlled way. And then you can propagate that through the constraints and somehow, but I think that white P wrote something about it, but I'll have to go look. I'll bring it to the, I'll bring it to the lecture tomorrow. Okay. And sorry and if you do that for arbitrary targets I would expect something similar that maybe just say that one more thing about that when you change these stability conditions in this way maybe you have a more complicated way but the way I'm talking about it's changing Gina zero stuff. So you have kind of control about what's going on. I have a whole question in the chat and I think we're going to have to finish afterwards. The question is, what properties of invariance. Do you expect for the code schemes of panels on of non torsion sheaves. So I would, I mean, I would guess the rationality would hold also. In terms of exact formula so yeah that the general question the general claim I would make is the rationality that that generating function is is still true. I don't know if that's true or not we have a little evidence I don't think we have much evidence, doing So the question is in rational and final cases turns out to be harder than in the general type cases. And one finds out also in that in the Vafa Witten world. And somehow the philosophical reason for that is that, and well my view is that the curve counting and rational services is very rich and complicated while the curve counting on surface of general type is very simple. So I don't expect they'll be as easy formulas to find or as as as universal formulas to find in the final cases, but I would expect the rationality at least. Thank you very much.


Hey friends, tonight we are hanging out at Disney's Animal Kingdom. I wanted to come out and make a little update video for the new year 2023 and then later on we have dining reservations at Jicco, which is one of the highest rated restaurants in Walt Disney World and fun fact I've never eaten there before so I figured we come on out, ride some rides, eat some food and have a beautiful Animal Kingdom kind of day. Any hoos, let's go do this It is a beautiful day to be hanging out at Disney's Animal Kingdom 81 degrees out the skies are crystal clear blue. I don't think we're expecting any rain and I'm excited to have some fun Like I mentioned earlier, we had dinner reservations at Jicco over at Disney's Animal Kingdom Lodge And I have never eaten there before all of my foodie friends tell me this is their favorite restaurant So I'm very excited to try it and after I eat there tonight I'm only two restaurants away from being able to say I've eaten at every single restaurant in Walt Disney World I have Victoria and Elberts and Montserrat Paul and I'm gonna be able to have that title Looks like we're just in time for wing it encounters and it's gonna be really pretty to see all the Macaws flying in the sky I Jeannie plus is $15 today, but I really don't think I'm gonna need to buy it Animal Kingdom I feel like it's one of the parks where I don't think it's worth it to buy Jeannie plus I mean unless you want to do avatar flight of passage, but I did notice that that's not even available So no point in buying the Jeannie plus if I can't ride that ride when all the other attractions have like a 20 minute wait Before we make our way into the park I wanted to stop at the island mercantile because they just released a whole entire line of awesome like vault collectible merchandise and Sometimes when you come to Animal Kingdom, it's more in stock here because people wouldn't think they come here and they do They have some great items. Oh, I am so excited right now Take a look at this lots of 25th anniversary Merchandise and I am obsessed with it. Look at the Kate Castle hoodie right here This is beautiful. Look at the backside of it. It's time to remember the magic. This is amazing. I love this so much I definitely think I should get it. They also have a 25th anniversary like bomber jacket. Look at this This stuff is so cool. Look at Simba on there. I love it and then the hat I think I'm gonna get this hoodie. This hoodie is $69.99. I think it's coming home with me. I Ended up buying the Kate Castle hoodie. I had to I was in love with it I'm sure it's gonna sell out pretty fast and because we're gonna be going into the park I don't want to carry it around with me So I'm actually leaving in here and I could pick it up when I leave They give you a little yellow slip and you return and you can pick it up before you leave So you don't have to carry it around the park Like I mentioned before it's not that busy in the park today Expedition Everest has a 20-minute wait Navi River Journey has a 20-minute wait I think we're gonna head down there first to ride that and then a flight of passage is 95 minutes It's tough to be above 10 minutes. This is like a really nice day dinosaurs a 10-minute wait.  