Okay, so I'd like to thank Maxim for organizing this this day, which is very interesting so So I'd like to talk about a project the which we're we've been working on for a little bit of time. Oh, sorry Sorry about the computer thing here Yeah, so we've been working on trying to understand the Gaioto more night ski Considerations at least some of them With Ludmille Katsakova Alexander Noel and Panov pondi as well as other people in Vienna So it's a pleasure to be able to speak here about this Okay Okay, so we wanted to understand the spectral networks of Gaioto more night ski from from the perspective of Euclidean buildings So this is what we'll be saying today And the idea is that this should generalize the trees would show up in the SL2 case So and luckily for me I didn't actually include any discussion of the SL2 case, but luckily for me we've discussed the SL2 case considerably already this morning and Well, this was maybe not emphasized But what you can see is that in the SL2 case we're talking about a foliation defined by the real part of the one form and When you have a foliation that corresponds to a tree which is the space of leaves of the foliation so Now the maybe the more somewhat more long-term goal, but this is not really my zone of expertise is to understand a relationship as was being discussed between the picture of spectral networks and the space modulite spaces of stability conditions in Particularly in the case of spectral networks there an important role is played by these transformations where where there's some jump between the in the topology of the spectral network and In the case of SL2 one way of thinking of that is that The space of the tree which is the space of leaves of the foliation has undergoes a mutation when the when you cross over a BPS state Okay, so so we'd like to thank Maxime and also another student and Vienna Fabian Haydn for important conversations about this Oh So I think I don't I don't have any pictures in the first part of the file here, but Then we'll do I have some other files with some pictures. So this is first just general Okay, so So let's take a Riemann surface X a vector bundle assume. It's has determinant one and we take a Higgs field I should maybe apologize that my round-of-fee is what Andy's non-round-of-fee was and my my the non-round-of-fee will become a The round-of-fee will become a non-round-of-fee anyway So okay, so let's take a endomorphism valued one form And let sigma be the spectral curve, okay, and we'll be assuming that that's reduced here Okay, so we have a tautological form and we just think of that as a multi-valued differential form on X Okay, so we can write locally phi is phi one up to phi R And the trace condition trace equals zero the determinant zero for the connection trace zero for the Higgs field Condition says the sum of these guys equals zero and the condition that the the Spectral curve is reduced is the same as saying that the phi I should be distinct for general points in X Sorry about the size of this. I was I didn't know you could do the full screen thing So let's just Consider the the divisor which is the locus over which the spectral curve is branched and let's let X star be the complement So the phi are locally well defined on X star Spectral curve is an atoll covering of X star Okay, so now we'd like to think of two WKB problems associated with this data The first is the Riemann-Hilbert are complex WKB problem So let's just take a base connection NABLA zero on e assume that that exists and look at a Connection NABLA t which is NABLA zero plus t times the Higgs field phi and t is supposed to be a large parameter So t is one over h bar here basically So then when we get a monoramy representation for the connection and Let's also choose a fixed metric on the bundle e so in this setup. We're just fixing a single bundle e So I mean I think Andy and Maxim were actually discussing a situation which is more more general in which The bundle e could also vary as a function of t and in fact even the curve and so on could vary the function of t Now we get a flat structure, which depends on t given by the flat connection So if we transport so if we take but now we've got a single bundle which doesn't depend on t And we fixed a single metric on this bundle, but now we have a transport function which transports back to the base point Okay, depending of course on a path. That's why x tilde is the universal cover of x So if we transport the metric By the flat transport back to the base point then we get a metric on the fixed base On the fixed the fiber of the bundle e Over the base point Maybe I should say We're fixing a trivialization here e x zero c to the r And we can think of this as a map to the space of metrics, which is this symmetric space Slrc module of s you are so we get a map from ht from x tilde into This metric space and it's row t equivariant So this is not necessarily a harmonic map, but it's supposed to be model It's supposed to make you think of a harmonic map basically as in Hitchin theory as we'll see on the next page and so for the complex wkb problem we'd like to understand the asymptotic behavior of the of the representation row t and HT or just the transport matrix itself as a function of t going to infinity So let's define things this way So let's take two points in the universal cover since we've taken two points in the universal cover There's a unique path going from one to the other Let's let tpq of t be from the fiber of e over the point p to the fiber over q be the transport matrix Now let's define the wkb exponent to be the limb soup of one over t of the log of the norm of the transport matrix okay, so The norm is being the operator norm with respect to the our fixed metric so this is supposed to be basically the highest eigenvalue of the transport matrix, which is to say Roughly speaking, this is roughly speaking the biggest term of this matrix And the our exponent is just this lambda one Now the the other type of problem so actually, you know this I think now that I heard Andy's talk I realized that we had a slight misconception about what you guys were doing you guys are actually looking at a holomorphic family of Connections depending on your parameter, which you're calling zeta Which which go to infinity in fact for z equals zero and z equals infinity Which are not actually exactly the same as this guy But anyway, we can do what we said here, too This is so this is not actually exactly what you guys are doing Well, in fact what you guys are doing I guess is more a little bit more closer to the to the complex wkb question Because he what's going on here will not depend holomorphically on the parameter t this time Anyway, so we can still do this anyway So Suppose that we have a stable Higgs bundle then we solve the Hitchin equations So we have the Hitchin Hermitian Yang-Mills metric on this Higgs bundle depending where we've multiplied phi by the number t Which is a large real number And let's take the associated flat connection and take the associated representation Okay, and this case we have a family of actually harmonic maps from the universal cover into the symmetric space So these are the maps given by the solution of the Hitchin equation And so one could also ask to understand the asymptotics of the solution of the Hitchin equation in this sense So we can define the transport and the And the exponent and so on But here we're using this metric which is which it also depends on t to measure the size. I Don't actually have a theorem about that for the moment anyway, but anyway, okay So now Gaiotto-Mornitzky explained that the that the exponent new pq Should vary as a function of p and q In a way so p and q are points on on x or on x tilde in a way, which depend is dictated by the spectral networks, okay? So let me maybe make a pause here to say, you know What does this basically say? so I'd like to just give it the following example So Gaiotto-Mornitzky say it first of all So this is kind of a classical but if we go from a point here for example to a point here, okay? So this is this is my picture of Of the spectral network, okay, so these guys are the Imaginary affiliations label something like one two maybe I'll probably mess up the labeling, but this is an SL3 case So there's labeling one two and two three, okay, and then we get some collision lines here If we go from the point p to the point q then Then their prescription basically just says that as we cross from here Once we've crossed the second one of these imaginary spectral network lines Then we can no longer calculate the transport from here to here as just a simple integral of one of these forms from p to q There's a jump happening here And the collision phenomenon is the same thing Let's call this The collision phenomenon says that the same thing happens, but it's a little bit more subtle here So if we try to integrate from the point p to the point q then we can say, okay, what's the What is the transport matrix going to look like when we go from the point p to the point q Okay, what happened here was that if we draw sort of the If we draw the foliation lines Those look like that Here we sort of clearly went over a mountain pass and then back down again Okay, and when you go over something and then back down again, then you have to count the back down again negatively basically so for example here you'd have to integrate P1 Up to here and then P2 here something like that Now if you look at the point p in the point q then we have an actual it we For example here We went over this mount into the upper part of this mountain pass, but we didn't actually go out of it because We got over to here and Similarly in for this guy p is already inside the the upper part of the the mountain pass here And then it goes down to the point q however, so what Gallardo more nights he say which is somewhat more You might say unexpected in this case is that actually you have to take into account these collision lines And since we went over a collision line and then back over the conjugate collision line That you cannot actually calculate the transport from p to q as just a simple integral of one of the forms from p to the q Whereas you would be able to from here to here or from here to here okay, so So that's a just a brief roughly speaking Rough explanation of what I mean by dictated the word dictated in here Now here's a remark, so we this was what Maxime was discussing in the complex