for the invitation. So the last thing about this web seminar is that you have a much larger audience and you would have usually in the seminar. Before I start with the talk in earnest, let me indulge a little bit in the past. So what you see here is the top part of the title page of my very first published paper. I can see it was fairly short, six pages. You can see my mouse cursor here. And it appeared in Arifta Mathematik in 1992. So I wrote it while I was still an undergraduate in Munich. Cornelius Greiter, who was an assistant at the time, pointed me to this problem. So you can see it's about iterates of polynomials. You have polynomial x squared plus a with an integer a, you look at the iterates of this. And the question is about the Galois groups of these polynomials. And there was a conjecture I think by Odoni who conjectured that the Galois group, when you take the constant term to be one, so iterates of x squared plus one, it'll always be the maximal possible Galois group you can have for iterated quadratic polynomials. Concretely, this means that the nth iterates should have Galois group, the pseudo two subgroup of the symmetric group acting on two to the n letters. So two to the n roots of the nth iterates. And so Cornelius thought that I might be able to do something about this and turn out that he was right. So I did this. And so you can see the main theorem of the paper. So I was able to show that for, well, infinitely many integers, say integers that are positive and one or two mod four or negative divisible by four and such that minus a is not as clear. These Galois groups really are the largest possible ones. And of course, this includes the case that a is one. So the original problem. Now, at the time, this looked like a kind of isolated curiosity, maybe. And so the paper didn't have much of an impact at first. So I think got its first citation 10 years later, 2002. Um, but then, um, yeah, so two things happened. Well, yeah, the one is that people started to look at, um, I will read a lot of our presentation. So that's a term point, I think by natural Boston, you look at, um, how the Galois group of the rational numbers, for example, in this case, um, acts on rooted infinite tree. So in the case of these quadratic polynomials and the iterates, you get a natural action on an rooted infinite binary tree. And then you can look at the actual the Galois group. Um, and so one can phrase the theorem, I prove there that for, for these values of a, what you get is a subjective representation. So the Galois group gives you the full automorphism group of this infinite binary tree, which is maybe an interesting result if you look at it in this way. And, um, also there's this field of arithmetic dynamics that's become fashionable. Things was pushed down to some extent by Joe Silverman, um, which studies arithmetic properties of iterates of maps of an algebraic nature. So for example, polynomials and of course, questions about Galois groups or irreducibility, which is what we will look at in this talk, they naturally fit in there. And so this little paper of mine has been picking up citations at ever increasing pace in the last 10 to 15 years. And so last time I looked, and this was yesterday was just one citation short of making the list of my top five most cited papers, which I think find rather surprising. The nice thing is that what I did back then actually has some connection to things I'm doing now. And so what I'm doing now is mostly related to questions about rational points on curves and hyperliptic curves in particular. And, um, yeah, we'll see how hyperliptic curves come into this picture. So let me switch back to the present, um, and sort of define the notions that I'm going to use. And I want to say that if you have questions during the talk, please do not hesitate to raise your hand virtually or write in the chat. And then I'll try to answer them. So we want to look at iterates of polynomials for the polynomials, um, in particular. And then I mean, easy to see that up to some, um, FI linear transformation to conjugation by FI linear transformation, you can always write polynomial in this form, they just have x squared plus a constant. So in my paper, I used a for the constant. I'm switching here to C because it's of more usual notation. And I now see to be any rational number, not just an integer. So fc of x, this is polynomial, and then I can define the iterates. So I start with the zero iterate, which should correspond to the identity map. So it's just given by the polynomial x. And the first is the polynomial itself. And then the second I obtain by plugging in the polynomial into itself. So I take x per plus C or squared plus C. And then I continue. And in general, I have the n plus first iterate of fc. Well, there are two ways of writing this down. I can sort of plug in the nth iterate into fc or plug in fc into the nth iterate. And of course, I get the same result, but sometimes it's more helpful to look at it in this way. And sometimes it's more helpful to look at it in this way here. So I get a sequence of polynomials. And of course, the degree doubles in each step. So the degree of the nth iterate will be 2 to the power n. And then, of course, I can ask questions about these polynomials. So the question I was dealing with in this paper was about the Galois group of these polynomials in the sequence. In this talk, we will ask a kind of weaker question. So the question will not be whether we get the largest possible Galois group, but just whether the Galois group extensively on the roots or put another way, whether these polynomials are irreducible or for which values of the parameter C and which indices n is this nth iterate irreducible. So if I fix C, then I get this sequence of iterates of fc. And then looking at this way of writing the n plus first iterate, it's pretty clear that as soon as the nth iterate is reducible, then you have a factorization. And this factorization is inherited by the next one, since we just plug in something into the polynomial. And so there are basically two possibilities. One is that all the polynomials are reducible. And the other one is that they are reducible up to some point. And then the next one is they are irreducible, sorry, up to some point. And the next one is reducible. And then it has to stay reducible. So there's this kind of dichotomy. And well, in the case that not all the polynomials are irreducible, of course, one can ask at which point they stop being irreducible and become reducible. And then if you look at examples, you may come up with this conjecture. So as far as I know, this is compatible with all the evidence that's available. And this is sort of all the best possible conjecture that you can make in this context. And it says that if you only require the second iterate to be irreducible, then it will never happen that at some point later, in the sequence, the polynomial becomes reducible. So if the second one is irreducible, then all the later ones have to be irreducible as well. So why is this best possible? Well, it definitely does occur that, well, the first iterate, so the polynomial itself is reducible when you take c to be a negative square. And of course, you can factor fc. And then all the polynomials from the first one will be reducible. But it's also possible that the first one is irreducible and the second one is reducible. And it's not too hard exercise to work out for which ways of c this occurs. And this is this remark down here. So the first case is when the first iterate, the polynomial itself is already reducible. And the second case is when the second iterate is reducible, but the first is irreducible. So there's an infinite family of parameters where this happens. Now, but conjectually, these are the only two possibilities. And then as soon as the second one is irreducible, it should stay that way. Okay, so that's the topic. And maybe to take out the suspense immediately, I will not prove this conjecture in this talk, but I will improve what is known about it. Okay, so to be able to say something about this, we need to be able to detect if at some point in our sequence, we switch from irreducible to reducible. And to do this, we have this lemma here. So if we are at the point in the sequence where we have an irreducible iterate, so the end iterate is irreducible, but the next one becomes reducible, then this implies that the constant term of this next iterate has to be a square. So that we give a proof because it's not so hard and every talk should contain a proof, I think. So the first thing to note here is that our n plus first iterate is an even polynomial because we plug in x-vector c into the nth iterate. So it's a polynomial that only depends on x squared. And so if I think of the factorization into reducibles, and I substitute minus x for x in the factors, then up to changing the sign of the degrees odd, I get another irreducible factor of the same polynomial. So I get an evolution on a set of modic irreducible factors of my polynomial, the n plus first iterate. Now I can look at fixed points. So if I have a fixed point of this evolution, then there are two cases, the polynomial that's fixed is an even polynomial or it's an odd polynomial. And it's an even polynomial. Then it's a polynomial that depends only on x squared. So I can also write it as a polynomial that depends only on x-vector plus c. And then I have a factor here that's a polynomial x-vector plus c. So this gives me a factor of the nth iterate, which by assumption does not exist. If my fixed point is a proper factor, which it must be because I assume that the n plus first iterate is reducible. So I'm looking at the irreducible factors, and so the whole thing cannot be an irreducible factor. When my fixed point is odd, the only odd irreducible polynomial is x itself, and then for degree reasons it follows that x squared