Welcome to the Young Scholars Initiative. We have thousands of members worldwide and we are developing the future of economic thinking. We invite you to join us in shaping it. So here's how you can use YSI's resources to pursue your curiosity and to create the stimulating environment you always wish you had. YSI has more than 20 working groups, each of which consists of young scholars that share an interest in a particular research area. There's working groups on inequality, on gender, and on sustainability to name a few. And each are grappling with challenging questions of our time. Make sure to join some of the groups that interest you. As a working group member, you will get notified about all the projects that your peers are setting up. You can join live webinars with experts in the field and register for workshops, convenings, and other in-person events. Oftentimes getting selected to present means that YSI will give you a travel stipend to do so. With each YSI project, we involve and enrich our intellectual home. We shape it to be inclusive, collaborative, humble, respectful, and non-partisan. A playground of ideas. So once you're familiar with the community, we encourage you to run projects too. Pursuing your own curiosity and economics, you can raise your own questions. You can arrange the lectures that you'd like to hear and organize the workshops that you'd want to be a part of. When you have a project idea, share it with a working group coordinator. Coordinators hold responsibility for the group and can tell you how things would work. They can make you a working group organizer and help you submit your project proposal to the YSI management team. Once approved, the project is yours to run. So join YSI now by registering on ysi.inateconomics.org because your questions need to be asked and your curiosity is worth pursuing. And because together, we can create the economic thinking of the future.


All right, it's time to get started with this session. So we have two speakers in this session talking about Dirac materials. The first is Oscar Wafak, who will tell us about twisted biographies. And I believe the next talk is closely related. Jet Pixley will give the second talk about some closely related physics. Thank you very much. And thank you very much to the organizers for the opportunity to speak here. This work was done at the National High Magnetic Field Lab in Tallahassee and Florida State University and was supported by the US National Science Foundation. Jian Kang is my collaborator. He's a postdoc at the Magnet Lab. And this talk will be mostly based on this preprint, which was recently accepted for peer-reviewed publication. So let me start by describing the system. Imagine two monolayer graphene sitting on top of each other. And then you twist one of them by an angle theta. If this angle is small and commensurate, you will generate a new periodic pattern on top of the underlying atomic periodicity of each individual monolayer. And this new periodic potential will then reconstruct the plane waves where the valence or conduction electrons of the individual graphene, the famous Dirac cones. So because of this reconstruction, there will be new band openings. There will be new smaller Brillouin zone, because in real space, this is a larger unit cell. And now we can talk about filling these mini bands in these so-called Moiré Brillouin zone. By the way, this pattern is called the Moiré pattern. So as you can imagine, if this angle gets smaller and smaller, this pattern gets larger and larger wavelength. And in principle, we are now in a qualitatively new regime from the original monolayer graphene physics. Of course, this field has seen remarkable development since the announcement at the APS March meeting by Pablo Herrero-Harrero group from MIT of rather remarkable effects. So let me just walk you through the results and then try to put it in a perspective of what I just discussed. So on the top graph, the x-axis is the carrier concentration measured relative to the neutrality point of graphene. At that point, the conductance shows a dip, but it doesn't go to zero. So we are dealing with possibly a semi-metal there. As we increase the carrier concentration, at first the conductance goes down, but then in a small window of carrier concentration, there is a sharp suppression of conductance and actually an insulating behavior unless you look at very, very low temperatures, where in some of the samples, it can become superconductive in some others. It's not, and it basically becomes insulating. And then if you dope near this point, superconductivity comes in. And if you dope too much, you empty some band because you enter into an insulator. Now, similar effects happen on both the electron and whole-doped side of this system. So experimentalists argue that the insulating state that happens between the neutrality point and this point, where, by the way, we would have four electrons in the unit cell that span this doping range, that this insulating state happens at commensurate filling. If it happens at commensurate filling, presumably this is happening because of some strong correlation effects. And the natural way to think about this problem is then to start thinking about the interaction energy as being the dominant term in the Hamiltonian, ask what kind of an insulating state that would stabilize and then turn on the hopping as a perturbation to see which one of the multitude of insulating states, many body insulating states, get selected. So the natural first step in a theoretical analysis would be to understand what kind of Vanier states we can write down for these four bands and then start thinking about what kind of insulating many body states we could generate. So this talk is about the first part of it. It turns out that the Vanier states are highly non-trivial and there is some debate about exactly how to construct them. So I'd like to walk you through those arguments. This talk will, at this point, have nothing to do with strong correlations, although it is motivated by trying to understand the strongly correlated behavior. OK, so how should we think about this problem? Imagine that the blue is the large, brillant zone of the top layer. The red is the large, brillant zone of the bottom layer. They're twisted relative to each other and therefore the Dirac cones are not exactly on top of each other. Now, if the twist angle is commensurate, we will have a new brillant zone, the more a brillant zone. And these two Dirac points will fold into the corners of this new brillant zone. And similarly, the opposite, so this is coming from one valley and similarly from the opposite valley, the Dirac cones