And I'd like to start by thanking the organizers for inviting me to this program. This is my first time here and I've really enjoyed myself. So let me start by giving you the rudiments of symplectic geometry needed to state our result. And of course I have to tell you about the classical symplectic non-squeezing result of Gromo. So starting slowly, what is a symplectic manifold? So a symplectic manifold is a manifold that is even-dimensional, in doubt, with a symplectic form. So what does this form need in order to qualify as a symplectic form? It needs to be closed, non-degenerate, alternating to form. Now the canonical example of a symplectic manifold is the following. Canonical. It's nothing else but r to the power 2n. So a point z in here will write it as x, p. And we view x, write it both x and p are elements of rn, and we view x as including the positions of n particles in r, and p, the momenta of those particles. Or if you want the position and momentum of a single particle in rn. For example, if n is 4, it could be dealing with one particle in r4, two particles in r2, or four particles in r. What is the canonical symplectic form? Is dp1, which dx1, plus dpm, which dxn. And it's clearly a two-form, which is closed, non-degenerate, and anti-symmetric. Now this symplectic form is the one that is responsible for Hamilton's equations in their traditional form. So if h is a Hamiltonian, let's say, from r to n to r is a Hamiltonian, then the flow generated by this Hamiltonian is defined as follows. Omega of dot z dot, the dot above the z stands for the time derivative, so z is a function of time taking values in phase phase. This is the differential of h at z, and you put over here whatever you put over here. Now if you decode what this means using the definition of omega, you recover exactly Hamilton's equations. xj dot, the xj entry of z is del h del pj, and pj dot is minus del h del xj, so Hamilton's equations. Now one can rephrase this canonical example as follows, so rephrase. We can regard r to the power 2n as cn, and we write the point z in here as x plus ip. And we rewrite to the canonical form as follows, omega of z and zeta is minus the imaginary part of the inner product between z and zeta in a product in cn. And our inner products are c-linear in the second entry as per Dirac's convention. So let me write this down, this is minus the imaginary part of the sum of zj bar zj. Now the advantage of rephrasing the canonical example in this way is that it easily generalizes to a symplectic Hilbert space. So what is a symplectic Hilbert space? This is a Hilbert space H, which is a complex Hilbert space endowed with a symplectic form defined by a minus the imaginary part of the inner product between z and zeta, where now the inner product is in H. This is the setting in which the non-linear Schrodinger equation can be seen to be Hamiltonian with the underlying symplectic Hilbert space being L2. I will tell you more about that soon, for now let me just retreat to the finite dimensional setting and explain to you what the simply non-squeezing result of Gromov is. Well Hamilton's equations have structure, so let me write here, Hamiltonian flows preserve the symplectic form omega. And to introduce one piece of vocabulary that is they are symplectomorphisms. So what is a symplectomorphism? It's a defiomorphism that preserves leptic form. So how do we write that down mathematically? What does that mean? It means that if you look at the Hamiltonian flow at some time t, then the pullback of the symplectic form via the Hamiltonian flow at time t is just the symplectic form, and that is true for all the times t. It is convenient, perhaps more enlightening, to write this down in the integral form. So with integral form, we can rephrase this relation like this, here are two copies of cn and the Hamiltonian flow at time t takes one to another. Now if you have a two-dimensional surface s over here, we take every single point on the two-dimensional surface and we flow it according to the Hamiltonian flow at time t. So what we recover is another surface like this, and this relation over here is nothing than the following. The integral over s of omega is equal to the integral of omega over the image of the surface s under the Hamiltonian flow. Now what does that mean for us? If the complex dimension n is equal to 1, then omega is a volume 4, right? It's after all a non-degenerate 2-4, then omega is a volume 4. So what we see in this case is that Hamiltonian flows preserve, so the area of s is the area of s under the Hamiltonian flow at time t. Now if the complex dimension n is strictly bigger than 1, then omega, which omega n times is a volume 4 as a non-degenerate 2-n-4. Now because Hamiltonian flows preserve the symplectic form omega, they are going to preserve the volume 4. So what we recover is the following theorem of L'Uville, which says that Hamiltonian flows preserve phase space volume. So this leads us to the following natural question. This preservation of volume, the only obstruction for the existence of a symplectomorphism. So let's write it down. Question is preservation of volume, only obstruction for the existence of a symplectomorphism? So what do I mean by that with a picture? If I have two blobs, so here is blob number 1 and here is blob number 2. Now I know that the volume of blob number 1 is equal to the volume of blob number 2. Does there exist a symplectomorphism between the two blobs? Now if, so Moser proved that if I have two blobs of the same volume with the property that there is a diffeomorphism between them, that is they have the same topology, then there is a diffeomorphism that preserves volume. Now if the complex dimension n is equal to 1 and I have two blobs of the same volume and a diffeomorphism between them, then Moser tells me that I have a diffeomorphism that preserves the volume, that is the symplectic form. By definition that is a symplectomorphism. So I mean the volume already is preserved? Yes, that means you know mapping every single patch in a volume preserving way. So you say that there exists then? If there is a diffeomorphism then there is one that preserves the volume, well on surfaces. So the answer in this case is yes and it is due to Moser. If however the complex dimension is strictly bigger than 1, then the answer is no, right? And this is Gromov's symplectic non-squeezing theorem. So let me tell you what that is. Okay, so I can't, that is too far from me to write on. And then I will never reach it again. So Gromov theorem due to Gromov, what does it say? Well, imagine that we have a ball centered at z star of radius r. So this is a ball in Cn. And imagine that we have a cylinder, okay, which I'm going to write like this. Cr of alpha and l, okay? So the parameter r is going to be the radius of the cylinder is positive. Alpha is going to be the center over here is a complex number. And l is an element of Cn, which is normalized to have length 1, okay? So what is this cylinder? Well, it's the points z and Cn with the property that if I take z in a product with l minus alpha, this is smaller or equal to r, okay? So what does that mean? Well, l determines a subspace of one complex dimension. Or if you want two real dimensions. So what do I do over here? I take z, I project it on that subspace and I say, well, it needs to leave within r of this center over here, okay? All right, then now imagine that I have a simplectomorphism, phi, okay? Now, if phi of the ball, right? If phi takes the ball inside the cylinder, then necessarily little r is bigger equal than capital R, okay? So in other words, a simplectomorphism cannot squeeze a ball inside a thinner cylinder despite the fact that this has finite volume and this has infinite volume, okay? So this is the point where people like to draw the analogy with a symplectic camel, which cannot squeeze through the eye of the needle, okay? Now, this is not if you have seen Gromov's theorem before, this is not exactly its traditional formulation, right? To obtain the traditional formulation, you simply take L to be E1, the first vector in the eigen basis for Cn, right? Because in that case, the cylinder