has to be a factor, and then I can argue in the same way. So whenever I have a fixed point, I get a factorization of the nth iterate, which I don't have by assumption, so there are no fixed points. If I have no fixed points of my evolution, then all the orbits have size two, and so I can pit one representative of each orbit and take the product, and I call this h. And then the other percentages I get by substituting minus x for x. Take the product, I get h of minus x, so one integer thing here is one has to note that the degree of h is even because the degree of the nth first iterate is at least four, power of two that is at least four. So I get h of minus x and not minus h of minus x here. And so the actual is that I can write this polynomial as the point of h of x and h of minus x, and then I can just plug in zero to get that the constant term of f n plus first iterate is the square of the constant term of h, and this proves the lemma. So I have a way of detecting when I switch from irreducible to reducible, but I should add here that this is only one direction, so it's not an equivalence, but so it gives me a necessary condition for switching from irreducible to reducible, and therefore a sufficient condition for staying irreducible when I turn this around. So here's an example that shows that it's not an equivalence. If I take for the parameter c, the value one-third, and look at the second iterate, I get this polynomial here, which has square constant term but still is irreducible. Okay. So if we use this to show that, well, if the second iterate is irreducible, then some others also have to be. So what is known about this? So we have to look at the constant term of these iterates. That's why I introduced this notation. So capital A n of c is the constant term of the nth iterate of fc. So starts with zero. I mean, the zero iterate is just x has no constant term, then f itself has constant term c, and then I get c squared plus c, and so on. And I mean, I can define it by this recursion here. And then the question is, well, is this, or for which c is this a square? And so the first, well, I will assume that the second iterate is irreducible. So the first interesting case is when n is two in the lemma, which means I have to look at the constant term of the third iterate, which is a three of c. And yeah, it's, I'd say something about this, what, in a few seconds. So one can show that a three of c can be a square only when c is zero. Of course, when c is zero, then all the a's are zero and are squares. And a four of c's is square only when c is zero or minus one. Yeah, when c is minus one, then all the even ones are zero and therefore squares. Yeah, how can one see this? So for a three, I mean, from what I've written here, you can see that a n of c is a polynomial in c with integral coefficients of degree two to the power n minus one. So for a three, I get the polynomial of degree four. And so this equation here defines the genus one curve. And it has special points, so it's an elliptic curve. And then one easily can determine that it has rank zero and five is racial points. And so I mean, it has three racial points, but two of them are at infinity. And so the only racial point is not infinity is the point that corresponds to c equals zero here. For a four, we get the polynomial of degree eight. And so we get, I mean, this equation is y squared equals the polynomial x defined hyperbolic curves. So when the degree here is eight, then this is a curve of genus three. And yeah, one can look at the Jacobian variety and the being variety of dimension three in which I can embed the curve. And so the group of racial points on this being variety is a finally generated to be included by model A. And it turns out that this group is finite. So there are only finally many racial points on the Jacobian and embedding the curve. They only find the many racial points on the curve. And I can find them by looking at the pre-images of the finite many points on the Jacobian. And it's not too hard to figure out that the only points here are given by these two x coordinates. So this is in a paper by Rave Jones, Nate Hins and a few other people. And at some point I also joined as a co-author. So I mean, you can find it in the archive and it also has appeared by now in the New York Journal of Mathematics. So this implies that when we have the parameter c such that the second is irreducible. And I should mention that when c is zero or minus one, then this is definitely not the case because then fc itself or f zero is x squared and f minus one is x squared minus one, they are both reducible. So in this case, all the polynomials are reducible. But if it's irreducible, then c is not zero or minus one. And then by the lemma, or then I mean, this proposition tells us that a three of c and a four of c both are non-squares. And then by the lemma, this implies that third and the fourth iterate are also irreducible. So it can go from two to four. And the main thing I want to do in the talk is to go one step further. So from four to five. And this basically means that I have to figure out for which there is of c a five of c can be a square. And the result will be that this only happens when c is zero. Again, now similarly to a three of c. Okay, so that's the goal. Now a five of c is a polynomial of degree 16, but it has no constant term. So maybe I just go back quickly to this slide here. Obviously, I mean, fc c squared plus c and then the square of something that ends in c plus c will always have zero constant terms. So it has a factor of c. And there is more convenient to work with an equation for the amplitude curve for the polynomial on the right hand side says odd high right hand side has odd degree. And so I basically just turn it around replaced a polynomial capital A five by lowercase a five bridges x to the power 16 times capital A five evaluated at one over x. This is now a polynomial of degree 15. Because the constant term of capital A five is zero. So the degree drops by one. And then the question is equivalent to to this. So I mean the points where C zero for y square equals a capital A five of x now move to infinity and the points that were infinity before now now move to zero. And so I have to point that infinity on the curve that comes from the capital A five. They move to zero. So they will be there. And then the other point moves to infinity. And what I want to know is that there are no other points. Yeah. So if my lowercase a five I can write in this form, I mean, it just comes from the recurrence. And so I'm looking at this curve, which I call a C was that equals to a five of x, which is this polynomial test degree 15. So you get, you know, seven here. And I want to know what the rational points are on this curve. The usual way of of trying to do this is to look at the Jacobian variety again. I mean, I've told you that for a four, they have this three curve and it's a common variety and it turns out that the group of rational points on the Jacobian variety is finite. Um, in general, when the rank of this group is finally generated within group of rational points on the Jacobian is strictly smaller than the genus of the curve, which is the dimension of the Jacobian, then you have a chance of applying Shabuti's method. Shabuti proved more dense conjecture that they're only finding many rational points on the curve of higher genus. In this special case, when the rank of the model of a group of rational points on the Jacobian is less than the genus. And one can make this approach at least say effective in practice. But this requires some some knowledge of the model of a group. So the first thing one has to do is one has to figure out what the rank is, or at least show that it's smaller than the genus. And the way of doing this is that one finds an upper bound for the rank by computing a certain suitable Selma group. And for hyperliptic curves, for the curvature of hyperliptic curves, one can compute the two Selma group relatively easily. And so in this case, if we do this, we get an upper bound for the rank, which is two. So we know that the rank of the group of rational points on the Jacobian is at most two, and two is less than seven. So in principle, Shabuti's method should work. When there are general results, it would allow us to deduce a bound on the number of rational points from this information. But this will not be enough. I mean, what you want to show is that this curve has only three rational points. They are the one point infinity. And the two points, when x is zero, then I mean, the right hand side simplifies to one. And I have two points, one with y equal to one, and one with y equal to minus one. And I want to know that there are no other points. But the bound that I get just out of knowing that the rank is at most two will never be three, but always be larger. So to really get at the points, I would have to know what the rank is, not just an upper bound. And then I would also have to know the right number of independent points in this group to do the computations I have to do. But the problem is that I cannot find points of infinite order in this group. So I mean, yeah, this is a seven-dimensional object. And searching for points there is not so trivial. And also, I just have this upper bound. So it's perfectly possible that the rank is actually zero. But I have no good way of proving this. So I'm sort of stuck here because I cannot go further and apply the method in the usual way. So that's what I mean by standard shabu-t techniques here. I also have to do something different. And this is this so-called Zener group shabu-t method that showed up in the title, which I will explain on the next slide. But first, I want to note that, I mean, the three points we know on the curve, the three racial points, the point infinity and the two points with x squared at zero, they all have the property that