So we're gonna head over there ride that first then probably a little Everest and just kind of enjoy the park And see if there's any new snacks new snacks. I know we're going to dinner later But I wouldn't mind trying a little lunch, you know Seriously with wait times like that It's absolutely pointless to buy the genie plus I feel like especially at Animal Kingdom and now we have made it to the Valley of Moira. You can hear all the different animals. I love coming in here. I Really wish we can ride flight of passage. I mean, that's a fun ride But I don't think I'm gonna wait 95 minutes if they had a lightning lane that I can purchase I would probably buy it right away, but they're just not available But Navi River's still a fun ride and I wouldn't wait more than 30 minutes for that ride. So 20 minutes is perfect It's also gonna be interesting to see if that's a true wait time 20 minutes and it's 148 So we're gonna head on in The reason I say I wouldn't wait longer than 30 minutes I feel like this ride is a fun ride, but it's very short and I can come all the time I mean, obviously if you're on vacation You're gonna want to wait as long as possible because when are you gonna come back on vacation? But the only thing I really love about this ride is the animatronic at the end. I Do have to give a credit for having an amazingly themed queue. Look at this. I Love all of Pandora. I think it's just really breathtaking and it's fun to just explore all the different like designs and Reasons why they do certain things. It's beautiful in here That was incredible. It was only a 12 minute wait. It said 20 and then the being 12 Our lucky mode of transportation right here. Look at that That was a fun little boat ride and I just loved the music in the animatronic at the end like the shaman She is so like majestic looking the way she just swirls around her arms it's hard to believe that that's like not real like it looks so lifelike and It's definitely probably my favorite animatronic out of out everywhere Now that we got to ride Navi River journey I kind of want to make my way over near expedition Everest as you guys know It's probably one of my favorite rides here in the park and it's got a super low wait time and even lower because we can do the single rider line As I start making my way over to expedition Everest I do want to stop at some of the little eateries or the food stands because they always have rotating specials Flametree barbecue always has a new cupcake that they have the smiling crocodile has new tacos And yes, there is a little food stand called the smiling crocodile and I think it's so awesome Right here is the smiling crocodile look it's got a smiling crocodile on top it always reminds me of Bobby Boucher and they always have like little tacos here pulled pork street tacos street corn tacos chicken street Tacos and then also an island sunrise, but this usually closes at like 3 p.m. So you have to be early to actually come here. I Hope some of you got the Bobby Boucher and the smiling crocodile Well, technically an alligator in the movie reference because I think that is so funny And then over at the flametree barbecue They actually have a sunrise cupcake and I had to get it I had to see if it was any good normally I try all the cupcakes here They might have had this on the menu before but it is pretty it's a chocolate cupcake But with vanilla buttercream on the inside and it has like a sunrise Kind of look to it's got sunrise sprinkles. Look at you can see Timon and Pumba and Simba actually walking across the Sunrise.  I honestly feel like he's gonna come and try to grab this cupcake. Look at them Do you like cupcakes? Now we don't feed the wildlife. Oh, he said no. He said no. Thank you And then we're gonna cut down the center here and we'll see the little vanilla cream on the inside. I Did not do that gracefully. Oh boy, look at that It kind of just tumbled it fell apart big time. That's the cupcake All right, let's grab a little bit of the icing a little bit of the filling All in one bite Yeah Not the best cupcake I'm attracting more birds here. Oh There they come you got the big daddy over there Big daddy over there. Sorry. You can't have chocolate. It's not allowed Gracie can't have chocolate. You can't have chocolate Honestly, I just always feel like the icing to cupcake ratio is over Proportionized like there's way too much icing on top of this cupcake I need a little thin layer of icing and then a nice little cakey moist cake Enough for the cupcake business though, let's cut through dino land as we make our way over to expedition Everest I mean if dinosaur has a short wait time over to mind riding that it's one of my favorite rides But also I just want to check out the whole dino land at USA area Obviously something big is coming over here, but we just don't know yet what it is As you guys are aware, they demolished the primeval world and there's speculation of a new land coming in the animal kingdom Some people say Zootopia some say incanto or cocoa. I mean, I Like dino land, I feel like it can stay I think they can get rid of the Chester and Hester area and keep dinosaur the ride and still have a new land And that's what I hope that they do Like if they just keep this area right here as the entrance to go into dino land and then you