becomes, let me write it like this. It's the points Z in Cn with the property that X1 minus the real part of alpha squared plus P1 minus the imaginary part of alpha squared is more or equal than r squared, right? This is the way Gromov's theorem is typically written. However, the two formulations are entirely equivalent, so they are equivalent. So how do we see that? Well, because L is normalized to have length one, there is a unitary map that takes E1 to L, right? And unitary maps preserve the inner product, which means that they preserve the symplectic form. So if there is a symplectomorphism between the ball and this cylinder, they compose and get with the unitary map. You get a symplectomorphism that maps the ball inside this cylinder. And the other way around, okay? Now, before I move on to the Adelaide setting, I would like to give you a few examples of squeezing, okay? They are not going to be counter examples to Gromov's theorem. They just serve to better appreciate the hypothesis of Gromov's theorem, okay? So examples of squeezing, all right? So there are really two remarks that I want to make over here. The first remark is that one might imagine that the reason why one cannot squeeze the ball inside the thinner cylinder is because for some reason, the ball is very fat in the x1, p1 direction, okay? However, that is not the case, and here is an example, okay? That is not the right way to think about it. Let's consider the Hamiltonian H to be minus p1 x2 plus p2 x1, okay? Then the flow generated by this Hamiltonian, you can write down Hamiltonian equations if you want in the following form. If Z is the flow, then Z dot becomes 0 minus 1, 1, 0, apply to Z, okay? So this is nothing else but a rotation, right? So Z of t can be written explicitly as cosine of t minus sine of t, sine of t, cosine of t, apply to the initial data at time 0, okay? So why is that good news or bad news depending on how you want to think about it? Well, if you take as initial data, right? If you take your initial data lying inside an ellipsoid, which is very fat in the x1, p1 direction, but thin in the x2, p2 direction, and you flow it for time pi over 2, okay? Then what you recover is now a cylinder which is very fat, right? You basically just rotate it. It's a cylinder which is very fat in the x2, p2 direction, but very thin in the x1, p1 direction.  So the reason why one cannot squeeze the ball inside a thinner cylinder has nothing to do with the fatness of the ball in the x1, p1 direction, but rather the whole property, right?  Now the second remark I want to make is the following. Remar number two is that it's essential to use conjugate coordinates when defining the cylinder, an example of which are x1, p1, okay? Example, okay? How do we see that? Well, you could ask, could it be possible to prove Gromov's theorem if we choose as coordinates x1 and x2, the positions of two different particles? Well, the answer is no, and here is an example. So cannot choose x1, x2. How do you see that? Well, simply consider the Hamiltonian h, which is minus x1, p1 minus x2, p2, okay? Then what is the flow for this? Well, it's sort of independent in x1, p1 from x2 and p2. So let me just draw a picture. What happens over here? Well, the flow looks like this, right? If you take an initial data, this is how it flows. So in particular, you see that it squeezes the x1 direction. If you start with the initial data in a ball over here, then as time goes by, it's going to move into sets like this, okay? And similarly, in p2, x2, it's exactly the same picture, okay? So it squeezes the x1 in the x2 direction. So you cannot prove Gromov's theorem with these coordinates for the cylinder. But then you can ask, well, what if I use the position coordinate of one particle and the momentum coordinate of another particle? Could I have squeezing then? I mean, could I prove no squeezing then? And the answer is no. So cannot choose x1, p2, let's say. Simply take the Hamiltonian to be, well, leave it at the same in the x1, p1, but reverse time for x2, p2, right? Then it squeezes x1 for exactly the same reason as before. But it does this in the x2, p2 direction, right? So if you start with a ball of initial data, then as time goes by, it's going to squeeze the x2 direction, the p2 direction, okay? All right, so that's what I wanted to say about squeezing. There are no questions, I would like to move on to the NLS setting. The non-linear Schrodinger equation that we will consider is the following. i times the partial derivative with respect to time plus the Laplacian of u is equal to absolute value of u to the p times u. The power p is assumed to be positive. And a solution to this equation is a complex value function of time and space. Time is going to be real, okay? But we can pose this equation either on Euclidean space or on the d-dimensional torus, right? This is rd mod zd. And the dimension d is assumed to be big or equal to what? Of course you can pose NLS on other manifolds, but for us this is enough for today, we're only going to consider these two cases. Okay, now this equation is Hamiltonian, okay? The Hamiltonian h of u, the integral of half of the gradient of u squared, plus 1 over p plus 2 u to the p plus 2 dx, okay? And the symplectic form with respect to which NLS is seen to be Hamiltonian with this Hamiltonian is the following, omega defined on L2. Let's stick to rd for now, cross L2 of rd with real values omega of u and v, right? So remember what it was? It was minus the imaginary part of the inner product between u and v in the underlying symplectic Hilbert space, which for us is L2. Our inner products were selenia in the second entry. I'm going to move the bar from u to v, and I'll get rid of this minus over here, right? So I'm going to write this as the imaginary part of u v bar dx, okay? So what do I want? Well, I want a symplectic not squeezing result for NLS. Let's state it, so we want the following fact. If we have a ball, b, centered at z star of radius r, leaving inside the symplectic Hilbert space L2, and we have a cylinder, cr of alpha and L. R is positive, alpha is a complex number. L is an element of the Hilbert space, which is normalized to be one. And what are these? These are the functions z in L2 with the property that the inner product between z and L, right? If you project z on L, then it lies within a disc of radius R around alpha. Okay? We want to prove that if we're looking at the NLS flow, a time capital T, let's say T is some positive number, and this flow, a time capital T, maps the ball inside the cylinder. Then necessarily, little r is bigger equal than capital R. Exactly the same statement as before. Now, we see that even by stating the problem, we immediately uncover an issue that needs addressing. Namely, what are we doing? We want to say that for all initial data in this ball, we can define the flow up to time T, right? And this was just some arbitrary ball, so the question arises, when is NLS well-posed on L2, right? Well, nowadays we know the answer to that question very, very well, right? So NLS is well-posed on L2 of R2. Precisely when P is more or equal than four over the dimension D. So how does that work? Let me just quickly review this. If P is strictly less than four over D, then the problem is subcritical, right? And using contraction mapping together with strict estimates, one can construct a solution, right? Locally in time, with the time of well-posedness being bounded from below by a negative power of the L2 norm of the initial data, okay? So using this, an iteration working immediately construct globally in time solutions, in particular, the flow is going to be defined for all times T, right? So how do you do that? Well, you start with your initial data at time T equals zero. And you run contraction mapping to the other strict estimates and you solve the problem, right? You construct a solution up to some time of well-posedness over here, okay? But now when you sample your solution over here, it has exactly the same mass, right? Exactly the same L2 norm as at the original time, right? Because mass is conserved. So you start your contraction mapping over here and you solve