So today, what I'm going to do is I'm going to construct the metric on square root 8 thirds, leave of quantum gravity. And the way that we're going to build it is we're going to construct a variant of the Eden model in the continuum using SLE6. So SLE6 on square root 8 thirds, leave of quantum gravity. OK, so before I do that, let me just quickly review the discrete story and try to motivate you why you should guess that any of this is going to work. So let me begin by supposing that we have a triangulation of the plane. So this is the uniform infinite planar triangulation. And I'm going to let t be the root triangle of the UIPT. Then what you can imagine doing is you can imagine performing an Eden exploration starting from this marked triangle. OK, and this is going to be in the dual graph. So how does that work? So you imagine that you have your initial triangle, t. Let's say that's t. And then each stage, what you do is you look at the cluster that you've made. You pick an edge at random on the boundary, and then you just explore the opposing triangle. So maybe the first triangle you see is this one, because you picked this edge. And then maybe in the next stage, you pick, let's say, this edge. Then maybe you pick this one. And then every once in a while, you might see something like this. So if I pick this edge, I can have a triangle come all the way around and cut off a big bubble, et cetera. And because I'm starting off with a uniform infinite planar triangulation, the holes that I cut out here, these are going to be conditionally independent triangulations of the disk given their boundary length. And it's also going to be true that the boundary length of the infinite component, this is going to be a Markov chain. This is a Markov chain. And you can write down what the transition probabilities for this Markov chain are. OK, so this is one very natural model. And another very natural model that you can do on the UIPT is site percolation. So another natural model is you can do site percolation. And so here, you start off with a triangle. Let's say your root triangle. And I can imagine that I've prescribed some boundary conditions. Maybe this is black, this is black, and this is white. And then I can do a percolation exploration. So starting from here, I reveal this triangle. And maybe I see something that looks like this, in which case I'm going to go this direction. And then I reveal this triangle, et cetera. And then just like before, sometimes I see triangles that do something more complicated. So I could see something that, see a triangle that does something like this, et cetera. And the point is that the holes which are cut out by the percolation exploration, these have the same law as the holes cut out by the Eden exploration. And not only that, but it's also true that the boundary length of the infinite component, this evolves in the same way. So this evolves in the same way as in the Eden model. And so the way that you want to think of this is that somehow percolation and the Eden model, they're exactly the same, except the only difference is that when you do a percolation exploration, you just have to keep track of where you're going to explore from next. So in some sense, the Eden model, this is the same thing as the percolation exploration, except for you sort of forget where you're going from at each stage. So the Eden model is the same thing as percolation, except at each stage, you forget from where you were growing from. You resample it uniformly at random, and then you reveal another triangle. OK, so that's sort of a very simple observation. And it's also very natural to expect that the Eden growth, this at large scales, this looks like a metric ball, looks like a metric ball growth. And this was, in fact, proved in a work, relatively recent work, of Nicola Currian and Jean-François Lagal. So this was proved by Currian and Lagal. OK, so what are we going to do? So we're going to use this intuition, but in the continuum. And what we're going to do is we are going to construct the metric for Li-Ville quantum gravity using the continuum analog of this discrete picture. And so the continuum analog of percolation is just going to be in SL-E6. And the planar map is going to be played, the role of that is going to be played by the Li-Ville quantum gravity surface. So we're going to construct this metric by reshuffling an SL-E6 curve. OK, so that's the general idea. So before I jump into the exact construction and the proof of the metric property, let me just do a very quick review of some of the things that we talked about last time. So let me remind you of some of the objects. So when we talked about Li-Ville quantum gravity spheres, we worked on the infinite cylinder, which is just the product of the real line and the interval from 0 to 2 pi with the top and bottom of the cylinder glued together. And the law of the distribution, which describes the Li-Ville quantum gravity sphere, is very simple to sample from. And how do you do that? So you do it in two steps. So first, you take its projection. So you take the projection of H, which is going to be the random field, onto the space of functions, which are constant on vertical lines, to be given by 2 over gamma times the log of z, where here z is a Bessel excursion of dimension 4 minus 8 over gamma squared. And once you form this process, you then have to reparameterize time appropriately. And the way that you do that is that you take its quadratic variation to be just given by dt. So it has constant speed. So to have a quadratic variation with constant speed. And this describes one part of the distribution. And the other part of it is what you get when you project onto the orthogonal complement and the projection onto the orthogonal complement of the space of functions which are constant on vertical lines, which happens to be the space of functions which have zero mean on vertical lines. You just take that to be the same as a Gaussian free field on the cylinder. And that's the construction of the Li-Ville quantum gravity sphere that we're going to work with. And let me just emphasize a couple of points about this construction, which are going to be very important for what we're going to do in a moment. And that's that, first of all, the points which lie at plus infinity and minus infinity. These are special points. They're marked points. And they turn out to be quantum typical. And what do I mean by that? This just means that, in other words, if you pick two new points, so if we sample x and y from the quantum measure, mu sub h, so this is the area measure associated with our sphere independently, and then we perform a change of coordinates, which swaps these two points with plus and minus infinity. So then we perform a change of coordinates, which swaps plus infinity and minus infinity with x and y. Then the result is going to have the same laws that we started with. So then the resulting field is left invariant. And this is true, actually, modulo one small thing. So this is true modulo horizontal translation and a rotation about the line from minus infinity to plus infinity. And the reason that this is the case is that, or one way of seeing it, it's not at all obvious from the definition that these points, I mean, they're obviously special from the definition, but it's not obvious why they're special. And the reason that this is the case is that the way that we constructed this sphere to begin with was by somehow pinching a bubble off a quantum cone. And the special point in a quantum cone is somehow you should think of it as a quantum typical point. So really, these are these spheres. They're doubly marked because they have these two special points at plus infinity and minus infinity. So the way that we like to think of this object, which it's a sphere, and it's parameterized by the infinite cylinder, and it's described by the distribution h, and it's marked by the points plus infinity and minus infinity. This is an embedding of a doubly marked quantum sphere. And if we don't want to specify how this is the particular choice of parameterization or field, sometimes we just write s for the quantum surface and then the two points x and y that we are marking it with. OK, and then the other thing we talked about is what happens when gamma is equal to the square root of 8 thirds. And we explore our sphere with an SLE6. So I'm going to let eta be a whole plane SLE6 from plus infinity to minus infinity. So the picture is that you have your cylinder that looks like this. Here is, sorry, I'm going to make it go from minus infinity to plus infinity. So here's the cylinder. And you can then draw your independent SLE6 on top of it. So you have this curve. Sometimes it wraps around, et cetera. And so I tried to explain last time that the holes that this SLE6 is cutting out, these have a very special structure. These are quantum disks. And moreover, they are conditionally independent, given their boundary length. So just like in the discrete story when we explored the UIPT with percolation, OK. And moreover, the point which contains plus infinity, that's special.  