Okay, so we can draw them. Okay, we can draw two of them, because I don't know how to draw P4 on a blackboard. And you have, for example, if this is x1, x2, y, x1, x2, x3, y12, y23, x1, x2, ya, maybe. Then the triangle associated with this graph will be, we have vertices, x1, y plus x2, x1 plus y minus x2, minus x1 plus y plus x2. Okay, one thing that you notice is that, for example, the midpoint between these two vertices is just x1 bar and the midpoint between these two vertices is x2 bar. And here is y bar. Instead for this guy, you will get this type of figure. This is for the loop. Okay, where the vertices will be given by, so remember always that you have these midpoints. And one thing that you will see that if these vertices x1 plus ya minus x2, and this is minus x1 plus ya plus x2, this point in the middle is ya. But then here you will have. This x1 is shared between this ledge and this edge, so it has to be common to triangle in the sense. So what you have is this type of picture.  And this other point will be x1 minus yb plus x2, in such a way that again this midpoint between these is x1 bar and this other midpoint is also x1 bar. Okay. So but in any case, knowing the vertices and asking to take the convex hull will determine this picture. In any case, drawing them is not something really fundamental because this is actually the only two examples that I can draw on the board. So for the rest just throwing them won't buy me anything because already for this, I'm going to go to before then. If you want to study the box diagram, this is going to live in P7 and so on. So making the drawing is not very useful. For this reason, so the main important message from the lecture yesterday that there is this correspondence and then which, as we learned from Julius lectures, each polytope as associated canonical form, which I will write it like this. So this is the standard measure on on the project space and this coefficient that is typically called canonical function is precisely what gives you the web function integrand that we are interested in. So you want to study this object, because you know that once you compute and once you relate the combinatorial properties of the point of to the properties of this function, we will have information about the properties of the wave function. Okay. I think this is the main message from yesterday. Questions about on this point if you have any of you want to clarify to get something clarified. Okay. Let. So, now, we want to understand both how to compute this canonical forms because is the way that we can compute the the wave function, but also how to structure the information without substantially do any any computation. Okay. And when when I said this, for example, what I mean is that one thing that we learned in the study in the analysis of the analytic structure of the function is that it has some poles. And this poles has a physical meaning that is typically given by either the the amplitude of the process if you go if you look at the total energy pole, or a factorization between lower point amplitudes and lower point wave function. If you look at the other the other poles. Okay, so the first thing we can ask is how we are supposed to see this information, which is the basic possible information that we can extract. So how can I know about this information, but just looking at the graph. Okay. Now, let's start with the simple example that maybe it's too naive, but let's let's let's do it in any in any case. So, the simplest sample as usual is discussed. Okay. So we know that this graph has the political process to this graph is a triangle with vertices, which live okay which lives in its own. And whose vertices are being by x one plus y minus x two minus x one plus y plus x two and x one minus y plus x. Okay. Now, what does it mean. So one thing that. The important thing is that this is also something that Julia mentioned is that if you take residue of your canonical form with respect to a boundary. You obtain another canonical form. Let me call it to a prime, which describe you. Which is associated to this boundary. What this mean in concrete terms is that typically, so in p2. So if you take this, this, this object, you can a boundary what is given is given by the intersection of the this poly top with a line, which contains one of the sites or more generally on the of the facet. Okay. So what we have to worry is that that is, you have your poly top. You have to intersect with an hyper plane, this intersection doesn't have to be zero and it has to fully contain the this poly top so the canonical form of of of the boundary will be given by just the one associated to this segment. Okay, will be that the one associated to this segment. So what this means is it means that whatever you have on the residue associated because remember that this corresponds to the poles of the canonical function. So whatever structure, whatever you get on the residue it's in a sense the consequence of the structure of the vertices that you have here so the first question that you need to answer is that what's the poly top associated to that to the particular boundary that you are interested in. So for example, we are interested in the boundary x one plus x two equals zero. Okay. That means that I need to consider a vector, which is a covector really, which is as coordinates 101. Okay, you want to answer the question about, okay, what's this intersection. What's the point opposite to it and what what are its properties. Okay, one thing that you realize is that so what does it mean so you have what does it mean that so you have to answer the question, which vertices of the cosmological point up are on this intersection. Okay. What does it mean that a vertex is on on on on on this surface this means that if I take, if you take, if you if you take the vector corresponding to the vertex. And you do the product with the covector, which identifies the side of triangle or the facet, this has to be zero. Okay. Why because it has to satisfy this condition. Otherwise, if it's not because the vertices will be on the positive side of the half plane. Identified by this by this line, then this product, this product, it is, it is to be positive so this if vertex is on this intersection, and this condition if the vertex is not on, please stop me if this you're getting if this is not very clear. It's important to have things clear because otherwise, you know, I will get exponential exponentially more incomprehensible. Very good, because in general, the way that you define this is that an arbitrary point inside the, the the polytope can be parameterized as a sum of coefficients with the vertices. Okay, and this coefficients ought to be positive. So what you're saying is that when I draw these lines. Okay, that's correct that yeah that's it's the because yeah these objects are going to be all the time positive in the sense of generation notion of convexity. Okay. So one thing that you assume is that you are taking precisely these planes and the orientation of these planes are such that this is the positive region. And these are the negative regions for these planes. So one point to be inside, need to satisfy this type of condition. And therefore, when you take this plane, you saw an arbitrary vertex as satisfy this type of condition either it has to be on the some facet, or as to be inside, and this tells you that it's inside. So on the positive region. Okay. Very good. Um, so in principle, you will okay in in this example is kind of of trivial but in principle, to answer this question, if I give you this graph, I can keep going arbitrarily complicated. You know, solving this conditions can be kind of cumbersome. Because I can give you a plane that I don't know which, for example, this is x one x one x two x three x four, etc.  And then identify which vertices are on what which vertices are not might be kind of not you do it but it says can be a little bit complicated. So, one is to develop a smart way of solving, you know, these conditions. Okay. So, an important point is that all the vertices as I recall in their cup has this this form. And here for all the ads. Okay. So, and the precisely is this triple is associated to the same to the same edge. So, one thing that one can do is saying that if this vertex is not on the faucet, which mean if this condition is valid. So, let's take the edge, which is associated to this triple that I'm looking at, and I put an X on it. I do, I introduce a marking if this is positive. If, and so this is where the site SC and this is site as primary. If that was so instead is positive. This is going to see as priming. And a marking close to the vertex SC, where here you have the positive number. And finally, for the other one, you might guess that I put a marking to identify this condition so this condition tells you that this vertex. The marking means that this vertex is not on the on the faucet that you might be considering and so on. Why this is useful. This is useful for for an important reason, the important reason if you remember from. I think two lectures ago but also I recall yesterday, there is a relation between graphs and sub graphs graph sub graphs and singularity, which means therefore boundaries of all these objects. So, for example, what we know from this condition is that the singularities are given by a sub graph was the singularity X one plus X two equals zero. This one is X one plus one equals zero. Okay, now we want to know which point of correspond to the intersection of this guy, we will have this line with our original point of this line with the original point of and so on and so which vertices among this tree are are there. So it turns out that you can make an association between this graphs and this marking so solves this condition of which one said actually are not on the faucet and which said are on the faucet and the prescription. It's just that when you have you consider in a sub graph such that the edge is containing it. Then you have to mark the one inside, which mean that in this case, the vertices, which will be on the this this this faucet will be just the two and three, because remember this I'm saying that I'm excluding the one. Here instead you mark close to the vertex which which corresponds to the, which is, which is contained in the in the sub graph. So you're excluding that soon. So this game by the tree, that one. And here the same you're excluding that tree, then you see that the the point of associate to this to this boundary is