Welcome, and thank you to everyone that has joined us today for another episode of the nonprofit show. Julia Patrick is here, the CEO of the American nonprofit Academy. I'm Jared Ransom, your nonprofit nerd CEO of the Raven Group, and we are constantly in awe and appreciation. We have so much gratitude to our presenting sponsors. Each and every one of these companies are dedicated to the nonprofit sector, dedicated to each and every one of you as leaders doing your part in the world. So please do check out our presenting sponsors. Again, we are so grateful to have their continued investment not only here in these episodes here at the nonprofit show, but truly around the globe in the sector at large. So thank you to our sponsors. And thank you to our guest who has joined us today. For those of you that joined us in the Chitty chat chat. You heard a little bit about Doris Vaught and the CPA journey that has taken her to become the president and CEO at Lifewell. So I would like to welcome you Doris and say thank you for joining us today. Thank you, Jared and Julia. I'm really happy to be here and excited to share my story and the story of our organization. Well, let's start there. Talk to us about what Lifewell does. Lifewell is not surprising a nonprofit organization. We provide mental health services and physical health services to people who are struggling with a variety of issues in their lives.  And many of our members that we serve are homeless, many of them have limited resources, limited income. And they also have limited social networks and a lot of cases. So what we really do for people, I think our vision kind of sums it up best as we offer hope, health and healing. So we really are here to help people live their best lives possible and to provide them with the tools and resources to do that. Wow, amazing.  So I want to start by saying thank you. Thanks for your service, your team service and all that you've continued to do throughout such a tumultuous and uncertain time.  And so when it comes to being a leader of this organization as a CPA, you've been with life well before so you didn't just start as their CEO, you spent some time in, you know, in in the trenches or you know really in into the programs as well. Tell us a little bit about that Doris. Yeah, I've actually been with the organization for about 20 years. Wow. Not even base. I've seen, I've seen our organization evolve significantly during during that time period. So it's been really fun. It's been really exciting for me to see what the organization has become and to see what it can still become in the future. I started 20 years ago I started as a financial consultant, and I had kind of my own little business that I ran with consulting and worked with a number of nonprofits, and I, I really fell in love with this organization. And at the time it was very small. I had about 60 employees, the annual budget was about $2 million. And the organization had contacted me because they needed some help in figuring out, you know, how they were going to be able to keep the doors open the resources were very very limited. So, that's how I started my journey. And over the next, I guess, I consulted for about three years and then became an employee in 2004. And in 2011 we, we went through a merger with another organization, the organization that I started with was called New Arizona family, and we merged with another organization called triple our behavioral health. And it was really at that time that we were able to when we combine those forces we were able to become one of the larger behavioral health organizations in our area, and really gave us some substantial opportunity for us to grow our services and expand. So, I prepared for a number of years for the CEO position we had my predecessor was with the organization for a long time and was retiring. So we knew that that was going to happen. So, you know, I never ever felt like I was going to have all the answers when I started because I had watched how difficult the job is. I thought that I was, you know, pretty prepared for anything. So, you know, as luck would have it. I was about eight weeks into the job before, you know, COVID really kind of took over. And you're still here for not saying this isn't what I signed up for. Yeah, it was, it was, I knew that I was going to be faced with significant challenges as CEO I think you know that's just part of what you expect. I didn't really expect that you know my first major decision would be you know a world emergency a pandemic so remember the feeling very well of thinking, okay, you know, this is your decision, you know, and you have to make changes and you have to do it very very quickly. So, well let's talk about that that quickness and we've heard. And Julia I'm sure that this is a word we've heard quite a bit is, you know organizations have really become nimble, and I can imagine Doris because of your 20 years of experience within life well that has really helped you to navigate change with quickness. And I was, I was impressed with really, you know how quickly telehealth and that increased I felt like it was there but it really increased at a dramatic during, you know, early 2020 I would say, but talk to us about you know the quickness that you've been able to lead with your organization as well as getting your board on board for these quick responses. Yes, we had to make changes very very quickly and we embrace telehealth we. The first thing that we did was take a step back and say okay you know this this is serious we need to make sure that that our employees and that our our clients are as protected as possible. We knew it was it was vital for us to be able to continue delivering service to services to our clients during this time so we knew that we were hoping that our doors wouldn't be closed and we would have the opportunity to continue. We shifted our services and our outpatient clinics where we really operated on skeleton crews at the site and tried to send as many people as possible home to do telehealth and to do telephone services with our members was easier for our admin staff because you know their positions kind of naturally lend themselves to that, but we do also have some residential treatment programs and some community living programs where we still, you know, had to be fully staffed at sites and we had to go, you know, visit clients in the community. So really making sure that we were as safe as possible so the first thing we did was you know try to figure out how we get the PPE how do we get extra cleaning services, you know how do we, how do we find Clorox. Yeah. Oh my gosh, I mean like so interesting because you're looking at your clients you're looking at the community you're looking at a health issue. I would imagine you're also looking at your budgets, knowing that. Oh my gosh, all these things that we're doing me you know we hadn't planned for and as a CPA I would imagine you put that hat on very quickly. There was there was a lot of stress around that because even before we were able to have the expansion of telehealth services and fortunately here in Arizona we have great support from our funders and great support from our you know our leadership so they did make moves very quickly that allowed us to do the telehealth services. But you know we didn't know how that was going to go because our staff had never done it are some of the members really don't like services by video so we were very unsure and we were even unsure as to you know the the level of services that we were going to be able to provide so immediately you know kind of saw probably our revenues going to take a deep dive. So how do we best protect ourselves and we did. One of the things that we did very quickly was apply for the paycheck protection program. I was going to ask you about that. Yeah. I'll talk to us about that a little bit because you know again having a CPA in the driver's seat. Oh my gosh. That made your board look like super geniuses. I mean talk to us about that. We had a long standing banking relationship with the bank that we had worked with and they were a large bank they had always been able to meet our needs and very happy with the services that they have provided. However they struggled with the paycheck protection program. And we found ourselves not able to get alone through the bank that we had a relationship with and finding out kind of along the way that you know kind of knocking on doors with other banks literally in some cases saying you know we have we have a need or is there any way that that we can get alone and fortunately we found one in the community who was willing and able to help us even though we didn't have a banking relationship. So that was that was definitely a learning process for us you know making sure that you don't keep all your eggs