if I embed the curve into the Jacobian in the usual way here, where we take the point infinity as the base point, then, well, of course, the point infinity, the base point will map to the origin. So it has order one and the other two points also have finite order and the order is odd. I mean, it's 15 as one can figure out here. So we have three points of odd order. And in fact, these are the only points of odd order on the curve. Okay, now, what's the idea for the Zener group shabu-t? So I mean, basically, we don't have points in this monorail group. We do have the Zener group. And so basically, we would like to use the information we have. So the Zener group has a kind of proxy for the monorail group. And so we look at the following diagram. So for the time being, I mean, you don't really have to know what the two Zener group is, only that there's a map like this. So I mean, they have the curve, the racial points on the curve, we have an embedding of the curve into the Jacobian given by mapping the point infinity to zero. So this also sends racial points on the curve to into the racial points on the Jacobian. This is a group. So I can take the subgroup that consists of doubles of elements and look at the quotient. And then the Zener group, the two Zener group, there's an injective homomorphism from this group into the Zener group. So the two Zener group is a finite, a billion group that's killed by two, so vector space over the field of two elements. And so this is just the f2 linear map here. Then the next step is to look at the same thing over the two added numbers. So of course, we still have the embedding of the curve. We have the group of q2 racial points on the Jacobian, and so this quotient. And the Zener group, the two Zener group has a natural map into this quotient. So not just for q2, but for every completion, but we only needed, every completion of q, we only needed for q2 here. Now, the q2 points of the Jacobian form a two-adic league group. So I mean, I can do analysis over the two-adic numbers. And this two-adic league group has a normal map that I can arrange so that it has imaged the two-adic integers to the power g, g is the g is seven, in our case. And then I can, I mean, I can take the reduction module two of something in here, I get something f2 to the g, and sort of completing this query here, I get the kind of reduced logarithm map on this quotient. I'll just take it pretty much here, take a look at the reduction. So if we work with this diagram, and so the sketch of the proof is like this. So we need to know that this map sigma here is actually injective, which we can check because we know the Zener group, so we can compute it. And then our goal is to show that the set cq odd that I defined on the previous slide, which consists of the three known points, almost all of the set of virtual points. So I want to show that if I have a virtual point, it's not one of these. I get the contradiction, the point cannot exist. So how can I do this? Well, I take my point p here, I map it into the Jacobian, so this is i of p. In the simplest case, this point maps to something non-zero here, which means that it's not divisible by two, and then it will map to something non-zero here because it's injective, and also here because sigma is injective, and I get something non-zero here, hopefully, and I can check that what I get here is a lot in the image of stuff that comes from the curve down here. That's the idea, but in general, of course, it's possible that my point is divisible by two, and then I would get zero here, and this would not be sufficient to compute something. So what I do is I divide the point by two as many times as possible, so n is maximal, and this makes sense because the only points in the Jacobian that I can divide by two arbitrary often are the points of odd order, and these are the points I have excluded here. Yeah, I mean, I didn't write it on the slide, the previous slide, but these three points are the only points on the curve that map to something of odd order in the Jacobian, but any other point will not be a point of odd order on the Jacobian, and so can be divided by two only finite number of times. Now I can play the game I sketched before with Q, so Q is the point here, and basically I map Q from here to here in the two possible ways, going this way or this way, and then going this way, I see that it has to come from some element in the same group, so it maps to something that's in the image of this map, composed with sigma. On the other hand, yeah, so that's what I know, and the logarithm of Q is the same as two to the minus n times the logarithm of I of p, because the logarithm is the linear map. And now to get a contradiction, I consider points, Q two points on the curve, map them over here, write them in this form, and then figure out what they map to here. And if I can show that no matter which point I take here, which is not one of these points of order, of course, ends up here in some element that's not in the image of this map, then this shows that, well, since every rational point also is a Q two point, this shows that these rational points cannot exist. Now, so that's the idea of the proof, and so we use the two same group here, which captures some information on the model Baye group, to replace the model Baye group, which you don't know explicitly enough in the Schabutzi argument. Okay, so I have to say a little bit more about the same group, and how we compute them. So we have this polynomial of degree 15, which is on the right hand side of our curve equation. It turns out to be irreducible. So I write theta for a root, and then I take the number field that's generated by this root, which is a number field of degree 15. Yeah, then for every field extension of Q, which I call L here, I can look at the rational points on the Jacobian and this quotient, which we have seen for Q and Q2 on the previous slide, and then there's a certain map, which I can write down, which I will tell you how I can define it into, well, K is the number field here. So I take L tensor K, and then the multiplicative group modulus squares. So this does not have to be a field, but it will be the direct product of fields. And the point is that the two cinema group can be identified with a subgroup of this when I take Q here. So then I just get K. It's sort of L tensor K, so subgroup of the square classes of K, such that when I take the obvious map to the same thing, when I take QV here for a completion of Q, so I take K tensor QV over Q, the square classes in that I have this map over QV. And then the cinema group, there's the subgroup of this consisting of elements that map to the image of this map for QV, so that I sort of locally in the image. Now, I mean, this is a condition for infinitely many completions. And so if I want to compute something, I have to cut it down to something finite. And the way to do this is to say that the elements in the image or the elements that belong to the cinema group have to be unarmified in a certain sense at most primes. And in this case, so because the discriminant of this polynomial is odd in square three, and the class group of this number field is trivial, as one can show by computation, one gets that the two cinema group actually sits in the square classes that come from units, also set of just elements of K. And the only condition that I have to satisfy then is the, essentially the condition that comes from the completion at two and at infinity. So if I remember the previous slide, one thing they had to check is that the map sigma from the cinema group to the group jq2 mod 2jq2 is injective. And this is something they can sort of check on the right hand side of these maps. So we know the cinema group is contained in here. And if I take q2 here, I get a similar thing on the right hand side, which is just this ring. So two electric integers. And then I have to join theta take the units when you lose squares. And so I can just check that this map is injective, that's a finite computation. And this implies that sigma is injective as well. Okay, but I still have to figure out what my image is of a q2 to really determine the cinema group. And so maybe that it's a little bit technical, but that's a way of writing down this map pretty explicitly given a certain value of a presenting element of the Jacobian. So maybe I don't go too much into detail here. I mean, if you want to know more, you can ask. But so one, one thing I have to do is I have to figure out what the image is of jq2, or 2j of q2 in this group here and z2 replacing L here. And so basically what I need is the basis of this group here as a vector space over f2. And because the curve has a fairly special shape, it's possible to write down such a basis. So if I take this polynomial, it's roots as x-coordins of points on the curve, and then pick suitable y-coordins and take the images of these points in Jacobian, and then I take their sum, I get the point in jq2. And if I do this for all these polynomials, I get the basis of this. So it's possible to show this by some very explicit computation. And this allows me to figure out what the image of this map delta q2 is and therefore what the two-seller group is. And in this way, I see that the two-seller group has rank two, and this gets really bound on the rank. But I also have an explicit representation of the two-seller group that I can use for computations. Okay, so they have this, and then they have this logarithm map. So we need to be able to compute this. And by the logarithms, the logarithm is obtained by integrating differentiations on the curve or invariant one-forms on the Jacobian. And so one way of doing this is that we define the logarithm by integrating differentials from this point infinity to the point p. So one needs a periodic integration theory for that, which does exist. And which I cannot go into further