just the identified by the vertices. It's that one and that two. And this is the case because this is if you were to solve it iteratively and algebraically this conditions. This what we'll find but now, if you have these rules. I can give you whatever arbitrarily complicated can be if you are asking the question about what are the vertices so which are on the on the on the boundary. Then you will correspond to the full graph. Then you know that you have to exclude all the ones which are contained in the middle. So you know that the object which describe this faucet and so which describe this cut in amplitude is given by all the set of this that to the tree for all the years. Okay. So, knowing this rule, I give you a graph, you look at it, use this marking, and you extract this information. Okay. So you know that, and then the canonical form for this guy. So for the incremental form for the intersected with this. Let me call it. Let's say that I call this the sum of all the access. It's just the flat space amputee to the associate to the graph. Okay, let's say the function so I won't bother to write the dimension. Okay. Now, if you know this, then you can start to ask. So, what you know, is that a scattering amplitude from this graph point of view, it just identified by all the pair of vertices that are associated to this to the end points of the of an edge. So this, this type of z two and three. So knowing this, we can ask the question about, okay, what's happened when I look at other facets, and actually what's happened for example here. And this is actually what we did with the Neema some some years ago was to ask the question, okay. We have a way function, the way function is not Lawrence invariant and it doesn't have a manifest metarity because it's an object which is defined on a space like surface. But we know that if we take a flat space limit, we get an scattering amplitude that can be Lawrence invariance and is unitary. How do we see this properties in particular, for example, if you're asking about the flat space unitarity. How do you see that from a way function. You get the when you get to the total energy pool, the flat space cutting rules, for example. Okay. And then that boils down in really considering this this facet and then considering other boundaries on top of this facet. Okay. So question about this point. Yes. Yes. So when you started this whole competition, so you want to do the way function, and then you separated the part which you call like the universal. Yes, like the flat base. That's correct. Okay. For sure. Actually, when you look up look a loop graphs, this gives you a different representation because in when I started from the way function and having the object defined on a space like surface. When you consider the loop ampoules you have just depends on the special momentum you don't have a time part. When you take the flat space limit, we will have we will have an object in principle, which still has just the special momentum and doesn't have the L zero component so you will you might say that. Yeah, that might be something wrong because you want to see something which is actually Lorentz invariance which has this propagator one over p square in the in the integral. This, I think I mentioned it answering the question of somebody in the past two lectures, but what's happened is that, yeah, for scattering amplitude if you take the simple example of example of a bubble and it's cutting amplitude typically you will write that you have. Let me keep the notation small d for the special dimension special direction, and you will have something of this type of square. Plus, I have to. L minus p square. Plus, I have to. While I will tell you that, if I consider the way function, and I take the flat space limit. What you will have is something like this. This is why a YB x one x two. And then you have why a plus will be squared minus x one square. Okay, here I'm forgetting but the epsilon, but bear with me, and here you have why a plus maybe. Okay. So, you will say that this don't look to has like the same object. However, where again this for example my y is modules of L and YB is modules of L minus p. However, what's happened. I can take this object. I can split it into the special that a special part and the time part. The time part goes from one's infinity to cost infinity, because this are Cartesian coordinate. I do the same here. Okay. So I just split these things up. Now you can recognize this this is our way a square. This is what be square p zero is the modules of the external state there so the sum of the modules external state of the vertex so this is my essay x one. Okay. And now you can do the integration, you can choose in this case to contours, you pick the one that you like. You put things together, and you get this formula. So, obviously, if you start with a set up, which is not a rinse invariant. And even if you take a limit in which you land on something which is a rinse invariant you might should not expect that Lawrence invariance is manifest, so you don't get this type of propagator. So this doesn't mean that it's not manifested this mean that you have to see how to understand that that you're you're actually getting the right result. And so this type of representation actually, there are several of them. Okay, and goes under different names depends on how you split this up. So it's called look to do it, or cause representation, or, yeah, or, or even in this case, you can get the whole fashion, personal theory. There are, there are a number of us. Okay. So, um, That's correct. And I guess the question was down to actually integrate. Well, no, no. I mean, it does give you some information, for example, that there was this at arbitrary loops discussion representation was just conjecture. Okay, there was no, and in a paper with a collaborator of mine. His name is William Torres Bobadilla. We actually proved it at all loops using this technique for for scattering amplitudes. Okay. Because what you do if you want to, if you want to prove some any statement about scattering amplitude. What you have to do is take your arbitrary complicated graph for which you want to know something in scattering amplitude. Eliminate all the vertices that you don't want. And ask this question there. So really the techniques are really are useful. And also another thing is that we could prove on scattering amplitudes was having a combinatorial proof of some relation. When in the sense so some some relation as important that as I explained yesterday are the statement that partially overlap that double discontinuity over partial overlapping channel in the physical region vanish. So in a sense, one, I don't know if my nice thing or not, is that in a sense of the fact that you look at when you do the analysis on this object, the fact that you restrict yourself to the physical region is kind of seems to be in built because for example, it is already been known from the from the sixties, if I take for example the box, the double the partial overlapping channel women doing double cut in the essence each other. Okay. So, one will think that because of this system relation that this has to be zero, but actually can discover that that's the there is a non zero double discontinuity, but typically you have access to it going outside of the physical region. So, and I, for the moment, in this picture. I don't know how to see this part. Okay, very good. Okay, so, okay, obviously, you know what you don't want to do is. I give you that arbitrary complicated graph, you land in expression like this, and then you can ask the question how do I know that actually that's really Lawrence invariant. You don't want to take the integral for this, and then do for all the LZ. Very good. So, actually, that's actually something that we already proved that there is. So, a contour integral representation for the canonical form of a poly top. Okay. There are some numbers that I never remember in front so I just put this quick go that you can represent the canonical form so what gives you this object here as an integral over some variables one for each. Okay, this new is the number of vertices of the political. So you see in the in the case of the, the amplitude, which correspond to this you are a race you are eliminated one vertex for each side so if the total point of love three any vertices. You will have to any vertices or let me put here just to any, and then here you have this is the dimension of the space of the fine space. Okay. So, you can actually use this representation. And there is a number, typically, if you are in the case of loop grad items, this number is bigger than the number of dimensions of your delta function. This means that this delta function or this delta function will fix a subset of this scene. So that you fix you have some arbitrary choice, you fix them, and you are left precisely with the L of this variables, and then you get some rational function here. And so now that these are precisely the zeros up to some shift that you have to go. Yes. Yes. Actually, you can get even representations that I do not think that are in any quantum filter literature because, for example, Sierra from this this integral. Yes, yes. Yeah, you get the first version here you get this and you get all that the one that she mentioned. What I wanted to point out is that, if you do this integral, you typically pick two poles. Okay, so you get a sum of two contribution you can get you can close the contour in the upper of plane on the lower of plane so you get to the representation to different some here I wrote it just as a single piece. You get something, some function expressed in so sorry partial fraction. Okay, from the point of view of the of the political this means that you're doing a triangulation. So, answering the question, which representation do you have of your skating amplitude of your way function boys dying to see to study the triangle, all the possible triangulations of the political. There are some of them, which has nice diagrammatic rules, some of them does, which doesn't. And, and there are. Okay, the ones that have been studied in in this cosmological political picture are of two type. One substantially, which correspond to more or less doing this integration, we will be like the triangulation that you might have in mind like, I give you a square, and I can draw either this line, which will represent the spruce pole, or I can draw this line, which I'm drawing this precisely because in this case, this