in one basket. We worked out and you know fortunately that that relief funding was there and available, and we've actually gotten some other relief funding through through Medicare Medicaid and other sources so it's been a life saver. You know truly without it we would have had to have taken some drastic measures that would have impacted our organization for a long time. So we're grateful. You know, this last two years I feel like we keep saying two years even though you know months continue to go by, all of us have had some insurmountable, you know, hurdles to jump. I cannot imagine because one of the things we've talked about almost weekly on the show is really about the return on relationships and as you just said Doris. One thing you learned is not to put all your eggs in one basket so I can imagine, maybe the fear and the stress when you find that your financial institution is having their own challenges. They have faced their own insurmountable, you know hurdle to provide, you know, these financial needs because at that point it was really impacting the workforce it was impacting the bottom line. And I can imagine maybe some flashbacks to when you started working with life well that says, Oh my gosh, are we going to have to consider what this does and what this means to our service and to the programming. So I cannot imagine what that was like but to Julia's point, you know, hurrah for the board because if anyone can navigate that it is certainly a CPA. Well that I think the board was really has really been helpful throughout this whole time period, really just allowing us to to lead the organization and make those quick decisions. They've been tremendously helpful. I heard from some of the most successful leaders and organizations that they are the ones that have the board that back them up and say absolutely whatever you need let's get in there they're rolling up their sleeves they're really leaning in to the obstacles and helping to provide their thought leadership to provide solutions right and one of the areas where we're all looking for solutions right now is the lack of workforce. And I'm curious if you would be willing to share how you have navigated the shortage within the workforce and as our labor market can continues to shift. How are you leading through this obstacle. Yeah, you know that was the second big surprise of covert I think, you know, first trying to figure out how we keep people safe and keep operating and then, you know, the number of people that have left the workforce has, I've really never seen it like this with you know, we, we, you know, publicly funded behavioral health organization, you know, we always struggle with resources, even though our state recognizes the need for mental health services and puts a lot of effort toward making sure that those services are provided. We always have more need for our services than there are resources. And so, you know, because of that, and some of the reimbursement rates, we're limited by what we can pay people in these, you know, critical roles in these very valuable positions kind of all the way down to our professionals case managers all the way up to our psychiatrist and medical staff. So we had shortages before the pandemic it was something that we as an industry we're trying to figure out. But the pandemic has made it 10 times worse.  So I'm happy that you know it's larger than life well it's larger than our industry I know that our state leaders are taking on workforce issues and trying to to see how they can develop plans with the universities to develop career tracks and also, you know, looking for information from us on on rates that are sustainable going forward so I think there's there's a lot to figure out but but we're finding our way through it. Well, let me ask you this. I mean, I feel like this labor issue, the great resignation, the aging of our leadership population is almost like another wave of the pandemic and something that is going to put us into a topsy-turvy place for quite a while and I'm wondering. And again, this is get out your crystal ball and shine it shine it out. But do you feel like we're at the beginning of it we're in the middle we're on the downside do you have any sense of where we might be in this journey and that's a big question. Yeah, I hope we're at least in the middle of it. I can't imagine that it would get much worse I hope. So, but I don't think it's a short term fix I think you know we're going to be struggling with it for a while. And some of the things that we are doing as life well. We've actually taken a pretty deep dive into our workforce issues and we've hired an outside consultant to work with us. We've done surveys of our staff so I think you know the kind of the first things that we did were immediate fixes that we can put in place. As you see, you know, salaries rising in the community, we had to follow suit somewhat. So we, we put some recovery pay some temporary measures in place and we also put some permanent changes in place but we know that that's not the only issue and it's not the only solution. This deep dive we've surveyed our staff we've spent a lot of time compiling the results and really starting to focus on how we can make the changes, you know within our resources how we can make changes that really people are asking for. I think it, you know, it really starts with listening so we've really been listening to what our staff have to say, and we know that the jobs that we asked them to do are very difficult jobs. Yeah. So what can we do to make it better you know we know that we can do more training we know that we can, you know, offer like discounts for childcare, things like that so. But we're, we're really being thoughtful about it and I think it's going to take us another year before we're able to really see changes from the things that we're going to implement. But in the broader picture, you know, I, what I see and what I read is, you know, you know, workers are really, you know, in control, you know they're in short supply and it's a time where I think organizations really need to listen and see how they can better meet the workforce. I think one of the things I continue to hear and I know you had shared with us Doris about the, the demographics of those that you serve, you know, so severely mentally ill behavioral issues. Many of the workforce itself is also experiencing this on their own right at a different level. So we've heard often about fatigue, we've heard about burnout we've heard about, you know, alcoholism spiking and there's just so much going on as we all continue to carry the weight of this uncertainty, you know, and so I can only imagine that not only your staff but all staff, the entire workforce across the nation, you know, we're all struggling with their own challenges and coping and resources. I even heard on the news lately, you know, therapy and counseling has gone up significantly, so much so that the waiting time to get on to, you know, a counselor or therapist books, sometimes is not until next year we're talking January of 22. And so that's, that's really trying and really, gosh, it's just heavy right, it's really heavy for so many of us. But as we talk through this there are silver linings there are good happening. And so I would love for us to take a look at how are you recognizing opportunities when they arise and what have they been. Well, I think, you know, it's far from business as usual. As usual anymore. Right, define that. And that in and of itself, you know, has been an opportunity for us to, to again, you know, take a step back and kind of reevaluate, you know, why are we doing things the way that we're doing. We're going to have ongoing flexibility for staff where we're able to, where their positions allow to continue, really a hybrid model where some days in the office some days at home. We've, we've offered some like schedules, you know, not everybody has to work five days a week we've offered the four 10s and we've even done some, you know, in a two week period where you work nine days. So just offering different options to people and I think we were never really able to do that because because of the telehealth really was not at the place where it is now so as we staff up as we get our organization healed a little bit I think those will be ongoing opportunities that will be able to offer more and more. There's a lot of relief dollars out there available so just trying to figure out how we as an organization might be able to tap into those resources. There's grant opportunities. More dollars from the American rescue plan that will be coming down. So we're looking at our facilities, you know, are we using our facilities to the best use. Do we have too much space now that you know people are working at home. Do we consolidate so lots of things I