It's sort of a special quantum disk. And the reason is that it somehow has to contain this extra marked point. And the way that that shifts its law is that it's just going to be a quantum disk, but it's weighted by its area. So it's size biased by its area. So we have this exact description of what's going on here. One other thing is that the process which describes the boundary length of the component containing plus infinity, this is going to evolve as the time reversal of a 3 half stable levy excursion, which has upward jumps, so with only upward jumps. So what do I mean by that? So the levy excursion itself has only upward jumps. But after you time reverse it, you get something which has downward jumps. And these jumps just correspond to the downward jumps of this time reversal. These just correspond to the boundary lengths of the holes that you cut out by the sl e 6. OK, so that's the general picture. And so what you get is the following thing. If you take a quantum sphere when gamma is the square root of 8 thirds and you explore it with an sl e 6, an independent whole plane sl e 6, then this produces for you a 3 half stable levy excursion plus the collection of quantum disks that correspond to the downward jumps. OK, so there's some way of going from this picture to this picture. And it's not at all obvious from what I've said so far, but it turns out that you can actually go in the other direction. So it's also possible to go this way. And what this means is that if you just observe this levy excursion and the quantum disks, as well as how they're oriented, then you can measurably recover the other picture. So these two things are actually encoding exactly the same amount of information. OK, so in other words, if you're given either structure, you can measurably recover the other. And this last point is something which I didn't say anything about, and unfortunately due to time constraints, I won't be able to say anything about how one constructs that arrow there. OK, so it turns out that these two objects are just exactly the same. And so what this means is that we have two ways of producing quantum spheres. We can either use our levy excursion or we can use a vessel excursion. There's actually a third way of producing a quantum sphere. Somehow using correlated Brownian motion, there's another excursion measure. But also due to time constraints, I'm not able to discuss that third construction. OK, so let me just mention that we have sort of two natural infinite measures that produce for us quantum spheres that we've talked about. So method one, this is where we describe the distribution, which corresponds to the field representation of the sphere. And this is based on a vessel excursion of dimension 4 minus 8 over gamma squared. And when gamma is equal to the square root of 8 thirds, this is just an excursion of a one dimensional vessel process or just Brownian motion reflected at 0. And the second method is you can just use a 3 half stable levy excursion. And what the 3 half stable levy excursion is encoding is it's encoding the quantum sphere plus the SLE6 on top of it. And this third construction, which I'm not going to talk about, somehow encodes the sphere. And then in addition to that, you have a space filling version of SLE6 rather than just SLE6 itself. OK, great. OK, so that's kind of the background, just reviewing what we did before. Now let me give you kind of the very high level idea for constructing this metric from these tools that I've just described for you. And I just want to emphasize a few things before jumping into the details so that there are a few things about it that hopefully you'll appreciate. So what we're going to do is we're going to build a growth process. And this is our QLE 8 third 0. And this is going to be built out of SLE6 on a square root 8 thirds level quantum gravity sphere. And this growth process, what it's going to be is it's going to be our candidate for describing metric ball growth. But it's not obvious from a mathematical perspective that this really is describing a metric. And then what we have to do is we have to check that the growth processes that we've defined, these actually do correspond to the growth of metric balls for some metric space. OK, so that's the general strategy. Let me make a few more kind of clarifying remarks about what we're going to do in just a moment before we actually do it. Number one, the metric that we're going to construct in just a moment, this is only going to be only defined, at least to start off with, on a countable dense set of points in our sphere. And these points, they're actually going to be, they're just going to be conditionally independent points. So IID points, x of i, which are all going to be chosen from the measure, so the area measure associated with this surface. And then it takes quite a bit of extra work beyond just checking the metric property. To show that, number two, the space that we construct is going to be, to show that the metric continuously extends to a metric on the whole sphere, so to a metric which is defined on the whole Lieveville quantum gravity sphere, which is homeomorphic to the Euclidean sphere. And it also requires quite a bit of extra work and another set of ideas to show that this metric space is equal to the Brownian map. So in the construction, if you're an expert in the Brownian map, you'll see some of the properties of the Brownian map sort of emerge, but proving that this is actually exactly the same as the Brownian map requires one to understand some additional properties which won't be obvious from this construction. And then finally, in the construction, we're going to show that this metric space is a measurable function of the field used to construct the Lieveville quantum gravity surface. But it's not going to be obvious that the Lieveville quantum gravity surface is a measurable function of the Brownian map structure. So there's also an extra argument to show that the metric space structure by itself determines the Lieveville quantum gravity surface that we started with. OK, and so these are all sort of separate issues. And what we're going to do today is we're going to focus on this one here, just constructing and making sense of the metric. And then if anybody's interested, I can explain later some of the additional ideas that go into establishing these points. Excuse me, Jason. Can you recall here, please, when you have the SLA6? How do you measure the boundary length? Oh, so the way that the boundary length is defined is that you can always apply a conformal transformation, which takes you, let me draw a new picture here. So you have your SLA6. You can always apply a conformal transformation, which takes you from the complement of this disk to, let's say, the upper half plane. And the field that you're going to get when you apply this conformal transformation is going to look like a free boundary Gaussian field on the boundary. So it has a well-defined notion of boundary length. Yeah, so in fact, I mean, what's very important is that the unexplored region is always a quantum disk, a levial quantum gravity disk weighted by its area. And these types of surfaces always come with an intrinsic boundary length measure. So you can always talk about the length of the boundary. That always makes sense. Right. But where it comes from is just the fact that these distributions arise from the free boundary Gaussian free field. And so you have the bicoformal transformation, then you have some multiplicative factor in the control zone. Yeah, that's right. There's always the correction term. That's right. So whenever you apply a change of coordinates, you always have the correction term. And that's just the right correction term so that areas and boundary lengths are preserved in the right way. Right. That's right. OK. Right. So let me write. OK. So before I do the metric construction, I have to emphasize one other point, which is very important to keep in mind when one works with these objects. And that's that there are, at this point, sort of two natural, well, let me just say, two parameterizations, which makes sense to talk about for the SLE6 on the square root 8 thirds Liouville quantum gravity sphere. So you have your SLE6. It's going from minus infinity to plus infinity, like this. And one kind of very important question is, what is the right time parameterization to talk about for this SLE6? And there are sort of two of them that make sense. So you can talk about, number one, the capacity parameterization as seen from plus infinity. That's one possibility. And the other one, which is going to be the one which we actually care about, is the time parameterization, which is associated with the 3 half stable Levy excursion. OK. And the reason that this is the one which is sort of natural to think about, the second one, is that this is just the one which corresponds to the continuum analog of what you would do if you were exploring a percolation exploration in the discrete one triangle at a time. So it's the continuum analog of the parameterization which corresponds to revealing one triangle at a time. OK. So when we talk about parameterizations for SLE6, we're not using the capacity parameterization that came from Schramm's original introduction of it. We're rather talking about the time parameterization which comes from the boundary length process. And the reason is that this is sort of the natural one to think of this somehow as being percolation in the continuum on Levy quantum gravity. And in a moment, there's going to be a third time parameterization, and that will be the one that we actually use to describe the metric. So now what I'm going to do is I'm going to start to construct the metric. And in the construction, it's very important to keep in mind the basic properties that one has for SLE6 on Levy quantum gravity, because the metric is going to have essentially the same, well, the right analogs of these properties. So how do you do that? I'm going to fix a positive parameter delta that I will eventually send to 0. And I'm going to define my approximations in the following way. OK, so here's the picture you should have in mind. So we have our sphere, which is described by the strip. And our process, I'm going to define my QLE process, which is going to go from minus infinity to plus infinity. And the way you should think about this is this is like a metric ball growth, where I'm only going to keep track of the component containing plus infinity, and the process itself is growing from minus infinity. OK, and so what do we do? So we are going to, number one, we're going to draw delta units of Levy process time. Sorry, let me say it this way, delta units of SLE6 with Levy process time, so the intrinsic notion of time, starting from minus infinity. And so I'm going to get my chunk of SLE6. And then what do I know? I know that the unexplored region is going to be a quantum disk weighted by area. And then I also know that the tip of my SLE6, this point turns out to be uniformly at random on the boundary of this component containing plus infinity. So what I can do is I can then resample the tip, so the location of the tip from the boundary measure. And then I can repeat. Just keep repeating this process. So in the next stage, I might, after resampling the tip, it might go to, say, this point here. And then I'm going to draw a bit more SLE6 from here. So maybe it does something like that. And