the the point of which represent the amplitude is precisely square. So, these two representations will begin by doing this, this type of of Nevertheless, there are also others, and so and this add the spruce spruce boundaries recalling in the P2 each line correspond to a denominator so you get this also gives you boundary but given that belongs to both cases upon the sum, this is going to be this this thing is going to be cancelled. However, you can do also another thing let me draw the triangle, the square like this. You can also make a triangulation through these points. Okay, and the specialty of these points is that are they identified by the same plane that identify the facet. This means that if I write this square as the if number one two three and four as one for a minus two three a, I'm always keeping the same lines so I'm not adding spruce singularities. So, I'm having a partial fraction of my integrant, which doesn't have spruce singularities. And this type, for example, in this simple case, this precisely I mean for the bubble, you don't have too many choices. So, one of the, so you can triangulate through a or through B, one of them gives you official perturbation theory. And the other one gives you what is called the causal representation. So, in this case, one gives you the fine man. Actually, this is too. Yeah, the one gives you the fine man representation the one that you will get just from the bulk, the using the explicit form of the propagator, like the advanced part plus the retarded part. While the other one is something else that I don't know how it's called. A bigger graph, obviously bigger it is more triangulation it has, and so more more more way of representing you have. Okay. So, in a sense. So one can say that representations of either the way function or the amplitude are given by triangulations of the full cosmological point for wave function. Or the intersection of political logical point up with the relevant hyperplane, which typical this we call it just not to repeat all the time I do the intersection blah blah blah. So, when I say that one can prove. Cossky rules from a combinatorial point of view or one can prove stem relations for amplitudes or actually for anything from a pro combinatorial point of view, both these problems map in the problem of how the two different singularities talk to each other in the sense that you see in this case of the square. Okay, you have here some linear condition here and what you have another linear condition condition you have four linear conditions which identified the, the, the square. The vertex of the square is given by the fact that the intersection, let's call this hyperplane one, hyperplane two, hyperplane three and hyperplane four. We'll begin by the fact that your political intersected with this plane intersected with this with the, with the, with the, with this other plane is not empty in codimension. In the sense that you can mention to means that I mean posting to condition one or two. Now, instead, when I look at these points. Precisely, this, they do not satisfy this condition, because they, they, the intersection of these two planes is outside of the point up so you can say that the intersection between, for example, plane three and four. Doesn't lie, it's actually empty in codimension to, which means that you're taking a double residue, which is zero, which is non zero here and zero zero here. So, if you want to ask the question about the category rules, you have to ask this type of question saying that one plane is the one that gives you the total energy pole, and the other plane is whatever other singularity which is both on the amplitude and on the scattering facet. Instead, if you ask him about the stamina relations in the for flat space here actually should be substituted directly with the scattering facet, and these are as to be planes which identify partially overlapping channel, or maybe you can do some general analysis say there's any pair of of other planes which correspond to subgraphs. And then you can use this graphical technique to actually answer this question. Okay. So, the way that actually I can, given that I have like 13 minutes left. So, I'm going to tell you how the flat space unitarity will work, because then this is more or less the same the analysis is a little bit complicated more complicated but the idea the underlying idea it's, it's the same. Yes. So, as Lorenzo was saying, this as such, so if you really then end up computing the canonical form gives you the, the, the way function integral that is like the flat space with function. Yes. Not, I mean now because he asked me something about the discarding amplitude I'm making the parallel also discussing amplitude but in principle if you consider full point of gives you the the integrand for the for the for the way function. Well, in the center space. Okay. Okay, in the center space. In principle, as. So, let's go simple. Let's consider the this option principle you have an integration from some big x1 cost infinity of the canonical form and this is what the cosmological point of the gives you. To this integral, you might, sorry, this is the one, as I said, the yesterday this alpha encodes, not only the cosmology are in if you're in the seat or something else, but also encodes the number of point of the interaction and the dimensions, the special dimension of where you're in. So, you can, for example, if you choose some information like this in such a way for example if I were to choose to do the computation in the center one plus three. But with the fight to do for this is actually completely conformal so you don't have any integration and cosmological point of already will give you the final result.  So, just, for example, I keep the seat in one plus three. And then, I have a fight with you. This alpha is one, so you get justice integration. And in this case, you can, then, I don't have time to go into detail, but there are operations that you can do on the polytope to actually extract what are called the symbols which means you means that you can extract the type of function that you will, you will have doing some projection operation on the point. So you still have the possibility of extracting information about the integrated function. Now now these symbols, which unfortunately I don't have with the nine minutes I don't have to go through but gives you tells you what are the branch cuts and the type of function that you have and typically this technology to work if you have stuff like poly logarithms, so this integrate to dialogue. Okay, and typically, they gives you the answer up to an ambiguity that here you can fix requiring that the limit from why going going to zero of this sorry. Is zero. Okay, so the, so that can be, you can be also fixed. So, for, so this is the case which has been more extensively studied. And obviously, if you start to increase the powers here, if you start to increase the powers of the outside here. Then you have to be more careful because you might run into IR divergences because which are divergences at infinity, which are infrared just because this x, this axis going to infinity really correspond to the time going to zero to the boundary. There is this representation that is this strange thing that sending some sort of things that looks like energies to infinity is infrared and not higher. It's infrared and not UV. And so you have to be more careful with that that's the reason for which so far, more general things has been a little bit has been a little bit less study. So, if instead this power is negative. Actually, you can trade, you can trade the integral for derivatives so it's actually very much simpler. Okay. So, in a sense, you once you know this information. So, this is where we know that everything is doesn't have divergences, you can actually do operation on the poet ops to extract the integrated function. There are the other case so that a simple case where you get to the answer by derivative operator just means that you know the answer. They're just on the operator that you know what it is that you act on it so in a sense to the cosmological point of will determine the final way function. Finally, there is what is the more complicated case that when you have this some sufficiently high power here. It's the cases in which you can get the infrared divergences and so you have to that this is the problem in that you have to understand how how to treat them. And these are divergences are the one which will correspond to this secular divergences that are given by the cut off or at the boundary. So, in a sense, it's true that once you do all the analysis, you are saying something about an integrand, but then you can still extract the information about the integrated with function. And in the cases that where you have, you know, you can, you can argue that you have polylogs, actually you can even all the inform all the statement that you can make on the multiple residues can get all the statement on the multiple discontinuities so you actually know not just the analytic behavior of the integrand, but also the analytic behavior of integrated function. And that this is typical the case for this axis integration in this in this cases, what we don't know yet is what's happening when actually you do the loop integration because this is has been. So, in the case in general is not very clear which space of function you have to understand and so you really have to study how the singularity of the integrand get crossed by the counter integration to make any statement about the this is continuity of final all the final answer. And I think there are four minutes left so prefer that you keep asking question rather me adding information. Yes. So, at least techniques, maybe not so much useful anyone in the case was, you know, what you can do. Well, well, it depends because for example, for if you take five to the four and in the one press three dimension or depends what you want to compute. Imagine that you want to compute even a super in function, okay. This you can put yourself this is the free result. Then you start to add this. And you can have, you can go to two loops, and so on. But this is precisely when you will start to see secular divergences. And so you might ask actually ask the question if this picture is useful to understand the the meaning of desire divergences how to deal with that if you can resummit or not, etc. And well, this is actually what I'm working on with my PC student and for the moment the