think that that we're able to do that are really setting us up for for growth in the future. Have you seen technology. One of the things we continue to talk about is really the advancement of technology. I've said Julie and I have said that you know of coven the pandemic and hit even two years ago or three years ago we don't know that we would have been equipped in a technological standpoint to really provide the services. Now your services are also with this telehealth so how has technology been an opportunity or has it been a challenge. It's both. Yeah, yeah, I mean we've seen huge opportunities with telehealth and I think you're right. I think if this had happened a few years earlier, we probably wouldn't have had enough laptops to provide our staff with to work from home, you know at the time that we needed it. So, I, I credit our focus on technology. We've always been an organization that's focused on advancing technology. So I think that was part of what prepared us for that but part of it was also just timing, and we had just ordered a lot of laptops to use in our training for our new hires. And so we were able to pull all those laptops out get them set up, and our it department was was great but the fun one of the funny stories. During the time was, you know I was sitting with our chief information officer and we were going through all this and he's like okay okay yeah we can do that we can do that. And he's like, by when. Yesterday. Right by yesterday. So, so that was that was just a fun story but then on the for our clients, I think there's both opportunity and challenge with with it on that side as well some of our, our members and our clients have really embraced the telehealth and telephone services, you know they don't have to worry about transportation. They don't have to take a half a day, you know to come into the office and have an appointment. And then others, you know because we do serve members with limited resources, many of them don't have the technology available to them they don't have the laptops they don't have internet. So, there's still, you know, there's still things we can do you know we can have those members come into one of our clinics, and see one of our staff that is working remotely but it has presented some challenges and that way. You know this has been a fast, pardon me fascinating conversation on many levels, I think your personal personal journey of leadership and your education and the way the lens with which you lead, I think that's fascinating. But I think the thing that's so riveting is that even a leader with 20 years experience within the organization, eight weeks before a calamitous situation. It's fascinating that you have been able. It seems like successfully manage this. And it's the hits keep coming as they say, you know, like the labor when we were talking when we first started almost two years ago. Everybody was like the experts that we had on were like, don't let go of your job because you'll never find another and I mean, isn't that true Jarrett. Yeah. So many things so it's been really exciting to hear your perspective and your success. What an amazing thing I think we need to sit down and figure out where we can have you back on and like another eight to 10 months to figure out okay. So what has also come about. What are some of these things that occurred, because we will have those hopefully those recovery dollars coming back in. And again, I feel like that's going to really put a whole new layer of opportunity and stress on organizations. You know, and that's still being defined and everything but still I think built back better it's going to be a huge, huge. Yes, and it's, it's kind of funny, even with that, Julia, because there are lots of opportunities out there but being short staffed it's hard to you know, find the resources to apply for those opportunities so it's definitely it's definitely a juggling act but I'd love to come back I really enjoyed speaking with you this morning. Well, it's just been great and you know I have to say Doris, when I know there are leaders like you in my community. I'm doing the hard work during the work that I could never do, nor even understand it is it like makes me realize how blessed we are to and loving loving the work that you do with so much passion is. It's hard to come by but it definitely shines through with you so thank you for all that you do and continue to do your leadership is needed during this time so. Thank you for spending some of your very valuable time with us today and to join us here on the nonprofit show. Thank you very much I appreciate it and just a quick shout out to my team members and all of our team members here at life well because they really do the hard work that makes a difference in people's lives so. So we're very proud of the work that we do. It shows and you know it's it's good to work, it's easy to work hard harder when you have a good leader. I think that's one of those things. Hey we've had a great great episode, I always love all of our episodes Jared, but today I'm really engaged in a new and different way of thinking about leadership. It's been fabulous. Again, we want to thank all of our partners in our new sponsor this without their buy in and commitment to us, we wouldn't be able to have these amazing conversations so we hope you've been as inspired. Thank you so much for today, as you are hopefully every day, but this is one of those times when we need to remember some of these lessons that we've talked about. Wow, Jared, I love love love these conversations you know I'm a leadership geek and so this has been great. It's been wonderful. Thanks again to everyone that has joined us I hope you will come back and join us again tomorrow. Until then, we will sign off the way we sign off each and every episode by asking you to please stay well, so you can do well. Thanks for joining us and we'll see you tomorrow.


Ok, alors, je vais expliquer très rapidement ce qui est la motivation de mon talk. Donc, c'est un joint-work avec Katya Konsani et très recentement, on a fait une découverte dont j'avais été sorti, en tant que 96 ans, j'ai trouvé un espace non-commutatif qui m'assurait d'écrire, si vous voulez, les formules explicites de Riemann Veil comme des formules de trait. Et, bien sûr, cette fois-ci, j'ai espéré que ce set-up donnerait un peu d'informations sur les zeroes de la fonction de Riemann. Mais, après un peu de travail, j'ai trouvé que, en fait, ce qui s'est passé, c'était comme en physique, c'est-à-dire que les zeroes, qui n'existent pas sur la ligne critique, manifestent eux-mêmes aussi dans un moyen positif par les résonances. Donc, gradually, I realized that there was not enough algebraic geometry in the set-up and that what was missing was, if you want, a kind of a very elementary, fundamental algebraic geometric structure that would be underlying the same non-commutative space. And so, with Konsani, we found, at some point, a few years ago, this was in 2009, that, in fact, it was possible to understand the counting function for this hypothetical curve, if you want, and that there was a lot of consistency going on. And that was actually involving the limit when Q goes to one of geometries of FQ and ideas, which have been developed since a very long time, in particular by Jacques Tietz. But, OK, this was still very, very far from anything concrete. And very recently, we found what I hope to be an answer to this conceptual question. And it turns out that the answer involves two fundamental notions, which are not completely obvious. The first notion is the notion of a topos, which is due to gothandik. And the second notion, which will play a crucial role, is tropical geometry or, if you want, characteristic one. OK, and they will be involved, in a way, which I will explain right away, but which is, as you will find out, I mean, if the answer was very complicated, it would be worth nothing. And so, I will explain the answer. And the answer is remarkable by its simplicity. So I will explain what is this arithmetic site. So it has, this arithmetic site is, as a space, it's a topos. So I remind to you that gothandik generalized topological spaces by the following idea, that by the idea that, in fact, you can encode the topological space almost entirely by the category of sheaves of sets on the topological space. OK, so in other words, when you have a topological space, of course, you can consider sheaves of sets, just sets, that's enough. And they form a category. And as gothandik noticed, the category of sheaves of sets, not only, of course, gives you back the topological space, but in the right manner. For instance, if you take the topological space, which is obtained from a variety by looking at the Zarisky topology, and not knowing about schemes, when you will compute the corresponding