it for another time of well-posedness. And you keep on going, right? Constructing a global solution. Now, if, however, the power P is four over D, then the problem becomes critical, right? And in this case, the time of local all-posedness for which you can solve the equation is a function on the initial data itself, okay? The scaling symmetry tells you that this time cannot depend on the L2 norm of the initial data, okay? In fact, it depends on how concentrated the initial data is with the intuition that the more concentrated the initial data, the shorter the time of existence. So this naive procedure over here is not going to allow you to construct global in-time solutions because maybe with every single iteration, the solution gets more concentrated so you can only solve it for shorter months of time. And this naive procedure will not help construct global solutions. So in this case, global well-posedness is a deep result of Dotson. And this is the following thing. For an arbitrary initial data, U0 in L2, there exists a unique global solution, U, to the mass critical NLS. So I del T plus Laplace in U is U to the four over D times U. The initial data is U0, okay? And this solution obeys uniform spacetime balance. So the integral over R, the integral of Rd of U to the power 2d plus 2 over d dx dt is bounded by a constant that depends only on the mass of the initial data. Okay? So in particular, this theorem allows us to consider flows for arbitrary amounts of time. Okay? Yes, this is Rd. Thank you. Thanks. Okay. Okay, so to make life interesting, we're going to consider the non-squeezing result for the mass critical problem, okay? So we're going to work with the mass critical NLS and to keep notation simple, we're just gonna look at the cubic NLS in two dimensions, which is mass critical.  Okay? So from now on, NLS for us is just the cubic NLS on Earth. In the theorem, I want to talk about the following. This is joint work with Rowan-Killip and Xiaoyi Jiang. And it says the following thing.  So R, capital R, these are positive numbers, finite. Z star is an element of L2 of R2. L is an element of L2 of R2 of length 1. Alpha is a complex number and T is a positive number, right? Positive time. If the flow, a time capital T of the ball, this is the ball in L2, is mapped, right? If this flow leaves inside the cylinder, then necessarily, little r is bigger than capital R. Okay? Now, the incentive for us to consider this problem came from our attendance of a talk during the full semester at MSRI, last form, a talk in which Dana Mendelssohn presented her symplectic non-squeezing result for the cubic Klein-Gordon equation on the three-dimensional torus. Okay? At the end of her talk, there were several questions from the audience about the existence of a symplectic non-squeezing result in infinite volume. Okay? And at that time, indeed until this theorem, all existing symplectic non-squeezing results were in the periodic setting. Okay? And we're going to see that there is a good reason for that. Okay? So, that got us interested in this problem. Particularly, we're wondering whether there is an increasing obstruction to proving a symplectic non-squeezing result in infinite volume, or was that just an artifact of the methods used that far? Okay? So, let me quickly review a little bit of history. Okay? The very first symplectic non-squeezing result for a Hamiltonian PDE is due to Cuxin, okay, who proved symplectic non-squeezing results for flows of the form linear part, where the linear part is assumed to have discrete spectrum, and plus a smooth compact perturbation. He offered examples of such flows on tori. Okay? Then the next entry in our history is Burgan. He had two contributions. The first one was a paper in which he gave more examples of flows that fall on the Cuxin's framework, and another one in which he proved non-squeezing for the cubic NLS on the torus, okay, which does not fall on the Cuxin's framework. Then the I-team, Kolyanda Kils, Tafilani, Takaoka, and Tau, proved symplectic non-squeezing for KDV on the torus. Recently, there is a paper of Hong and Sunsik Kwon in the audience, who re-proved this result of the I-team dispensing with the use of the Miura transform, and they also proved symplectic non-squeezing for a system of coupled KDV equations, again, on the torus. We have a result by Ruben Gu, who proved symplectic non-squeezing for the Benjamin Bonamahoni equation, a close relative of KDV, again, on the torus, by proving that this equation falls on the Cuxin's framework. And then finally, let me just write it here, we have a result by Mendelssohn for the cubic Klein-Gordon equation on the three-dimensional torus. Now, Mendelssohn's result is a critical result in the same way that ours is a critical result. What do I mean by that? The regularity needed to define the symplectic form coincides with the scaling critical regularity for her equation. Now, in that case, we only know local well-poseness for solutions in the critical space, with a time depending on the profile of the initial data. So in order to prove her symplectic non-squeezing results for arbitrary time t, she assumes that the cubic Klein-Gordon equation is globally well-posed with uniform spacetime bounds. She assumes more than that. She actually assumes that various frequency-tragic versions of the equation are globally well-posed with uniform spacetime bounds. That assumption is stronger than the initial assumption. Well-poseness for the frequency-tragic equations implies well-poseness for the cubic Klein-Gordon. But the reverse is not true. I'll tell you more about that later, because it actually has a bearing on our theorem as well. Okay? All right, so what is so special about this periodic sitting? Yeah. To formulate the problem, you don't need global well-poseness. Well, you say, give me a time and I want to flow a ball, an arbitrary ball. If you take arbitrary, okay. Right? I mean, you want to imitate Gromov, right? You want the result as general as possible. You can always restrict this. Well, we're going to see soon why that is not the case in the critical case, because scaling tells you that you're going to have to do it uniformly, globally in time. I'll point that out when I get there. Okay? It's because you will need to prove stability of your finite dimensional approximation. In order to do that, you're going to have to work globally in time. No question? Okay. So what's special about the periodic sitting? Very briefly, right, in the periodic sitting, what one can do is take the solution and express it as a superposition of plane waves, right? So this is just the sum. Let's say periodic setting td. I'm summing over k's in zd, u hat of tk, e to the ikx. Okay? So if, right, I mean, if one takes the solution and truncates it to finitely many frequencies, then one obtains a finite dimensional Hamiltonian system, right? So truncating two frequencies, smaller or equal than n, either by using a sharp projection in the case of Burgan or a nice smooth projection in the case of the i-team, right? One gets a finite dimensional system. But on finite dimensional systems, Gromov's theorem applies, right? So Gromov is going to tell me that no squeezing holds for the finite dimensional system. So all that one needs to do is prove that solutions to the finite dimensional system are a good approximation to the solutions to the full equation in order to deduce that non-squeezing the A implies non-squeezing here, okay? All right. Now moving from the periodic setting to the Euclidean setting, it's not clear how one should define a finite dimensional approximation, right? In fact, that was one of the key challenges that we had to overcome, right? How do you define a finite dimensional approximation, right? Moreover, because the Laplacian or Euclidean space has absolutely continuous spectrum, one cannot even find a finite dimensional subspace of L2 that is left invariant even by the linear flow, right? All right. So what did we do? Let me just sketch the proof. We argue by contradiction. So assume that we have parameters as in the theorem, but we take little r to be strictly less than capital R, and the flow of time