will fold into the same point. OK, now, because we have some interlayer coupling, this will introduce band hybridizations and therefore we'll reconstruct the band structure and for some special values of these angles for the experimentally, for the realistic coupling between the layers, it was predicted theoretically some time ago before this was discovered that some of these bands that you generate can become very, very thin in energy. And there are so-called special magic angles where some of these bands get very, very thin and I will try to make this precise. And so once you have a thin band, you have these narrow bands, you would expect that the correlation effects, the many body effects play some important role. OK, so that's sort of the basic playground in which I would like to discuss the physics. OK, and so in terms of those bands, well, there are, not including spin, four of them. But this would be the top of that four band structure and this would be the bottom. So these two are presumably the trivial band insulators and everything in between, if you have an insulator, should be interesting. It should be due to many body effects. So this in particular would be happening for two electrons or two holes per Moray unit cell. OK, so strictly speaking, that's actually quarter-filling. OK, so to build these vanes days, what we consider is the following structure. We started from a monolayer, let's say, described by the red, which is the bottom layer, and then the blue, which is the top layer. And then we twist the top one by some angle theta over 2, the bottom one by minus theta over 2, and then we introduce interlayer coupling between them. Now, we start from a situation in which the atoms here were in registry and we twist about this position. And we ask, what kind of point group symmetries does this new crystal have? Well, it's periodic because the angle was chosen to be commensurate. It has a three-fold rotational symmetry around the stacked site, the carbon registered site. It also has a two-fold rotational symmetry, which is perpendicular to this main axis. So imagine rotating in the plane of the two monolayers by 180 degrees. It will bring the structure back into itself. And this generates the rest of the group, which in this case, the point group is so-called D3. OK, so as I said, we start with an end with a situation where the so-called AA sites are in registry. Now, we can get to the end of this unit cell in two different ways. We can either track the blue lattice and the triangular unit cell vectors of the top monolayer as well as the bottom monolayer. We have to come to the same point. And so this commensurate structure is therefore described by how many times you go along each of these legs. And so a pair of integers m and n defines this commensurate structure and the twist angle. So we will be referring then to this pair of integers. So if these are some small numbers, well, then your unit cell will be small. These are large numbers, a little bit offset from each other. Then your unit cell can be actually quite large. So many of these band structures have been calculated before the discovery of the correlated effects. And the band structure that we calculate from the model that was published, for example, in Moon and Koshino's paper. And it is essentially a tie-binding model for this large unit cell where the hopping amplitudes, which can be within a layer, either top or bottom, or between the layers, are described by a two-parameter family of functions. These are fixed experimentally, as well as the rest of the coefficients, to match the LDA calculation for the large twist angle, so small unit cell, where the LDA calculation can actually be done. So this tie-binding model has been calibrated. Unfortunately, the LDA calculation for the realistic magic angle is too costly. And so the best that one has to actually compute the band structure microscopically is to fit these tie-binding coefficients for larger angles to LDA. They work there.  So that's what we did. We used these to compute the band structure. And as you can see, the progression, for example, is shown in their paper of these angles. And the set of integers that they show is shown here. So once you have these larger integers, you can see there's a new pattern, clearly new pattern that emerges. OK. Now, it turns out that the experimentally observed, yes, please. Not be able to extrapolate this thing, right? Or maybe at least? Well, they're apparently there. So experimentally. OK. So now, experimentally, it turns out that the magic angle is close to 1.06 or so as measured. And if you use this model to calculate the band structure at that angle, the closest commensurate integers, simple integers, to that would be 30 and 31. It turns out that for 30 and 31, this model gives band structure, which indeed gives four bands. But these four bands are not separated from the rest of the spectrum, I think, from the bottom. And so as a result, one doesn't actually get an insulator, as one should, at the bottom of these bands. Now, people who add the relaxation to this claim that, in fact, that does produce the gaps from the top and the bottom, but we were just simply taking the un-relaxed model. Now, if you want to understand the qualitative effects, like the shapes of the vanille orbitals, et cetera, well, I think you are justified in taking a pair of integers, which is large enough, although not necessarily exactly at the magic angle, where this kind of model does give you bands that are separated. And that turns out to be true for 25 and 26. So this unit cell contains now something like almost 8,000 sites, not quite the 11,000 sites that are there for 30 and 31, but close enough. But qualitatively now, you can reproduce these gaps. So the band structure that we get from here looks something like this. And I will discuss its details in a moment. Now, if we were to zoom in at the gamma point, which is the center of the Moray-Brillant zone, we would find that the two top energy states are degenerate. And if we check how they transform under the three-fold rotation, they form a doublet. They transform as an E representation of the D3 group. The same is true at the bottom of this band structure. On the other hand, if we zoom in to the corners of the Brillant zone, the Moray k and k there, we see one doublet. And this indeed transforms as a doublet under rotations. But then we see split pair of bands. And these are one-dimensional representations, A2 and A1. Now we have a setup that we can try to build our Vanier states for. We have a gap that separates our four bands from the top and from the bottom. And so in principle, we should just be able to