here. So this defines my logarithm on points that are images of points on the curve. And then the image of the curve generates the Jacobian. And so I get it on a Jacobian by extending it linearly. Okay, now, I mean, our goal is to show that the three points we know are the only three rational points. And the way we will prove this is we will look at each of these points and then show that there's no other point that's too adequately close to it. So we begin with this point p0, which is the point 01. As I said, it's a torsion point of odd order even. And so the logarithm is 0 because the logarithm is a homomorphism into a torsion free group. Now, near this point, we can expand this map log as a tuple of formal power series in the uniformizer at this point, which we can take to be the x coordinate. And then I mean, looking at this, one can show that one can determine the ways of convergence of this power series. And so this process converge when the two edit valuation of the x coordinate is a little bit larger than zero. So definitely for all rational numbers, that whose numerator is divisible by two. And using this, because the points that are involved in this basis of jq2 over 2jq2 that you saw on the previous slide, they all satisfy this condition here. They can just compute the logarithms by evaluating the power series. And this way, we can determine the basis of, so I mean, the other way I have to find it, the logarithms maps into q2 to the power g. But by evaluating the logarithms at this basis, representatives of this basis, they get a z2 basis of the image. And this allows us to identify the image of the logarithms with z2 to the g. So now because the points that are close to p not in the two edit sense. So I compose the logarithms with isomorphism in my head in the previous slide to get the version of the logarithm which I now call log prime that surjects onto the z2 to the g. So that's the logarithm that was in the diagram on the slide before. And yeah, then I have to figure out the image of this map log bar composed with sigma. And I can just do that. And taking the suitable basis of my space here, I have these two explicit generators. So this is an f2 vector space, g dimensional, the fg tuples of elements of f2, and both these generators have the property that the first and the last coordinate are 0. And so this is the only thing that we need to remember here. And then, yeah, so computing this power series that gives this modified logarithm, so it's the ocean logarithm, and then I multiply by some matrix. You can compute this and we get this up to o of two squared terms. So we have a t squared here and then everything else has more powers of two and higher powers of t. So this is, I mean, I'm looking at points that are we'll have the same reduction as p0 mod 2, so they have an even x coordinate. So I write this as two times t. And so, well, I have to work out what the images are for all these points. So when t is odd, then I have something odd here, something odd here, and everything else is even. So I get this point if I reduce mod 2, 1, 1, and then zeros, but it has a one in the first component, so it's not in this image. And then t is even and nonzero. Then, whether I have this t here and everything else in most of these t squared, or two times t, or actually four times t. And so this term here will be the one that has the smallest triadic variation. And if I divide by the highest possible power of t, this gives me something odd and everything else is even. And so I get this element, and I reduce mod 2, and still has a one in the first component, so it's not in the image of the same apple. And this shows that there's no rational point other than the ones with x coordinate zero that has positive triadic variation of the x coordinate. So that is with these points, and then we do basically the same kind of thing with the pointed infinity, and then we still have to look at points that have odd x coordinate. So let's do that, but it will be, I mean, yeah, so for the points infinity, it will be very similar. So by this, I mean points that have negative triadic variation, so the x coordinate has a two in the denominator. I can write down a uniformizer at this point infinity, which is this here. And then again, write down my log of the midterms of power series in this uniformizer. And well, this gives the following. So I get stuff that involves of higher order things. And then the last coordinate, I get minus t plus 2t squared plus something divided by 4. And so this makes it pretty clear that regardless what t is, this last term will be dominant. We have the smallest triadic variation. And so if I divide by power of 2 and reduce, I will always get this element. And this is the one in the last component and therefore will not be in the image of the same apple. And so by the same reasoning, this shows that the point infinity is the only universal point that has x coordinate with negative triadic valuation. So this leaves points where the valuation is zero. And then by looking at the equation module eight, you see that this can only be the case when the export is going to minus 3, module eight. And all these points give the same image under delta q2, which is not zero. And this is not in the image of the same apple. So this shows that these points also cannot exist. And this proves what we want to prove. So the only point coordinates are the points with x coordinate zero. The only point with x coordinate has even denominator is the point infinity and there are no other points. So this finishes the proof that A5 of C is a square or even C is zero. And this extends what we know about the conjecture from the beginning by saying that if the second iterate is irreducible, then the fifth iterate is also irreducible. And now we want to go a little bit further. So what about the sixth, seventh, and so on. And yeah, so one, one idea here is to reduce the question whether A n is a square, the question whether A m is a square for a divisor of m. So then I have a divisor of m, and A n of C is a square, then A m or minus A m has to be a square. And if m and n have the same parity, then you can get rid of the minus sign here. So this allows us to reduce the case n equal to six, basically to well n equals three, and then the combination of n equals three and n equals two. So if I take n equals six, n equals three, then A six is a square. I know that A three is a square or minus A three is a square, but A three is a square when even C is zero. And for minus A three, I also use that A two of C has to be a square and combining these two things, I get a genus two curve of rank one. And it's by now pretty easy to figure out what the rational points are on such a curve. And I mean, this gives the point minus one here. The next one is seven. So seven is a prime number. So they cannot use divisors of seven. So we have to work with this curve, y squared equals A seven of x directly. But in this case, it turns out that the two same group is trivial, at least assuming the generalized Riemann hypothesis, because the polynomial on the right hand side of the curve now has degree, well, if I take the little a seven as degree 63 is irreducible. So I need the class group of a number field of degree 63. And this is only feasible assuming grh. Now, but otherwise, I mean, if I assume this, the seven group is trivial, and therefore the group of rational points is finite. And we can argue in the same case as for A four, if you remember, so we have just finding many points on the Jacobian, which we can pull back to the curve and see that this is the only possibility here. Yeah, eight and eight, nine, we can deal with using three and four or four and three. And then 10, we have to look at the case that we have that minus a five is square. And so we have to look at this curve, which is a quadratic twist of the curve we have been looking at as a morphic or to it over to a joint I and I mean, we can use that to do something similar, which I will not go into here. What the upshot is that if you assume that the second eternal is irreducible, what we've done here, so this and this implies that the sixth eternal is also reducible, and then assuming your age to get to seven, we can even go up to 10. So the 10th eternal is irreducible. And then we're going up further. So one would have to look at a 11. But this is a polynomial of degree 1024 or 1023. If I, if you have turned it around, to get something of odd degree, and it's irreducible. So we would have to work with a number field of degree 1023. And this is completely impossible, at least with the current technology. So this is what we can do regarding this conjecture. So I guess I should finish, but let me just mention this that the same kind of method does work for curves or more generally for curves of this form. So if you remember the expression we had for the right hand side, which was x to the 15 plus x to the seven plus blah blah blah square. So we had a square here of lower degree than this. And the concentration was one. So we only needed its odd because we do things too adequately. Then one can try to use the same kind of approach to show that the curve has only the obvious three points. And for this, it's essential that one can write down this basis of this group here. And just as an illustration, I've implemented this and then run this on lots of curves. For example, I've taken curves of genus five with coefficients of h between minus three and three and h of zero or h of one even, positive leading coefficients efficient of h, sorry. And then while you can see that most cases the method is successful, it shows that these are really only points. And then the remaining cases split well roughly evenly into curves that actually have more points and then curves where it's not successful. Okay, but that's that's all I wanted to say. So thank you very much for your attention.