answer is that yeah that there is some utility of this techniques for this. Okay, so again, obviously, if there are things that you know to compute another way and it's easy, probably there is no much point in just looking at this thing. However, the reason for which for me is also useful to extract information, simple information that you can expect in other ways it says to check that there is no loophole in what I'm doing. And so you really have to extract some of the information also for consistency of the full picture. No, it's not. It's not. What is not the pointer is not the point. Point doesn't point. Point doesn't point. I have to press it. Okay, very good. Anyway, we don't have particular rush on time. There is no much right on time. Okay, if I can grab 10 minutes, I'll be happy. Thank you. Thank you. No, no, no. Okay. Okay, okay. So we are very happy to have the power. Okay, thank you. I would like to thank the organizers for the opportunity to be with you and I felt obliged to extend a little bit the title of my talk giving the conference and in particular the input that have been shared over the week. In this talk, I would like to try to define a computational tool which can be applied from particle physics to any to compute any sort of cerebral that meet the dramatic approach and in particular to special functions that appear that are very ubiquitous in physics and not totally. Be based on development of intersection theory. So, when we start considering the evaluation of magnetic field in electron class electric magnetism. It is a law to define the magnetic field the flow through a surface. And in particular, if the geometry of our circle or our circuits are complicated, one aspect this calculation to become complicated. Actually, thanks to geometry, even the calculation of any field into from a configuration that is highly not really I can become extremely simple, and in particular can be broken down in individual contribution who's coefficient comes from geometry. And in particular, I'm talking about the linking number, which is the number of windings of the basic circuit around the boundary of the surface. So this linking number is the main building block and main building ingredient we are going to elaborate on so reading classical dynamic essentially we are in presence of the composition in terms of basic functions whose coefficients are dictated by geometry. How this is related to a famous integral calculus we have already been listening to several talks over the weekend about the application of integration by parts identity is two dimensional evaluation of dimension regulated integrals and we know that although in introduced for particle scattering nowadays their application to effective field theory true effective field theory to gravity and we listen to several talks to cosmology allows to consider them really as a as a syntax as a language rather than as a toolbox for calculation rather than as a function of tools that are limited to a special to a special problem. So the basic part of the and the basic application of integration by parts identities. We find them in the decomposition of amplitude in terms of what we can now define as a basis of integral or even to the compose or to derive differential equation for the master in the as themselves or in terms of differential shift functional the current relation. So the outline of my book is based on trying to convey the underpinning mathematical structure that connects. Interest and I will try to share what we understood so far, meaning how this can be controlled by the Ram twisted theory and particular intersection number, which I will just recall I don't have a limited number of limited time to go through three different methods that we propose to compute intersection number and that I'm not satisfied yet with them with them and eventually go to application range from basic functions to find an integral correlation function which are important from quantum field theory but also for cosmology and on a special integral which came into view during this week and which we applied successfully. This theory on. So let me start directly to say what we are found in particular when the integration by parts identities were proposed and have been used for 40 for 40 years, they were suggesting the existence of a vector like structure for Ferman interval, but this was only only only a suggestive a suggestive consideration in order to define really a vector space you need to find an internal scalar product between integrals in order to enable the decomposition and eventually you want really to consider a completeness relation among the element of the basis. So, the two big question that we answered was, what is the dimension of the vector space that the talk will be called new, and what is the scalar product between the integrals. And to answer to this question, we relied on intersection theory was proposed for so called twisted period integrals. So any integral that is the pairing between a twisted cycle and a twisted co cycle with the property that the function multivariate function that appear in twisted cycle, vanish on the boundaries. Is extremely good candidate to apply in the section theory, and what the main goal became then our research is to try to fit as many examples we know from physics on mark into these form, so that we could eventually benefit of the geometric property deriving from this representation. The, in particular, since integration by parts identities come from the notion of the vanishing of a total divergence. We started, I mean these integrals obey integration by part identities, in particular, any integral of a total differential of M minus one twisted form can be shown to vanish upon the introduction of a covariant derivative. So where, in particular, giving you namely the definition of the twist log you acts like a Morse function, and this omega which is a pivotal concept is the deal of view will be characterizing the full geometric property of the integral and of the space. So what happens that owing to the vanishing of total differential one can easily identify integrals that although that integrals that give the same results, although they integrate differ by total surface time we will say in this language by a covariant derivative term. Stokes theorem allows us to also obtain the same value of the integral if we modify the boundary this time instead of the integral and leaving the expression of the total integral still an alternate. So we have the possibility to define homology at the homology space of the closed form model on exact and they let me say do a representation they do a space in terms of boundary. Now beside the twist dictated by you, we can also consider you inverse. And in this way we can define other two spaces which are related to a dual covariant derivative, where omega is replaced by minus omega. We are in front of so we have in particular four spaces of co homology dual co homology and then homology and dual homology, which contains elements that can be paired up together to form integrals dual integrals intersection number for cycles, which are the natural intersection of Gaussian in numbers where intersection number between integration contour and what matters for us. And what we elaborated on is instead intersection number for twisted co cycle or better differential form regulated differential form in this vector space, we can imagine that somebody gives us the dimension of the vector space, then we can introduce co homology of the co homology group and dual co homology group and define identity metrics resolution of identities, which, which are built in terms of intersection matrices between the basis, which play the role as the metric in this space. And we can do it for forms or for contours and for the rest of the talk we will more focus on these property of forms. So the resolution of identity in co homology space is what controls the linear and quadratic relation or the relation that appears among the integrals which are the composition formula differential equation and higher and shifted dimensional relation. That comes when imagine that in this representation you have a new master integral where new is the number is the dimension of the co homology space characterized by a given twist you and characterized by which is common to all the integral of the of the differential form and the goal is to decompose any integral in the in terms of master integral. Then these can be done by exploiting the decomposition of the differential forms, instead of working at the integral level, we work on the differential form. Here we have a differential form belonging to the co homology class, we apply the identity metrics, and we obtain directly the composition in terms of our basis of forms, where the coefficient can be computed by intersection . So this formula for us will replace the integration by parts identities and allow us to project directly any integral on an integral basis working on or using the intersection number for differential forms. So now we can really now that we have an inner product we can really speak of having defined the vector space of a feminine but more generally or any twisted integral. And the main and the two main ingredient is that the number of master integral will correspond to the dimension of the co homology group and the intersection number give us the probability to define the inner product. So how then one recognize that that feminine integral cannot admit a treatment can be treated in the language of intersection theory where any feminine integral beside the momentum representation cannot meet also parametric representation. Among the various parametric representation, the various parametric representation are usually characterized by a certain graph polynomial raised to an integer power. And what we would like to find is some representation in which the integration by part identity holds. The bike of representation as this property and the bike representation is an application where the integration variable are the denominators of the integral and where the twist is the grand determinant of external internal moment. So for instance if you have a one loop nonagon or a two loop box, they both are represented by a twisted period integral with with nine integration variable. And for this it was known that in this representation, one can derive integration by parts identities. So among the three special application which we are interested usually in the different in the