construct such exponentially localized Vanier states. Now, before you computed the Vanier states, you looked at the block states. And from the block states, you computed the local density of states, let's say for positive energy. You would discover that this local density of states is peaked at a series of triangle lattice sites, the Moray triangle lattice sites. So this really suggests that the Vanier states should be centered around the triangle lattice sites. So perhaps you have some type of a triangle lattice Hubbard model or something like that. And there are people who try to argue for this kind of reasoning. I will argue to that this is, in fact, incorrect. This was also pointed out by others. Let me present the argument for that. So let us assume that we indeed have four Vanier states associated with the triangle lattice sites. Now let's say that these Vanier states are chosen to have a corresponding site symmetry of the D3 group. So if there is a point group operation that now acts on these Vanier states, there will be some unitary matrix which transforms between our set of four states that depends on which symmetry we picked. Now if we construct block states out of these Vanier states in the usual way and then act with this symmetry operation, we discover that if we pick triangle lattice sites, then the block states would have to transform the same way, this matrix U has to be the same, as the Vanier states at high symmetry points, the points which get mapped onto themselves due to these point group operations. That turns out to be a problem. Because once you fix your Vanier states, you fix this matrix. And what this is telling us is that at high symmetry points which map onto themselves, let's say under three-fold rotation, such as the gamma point and the corners of the Brillouin zone, this matrix has to be the same. But I just showed you that at the gamma point, we have two E representations, two doublets. And at the k point, we have one doublet and two separated states. So this is, therefore, inconsistent. So as I said, the representations at gamma and k, according to this, would have to be identical, but they are not. So this suggests, oh, and the similar argument was presented by Liangfu's group, and then Centel and Ashton Vishwanak's group before, although it was a slightly different argument. And so this suggests that we should not think of this as a triangle lattice, despite the fact that the local density of states is peaked on the Moray triangle lattice, but rather as a honeycomb lattice, the dual lattice of the triangle lattice. Now it turns out that this argument doesn't quite go through the same way if we pick those sides, because now under some of the point group operations, let's say rotation around this point, you see some of these Vanier states, they change position. You're no longer rotating around the triangular side, and they go into different units. That's why under this operation, there's an additional shift. The shift depends on which atom you're considering and which point group operation. That's why R prime depends on G and I. So now if you follow the logic I presented on previous slide and ask how do the block states transform, you discover that they don't actually quite transform the same way. There's an additional phase factor that you pick up. And this phase factor depends on this shift R prime. Now notice that this does not matter at gamma, because at gamma K is equal to zero and this factor is just one, and so at gamma, the block states should transform the same way as the Vanier states. And I told you that the block states are two E's, and therefore we should construct our Vanier states in such a way that they transform as E under the point group operation, the site operations. But because at the K point, this factor does no longer disappear, it's present, we have the desired results that the representations at gamma and K need not be identical anymore. Okay, so to construct our Vanier states, we're gonna focus on the gamma point where this factor is not present. And because the threefold rotation around the Z axis at the AA registered point does not change the sub lattice of the honeycomb lattice, the Vanier states must transform the same way as the block states. And so this means that U must be diagonal for G that is C3, so for the threefold rotation. Okay, and so now let's imagine that we apply C3 here and then ask what happens. Well, let us pick the one one element of our diagonal matrix to be a phase factor with two pi over three rotation. So that would fix the transformation properties of our first Vanier state, W1. But once we have W1, you see, we can then apply time and resource symmetry and that doesn't change the site, it will generate a new state, let's call this W2. And because under time and resource symmetry this phase factor will change a sign that gives us the second element of our diagonal matrix. Now to get W3, this is generated by a rotation around this dashed line. Well, we're just gonna be taken from one honeycomb sub lattice, more a honeycomb sub lattice to the other. It also tells us exactly how that new Vanier states, W3, should transform under threefold rotation. That's fixed for us, this phase is now fixed for us. And similarly, or lastly to get the fourth Vanier state we just apply time and resource symmetry to W3 and that forces this phase factor to be this. Okay, so now what do our block states look like at the gamma point? I'm plotting the absolute value squared for the top energy state at gamma which transforms under rotation such that it picks up a phase two pi over three. This is separated into the top layer and sub lattice A and B and the bottom layer and sub lattice A and B. So it has this peculiar, if you wish, propeller-like structure. Similarly, but it's smooth other than that. If you similarly look at the block state below in the doublet below which transforms as epsilon, also has rotation eigenvalue with the phase two pi over three then this side, I'm sorry, yes. This side is the A-A side, the corner here. So that would be the triangle lattice. Same, yeah, this is a triangle lattice. Yes, thank you for your question. Okay, so this is what the block states look like. Now we're gonna use these block states, this is just to give you a feel for what they look like. We're gonna use these block states to construct our Vanier states. Now the method for constructing Vanier states was developed by Marsari and Vanderbilt and to optimize their spread and we just simply follow that method. Perhaps I can walk you very quickly through how this works. Remember the Vanier states are related to block states by Fourier transformation. Now numerically, when