topos of sheaves of sets, and when you will look for the points of these topos, you will find the points of the scheme. Now, but gothandik understood that somehow, if you want, there were more categories of sheaves of sets than topological spaces. And among them, in particular, and this is evident when you read the SGA4, there are, if you want, toposes which are obtained as dual of small categories. So you take a small category and you take all possible functions, contravariant functions, from this small category to the category of sets. It turns out that the corresponding category has exactly the same properties as the category of sheaves of sets on the topological space. And this is what we shall do here. So here, the small category that we shall take is amazingly simple. It's just a category which has only one object. But the endomorphism of this object is the semi-group of non-zero integers, and they are multiplication. So this is a small category. And at the end of my talk, I, sorry. You wrote any process usually positive? Yeah, positive, positive. Yeah, sure, sure. When I, yeah, my integers are always positive, OK? So it's a very, sure, sure. OK, so it's positive, positive integers. So I mean, what is an object? I mean, this topos is the topos of functions from this category to sets. But of course, since it has only one object, it will give you just a set with an action of n cross. And at the end of the talk, I will explain the link with non-commutative geometry of this space. I mean, this space is, in fact, exactly the absolute point in non-commutative geometry. So it has a structure sheaf. And it is a structure sheaf which involves tropical geometry. So what is a structure sheaf? The structure sheaf has to be an object of the topos, because it has to be like a sheaf on topological space. And as an object of the topos, moreover, it has a structure. And that structure is a structure of a semi-ring of characteristic 1. What is it? It is all positive or zero integers, where you add infinity. So you adjoin infinity. And the operations are the following. The first operation is the infimum. So if you have two integers, one of them is a smaller. So this is what we replace if you want in this semi-ring, the sum. And the other operation is a plus. So it is a plus which plays a role of the product. So you should think of elements. Mentally, you should think of elements of this n-bar as being powers of a variable q. So you should think of them as being q to the n.  So you should think of them as q to the n. All right. So these are the two operations. These are the two operations. The first operation is infimum. And the second is the sum. And it's pretty obvious that when you act on this by multiplication, you preserve the two operations. It's compatible with the two operations. Because, of course, if I multiply two numbers, if I take the infimum, it's the same thing as taking the infimum of the multiplied and the same for the sum. So what do we have? We have a topos with a structure shift. And we define this arithmetic site as being this pair. So the topological space, which is this topos, and a structure shift, which is viewed as a summary ring in the topos. Do you like to add 0 to n bar? No, because 0 is coming from plus infinity. You see, when I say it's q to the n, q is smaller than 1. So infinity will give me 0. So it's non-infinity. Right. Yeah, yeah, OK. So no, the unit correspond to n equals 0. You see, q to the 0 is 1. No, you want to add q to 0 as well, you know? No, no, but when I add n, n, n I include 0. This is a French notation. You see. Yeah, n includes 0. OK. Right. Sure, sure, I know. It is multiplicative. No, but the semi-group operation is also non-infinitation. No, multiplication, of course. But it's positive numbers. Positive numbers. Strictly positive, OK? But the other n is not strictly positive. No, sure, sure, sure. No, that's not standard. Yeah, OK. But I mean, I like to confuse people because then, you know, they understand better, OK? So, OK. So let me write this, OK? So my topos is n cross, and this contains 1, 2, et cetera, et cetera, OK? And the structure shift is n bar, and this n bar contains 0, 1, 2, et cetera, and infinity, OK? And here we are. And they are both countable, OK? So that's the structure. This is the structure, OK? And you should think of n bar as being Q to the power n. OK, of course. All right. OK, so now what we are going to find, the first thing that one finds is that when you look at the points of this topos, you find a non-commutative space. This is the first surprise, because it doesn't use the structure shift. It's really only the topos. And there is a definition of a point of a topos, which is very simple. It's due to Grotendick. What is the point of a topos? It's a geometric morphism from the topos of sets. So the topos of sets is the simplest possible topos. You take all sets, OK? That's if you want, it's functors from a point to sets. But I mean, this is all sets, OK? So that's a topos. And a point in general, a definition of a point of a topos, it's a geometric morphism from the topos of sets to the topos that you consider. So here we consider geometric morphisms to this topos. And this data. What is geometric morphism? Well, a geometric morphism, you see, when you take two topological spaces, and if you take a continuous map from X to Y, it induces two maps on shifts. It induces a push forward and a pull back. And they are adjoint functors. And moreover, the interesting piece is a pull back. It's not a push forward, because a pull back has an additional property. Not only it's commuting with arbitrary inductive limits because it's an adjoint, but it also commutes with finite projective limits. And that's a crucial fact. That's an absolutely crucial fact, because it's what governs everything in shift theory. The fact that you commute with finite projective limits. OK, so that's what you have here. And so in particular, this is what? Inductive limit geometric morphism. So a geometric morphism between two topos, it's a pair of adjoint functors. A pair of adjoint functors. OK, which go in and such that the pull back functor is actually exact. It's a pair, not one functor, but a pair. No, it's a pair. But it turns out, by general theorems of category theory, that you only need to know the pull back part. It determines the push forward part. So here, if you want, you are giving your own interests indeed in this pull back part, which is left exact. And it doesn't take much work to determine this pull back part by the following. So you don't need to know the pull back part everywhere. It's enough to know this pull back part when you're restricted to the composition with a unedited embedding. Because there is a unedited embedding, as usual, of the original category in the topos. And when you compose with a unedited embedding, what you find is that you find something which is quite simple. Namely, you find really a functor from this small category n cross to sets. But it has to satisfy an additional property which is a flat, which is to be flat. So what does it mean to be flat? It turns out that the first thing that it means is that the corresponding set is non-empty. And the second, and two other things, I mean, the second thing is the following, is that if you take two elements in the set, they will have a common root. So if you want, they will be written as the application of this functor to the same element, but with, of course, different values of k and k prime. So this is the first property that you find. So and the second property that you find is that, in fact, the action of n cross on this set x has to be free in the sense that if the same element is sent by k and k prime to the same target, then k n has to be equal to k prime. So these are the two properties. This follows from the definition of flat functors. You can read it in SGA. You can also read it in a book which is much more accessible, which is a book by MacLennan Mordech on Topos Theory, which is more modern in a sense. But I mean, it's all in SGA. OK, so I mean, this is what you have to look for. You have to look for such flat functors. And what you find out after it's really an exercise, so what you find out is the following. You find out that this category of points, so the flat functors, if you want, of this Topos n cross, is, in fact, canonically equivalent to a category of totally ordered groups HH plus, which are groups of rank 1. So if you want, they are isomorphic to non trivial subgroups of Q, and they are ordered, so they are subgroups of Q, so they inherit the order of Q, of course. So they are ordered groups, totally ordered groups. And the morphisms between two such ordered groups are the injective morphisms of ordered groups. And how do you see that? How do you see that? You see that as follows. I told you