capital T maps this ball inside the thinner cylinder. We like to derive our contradiction. So what are we going to do? We are going to use a frequency truncated large box approximation to our NLS, okay? So I'm going to choose parameters, frequency scales going to infinity, and spatial scales going to infinity, and I'm going to consider the following finite dimensional system. Let me call it NLSN. I del T plus Laplacian, UN. What do I do? I insert a projection in the nonlinearity to frequency smaller or equal than NN, and I truncate every single copy of my solution to frequencies smaller or equal than NN to make sure that this is Hamiltonian. I'll write down the Hamiltonian in just a second, okay? Where do I pose this problem? Well, T is going to be in R, but X is going to be in TN, which is R2 mod ln Z2, okay? So I'm posing this problem on ever larger tori. I take more and more frequencies. I project on higher and higher frequencies, and I pose the problem on larger and larger tori. And I'm taking my initial data, UN times 0 to be U0N, a function in HN. Okay, what is this? These are the functions f in L2 of this large torus with a property that they do not have frequencies larger than 2N. Okay? So I have made it a finite dimension. Okay? And as I promised you, this is a Hamiltonian system. Let me write down quickly the Hamiltonian. So the Hamiltonian is HN of UN is half of gradient of UN squared plus 1 over 4, the projection to be considered smaller than NN of UN to the power 4. Okay? Now, because this is the finite dimensional Hamiltonian system, Gromov's non-squeezing theorem applies. So one can find witnesses to non-squeezing. Okay? So I can find U0N, okay, living in the ball centered at z star of radius R where this ball is in HN, such that the solution corresponding to this initial data to NLSN at time capital T lies outside a cylinder. Okay?  So this is bigger than, let's say, R plus R over 2. Now, you see that there is a little bit of fudging going on there, right? Z star and L are just some elements of L2, right? So in particular, there's no reason why z star should live over here. However, I can take z star and I can project it to frequency smaller or equal to NN. And as little N goes to infinity, I'm making a smaller and smaller error by the monotone convergence theorem, right? And at some point, because I have a positive distance between little R and capital R, if I draw that down, okay? At some point, that error becomes acceptable, okay? So I'm not going to write a projection over here. Just bear it in mind that there is one to make things perfectly true, okay? Similarly, over here, right? By making an acceptable error, I can replace L by a compactly supported function and as little N goes to infinity, it will eventually live in the domain of one of the ones that are of the torus, TN, okay? So what do I want to do? I want to take these witnesses to non-squeezing and I want to produce a witness to non-squeezing for our equation, the cubic NLS on the torus, on R2, sorry, okay? So what is the strategy? The naive strategy is the following. Take your initial data. It's mapped by NLS N to UN at time T and let's take weak limits. I recover U0 infinity over here. I recover UT infinity over here. And let us assume that we can prove somehow that these two solutions are related by the NLS flow. Okay, so what do I mean by that? That this function is the solution to the cubic NLS on R2 with this initial data at time capital T, okay? Now, if I can do that, then I am done, okay? And the reason why is because the complement of the cylinder and the ball are closed under weak limits, okay? So these being witnesses to non-squeezing will imply that this is a witness to non-squeezing, okay? All right, so can I make this strategy work? Well, if you think about it for a few minutes, you will see that even the simple step of passing to the limit and extracting those weak limits over there is a little bit controversial. And the reason why is because every single element in the sequence and the limit itself live in different Hilbert spaces, right? The ones above live on L2 of the torus and the other one lives in L2 of R2. So how do I pass to the weak limit? Well, what can you do if you have a function on the torus? How can you embed it to be a function on the plane, right? Let's say that this is TN and this is our initial data U0N. Well, what do you do? You cut the torus, you unwrap it, and you embed the solution in the plane, right? However, you have to be very careful where you cut the torus, okay? You cannot cut the torus at a point where let's say that the initial data has a bubble of concentration, a bubble of mass.  And the solution to the equation with one bubble on the torus looks nothing like the solution on the plane with two half bubbles, okay? So you're gonna have a lot of trouble proving that the weak limit is gonna be a solution to NLS, okay? So because of that, we have to cut at the point where the solution doesn't have a lot of mass. In fact, for the same reason that I just explained right now, not only do you have to cut it at the point where the initial data doesn't have a bubble of concentration, but the solution up to time capital T doesn't have a bubble of concentration for exactly the same reason, okay? So where to cut? Well, choosing the torus to be sufficiently large, right? By taking ln to be really, really large and using a pigeonhole principle, I can find the region on the torus where the initial data has tiny mass, okay? And that's where we're gonna cut, right? We're going to prove that we have an almost finite speed of propagation for the equation, right? Because the initial data doesn't have a lot of mass over here, the solution up to time T is not gonna have a lot of mass over here. In order to do that, we're gonna take the torus to be really, really large compared to the frequency scale ln, right? In particular, ln is gonna be much larger than ln times T so that the solution doesn't have time to wrap around the torus and put a lot of mass over here, okay? Are you cutting in fuller space or in physical space? Physical space. But you can have huge velocities. Well, the velocity, I have the projection. Remember, there is a projection to frequency smaller than ln in the nonlinearity. Eventually, we go to infinity. Yes, but ln is gonna go to infinity much faster, right? That's the whole point. ln is gonna be much, much larger than ln times the time T where you wanna prove non-squeezing, okay? All right, so that takes care of the weak limits, okay? Now, the next issue we have to deal with is stability. This is where I owe Sergio an answer, okay? So what do I mean by stability of the finite-dimensional approximation? I mean that as you increase the quality of your approximation, you get uniform spacetime bounds for the solutions, right? And that is gonna give you a fighting chance to prove that the limit is gonna be indeed a solution to NLS, okay? Now, this turns out to be a very hard problem because of criticality. And to make the discussion a little bit simpler, let us consider stability for this equation where we dispense with the geometry, right? Rather than, you know, asking for this equation on a torri, let's just consider it on the Euclidean space, okay? Even then is non-trivial, okay? So let us ask stability for NLSN but posed on R2, okay? Can we solve this problem? Now, because this equation has a scaling symmetry, you can use that scaling symmetry, and you will see that stability for NLSN on R2 is equivalent to uniform spacetime bounds for the following equation. I can replace all the NN by one, okay? Of course, in doing so, there is a vestige of the scaling that I just used, and that is that now I have to find uniform spacetime bounds globally in time, okay? Now, can we find uniform spacetime bounds for this equation globally in time? Well, if you look at Dotson theorem, he dealt with a cubic. What we're doing here is adding a projection to low frequencies. Surely you should be able to prove uniform spacetime bounds for this equation. However, it turns out that this is a strictly harder equation. Uniform spacetime bounds for this equation actually imply