you compute the block states, there is nothing that fixes the phase of the block state at one k point and the nearby k point. And so as a result of that, numerically or in practice, this block state will have a random phase in k space or at least that phase will not necessarily be smooth. And that presents a problem because when you Fourier transform that non-smooth function in k, you will not necessarily generate something that is nice and localized in real space. So you would like to make this block state as smooth as possible in k because that would then produce as localized over Vanier state as possible. Okay, so the way this is done is that you start with some trial functions. In the case of an isolated band, just a single trial function. That's usually something that has some nice smooth peak and but it should have the right symmetry of the band if you'd like to reproduce that. And then you project that onto the Hilbert space at a particular k point, spent by your isolated band. This is done this way. This defines your new basis which lives in the correct subspace of the block state you are interested in and has an additional advantage that it is smooth in k because this projector is smooth in k. So if you have a random phase here, well, the random phase here exactly cancels that. That defines our, at this point, complex k dependent term and to normalize it is trivial if you have a single band. Okay, now in the case, this is not our case because we have four bands. These four bands are sort of intermingled but they're separated from the rest of the bands by a gap. So instead of just taking one of them, we're gonna consider all four block states. We're going to find a linear combination of them and we try to make this linear combination as smooth as possible in k space. Okay, so there's a procedure for this. You have to find the single value decomposition of the overlap matrix of these states and from there find the square root of an inverse of this but this is a well documented procedure, okay? So as long as this inverse exists, in other words, none of the single value decomposition eigenvalues vanish anywhere in Brillouin's zone, then this procedure can be done and there is no obstruction to building these binary states. Okay, so this is what we follow. To get a good trial state, we construct them as following. We just take these block states and then we imagine them centered around the dual, the honeycomb lattice sites and we're just simply going to take a wave function that lives in here on the top layer in sublattice A. So from the top energy state and from the bottom in sublattice B and then mix in a state that lives in the different energy, the bottom doublet, but they're both coming from the same C3 eigenvalues. So that now our block state, even though it's mixed in energy, sorry, now our Vanie state, even though it's mixed in energy, transforms under rotations as it should. It should pick up an epsilon and should be an eigenstate. And so then we mix in these two together that gives us our trial state. We fit into this machinery, go through the single value decomposition and outcome the Vanie state after the so-called projection method. Now one can do a little bit better than that. One can try to then optimize these unitary matrices that you still have a freedom of acting with the unitary matrix and try to get the spread of these Vanie functions to be as small as possible. This was developed by Marzari and Vanderbilt, the so-called maximally localized Vanie functions. And you can do that, you will achieve about 20% improvement in the spread. Now, what do these functions look like? So remember, they're all centered on the honeycomb lattice. So the dual lattice of the triangle lattice. But interestingly enough, they all have three peaks which are centered approximately on the triangle lattice. So this immediately explained how you're going to recover the local density of states from something that is centered at the honeycomb lattice, but while the density of states is peaked on the triangle lattice. Okay, and I can tell you more about these if you are interested, but once you build these Vanie functions, they live in the correct subspace of interest in these four flat or nearly flat bands. And then you can sandwich the original Hamiltonian in this basis and compute your type binding Hamiltonian from that, yeah. Yes, this shows the triangle lattice. Just consider the middle of these triangles. That is the honeycomb lattice. That is the more honeycomb lattice. Pretty close to that, yes. Yeah, triangle lattice. The block state. But this is just a gamma point. Remember, you sum over all Brillouin's zone. So once you project it and then do the integrals with the projectors throughout the Brillouin's zone, this is the structure you pick up, see? So it naturally explains the peaks in local density of states and it doesn't suffer from this issue of the symmetry mismatch. Okay, so the story here, therefore, is once you pick W1 to be the eigenstate of C3, you can generate W2 by time reversal, W3 by C2 prime, which is the twofold rotation I discussed, W4, and by time reversal of this one. Okay, so all the symmetries that I discussed, the exact symmetries of the D3 structure are therefore represented locally in these Vanier states. And so how well can you do with this? Well, this is how well it reproduces the original band structure, just from the tie-binding model of these Vanier states. The blue is computed from the Vanier-based tie-binding and the red is the original tie-binding problem with almost 8,000 sites. Okay, okay, now I hope I have another hour to discuss the relation to the obstruction that has been discussed in this field. And so let me just tell you how this is related to this. So let me start with the claim. The claim is that if the system has a twofold rotational symmetry around the z-axis followed by time reversal symmetry, and then in addition to that, it has a valid conservation symmetry. Then there's an obstruction to prevent, to constructing exponentially localized states, exponentially localized Vanier functions which are symmetric in areas which transform locally under the operations of the crystal. Okay, so by symmetric I mean represent the crystalline time reversal and valley symmetries locally. So the question is how are we avoiding this obstruction? So this argument has been quite insightful argument has been pointed out by Senkel and Vishwanak collaboration. So where does this come from? How does one understand this? So imagine for just a second that we are going to conserve the number of particles in a particular valley. So now you twist the two layers and so naively because