just in the page before that if I have two elements, x and x prime, they can be written of the form p of k times z and p of k prime times z for the same z. So you define the addition to be simply p of k plus k prime times z. So now, you see, you add an action of the multiplicative group. Now you use the addition in the integers to define a sum. And what you will find out is that it is independent of the choice of the element z that you have here. The independence of the choice of z follows, because if you have another element, put another element below the two of them, and then it will work out. It will show you that you have the same sum. So the sum is well defined in the set x. And what you find after a while is that your set that you are considering is not, of course, the full group, but it's the strictly positive elements in the group H. So and also because of the fact that every finite set of elements can be written as the image of a single element, what you find is that this group is, in fact, an inductive limit of rank one groups, I mean of groups of the form z. So then you find easily that it's a subgroup of q and that it's ordered. It's ordered because you have the positive part, OK? So that's the first observation. When you say it's isomorphic, is the group to which it's isomorphic really have the scaling? In q, yes, up to scaling in q. And we are going to see that in the exam. So you could have an easier description, forget all this talk. It's all about the order, it's in q. Yeah, exactly. Exactly. That's perfectly correct, OK? And this is what I'm going to use now because indeed, now I have an Adelic description, which is very simple, of rank one subgroups of q, which is the following. It's easy to see that any rank one subgroup of, I mean any subgroup of q, non trivial, can be written in the following way. It's a set of elements of q such that aq belongs to ziat, where a is an Adel. OK, where a is an Adel. So you find an Adel. This Adel is not quite unique. It's unique only up to ziat cross. But it's easy to see that this is what you get, OK? And because of that property, that a subgroup can be associated to an element in this quotient of aqf mod ziat cross. And as you were saying, we are looking at them up to isomorphism in q. So in fact, we get a double coset space, which is the quotient of the Adels by q cross plus and by ziat star, OK? So I mean this was the first int. And these are finite Adels, of course, OK? These are finite Adels. OK, so this was the first int because this is an uncompetitive space. This is not an ordinary space. What do I mean by an uncompetitive space? I mean a space where if you drop the uncountable axiom of choice, you will not be able to construct a bijection with a continuum. So it's a set for which if you want the cardinality is actually larger than the cardinality of the continuum, the effective cardinality, OK? And of course this set, I knew very well. So in naive sense, we need certain actions as kind of transitive, OK? It's ergodic, ergodic, ergodic, ergodic, OK? I mean the fact if you want that the action of q plus cross, sorry? Classical to usual space, of course. But there are ways to use space. Sure, exactly. So this is not usual space, by any means, OK? So that was the first int. But of course, I mean when I am dealing with this, I am just dealing with the points, OK, of the... So as I said, you know, it follows from this fact, I mean it follows from this fact that any non traversal group can be written this way and that, you know, you have isomorphic points if and only if they have the same class in the quotient, OK? So the... But one remains a little bit unsatisfied with this because this is not the space which I had found for the zeta function. The space which I had found for the zeta function was not this one, it was a space without the finite adults, with all adults. Why it is very important to have all adults? Because then you have a flow which is the action of r plus star which will commute with everything. And you need a flow, of course, OK? So, I mean, at that point, it's a little bit... It's just saying, you know, there is a hint of non-commutativity, but I mean that's all you can say. OK, so the next point now is really to let into play a fundamental semi-field which is not B, but which is a semi-field which has been used very much in two different frameworks, I think. It has been used in the Russian school of decontisation and it has been used in potropical geometry. And this semi-field, let me denote it multiplicatively. I much prefer to denote it multiplicatively. So it's r plus max, what is r plus max? It is, if you want, zero infinity. It is zero infinity, I do not include infinity. OK, I use on zero infinity the usual multiplication. So it's the usual multiplication. OK, and I use for the addition the max. So now x sub y, OK, this is the max of x and y. OK, so let me convince you in a few words, summarizing a few of the discoveries of the people on that why this is a very interesting object, even though it looks very simple. The thing that convinced me that it is a very interesting object is the following, is that if you develop the theory where you don't think of real numbers and so on, but you replace by r plus max, then you can wonder what is Fourier transform. And when you do that, you find the Legendre transform. So that's really a beautiful thing, because normally the Legendre transform, we know it's very useful, but we don't know where to put it, in which draw to put it. And it's just Fourier transform in this setup. So moreover, moreover, this... It's actually discovered by the police ministry. It's much over, yeah, exactly. Yeah, sure, I'm sure, I'm sure that one. It's a climate, exactly. Sure, yeah. But in fact, what I would suggest actually to a student to do the following, I mean, this language of r plus max is the correct language for thermodynamics. And somebody should rewrite all of thermodynamics using this r plus max. Somebody's looked down for lecture, essentially, it's done, yeah. Yeah, okay, but sure, but I mean, somebody can do it with a lot of care. Fine, fine, this all goes in the right direction, okay. So not only... But of course you have one very important property, okay, which is that this r plus max contains what will play the role in this theory of the finite field of q for q equals one. And I mean, this cannot be simpler. It's just formed of zero and one with the rule that one plus one is one. And this is what we call characteristic one. Characteristic one is not one equals zero. I mean, that would be bad, okay. But it's one plus one equals one. Okay, so that's what you have here. That's what you have. And moreover, okay, it's an exercise also to show that this my field doesn't have any finite extensions. But in fact, so it has this infinite extension. And moreover, in that extension, which is r plus max, this is the beauty of the thing, you have a one parameter group of Frobenius automorphisms which are given by the map x goes to x to the power lambda. So if you look for the group of automorphisms of this my field, what you find is that for every lambda, every lambda you have this Frobenius automorphism. And moreover, they form all the automorphisms, okay. And their fixed point is equal to b, okay. So you see this situation is to be sort of as a parallel situation of what happens, of course infinite characteristic when you take the algebraic closure of fq, okay. So now the next step, the next step is to consider the points of our gadget on this r plus max, okay. On this r plus max, okay. And when you do that, actually, so this is what we found.  