Hey creatives, welcome back to the channel. I am Amanda. In the previous episode of composition and framing We broke down the commonly compositional framing used by filmmakers. Now that we've gone through that vastly, the next question becomes How should we compose the subject in each frame to fit into the cinematic framing? Welcome to the part two on compositional framing, guest snarking as we journey through this learning curve together. Compositional framing is a delicate dance of arranging elements within the frame to elicit emotions, tell stories and immerse the audience. It's about more than just capturing a scene. It's about crafting a visual narrative that speaks directly to the heart and soul. Every shot is a canvas and every composition is a brush stroke that paints the story's essence. The arrangement of characters, objects and space within the frame is a language through which we communicate emotions, tensions and relationship. So first off, we're going to begin with one shot. A one-shot or a single is a compositional technique that focuses exclusively on one character within the frame. As the lone character remains the primary focus, the technique emphasizes the character's facial expression, body language, reactions and can be used to convey a character's isolation creating a surreal and immersive connection. Between the viewer and the character, single shots are used in various ways imaginable. Most single shots are more frequently medium shots or close up, but in cases when they are wide singles, it's mostly compelling showing emptiness. Let's delve into the concept of a single shot with these movie references. This movie was directed by Sofia Coppola in 2003, lost in translation. The single character's shot was employed brilliantly to capture the emotional isolation and introspection of the main character. The film contemplative tone is accentuated by shots that linger on your faces allowing the audience to witness their unspoken thoughts and the complexity of their feelings. Through single character shots, the film conveys and disconnects the feel in a foreign environment and their genuine connection with each other. And this next movie was directed by Alejandro Gonzalez, the Revenant. The single character shot plays a vital role in immersing the audience in the harsh and unforgiving wilderness. The film's protagonist Hugh Glass is frequently shown in single character shots that captures his struggle, determinations and raw emotions. This shot intensifies the physical and emotional challenges he faces emphasizing the resilience and the relentless nature of his journey through the movie. Our next point is the two shots. A two shot involves framing two characters within the same shot, often side by side or facing each other or in any other form needed for the scene. The characters are usually filmed from the mid chest or visually in a way that highlights their interaction, relationship or dynamics, which can also show harmony or disharmony depending on the scene. Having the characters with the main focus of the shot while the surrounding environment can also play a role in conveying the scene's contest. In this film, The Piano, directed by Jane Campion, there's a memorable two shots that beautifully illustrates the complex emotion and relationship in the story. In a pivotal scene, Ada and George find themselves in a lush forest clearing. As Ada plays her piano, George listens attentively. The camera frames them in a two shot with Ada sitting at the piano and George standing nearby. This composition captures their emotional connection. Ada's music expressing her innermost feeling and George's attentive presence revealing his growing affection for her. As the scene progresses, Ada and George unspoken bond becomes palpable. The surrounding nature adds the sense of intimacy and vulnerability. As the character connection deepens beyond words, with that being said, the piano exemplifies how this technique can convey intricate relationship on spoken emotion and pivotal moments without relying solely on dialogue. As most subjects or elements are added to the frame, the name of the population size changes accordingly. And over the shoulder shot occurs when the camera is placed behind the shoulder of the character. The character's head and shoulder are seen in the foreground and are used as a framing device for the shot. While also revealing the face of the character they are interacting with, it serves to create a sense of spatial relationship and engagement between the characters. Over the shoulder shots can be used to suggest tension, intimacy, desire, hatred, imprisonment or conspiracy. It depends on the storyline and staging. In this movie, The Social Network, there's a compelling use of the over-the-shoulder framing during a pivotal conversation between Mark and his lawyer Sean Parker. During this scene, Mark and Sean are engaged in a conversation about the potential of Facebook. The camera is positioned behind Mark's shoulder, focusing on Sean in the foreground. Mark's profile is partially visible and the camera captures Sean's face as he speaks. This over-the-shoulder framing achieves several effects. The audience feels as though they are part of the conversation, observing Sean's expression while hearing his dialogue. The viewer can closely analyze Sean's expression and body language as he talks, providing insight into the motive and emotion. The framing highlights the power dynamic between Mark and Sean, emphasizing Sean's charismatic presence. And now, we move on to the POV shot, which we call Point of View. A Point of View shot is intended to present the subjective view of specific character. The camera lens is physically placed at the eye level of the character, which POV was seeing and the way the audience sees what the character sees. By adopting this character's viewpoint, the audience gains insight into their thoughts, emotions and reaction, making them an active participant in the narrative. In Alfred H. Cork's iconic film cycle, there's a famous POV shot during the shower scene as Marion takes a shower. The camera's perspective switches to a POV shot from behind the shower curtain. This viewpoint aligns with the audience, with the vulnerable position of the character intensifying the suspense and terror of the moment. We witness the menacing figure approaching through Marion's eyes, creating a visual and chilling experience. The POV framing technique is a potent tool in cinematic storytelling, inviting the audience to experience events as if they were the character. H. Cork's mastery of his technique contributed to the lasting impact and visual nature of the scene, showcasing the power of cinematography to evoke genuine and emotional responses. Insert. An insert in composition framing is a cinematic technique where a brief shot, often a close-up, is inserted into a sequence to focus the viewer's attention to a specific detail, object or action. This shot provides contextual information and emphasizes emotions or adds depth to the storytelling. Director Alfred famously said the length of a film should be directly related to the endurance of the human bladder. Insert. Shot supports this notion by conveying essential information efficiently without unnecessarily prolonging scenes. In the movie The Shining by Stanley Kubrick, he utilizes the insert shot to create an unsettling atmosphere in one of the iconic scenes when the Torrents discovers Jack's manuscript, filled with pages repeating the same sentence, all work and no play makes Jack a dull boy. The camera captures an insert shot of the manuscript, emphasizing the repetition and the descent into madness. This insert shot becomes a visual representation of Jack's mental state and adds an eerie dimension to the horror of the film. An insert shot can carry symbolic meaning or context or subtext, revealing characters' motivation or thematic element. Insert can build suspense by focusing on a key object before revealing its significance, creating intrigue and engagement. Central framing. Central framing also known as centered composition, where the main subject is placed at the center of the frame, is often motivated by aesthetics and visual storytelling. This technique can create a sense of balance, symmetry and visual harmony that appeals to viewer's sense. It's a deliberate choice that can convey certain emotions, theme and messages, depending on how it is used within the context of a film. Center composition aesthetic appeals lies in its ability to draw immediate attention to the subject. The centered placement