WKB case we can actually View this transport function in terms of resurgent functions and to that we take the Laplace transform So Laplace transform is the integral from zero to infinity of the function of t times e to the minus zeta t Also, I'm again. Sorry because my parameter zeta here is Zeta is not the Andy's zeta T here is Andy's zeta anyway So here we have a homomorphic function to find it for for large values of zeta And it emits an analytic continuation with infinite but locally finite branching However, that's not what I'm going to talk about today We can just well let me just say we can describe the possible locations of the branch points And it turns out as far as I can tell at least this can This description is roughly speaking compatible with what Gaiot and Marneitzky and say and I think it's relatively precisely compatible in the sense that I think we can show that That we get exactly the integrals from points line over p to points line over q along the spectral curve Of course the question is you know, which actual integral you get and So I don't have a way of saying that in general T and the one over t at the same time That's a slight difference, but I don't think it makes too much difference So I I think we can deal with a one over t in the in the perturbed It essentially comes down to putting a one over t in the perturbed piece of the Of the connection But anyway, that's not what I want to talk about today Today we'll look in a different direction, okay? So in particular, so I Mean one question is you know, why is it this particular thing? For example integrate phi 1 up to here and then phi 2 to here or here We integrate something like phi 1 up to here and then phi 3 to here or something like that Why do you get this particular value of the exponent instead of some other one? and so so we would like to look at that in terms of harmonic maps to buildings Let me also give a Sort of a first look as to why Why buildings might be involved? Well among other things you could really see what was happening in terms of the the The harmonic map to the tree in the case of SL2 so that already suggested that buildings would be involved But we can draw a picture. We can draw the following picture Of this our first the first thing I said there map going from p1 to q1 by going over a mountain path Okay, so let's draw just this the lower part of the function. Let's draw that Like this, so here's our mountain pass, okay? And say we we had a point p on the front sheet and the point q on the other On the back sheet and let's suppose that the other differential the 2 3 differential is roughly speaking constant between the point p and the point q Okay, now if we don't want to vary the the 2 3 differential too much our curve in principle shouldn't really vary too much from the straight line joining p to q so In terms of the and so this will be the function of the one two Function the real part the total variation of the function one two along this path is actually sort of too big It's not what we get from the prediction However, the prediction tells us that what we should actually do is go up to this point here and then back to here That that's the exponent that you get from the point p to the point q And we can see that that's what you get by looking at the distance in the building so we map the the remount surface to this building which is just a Three-legged tree thing cross an interval The distance from one point to the other in the building so the point p That whole part goes to the front sheet and that whole part goes to the back sheet Okay, so they're not on the same sheet of the tree of the building Okay, to go from one to the other we just have to go up to here and then back so this picture was sort of a basic picture which coincided with the the prediction that you get Just in the simple case where we were not looking at collisions or anything yet and so our goal with the With the people in Vienna was to try to understand what's going on with the conjecture that in fact Our our wkb exponent is actually going to be the distance in the building between the two points That's going to explain the sort of cancellation phenomenon if you sort of try to integrate the differential equation By little pieces from p to q you're going to get a much bigger answer than the correct answer The cancellation phenomenon Sorry to turn the cancellation phenomenon says that a lot of those pieces cancel out and the real exponent is just integrating up to here and back Because So that that's the subject of the talk Right, okay, so here's the so here's the basic idea. So this is a wrong basic guide. I mean I at least I don't know how to make it correct basic idea, but it's the basic idea So the idea so let's look at the field of functions of germs of functions on our germs that infinity of functions are Let's sort of a suppose that we could define a valuation given by the sort of exponential growth rate of the function Then our family of monoramy representations constitutes a map from pi one x x zero into SLR of the field K So if I want to act on the Brouhatt tits building and We could just try to choose an equivariate harmonic map from the universal government to the Brouhatt tits building following a Grumov shun Unfortunately, it doesn't seem clear how to make this precise because of the fact that We can't really define a field of functions because yeah, if you take a function like sinus sine of x That's supposed to have exponential growth rate equal to exponent equal to zero. It's supposed to have a growth rate constant However, of course if you take one over a cosine of x it's got lots of poles So you can't really say that it has a bounded growth rate So so at least to me at least it doesn't seem clear how to make this precise But that's kind of the idea. So luckily turns out the and and power O has actually developed exactly this type of a theory Based on work of Kleiner and Leib So it turns out that we can just apply their work basically Maybe some slight modifications are necessary of course So the idea is to look at our maps HT as Being maps into a symmetric space, but let's rescale the distance on the symmetric space by dividing by t Then let's take the Gromov limit of the symmetric spaces with their rescale distances That just means take the symmetric space and sort of look at it from really far away Okay, of course, it's it sort of since it's kind of infinitely big even if we look at it from far away It's still infinitely big and as we take those sort of the limit of that you get exactly the Some kind of building modeled on this LR Brouillard hits building You get a building It's an R building. Yeah, or even worse, perhaps I mean it's some kind of building and the limit We don't actually need the definition of I mean we'll have a definition here, but which will be concrete so We don't need to worry about that kind of thing But anyway, so their construction depends on the choice of an ultra filter. I don't even know what an ultra filter is but Should should one day. I hope to know what it one is, but any in case Roughly speaking it means choose a sort of compatible sequence of choices of subsequence whenever you need one And then you get a limit which is denoted cone Omega. In fact, it's cone Omega of the of the building I think you have to you have to fix a base point in there and so in Perot's paper she discusses this limiting building cone Omega and and says that you have a limiting action of the group pi one on the building and It turns out that you can Equally well get a limiting map h Omega from the universal cover into the into this limiting building and the importance for us This whole situation is that the distance between two points in the in the building The distance between the images of two points is exactly what we want Which is to say take the distance in the symmetric space between the two points Rescale by one over t and then take the limit, but that's an ultra filter limit So that's that means it's a limit of some subsequence which is Yeah, sorry That's the limit of some sec subsequence which is sort of cleverly chosen you might say Now in fact we can be a little bit more precise because on the building. So let me just say this is a The limiting building is a building model on the the affine space for the SL2R For the SLR brought its building was which is just r to the r-1 Yeah, so I'm gonna discuss that a little bit in fact, that's a good point But but for now, but let me just say this first So in fact on the building there's several different ways of measuring the distance So the perhaps standard way is to measure the Euclidean metric Let me just say the way to measure distance between two points in a building is that one of the axioms of a building is any Two points are contained in a common apartment Apartment is just an image of a map from here into the building So we take our two points we put them in a common apartment and just measure the distance in the apartment between the two points Now depending on what kind of distance you put on the apartment you get a distance function on the building The standard one is just the Euclidean distance Square root of the sum of squares of coordinates That's not actually a very good choice for what we're what we're interested in There's a better choice, which is the finzler distance The finzler distance is the log of the operator. Sorry the finzler distance is just the The maximum of the coordinates Of the variation of the coordinates And then uh more refined which I'll say on the next slide is the Apartment you take finzler distance. Yeah, I guess they just co-operate the wire coordinates. Yeah. Yeah And you can actually combine these all together the different coordinates all together and get what's called a vector distance Which basically means uh take take all the coordinates but arrange them in decreasing order And these guys all come from limits of the corresponding distances on the symmetric space The Euclidean distance comes from the usual distance The finzler distance comes from the guy which is just the log of the operator arm of the matrix And the vector distance comes from the full collection of dilation exponents between the two metrics On the symmetric space And in fact, so what we're really interested in uh to get kind of a full information About the wkb problem Is the vector distance so in the affine space so the r Affine space is the set of points in r to the r who sum is