make a right to go to dinosaur They have all of this to play with over here to actually have a new themed area Honestly I am all for a new themed land here and as much as I love dino land USA I can care less if dino Rama actually disappears. I'll be honest I mean, I think it's a fun little area, but right now it's just empty This is the area that I'm talking about Primeval world used to be on the other side of that big blue fence But they just demolished it all and now it just looks like an empty parking lot It literally looks like we these are parking lot spots right here. Look at this. It's like, yeah, we are in a parking lot I think it's always been that way though. It's a part of the theming of the area over here I'll be honest. I think that's really cool that they have the parking spots actually Inside the dino Rama area. It kind of fits as like, you know a hometown like Carnival or fair where it's usually in a parking lot and since we're over here I think maybe we will play a game or two. You have to buy a little game ticket There's six dollars a piece and then you can go play a game to win a prize or possibly win a prize. Oh Let's do a fossil fun the water gun game I'm doing it We are currently at three players We are playing for a little bit you little mini or baby triceratops Oh, that's what I was looking for one of the iguana Don That one is a single player game comment crashers and all prize options are available there However, it is a game of chance. I'm gonna do it. I'm going for the iguana Don Thank you Alright, well scratch the water gun game You need eight players to try to get the iguana Don and that's what I want from the ride of the dinosaur So instead they do have the iguana Don over at the ball game We have to throw the ball in the colored circle. So I think we're gonna try that one All right, we're gonna try for the iguana Don. We need a yellow or a white to get the iguana Don That did not I thought it was gonna bounce a little bit. I really thought it was gonna bounce around a little bit Oh Nope, no luck here. Oh, no, yeah, they're not bouncing much Oh Access denied man Okay, that was just the first try six dollars I decided to get another ticket and I'm gonna try to bank it now I'm gonna try to bank it because my first one I just kind of like threw it right into a hole. It didn't even bounce I Came back for more. I Feel good about this time. I have a new strategy Yep, I'm gonna try to bank it All right, here we go. We're gonna try to bank it off the wall. That did not work That did not work at all. Oh, I thought it was flying down there. Hold on Yep, we got to formulate a new plan. It's not working Oh come on get in there. Oh, it almost went to the blue one. I get one extra. Here we go. Here we go Now I Tried Access denied I failed the mission and I did not get the dinosaur. I lost $12 and I think I'm walking away. I'm walking away Technically if I really wanted to get that dyno I can go back and buy eight tickets for six dollars a piece and then kind of buy out the water gun game and Then just have mine as actually the only one that wins that way. I can assure a victory But now we have made it over to expedition Everest and we're gonna ride the single rider line see how long it takes Hey friend and Enjoy a nice little ride. Let's go get that yeti Look at that single rider. It's at a 20 minute wait, but I might as well just ride it as soon as possible honestly, I think on The only one that's gonna be waiting Yeah, look at that fancy That was a solid one minute wait here comes our train I Would like to walk up them steps one day every time We made it back with us not that dyno expedition Everest always just puts me in a happy mood and look at that It's still a 20 minute wait after we got off and honestly I feel like that 20 minute wait was probably like a 10 minute wait and Yeah, now we're gonna keep moving along. It's been a great day so far. I'm really having the best time ever at Animal Kingdom I Thought about going to ride Cali River Rapids But I don't think I want to be a soggy bottom boy when I end up going to Gecko You know what I mean? It's kind of a high-class restaurant there and wet shorts probably doesn't fit the dress attire or the dress code Since Cali River Rapids isn't in our agenda today I wanted to come over and at least check out Kilimanjaro safaris See how long the line is we might be able to ride that if we have time and Then kind of complete our full lap around Animal Kingdom Kilimanjaro safari says 30 minutes, but I have a sneaky suspicion. It's not a 30 minute wait Honestly, if all of these rides have a 20 minute wait, I feel like Kilimanjaro safaris is gonna be like Less than 30 minutes. I mean it has to be honestly, I could say probably 15 minutes 15 minutes tops It's also important to point out the safari Closes at five o'clock the last safari goes out at 5 p.m. Because all the animals got to go to sleep They got to go to bed at five o'clock. They're like me Up here comes our safari vehicle Simba one Alright guys on the right side of the truck watch your hands arms. They speak