naturally focuses the viewer's gaze, allowing them to engage directly with the subject without distraction. However, central framing can reveal power dynamics in a scene by using the centered placement of character or object to emphasize dominance and authority, and that visually establishes the dominance over the scene. It can also be used to depict vulnerability or powerlessness. Stanley Kubrick's film, A Clockwise Orange, features central framing to create a sense of unsettling symmetry. In the infamous singing in the ring scene, Alex commits a brutal act of violence with an eerie calmness. The camera remains static as Alex, with a sinister smile, assaults his victim. In other contexts in the film Moonlight, directed by Barry Jenkins, central framing is used in ways that captures vulnerability and intimacy. In a pivotal scene, the young protagonist, Karen, sits across one who becomes a father figure to him. This camera centers on Karen's face as he listens intently to one's word. Central framing is a versatile technique that can evoke different emotions and serve various narrative purposes. While it can provide stability and balance, it can also create tension, highlight emotions, or emphasize the importance of a particular moment. Now, Rule of Space, the rule of space also known as Principle of Directional Movement, is a cinematic guideline that involves leaving visual space within the frame for a subject to move or to look into. This technique creates a sense of visual balance, direction and anticipation by providing room in a direction of a subject's movement or gaze. The audience's attention is drawn in that direction, enhancing the narrative and conveying a natural flow. In the movie Lost in Translation, directed by Sofia Coppola, there is a pregnant use of Rule of Space using a conversation between two main characters, Bob and Charlotte. In the Tokyo Hotel, in the scene where Bob and Charlotte sit on a window sill, looking out at the city lights, the camera frames them with the space to the left of the frame, allowing the audience to see the city space. The character's gaze and body language direct attention to the cityscape, leaving room for the viewers to explore the environments they're observing. The space to the left of the frame balances the composition, providing a harmonious arrangement of the characters and their surrounding. And by framing the characters with room to look into, the audience's gaze is drawn to the same point of interest, enhancing the scene's emotional impact. Roger Dickens' word reminds us that cinematography is an art of intuition and experience. It's about translating the feeling of intention of a scene into visual elements that resonates with the audience. Every decision about framing and composition and perspective has a profound effect on the cinematic experience, creating a symphony of emotion that unfolds on screen. And that's all on our topic today. If the topic is not clear enough, please feel free to ask questions in the comments down below. If you have enjoyed this video and found it educational, give us a thumbs up, hit that subscribe button, and don't forget to ring the notification bell to stay updated with our upcoming content. Until next time, improvise, adapt and overcome, and I'll see you soon. Bye.


There's one thing in this world that I don't want you to be. Today, we're going to be talking about how all of the limitations that you place on yourself, all of the things that you think are real are completely optional. All of your limitations are optional. And I'm going to start off with a story. Back in 1954, and before 1954, for all of human history, I guess until time came in and people started recording time, they thought it was impossible for someone to run a mile in under four minutes. So years and years and years and years and years people have been running miles. No person had ever made it under four minutes. They thought it was literally impossible, physically impossible. There were actually doctors that came out and said that a human heart would actually explode if somebody were to push themselves hard enough to run a mile in under four minutes. So nobody did. Why? Because there were all these people that were in white coats. There were supposedly quote-unquote professionals. They were the doctors. They were the ones that quote-unquote know it all. So of course we can't do it. And then so what happened? Someone comes out and says something and the mass amount of people in the world go, okay, I believe that. And this limitation was placed on all of mankind that actually paid attention to it. Everybody thought that it was impossible. Except for one person. There's a guy named Roger Bannister. And in 1954, Roger Bannister ran a mile in three minutes, 59.4 seconds. The first person ever recorded to run a mile in under four minutes. And what did he do? He showed everybody that running a mile in under four minutes was possible. And once people saw it was possible, they stopped limiting themselves. So they went, oh, this is possible. That means I can do it. Here's the craziest part about it. Nobody had ever been recorded running a mile in under four minutes because they thought it was physically impossible to human heart explode. It was not going to happen. So why would they even try? Right? Within two years of Roger Bannister breaking the record and running a mile in under four minutes, 300 people ran a mile in under four minutes. Now let me ask you a question. Was it in those two years that Roger Bannister ran the four minute mile that humans just magically evolved? They became a great species. They became more superior. They became faster. They became stronger. Their lungs grew to a larger capacity. Did humans change? No. What changed? Their perceived limitations changed. They had the ability the whole time. But the problem was their belief that it was not possible was the thing that held them back. So think about that for a second. Before I go any further in this podcast episode, how many times have you held yourself back because of perceived limitations? How many times have you believed something simply because somebody in a white coat told you that that was the truth? How many times have you listened to someone who's older than you because they said it was a truth? Right? There's another example. For all of human history, people look up in the skies and they'd see birds flying and they're like, I wish that humans could fly. And that was impossible until the Wright brothers proved them wrong. People thought that putting a man in space was impossible until the Russian people put their very first person in space in 1961. People said that it was impossible to land on that big old thing called the moon until 1969 when Neil Armstrong proved them wrong. And you're telling me that in 1969, we successfully sent a rocket ship 238,900 miles away to a rock that's orbiting the earth at approximately 2,288 miles per hour and then got them to take off from that freaking moon and land back on the earth safely with something that literally their technology was nowhere near what you carry in your pocket and you can't accomplish your dreams? Really? Does that make sense to you either? Because in my head, that doesn't make any sense. What do all of the four-minute mile, the Wright brothers, Russia, Neil Armstrong, what do all of those have in common? One thing, they were impossible. They were impossible. And they were only impossible until they were done. And then once they were done, what happened? People clicked over and they went, Oh my God, this thing is possible. So I've got a question for you. What is holding you back from following the life that you really want? What are the limitations, the glass ceilings of perceived limitations that you're placing on yourself the same way that every one of those was a perceived limitation? It's just they hadn't figured it out yet. Right? The four-minute mile, perceived limitation. That perceived limitation was people saying that it was impossible, physically impossible for a human to do it until, oh, actually it's not physically impossible. We were wrong. It's physically impossible to go into space. Oh, we were wrong. It's physically impossible to get to the moon. Oh, we were wrong. Right? How many times have things people then, oh, we were wrong. What is the oh, we were wrong that still exists in your world right now? What are you telling yourself that is not possible? What limits are you placing on yourself that are completely false?  