zero so it's r to r minus one The val group The val group is just the symmetric group acting by Permitting the coordinates And the vector distance between the origin and a point is Obtaining just by reordering the point so that they're in decreasing order It does satisfy triangle inequality If you if you I mean you have to interpret I mean you add term by term. Yeah, yeah Now so so the vector distance it just means put our two points in a common apartment then use the vector distance in that apartment Now it's supposed now what's the corresponding thing on this metric space Let's just put the distance define the distance the vector distance between two different metrics to be lambda one up to lambda k Where basically one metric is equal to e to the lambda i times the other metric on an adapted common orthonormal basis Now somebody will probably say but that's not a distance function or something like that We're only interested in what's going on in the large Ah, sorry As it's more like one of exponents yeah, it's collections not one numbers. Yeah, it's the exponents. Yeah Okay, so in terms of transport matrix matrices the first exponent the first distance is just the log of the the norm of the matrix the operator norm of the matrix and We can just use operator arms to get at hold of all of these guys because we can Take the operator norm of the matrix on the k-th exterior power. That will be the sum of the first k exponent So this is just to say that the looking at this vector of guys is just basically the same as looking at the At the operator norm of the transport matrix So we can just basically think of the lambda one distance and that's the fin sort metric So we're only interested in these distances on the symmetry space in the large as they pass to the limit after the rescaling And the distance and the rescaled distance is just For the finseler guy is just one over t times the log of the operator arm So let's define the ultra filter exponent to be uh The same thing as the limb soup except the limb soup is replaced by the ultra filter limit So that means it's the limit of some subsequence In particular the limit of a subsequence is going to be less than or equal to the limb soup The ultra filter limit means means we choose a some cleverly chosen system of subsequences And and so that'll be less than the limb soup and so now the question that maxime was asking is is this really the same as the limit So that will be the the case in some cases So the first observation is that if we fix the choice of points p and q Then we can choose some ultra filter such that the ultra filter limit is equal to the limb soup for that choice of p and q We just choose a subsequence that calculates the limb soup and then subordinate the The ultra filter to that subsequence Uh, what's not clear is whether we can really do this for all pairs of points p and q at the same time In the example that we're going to treat that actually will follow api posteriori Api a priori no Well, of course the second observation is that suppose that the limb soup is actually equal to a limit then of course it's the same as the ultra filter limit So this is the case for example when we use the local wkb approximation And uh, it would also apply in the complex wkb approximation for generic angles of the of the For after we twist by a generic angle If we knew that the laplace transform didn't have essential singularities. So that's uh So, I mean, uh, uh, I don't in general know how to prove that the laplace transform Of this guy's doesn't have essential singularities And if it had essential singularities that would be like sort of having several different singularities all glued together at the same point So it'd be sort of like having a uh wall also And then then you might just get some kind of weird behavior If the laplace transform doesn't have essential singularities And if you twist things by a generic angle so that there's only one Singularity of the Laplace transform which has biggest real part Then everything is governed by the asymptotic expansion at that real part and the asymptotic expansion has a leading term And so you can just see that the The the limb soup is actually equal to a limit So in fact, so so we pretty much expect that some other way you will be able to prove that that will be the case okay, so now uh Our main remark here is the just the classical wkb approximation which says the following thing I mean it says more than this but uh at least it says this I guess Uh, which is suppose we have a short path in the x tilde star That means x tilde but not going near the the singular point And suppose the path is non critical So non critical that means that if we pull back to the path the real part of the differential forms Then they're all distinct all along the path So that means that we have a reordering of the differential forms and up to the reordering We can assume that the first one is bigger than the second one and so on In that case the just that's just complex wkb theory is is my understanding goes Uh, that's just sorry. That's just classical wkb theory From like 1920 or something like that And in that case you can really describe the dilation exponents for the transport matrix And that we exactly get the Approximation like that So these lambda i's are the here Sorry again, the notation is not quite the same as in the other talks the lambda i the what was called lambda i before here was Is here fi and here the lambda i's are the integral of the fi so and so now given Let me say that uh Here this is supposed to be really a sort of a fairly good approximation which is to say The the dilation exponents Grow like lambda i but not not last basically And the corollary is that we have it in the limit. We always have the the distance in the cone omega It's just equal to this family of Of exponents given by the integral And that's the statement for the complex wkb problem I think we conjecture that the same should actually be true for the hitch in wkb problem So and in fact, this was a pre this is a misconception about What you guys were doing which is I thought that you guys were actually claiming this statement But in fact you guys are actually doing more of a complex wkb problem even though it has this 1 over t term also but It's still it's holomorphic in t So I guess you guys aren't really claiming that as far as I can tell Because in hitch in case we're we're actually solving the hitch in equations But we think that that should also be true in the hitch in case Okay, anyway, uh corollary in the complex case and also of course in the hitch in case if that turns out to be true Is if we have any non critical path even not even a long one not necessarily a short one Then if we use the map to the building Our path is going to map everybody into a single apartment in the building And the vector distance in the apartment is just given by the same integral. Okay remember these lambda i's are These lambda i's are the integrals of the real part of the forms fii So this is just from the classical wkb approximation. We can get sort of a Fairly good control already And the the passage from short paths to long paths is as you were saying the triangle inequality for the vector distance The case of equality and triangle inequality actually tells us that if we have three points Such that you have the sum relation for the vector distance Then the three points are are all three in the same apartment Usually in a building any two points are contained in the same apartment But three points are not necessarily contained in the same apartment If we have a relation like that, then they are contained in the same apartment And in fact, they're in opposite chambers centered at the middle point Maybe let me just draw The thing I mentioned here The inside and what does an apartment look like? We have three different families of reflection hyperplanes You know one two two three and one three um And the convex hull for example The the the segment in the finseler distance from this point to this point for example is something that looks like that So uh, so these are actually sort of the good segments to look at in the building The actual Euclidean segment is not all that useful in this context The the the real segment between this point at this point just the convex hull of these two points But in the finseler distance, uh, that means we go like that that means any path that looks like that is sort of non critical And we'll calculate the correct finseler distance or the vector distance So now the corollary is that in fact this this actually implies that our map from x tilde into cone omega Is a harmonic phi map in the sense of grommon shown What does that mean? That means that any point In the complement there's a discrete set of singular points Which could which of course contain the singular points we have already they could also in principle contain some kind of other bad points but discrete In the complement of that set then any point has a neighborhood which maps into a single apartment in the building And furthermore the map to the to the apartment is given by just integrating the one form So I think this is uh, this is seems very similar to why andy was saying that we have sort of local systems of a fundamental solutions on the on the sector I don't think that we can go from this statement to to what you were saying You probably can go from what you were saying to this statement on the other hand But anyway, so okay, so this this finishes what we can say about sort of the general situation Which is that we get this harmonic phi map h omega, but which depends on the choice of ultra filter The the the exponent that you get using that map namely the the finzler distance Is smaller or equal to the wkb exponent And furthermore we can choose the ultra filter so that that that holds the equality holds for at least one choice of p and q And also equality holds in the local case. That's to say when p and q are close by Now we actually expect that one should be able to choose a single Omega which gives the building which works for all p and q or maybe almost all p and q or something like that Sort of the set of points the set of bad points The set of bad values of t which you might have wrongly taken in your subsequence Should have sort of small enough major that for the different points that they don't intersect too much That they don't cover everybody too much And this is just so you start