the two direct cones in the Morel-Brillant zone come from nearby valleys at small angle. In the one valley problem, their winding numbers should be the same because the winding number of this is presumably the same as the winding number of this on the other layer. And so in the band structure if you now get the two direct cones for each valley separately, then they should have an equal winding number. This is not the situation for each monolayer. In each monolayer the winding number at the corner of the Brillant zone k and minus k are opposite. But here the claim is that they are the same. I just made it clear that this is just a very naive argument but it is a lot more subtle but it's actually correct conclusion. So now remember for this to work you need to have a twofold rotational symmetry around the z-axis. That is true if you combine the twofold rotation axis that we have present in our structure and then an additional twofold rotation axis that's perpendicular to that but still in the plane of the bilayer. So it's only if we have this additional C to double prime combined with C to prime that the twofold rotation is present. Now to establish that the winding number is the same is not still enough to establish obstruction. For that one as they have done argued that in some open region in the single valley band structure that surrounds the corners of the Brillant zone one can define a K smooth basis on the two narrow bands and that in this basis the structure of the effective Hamiltonian has to look like this in other words it doesn't contain sigma three matrix. And then what one can ask is because it's now a smooth field how do the components of this field wind as you go around these two point. And in the presence of the additional twofold rotational symmetry the point in the middle between K and K prime so the endpoint turns out to have block states which are eigenstates of this twofold rotation symmetry with opposite eigenvalues. With that it is enough to show that N one under a mirror sorry under a twofold rotation operation that goes through the end points and therefore takes one to the other one direct point to the other under that operation it turns out that N one does not change sign so the X component of this phase field will stay but the Y component changes sign. So if you were to plot the zeros of these two functions under this symmetry it turns out that they would have to have the same winding number. Now so these two components of the argument not only force the phase field to lie in the XY plane but it also shows that the winding numbers are the same and there are but there are no periodic functions N one and N two that you could now extend throughout the entire Brillouin zone which would have two same winding numbers in the entire Brillouin zone but only two. In other words you would have to accompany that with opposite winding numbers. And this is the basis of the essentially a proof that there has to be an obstruction to building exponentially localized vignette states provided that the value symmetry which we needed to define this and C to T are good symmetries. Okay now it turns out that in the D three structure the C to the prime is not an exact symmetry and because of that the C to T neither is C to T and so therefore our if we did have valley number conservation then our effective Hamiltonian would have to be two by two but it could contain sorry about this typo but it could contain the sigma three matrix and that immediately tips this vector out of plane and there's no longer any obstruction. The obstruction only works if this is in plane. Moreover because we are on the lattice the U1 valley symmetry is not exact so there is some mixing between the two different valleys. So neither the valley symmetry is exact in the microscopic construction that I showed you nor the two fold rotation symmetry followed by time reversal along the Z axis are exact and therefore there's actually no obstruction. And so in practice when we look at our Vanya states after the projection but before the but before Vanya 90 which is essentially the maximum localization we find that a significant portion of the wave numbers that compose our Vanya states are actually coming from one valley but there is some spillage into the other valley in this case it's something like nine to one and if we then maximally localize it then it's something like five to one never do you have a perfect valley polarization. I like to connect this to the so-called obstruction issue that has been discussed by Soryanov and David Vanderbilt in the case of 2D topological insulators. So it is known that there is an obstruction to building exponential Vanya states if you have a separated band but that band has a finite churn number. Now Z two topological insulators do not break time in symmetry so they don't have churn numbers. So the question was is there any obstruction in this case? The immediate answer is that there shouldn't be because you don't have churn numbers but it turns out that if you try to build exponentially localized Vanya states in the 2D topological insulator with a Z two index that's odd and you require your Vanya functions to respect the time of symmetry locally then there is an obstruction. So you have to relax this condition that the two Vanya states come in Kramer's pairs. And that's because loosely speaking the basis that composes your Hilbert space at the two different K points is completely orthogonal. And so it's actually quite related but there is a way around it that they proposed which is that you simply pick up something that does not, you pick up Vanya states which don't go into each other locally they're not Kramer's pairs under time overs. Okay so I think I don't have enough time to talk about what happens to this band structure as you start twisting the angle closer and closer to the precise magic angle. They're actually quadratic band touchings that in itself has many body instabilities but let me just stop here and flash my conclusions so thank you for your attention.


Leia Black was born in 1956 in Waco, Texas. She's not just a reality star, she's a powerhouse mom, wife, beauty, and businesswoman. In the 80s, she launched the sudden youth skincare facial, and recently she stepped into the CBD beauty skincare world. But that's not all. Leia's journey includes two decades in the beauty and fashion, a stint on the Real Housewives of Miami and appearances on various talk shows. Plus, she's the brains behind the star-studded Black's annual gala, raising millions for troubled teens through her consequences charity.  From her hard-working parents, she learned the difference between today's era from before.