Dotson's theorem, but the converse is not true, and the reason why is because one can embed solutions to Dotson's, to the usual cubic NLS into the class of solutions to this equation by using scaling, okay? So then you say fine, you cannot use Dotson's theorem directly, how about the proof? Can you rescue the proof? And the answer is no, and the reason why is because these projections over here destroy the Morowitz monotonicity formula, okay? The reason why is because they do not commute with the weights in the Morowitz formula, okay? And there's no reason why the commutator should be small, okay? So, right? The previous authors also had to face this stability problem. Why was it easier before? What did they do? Well, in the subcritical case, right, let's say these results, well-poseness of the original equation actually the same method implies well-poseness and space-time bounds for the frequency truncated equation. Basically, because one has the luxury of a holder in time, okay, when deriving well-poseness. And what about the other critical result, right? What did it Mendelssohn do? Well, she assumed that the frequency truncated equation is globally well-posed with uniform space-time bounds, right? This is one of her assumptions. So, what do we do? Well, we developed a method to prove uniform space-time bounds, okay? Provided one modifies these projections slightly. So, what is the modification? Well, they are no longer the usual little or fairly projections, but rather they are projections that decay very, very, very slowly, okay? What is the advantage? Well, there are two cases you can consider. Either the initial data is very localized in frequency or it is supported on a very large frequency band, right? If it is very localized in frequency, then it meets these projections over here basically like coupling constants, because they vary very slowly. And because Dodson gives us uniform space-time bounds for the cubic NLS, we have space-time bounds for the cubic NLS with a coupling constant, okay? Now, what happens if the initial data is supported on a very large frequency band? Well, if the initial data is supported on a large enough frequency band, then by the pigeonhole principle, somewhere in that frequency band, a region where you have very tiny mass, and I split the initial data into two bubbles, right? One to the left of the tiny mass area and one to the right of the tiny mass area, okay? Now, we use an induction on mass argument, right? How does that go? If, right, once I have split, I split my initial data into two initial data, each of them is going to have mass strictly less than the original mass I started with. So I can solve the problem globally in time with those two initial data, and I have uniform space-time bounds. Now, because the two initial data were well separated in frequency at the initial time, right, you can prove that the interaction between the two global solutions are weak, and while the sum of the two global solutions is not a solution, it is almost a solution. So perturbation theory is going to give space-time bounds for the solution to this equation in that case, okay? So induction on mass when it's supported on very large frequency band, and simply Dotson theorem when it's very localized in frequency, okay? All right, so that's what we do. We come over here and we modify this projection simply by rescaling, okay? So what we did so far, we obtained stability for this equation but now posed on R2. So what do I have to do? I have to take the solutions that live in the plane, I have to wrap them around the torus, okay? And I have to prove that they are approximate solutions to that problem, okay? Let's put a deal over it, okay? Now, in order to have a perturbation theory good enough to do that, that means that I need to have a perturbation theory in critical spaces, which means that I need striccate estimates at the critical regularity, L2, right? Critical striccate estimates. However, Burgan tells me that they fail. There are no striccate estimates at the critical regularity, okay? So what do we do? Well, we prove striccate estimates at the critical regularity but they are again adapted to our parameters, ln and nn, right? If we take ln to be much, much larger than nn then the solution doesn't have time to wrap around the torus and you can prove striccate estimates in that setting, okay? So striccate estimates. If you want critical striccate estimates on the torus, okay? All right, so I'm a little bit out of time. Can I have two more minutes to finish? Yeah? All right, so the very last thing we have to do is we have to prove that these two weak limits over here are indeed related by NLS, right? We basically have to prove that a weak limit of solutions to NLS is a solution, a strong solution to NLS, right? So what we need is well-poseness in the weak topology, okay? Now, in the subcritical setting, well-poseness in the weak topology goes back to work of Cato. In the critical setting, the only result that we were able to find in the literature is a result of Bahou'i in Gerard who proved weak well-poseness for the energy-critical wave equation, okay? And in order to do that, they used the concentration compactness principle they developed for that equation. Now, let me just one minute explain to you why you need something as heavy as concentration compactness to do that, okay? So consider the following scenario. Let's say that in frequency you have, right, initial data looks like two bubbles. One supported around zero and one supported around Nn. Nn runs to infinity. Then you pass to the limit, right? The weak limit. All you recover is this initial data that lives at the origin, right? So, weakly this converges to zero infinity, okay? Now, in the subcritical setting, the high frequencies are weak. In particular, they are small in all the relevant norms. And it's not hard to prove that the solution with this initial data converges to the solution to this initial data weakly, okay? Now, in the critical setting, all frequencies are equally strong, right? So, in particular, the solution can... sorry, the initial data can have half its norm over here and half its norm over here, okay? So, in order to prove that the solution with this initial data converges weakly to the solution with this initial data, what we have to prove is that asymptotically the solution to the problem with this initial data is the sum of two solutions, one with this initial data and one with this initial data, okay? And this is precisely what the linear profiled decomposition does, okay? It gives you an asymptotic principle of superposition for a nonlinear equation, okay? Now, we also use concentration compactness to do that, but in our setting it's a little bit more complicated because there's also a change of geometry, right? This problem is posed on the torus while the problem we want to solve is posed on the whole plane, right? So, there's a change in geometry. And, of course, there's a change in equation because the Laplacian on the torus is nothing like the Laplacian on Euclidean space, okay? So, thank you. Sorry for taking so long. Any question, comment? I have a comment of no one minds, which is there's connection with your talk and with the talk on Monday afternoon of Law and Tonan whose one of his corollaries was growth of so-called norms. That's right. That was actually the original motivation for Cookson to prove his result, right? He argued that the non-squeezing theorem measures in a way how weakly turbulent a flow can be, right? It proves that the energy cannot fully evacuate the low and the middle frequencies, right? It cannot run to high frequencies. It can for one solution, but it cannot do it uniformly on both. Exactly. Not uniformly on both. Okay? Actually, can you allow me to do this? So, what is exactly the connection between the non-squeezing theorem? Weakly turbulent? I can allow you to make my comment. It's take a ball in L2, and you'd like to know, you'd like to say, can I uniformly make the Sobolev norm grow? Yeah. So, that's like saying, is there a map from this ball to the cylinder where the cylinder is the projection onto all high frequencies less than little r, but you allow the low frequencies to be, sorry, all the low frequencies less than little r but all the high frequencies to be free.   