with a Higgs bundle and from the and Just from the Higgs bundle you get this harmonic map or does it need? well, yeah So the statements i'm making here are for the case of the complex wkb So here we're taking a Higgs bundle and an initial connection nav was zero and we're looking at the case The complex wkb is the Something like that and so this and this harmonic map so it depends not not only on the not only on the Higgs bundle But also in this case h is just fixed Right, I mean it doesn't depend only on that phi there. It depends also on The harmonic map that you get doesn't depend only on five, but it also depends on Nabila. Yeah, it depends on Nabila. Yeah, that's true. Yeah So for example, it might be that it has different behavior dependent for a special choice of Nabila not And then We we will we would sort of Imagine that we get a better control on the exponents if we choose a generic value of Nabila not basically And so so so what i'm saying now is is for this case because this and this is the case in which we know how to do the local wkb approximation So one could hope that that for the actual Hitchin equations, but where we take So we're doing this case We would hope to be able to treat the case that we take t We'd hope to be able to treat the case where we take This considered as a Higgs bundle and solve the Hitchin equations for t times phi and then Look at that or I mean you guys are already talking about more general things where you let the bundle vary the the space vary everything I don't know what What to say about that. I mean of course you hope in the abstract to to be able to have a statement for that kind of situation also but but What we're saying in principle is for the for the complex wkb question Okay, so so now we'd like to analyze the harmonic phi maps in terms of spectral networks And the the main observation is just to note that the if we take the reflection hyper planes in the building So that means the the the hyper planes in the building along which the different faces are joined Okay Those guys are just everybody that's parallel to one of these three guys in the In the apartment in the sl3 case Now if we pull those guys back to to x tilde By our harmonic map Then we just get the imaginary affiliation curves, which therefore include the spectral network curves Because the reflection hyperlamp planes in an apartment in our case in the slr case are just the equations Equations x ij equals constant where x ij is the difference x i minus x j And of course these pull back to curves in x tilde which satisfy the equation a real part of phi ij equals zero So these are exactly the curves the andy was talking about So these are the equations for spectral network curves Of course the spectral networks curves are the special ones that you get by sort of Starting out from singular points and then doing collisions and stuff like that. So So so of course what we're trying to do is to understand geometrically Why these collision type guys show up Okay, so uh, so that that sort of finishes what we can say in general Uh, so then we decided to look at the original berk nevins roberts example because we saw one of you guys in paper You said that this was the original example The japanese guys referred to somebody else also niki shima or something like that But we didn't check up and not yeah, but anyway, uh, so uh, this is seems to be the first example Where you actually see a collision of spectral network curves Uh, this was the the pretty much the original place So we looked at their paper and in their paper, they don't actually have a parametrized family of equations So the hypothesis is that they're actually setting h bar equal to one Which physicists tend to seem to do and in fact maxim also did I notice no one in your slides So if you look in their paper, but you say okay, actually they were setting h bar equal to one So we should put h bar back in of course that you want to put it in in a way, which is homogeneous Uh, you know going with the derivative so and then we just take uh Remember for us, uh T is one over h bar. Okay, so h bar is one over t and again, we don't care about the i's either the square root of minus one so somewhat conjecturally the the thing they're really talking about in their paper is the family of differential equations Which looks like this basically then we use the companion matrix and so on So instead of a single instead of a third order equation for one function We transform that into a matrix equation for three functions And you see that the you got a hinge field And the spectral curve for the hinge field is just given by the Thing that you would get from the equation by replacing the derivatives by the cotangent bundle direction So y is the coordinate on the cotangent bundle direction x is the coordinate on our Riemann surface And up to maybe a sine error or something like that The the spectral curve is given by the equation y cube minus three y plus x Equal zero. Okay. Now the problem here is I wanted to also have the other Okay Yeah, so here's just a real drawing of the spectral curve Okay, this is just the real points of the spectral curve here. This is x and that's y The nice thing about this polynomial I'm not sure how they found this polynomial It has the property that if you take so the the image is in x of the two branch points are plus or minus two and if you take the pre-images on The curve in terms of the white coordinates of these points you get plus or minus one which are the branch points and plus or minus two So if you want to do some calculations by pulling back to the spectral curve itself and using that as an equation The the formulas are not too terrible So there probably are not all that many polynomial. We also have that property, but Point of their example that it's just they can treat it explicitly. Ah, yes Well, they didn't say how they calculated They didn't say how they calculated anything in their example, but maybe that's the way they Yeah, exactly But you know, I don't think that matters too much for what we're doing in fact, uh, but in any case so so they As we'll see, uh, we'll see here They sort of gave an explicit description of this collision phenomena. In fact Okay, so our differentials p1, p2, p3 are all of the form yi times dx for y1, y2, y3 the three solutions of the curve And of course we have branch points in x which are two and minus two as we saw from the picture And the imaginary spectral network is as in the following picture Let's see So this is the this is the spectral network A modulo maybe a rotational question or something like that, but so here we have the two The two singularities minus two and two they Sprout at three lines. So this is like in Andy's videos, but this is just a fixed picture Extremely basic, but these two lines up here go up here and they collide at these collision point And great and send out a collision line here And similarly in the other direction we get a collision line going down here So there's two collision points So let's just sum up what we'll see in the picture. So the first thing is that we have two collision points Which lie on the same vertical line Now the next thing which we'll see in the picture in a minute the spectral network curves divide the plane into 10 regions There's four regions Maybe I should write draw the picture Let me just draw the picture as a reminder Okay, the there's a sort of vertical spectral network line Which actually continues on the other side and it's the same for the two curves. I mean, that's probably not a general phenomenon Okay And the spectral network divides everybody into into 10 regions Eight regions on the outside of here one two three four five six seven eight and then two interior regions Okay The interior regions will be colored yellow in our picture. So Those are some yellow top here so These are the yellow regions We're not we're not so good for time Maybe I'll go faster here. Okay, so arguing with the local lwkb approximation We can actually conclude that each region is mapped into a single vial sector in the single apartment of the building And the interior square this yellow guy maps into a single apartment With a fold line along the caustic The caustic is this is a curve like that That's the curve where the where the differentials cross over And in doing this discussion, we just use the axioms of the building As they were discussed by perot and the other people That's our yellow region. That's the image of the yellow region inside the apartment But you can see it folds over itself And the fold line is not it's not it's not this guy the fold line is the caustic. So These two points actually go to the same point So let me just So now just arguing by the general principles of maps to buildings what we see Sorry, no, there's one further piece of data, which is that it turns out in this case that the images of these two points Are the same point in the building So the two collision points go to a single Point in the building and everybody goes to sectors which start from that point so Those are the different sectors And then those guys go to so we can make a graph of the sectors So this is a this is that this is the actual this is the spherical building Which corresponds to the graph of the sectors in in a if you if you have an affine building But you have a bunch of sectors which all start from the same point Then you can just represent that as a spherical sl3 building Which is just a graph The segments of the graph correspond to the sectors in the same way And they're arranged like that So let me just quickly finish the so the the the remaining part of the argument says that uh Says that once if you have a configuration like this and you try to put it inside a building then that amounts to trying to put the This graph into a spherical building A spherical building is something that's sort of covered by hexagon And no two points are allowed to be distanced more than three apart So every time you have a segment of length four you should you should join it together So for example, we should add a segment of length two joining this point to this point To to close up this these four into a hexagon Pretty much any two pairs of opposite points should be joined by a segment of length two We've already done this with the yellow guy Uh Now when you do that What basically happens so if we remember our sorry, so the The collision phenomena in the bnr example comes from Taking a