then the tip of this SLE6, let's say at this point. And then I'm going to resample it from the boundary measure. Maybe I get something here. And then I draw more SLE6 again, like this. Here's its tip. And then you resample the tip, et cetera. And the point is that this is just like first passage percolation. The only difference is that rather than revealing the surface one triangle at a time, I'm revealing it by these chunks of percolation exploration. So this is just like the Eden model. This is an Eden model growth, at least in spirit, except my chunks that I'm adding, these are somehow pieces of SLE6, rather than somehow corresponding to a single triangle. And the point is that we don't know how to draw single triangles at a time in the continuum, but we can do these SLE6 explorations. And the reason that you would expect this to sort of work is that somehow it shouldn't matter what types of chunks you're adding at each stage, still the final answer should approximate the metric ball growth. Now, the reason for doing this construction in this particular way is that these approximations have a number of very special properties. And this is why the whole thing works. And that's that we know that, number one, the holes which are cut out by the exploration and the law of the unbounded component, including its boundary length, these things have, they're the same as what you get when you do SLE6. So they have the same law as in an SLE6 exploration of a levo quantum gravity surface. So we know kind of exactly what's going on, at least in terms of the holes, the boundary length process and the unbounded component of this process, regardless of how we chose delta. And so how are we going to define QLE? Well, at least initially, our candidate for metric ball growth is just going to be what you get when you take a sub-sequential limit as delta goes to 0 of this construction. So what's happening when delta goes to 0 is that you're just re-randomizing your tip more and more and more frequently. And because of how it was constructed, these two properties are also going to hold for the limit as delta goes to 0. So we're going to know that the boundary length, at least at the moment, evolves as a 3 half stable levy excursion up to time reversal. We also know that the holes are going to be quantum disks and the unbounded component is going to be a quantum disk weighted by area. So properties 1 and 2, these are going to hold for the sub-sequential limit. OK, and then let me also say that there are also some things that you could be worried about here. So number one, we took a sub-sequential limit and not a true limit. And you could ask whether or not the metric that we're going to define is going to depend on the choice of sub-sequential limit. It turns out that it's not going to. And sort of after the fact, much later, it's possible to show that this really is a true limit. And then the second thing you could worry about is that the construction involved additional randomness beyond the Leaville quantum gravity surface. So we had our SLE6s we were drawing. And we also had these points that we were re-sampling on the boundary of our growth. And you could worry that somehow this extra randomness is going to persist in the limit and not give you something which is measurable with respect to the Gaussian free field. But this also turns out not to be the case. And that will all come from the proof. So it really is a well-defined metric that is measurable with respect to the Gaussian free field and does not depend on the choice of subsequence. And as I said, sort of after the fact, it's possible to show that this really is a true limit and not a sub-sequential limit. OK, right. OK, so the notion of time that comes with this limit, the sort of a priori notion of time, is the wrong one because it's the one that comes from the SLE6. So this comes from the Leaville excursion. This is the wrong time parameterization for the metric. So this is not the correct time parameterization for the metric. Time equals distance to the root. In the metric, time equals distance to the root, right? That's right. That's right. So the notion of time that our growth process is currently equipped with is the one that you would get if you were doing the following thing. So let's say this is your root triangle and your UIPT. And let's say you did the Eden model where in each unit of time, you revealed one triangle. So that was one unit of time. Then the next unit of time, you did this. Maybe the next unit of time, you did that, et cetera. So that's sort of the wrong notion of time because when the boundary length of your cluster is getting longer and longer, you should be revealing more and more triangles because each unit of time, you're supposed to be revealing one distance one. So all of the triangles which are adjacent to your cluster. And so you have to make a time change. So again, in the Eden model, with the correct time parameterization, you add triangles to your cluster at a rate which is proportional to the boundary length. And so we want to make the corresponding time change now for our QLE. And so if x sub t is equal to the boundary of the complementary component, which is containing plus infinity when we're doing our QLE growth, then from the construction I've given so far, this is just given by the 3 half stable levy excursion up to time reversal. What we want to do is we want to make a time change so that we are in fact growing at a rate which is proportional to its boundary length. So we're going to change time by, we want our time to be, let's say, s of u, which is the first time t that the integral of 0 to t of 1 over xs ds is bigger than u. So that's the time parameterization which says that we're now growing at a rate which is proportional to boundary length. Is that the natural thing to do? And so what happens when you do that? So before the time change, x sub t was a 3 half stable levy excursion. And after the time change, so x sub s of u, for people who know about these things, it doesn't actually matter what the definition is. But this is going to be a continuous state branching process with branching mechanism u to the 3 halves, up to time reversal.  In so particular, the distance from minus infinity to plus infinity, in terms of levy process time, it's just going to be equal to the integral from 0 up to, let me say, xc of 1 over xs ds, where xc is the first time t that xt is equal to 0, the first positive time that it's equal to 0. OK. All right, so now I think I'll maybe we'll just take a short break. And then in the second half, we'll come back and prove that this defines a metric space. OK, so maybe I'll start again. OK, so I just defined the metric ball growth that goes from minus infinity to plus infinity. How are we actually going to extend this? So we're going to assume that we have points xi. And these are going to be iid from the quantum measure, so from the area measure on your sphere. And what we're going to then do is given xi and h, what we're going to do is we're going to sample conditionally independent QLEs from each xi, so from each of the xi. And actually from each xi to each xj. And then what we're going to do is we're going to set the distance between xi and xj. This is just going to be the amount of time that takes the QLE from xi to absorb. OK, and now I've defined this function, which is sitting on top of the level quantum gravity surface marked by these points. But as I said before, there's in principle a lot of extra randomness here, because I have, well, in the definition of QLE itself, there's a lot of extra randomness. And beyond that, I have sampled a bunch of conditionally independent QLEs given my Gaussian free field and the collection of marked points. So that's something to keep in mind. OK, so what we're going to do now is we're going to show that this d is a metric on the points xi. And so to do this, there are three things to check. Well, actually, two things. We're going to show that d is symmetric. So this is something which is not at all obvious if you think about it, because let's say in the picture where I have minus infinity and plus infinity, what I have done is I've sampled two conditionally independent QLEs, one which is going in this direction. And I've defined the distance from here to here to the amount of time it takes this guy to make it to infinity. And the distance in the reverse direction is defined with a conditionally independent QLE going in the other way. And from the way things are defined, it's not so obvious actually that this is going to be something which is symmetric. And then number two, we need to show that it satisfies the triangle inequality. And actually, the proof of part two is going to somehow come from one. And the proof of one will also imply three, which is that the metric d, this is determined by h and the locations of the xi and does not depend on the subsequence. So it's not going to matter which sub-sequential limit we chose, and that will come from actually just the proof of one. So really, the heart of the whole argument is going to be this proof of symmetry, because once we have that, that will sort of give everything. OK. Excuse me, why would they all be conditionally independent? I've just chosen them to be. Right, but what's the motivation? Because that's the most natural way to couple things together when you don't understand their joint law at all. If you're going to put them in the same space and you don't understand how they're supposed to be related, just take them to be conditionally independent. It's also crucial for the proof, too. It's very important. OK, so now I'm going to explain the argument for symmetry. And so how does that work? So I'm going to imagine that I have, in general, my quantum sphere, Sxy, and it's marked by the points x and y, which are quantum typical. OK, and I'm just going to focus on the QLE, which goes from x to y, and the one which goes from y to x. So gamma, this is going to be the QLE, which goes from x to y. And gamma bar, this is going to be equal to the QLE, which goes from y to x. And here, just to emphasize it again, gamma and gamma bar, these are going to be conditionally independent, given the underlying surface. And what we want to show is we want to show