context of feminine integral is how to derive the menstrual recurrence relation, where in terms of twisted form language this become just a problem of the composing of the composing an integral characterised for instance a master integral characterised by EI its own differential form whose integrant is multiplied by no power of the twist function. So we can choose in particular a basis of other masters where the arch which are characterized by different powers of the of the twist and apply our decomposition method. For this special base choice, we can cast this decomposition as an equation of vanishing equation and recognize that the insertion of the grand determinant under the integrant sign actually what they do they generate shifted dimensional. In a similar fashion one can derive differential equation, namely the action of an ex derivative with respect to external variable on to a twisted integral generate a differential, generate a differential form in which one introduced a new covariant derivative, where this time you take the delog derivative with respect to the external variable itself. Once you generate this form, this form still belong to the same n-comology group and can be decomposed as any other differential form, and therefore the one derive directly the differential the faffian differential matrix obeyed by the master integral by projection technique in terms of intersection number. A third equation which is less known to people working on the evaluation of faffian interest but it's more familiar to people work on the mathematical side on GKZ system is the so-called secondary equation in which one can show that the intersection matrix is solution of a matrix differential equation ruled by this omega check, the omega tilde, which are the matrix differential equation of basis and dual basis. So, once we have these these understood that in order to decompose an integral and amplitude or to build a differential equation or regarding the relation for our integral can be done in terms of intersection number, then it matters how do we compute the intersection number. So, something from the most function log of you, as I said, the dimension of the homology group can be computed by counting the critical point of this function, namely the number of solutions of omega equal to zero. Omega also contains information about the poles, so omega contains the geometry, the full information geometry associated with this integral, and the intersection number indeed exploit. Very technical point for the screen sharing. Thank you. Thank you very much. The intersection number can be computed in two moves. Ultimately, it is computed as a residue, computing residue around the poles of the Morse function, but the function of which we have to take residue on is built on the form on the right form. It is not on the left, but rather on a different on the solution of a differential equation of a protein, which we call potential. Nama omega psi equal phi life. So we first have to solve these differential equation around the poles, then plug the solution in this product, and then compute the residue. The formula is valid for intersection number, namely for integral, which are one fault integral. The problem is to extend the evolution of intersection numbers to higher higher number of integration variable. And in the case of the logarithmic form, the intersection number can be computed via global residue theory. Or very efficiently we discussed once with Julio, and this can be done also by means of companion metrics which eventually avoid to go around each pole. The main problem is that we want to apply this technology to interest, which not necessarily have these beautiful mathematical properties like being differential logarithmic, I would like to apply to generic meromorphic form. So the one approach that we try to apply is to consider the integral as a vibration, so as an iteration of multiple integral function, and apply the composition in terms of master integral one layer at a time from the most internal layer to the most external, is recovering the full decomposition. Thanks to this idea, we were able to propose an iterative method to compute the intersection number of n forms, like skip the details, but again to derive the formula of intersection number for n forms in terms of intersection number of one forms exploit the identity resolution metrics for the internal states. For instance, let's see an application to a rather known, first, probably non trivial mathematical function, which is the three F2 a per geometric Gauss integral. This integral is characterized by the twist, which is built out of, which is built out of product of a polynomial simple linear, linear term so would say in a plane arrangement and from the definition of omega, we can count that there are three master integrals for this class of function and two master integrals for the internal layer, and we can build either a grid relation of Gauss a per geometric type, which would be the analog of integration by parts identities for this class of function, or differential equation, applying intersection theory and for instance, for this choice of basis, we can even derive a canonical system of differential we applied this technology to to one loop, to one loop integral here you can see for instance the masterless box function which can be decomposed in terms of master integral, and the most recent application to the enthusiastic contribution of very clever PhD student who were able to consider a planar and non planar planar and non planar for point function hexagon masterless hexagon or with one mass hexagon even going to have the guns. As I said, we have developed more more. We have benefited from the mathematician contribution, we are also helping them providing computational tool. Now we have another way of computing intersection number, we generalize the, the, the univariate representation. And in a very recent paper we benefited of the construction of the module theory to build the matrix differential faffian omega, omega check, we can solve the secondary equation finding directly the intersection matrix as a solution of a matrix differential and then we can build directly the coefficient of the composition by matrix multiplication, essentially by solving twice the secondary equation for specially intersection matrices and and deriving the coefficient of the linear decomposition as the last row of this product of matrices. So, beside the mathematical, let's say that the interest in developing efficient computational tools for intersection number, we tried also to apply it beside the final interest. And in particular, it began like a joke to apply it to certain class of function in quantum mechanics for instance and quantum field theory. We focused essentially on functions or internal function that have this twisted period form. And eventually, if the original definition is not regulated, we introduced a regulated twisted, so to have a full regulated twisted period integrals and applied the composition formula in terms of master integrals. So, from the composition formula of the differential forms, we, as said, arrive to the composition formula of the integral itself. And among these we identify this class of example we identify two paradigmatic case, one are the orthogonal polynomial relation, and the other one are the matrix element in quantum mechanics. Both these type of integrals, as we see, will be computed simply as a problem of the composing one form in one master integral. So, the dimension of the homology group is either directly one, and therefore we have a master integral, or as we will see in the case in which it is two, the coefficient of the chosen second master integral will turn out to be zero by an intersection number. We were able to recover all the orthogonal relations for Lager, Le Jean, Cheviger, Gegenbauer and meet polynomial, simply remapping the canonical integral form that we learned during the introduction to quantum mechanics course into regulated twisted period integral, and the differential form is built out of the product of the polynomial themselves. And for instance, for each of these case, for each of these type of polynomial, we identify the twist, the dimension of the homology which turn out to be always one except for their meet polynomial, which is a Gaussian, regulated Gaussian integration measure. By intersection number we define and find the C matrix, which is a two by two matrix only in the last case, in all other cases it just contains one element. This is the regulator choice. These are the master integrals, and these are the coefficient of the combination. From orthogonal polynomial, the analysis to matrix element of harmonic oscillator, and then as we will see of hydrogen atom, we know that the wave functions in the definition of the wave function in the quantum harmonic oscillator can be built in terms of a meet polynomial, and therefore we mapped the canonical, the standard, I would say, definition of integral in quantum mechanics in the twisted period integrals, and we applied the composition algorithm that I showed you before, and we were able to recover by intersection number all the standard result which usually are obtained by direct integration. We did it also on the radial wave function, we applied also to the radial wave function harmonic oscillator, and together we said we enjoyed reproducing some formula in the Landau book. Yes. The contour is still from zero to infinity, stays in zero to infinity, but I use it as a U, so this is what we are going to do, but I use it as a U, so this is what comes out of the definition of quantum mechanics. quantum mechanics then I built this inside the radial inside the radial wave function you already have this twist okay this is non-regulated so I multiply it by z to to draw and then I take draw equal to zero at the very end so at this point my integral is regulated between zero and infinity I create a multivalent branch cut locally in order to apply in order to apply intersection theory I don't do integral I compute the intersection number but I need regularization thanks for the question in order to be able to apply so I slightly modify my integrand and then if you if you want this is one way of seeing but I don't want to see like in a sense it's a one way of interpreting I want to say that I want to see it that once I have a regulated integral okay and this and the important thing is that you vanishes on the I build a twist that vanishes on the