point p here and taking a point q here we're looking The phenomena the collision phenomena says that if we try to do the transport matrix from the point p here to the point q here That we can't get the exponent just by doing an integral of one of the forms along this path You have to integrate this you have to integrate the form like this You have to integrate to a collision point when you go to a collision point you have to go like that then like that And then like that basically That amounts to doing a phi one here And then phi three all the way over to here. You can't just do phi one phi two or phi three along the path That's the collision phenomena And what that translates into saying here is that these two Is that sorry that these five Adjoining paths one two three four five Cannot all be in the same apartment Okay, that's to say a path like this guy He's not going to go into a single apartment in the building And you can actually just see that from the axioms of a spherical building Uh Using additionally the the statement that you got from the local wkb picture, which is that as you cross over If you look at a point for example here When we go from here to here The image of these two sectors inside the building is not supposed to fold like that The local wkb thing tells us that in a neighborhood at this point We go into a single apartment and the the map into the apartment is given by these linear forms So we're not supposed to fold If you're not allowing anybody to fold along the the octagon that you have to start Then you're not then you can't put those five guys into a single apartment Basically because uh let's see The maybe the well, let's see. Yeah, if you if you try to put these five guys into a single apartment then then Uh, if we get this guy and this guy in the same apartment then that would mean that there would be a another sector going directly from here to here But then we would have a path of length four We'd have a sorry a cycle of length four You're not supposed to the building is supposed to be negatively curved So you're not supposed to have only four sectors that join together so, uh If we try to if we try to put these five sectors into a into the same apartment these four sectors are in the same apartment That looks like this actually One two three four if we try to put the fifth sector into the same apartment then Then then then there has to be only a single remaining sector joining from here to here Okay, but then we would have one two three and then this remaining four sectors So that can't happen in a building And that's basically the phenomenon which forces these two guys to to not be in the same So let me just finish with this drawing This is sort of the image. This is what the image of this guy looks like um That we have we have two basic apartments that are joined at a think of joining two joined two sheets of paper together along something which is a union of two sectors the The image of x goes into this sub piece of the building Then let me just this is what it was all said here The conclusion is we've got a universal building with a harmonic fmap Sorry from x tilde into the building Such that for any other guy it factors through our universal guy And in our example we furthermore have the property that the factorization It's not necessarily an isometry on the on the universal building itself But it's an isometry on the part which contains the points in x basically So the distances in c between points in x are the same as those in the universal building That coupled with the Remember that we could choose us an omega depending on p and q which gave us the right dilation exponent Since now we know that the the distance is independent of the choice of omega So now the wkb dilation exponent is calculated as the distance in the building b this universal building b phi So that's what we're hoping to say is the generalization in the higher rank case of the fact that in the sl2 case You can look at the tree which is the space of leaves of the foliation and the In some kind of under some generosity hypotheses such as the spectral curve being being smooth Um The the wkb exponent is equal to the distance transverse to the foliation in the space of leaves of the foliation So the hope that this is kind of only a vague hope, but the hope is that this b phi We would conjecture that something like that will exist in general. We've only constructed that in this example We'd hope that something like that would exist in general And that that would be some kind of generalization of the space of leaves of a foliation Uh in our case, we're doing sl3 The building the pieces of building have dimension two and our remount surface has dimension two so Most of our sectors actually cover their corresponding apartments except of course these yellow one which only covered a small piece of the apartment um Of course in the higher rank case The remount surface still only has dimension two and we're mapping it into some higher dimension building higher dimensional building We don't have any idea right now what What kind of geometric description you could give of the points of the building which are not In the image of of the curve So I think that's a main question is to if is there are some intrinsic description of the points of the building Okay, so I'll stop here In this cubic example, if you would do the Fourier transform easily solves the equations Then you have to see all this phenomenon just in the Fourier transform Well, I think that that's what they say in the paper. They give the explicit I mean they give the explicit Uh description of the solution and you can see that something that there's a stokes phenomena along this line Which will become a phenomenon of the actual integration. Yeah Because it's just for an integral of the simple So right so so we're trying to say that this stokes phenomenon on the collision line Uh sort of has to come about because of the fact that we're mapping into a building basically It didn't understand something that in standard complex wqv theory you never have to worry about taking the limit on an ultra filter So what what is the complication you have? Where does it come from that you have to to worry about that? Well for one thing, um We were just applying this totally general theory for what happens when you take the limit of some Some symmetric space maps into symmetric spaces where we don't necessarily know anything about that Um, so for that general theory you need to use the ultra filter Then you can say how does that apply in our case? um in our case the The the wk the the the function is going to look something like you know, maybe e to the a1 T plus e to the a2 t or something like that right If the real part of a1 and the real part of a2 are the same Then this might look like something like e to the i t Plus e to the minus i t which is just cosine of t or something like that This this function is equal to zero for a sequence of points going to infinity So if our ultra filter sort of had the wrong choice and we chose the points where cosine of t equals zero Then we would get zero rather than one. I mean that wouldn't be so good That doesn't need an ultra filter if you're extraordinarily Elaborate right so here you would say okay, let's just choose you know Some points which are kind of generically chosen so that uh the the You know cosine of something or there's approximately usually equal to a half or something like that um The problem is that I I mean I don't maybe you guys know how to do this But I don't know how to prove that the Laplace transform doesn't have an essential singularity If you have an essential singularity, it's kind of like adding up a whole bunch of terms like this infinitesimally close to each other So, uh, I don't I think you probably have no control whatsoever on what the function Uh looks like how many zeros there are of the function So I think you could hope to try to say that sort of the the places where the function goes to zero Is sort of too small kind of have smaller and smaller major And then try to argue that way or something like that or try to prove that this Laplace transform doesn't have essential singularity But uh But so excuse me so two questions if I mean so is this theory what you've represented is it about uh The uh stalks lines or is it actually about the solutions of the differential equation? So, uh, because does it actually prove that there is a stalks phenomenon? You know, um study the uh, it's only going to It's only going to tell us what's uh, it's only going to tell us something about this exponent Which is the limb soup of the size of the of the matrix At least uh, as it's currently done here. I mean, uh, of course, you would hope that that maybe some Uh, some generalization or something like that might really give you the exact solutions Or in some way or something, but I mean, I don't see how to do something like that In the a priori, we're just talking about, you know, what's the biggest, uh, what's the biggest exponent here? um I don't know if the answer is the question so uh, so, you know We're saying that you can get uh, you can write down the the vector of dilation exponents We're not saying anything for example about what type of uh What type of properties that the bases the diagonalizing bases would have All right second question Do you imagine that there should be something like this also first? I suppose I want the second exponent Uh, maybe this question the you mean lambda two No, I mean uh, no, we're claiming that this is true for the vector distance in fact No, yeah, I think that's just because I mean you can think of that as just saying let's just take the exterior powers of our The vector distance is given by looking at the exterior powers of our original system and just looking at the at the operating norm But uh, and I think but I think this is explicitly stated in perro's paper that the vector distance on the symmetry space Maps to the vector distance in the limiting building


Without knowing the sweetness of love and the overwhelming nature of the divine, you're here to touch life. If you live in the barren lands of just your thought and silly little logic, which doesn't solve anything, but through the prism of your thought, if you look at the world, if you look at creation, the whole creation looks like a problem. If you look at the world through the teardrops of your love and ecstasy, the whole world is a blessing.