that the amount of time it takes this QLE to go from x to y is the same amount of time it takes its time reversal, not its time reversal. It takes gamma bar, the other QLE, to go from y to x. And so how is that going to work? I'm going to define a law theta. And this is the law of S, x, y, gamma, gamma bar, and some additional randomness, u, where here u is going to be uniform between 0 and 1, independent of everything else. So that's what we're going to do. And the way that we're going to show that these two distances are the same is that we are going to define two new measures. So I'm going to define d theta. So theta x to y is going to be the measure whose radonichodium derivative with respect to theta is going to be the distance from x to y. And I'm going to define my other measure, d theta from y to x. This is going to be the measure whose radonichodium derivative is dyx. And because we don't know symmetry yet, it's not clear that these two measures are the same. And what's our goal going to be? Our goal is going to be to show that theta of x to y is equal to theta from y to x. And once we show that, then this has to be equal to that by the uniqueness of radonichodium derivatives. So this is what we want to do, because this then implies that the distance from x to y is equal to the distance from y to x. And as you're going to see, this is going to somehow boil down to the reversibility of whole plane SLE6. And I should mention that parts of this argument are actually based on another work, which is due to, so this is based on or related to a strategy developed in another work by Scott together with Sam Watson and Hao Wu, where they construct a metric on CLE4, somehow using an argument which is related to what I'm going to show you now. OK, so this is what we want to do. And there are sort of two steps to proving this symmetry statement. Here's the first one. So I'm going to define a stopping time tau. And tau is going to be a uniform between the distance of x and y. So it's just u times the distance from x. The amount of time it takes for the CLE to go from x to y. And I'm going to let tau bar be the first time t that gamma bar of t intersects gamma tau. So here what I'm doing is I'm drawing one of the QLEs up until a typical time between its distance from x to y. And I'm going to draw the other one until it hits it. And similarly, I'm going to let sigma bar, the u times, the time it takes to go from y to x. And I'm going to let sigma be the first time t that gamma of t intersect gamma bar sigma bar is not equal to the empty set. OK, and what is the first step? The first step is to show that, and this is really the heart of the argument, it's to show that the theta x to y law of s comma x comma y, gamma up to time tau, and gamma bar up until it hits it. This is equal to the theta y to x law of the same thing, except for its gamma up to time sigma, where remember sigma is the first time that gamma hits gamma bar up to time sigma bar. So we're first going to check that. And then there's a second step, which one also has to check. And that's that you need to show that the theta x to y law of gamma and gamma bar, given everything that we've seen so far, so s, x, y, gamma up to time tau, and gamma bar up to time tau bar, this is equal to the theta y to x law of gamma and gamma bar, given everything but with sigmas in places of tau. So s, x, y, gamma up to time sigma, and gamma bar up to time sigma bar. Because we checked those two things, then the two laws have to be the same, and therefore the radianicademe derivative is the same, and therefore the distance measured from the left is equal to the distance measured from the right. OK, so due to time constraints, I'm just going to focus on the first step, because the second step is actually very believable if you think about it, because this is just something about the sort of Markovian nature of this construction. Because we took everything conditionally independently, these are all sort of stopping times. It's natural to believe that these conditional laws should behave well. And it's sort of a technical exercise to check that this is all the case. So somehow all of the meat is in the first step of the argument. And so as I said before, the way that this is going to go is we're going to reduce step one somehow to the reversibility of whole plane SLE6. OK, so that's the idea. And now let me remind you that we have three, well, I've described one so far, but there are sort of three natural measures that we're going to consider on quantum spheres, on square root 8-thirds,劉ville quantum gravity spheres. And somehow by understanding these measures, this symmetry statement is just going to pop out. So the first one is the one which is induced by N, where N is the excursion measure for a 3-half stable levy excursion. OK, so if you have a 3-half stable levy excursion, then you can produce a quantum sphere with an SLE6 just by associating with each jump a conditionally-independent quantum disc. So there's this measure. The next natural one is you can take the one which is induced by taking the measure N. And then what you do is you add to it another variable T. OK, so there's a second variable here, small t. We're here dT is Lebesgue measure. T is equal to the length of the levy excursion, capital T. OK, so this second measure, all that I'm doing is I'm taking my levy excursion and I am biasing it by its length and then picking a typical time between 0 and the length of the excursion. So in this second construction, the marginal of the levy excursion, so of x, this is just given by just taking N and waiting it by its length. So it's just a size bias levy excursion. OK, and then the third possibility, the third measure is the one which is induced by the measure 1 over xT times the indicator of the event from 0 up to the length of the levy excursion. And then dT again is Lebesgue measure and this is the levy excursion measure. Right, OK, and so here in this third construction, the marginal on x is just N weighted by the integral from 0 to T of 1 over xS dS. OK, and so the way that we think of these things is that here you're just waiting by somehow the length of the levy excursion in construction 2. And in construction 3, if you think of this levy excursion as somehow being related to the QLE, then here you're somehow waiting by the distance. It's somehow a distance weighted construction. So this waiting is like waiting by the distance between two typical points on the sphere. OK, and let me give these measures names. So this one, this measure sometimes we just refer to as M2 SPH because this is describing a sphere with two marked points. This measure is a sphere with two marked points where we've weighted it by the amount of time it takes the SLE6 to go from one point to the other one. And then the last one is like it's a measure in a sphere with two marked points where somehow you're waiting it by the distance as measured by the SLE6 to go from one point to the other point. And what do we want to show? We want to show the following. So let's say this is my strip. Then if I produce a sample from M2 SPHD, so this consists of a sphere together with an SLE6.  So eta here, this is the SLE6. It's going from minus infinity to plus infinity. And I've drawn it up to the special time t that's associated with this measure. And then I'm going to let eta bar be the time reversal of eta. And I'm going to let t bar be the first time that eta bar hits eta up to time t. So I have this picture where I have drawn one of the SLE6s, sorry, the SLE6 up until a typical time. And then I draw the other, I draw its time reversal, eta bar, up until it first hits eta. So this right here is eta of t. And this point down here is eta bar of t bar. And so what do we want to do now? So if I let x be the surface, which is separated from plus infinity by eta up to time t, x bar is going to be the surface separated from minus infinity by eta bar up to time t bar. And then b is going to be what's left. So it's the remaining surface. So in this picture, the stuff cut up by the white curve corresponds to x. The stuff cut off by the purple curve corresponds to x bar. And the stuff which remains is just b. And what we want to do is we want to show that, so our goal is to show that the law under this particular measure of x, comma, x bar, comma b, this is equal to the law of x bar, comma, x, comma b, OK? So we want to show that if you look at the surface cut up by the first SLE6 drawn up to the typical time, the second one until it hits it, then you can't tell which one is which. They look exactly the same. And the reason that that suffices to finish the proof, the reason that this suffices is that you can take this picture. So let me copy it down here. Save the same picture I had a moment ago. So here is eta up to time t. Here is eta bar up until it hits it. So this is eta bar of t. If you take this picture with the two meeting SLE6s, and then you reshuffle it, so you perform the tippery randomization procedure, then what you're going to do, what you're going to end up with are not two meeting SLE6s.  You'll have something like a QLE growth gamma up to time tau. And you'll end up with an eta bar will turn into gamma bar up to time tau bar. And if this picture is symmetric, then what you get when you perform the reshuffling operation down here is also going to be symmetric. And then this will then prove exactly what we need to show what we needed for the first step. So really the heart of the problem is just to understand this sort of strange symmetry statement for SLE6s on a levopontum gravity sphere. And so where does it come from? It's going to come from, again, the reversibility of SLE6. And why is that? So for a moment now, let's prove an analogous symmetry statement for not M2SPHD, but M2SPHW. So in this measure, what you have is you have a quantum sphere. And on top of it, you have your SLE6, which is going from minus infinity to plus infinity, like this somehow. And the measure is what you get when you take this picture under the original quantum sphere measure and you weight it by the amount of time it takes the SLE6 to go from minus infinity to plus infinity. So what you see here is SLE6 on a quantum sphere weighted by the time to go from minus infinity to plus infinity. And in addition to that, you have a time t, which is uniform between 0 and this time. So the structure you have is you have this sphere. On top of it, you have an SLE6. And then you have a time which is chosen uniformly from the amount of time it takes for the SLE to go from here to here. So you have a marked time. Let's say it's this time. And the point is that this picture here is completely symmetric because SLE6 is reversible. And the time was just chosen uniformly from the amount of time it takes to go from one side to the other. So in particular, under this law, what do we know?  