boundary okay this is the key property I don't need even to know where it is or what this is explicit form it's sufficient to say a form it's like integration by parts and ends when we derive them we don't care at all about where is explicitly the the integration domain okay here instead the integration domain the integration path is connected to the definition of the regulated twist and for instance with z to power rho I regulating the integration from zero to infinity for this and then we went a bit to study correlator function in particular for quantum fusion theory it was proposed already by vansel in in 2020 and also concept with tau function this because correlation function by definition okay can be read as a coefficient of the decomposition of the numerator with respect to the denominator okay so the the green function or a correlator function they are coefficient of a linear decomposition of a twisted period interest once you make this connection explicitly then you have assumed the tools for computing it and so for instance we start with a toy model phi to the four where the free action contains the quadratic term and interaction term the quartic terms and we mapped each fields to just a coordinate in order to to to derive a toy model with one-fold representation and the calculation then of g n of the endpoint correlator function simply boils down to the composition of of the form z to power m dz in terms of a basis function when the basis function is just one always integrand equal to one okay and for instance so in order to decompose to apply the decomposition of this problem we used exactly the same twist as in the hermit polynomial with the dimension of the homology space being two the choice of master integral are one and one by z and it happens that the coefficient of one by z it tends to be always zero so by intersection number we are able to derive this uh uh known relation for for classical uh uh phi to the four theory and then you can ask uh how how can you compute uh uh the the green function in uh for in the full theory for instance by perturbation theory and so in perturbation theory you can expand uh uh the integral on in the numerator in and define the various endpoint functions in terms of coefficient with respect to the same master integral as before and compute the correction this would be the next leading order in epsilon where epsilon this time is the coupling in this theory simply by intersection number you not only can apply this method for in perturbation theory but then you can also attempt to solve uh uh uh the exact theory and if you consider then the full action as exact then what happens that your twist really generate an homology group which is a larger dimension instead of having two master integral you have four master integrals this is the matrix differential and then you can again find relations between endpoint function and an independent number of endpoint green function the business goes on and and and we applied it to this non trivial uh case proposed in earlier nineties by itsinson and zuber the to to compute the coefficient of the tau function and we did not have to change much again it's again the same structure that we saw for the mid polynomial we could recover an independent alternative expression of the results by intersection number then before concluding let me just mention an inspiration that came during during this week certain point in the in the talk of p mantel it was shown this integral brilliantly explained how to derive differential equations of this integral and it was captured in court by by the the the the the the number of boundary or the number of boundary region it was speaking about and and it was too suggestive not to attempt to to redo the analysis in the context of in intersection theory so we went from these uh uh uh ben in casa uh like uh uh like uh integral to its twisted regulated form okay where we consider the numerator and then the product of linear factor here raised to a regulator gamma at the end of the day we take gamma equal to zero out of these twist we derive the the the number of critical points and therefore the number of master integrals there are two master integral in the internal layer three master integral in the external and we chose them to be the product of this denominator any alternative choice is okay but with this choice this is the sorry the intersection matrix obeyed by this differential form with this choice we could derive differential equation and for instance they admit a canonical form and you can see they are epsilon factorized they contain only simple pole and this structure as paolo so mentioned before and it's not stretch it's suggestive okay so i i i can't my conclusion to say we started our analysis from integration by parts and then this explicit calculation of of internal related scattering amplitudes so in a pure quantum field theory context we define our integral these integral are fully characterized by graph polynomials the from the graph polynomial and then the geometry of the space uh we could define a vector space generated by master integrals and this integral obey integration by parts and then this differential equation and also quadratic relation which i had no time to explain but the underpinning structure which as i say is common to many other theory is that in differential algebraic geometry you can define a wider class of integral or moto gelfand g k z per geometric integrals where the graph polynomial is replayed by any multivalent function described by the fine variety connecting to it what the vector space role is played by the homology group and the basis can be either classified in terms of independent integration contour or as we so as in the independent differential form so the integration by parts are continuity relation differential equation are farfian system and and and and there exists also quadratic a remand period relation so there is a a full one-to-one matching so i conclude my my talk by saying that we believe to have implemented we have found an underpinning important structure of generic mathematical function which are of interest in physics as well as in mathematics and the the the key concept here is to find the isomorphism with with the ram twisted group so and in many disciplines i'm enjoying this you can find that functions can be mapped into this into this context so i like to close this this seminar by saying that i hope that this new insight can offer some new connection between physics geometry and and statistics and that this tool can be of benefit for our communities thank you very much oh thank you actually usually integration by parts identities the choice of the basis is an option that is a posteriori namely you first decompose it right then you know and check how many master in there you have and eventually then you can change your basis to to have here the knowledge of the notion of the basis is an input is a priori because when you start the twist okay from the you function you get to know what is the dimension of the homology so you know really beforehand what is the number of master in then you choose them you choose and you build the the the the c matrix namely the intersection matrix in order to know if your choice is a valid one you want to this matrix to be invertible so you check the determinant and if it is vanishing means that your basic choice is not valid so then you choose the basis and once your your basis is a valid one you apply the composition at the end of the day because if the choice that you make is the same as the one that your colleagues has done in momentum space the coefficient are exactly the same as the ibp the advantages of this method is that is that the you don't have to go under the huge system solving it's a conceptual advantage because i want to say that there has been a tremendous work done in improving the system solving solution nowadays therefore you still these methods are canonical in the way that are faster and to find the relation but slowly with this improving so i don't think yet we have the best argument for computing in the section number already with one we propose we have we manage to to to really make a lot of progress in computing time okay and therefore my hope is that by finding the the best algorithm in the section number and using also finite field reconstruction method it helps so much solution with ibp we can be at least competing i'm more really very happy i think the differential equation for the heptagon at one loop i don't know was not derived recently was sold the exagon at one loop so we are getting there okay and and i see that we need so we don't know well enough the the mathematics namely the formal theory is there but to translate it on something to the computational efficient it requires some effort and so that's why we are putting our effort okay being the last speaker of today let me know at least the last external speaker let me take the occasion to thank the organizer for the wonderful workshop that they put together okay we did a summary so yeah i think it was a very very nice week thank you very much for coming i learned a lot especially over the cosmology time and hopefully each interaction will continue and would eventually emerge to some new insight the program doesn't end we still have informal discussions in the afternoon so i don't think it's over yet nothing to add for me on that point however there's a nice thing we still have on the on our agenda which is that the the icdp has a nice feature of awarding some small prize and this is for we had this poster presentation which i think was also very successful but you know many many discussions and so there's a poster prize awarded by the icdp through the organizers i should say that it so happens that several of the posters have some kind of connections to my institute so so i think that the house of the nima should you know have the believe on this that i am so i happen to have here a little diploma a poster poster prize diploma but i think it's appropriate yeah it's not on the zoom or it's four a.m. in the morning so this evening but after much thought we decided to award and the prize for the best poster to my team so i go ahead okay so so that's it have a nice rest of your of your visit here and yeah hope to see you soon some other occasion thank you thank you so by one of them . . . . . . . . , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , 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Hello, my goal today is on setting up tests. Only for years. I don't know why that's so funny to me. My goal is to kind of suss out all of the things that need to be set up and some of the concepts. So, that's it, that's what we're doing today. Pretty simple.