As to why you can't create the family that you want to? As to why you can't create the business that you want to? As to why you can't be a successful artist? What is it? What are you afraid of? What's holding you back? What are the perceived limitations that you're placing on yourself? Is it the fear of success? Are you afraid of what would happen if I were to succeed at such a high level? What would be required of me?  So many people are afraid of the fear of success. Is it the fear of failure as to what if I go after this thing that I really want and it doesn't work out? What if I go after this thing and people make fun of me when I fall on my face? Is it the fear of uncertainty as I don't know what's out there in the world? I don't know what's going to happen to me when I go for this thing? Is it the fear of making a mistake? Is it the fear of other people's judgment? Is it the fear of other people's opinions? Is it the fear of disappointing somebody else? Is it the fear of stepping out and following your dream and creating this dream business because you don't want your parents to come back at you and say, hey, why the hell would you do that? We spent $60,000 for you to get your engineering degree. What is it? Are you afraid of disappointing yourself?  Are you making things up in your head? Are you taking the safe route? What is it that you're doing? Are your dreams so big and so vast that it paralyzes you thinking of all of the things that you must do in order to get that? What is it? Are you making up your limiting beliefs? Are you making up your excuses? And then once you think of those limiting beliefs and think of those excuses, you're believing them to be truths. You're believing those things to be impossible. Are you saying that you don't have enough money to do that? Are you saying you don't have enough time? Are you saying that you're too lazy? Are you saying that you don't have a car? Are you saying that you didn't grow up in the right part of town? Are you saying that you don't have mentors? Are you saying that people from where you're from don't succeed? What is it?  What is it? Because it's all false. See, we all have near unlimited ability between us and in us, but we restrict ourselves in the potential that we have because of limitations that we put on ourselves, the excuses that we make up as to why we're not living our dreams so that we can play small, why we can't reach our dreams so that we continue just to sit on the couch and play around on Instagram because I'll tell you what, it's a lot easier to sit on the couch and just mess around on Instagram than it is to actually go after your dreams. Why? Because you're going to fall on your face over and over and over and over again going for your dreams. But guess what's the beautiful thing about it? You eventually will succeed. The creator of Honda says success is 99% failure. You only got to succeed once. You've got to look at your limitations and say, yeah, those are false. I can see how false those things are. We have to rise above the limitation. We have to break fear free from our fears, from our excuses. When are you going to stop telling yourself that you can't do that? When are you going to stop telling yourself that something is completely impossible? Because when we remind ourselves that something's impossible, then we give ourselves an excuse as to why it does not go for it. Because once again, it's easier not to go for it. But we remind ourselves that something is possible, that it is within our reach, that it is going to take more of us. Is it going to demand more from us to get there? Absolutely. But are we going to be able to get there? For sure. When it comes down to you, it's always, has always and will always be in your head. You're not at battle with anybody else. You're not competing against anybody else. You're competing against the person in the mirror. You're competing against the the eight inches between your two ears. That's all you're competing against. It's you. It all comes down to you. It comes down to your beliefs. It comes down to your fears.   It's not your current situation. It's not the fact that you don't have the right circle of influence. It's you. It's always been you. It will always be you.  And you go, this is, I'm the one to blame as to why my reality is the way that it is. We need to reprogramming of our minds. It changes with the story that we're telling ourselves. And so I want you to start to think about that. What is it that you're doing that's holding yourself back? There's one thing in this world that I don't want you to be. That's realistic. I don't want you to be realistic. Do you want to know why? Because realistic doesn't exist. Realistic is fully 100% subjective. You make up what is realistic. So that realistic thing, like it's a glass ceiling. And what's interesting is as you work harder, you know, for some people that might be listening, it might be your goal to make a hundred thousand dollars, but that's so unrealistic for where you are right now. And I get it. That's unrealistic. But that, that hundred thousand dollars is a glass ceiling. It's not a real limitation, but it's a glass ceiling. Here's what's interesting about it. Once you push yourself and push yourself and push yourself and you don't pay attention to your beliefs, and the beautiful thing about, about the beliefs that you have is you don't have to believe in yourself in order to take action. So if you have this glass ceiling that's saying, not good enough, not good enough. I can't do it. I can't do it. I can't do it. But you just work and work and work and work and work and you eventually do it. You break through that glass ceiling. What happens? You create a new glass ceiling and you go, oh, I just made over a hundred thousand dollars. You probably won't make less than a hundred thousand again. Why? Because you now have a new belief of what is quote unquote realistic for you. So the last thing I want you to do is be realistic because there's no such thing as realistic. A four minute mile once wasn't realistic. Going to the moon wasn't realistic. All of those things weren't realistic.   Why? Because back then it wasn't realistic. Why? Because it had never happened.  So I want to leave you with a couple of questions to help you kind of dive into this and to get a little bit deeper into this. So I want you to jot this down. Get a pen and paper real quick. Write this question down. What limitations are you placing on yourself? What are they? What are the limitations you're placing on yourself? Write them all down. There might be four, there might be five, there might be 200. I don't know what they are. What are they? Next question I want you to ask yourself is this, why are those limitations false? So instead of looking at the limitations and believing them as limitations, let's flip it on its head. Now that I'm looking at those limitations, why are those limitations false?  And you can say, why is that limitation? Why is that belief false? Well, because I'm sure there's other people in the world who have come from the part of town that I'm in or something equivalent to it and succeeded. And if someone else has done it, that means I can do it. If Roger Bannister can run a four-minute mile, I can too. A few hundred people realized that two years after he had ran the mile. Right? So why are your limitations false? Why are they not true? And then last but not least, the last question is how can I overcome these limitations? What do I need to do? What actions do I need to take? What systems do I need to set in place? What habits do I need to create? What life do I need to create? How do I need to change my environment around me so that I can overcome these limitations as often as possible? Maybe I set up some sort of way where whenever I start thinking about this limitation, what do I do? I have something else that I do instead. And I develop a system, a routine, a habit, action, something that I need to do to get myself past limitations because none of those limitations are real. They never happen. They never will be. The only thing that holds you back from creating the life that you want to is you. It always has been. It always will be. But the beautiful thing about that as well, the only thing that would propel you forward into creating the life that you want to is you. So you can either look at your limitations or you can look at your opportunities. The choice is yours. Hey, thanks so much for watching this video. If you want to learn even more about Master Your Mind, click right here and watch this video as well.  You are. Failing at something is better than doing nothing.