You see when you do scheme theory, you will find out easily that when you are taking the points of a scheme over a local ring, essentially what this is given by, it's given by a point of the scheme because after all the local scheme, if you want a scheme corresponding to the local ring, we'll have only one point essentially. So it will give you one point of the scheme but it will also give you a local morphism which goes backwards. So which goes from the stock at the point in the other way, okay. And so, but I think I didn't talk about the stocks. So let me tell you very briefly what are the stocks of the structure shift. This is very important for what I am going to say now because you see, the beauty of the theory of topos is that the stocks, so what is the point of a topos? A point of a topos is really the function that gives you the stock of a shift at that point. Okay, you should think of it this way. And the beauty of the function which gives you given a shift, the stock of the shift at that point is that this function commutes with inductive limits, arbitrary, but it commutes with finite projective limits. And the fact that it commutes with finite projective limits is crucial because it tells you that, for instance, if the shift as a structure of the wiring and so on, the stock will also inherit the same structure. Because what you can say, for instance, is that the structure is given by a map from the product to the thing and so on, that will continue to work for the stock, okay. So we know, a priori, that the stocks have to be some irings, okay. And we have to discover what they are. And we discover what they are by the following theorem. So the theorem, we know that the points, so the points of the topos are associated to subgroups, okay, to ordered groups, I mean, so abstractly to an ordered group. And so it turns out that when you compute the stock of the structure shift, what do you get? Well, you take this ordered group and you make it into a sum wiring. And how do you make it into a sum wiring? Well, exactly in the same way as what we were doing for the integers, but you change the sign. So you use the max now, not the mean, and you add not plus infinity, but minus infinity, because, okay, this will fit with standard notations. And so what you do is you take your ordered group, you add minus infinity, use the max, okay, and use the plus as an operation for the product. But of course you should think of it multiplicatively. So you, exactly as I was explaining for the case of N, okay. So what do we have? We have now a very good understanding of the stock of the shift. And we get, in that way, this is not difficult to see, you get sort of all possible, under quote algebraic extensions of what is called Zmax. So Zmax is a semi-field. It's very used in the tropical geometry. And in that way, you get all possible algebraic extensions, okay, and the maps, the relative maps are the maps of semi-fields. And of course, there are two things which are important to say at this point. There are two things which are very important to say. The first one is that you have a semi-ring, we add the semi-ring and bar. What is it? Well, it's just the stock of the shift over the simplest group which is Z, okay. So this is quite simple to see. But there is another thing which is quite important, you know, which is that at some point, we shall need to have what in geometry is the field of functions, you know, of the function field. And when you look at the various definitions of the function field, you find out that there is only one which works in this case. And this is the stock at the generic point. There is no other definition. And it's good that Grotendie was insisting on this property because here there is a generic point and the generic point corresponds to the subgroup which is Q itself. Because any other group will go inside. So this is the generic point. And so there will be a stock at the generic point, okay. So that's where we stand. And so now we can look at the, now because we have the stock and that the stock is a semi-ring, we can find out, we can compute what are the points over. And for simplicity, we prefer to take the points over the maximal compact subring of R plus max. But I mean it would be easy to know what it means if you don't take that. And what's the, I mean just, you have some quotes, why not? Sure. What's the semi-ring associated to the generic point, is it? No, it's Q, no no. It's Q, but you have to think of Q as being the exponent of a variable, okay. So it's like arbitrary roots of a variable, okay. Rational roots, okay. Okay. So what we find is the following. We find that when you compute the points defined over this R plus max, which is the maximal compact subring in fact, okay. Then of this tropical semi-field, then you find exactly that it is in bijection with a non-commutative space which I had considered, okay. Which is the coefficient of the adult, the full adult now divided by Q cross and divide by the adult cross. And the division by the adult cross when I had, we shall see what is its relevance later. I mean its relevance is to select among the various cell functions with grosses end-characters, the remand data function. This is why, this is exactly what one needs. Okay, and moreover when you look at the action of the Frobenius automorphisms, which are automorphisms of this R plus max, so they are exactly the way you do the computation when you do algebraic geometry. So when you look at the action of the Frobenius automorphisms on these points, they correspond exactly to the action of the del class group moduloziat cross. So it's just multiplicative numbers on this non-commutative space. And when you look more carefully, you find that this non-commutative space, so the space in question, is disjoint union of following two spaces. So there is a first, the one that was coming from the finite adults. And when you look, this corresponds to the points here, which are not only defined over R plus max or over the maximal compact subring, but which are in fact defined over B. So this is very natural, you see. You have, and they correspond to the fixed point of the Frobenius. And that's exactly what happens in the del class space. So they correspond exactly to the fixed point of the Frobenius. And then you have another piece. And that other piece corresponds to those points which are not defined over B. And they are in fact in bijection with rank 1 subgroups of R. So I mean it's a fact, which doesn't take much thinking, that if you look at the space of rank 1 subgroups of R, it's a non-commutative space. Okay, by rank 1 subgroup, I mean groups such that any two elements are proportional by an integer. Okay. And I mean it's pretty much like a space of leaves of affiliation. I mean, because you cannot, you cannot find a canonical element there. So you, and the weight, so the weight occurs if you want the weight relation between this and adults is very simple. It occurs by this simple formula where you have a simplification by q. Okay, so that's where we stand. And now what we want, of course, we want to relate this to the Riemann zeta function. And then, okay, this is a relation with previous work. And so in order to do that, because the counting is extremely delicate. So what we do is we now consider the space of points and we consider the scaling action, which was given by the action of the Frobenuson points. And now we consider complex valued function and we count the number of fixed points by taking traces of operators. Okay, so in other words, we consider this group acting on functions on the adult space. And what I had shown, okay, and which was then refined in particular by Ralf Mayer and so on. What I had shown was that, okay, one could understand the explicit formulas in terms of a trace formula. But the important part was the following. The important part was that, in fact, you can compute the trace of integral of these scaling operators and they are given by this expression in terms of adults and adults and so on. And in this expression, you can see immediately that you have exactly the terms that Vail introduced when he gave the conceptual understanding of the explicit formulas. What is the measurement about? You say, what is the measurement there? Which? We integrate the measurement on the right side. What is the measurement? Yeah, you mean? This is the U over U. This is the U over U. I am working with functions of U because the problem is depends on a real number. So I am iterating over, okay? I will explain more about this, okay? Okay, so in fact, we apply this formula and but we apply it to functions which are invariant under Z at star because to get the Riemann data functions, you have to take the trivial grossen character. So we take functions which are invariant under Z at star. We take that function and what it does, it gives us a distribution N of U on U belonging to one infinity which is counting the number of fixed points. Of course, in a very subtle way of this flow, of this Frobenius action on points, okay? And now, the first result that we have is that this counting distribution of the Frobenius action, in fact, it's given by the following formula.  