the doors gonna be slide and closed And guys look above you up there. There's a game spotting out to help you identify the animals We might find out on the reserve. Hope we get a chance to see for those. That's not a sauna All right guys. Let's say a Harambe twin day, which means let's go. Let's go Wonderful Everybody welcome to Kilimanjaro safaris My name is Travis and I'm gonna be your game driver for your safari through the Harambe wildlife reserve A few rules before we get started get rolling boy. I know these are massage around because they have an irregular shape pattern As opposed to like a reticulated Rothschild or even newbie and giraffes which have all blocky square or rectangular pattern Many people view there were different types of giraffe. I didn't know there was different types of giraffe My Oh, yeah, got a couple up there Don't never let a giraffe lick you Nasty this can stick your foot in the mouth, which is there because they have some mean I'll give you an example see the bushes on your left here. These are acacia giraffe love occasion But look at the branches closely. They're covered thorns But a giraffe will actually wrap its tongue around one of those 40 branches now the discus foot in his mouth will help protect his tongue Wow, look at all the large animals you see back there those are Patterson's elan Patterson's elan are the largest animal I miss when the bridge used to make noises like a squeaky noise Yeah Communicate using a low rumble that's actually below the ability of humans to hear it called infrasound An elephant can hear it at the 10 miles away through its feet actually feel the vibration of the rumble in the ground And then let me introduce you to those driving animals They don't like me You run 45 miles an hour and all they do have we exactly what they're doing on there You know just kind of hang out lay around and sleep 20 plus hours a day. What do you write white rhinos? Why right? Let's get their name from the African word fight meaning I think in charge of speeds, but the 35 miles an hour I've really bad eyesight, too. Don't ever sneak up on when I always don't know you're coming make a trip There's a body bar in front of us in there. But you're not very extinct in the wild. Montabac There's a bonzabac Unfortunately 150 years ago, they're only like 20 boxes from the planet That was a great Safari Travis was amazing. He was one of the best safari guys I think I've ever had his voice just fit perfectly in the role too and I got to see lots of cool animals It's probably the most I seen of the lion and it was also really exciting to see the bonzabac It's kind of funny because you guys caught me in action of getting excited to see the bonzabac Well, that was great success and now I think it's time. We start making our way to dinner We had a fun day today. I had so much fun low-weight times everywhere. We went It was just a chilled day and now it's time for a very nice dinner I Just walked all the way out of the park and realized I forgot to grab my cake hoodie my 25th anniversary cake hoodie So now we're gonna walk all the way back I'm so glad I got it when I did because when we came back, they're all gone. Look at that They are absolutely all gone. Wow There we go. We have secured the package. I love this hoodie. I'm so excited I think that is such a cool added benefit to be able to pick up your packages So you don't carry them around but I almost left the park without it I did put my name and phone number so I'm sure they would have called me if I did forget it Here we go We have made it over to Disney's Animal Kingdom Lodge and if you guys ever want to come visit the Lodge from the park You can just take a bus. They have bus Pickup locations right outside the park and it's literally only two minutes. I Have to say this is probably one of my all-time favorite Disney resorts. Look at this Animal Kingdom Lodge is just so beautiful We're gonna head down to the restaurant and scope out the situation a little bit I wanted to try to see look at the menu and then go over some more details and facts about it I'm sure lots of people just walk past Giko because Boma is right here and everyone knows Boma because of its you know Famous buffet, but Giko is literally right over right over here This restaurant is classified as a hard-to-get reservation. It's only open five to nine thirty five days a week I think it's closed on Tuesdays and Wednesdays and like I said, it's right next to Boma So like I mean I love Boma. I love the stews in there. It's a buffet serving This is a little bit different to table service and a little bit upscale, you know Here is a look at the menu and it's got some pretty interesting entrees. It's got the oak grilled filet I heard that the mac and cheese was really amazing with this. That's around fifty four dollars They've got a nice short rib dish. They've got some cool stews It's a statable fish a lamb shake It looks like the most expensive thing on this menu was fifty four dollars and then they've got the Giko salad some seared scallops Oh, there's the boar tenderloin. I