Then by the reversibility of SLE6, this whole picture is symmetric, we have that the M2 SPH w law of y, y bar q is equal to the MSPH law of y bar y and q. So all that this is saying is that in this picture here, you can't tell in some sense in which direction this was drawn. And this is just because you can't tell in which direction in SLE6 was drawn. OK, and so to finish the proof now, we want to deduce the corresponding thing for the law M2 SPH d. Now remember, in the law for M2 SPH d, we ran the SLE6 up until the special time t, and then we ran the time reversal up until it first hit the SLE6 in the first direction. And the difference between the two laws is that in M2 SPH d, we have this extra factor 1 over xt appearing. And somehow we want to make that appear under the law M2 SPH w that we had over here. OK, and so how is that going to happen? So the idea is very simple. So what I can do is I can assume that I'm working under the law M2 SPH w. And I draw my SLE6 up to the special time t. And then I can condition on the event that this time is actually a cut time. So if I condition eta of t to be a cut time, then what's going to happen is that that's like conditioning the time reversal to first hit eta exactly at its tip. So then this becomes eta bar, first hitting eta at exactly at its tip.  So then yy bar q will still have the same law as y bar yq because I took a measure which was symmetric and I conditioned it on a symmetric event. OK, but the point is that the probability that this event happens is just going to correspond to the inverse of the length of eta at time t. So this law, it turns out, is exactly equal to the M2 SPH d law of x, x bar, and v. And again, the reason for this is the following. Let me draw the picture again. So let me just explain that in a little bit more detail. So if this is the strip, then how would you make it precise the event that you're conditioning on something being in a cut time? Well, you imagine you draw the first process up to time t. And then you can condition on the event that the time reversal hits the first process in some interval, like it could hit the interval from here to here of length epsilon. So if that's an epsilon length interval, then the probability that this guy is going to first hit eta exactly in that interval is going to be of order epsilon over xt. If xt is the boundary length, then epsilon is the length of this little thing. And so when you condition on a time being a cut time, it exactly is going to introduce the extra factor which allows you to go from NSPHW to MSPHD. And so this symmetry statement falls out. But of course, there's quite a bit of work that goes behind checking that this is actually the case. So one has to be careful with this conditioning argument. But that's sort of the main idea behind the proof. And this is sort of the heart of the argument that this defines the metric. OK, so let me just finish in the last few minutes by explaining a few more things. So at this point, what does one have? So you have a metric on a countable dense set, so a sequence of iid points on a square root 8-thirds level quantum gravity sphere. OK, and what one wants to show to sort of finish this program is to show that this metric somehow extends continuously to something which is homomorphic to the real sphere, and it is the Brownian map. And then we just sort of very briefly explain in the last few minutes how one checks these things. So to show that the metric continuously extends, there is a particular trick that we've used many, many times, a number of times in these papers, which allows one to control the size of QLE. So you have to control the size and shape of a QLE growth.   And that's that if you're working on, let's say, a quantum sphere and you grow a QLE up to a given time, then the unexplored region here is going to be a quantum disk. And one way of thinking about that is that this just tells you that if you apply the conformal transformation phi, which takes you from this complement to, let's say, Euclidean disk, then you know that the field H composed with phi inverse plus Q times the log of phi inverse prime, this is a quantum disk. And let me call this H tilde. And if you somehow want to control how rough the boundary of this QLE is or how big it is, basically, one way of controlling it is trying to understand this conformal map here. And so what you can do is you can just solve, you can use this equation and solve for the log of the derivative. So you have this change of coordinates formula where you know that this comes from a quantum sphere, that comes from a quantum disk. They're coupled in some complicated way. But anyway, it's just going to be true that phi prime is equal to, well, you can solve for the log of phi prime. So Q times the log of phi prime is equal to H tilde minus H composed with phi inverse. And because this is related to a Gaussian free field, and that's related to a Gaussian free field, you can use this to bound the behavior of this map and argue that these QLEs are actually not going to be too large. So this constrains the behavior of phi prime, and that constrains the size of the QLE. And arguments of this type are actually good enough that you can show that the metric space defined in this way does, in fact, it continuously extend to the case of the sphere. And once one does that, then one can also show that this metric space is not only going to continuously extend to something which is homomorphic to the sphere, but it's also going to be geodesic. It's a geodesic metric space. And essentially, to show that the metric space is actually the Brownian map, what one has to do is somehow understand how these geodesics interact with each other. But since now, well, there isn't very much time left. I won't say anything about that. But if anybody's interested, I'm happy to discuss it offline. OK, so maybe I'll stop there. You don't have a responsible person, so I guess I'll ask. Any questions? So where's the Brownian map? Yeah, the Brownian map. Oh, where does it actually appear? So this metric space is the Brownian map. It's the same metric space as the Brownian map. It's not obvious that it is the Brownian map, because the way that the Brownian map is defined is somehow not obviously appearing in this particular construction. But the way that you can see that this is the Brownian map is you try to understand how geodesics in the Brownian or sorry, in the in Leaville quantum gravity interact with each other. So you imagine you've grown a metric ball, which goes from minus infinity to plus infinity. This is your field metric ball. And what you can then do is then you can draw the geodesics, which go with some kind of tree of geodesics going from the boundary of this ball back to minus infinity. And what you can do in both the Brownian map setting and also in the Leaville quantum gravity setting is you can understand the law of this geodesic tree, not only at a fixed radius, but you can understand what happens as you move the boundary of the metric ball back in time and get closer and closer to minus infinity. And essentially to show that a metric space is the Brownian map, you just have to check that this geodesic tree has the right sort of behavior. And it also evolves in the correct way when you move the boundary of the metric ball back to minus infinity. And then you also just need to know that somehow this property is invariant under the operation of sampling these two points. And once you check that, then it's possible to show that it is the Brownian map. So those two properties essentially single out the Brownian map. On the Leaville quantum gravity side, somehow where these properties come from is from the sort of construction itself of QLE. So roughly speaking, you need to understand how these boundary lengths between geodesics evolve. And what you need to know is that they evolve as independent continuous state branching processes as you move the metric ball back in time. And where that comes from is in the approximations to QLE, basically what you're doing is you're somehow attaching little pieces of SLE6 if you want to advance the metric ball either forward or backward. And when you attach a little piece of SLE6, let's say to here, what happens is that this interval is going to get a Levy process increment. And it's going to be independent of, it's not going to depend on these particular intervals. And so when you draw these little SLE6s in, the different intervals are getting independent Levy process increments. And somehow you can extract from that that the boundary lengths really are evolving in the correct way for it to be the Brownian map. So you don't see the Brownian snake appear exactly, but rather you see this sort of geodesic tree structure, which is built into the Brownian stake appearing in Leaville quantum gravity. Yeah.