This would correspond to projective space minus the derivative in the sense of distribution of the sum over the non-trivial zeros of data of the order of the zero multiplied by U to the rho plus one over rho plus one. And if I was careless, if I would differentiate each term, I would get exactly the formula that you have for finite fields, for function fields. Namely, I would get U to the power rho, okay? But I cannot do that. This sum exists as a function, okay? But it doesn't exist when you differentiate. It exists only as a distribution, okay? So that's what you find for this distribution, N of U. The second thing that you find is that it's positive. It's a positive distribution for U belonging to one infinity. There is a notion of what is a positive distribution. And the most surprising thing is that its value at one is minus infinity. You see, I have to explain why this was needed and why this is something which was looked totally impossible. Why was it needed? It was needed because it's known that the value at one of a counting function in the case of function fields is the other characteristic of the curve, say, okay, okay? But yeah, okay, imagine whatever curve this is. It has certainly infinite number of dimension for H1 and H0 and H2 are of dimension one. So, I mean, it's totally improbable that the other characteristic would be something like zero or something finite. It has to be minus infinity. And so, this is what happens. This is what happens when you compute this function and I can show you graphically what it does. So, graphically... So, again, what's the meaning of this function? What's the meaning of this function? It's counting this point, so what is N of U? Yeah, N of U is, as I said, obtained by the trace formula, okay? So it's a distributional trace of this operator, yeah, which is computed there. So, it's a distribution, okay? That's where the analysis comes in, of course, you know. It's not counting in a naive manner, of course. Yeah, but what is it counting in a naive manner? What? In a naive manner, what you would do is... Okay, this is, again, a subtle point. I mean, what you would do, you would take the points of this topos and so on, okay? And you would count how many fixed points you have for theta of U, okay? Exactly. But now, you see, there is a big, big difference with the case of function fields, which is, in the case of function fields, the Frobenius, there are intermediate subfields. And you can look at the points with values in the intermediate subfields. There is no intermediate subfield here, okay? And this we knew for a long time, okay? So, I mean, the count has to be global, okay? It cannot be by looking at the points of our intermediate subfields, they are known, okay? So, okay, so this is what you get. And the graphic of the primitive, okay? Because, as I said, you know, it's best to think of it as a derivative, this distribution. When you draw, when you plot a graphic of the primitive, what do you find? You find that you can approximate it by not taking the sum of all zeros of theta because this is impossible, but by taking the sum of finally many zeros of theta, and then you get this approximation, which is wiggling and so on. And what you find out is that, in fact, this function, j of U, it doesn't look on the graphic, but it shoots down to minus infinity and it has a finite value at one. So, when you take its derivative, the derivative is minus infinity at one, but it's positive everywhere because the function is increasing, okay? So, for U positive, it's positive. All right, so this is what we have, but now there is a question, which is the troublesome and which we have to answer, which is that, is it possible to associate to this counting distribution a surveil type theta function because this is what happens in the case of function fields. We really have a surveil formula. And for that, we start with the formula which had been given by Christophe Soulet. And what Christophe Soulet had done, he had considered geometry characteristic one and he had found that there was a way to define a limit when Q goes to one of the assay-vail formula. And his technique was the following. His technique was that you take the assay-vail counting formula, okay? So, this is a generating function N of Q to the R is supposed to be a number of points over the field corresponding to Q to the R. So, you take this function and Soulet had the idea of taking the limit when Q goes to one, okay? He has to renormalize this limit. He has to multiply it by Q minus one to the other characteristic. And then he takes this function but he couples the two variables so he put T equals Q to the minus S, okay? And he computed it for, for instance, projective space. For projective space, the counting function is N of X. If you take N-dimensional projective space, it will be one plus X plus X2 plus X to the N. So, it's easy to compute and you find the right answer, the expected answer, okay? Which is a rational function of S. Okay, so, this is what Soulet was proposing but you see immediately that we have a big problem with the formula of Soulet because our characteristic was minus infinity, okay? So, at first, when you try to apply this formula naively, you meet a wall, okay? And because N of one is minus infinity. And, however, we have solved this difficulty. We solved this difficulty in the following way. What we have found, you know, moreover, if you take the definition of Soulet, there is no good idea why you should have a limit to this formula. I mean, this formula could very well not have a limit. Now, what we found is that this formula is, in fact, the Riemann sums for an integral. And how does this go? Take this formula and imagine that you want to eliminate this Q minus one to the chi. How do you do it? You take the logarithm. The logarithm will put this term as an additive term. This term is independent of S. So, when you differentiate with respect to S, it will drop out. Okay, so what you have to do is to take d by dS of log of zeta N of S. If you do that, the log will counter the exponential and you will just get this sum here. Imagine that I take this sum and I replace T to the R by Q to the minus S. And I differentiate with respect to S. Then there will be an overall minus sign. And there will be, the R will drop out because it was in the power. And I claim I will get the Riemann sums for a Riemann integral. Okay, because what you get now, as I said, you know, you take the derivative with respect to S. The R will drop out. You will get a log Q. You will get a log Q. And in fact, what I claim is that these are exactly, of course, Riemann sums for the integral which is the integral from 1 to infinity of N of U times U to the minus S because of this times dU over U. It is a multiplicative armature because you have the log Q. Okay, so that's what you get. And then, because of that thing, now you can completely eliminate the problem that key is minus infinity. And you can say that if you have a counting distribution, you can associate to this counting distribution the asservé zeta function obtained as a limit when Q goes to 1, but this is better to take as a definition. So if you want, it's a distribution, so if there is no Riemann sum, it's something which is more subtle. But still, this integral will make sense and this will define the logarithmic derivative of the corresponding zeta function. So this is really the definition if you want of the asservé zeta function. And it combines the idea of Soulet with this fact that, sorry, that we have the Riemann sums, okay, for the, for the... Can you make a, you define the function itself? So this is up to a constant, yeah. No, I don't know, no, no, no. Yeah, yeah. Well, I didn't, I didn't really try. I didn't really try because, yeah, one would have to, I thought a little bit about it. It would be good to actually do it, you know, up to a scale, I mean, to find it up to a scale. It's like a finite value. Okay. Sorry. So for Colton again, and if you, because I didn't understand the definition, but some transform of the view is an ordinary function, it's a part of you, when you have a known function. Yes. So, no, but that's exactly what I'm going to say now. So what I'm going to, what I'm saying is that our end of view, it's exactly what you are asking. Our end of view, when you do this formula, what you find exactly is the logarithmic derivative of the complete Riemann zeta function, okay, with the other factor, with the factor. That's, that's exactly, okay. So we take it, we do it, okay. And so we can state that the RSEV zeta function of the arithmetic site is this complete Riemann zeta function where it has also the factor at infinity. And it's easy to see also that the arithmetic site is a complete, I mean, it's complete. Why? Because you can look at global sections of the structure shift, you find they are known. Okay, so it's not missing a point or anything like that. It's a complete, it's a complete thing. Okay, so now of course we are only at beginning and what we want are the Frobenius correspondances and we want a square of the arithmetic site, okay. And the square over what? Well, the square over this B, okay, over this spec B, if you want, which is playing the role of, okay, you know, a very, very, very basic, some people call it F1, but okay, I mean. So that's what we want, we want the Frobenius correspondances. And the main result that we found is that even though the arithmetic site is countable, you see, because the arithmetic site is countable, you could totally give up the idea that there would be a one-parameter group of automorphisms. It cannot be, things cannot move