This video walks through the solution to lesson for practice exercise a Where you have a text file with some coordinates that looks like this And you need to loop through those coordinates parse them out and then create a polygon out of it to create the shape of a US state So this is a very basic list of coordinates. There's no header here. So we're just gonna blast right through it As if it's a CSV file with two columns and no header In order to do this, we're going to use art pie, which we import in line three and The CSV module for Python, which we import in line four In lines eight and nine We set up some variables referencing the files that we're going to work with in the script So we're working with the mystery state shape file that was provided for you The the shape file is in the WGS 84 geographic coordinate system However, there's no shape or polygon within the shape file yet You're gonna add that and Then line nine references the path to the text file containing all those coordinate points In line 10 we do a little bit of work to detect the spatial reference of the mystery state shape file We're gonna use that later when we start creating geometries This pattern of using the describe method and then getting the spatial reference property should be familiar to you from your work in lesson two In line 13 we open up the text file that has the points So the first parameter with the open method is point file path, which we created in line nine and Then the second parameter is our in quotes, which just means we're gonna open in read mode We're just gonna be reading it. We don't need to write anything to it Then in line 14 we pass that file into the CSV reader constructor so that we can create a CSV reader object That will let us iterate through all the rows in the file in an easy fashion Before we start doing that though in line 18. We create an empty arc pie array object We know in this case, we're just gonna be creating one geometry It's going to be one polygon with one array So it's okay to create that array before we start looping And actually we don't want to create it inside the loop because then we would be creating multiple arrays And we just need one of them in this case In line 21 we actually start looping through the rows in the file each row is going to be represented here by a variable variable called Quartz and so in line 24 We create an arc pie point object And when you create a point you need at least the longitude coordinate, which is the x value and latitude, which is the y Now why we while we typically say latitude longitude when talking about coordinates It's important. Remember that latitude is the y value and longitude is the x So when we're creating a point object, we want to make sure that we supply the longitude first then latitude Now going back to our file, you'll see that in the first column or column zero That's the longitude. So we have Quartz Zero in square brackets in our code and then the next piece is the latitude or the y coordinate And so that's the column with the next one. So we use Quartz one in square brackets for the y Once we have that point Once we have that point We can add it into the array And we just keep doing that cycling through the file until we've added point after point Until we've gotten to the end of the file Now in line 30 The loop is exited Which we know because that line is not indented And we're going to take the array that we made and make a polygon object out of it So we use art pie polygon to refer to the polygon class and the things we pass into the constructor in parentheses are The array itself and then the spatial reference for this geometry, which we retrieved back in line 10 So we've got the polygon all set up But we haven't added it to the shape file yet in order to do that We're going to use an insert cursor in this case. We just use a very simple with statement to create the cursor The insert cursor needs a couple of parameters in order to get started It needs the feature class that we're going to modify Which is referenced by the variable shape file Remember we created that back in line eight and then there's this couple of fields That we're going to modify in our case. We're only going to modify the geometry We're not going to modify any attributes So the way that we supply this tuple is it just has one item or token called shape with the at sign And then our line 34 actually inserts the row and it inserts just the polygon geometry nothing else as we describe When this is done, we should be able to go into ArcGIS pro and verify that we indeed have a polygon inside of that shape file Now one little quirky thing to note about this code is in the last line Where I supplied my tuple of values Corresponding to the tuple of fields associated with the cursor When you're supplying just a single value, it's necessary to follow that up with a comma If my cursor was defined with more than one field say two fields Then my values tuple would not need to end in a comma It's only when specifying just a single value that the values tuple needs this trailing comma Now up through ArcGIS desktop 10.6 This was the way to create polygon or polyline geometries At 10.6 it became possible to create these geometries Using a list of coordinate pairs Instead of an array object So here is an alternative approach to Creating our mystery state polygon Down through line 14. Everything is the same Then on line 18 Note that I create an empty a new empty python list rather than a new object of the arc pi array class Then inside the loop instead of creating an object of the point class and adding it to the array I simply use the list append method to add a tuple Representing the current coordinate pair Finally on the last line of the script I go on to plug my list variable into the insert row statement instead of an array variable So that's a little bit simpler approach to creating geometries and you should feel free to use it if you prefer Though either way will work and you'll definitely see the array of points approach if you're looking at pre 10.6 ArcGIS desktop geometry scripts


Time to go to the Vrables Trader. Dave Beganis is going to show us Will Leviss' fourth touchdown pass, where he basically threw Nick Westbrook a quinae open. Show us, Coach. I love the Vrables Trader. The Vrables Trader is really solid here. The first thing I want to do is show you. We're an 11 personnel, three wides, one tight, one back. Now, what we're going to look at here, we're in the slot to get rid of this. Right there, hit that again. This right here. Yes. All right, now, here's what we want to do. Where's my free hand right here? This is the dude that's going to get the ball butt. These guys right here, they influence what's going on. I want you to watch the coverage starts to roll outside up here to the top. It rolls outside up here to the top. Nick Westbrook a quinae is in the slot. They are playing man-to-man, but the safety in the corner are going to lean to this side. So it's going to put the deep safety man-to-man on Nick Westbrook a quinae in the slot. Are we ready? Now, watch it, because they're going to bring some pressure. And here we go. Let's get this all out of here. All right, here we go. Stop this thing right here. OK, now, here's who we're looking at right here. Right here, you can see, all of a sudden, Nick Westbrook a quinae has put a nod move, a nod outside and go back inside to the post. You can see everybody is caught up man-to-man here. This man right here is over the top out here for this receiver here. And look at the pressure on Levis right there, even before he's thrown the ball. He's going to get belted. This right here is key to this. As he moved outside to the outside, I want you to watch at the launch point, he has got nothing to step into. This is all arm. And it's really all arm from here to here. All right, hit it, coach. OK, are we ready? I don't need to hit that again, do I? OK, now here we go. Just roll it from the top. Let's roll it. Why do I keep hitting that, actually? Just hit the top one. Now, let's go. There you go. Boom. All the way, let's watch it all the way across the field. Now, let's take another look at this thing as it starts. We got a little white shoes here in the end zone. All right, let's let it run here. And I'm going to show you how far this ball travels. It's a 33-yard touchdown, but it goes 55 yards in the air. 33 yards from the launch point back here, but this thing travels 55 yards in the air. And the thing that's beautiful to watch is he leads him with 55 yards with a half-armed toss, which is really, really pretty impressive.