Hey guys, Thunder E here, you guys asked for this video and here it is, the P40 Pro from Huawei versus the Galaxy S20 Ultra versus, yes I added a few more, the LG V60 ThinQ and the iPhone 11 Pro Max. Now we're starting off with the front facing camera, we're going to switch to those other two devices, all shooting at 4K60, and then we will go ahead and record some more, so let's switch. So in terms of audio, both phones, the P40 Pro and the S20 Ultra, do a good job with audio recording, now we're looking at the rear camera's 4K60 and you can see the S20 Ultra has better stabilization than the Huawei P40 Pro. When it comes to colors, especially in the horizon, the sky, the P40 Pro is much punchier, the S20 Ultra is a bit more muted and more balanced I would say. Now in terms of this backlit shot here, I like what the S20 Ultra has given me, but in general both are still doing a good job, though stabilization is much better on the S20 Ultra. Now let's switch over to the V60 ThinQ and the iPhone 11 Pro Max. Alright so now we've switched over to the LG V60 ThinQ and the iPhone 11 Pro Max, you're going to see how of course they handle video walking with the front facing camera 4K60. Now the LG has three different audio modes, what I'm using now is called Voice Bokeh which focuses on my voice directly, so it should sound better than any of the other devices, but now we're going to walk and finish this and then switch to our rear cameras for both devices. So moving over to the rear cameras are both the iPhone 11 Pro Max and the LG V60 ThinQ. The first thing you notice is the iPhone stabilization, it is really really good compared to of course what you're seeing from LG here, which is actually quite shaky. In terms of color tones, they look quite similar in terms of color tones and similar to the environments that I actually shot it in. Now in terms of stabilization, I ranked the iPhone number one, the Galaxy S20 Ultra is number two, the Huawei is number three and the LG V60 ThinQ comes in at number four, but you've got to just be impressed with how well the iPhone is stabilized as a video camera. Now when we look at shots from the front facing camera of all these devices, the Huawei P40 Pro here at the S20 Ultra with its portrait mode do a really good job with edge detection. Also skin tone is really good, though the Huawei looks like a more polished version of me. I do like what the Galaxy is still doing right here. Shirt colors are also quite similar to what the shirt color should be. LG, the shirt color is off, also my skin tone is more orangey and there is some edge detection issues at the top of my head. The iPhone keeps the shirt color image, the shirt color correct as well as also my skin tone to match. Edge detection is pretty good all around with just a few places here and there if I was going to nitpick. Now as we transition to this image outdoors here, the Galaxy tends to darken me quite a bit but its edge detection is pretty solid because my finger isn't blurred out while the Huawei P40 Pro is blurring out my finger but everything else seems solid, the colors of the Huawei actually spot on right here. Now moving over to the V60 and the iPhone, both of them have brightened up the image quite a bit and you can see the edge detection issues with the finger as well but again the whole image is brightened. I do prefer the V60 over the iPhone, what the iPhone is doing here is kind of fuzzy in the sense. Now this is more of a backlit shot and you can see again the darkness and the contrast the ultra brings into the shot though it still keeps a nice focus of blurring the background. The Huawei does a good job with me as well as also my background as well but it doesn't blur it out as much in this shot. Now the iPhone and the V60 have done something completely different, they tried to brighten out the image while the V60 has darkened me quite a bit and also my shirt. The iPhone has kept the shirt color intact but it's made me a bit more orangey and everything in the background is just super bright and blurred out. So in an era of multiple main camera sizes the Huawei here in this shot does a good job with this background bokeh but I like what the S20 Ultra is doing, 108 megapixels compared to 50 of course binning and all that fun stuff. Now when you look at the LG V60 here it's a bit brighter and I do like what the iPhone is doing here because of course the iPhone is giving a better representation of my skin color as well as also the color of my shirt so that's something I think you guys should just pay attention when you're taking photos with these cameras. Now this is a shot we'll get back to the P40 Pro and the Galaxy, they both look great though the color of the S20 Ultra is closer to the exact color of this setting itself. While the LG V60 think you here is much brighter and the iPhone has gone into this blue tint though they all look good but very different color palettes that all these cameras are bringing to the table especially for this shot which we'll see later with some low light shots in this video. Now here's an ultra wide shot, this one thing the Huawei does, it tends to punch up the colors I do kind of like it with the ultra wide because you're trying to capture a lot of a landscape the Galaxy is a bit more muted here in this shot especially this view of Brooklyn the iPhone is also quite muted and the LG is a bit pumped up not too much though in terms of the the color palette but all look good in the ultra wide capture so moving to this main shot here both the Huawei and the Galaxy look pretty solid the Huawei colors of course a bit punchier but that's fine we're gonna look at this path here straight down with all cameras as well as the V16 the iPhone 11 Pro Max here both of them are a bit on the wash have a bit more washed out look but the colors still look fine now we're going to go into zoom first we're going to a 2x zoom for the LG V16 the iPhone you can see is a bit much brighter you can see more of that path in between and this is the maximum zoom you can get without going to your digital zoom but we're gonna check out a 5x digital zoom so we can match up with the other devices so with the V60 thank you it's a bit brighter you can see the two buildings in the back quite well that ends a bit more focused than what the iPhone does with its 5x digital zoom I prefer the V60 image it just feels a little bit more sharper now when we go to the 5x of the Galaxy and the Huawei the Huawei has the natural darkness which you would see because of the way the Sun is at this point in time the Galaxy brightens it up so you can see more detail with 5x zoom here and when we move into 10x zoom though the Huawei I think is just a sharper image even though there's more shadow the the Galaxy is sharp but it's a bit faded on the building on the right so you have that in there now the Galaxy can go into 30 in and of course 50 in 100x zoom and while the Huawei goes to 50 now when we switch over to the front-facing camera at night now this is quite interesting the Huawei and the Galaxy you can clearly see more of me on the Galaxy than the Huawei the Huawei image is very noisy and when we get into more lighting situations the Galaxy lights it up a bit more and just the imaging comes out much better on the Galaxy now using the front-facing camera with the V60 in the iPhone surprising with the iPhone the iPhone is completely dark the V60 is noisy but you can clearly see me you can certainly see me wear the Galaxy buds a much better image overall but the iPhone once you start getting in more lights the iPhone basically of course looks a bit better right there now with the rear cameras for both devices here we have the Galaxy now again this is post update so I've got autofocus and you can see better focus stabilization of course is not as smooth but still good the Huawei is brighter but very noisy image overall again this is all recording at 4k 60 and then when we move back to the LG V60 in the iPhone 11 Pro Max the iPhone here is more stable though the image quality I think though brighter not as sharp as the V60 V60 is not as stable as the iPhone and the imagery also though sharper but not as bright as what you have okay what about night mode photos with the front-facing camera now the iPhone doesn't have a night mode so this is a shot here with the iPhone is quite grainy but you can clearly still see my face now when we move over to the other three phones the Huawei the LG the Galaxy the Galaxy has the best image it's less grainy it's sharper it's brighter you can see my face the colors of the two jackets I'm wearing the LG V60 is the second best though the color of my skin is more orangey in look while the Huawei is probably the least out of all three but still a good image you can still see me clearly but it's much darker of all three images here when it comes to low light selfies now trying to recreate a club like scene we have some pink light coming in and you can see the Huawei is kind of dark in my skin tone especially around my face and also dark in my shirt while the Galaxy has kind of brightened the shirt up and added and fused that pink light all around giving you a much clearer image I think overall with between these two and then looking at the V60 it's really taking a lot of that light in the the shirt color actually isn't bad compared to the iPhone where the top half of the shirt really looks deep blue and then you've got more pink infusion in my face as well making my skin tone look a bit reddish in general again it's a tough situation but wanted to see how what it would do in complex lighting and low light situations now this is a shot I mentioned earlier that we were going to get back to and we check it out in of course low light situations the both of both images look solid from both the Galaxy and the Huawei though the Galaxy represents the color of the building much better and less contrast and saturation as opposed to the Huawei which has that kind of shiny sheen to it still lovely images while the LG V60 very similar to the Huawei and the iPhone is similar to the Galaxy though the iPhone has brightened the sky and added a little bit more of a slight orange tint which isn't there it's just adding more of the yellow from the lightings around the city it's