was talking about this the other day I don't know if I'm gonna get that though and then a sweet potato grundit soup that actually sounds good All of this looks so amazing. So I'm I'm here for it. I am all here for it It is time to actually head on in it's our reservation time and I'm excited I'll show you a little bit of the restaurant. Also, if you didn't know Giko actually means the cooking place and they have the largest collection of African wines outside Africa and I'm kind of interested in getting that but they do have a Old-fashioned here that I kind of like looked at and I was like I got to get this it comes with a matted gas car Bourbon like a vanilla bourbon. So I'm all about the old-fashioned Here is the restaurant itself and it also has the largest hidden Mickey in doors And I'm kind of excited. They said it's over here by the stove or by the ovens. Oh, I see you already look at that Hi friends. How are you? I'm looking for the hidden Mickey and I think I caught it now You don't notice it until you see it. Look at that. That is a huge hidden Mickey Yeah, what are you guys cooking for back here? Yeah, how are you? Oh very fancy. Oh, thank you Yeah, are you guys making some of the appetizers here? Right now?  That's the most important thing A hot towel with rose water very fancy It's rose water that they put in this. Oh greet it with the hot towel with rose water. I I Decided I'm going with the filet because it has a bone marrow crust on it And I was told instead of getting the mac and cheese I can get lemon mashed potatoes And I've never had lemon mashed potatoes So I'm kind of excited to try those and I maybe have high expectations for this filet It also comes with a chocolate red wine glaze on it So this might be a winner a winning a winning filet Cheers The old-fashioned is phenomenal, I love it It's like the perfect thing for me and now we're gonna dive into some apps We ended up getting the wild boar tenderloin which looks really good, and then I think we got some scallops too Here is the wild boar tenderloin This looks actually way better than the wild boar sausage I tried a festival of the arts and it's got a polenta on the bottom there. It looks really good And then we got a bread service Look at that very fancy-looking bread service All right, here we go trying the wild boar That's so good, wow I don't know what that is on this boar, but it is so good I really do love it a lot totally tastes different than the boar that I had at the festival of the arts Like I said that was more like a ground-up sausage This is more like a filet and it is divine That's a fancy word divine A little giraffe bread and I like the texture already on it I That might be my new favorite bread Thank you, you're welcome. I'm gonna be honest. I never had a palate cleanser before So the fancy word is intermezzo Intermezzo? Intermezzo Mango sorbet with a tiny spoon Can you the castle I have a chair I have a chair Oh, it is very mango. It feels like dessert. Yes. I like this. I'm cleansing my palate Enough of the cleansing the palate business the food has arrived here is the steak and then we've got is this the fish It's grouper. Oh grouper very fancy. And then you got the short rib. I love that dish Look at that. It's a bone-in short rib too. I feel like I should have got the short rib Yeah, it looks good, but I'm excited for this. This actually looks amazing a little out of blur Oh, wait, there we go. And I will focus in yeah, this looks amazing. Look at that Very excited and that's the lemon mashed potatoes. Yes Speaking of lemon potatoes We're gonna dive in and they said that there was no dairy made with these no dairy That's kind of crazy olive oil. They said and I guess lemon some lemon Not often you get to try lemon mashed potatoes Very lemony very good. Those are honestly, I'm obsessed with these lemon potatoes We're gonna cut into the steak here. Look at that nice char Bone marrow crust in there. Look at that chocolate red wine demi-glace. This is actually I'm impressed Can this be a high-ranking filet for me? I mean It might be Here we go first bite Normally I'm not a filet fan, but I got to say I can't beat that crust the bone marrow It's just cooked perfectly and it's also cooked on an open fire too. I noticed all of their like Flatbreads are cooked on an open fire and then the grill is an open fire too So the char is amazing and the demi-glace here much too good. I think this might rank actually high for me That steak is phenomenal. I have to say I absolutely love the bone marrow crust on there I would rank the filet maybe in my top 10 like I know that's that's shocking But yeah, I mean it was exceptional and then the wild boar was good The whole experience was absolutely amazing now. It's time for dessert For dessert they for dessert For dessert they have a lot of cool kind of unique items. They have African drum beats They've got candied beats. I'm sure Dwight Schroet would love that. Honestly, I might want to get the beats But I'm too drawn to this Malva pudding they said it tastes like cinnamon toast crunch and it looks really cool When I seen it at another table there. So I think I might go with that. It's their signature item too.  