What's going on everyone Sam here with late in sports car driven 2022 punini baseball 3 and 2 10 box case number 2 random team style. Here's a look again at your random results As always you can find these on the website under the results tab as well as the link that's in chat guys We've got seven spots to go on our last case that has spots for three and two that breaks Not next but the case after number three you guys sold out in a blink of an eye which I Can't even uh, can't even tell you how uh appreciative I am. Hey, what's up chris? What's up chris? How you doing? 10 boxes. Let's do it. What do we got? Certified divisions break. All right. All right Okay, shane we'll get you north down to six six left Oh nasty Oh my god, that could be nasty. Uh, Sean it is Sean empty box Yeah, what's up my man? What's happening is it Sean? We've got paul goldschmidt 82 out of 99 Cardinals got hans kraus 13 out of 32 going to the fillies And then we've got sultans erin judge 4 out of 50 Very nice one there. Congrats. What's going on, brother? Um, let me see. I just want to see what this is numbered to Whoo, that's gonna be nasty. We've got a joseph gray going to the nationals. Congrats washington steven mc And then who's got this team? What a way to start brandon s We've got a redemption quad jerseys autograph hollow gold erin Boom What a way to start this case? Wow, that'll be numbered to 25 or less King kratz brandon Sick that is nasty The mvp right there We're on to number two Right. Uh, yeah, that's fine. Yeah, I don't want to take it out.  Yeah, sometimes it happens It's just like they do it in all the encased products when it's redemptions Uh, freddy freeman free out of 12 Dodgers on the gold Gold joe perez rookie for the astros 9 out of 12 And then we've got in a cunha junior 45 out of 50 That is going to the braves Sam we only have two spots left We only have two spots to go on Three and two baseball case number four Let's get that filled. Why not? We've got going to I believe this will be the milwaukey braves. This will be going braves Warren spawn eight out of 49 game used going to double check that make sure he goes to the braves Um, yes milwaukey braves So this reverts to the braves since they relocated to Atlanta So this reverts to the milwaukey to the Atlanta braves. So I'd say just probably tag that one. Yep And then we've got uh three color two color Gavin sheets rpa going to the white socks Six out of 10 nice one there. Congrats. Chicago south side Steven m a South side south side south side box number three Place to be is box number three Turned into a rapper overnight. I guess I did. Yeah, I'm dropping the south side tag line. He's a producer really he always drops that Yeah, oh dude This stuff is crazy Let's go, man mc sam mc sam on on the Mitch the beat right here Matt Olson 94 out of 99. That's a braves card There you go, Atlanta. Uh, henry davis on the prospect 60 or 89 out of 99 for the pirates Nice henry davis and we've got sultans 114 out of 150. Vladdy Guerrero jr We've got going to the fillies quad jumbo jersey briston stock nine out of 60 Which nine out of 60 and then if you flip it over It's still nine out of 60 Even one of one that's an ebay one of one. That's the ebay one of one reverse flip the card over right there Psycho uh player worn on that one Congrats fills and how about this go into alexandros s. How about a cow ripkin jr. Bat jersey auto One out of ten game worn and game used bat that is nasty Congrats alexandros on the iron man One out of ten the little hall of fame up there. Oh, so cool dude Box four give me some more Oh Oh I think we are I think we are One that's what it's gonna be one to go guys one to go Corbin burns 62 out of 99 going to the brew crew Tristan cassis that is uh 83 out of 99 that's going to the red socks on the prospect card And we've got sultans frank thomas 50 out of 150 Going to the white socks We've got Jumbo swatches ronald akunya jr. Seven out of 25 Player worn on the akunya. I believe that's a bat that looks like a bat Nice one there and we've got lone star swatches for the padres. That's jackson maryl three colored patch autograph numbered 23 out of 32 Wait, that's another ebay one of one two three three two if you flip the numbers around it's still 23 out of 30 Man, we are pulling ebay one on one. This is insane all over the place right now load it box five don't touch that beehive That's about as best as I can do on that I like onions that would not the chive As a bit of a reach I like my sports live. Oh, there you go box five. We are alive. There you go. There you go Waitin is live in the hive in the hive Oh, uh, that's goose goslin one out of two on the purple going to the