That is, you cannot uniformly send frequencies higher frequencies up. Just take L to be a character, and you get that statement. It tells you that the Fourier transform of the function at a frequency, right? The Fourier transform of the function at a frequency lies within a cylinder or not, right? Within a smaller circle, okay? So, you're interested in whether, you know, can you say something better about the resolution of a single frequency, even if you're willing to give up on all the other frequencies? And the answer is no, right? Not uniformly impulse. But you actually get the result on the... That's... Yes, for us, for us... Quantitativeism. Not more quantitative than what I wrote. Okay? But you can think of this result being on the plane as a statement about scattering as well, right? So, what does scattering intuitively tell you? This equation scatters, right? Dotson showed us that. Intuitively, it means that the energy believes any compact set. It just runs away to infinity. What the non-squeezing theorem says is that it cannot do this uniformly on both. Any other questions or comments? No, no. The wave maps, wave operators are simplistic. So, doesn't scattering... scattering reduce the question to a linear setting? That's not exactly the same one-squeezing, but... I'm sorry, what are you after? So, many of the symplectomorphisms to the... the wave operators. Scattering linearizes the problem. On a slimmer problem, you have non-squeezing. Because it's linear. So, that gives some non-squeezing. In terms of scattering for body, yes? For feet, yes. It completely distorts the ball and the cylinder. It disrupts the ball and the cylinder, but actually non-squeezing should not just be for balls and cylinders. You should have any set and symplectic capacity of the set. So, if it's a more general picture, which I think is not impossible. Well, it's not really clear how to define a symplectic capacity in infinite dimension. Right? I mean, I thought that the symplectic geometrists have been trying to do that for a while. But we have a... you do have some tools that you explain to us.  Any other questions? I'm going to send this speaker again.


Could the shortest receiver moss the tallest corner? Now Madden has an X-Factor ability called DoubleMe, which makes your receiver insane in single coverage. Let's push that to its limit. I made Jamar Chase five foot five, and I made all the corners seven foot tall. Then I had them run 1v1s. Weirdly, Burrow would randomly overthrow the ball. It was almost as if the receiver was so short that Burrow just couldn't hit him. But when they did connect, it was absolutely disgusting. The shortest receiver can moss the tallest corner. I just don't recommend it.


Hello, and welcome to today's stream Advert of day, Advert of code day six Sorry for being late. I was playing around with chat GPT making it tell me stories and the about Various things like the creation of Haskell But in the style of Tolkien All right, it's good fun Is it gonna replace people? Probably not, but I know that it will empower individuals to do a lot more with a lot less which is Not a bad thing Anyway, we're gonna do day six. We've been doing doing it every day. So a bit later today But maybe that's better. I don't know maybe more people are Available at this time. So we'll see see how people like it Yesterday we did some stacking And there was a lot of stacking actually, but most of it was just parsing and we had to Take this same This diagram here and kind of make it into something we could compute on Uh, so we did a lot of parsing and then the actual problem which is Insert into an int map folder boom But That's how it is sometimes um The old videos are up on the youtube. So if you want to see previous installments Uh, you see how people Write code in Haskell prefer to do that. Uh, I think the 2020 edition it was actually included in a paper Uh, which was like a study on how people write Haskell programs So We're doing good work here folks. Anyway, let's get started with day six Um Oh no, so we're on our way to an expedition to get star fruits. We're picking up fruits along the way But we're also getting them for solving tasks I don't know the lore Is a bit thin I feel but That's okay Right. So we're On our way to the fruit grove star fruit grove We get a handheld device And it's a communication system ah Maybe this is going to be like one of those um That kind of running things right It was a lot of fun last time Like I think in 2020 we had like this computer and we kept Maybe it was 2019 we had like a computer and we kept having to self modify the code and stuff. Anyway, so let's make sure we do it Extensive an insensible way today, right? um so I've had a significant experience dealing with Signal based systems Yeah, we did these Huh. Yeah, this is the int code. It was so much. I feel like I did it uh in 2019 I really definitely did the int code stuff and funny This is 2021. Yeah, I mean we did things in 2021 2020 we didn't do everything We managed to get quite far Oh, but 2019 apparently not all right Maybe I wasn't logged in at that point Because I feel like I Yeah, okay, I did it in 2017. I've been doing it, right? Anyway We're doing 2022 now Surely you'll have no problems. Let's do inspired by community broken He's a lock onto their signal. The signal is a series of seemingly random characters That device we use one at a time Okay, we We need to detect the start of marker In the protocol being sbs the start of a marker is indicated by a sequence of four characters that are all different The device will send your subroutine a data stream buffer And your server needs to identify the first position whether for most recently received characters are all different Specifically needs to report the number of characters from the beginning of the buffer to the end of the buffer And the first search for character marker Okay Okay, let's see um So I think this will like we're doing like a sliding window here, right? um And so either it's in the first four, right? Or and Okay, let's see Other first three characters will read There have been enough characters needed to find a marker. First time mark should first have the four characters received Making the most recent four characters and j q j because j is repeated. This isn't the marker The first time a marker appears is after the seventh character Arise once it does The last four characters you are j p q m which are all different in this case Your subroutine should report the value seven Because the first start of park packet marker is complete after seven characters may process Okay, so we have the an example That's out of here new file example Okay, and these are actually many examples, but We're ready to all the examples, huh? Okay Okay, so this is going to be easy to parse because it's essentially just a character stream, right read input File path to IO a list of stream Read input today is just gonna be Lines over read file read file Ah, but I need to do f map Because we're doing dot free. Okay Let's print out the example read input an example print gtao base three six And six dot chat We didn't save the example Okay, and we're not gonna have this line actually All right, we have the Arrays of strings Okay close others Now let's have the example right here Okay, and I want day six. Yeah. All right, so we have to check um First of all, we have to check how many there are. Okay, so wait, okay, so Because yeah, so value seven because the first startup argument I completed after seven characters have been processed. Okay Let's see, um Ask one takes a string and returns an int Ask one So we're gonna we're gonna process the initial for first, right? So C1 C2 C3 D4 and then the rest equals Okay, let's actually produce the first. Yeah c3 c4 Okay, so then we just say if let's import import In a dot set Okay, let's actually make this a bit different Equals task one prime A zero Task one prime a N in okay, then we actually match on them equals Okay, now we're just gonna be we're gonna do this