It's daily and Sunday, Los Angeles Times. It's 530 KECA, AM & FM, Los Angeles. If it's 3% you want most, open a savings account at Coast. Yes, Coast Federal Savings pays you more, 8th and Broadway, 2nd floor. So if you want earnings and safety, Coast Federal Savings is the place for your savings account. You see, beginning July 1st, your savings account at Coast Federal pays you 3%. And when you open a savings account at Coast, it's protected by federal insurance and over $50 million in assets. Find out about higher earnings, greater protection. Visit Coast Federal's banking quarters at 307 West 8th Street, just west of Broadway, or phone Michigan 4343 and ask for Ray Smith. That's Michigan 4343. Coast Federal Savings is the largest savings and loan association in the West. Remember, if it's 3% you want most, open a savings account at Coast. Coast Federal Savings pays you more, 8th and Broadway, 2nd floor. Safest and the best, largest in the West. Coast Federal Savings, C-O-A-S-T, Coast Federal Savings. The following program is transcribed from an earlier network broadcast. And now ABC brings you a special on the spot broadcast. Ladies and gentlemen, we're speaking to you from Cheyenne, Wyoming, a town whose residents woke up this morning to find that their city is changing its name. This town, capital of the state, is being renamed today in honor of one of America's best loved and respected characters. Since morning, a tremendous celebration has been in progress. A celebration which has moved from the downtown area to the grounds in front of the state capital building. We've set up special broadcasting facilities on the rotunda just outside the entrance to the state house, where we're right in the thick of things. And it's from this on the scene spot that we're bringing you this special ABC broadcast. From our position on the reviewing stand, we've been watching a parade of Gaelic room floats, horsemen, vans, cowboys, bugle corps, troops of Trottergaard Indians, and pretty girls with typical western costumes. The parade headed by the mayor and his guests has recently ended, and the governor and scores of other prominent citizens are waiting right here at the state house to receive the nationally-famous guest in who's honor the city is taking the new name. I'm going to turn the microphone over at this point to our special ABC announcer, Star Yellen, who will bring the mayor, governor, and Cheyenne special guests to the microphone just as soon as they appear. Take it, Star. Hi, everybody. Ladies and gentlemen, I hope you've been hearing me. This is a big, a serious celebration I've ever seen. Are we still on the air? I hope we are, because we've really got something here with all this terrific climber going on. We've got our microphone, but it's right here in the middle. Oh, I give up. I have to give up, so help me. Well, folks, anyways, you can see the crowds in the excitement out here today. You'll think this was the 4th of July, New Year's Eve, and DJ Day all rolled into one big, huge celebration. Yes, everybody seems to be here, kids by the hundred, cowboys from the range, Indians from nearby reservations, chief ranchers, I see a couple right over there, cattlemen, soldiers from Fort Warren, state and city officials, all timers who still remember Kit Carson and Calamity Jane. There are Sue Braves in tribal dress, the most colorful assemblage of western characters you've ever seen. They've been parading through the streets of Cheyenne for the past hour, and now they're gathered here on the fast lawn in front of Wyoming's beautiful state capital built. They're waiting for the guest of honor, the man for whom the celebration is held, and in whose honor the city is changing its name. This man is so famous that almost any boy or girl in the country could quickly identify him on site. Now, he and Mayor Nelson of Cheyenne are in the state house now, where they've gone to pay their respects to Governor Hunt. We've been told, yes, we have been told that both will be out in just a moment, together with the governor. Now, let's see, I think in just a moment here, we'll have him right here. Yes, I can hear the crowd now. Here they are. Uh-oh. Now, here's the mayor coming out of the state house right now. Mr. Mayor?  Uh, Mr. Mayor, let me say that never before have I seen such a genuine enthusiastic reception, given like that shown today. You know, I can hardly believe my eyes on the whole thing. Well, speaking from a fellow townsman and myself, Mr. Elman, I can't say I believe mine either. It's the biggest day for us since back when Wild Bill Hickok, Flammity Jane, Kit Carson, and some of the famous old timers used to blow into town. And don't think for a minute that there aren't half a dozen old timers standing right here in the crowd. Right this minute. Uh-huh. Well, we can certainly believe you, Mr. Mayor. You know, I've seen a lot of mighty colorful old timers here in Cheyenne, and of course, some pretty sharp young timers here, too, like yourself, if I may say so. I understand you're a veteran of the last war. You were a pilot in the Army Air Corps, weren't you? Uh, yes, that's right. I spent five years in the service, but about 20 months of that time spent overseas in the ETO as a photographic pilot flying B-25, 17, and 29. Uh-huh. Mr. Mayor, as a young timer, just how old are you, sir? Just 32. 32 a mayor of Cheyenne. Well, I'd say that you young timers are giving the old timers a mighty close race when it comes to doing things in a big way. Hey, hey, wait a minute. Wait a minute here. Let's pause. Look who's coming out of the State House. It's the man in whose honor the city is changing its name. He's walking over toward our microphone. Here he is, the Lone Ranger! And fellow citizens, in honor of the 15th anniversary of the Lone Ranger, a man known and respected throughout the West for his unswerving efforts in upholding law, order, and fair play, we today are doing something that no other town in the history of the state has ever done. By virtue of this authority, as Mayor, I have proclaimed the name Cheyenne, to be officially changed to Lone Ranger Frontier Town, and I have made the Lone Ranger Honorary Mayor. I will now read the official proclamation. Whereas the Lone Ranger has come to town, now, therefore, by virtue of the power invested in me as Mayor of Cheyenne, I do hereby change the name of this city to Lone Ranger Frontier Town. Simultaneously, I relinquish my rights and title of office and appoint the Lone Ranger Mayor of the City. In witness whereof, I have hereon to fix my hand and seal this 30th day of June and a year, 1948. Thank you. Thank you, Mr. X, Mayor. And thanks to the people of Cheyenne for this great honor at the beginning of the 15th anniversary of the Lone Ranger program. I want to say that whoever first referred to Cheyenne as the magic city of the plain certainly knew what he was talking about. Because I know of no other city which has retained quite the same atmosphere of adventure and hearty western spirit as you folks have. I also want to say right off that wild horses couldn't keep me from being here for your nationally famous Frontier Day celebration during the last week in July. And now, as official Mayor of Lone Ranger Frontier Town, I assume I have the right to make an official proclamation in honor of the Lone Ranger's 15th anniversary. Don't I, Mr. Mayor? You certainly do, Mr. Mayor. Very well then. As Mayor of Lone Ranger Frontier Town, I declare official thanks to my young friends everywhere who have made contributions to the National Society for Cripple Children. Contributions which will help less fortunate boys and girls walk, walk and play again. As you may have heard me say before, the best way to help make ourselves better is by helping others. All right, now I'd like you to meet two young friends here. Little Nancy Gore, age six from Southlake City, and Tommy Sykes of Grand Junction, Colorado. These two young friends together with other boys and girls here in the audience have been appointed special deputies to help Sheriff Tuck and me preserve law and order in Lone Ranger Frontier Town. Now, I want each of you special deputies and any other boy or girl who would like to be a Lone Ranger deputy to raise your right hand and repeat after me this pledge of allegiance. Are you ready, Rangers? I do solemnly pledge to place God and country above all else. Honor my father and mother or guardian. To be honest and truthful, to pledge my hand to the weak, to pledge my heart to the helpless, to pledge my life to my fellow men. Welcome, Rangers. You are the first of the official Lone Ranger deputies. And now for my second official proclamation as Mayor of Lone Ranger Frontier Town. I officially declare that the rest of this day be a town-wide holiday and you young official Lone Ranger deputies it's up to you to help to see that the day will be long remembered. And here is Captain Ozy Waters and his Colorado Rangers playing Old Susanna. Here's Captain Ozy Waters. And now, ladies and gentlemen, here come several of the Lone Rangers young followers and they're carrying a mammoth breeding card. This card is about five by eight feet in size, I'd say. It has been signed by practically every boy and girl in town. Deputy Jack Carleberg is going to present the cards of the Lone Ranger on behalf of the boys and girls of Lone Ranger Frontier Town. Jack? Ladies and gentlemen, in honor of the Lone Ranger's 15th anniversary, we take great, we take pleasure in presenting our new honorary mayor with this breeding card signed by his loyal followers. Congratulations, Lone Ranger, on your 15th anniversary and greetings from us all. Boys and girls, I want to say that this is all the nicest tributes I have ever received. Thank you, Jack. I know you boys and girls who have signed this memorable card. Thank you from the bottom of my heart. Uh, Jack, who would you say was probably the best known man in Wyoming? That's easy. The best known man in Wyoming is Governor Hunt. Everyone knows Governor. Yes, you're right, Jack. And here he comes now, the honorable Lester C. Hunt, Governor of the State of Wyoming. Thank you, friends. Thank you, Mr. Mayor. Thank you, Mr. Ex-Mayor, distinguished citizens and guests. I suppose now, just as soon as I get back to the Governor's mansion, that the children in my neighborhood are going to be after me to declare an official holiday just like the Lone Ranger did so easily. And if I thought that we were all going to have as much fun as we're having here today, well, I think I probably would. Seriously, though, days like this and we have time to relax and to join our friends and our neighbors who are a very definite part of the friendly life of Wyoming. But I mention the friendly spirit of Wyoming only for those in the radio audience who have never met us. I don't have to tell the hundreds of thousands of Americans who have visited us as tourists driving through our wonderful Yellowstone National Park or camping or fishing in our National Forest. These Americans who have been here know that Wyoming's friendliness is as generous and as picturesque as the