picked up now the ultra-wide shot here low lights both images look nice though the Galaxy is more representative of the colors in daytime and you can see more of the yard as opposed to that darkened feature set where it's still clear but very different on the Huawei P40 Pro while with the V60 thank you it's kind of added a bit more of a soft glow to it while of course the iPhone doesn't have low light capabilities in the ultra-wide camera hopefully that changes this year you can see it's a bit more faded on the V60 now this is of course the main camera lens and you've seen some of the similar features where I do like the way the Huawei looks in the background while the foreground the Huawei's colors are a bit off the Galaxy is much better and I like that that's the focus and it's much sharper with the Galaxy than it is with the Huawei P40 Pro now with the V60 thank you it brightens everything all the way through and I think the iPhone has the best background color palette while the foreground is brighter it's not as bright and see the Galaxy but it's still nice and it's a good composition image as opposed to what the V60 brings which is still good now our final set of shots here looking at of course this water tower we've taken a couple of times the Huawei P40 Pro image is really nice it showcases the fans issues the water tower the building the Galaxy is a bit more muted something that we used to be brighter maybe the update has changed a couple of things here and it's a bit fuzzy in the foreground while both the V60 and the iPhone 11 Pro Max do a really good job here again with the foreground the background while the V60 is really bright especially around the subject matter of the water tower and the iPhone's a little bit more balanced in between for the foreground the background images now when we look at zoom here the Huawei is brighter with the zoom at 5x the Galaxy a bit more muted but again both images I think I do like the Galaxy it's not as shaky image as what I have with the Huawei and then when we go to Tenex zoom this is where the Galaxy really shines you can see Tenex zoom on the Huawei it's really dark you can't see the subject matter quite clearly as opposed to the Galaxy where it's quite bright you can read the letter in you can see the water tower in clear full distinction so I have to say though this comparison has been quite interesting with the results you see here on the one hand the iPhone 11 Pro Max still does a good job all around but of course lacking certain aspects like no low light photography in ultra-wide and also the front-facing camera on the other hand you have the P40 Pro which comes in really strong with this imagery in terms of taking photos the front-facing camera does a really good job the rear cameras also do a solid job and the zoom also is quite impressive matching or getting close to what the Galaxy S20 Ultra brings to the table and ultra also is no slouch with its front-facing camera and its video occupants now LG on the other hand seemed a bit average for me all around but still solid maybe not as high as the other ones I did really good job with audio though especially in the voice recording with the front-facing camera as well as also video recording with the front-facing camera in low light conditions I think there are many parts where these smartphones do better than the other depending on what you're looking at but I think it all resides on what you want to use it for and that's why I'm going to ask you guys what did you guys think was the best smartphone for you here in this video if you have any questions or any comments let me know if you're looking to buy any of the items you've seen in this video use our links down below otherwise don't forget to like share subscribe and always enjoy your entertainment


I'd like to share some photos with you about deposition in Antarctica and in particularly in this area called the dry valleys which are at the edge of the ice cap near the McMurdo station where there's actually a bare desert land that is accumulating sediment. So we zoom in a little bit. This is Ross Island which is an active volcano. There's this beautiful glacier that's coming out and onto sea ice and so this is all sea ice and you can see it broken up into icebergs here. This area is much colder than Greenland and there's very little open water. In the summer you can get some streams and rivers flowing as well as open water ponds and the sea ice can break up but in general it's much colder. So we have flown from McMurdo to several places in the dry valley, Pierce Valley here and Wright Valley here. I'm going to show you mostly photos from Wright Valley and the flight and it really is quite different from Greenland in that it's again mostly frozen over and there are no plants at all. So this is a view looking south from the McMurdo station where it's mostly covered in ice and snow. Particularly this was a spring photo. It's still quite cold. This is called the Ferrara Glacier so the ice cap is up in the clouds here and it's filling the valley all the way through here. There's in general less sediment because a lot of the ice is coming off the ice cap and it's not melting as quickly as it is in say Greenland. When you get into the dry valley areas the ice cap again is back in the clouds and the Taylor Glacier is visible here. This is Taylor Valley but the air is very dry and so the ice sublimates it basically goes from ice directly into vapor and so the ice is flowing down but most of it just goes up into the atmosphere. There are also alpine glaciers that are from snow that are collecting on the mountains on either side and as you can see in this close part it's pouring down over the mountain and it has a lot of crevasses and breaks up and then it freezes to form glaciers coming down and these areas in here and this one back here are lakes. They have liquid water but they're covered with meters of ice and that ice is permanent. The edges around them can melt in mid-summer in part due to the heat of sunlight being absorbed by the dark rocks helping melt the ice but they do have that permanent ice cover and you don't get the open water lakes that you see in Greenland. So this is Wright Valley it's really quite dry and this is taken in the summer and there's enough melt water and enough absorption of sunlight by the rocks that you get liquid flowing water so this is the Onyx River. It's really just a stream but it's the largest river on the continent of Antarctica and it's flowing down and to the left in this image and it has a typical river form with a meandering and sometimes a braided channel. It flows into Lake Vanda so this is a picture I took standing on Lake Vanda looking up. The Onyx River flows through here into the lake providing it with its melt water. One of the nice features about Lake Vanda is it shows that really nice U-shaped profile. So when this valley was carved out by glaciers and when glaciers flow down a valley they have a wide erosive basis. The flow fills a lot of the valley like we could see with the Ferrara glacier early on and thus they're eroding all along the floor of the valley which tends to make them wide and flat bottomed. When rivers erode the erosive power is concentrated right in the stream golly and so they tend to erode the most there and tend to get V-shaped valleys. The fact that this valley is still U-shaped even though the largest river in Antarctica flows through it shows that the river has very little influence on the landscape compared to the ice. So this is looking in the other direction and from up a plateau so this is the head of right valley. There's a little ice covered pond and this is the edge of the Antarctic ice sheet supported by bedrock mountains and ice flowing off the top and reforming as glaciers down below but not enough to still fill this valley. There might be some ice under the debris right here it has the morphology that looks a little bit like it's been flowing as if it had had ice in it. So obviously there's bedrock around the edges but a lot of this valley is covered in in glacial till that was deposited when the ice was last in this valley and that till looks something like this. There are large blocks that stick out but a lot of it is a smooth surface with grains that are sort of a class that are sort of fitted together and you'll notice that this this rock has a very strange weathering profile to it and it's been a long time since ice was in this area and there's been a lot of wind blowing and the wind reworks the sand and abrades a lot of these rocks and then there's extensive freeze saw cycling particularly in the summer when the rocks can absorb enough heat to have very large temperature changes from day to night and that breaks up the individual crystals and particularly in pockets where there's maybe a little bit more humidity or slightly different changes in the profile and you end up with these very irregular weathering shapes. So if we look at the ground in more detail the till we can see that it is poorly sorted. So there's a lot of sand in here this is a cobble size grain as these are as well and in places where the wind where the till is protected from the wind the the glacial flower and the sand are still between the clasts and that includes if you dig down onto those flat surfaces at all you'll find a lot of very fine grains. So there are also places where the sand accumulates and this is one of them this from the one ripple crust to the other down here is about three meters so nine nine or ten feet and this is actually in granule sized sand and then as the you go further up in the image which is also uphill the ripple size gets finer and the ripple wavelength is smaller the ripples are closer together and smaller and the grain size is also smaller and the wind gets funneled into this valley and the very fine grains are too small to be deposited in this zone here but a little further up on the valley wall the wind is a little bit less strong and the and the coarse grains don't make it up there and the fine grains can actually accumulate. So these grains were all once part of the till that he wrote it out from other places in the valleys. So this is going back into right valley and you can see that a lot of the pebbles on the top are flat and that's from the abrasion of the wind grains flattening them off but we do have a lot of the big rocks sticking out with some really beautiful amazing sculptures created by the freeze thaw and temperature cycling and thanks for watching