crazy Stupid right um So this was position it actually starts at position four, I think Okay, so let's see here, uh, where S equals Let's import qualified Qualified as set let's say as is set dot from list C1 C2 C3 C4 uh So, uh, we say if length S equals So the size of the set is four. Let's see set dot size equals four Then we return and so um Basically if we have four unique characters Otherwise We do task one prime and plus one a C2 C3 C4 We we we you know, okay, and now let's write a task one prime Error End of input without marker I think this will work. Let me see print task one Uh print a map right crush 7 5 6 10 11 7 5 6 10 11, okay so That's a chunky boy Uh, maybe we're doing it super slow. I mean so we are doing it super slow well Okay, let's see because we should actually just kind of Remove from the set add to the set and So what do we know here? Okay, so Yeah, okay. Let's just see how how slow this is. I because I don't think it would be that slow If it is super slow, we change it actually Probably will be slow for task two. I don't know what task two is but But it usually gets slow. Okay, let's see if 16 25 is good All right, we got the first one now. Let's see if our Being lazy Actually didn't work out The start it's girl you're taking packets. We still isn't working. It's like other little messages. Oh my god Except because it's a 14 distinct characters rather than four ah now 14 We're not gonna do that right now. Let's see now. Let's rewrite this in a non lazy way Okay, uh Okay, so here we have the list And then this is gonna be this is gonna start off at Zero I'm just gonna start off at 14 Okay, um We are gonna say Case things set dot I had sets outside. Yeah So this set here, okay, let's say So what we should be doing is like not constructing the set every time right we should be Like adding the last one to the set And taking the first one out of the set and maybe the last okay. Let's let's let's not be so lazy. Let's say a Let's try and it's actually Yeah, okay, so see if we have to add c I see and see yes, okay now let's say So we start off with um Does we need to kind of keep track of the order of the letters? Okay, let me Uh, let's just see where That's equals take 14 What about a s comma r equals take 14 So we're gonna say a split at 14 Let me just see ggi a split at Three one two three four five Yeah, that takes three. Okay Okay, so this is gonna be the start and then we take we take 14 of them Let's see Task one right 14 equals Where let's call this Okay, sorry for being a bit confusing. I'm trying to kind of keep so we're gonna keep the track of the list It's gonna be like of length 14 and then we're always popping off the first one and And adding the latest one. Okay, so let's actually start at 13 here And then we are gonna start with We're gonna start with a set dot from List s and we're gonna start at we're gonna take 13 of them And we're gonna Give it the rest of the list And this is imp. Okay Uh, okay Ask one prime and A This is actually not gonna be s s Set up from list s Okay, so here we have the a In set In set charts we have the curve set And we have the next character and the rest of the characters Equals, okay, so pc pcs Previous characters previous characters set. Okay now We have a set of 13. Okay, so if c set dot member Curse set Now let's do it like this case if the c is the member so so adding the 14th character in um If it is a member Equals, then we return then we see when we we we return uh n plus 1 otherwise Except it was just 14 distinct characters So this is actually gonna be a Not set member right case a Not so if it's not in the set Then we return n plus 1. Okay, so otherwise it Ah, otherwise it's in the set otherwise it's a otherwise it's in the set so we um I think we're doing something wrong here, but let's see otherwise. It's in the set. So we do task one prime and plus one We'll do pcs Uh, and we append c to it bcs plus plus c And then curve set. Okay. How is it a set? Uh, that How do I uh remove from a set? How's everyone doing? Is this add and code hard to solve? and I would say that it's harder to solve than the ones we've seen so far See set up in a sense that like the input is actually Like it you don't have to parse it all right But uh, we're actually having to do some work. Okay. I think this won't work But um, but I want to see what it does 14 14 15 14 15 Yeah Okay, so we're doing something wrong. So first I was thinking like if we put it in the set Do we actually have to we can just keep putting in the set, right? I think we can just keep putting in the set Um I think yeah, let me just check Okay, we don't need any of this task and Curse set No, hey, it's all Uh, and then we have the C and Cs Okay, uh, if set dot size set dot size cur set equals 14 then uh Then we return n otherwise That's one prime n cur set set dot insert Wait It's insert. Yeah. Yeah C set out insert set Cs right because we want to see 14 distinct characters But we probably this probably won't work because we haven't we don't delete anything out of it uh Of course we need to do n plus one here 19 23 20 28 25 19 23 23 29 26 Okay, it works for the example But it doesn't work for anything else Yeah, we still we need to um, we need to remove the Ones uh that we previously added Okay, so if the set of the size of the set is 14 then we return n otherwise We insert this one, uh, and then a pc Set dot So now we we kind of we we remove that So we're keep we keep adding to the set It's just like the cs is Being empty equals Okay, I know said it's not a good message actually, uh, but uh Okay, so the set of the current set never okay, I think I might need to Okay, so it's a prime Yeah, so we like we insert and then check How's everyone doing today? have you tried Have you guys tried chat gpt? that uh Because it's uh, it's a lot of fun to play around with actually maybe I just I don't need the Maybe I don't need the Let's just do it the bad way. Okay Uh pc pc Yes, let's actually use a data.sequence here That's cool. So then we can Add to the beginning and end and it's good. Um Okay, so let me see no sets Import data Okay, um now So this is like a data structure that's it's not just the list you can always like operate on the front But for this one you can actually operate on both ends efficiently Okay, so uh We don't do c except from that we do um seek dot from list a s And then we just we don't need to seek for our For the rest ah Yeah, okay So here we have a and we have pc And it's this funny operator saying Okay, this is the front right, okay, so We do uh This is the front of the sequence Okay, so case uh escu and Okay, sequence. Okay, so case member I feel like we're doing uh too much here So here we will be That's not a workout the key things, huh? We just take the first s So pcs here Then we say R Say that 14 17 14 15 18 That's not at all 14 distinct characters rather than For so if we were doing this we're raised right we could just have like indexes and like okay, so Too early, right? So if the if the if the new one is not just go back to them So the thing is now we check but we don't check if all of them are distinct All right, so let's actually have the cursor here set up So we have the cursor here There's said What are you complaining about? This is supposed to be uh, no, yep Okay, so we have the cursor set. So we actually check a that size pcs plus I think maybe just the set dot in search Okay, so and I want to delete the So yeah, so I think we were like deleting before we were adding So now I want to I want to insert first I want to delete and then I want to add because I think before like we would delete kind of twice, right? PC Ah, but also yeah So the thing is that we were deleting But like maybe we'd seen it twice, right set dot from list or set So actually we're not gonna Do any clever things here. What is wrong here? Good much type a All right, 19 23 20 20 20 26. Okay, so the problem was that We were gonna be kind of adding and deleting from the set The thing is like if we'd seen a character like four times in the set Like the deletion of that character would delete it Like a lot of times, right? So this one works Like it was quite fast