The work of firefighters and emergency medical service personnel is physically and emotionally demanding. Emergency service personnel are routinely called upon to respond at a moment's notice, lift heavy equipment and patients, and risk their lives for the good of our community. The care and well-being of people and property are often thought of as most important. However, if you are experiencing musculoskeletal pain or if you become injured, it will compromise your ability to respond safely and effectively. The most recent National Fire Department injury data shows sprains, strains, and muscular pain are a problem for fire and EMS personnel. They account for the highest percentage of injuries, even outranking injuries due to burns, stroke or heart attack, smoke inhalation, and wounds and bruises. For example, 50% of injuries due to traveling to and from incidents, 42% of injuries occurring on the fire ground, 57% of injuries during non-fire emergencies such as EMS responses, and 59% of injuries during training activities. One of the ways to reduce sprains, strains, and musculoskeletal pain is by applying ergonomics. This video is intended for fire and EMS personnel to review ergonomic principles. We will talk about what ergonomics is, what musculoskeletal injury risk factors are, what ergonomic solutions are, and finally, how these apply to you. Ergonomics is defined as modifying the task or environment to fit the anatomical, physiological, and psychological capabilities of the worker. It is about the relationship between the worker, the task, and the environment. In this example, the firefighter is removing a ladder from an engine and setting it. Some of the worker factors that could affect this relationship are his or her age, height, upper body strength, experience, fatigue level, and stress. The task he is doing is handling the ladder. Some of the task factors include the ladder's weight, length, the method he uses to lift and carry the ladder, for example carrying it on the shoulder or at the waist level, and the number of people available to assist. Finally, the environment in this example takes place at a fire scene. It may be noisy, cluttered, and the ladder may have to be carried extended distances. Beyond this example, there are other worker, task, and environmental factors that arise with the variety of work activities that you do. Can you think of any? Injury risk factors are postures or activities that can lead to acute and cumulative trauma injuries. There are several musculoskeletal injury risk factors that fire and EMS personnel commonly face at work. Awkward postures are any postures that are not in a neutral or anatomical position, such as reaching with your arms overhead or away from your body, and bending or twisting your torso at the waist or neck. Excessive force means exerting large forces such as when you push, pull, or lift. For example, lifting the fan or handling patients. Repetition is performing the same motion over and over. It can be highly repetitive over a short period of time, such as when pulling down a ceiling or low repetition over a long period of time, such as the common activity of putting the gurney into the medic unit. Impact forces are large forces transmitted to parts of the body by contact with the ground or a hard surface, such as the force on your feet, ankles, and knees when you jump out of your vehicles. Excessive temperature occurs during very hot or cold conditions, which can compromise your soft tissues, such as when you're fighting fire in your protective gear or during your training drills. Whole body and hand-armed vibration result from activities such as riding in vehicles or operating powered hand tools. These risk factors may not result in immediate injury. However, if you increase frequency, duration, or intensity, or you combine the risk factors, this increases the risk of an injury or an eventual disability. An example of combined risk factors is when you are moving a patient from their bed to the gurney, which is a common activity that involves awkward postures and forceful exertions. Once you are aware of what risk factors are present, it is important to come up with solutions to eliminate or reduce the risk factors. In general, you want to reduce the time, frequency, or intensity of the risk factor. There is a hierarchy for basic ergonomic solutions. Some solutions are very effective at reducing the injury risk, but may be very involved while others may be less effective yet easier to implement. Oftentimes, the most effective solutions are not in your personal control and take an involved departmental approach to reduce the risk. At the top of the hierarchy for being effective is equipment and engineering to eliminate the risk, followed by job organization, then personal protective equipment, and finally, body mechanics. While you may already be doing some of these things, the following examples are meant to illustrate different types of solutions in the hierarchy. The most effective ways to reduce risk are equipment selection and engineering controls. Using appropriate ergonomically designed equipment or redesigning current equipment can reduce your injury risk. One example of equipment selection is the type of medic pack carried on calls. One type is this hard case. These weigh 12 pounds empty and must be carried away from the body due to their shape. An alternative is a soft pack with shoulder straps. These weigh only 3 pounds empty and can be carried closer to the body. Also, consideration should be given to the supplies that are carried in your pack. Reducing the supplies down to only essential items can decrease the weight that has to be frequently carried. Another example of an equipment selection is a lateral transfer device. Slide sheets are made of a slick material that reduces the friction during lateral transfers. This decreases the stress to your low back. They are lightweight, can be stored quickly, and are comfortable for the patient. There are a variety of styles available for your equipment committee to consider. An example of an engineering control is this bariatric ambulance. It uses a special gurney that travels up a ramp pulled by a winch mounted inside the medic unit. This moves the patient safely and quickly into the medic unit without using any manual force from the emergency service workers. These types of interventions need to be approached through the safety committee and may involve the survey of the department's needs. Also, equipment purchasing committees should take ergonomics into consideration. Line personnel can really provide valuable input to safety and equipment committees since they know what might work best. Changes in your job organization or work practice can also reduce injury risk. An example of a job organizational change is having four people lifting the patient instead of two or three people. Even with four people lifting, using proper body mechanics is still important. Are all these people using good body mechanics? Another example of an organizational change is two people carrying the fan instead of one. Both of these examples decrease the weight being carried by each person. A work practice change could consist of using a handle and stepping down when exiting vehicles instead of jumping. This reduces the impact forced to the lower body and decreases the risk of spraining an ankle. Even in the office, the way the workstation is set up can impact stresses on your body, especially the low back and risk. Organizational changes can be complicated and difficult to implement, but working with your station, you may be able to develop ideas on how to improve your work practices. Approaching the safety committee, training coordinators, and command staff with ergonomic ideas are effective avenues for making organizational changes. Personal protective equipment is lower on the ergonomic hierarchy. Ergonomic personal protective equipment promotes neutral posture or reduces forces to the body. Examples include lumbar rolls to keep your spine in neutral posture while sitting in the vehicle, doing paperwork, or reclining. During drills without live fire, knee pads could be worn to decrease the force to the knees, a joint prone to osteoarthritis. These are items that could be requested from the equipment committee. Finally, body mechanics is an option. This is the least effective protection from risk of musculoskeletal injury, but the one that you have most personal control over. Maintaining neutral spine while lifting, such as when you are lifting your air pack, rather than bending at the waist, is an example. Similarly, when raising a patient on a gurney, you should lift with your legs, keep the gurney close to your body, and your back in a neutral posture. The most important body mechanics concept for the low back is maintaining neutral spine. Remember to keep the three curves of the back, the cervical, thoracic, and lumbar curves. One way to practice neutral spine is to hold a straight edge against your spine. You should be able to slide your hand between your low back and the straight edge, even as you move. Proper body mechanics should be practiced in all your movements at work and not just when you are lifting heavy patients. This ranges from firefighting and patient handling to housekeeping at the station and physical fitness. Now that we have talked about the risk factors and potential ergonomic solutions, let's take a look at a few scenarios to practice ergonomic skills. For each scenario, think about why it is done this way, then identify the risk factors, come up with possible solutions, and discuss barriers to successfully implementing these solutions. These scenarios won't