misty falls that dropped from the sheer walls of Yellowstones. Thank you, Governor Hunt. Thank you, Duffman. And now, friends, I'd like you to meet another very distinguished man who is here for today's celebration. He is Mr. Leslie Perron, President of General Mills. Mr. Perron. It's a thrill to me to be here for today's celebration. He is Mr. Leslie Perron, President of General Mills. Mr. Perron. It's a thrill to me to be here for today's celebration. As a visitor, I can vouch for every word that Governor Hunt said about Wyoming's friendliness. Also, I want to add my congratulations, Lone Ranger, on your 15th anniversary, which starts today, and mentioned that grocers all over America plan to help you celebrate your anniversary throughout the month of July. I would also like to commend the boys and girls who listened to the Lone Ranger program. Many of these boys and girls sent in their pennies, nickels and dimes to the National Society for Cripple Children to help less fortunate boys and girls to run and play again. And I too would like to make a personal contribution to this worthy cause. Please accept my personal donation to the fund which the Lone Ranger program is giving to the National Society for Cripple Children. Thank you, Mr. Perron, on behalf of the National Society for Cripple Children. Ladies and gentlemen, boys and girls, Mr. Perron has just handed me his personal check for $1,000. This donation, together with those of the boys and girls who listened to the Lone Ranger program, will certainly aid the fine work which the society is doing. And all of these gentlemen, here's another very special guest. He is Mr. Mark Wood, President of the American Broadcasting Company, the network over which the Lone Ranger program is heard. Mr. Wood, if you please. Folks, as a tribute to the Lone Ranger program, at the beginning of its 15th anniversary on the air, the American Broadcasting Company wishes to make this award. No yield by these presents that whereas the American Broadcasting Company is desirous of giving recognition to the merit and the solid accomplishments of the Lone Ranger, as well as its fine value of entertainment. Now therefore, be it known that the Lone Ranger is cited and presented with this goal for distinguished achievement and radio. On the occasion of the 15th anniversary of its first broadcast by the American Broadcasting Company. Mr. Lone Ranger! My sincere thanks to you, Mark Wood, and to ABC. I am delighted to accept this award, not as a personal one to myself, Grace Beamer, but to the legendary character, the Lone Ranger, whose thrilling air adventures I portray, and for those men behind the scenes who have made the Lone Ranger program possible. I refer to the director of the Lone Ranger program, Mr. Charles Livingston, the chief of the writing staff, Mr. Fran Stryker, and to the producer and creator of the Lone Ranger program, Mr. George W. Trendl. These men are here from Detroit for today's celebration and on their behalf, as well as for myself, I wish to thank you. Thank you again. No celebration for the Lone Ranger could forget his faithful Indian friend and companion, Tato. In fact, the Lone Ranger from Long Association with Tato has demonstrated another American ideal, the way that thousands of pioneers lived on friendly terms with their Indian neighbors, helped them, and were in turn helped by the Indians. As indication of this mutual regard, the name of the great Sioux Charles. It is a pleasure for me to see the Lone Ranger, the white warrior, who think it is a wonderful thing to be in here to see him and to see him as one of our own men. A warrior in our Indian race of people, it's something to be proud of. We have wonderful boys that went to war, and they have done wonderful work. And the boys at home, the little boys, all watched the Lone Ranger and praising for the wonderful things that he has done, the kindness that he has showed to humanity. And I am very pleased on behalf of my Indians that they are proud to meet the Lone Ranger. On behalf of this thing, for the crippled children, I have worked on a march of dying. Nothing just grieves me to see these children. I wish them all luck. So someday, may they be rid of this year's paralysis. That's my one prayer to Almighty God, with sincerity. I thank you. Thank you for that wonderful greeting, Princess Bluewater. And I hope that you convey my own sincere greetings to the Sioux Nation as warmly. I will. Well, this day you are going We will miss your bright eyes and sweet smiles for the day you are taking the sunshine. That's right, and that's the way it was. Come and sit by my side here, you leave me. Do not hasten to bit me ado. But remember the Red River Valley and the cowboy who loves you so true. As you go to your home by the ocean, you never forget this sweet hour. That we spend in the Red River Valley and we'll always be together. And now, ladies and gentlemen, as a final epic tribute to the Lone Ranger, the great American poet Edgar Guest has written a poem which puts into simple heartfelt words the feeling and sentiment we all have for this great champion of friendship and justice, whom we are honoring today all through July on his 15th anniversary. Edgar Guest's poem is entitled Fifteen Years of the Lone Ranger and we have asked Governor Hump to read it to you now. Fifteen years since the first brave deed and Silver's glorious burst of speed. Fifteen years that have gone and come and the gallant Lone Ranger still writes on courageous and unafraid, loved and admired by the friends he's made.  Many are fathers with children small who wait as did they for the Ranger's call. With eyes to glow they sit about as high-hoes silver with him they shout. Fifteen years of riding hard with the faithful tanto standing guard. Fifteen years filled with dangerous grim but never a bad man mastered him. The good upheld in the evil state and always the law of our land obeyed. The Lone Ranger, friend of the poor and weak, the first to answer when the help they seek. Orson Ryder who takes the trail with high-hoes silver but we must fail. And always at night when the races run the boys and the pride of a good deed done. Lone Ranger, never a sane appears on that record of yours through fifteen years. Our boyhood has learned through peril and strife with courage to cherish our way of life. And because of your kicking our land will be forever the home of the brave and the free. The Lone Ranger pounding other hard riding off on another fifteen years of his famous adventure. Today's program brought to you direct from Cheyenne, Wyoming which was officially proclaimed Lone Ranger Frontier Town as mark the opening of the Lone Ranger 15th Anniversary which will be celebrated in grocery stores all over America during the month of July. The regular broadcast of the Lone Ranger program will be heard at the usual time tonight and will have come to you as soon as possible. Today's broadcast of the Lone Ranger program will be heard at the usual time tonight and will of course come direct from the city. My special request tonight's Lone Ranger program will retell the stories of how Tonto and the Lone Ranger first met and how the Lone Ranger got his name. This program was transcribed from an earlier network broadcast. This is ABC, the American Broadcasting Company. It is six o'clock at KECA, AM and FM, Los Andrews.


So, most large stars, more than eight times the mass of our sun, die in brilliant explosions called supernova. But some undergo near-death experiences, having a small eruption about a year before they eventually collapse and die.  But this shockwave, this near-death experience, will go out into the star system, hitting the other planets. And by hitting the other planets, it may actually affect the atmosphere and conditions of those planets, and therefore affect the possibility of hosting life. So by knowing the ranges of stars and planetary systems and what they end and how they end will tell us, is there life elsewhere out there in the universe.


وَأَقُولُ فِي القُرْعَانِ مَا جَاءَتْ بِيهِ آيَاتُهُ فَهُوَ الْكَارِيمُ المُنزَالُهُ وَأَقُولُ قَالَ اللَّهُ جَلَّ جَلَالُهُ وَالْمُصْطَافَ الْهَاديِ وَلَا أَتْ أَوَّالُهُ الحمد لله يا رب العالمين له الحمد الحسن والثناء الجميل وشد الله إله إلا الله وحده لا شريك له يقول الحقا وهو يهد السبيل واشدوا أن محمدان عبده ورسوله صلى الله عليه وعليه وعلى آله وأصحابه والتابعين لهم بإحسان إلى يوم الدين أما بعد إن شاء الله تعالى my beloved brothers and sisters in this contradiction I want to speak about the contradiction of الاستيزاء بلافع إثوبي فوق الكعبين من الرجال mocking the men who have their garments above their ankles and not giving any concern to the women who have their clothings which are above their knees you're mocking, you're laughing, you're ridiculing the men who have their garments above their ankles and you are fine and you don't say anything about the women who wear their skirts above their knees you don't say anything about it that's a contradiction the reason why the men having their garments above their ankles is an obligation from the Prophet ﷺ it is obligatory upon the men to have their garments above their ankles the Prophet ﷺ he said in the hadith to the صحابي he said إرفع إذا ركى إلى نسى الساقين the Prophet ﷺ commanded him he said raise your garments above your ankles raise it up to إلى نسى الساقين up to half your shin فإن أبيتة if you don't want that and you don't want to do that فألالكعبيني then do it up to your ankles وإياك and stay away from the Prophet ﷺ said وإسبال الإزار stay away from having your garments below your ankles فإنه من المخيلة for verily this is from arrogance والله لا يحب المخيلة الله does not love those who are against also the Prophet ﷺ he said about the women يُرخين الشبرا that the women she should have her garments down a shibr a shibr is a handspan يعني the women are told have your garment a handspan down so you go a bit down and then أم مسلمة رضي الله عنه she said إذا تنكشفوا أقدام but their legs are still going to show then the Prophet ﷺ he said فا يُرخينه ذراع لا يزيدنا علي rather what they should do is they should lengthen it a handspan they should actually lengthen it an armspan so the beginning of the Prophet ﷺ she said but the legs are still going to be seen he said okay an armspan so the women are being