Yes. Yeah, it's a mango Look at this Complementary put them in your purse And with that I think we are done here today Jico was amazing I absolutely loved it. The service was exceptional all of the little things that they add in there I thought were great like the palate cleansing and a little pineapple fruit tart at the end and My steak was phenomenal. The bore was good. The dessert was out of this world. The dessert was so good I was obsessed with it and even the bread service the whole meal was exceptional And it wasn't like that high-tier price either like it like $50 range But it wasn't like the $90 a person or you know monster of Paul's like $189 a person But I really enjoyed that it was great to hang out with Tim and Mary Ann and her family and They actually gave me this buffalo bills hat. They're from Buffalo. Even though I'm a Niners fan. I'm gonna root for Buffalo They're my AFC Companions there. I'm not the biggest Pittsburgh fan. I know I'm from Pennsylvania or stuff like that I'm always a Niners fan through to my upbringing there and Yeah, I hope you guys enjoyed the video. I enjoyed making it. We'll see you next time. Bye Just casually sitting here watching fireworks on my way home Look at that. Love it. Oh Wow red light Green light go I always like bringing my doorbell and Gracie freaks out. Let's hide over here What


Good afternoon Foundation staff, this is Dr. Theron Sherman. Today's briefing doesn't contain any objectionable material, so let's begin. Item Number SCP-6622 Classification Level 4 Containment Class, Esoteric Secondary Class, Archon Disruption Class, Dark Risk Class, Notice Special Containment Procedures Mobile Task Force Omega-43, Codename, Leave It To Beavers is tasked with both limiting public awareness of SCP-6622 and grounding 6622-B instances until nearby systems can accommodate them. Beyond this, containment is limited to regular feeding and care for 6622-A instances in high-risk regions. When possible, newly commissioned Foundation facilities are to be built in areas already serviced by SCP-6622 infrastructure. Description SCP-6622-A is a subgroup of American Beavers, genus Cantercanidensis, which builds and maintains SCP-6622-B. Only 6622-A instances will interact with 6622-B. Nonanomous Beavers will not approach. SCP-6622-B are beaver dams and lodges which generate electricity via unknown and abnormally efficient means. Each instance continuously outputs 6,800 megawatts, approximately the peak electrical demand of the city of Los Angeles. Presently, the Foundation knows of 12 distinct 6622-B instances. Power produced by SCP-6622-B is distributed into man-made electrical infrastructure via ferry rings, tree roots, and other naturally occurring pathways, entering said infrastructure at any point where it contacts fertile soil. Discovering the mechanism behind this transmission is a beta-level research priority. SCP-6622-A instances often dam at remote or inopportune locations. Mobile Task Force Omega 43 has partially incorporated 6622-B instances into the electrical grids of 96 cities, but gigawatts of power are wasted each month, both in preventing grid overload and the Foundation's inability to transmit more power to high-demand areas. Having better regulatory infrastructure for SCP-6622 is an alpha-level research priority. Attempts to establish SCP-6622-A captive breeding programs have failed, as has relocation of wild colonies, instances constructed on anomalous dams. Reproducing the anomaly in a controlled setting is an omega-level research priority. Discovery On October 11, 2020, Site 43 experienced catastrophic power overload. The first known SCP-6622 Lodge was later found on the Osable River, 5.8 km away. Over a period of 9 months, Genitorial and Maintenance Section Chief Amelia Torosian was able to convert the site to rely exclusively on SCP-6622 for electricity. Torosian's initial designs have formed the basis for SCP-6622 transmission systems worldwide, preventing an estimated 450 megatons of CO2 emissions since implementation. Afterward, Upon entering the Site 43 main power station on March 8, 2021, Chief Torosian found on her workbench a pile of dried mud and twigs cruelly molded into a heart. Torosian's request to keep the sculpture in her personal quarters was approved.


Construction costs would make up $220,000 to $250,000 of that, leaving just $50,000 to $80,000 for the section. Do you know what the average section price in Auckland is? Well, it depends on where the section is going to be. So that's a no. The average section price in Auckland is actually $300,000. I reckon they're in fantasy land. It's a nice idea, but they're dreaming. He says it simply isn't possible to bulk buy cheap land in Auckland. I don't know where you can get land for $50,000, maybe in Mosegill out of Dunedin somewhere. It's not realistic at all. I think it'll cost five times as much. It'll be cheaper to buy 100,000 caravans and put them on site and then you can take them away later.