nationals that'll revert washington senators to the nationals We've got cold wind on the prospects 83 out of 99 texas and full count frity freeman 20 out of 150 Going to the dodgers. We've got patch going to the rockies on the combo spot tim j Ryan belayed that is 30 out of 35 player worn I'm the belayed And then we've got rookie day one auto shame boss That is six 15 out of 32 We are sold out guys. Oh, that is awesome. Thank you so much Going to the rays joe v. Congrats, joe Uh, now we got to fill up luminaries is the next one Asha toko goes or asha tokyo goes don't quit your day job I dare you box six box six This is nasty We're rappers for years. Oh, yeah big time Sean and i are big time. Yeah Uh cool papa bell for the cardinals 19 out of 32 Jordan walker, uh 38 out of 99 Uh cardinals and sultans franando tatis jr Uh, I think that's some like gunk on there. I'm gonna send you guys an empty box and wrapper with the padres I don't know what that was all about Sorry about that padres. I am gonna make it up to you with this next uh, next couple of cards, so We've got going to the um pirates. That's ronzi contraris three out of five Get what uh, what's it called said Vaughn said three out of five on the ronzi And then we've got 24 out of 49. How about a cj abram's rookie jersey auto for the padres I think that makes up a little bit for it there james Nice one there congrats padres player warn That is wrong vaughn. I know vaughn just did this pretty dirty.  I know the chemistry isn't there anymore. Say sam is trash Ah Directed all towards sam you started it. It's true sitting here with a headache and i'm rapping johnny meyes 23 out of 25 matt brash one out of 15 going to the uh, mariners And brason stott 78 out of 150 for the fillies We've got chris bryant on the jumbo swatches. That's 40 out of 49 going to the Rockies gave more on that one Not pull the swag yet now Not pull the swag yet And then we've got seven out of 10 jonathan india Two-color patch auto on the second year congrats reds Reds is jeff n Player warn on the india I hear we can we'll just I guess probably be photos for the recaps now There's really no reason to do the jerseys Oh, that's cool I can see a lot of that that guy on that team warned that really Exactly. Oh game game use that man What if we have new listeners? I know jess shizzlem 11 out of 25 going to the marlins Nicoladolo 22 out of 99 reds and full count ronald acuña jr. 5 out of 150 Uh jerseys going to the reds. That is tony santian Uh 10 out of 35 Player warn and here you go mic D Going to the angels. How about a noa cinder guard 33 out of 35 Game worn jersey auto of noa cinder guard of thor Very cool. Congrats mic Now it's full timeless moments You got two boxes to pull timeless moments. That's what we got to do here zander bogart's 29 out of 32 for the red socks We've got kevin smith 4 out of 15. That's going to the a's And we've got sultans john carlos stanton 8 out of 50 Going to the uh yanks We've got paul goldschmidt uh 2 out of 35 and the full count swatches that is Game used game used bat Game used bat 5 way for the cardinals And then we've got rookie jersey auto. That is edward cabrera going to the marlins Uh 12 out of 25 player worn on the edward cabrera. Nice rookie jersey auto for you there the fish The fish of my end last box of the break guys Not really it's more of like a shadow box shame. Um that ricky henderson we pulled in the first uh break That's the timeless moments Um interesting kind kind kind kind kind rocking with it i'm rocking with it 10 to go on luminaries dual case random hit style That is paul goldschmidt 17 out of 32 We've got matt brash 77 out of 99 for the um Uh mariners and then we've got yachty 47 out of 50 all hand steam for the cardinals great photography on that And then we've got a jumbo swatches mookie bets 16 out of 20 game used bat congrats dodgers There you go la This one's cool actually Going to the royals we've got a quad jersey auto andrew benentendi That is two out of 49 yankees legend redsox legend royals legend andrew benentendi congrats kc All right guys, we'll do a quick recap just on our jersey car on our autographs We've got because it's all pretty much an all hit product Edward cabrera marlins. That is noa cinder guard, uh, angels jonathan india reds cj abrams padres shane bas for the race We've got jackson maryl going to the padres. That's cal ripkin going to the oryals. That's nasty, dude Gavin sheets white socks. That's erin judge on the redemption number to 25 or less And benny biceps andrew benentendi for the royals that'll do it for the break guys Thanks again for filling this up. We've got a couple more cases of this. Let's let's keep getting into it guys