Well, I mean, I think it's always about the Titans needing to be better. But it's a good football team. So when you make mistakes and know that any margin of error against a good football team is going to come back to hurt you, so it's probably a little bit of both. We'll consider everything we possibly can to win a game, Tron. So thank you. I think we struggled everywhere today. You're at the game, so everybody's quick to make it about one guy. But it's never about one guy. We've been down this road before. So when you get beat, there's a lot of things that went wrong. And certainly, we'll have to look at a bunch of different things. Well, you have to understand how critical it is to protect your quarterback. When a quarterback gets hit, nothing good can come of it. But we also have to be able to get open and be able to, there were times where we had some protection. And the protection was good. But then when you get down and you get behind late in the game and it becomes a drop-back game, that's not certainly not where you want to live against a team that has some of the rushers that they do. Well, making sure that we protect them, making sure that we use our chip help, making sure that we're given an opportunity for the quarterback to progress through. I mean, I think it would probably affected it. I mean, we had four wide receivers available for the game. So we lost one, and that kind of put us into a little bit of, you have to be prepared and use some 12 personnel and had CJ in there. Well, I mean, I think he's doing OK from what I've seen and what has I been able to see him at halftime and talk to him after the game. We expect that from Chalen just because I know what he's capable of. It's unfortunate that he couldn't finish the game. You said time and time. It's not about AJ. AJ had a good game. We can't give up big plays on defense. We tried to make him earn it coming into the game. And we clearly weren't able to. Was that a question buck or was that a comment? Good. You made a great comment. Yeah, we did. They made plays, and we didn't affect the quarterback. And again, they were able to come down with some plays down the field on some 50-50 balls. I thought early, we tried to cage him and make him throw from the pocket to his second read. And he was able just to work the vertical game. And so I think we started to rush a little bit and try to get him. We affected him at times, just not clearly enough. He's training us in a position. Yeah, you get a lot more leeway when you're playing the football, when you're looking back. And again, when you're not playing the football, we've known for a lot of years that your any contact is really going to be exaggerated, which it should be. So we have to play the football. We have to be able to be in phase and turn around and try to play the football, and then they're going to give you probably a little bit more leeway. Yeah, I'll have to look at the tape, but my guess is it was a double move. Christian had right to the space. And again, this crew is from all over. They just put this crew together today, so I don't know. I just know that I hear what they tell me sometimes about, yeah, the DB has right to that space. Did he take a charge? I don't know. But that's supposed to be, again, is he coming in from the side, or do we have that space? And then we're entitled to that space. I couldn't tell, and it didn't see enough in the replay. Why is that something you had? Yeah, that's over with. Yeah, I mean, what do they call it? Legal contact? Yeah, legal contact. Whatever. Why did they just end? I don't know. Again, these are all great questions you can go ask.  And I don't know. I mean, I had a very good year coming in. And again, he can scramble and throw from the pocket, did a nice job from the pocket, and was able to make it difficult. So we stopped to run game and just weren't able to make them earn it down the field.  I'm just concerned when we lose and disappointed for the players when we lose. Well, I mean, I think we're at a crossroads. I told the team, I think we're at a crossroads in kind of how we want to continue down this season. And we can't point fingers. We have to assume that each and every one of us didn't do a good enough job because we didn't. We have to come to work with greater energy and greater resolve to prepare, to win a football game against a division opponent.  It sucks getting beat the way we did. But we have to, we've got to make a decision. How much are we willing to invest and trust in what the coaches are doing, trusting what each and other players are doing? So I just think it's a critical time for us. Yeah, I mean, I've never been safe or secure in anything that I did or wanted to try to ever take that approach. I hope the team hasn't, I mean, coming on the road to the best team in National Football League. I hope we didn't take a safe approach. We just got to coach better and got to play better. We got to make some decisions and see what we can do to win a football game. Yeah, it was a huge play. Obviously, it took a big hit there and held onto the ball. It says a lot about the first thing when I saw him in the locker room at halftime. He said, I wasn't going to let that ball go. So it just tells you a lot about the kind of guy he is and the competitor he is. How much different, how much? Well, I mean, obviously losing trailer ends as a blow. And the shuffling we had to do, personnel wise and everything else just makes it tough. But yeah, anytime you lose one of your top guys, it makes it more difficult. Yeah, we're going to find out what we're about. I think we have a resilient group. We're tough and a bounce back from a lot before. And I believe we'll bounce back from this. Got to make the corrections that need to be made and come ready to go tomorrow and this week to win one game. At the end of the day, this one's done. It stings, it hurts. We're all disappointed. We're all frustrated, mad, angry, every emotion. You can name it, but have to be able to get over it and get ready to go in the next one. Yeah, just as a me as a player and as a leader, just have to go in ready to go with a great attitude and a focus and intention and just drive our guys to go and prepare, have our best week of preparation and get ready to go play the next one. It makes it a little tougher, obviously. We want to play a balance and that's no secret, I don't think. But whatever the case may be, if we're struggling to get it going, we need to make more plays in the pass game. And each game is going to be different, so have to find a way to move the ball. Yeah, I mean, like I've said, the ankle is feeling better steadily as the season's gone on since the injury. Just want to keep it going in the right direction and then have some opportunities to get out and get outside and use my legs a little bit. So it felt good to get moving a little bit. And I'm glad I was able to make those plays. With what? Obviously, we're just not doing a good enough job. I don't know exactly what it is. Can't tell you exactly what it is, but I'm going to need to be better. Yeah, everyone has a role in it. I do, receivers do, Office of Line does. We just have to be better all the way around. No, I don't think there's any reason to believe that. From what I've seen from our guys and our team, I don't think that. Honestly, I don't try to take a step back and look at what all is in front of us and where we're at in the standings and all that stuff. I try to go win a game each and every week. And that process won't change. And I know we have a division game at home this week. So I'm going to be a big one and need to go find a way to win. Resilience, when you face adversity and you find a way to stick together and push through and find a way to win. When you've seen it before, you believe you can do it again. I think it is execution. We're just not being able to string drives together. We get a couple of plays here and there, and then we stall. And then we just got to play better overall on offense and individually, self-evaluation, how I play. I've been playing each these last two weeks and just learning and come in. And we just got to be better, a lot better. You mentioned self-evaluation. No, like I always tell you, I'm a rich critic. So I can evaluate myself about everything and anything possible. But I got to be a better leader, small things, details, and making plays. I pride myself on being a playmaker for this team. I feel like I haven't done that these last two weeks. So that's been tough and adversity. But you just got to stay positive through it and continue to work and focus on improving. And then eventually, I think everything happened for it. But right now, it's tough. Just get two losses and haven't had success. So I think we just come together, stick together. These times need to be closer and be positive and be real with each other and know what we got to do to be able to win games and play this how we want to play. And I think that's what you got to do. We have two tough losses like this. I think leaders got to step up and just focus on every guy, focus on doing their job a little bit better, much better. I'm great, so good. We just got to be better. I mean, that's all I can really say. SEQ better. It takes all 11, all 11 less tied in. Know it's useless. I need to be better, speaking for myself. And I'll take a count on the building to that. And I will work as hard as I can to be better here and so on. So I'm sure every guy in that locker room and on this team, we've tried to do the same thing. Sit again? I mean, well, it's tough right now. And it's easy to point a finger at something else. But I don't think we guys in the locker room, we're not focused on that. It's about how can we get better? How can we do things to where we put ourself in position to win games?  And just letting that lead throughout the weekend going to Sunday. But I mean, it's tough right now coming up for loss. But thank you to try to stay positive as much as you can. While you guys showed up to execute, was it tough to watch somebody who could easily command them still in the locker room to execute so well? No, not at all. He's on that team now. Been on that team for how many games, 12 games. So I mean, that's been done and over with. He's got to play better as a team, and we need to execute. So I really focus on one person. It's about what we need to do and focus on ourself and how can we get better and come out and play better. Would that execute a kind of struggle that you guys have had in the Russian attack? Is there something you think is more what the opposition is doing, or is that internally a lack of execution?  I mean, they've been doing a great job. But you know how much pride we have in the run game. And I don't feel like we've been doing our best to have success. And we just got to get better. I mean, it sound like a broken record, but that's just what it is. I mean, we're still in the season. Got five games left, and it's time to focus on getting better. And like I said, I hope myself accountable for being better and do the best I can here and now.