instructed to have a long garment and the men are instructed to do what to have their garment high up we on the other hand have become different we mock the people or we look down at the people whose garments are below their ankles from the men if a man has sorry we look down at the man who has his garment above his ankles and we are fine with we have no issue with the woman who has her skirt above her niece fine no problem we don't say anything to her and it reminds me of the line of poetry which is there's no fascination the women have become men okay there's no fascination that the women have become men there's no fascination that the women have actually become men ولكن تأنيت الرجال عجيب what's fascinating is that the men have become females that's what's fascinating to be honest I'm going to stop there insha'Allah anything which I have said that was wrong and incorrect it's for me and Shaytan and Allah and his message are both free from it سبحانك الله و بحمده أشهد بلا إله إلا الله أستغفر كوة الله السلام عليكم و رحمة الله و بركاته كيف can you do a two second action right now that will give you a share of the reward of everything we're doing on this YouTube channel simple like this video and click subscribe why it will allow YouTube to recommend our videos to other users and imagine the huge amount of reward that could be waiting for you on the day of judgment if you did that with a sincere intention of spreading the Deen of Allah you'll be rewarded for every single person who benefits from one of our videos as a result of your like or subscribe that's an easy two second action that you definitely don't want to miss out on do it now click like and subscribe and don't forget to make that intention


Everybody forced here up in 22 23 prism collegiate draft picks basketball hobby 4 bucks number 8 random team style check the results anytime Elite sports cards calm. Let me get real GM up here in a second. We'll get started Yeah, 130 point calm. I'll drop the link here, too. We'll also kind of help But you're gonna want to just look at if you can't find the exact card on eBay Maybe higher number parallels or maybe other Michael Parsons autos out of lower numbered or maybe lower numbered autos from the same set That'll help you Kind of a determined value stuff with low numbered cards though But 130 point comm is a search mechanism where you can search all eBay sales and everything Fantastic Good luck everybody. Oh, yeah, Tara peak now. I keep forgetting eBay has that now Yeah, card ladders are good place to go you can look at Sales of cards sometimes, but if you want something for free 130 point comm is great. That's just been my go-to forever Silver a Trevor Kiehl's rookie was he Nick's Trevor Kiehl's dude. Look at this auto Tevin Brown's silver auto. There's no way Dude, there's you see this auto There's no way dude No, dude, I refuse to believe that he signed it like that. This guy's going Pacers I Don't rag on a lot of guys autos I know cursive isn't taught but man that is a small set of initials And if you're gonna do that at least make it take up the center the complete center of the sticker at least you can Pacers on that one. That's Pacers Fort Wayne Matt answers it Pacers G League rookie hyper Caleb Houston blue wave James. Oh James worthy That's pretty cool It's a 249 That's crazy. Like I've seen like initials now. Whatever. It's what's not ideal, but oh man That's another that's another step too far right there Silver Jordan pool silver auto is buddy Bayhime is he did you ever have NBA affiliation? Buddy Bay, I know that Pistons looks like Yeah, pistons. Oh, we saw the pistons Huh, I'm thinking of another Bayhime. There's a lot of Bayhimes. There's oh Jimmy. Jimmy Bayhime is a in Europe. I think I Gow in turkey. I heard the horn. It's hyper Patrick Ewing I'm like, oh man, I'm thinking of I'm thinking of Jimmy all these damn Bayhimes out here Silver rookie Kennedy Chandler a auto is Jordan usher usher usher. I'm sure that That gentleman has never heard that before Mmm, he was with Utah the summer league and he's in Europe now jazz on the Jordan usher Hyper Jerry West Purple E. J. Lindell solid one there number to 75 Jaden Hardy hyper hearties good pig. He was a second-rounder wasn't he? Silver Dominique Wilkins a blue auto Jalen Williams for the Thunder numbered 146 of 149 Hyper Chris Paul Hyper Rookie Orlando Robinson and a purple rookie of Keegan Murray nice to 75. There we go Kings NBA team color match I don't know if you get away with that one on eBay. I don't think you're gonna drop a lot of attention to the NBA color match Boy Herman. Oh boy Herman brought up camera radish Don't tell Sam. No, he doesn't That's Silver rookie Bennett Mather and pop Mather and silver got silver auto of Colin Gillespie Where's and is he blazers Gillespie a lot of Gillespie? Colin Gillespie Denver Denver Paul Pierce hyper hyper rookie Alondas Williams purple ice Jalen Williams rookie Number to 149 for the Thunder That's also Sam's favorite song guys. He loves the Imagine Dragons All right, it's not the it's just imagine dragons, right? Sam would know I'll ask Silver Julius Randall red auto fats Russell Fats Russell Let's see. I think he has no NBA affiliation Undrafted and then he's playing in Europe. It looks up Fats Russell to 199 Rookie hyper Peyton Watson snakeskin Scotty Barnes. That's pretty cool Eagle Raptors for a state Scotty Barnes right there Silver Tatum brilliance Yeah, he was a good second round pig illegal He better be playing now. They traded everybody for not everybody, but they've tried a lot of depth for Kyrie So silver Bosch up Blue ice Davion mince auto Davion mince. Come on mince All right, I'm gonna search then Davion mince. Let's see capital city go-go. That is the wizards G. Lee wizards 75 hyper Devin Booker hyper rookie was Monty Yang You'll have to talk to him about that I'm not the camera out of Shakespeare Summer day Donovan Mitchell She goes to the cabs red is a champaign auto. Is that a chinch and Pagney? He's a spurs now. I think he's a spur. Yeah, it spurs now. No Champagney, I think he does go by champagney. Can anyone confirm that any good hits? I mean, yeah, we've got some solid stuff hyper Keegan Murray. It's a good hits in hockey a Jang rookie red. I think he's thunder Champani Champani Champani Someone for sure told me of a champagney someone for sure told me that in chat one time But I was like Okay, sham wow Silver rookie Dyson Daniels we got a blue auto of Michael DeVoe Michael DeVoe Ontario Clippers would be LA Clippers Number two 149 Hyper Paul George blue Jalen Williams rookie again for the Thunder Thunder having a pretty good break here to 199 without it without having big chat anyway One of one Donnie B. Crumb Now laugh is too good silver shack Kai Soto Kai Soto base auto Kai Soto he went undrafted He's the Australian we played in Australia. He's ufa. That's Rams everybody at the end Filipino yeah, play in Australia. Yeah, no NBA affiliation. I'm Baji rookie hyper got purple eye so Han rookie There you go spurs to 149 and Hyper incident impact tie-tie Washington for the Rockets, right? Silver Ryan Rollins a silver national pride Wong Zee Zee. That's pretty cool I believe he was with who was he with the longest just want to know where this goes real quick I always think Mavericks in my head he It will go Mavericks He had two seasons with a few teams But the Mavericks he played 60 games with which is the most that he had so that's a tiebreaker there Mavericks on the Wong Zee Zee There's only six. Let's go. The illegal is making a joke right now for sure blue auto Isam Obli for the Cavs the 149 Illegal is definitely making a joke Bryson fell for it illegal hyper rookie Chet Holmgren But he's it's illegal snake skin John Morant My snake skin there racers Yeah, Bryson hasn't toppled it enough and dealt with illegal enough to know his game But I mean you guys are friends as your Eagles fans silver Anthony Davis Silver is Cole Swider is that how you say it swider swider Lakers on the Swider looks like spider with a W Jimmy Butler hyper Jaden Hardy hyper Jets those under the devils.  Who was that? What's he doing? Yeah Switter swider swider Sweeter could be sweet AJ grip and rookie silver. There you go another Wong Zee Zee. Wow I haven't seen any national prides at all. We have two silver Wong Zee Zee's in this break a Rocky cruiser or Croisier auto Rocky cruiser these are the Nuremberg Falcons And after North Dakota stay he doesn't have any NBA experience So random to everybody at the ends Jabari Smith rookie hyper. There we go and a Wendell Moore Rookie per rookie red to 299 silver chat Paloas Palo silver incident back will topple that one too silver rookie Jalen Williams a Silver David McCormack auto Think he has a team McCormack he is with beseeked us But he was with Minnesota in the summer league Timberwolves on that one hyper rookie Patrick Baldwin hyper Dalen Terry Jeremy Sohan Silver Russell Westbrook blue ice Croisier again or cruiser again random to everybody at the end 75 Hyper to Rosen blue wave Jalen Williams going to the Thunder Rookie to 249. All right last pack Silver rookie Jalen Williams Black and gold Blake Wesley to a five on the auto spurs on the Blake Wesley Shredding get ready Ice So I want to get to obviously a Jabari Walker shade on sharp hyper Mark Williams blue nice Duke color match to 199 and That will do it other solid draft break there. Let's see randoms to everybody here We're just going to order by parallel then makes more sense So four autos here Random to everybody top four spots in this in the random will get one of the cards here In this order top four one will get the cruiser blue Second will get fat trestle red third the cruiser base and fourth the kai so do in that order. Good luck Five and one six times again top four one two three four five and six So it's going to go in this order pelicans Celtics Rockets Nets Pelican Celtics Rockets Nets All right, so we had the Blake Wesley nice car there to five for the spurs David McCormick Cole sweeter Isaiah Mobley Michael DeVoe Julian Champani Davion mints Colin Gillespie Jalen Williams Buddy Bayheim and a Tevin Brown That will do it. Thanks again everybody Better parallels a lot better parallels and autos not what I'd say, but yeah, let's keep going. Thanks y'all