Username time 2 to 15. Okay. I mean, yeah, it took We're using all the commands here It took less than yeah 15 milliseconds. So that's just kind of Cheatsy boot time. Let's see All right Ugh That did take us longer than usual, right? but we did it And yeah, so I think like we were thinking about it wrong, right? We were gonna we would wanted to have this current set and then we wanted to You were I wanted to kind of add to it and delete from it, right? But uh Yeah, we should have been using something called a multi set, right? Which is also known as a bag. So then you Yeah, it's a set but So you kind of insert and delete into it, but it doesn't have an order And at that point just also it's also we could have just been looking at the PCS prime, right? Okay, well We did it And yeah, like I said, it's 15 milliseconds, right? So now we could have probably do this in a more optimal way, but it's It just seems fine, right? put example A6 All right, uh That's it for today We did have some viewers a sometime. I think uh a Some of my biggest fans are in the us very busy right now, so There was less chatting than normal. I think but I think I need to Because I put it on twitter, right? But I think like Last year I put it sometimes on Like the Icelandic programmers group and then I would get people Anyway, but also we should be on time, right? But uh, yeah, that was it for today and And you'll find me, uh playing chess after this or actually Yeah, probably have to review some homework from students. Anyway Thank you for tuning in And uh see you tomorrow for day seven, but you know, we've been doing this for almost a week now So I think, you know, we're doing quite well All right, but thank you for the day


Let us start. It is really my great pleasure to open this second day of what is our flagship conference monetary policy conference, I should say, called Ritching Science and Practice today, and I'm really very pleased and honored to announce that today's keynote speech will be given by Lars Peter Hansen who is the David Rockefeller Distinguished Service Professor of Economics at the University of Chicago and he is also a 2013 recipient of the Nobel Prize in economics shared with Eugene Ephama and Robert J. Schiller The methods that he has developed are widely used in empirical research in financial economics today and the Nobel Prize recognized his work on his work on the empirical empirical analysis of asset prices which has been used to test theories and models that have shaped our understanding of asset pricing and and there is probably no a graduate student of economics who has not used Professor Hansen's generalized methods of moments or GMM estimators his recent research attempts to quantify ways to model economic behavior when consumers and investors struggle with uncertainty about the future This work is highly relevant with respect to the uncertainty arising from climate change and the challenges resulting from it for for policy makers And I'm very pleased that Professor Hansen will share today with us some insights from his recent work on the challenges for central banks posed by uncertain climate change Uncertainty about how climate change interacts with the economy and the resulting macroeconomic policy consequences His observations are highly relevant for us as policy makers including central banks and supervisors And we share the views expressed by Professor Hansen and others that governments and parliaments have the primary responsibility to act on climate change At the same time within our mandate We at the ECB are committed to ensuring that the ECB takes into account the implications of climate change and the carbon transition in its monetary and supervisory policies To include climate change considerations in our monetary policy framework The governing council on the 8th of July 2021 Decided on a comprehensive action plan to include climate change considerations into its monetary policy framework And we have published an ambitious roadmap with concrete timelines for seeing a wide range of actions Among other things we aim to enhance the analytical capacity at the ECB in macroeconomic modeling to account for the impact of climate change on economic activity and inflation and on the monetary policy transmission Moreover, we have also set ourselves clear targets with respect to including climate change considerations in monetary policy operations in the areas of disclosure, risk management, collateral framework and corporate sector asset purchases Being conscious about the systemic risks related to climate change for the financial sector Climate-related risks will be an essential part of financial stability monitoring and prudential supervision of banks A few weeks ago, the ECB published its first economy-wide climate stress test concluding that firms and banks will be severely affected if climate change issues are not addressed And we have to act now Short-term costs for an orderly and swift transition to a zero carbon economy outweigh the costs over medium to longer term We will incorporate the experiences and lessons learned from this economy-wide stress test into the climate stress test of the euro system's own balance sheet and the bottom-up stress test by our banking supervisors Both stress tests are planned for next year Now, when they're telling you about all these things, I am aware of the skepticism that Professor Hansen recently voiced with respect to central banks and supervisors' ability to conduct these types of analyses And he is right to stress the highly uncertain relations between climate change, the macroeconomy, the financial system, and policy instruments And it is a great challenge to deal with this uncertainty when designing and revising our internal models and instruments As policymakers, we should not downplay the knowledge gaps that remain No one knows what combination of physical risks and transition risks will materialize in the upcoming decades We just know that a combination will materialize And to this, we need to adapt And there is no better way to kick-start this process than to start with imperfect data and analytical tools and improve as we learn more as physical and transition risks unfold Because the latter will not wait When we will embark on addressing the challenges emerging from climate change at the ECB, we will carefully listen to the evolving findings and insights of academics in both natural and economic sciences And I'm therefore particularly happy that we have today one of the leading thinkers in this field to share his thoughts with us I'm looking tremendously forward to an interesting talk and an enriching Q&A session And without further ado, and I've spoken much too long already Let me now give the floor to you, Professor Hans


What are some of the job and training opportunities that one might find in the shale industry? So workforce development is one of the key considerations of shale energy development that we've seen in the U.S., certainly in the Appalachian Basin and then beyond that. And the models that have been put in place, the assessments to determine what workforce needs would be, and then the models to be put in place to actually develop those same programs, either at an institutional level or also at a political level, meaning that there is often funding that's needed to make those programs operate. So all those are key considerations. We have found that about 70% to 75% of the workforce is blue collar, meaning that technical skills are key. So community colleges are a very big part of the institutional capacity to train workforce and will continue to be as we go forward. The remaining 25% is more of an academic type of process. So training the geologists and the attorneys and some of the environmental engineers and such are a key aspect of that as well. So collectively it's an initiative that's important to bring those multiple tiers of institutions together to develop the curriculum and also to make sure that they've done the proper assessments so that we're training the right number of workers in the right quantity with the right skill sets and bringing them into the marketplace as Mark is asking for that.