apply to all departments or all situations, but they present an approach for identifying problems and exploring ergonomic solutions. This first scenario depicts two people carrying a patient down a stairway using a small cot with handles. Why is it done this way? The risk factors are excessive force caused by weight of the patient, awkward postures associated with keeping the patient level, and twisting to see the stairs. A potential solution for a wide stairwell is using a backboard with four people or the same cot with four people. An even better solution for most stairwells, wide or narrow, is a well designed stair chair that places little stress on your body. This model is excellent since the patient does not have to be lifted and carried down the stairs since it rolls down the steps. It can also make the patient feel more comfortable and more secure. Examples of barriers to these solutions might be the lack of storage on the ambulance, or restricted funds for purchasing the equipment, or resistance to changing equipment and work practices. Are there other solutions or barriers? The second scenario demonstrates a firefighter rolling a hose during a training drill. Why is it done this way? The risk factors associated with this are awkward and prolonged postures to the back from flexing at the waist. Depending on the size of the hose, different techniques may be used that reduce the stress on the low back. For two and a half inch hose, the firefighter could roll the hose from a kneeling position, maintaining a neutral spine. For larger, four inch hose, a combination of bending and kneeling could be used to reduce the time spent in each posture. The barriers could include the perception that it is quicker to roll the hose while walking in a stooped posture. Another barrier might be a knee problem that causes pain when kneeling. What might other solutions or barriers be? In the third scenario in the station, this firefighter is performing a routine check of the medical pack. Why is he doing it like this? The risk factors are prolonged and awkward postures to the neck, back and knees. A solution might be to place the medical pack on a table or cart about waist height to elevate the work and allow the worker to stand upright while checking the pack. Some potential barriers include space for a table in the bay or the time it takes to get the table if it is stored or walk to the table if it isn't near the vehicle. What might be other solutions or barriers? The final scenario shows a common exercise performed by fire and EMS personnel to strengthen oblique abdominal muscles. Why strengthen these muscles with this particular exercise? The risk factors to the back include repetitive twisting and flexing of the low back which are awkward postures and forceful exertion to the lumbar discs. Solutions include exercises such as the bridge which keeps the spine in neutral posture while maximizing trunk stabilization, strength and endurance. A barrier to using this exercise might include resistance to changing an existing workout routine. Can you think of another way to safely strengthen the obliques? And what are other barriers to adopting new behaviors? For ergonomics to be effective in preventing musculoskeletal injuries it must be part of a comprehensive safety and health program. An ergonomics team can be part of the existing safety committee which is the appropriate vehicle for creating change. Your role is to identify and prioritize risky job tasks, assess the hazards, develop plausible ergonomic solutions and bring them to the attention of the appropriate committees. As a review ergonomics is the relationship between the worker, the task and the environment. It is fitting the job to the worker. Musculoskeletal risk factors include awkward postures, excessive force, repetition, impact force, excessive temperature and vibration. And remember, time, intensity, frequency and combining risk factors increases the risk. The solutions in order of effectiveness are equipment or engineering changes, job organization and work practice changes, personal protective equipment and body mechanics. Attention to ergonomics can help prevent injury, disability and help you provide the best care possible to your community. The long-term benefits of ergonomics can prolong your career and enhance the quality of your retirement.


From around the globe, it's theCUBE with digital coverage of Actifio Data Driven 2020 brought to you by Actifio. Hi everybody, my name is David Chang. I'm the Chief Product Officer and Co-Founder of Actifio. Extremely excited to have you join Actifio's Data Driven where it's the only event customer essentially is on the center stage telling you what they're using Actifio for. Excited to have you looking forward to the event.


It's my son Turner Smith. Oh, this is Lou. Come out there. Well, that's Jamie. Jamie. Mr. Turner Smith. Well, it's nice to meet you. Are we going on TV? Sandy? Yeah. Oh, we'll be going on TV. Good to see you. Good to see you. Oh. We'll get out of the way. I think they're in family picture. All right. Is that a TV? No, it's not. Well, they're live. It won't be on TV. Yeah, that one's a TV camera. Jamie, look over here. That one's a TV camera. You got a smile on it, right? Jamie Smith. Jamie Smith. There you go. Thank you. What about with the whole family? Oh, all right. All right. We'll be back in a moment. Are you filming right now? Yeah. What? You're still very fit. Just look back. Thank you. Thank you very much. President, we're going to get the grandparents in here now. Oh. All right. That's so great. Thank you. Yeah. Thank you. Thank you very much. President, we're going to get the grandparents in here now. Oh. Oh, thank you. Thank you. Jamie, thank you so much. Jamie. Jamie. Right here. Right here. Right here. Right here. Right here. Thank you. You took it apart. She learned it quickly. All right. It's great to meet you. It's great to meet you. Yeah. It was so great. Thank you very much. Well, thank you. Thank you very much. Will we see you shortly? Huh? That's right. It's great. OK. That's going to be great. Can you say hi?  Can you say hi to the boss? This is Greg. He's three and a half months old. He weighs about 15 pounds.  So he's going to be bigger than his sister. Should we go over to the side over here? It's fine what you want. Can you give the president a big smile? Here, Caroline, we'll put you in the center. Can you give him a big smile, Caitlin? That's a good smile. Hey, let's get one more pose from here. You don't want your hands in the middle. Greg, can you look at the camera? Use your hands all the time. There we go. I want to let Caitlin stand on the ground and we'll get out of here, Mr. President. You don't want to stand on the ground? You want your picture taken one time? Yeah, do you want to stand next to the boss and have your picture taken? Like a big girl? No, it's a girl. No, it's a girl. Thank you. Thanks a lot. Now, you know, of course, don't you? Day after tomorrow, we're going to be that tall. Well, as Mrs. Reagan says, tomorrow we'll be 20. My favorite story is about... And this is a really true story. Gala was having her first child in the hospital and thinking pretty rough and it was a pain. And there was a pretty experienced nurse there. And finally she turned to the nurse and she said, oh, she said, this isn't going to last long, is it? And then the nurse said, just about 18 years. That's true. We've got a lot to learn that, don't we? About being parents. Now, he slept through the night for the first time last week and then what, he's done it a couple of times since then? I can't believe that he likes it up there. She's breastfeeding and as you know, that doesn't hold him as long as the formula. So he'll wake up at three or four and want to eat and then wake up again at six or seven and not eat again. But, you know, the last week or so he's been sleeping through the night. That's nice. You want to go to your mommy? Can you tell him I said bye-bye? Can you tell him bye-bye? Yeah, he's going to give me a tour. Katelyn, can you say bye at the bar? Say bye. No, we'll get you some jelly beans on there. There's comic strip readers who know what I am. Right now, the one wants to, is it Kathy? Yes, the girl. But her friend is having her first baby. Well, I wish you'd see what they go through day after day, the routines. They've read all these fantastic things about the child and while he's in there. So now, she has the thing that she holds that plays music. And then she talks to everything. They're going to have a baby born able to go to school. Well, maybe it'll work. Well, we get out of the way. Say goodbye to Katelyn. We have jelly beans out there, mommy. There are jelly beans out there. We'll have them all over this pretty dress. She saw those. You want to take a handful out of there? Not a plus. Say thank you. Don't worry, I'll give you this. There. Not one at a time, Katelyn. Katelyn. Don't talk. Hi, how are you, Elizabeth? Do you want to put that back in? You're looking your screen. And look at Katelyn. Don't rest up. Hi, can you say hi to Elizabeth? She just, we went into the family. Oh, how nice. Katelyn, chew those things up good. Great. First time for photos. Hi. Good for him. Hi. Good photo. You want to spend those jelly beans out? We'll see you this weekend. Okay. Thank you for taking time. Well, I'm pleased to see you and me, and we'll introduce each other. Thank you. Let me introduce our leadership team, Dr. Ken Scholl. Nice to see you, President Reagan. Nice to see you, Mr. President. Mr. President, nice to see you. Nice to see you. Nice to see you. They are sure, President Reagan, and President of the State of California. Newport Beach, sir. Hold on. Hold on. Hold on. Hold on. I'm sure you'll be able to make many lines. Very good. I used to do my box. With the senators. I'm Pula Martin, the media's past President. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. I'm Pasta Shade. Back as all boom I banster pack. Good to see you sir. We all go over here by the desk. Do we need everyone on the right side? Can we ask, want you to step up with right side and balance it off? Absolutely Mr. Secretary, why don't you get in the picture too? All right. Well, listen it's nice to have a chance to also say thanks to all people that alwaysbolメ this sensitive year We're still one of the 1.8 but we certainly want to appreciate you taking time to meet with us today you know you've got a tremendous track record on housing this fourth consecutive year 1.8 highest housing levels I think of any administration that's ever had consistent in four years the less interest rates we've had in the last four years we really appreciate the job you're doing for housing we did have about three areas of importance that we felt it might help you take credit for the job that you've done first one was a housing priority national priority second we're a little concerned about the future of FHA and VA certainly aware of and the third one was a formulation of the future national housing policy in this country those are the three major areas and I'd kindly be remiss if I didn't say something about tax reform we got a couple concerns about that the transition rules and fairness on it and I'll tell you if I didn't mention that to you this afternoon I'd probably be slapped upside the head with a two before membership so we do have some concerns there and we just appreciate the opportunity to meet with you and John we're going and we would hope to sit down and talk with members of your administration further in detail about these we have prepared a briefing report that we're going to leave with out came more detail on those okay issues and we really appreciate the opportunity and want to thank you on behalf of all of our members under the 42,000 business firms and we'd like to extend an invitation to join us down in Dallas at our annual convention talked our board again you know you've had a tremendous record since you talked to us last time and we'd love for you to come down and I think Fremont Crow's Customs Festival provisions down there in the Anapole to accommodate you we'd love to have you come down with it January. There are 58,000 people who attend that so it's it's quite an opportunity and we're going to stress the bicentennial of the I've been a customer just in the last few weeks not building anything doing a lot of repairs and so what the ranch well we just came that helps our energy. Yes, as a matter of fact the same contractor that did the work on the original ranch house when we got there that built the house for the form and thereby he was in on this work so I hope he's a member of our association I can say this that having preceded Dave last year it made it very easy to register and have a growing association and the leadership has helped to make that come about. We sure appreciate the statement you made on Barbara Walder's show when she asked you what message you want to send to the average Russian when you said you know it was a single family home with a car and a garage and that's the right way we'd like to see you make more statements. I'm so frustrated. I've never heard of this country before. I get in here at the summit and I keep thinking of all the things I didn't see for the first time but then the frustrating thing I said, I said to myself, yes, everything that you've shown, you'd swear, was just perranged. And the other day that was reaffirmed. If you haven't read it, you wouldn't want to know about life in Russia. There was a book by John Barron, the Washington Bureau chief here of Reader Duchess, but the book is true. It's a factual book and it's simply called MIGPilot and this was that Russian pilot, he was a real top ace there in their air force that had just grown up believing increasingly to the government, their government lying to him and so forth. And he's the one that then flew the MIG-25 on to Japan and then turned over to Washington so that we could have the information. And he's now here, married to a girl in South Dakota. But the book is so fascinating because they write, as he's making his escape in the MIG, trying to fly to Japan. And before he gets to Japan, they flashback the book that's to his childhood. And now you see him grow up in Russia for all the things in their daily living. This great pilot got to this country and had to be taught to drive a car. But all the things that are in there and the things that led to his not leaving, but then when he came here, he tentatively did all kind of fly to Japan. So when he showed him a supermarket, he believed it was all set up just to show him that. Well, that statement that you made certainly sent a signal to the American public that you supported housing. So that's wonderful. We appreciate it. We need more of that plain talk about housing. Yes, we do.  Yes, thank you very much. Thank you, Mr. President. Thank you, Mr. President. I think you're going to get your tax bill. We're trying to make a little change in the context of the transition. We support the concept certainly. The objective of it. Okay, you're welcome. I'll show you after you have too much to dash in, everything you say will naturally be true. Thank you. Everything you're watching folks say to me. Thank you, Mr. President. Well, we're not done with the change. Let me get ahead outside again. Do before now. Thank you, Mr. President. We appreciate your position. Thank you for everything that helps. We appreciate your work. Thank you, sir. I look forward to seeing you back in California when the truth is found out. Next week we get back there in a few days.


Hi everyone, Joe for Jaspy's casebreaks.com coming at you with a two-box break of 2023 Topps Gilded Collection Baseball It's a division break and on sale as well. This is random division break number two All hardship obviously big thanks to this group Danny Alex and William for getting in on it. There are the divisions right there Let's roll it randomized names and divisions one and two three times one two Three easy all right one and two three times for the divisions one two and three Got NL West down to AL East William last Bob Mojo NL West Danny AL West Alex NL Central Danny AL Central Alex NL East Danny AL East Let's order these alphabetically Danny has all the ALT AL divisions There you go. All right. I'm gonna pause the video We're gonna see if there's any trades when we come back. We'll have the break as well So stick around BRB. All right, welcome back everybody. No deals were done. So that list right here remains the same So thanks everybody for hanging with me on a Wednesday hump day new release day the 25th And we've got it's a six-box case The final two box division break is in the store right now jasmine's case race calm. It's my reflection right there So we'll go one two three for the top two boxes and four five six for the bottom two boxes. Good luck boxes It's two so we'll go top two boxes So put that right there these bottom two boxes will do for next time for break Three you should be able to get this done tonight. We may as well finish off the case We'll put those two boxes right next to these back here Keep those back there Grab my knif Got enough supplies here All right. Good luck everybody Got caught on the edge of that box What are the first two box that were the first two boxes good? I feel like it was all right. I mean this is This is not bad Even though even the worst box is still pretty good for the product like this, but We've seen some incredible boxes as well. So let's see what we got in here Let's say one autograph box. Sometimes we'll get more out of here All right, we're gonna start off with a rookie of Brett Beatty 60 out of 75 and L East Alex Got Vlad Guerrero senior 84 out of 99 that's gonna go to the NLE east Alex Expo stuff goes to the Nats and the autograph behind Pudge is Leo perguero rookie auto 56 out of 99 pirates. That's NL central Alex Their divisions are only bad if they don't hit Ivan Rodriguez Pudge Yvonne Rodriguez that is 50 out of 50 Rangers AL West for Danny gold frame the gold frame is gonna be Paul Molliter 41 out of 50 Brewers and L central for Alex. It's only bad if they don't hit second box Bring the ghost right out there Taking a look at you. All right second box. Good luck everybody Quick two box break like I said another two box division break is in the store right now Jasper's case breaks on calm just needs six people to get a division each And then we're in breaks for us one auto Right here. We got a Tanner BB 161 out of 199 guardians AL central Danny There's a Corbyn Carroll rookie card one out of 99 That's gonna go to the NL West William Dimebacks in the World Series Corbyn Carroll World Series MVP It'd be good for the hobby and Behind a von Herrera 24 out of 99 for the NL central Alex is JJ Bledet rookie auto 93 out of 99 He may be a top five pick when he was drafted. This is gonna be for the AL West Danny He's originally in the Marlins organization and your final gold frame. Good luck everybody final gold frame will be Tony Perez 12 out of 50 Old school red going to the NL central for Alex who previously said NL central on It's only out if they don't hit Everyone else is saying about their divisions now So there you go gang two box random division break number two in the books. I'm Joe I'll see you next time for the final two boxes Jasper's case breaks calm