continuously. However, what we shall find is that if you think about correspondances, then it indeed has a one-parameter group of correspondances on the square, okay. And they are exactly the ones we want. So in order to do that, we have to find a square. And to find a square, we use the tropical geometry ideas. So we take the tensor product over B of this semi-ring and bar, okay. And to tell you what happens, you see, of course, you can always, as always, you can represent elements in the tensor square by tensor products of this type, because we are in characteristic one, there is no coefficient here. And what you find also very easily is that there are simplification rules. For instance, if I take, you know, if I take something like, sorry, if I take something like q2 times q3 plus q times q, well, I can cancel out this thing. And so it's equal to q times q, okay. So what happens, geometrically is the following. What happens, I will tell you the proposition later, but what happens geometrically is very clear, is that whenever you have a tensor, this is like q times q6, okay. So whenever you take a tensor, then all the elements which are in the yellow quadrant above it do not play any role. So in fact, the elements in the tensor square can be sort of as a yellow sets which are here. And these yellow sets, if you want the best way to think of them, is that they are hereditary subsets for the ordering of n cross n, which is the obvious ordering, namely that the ordering where a pair is larger than a pair if both elements are larger. So for every partially ordered set, you get a b module and just by this trivial operation of union, okay.  And then it turns out that you take this product ordering. And okay, then if you want, there is a statement which is not difficult to prove, which is that the tensor product is really the tensor product, namely it's uniquely characterized by the universal property that if you take any other b module and you take a bilinear map, then this bilinear map will factor through this gadget, yeah, through this tensor product. So there is a tensor product. There is a tensor product. And once you have this tensor product, you can begin, you can begin, but one will begin slowly, okay. We shall see eventually that the true tensor product will be related to Newton polygons. And so it will be a summary ring made of Newton polygons, okay. That's what we shall see. But this is just a preparation, okay. So this is just a preparation. And now that we have this, we of course, at first it's just a b module. Now we have to make it into a summary ring, but the summary ring structure is forced upon you because what you want is that the multiplication of these tensors gives you this property here. And what does it mean geometrically? It means that you take yellow sets like here and you add them, okay. So the sum will be the union. They will still be of the same form, but the product will be the addition of them, okay. And this will still be of the same form, of course, okay. So that's what you get. And moreover, you have a two parameter now, family of endomorphisms on this summary ring because you can rescale the power a and you can also rescale the power b, okay. And as we shall see, you know, there is a, we shall have to improve a little bit this. I will explain why. So this is the n-reduced square. So the n-reduced square is a topos which is a cross of n, okay. Two copies, I mean the product. And it has this as being the structure shift. But as we shall see very soon, this structure shift is a little bit like if you had a scheme and you would take an n-reduced scheme.  I will explain it in some detail, okay. So this is what is explained here. What happens if you want is that this summary ring is not what is called multiplicatively cancelative. It's like if it had zero divisor if you want. So because of that, okay, one will have to do a completely canonical operation on it which is to make it multiplicatively cancelative and we shall see how it works, okay. So this is not really an important point. Now what is really important now is to define the Frobenius correspondances and they will of course go down to the reduced thing automatically. And to understand the Frobenius correspondances we first understand the product. You see the product can be seen geometrically. Of course because n-bar was a summary ring there is a morphism of summary rings from n-bar times b-n-bar to n-bar. And what is it? Well, it corresponds to the product of course. When you do the product of this n-i and n-mi but because the sum n-bar was given by the if what you see is that this product is in fact given by q to the alpha where alpha is the infimum of n-i plus n-i. And this has a very clear geometric meaning. What it means is that you take a yellow set and you take the parallel to the second diagonal which touches the yellow set. You take the first one which does it. This is the product. So this is pretty obvious. But it will guide us to define the Frobenius correspondances. Why? Because now we have the product and we can compose the product with the two variable Frobenius on the both sides. So when we do that what do we get? You see, we take this FRNM which was raising qA and qB to some power. And what do we find? We find that in fact, the corresponding morphism can be written very simply. It can be written as the infimum not of n plus m but now of r n-i plus mi where i is a ratio between n and m. It only depends on this ratio. And the picture is like this now. Now we no longer take a line which was parallel to the second diagonal. We take a rational line, like the blue line here, and we push it up until it touches the yellow set. And that gives us the corresponding map. And what is important, it's exactly like when we do ordinary algebraic geometry, what is important in algebraic geometry when you define a correspondence is the ideal defining the sub-variative. Here it's a congruence relation defining the sub-variative. And the congruence relation is that two yellow sets are congruent if the blue line is the same. Of course, you can have many sets which have the same blue line. They will be congruent for that relation. So this is what is described here in great detail. And there is another thing which is quite amazing, which is that when you look at the image of this morphism here, you have to characterize the corresponding algebra. And then you find that it was already done by Frobenius. Then it's called Frobenius problem. It's a problem of knowing which are the numbers which are a combination with positive coefficients of two positive integers, or more in general. And there, the name of Frobenius is really important. So that's what you get. But now, you get immediately from this thing the extension to arbitrary positive real numbers. Because after all, this infimum of alpha ni plus mi can be extended to real numbers. And when you extend it to real numbers, what you find is that on finitely many elements, finitely many yellow sets, it doesn't see the difference between a real number and it's a diophantine approximations. But if you look at all possible yellow sets. I have a approximation by continuous change. Sorry, by the continuous fraction. By the continuous fraction. The best rational approximation, of course. You have to use that otherwise it doesn't work. But now, if you look at all possible yellow sets, of course, two different real numbers will give you different answers.  So there is this reduction. The reduction turns out to be passing to the convex cell. So there is a theorem which tells you that in general for summaries, you have a reduced summaries, which is obtained by reduction. And we computed it in our case. And what did we find? We find that, of course, if you take a convex set, it is multiplicatively cancellative. But what we found, which is more important, is that the homomorphism, which takes a convex cell, is exactly the reduction map, which is canonically defined. So this means what? And it has a universal property. So what it means is that the reduced square, which is really the important thing, what it is, it's a theory of Newton polygons. So the topos is the cross, is the square of this topos. It's a dual of n cross square. And the structure shift is a shift of summaries, which is given by Newton polygons. And what happens to see the small nuance with the yellow sets? The following is that when you take, for instance, this element, where I have this Q5 times Q3, this element will disappear when you take the convex cell, because it's inside the convex cell. But it was obvious already that when you take the Frobenius correspondence, I don't see the difference between a yellow set and its convex cell. Of course, because you were taking the half plane. OK, so that's what we have. And then the really important result are composition of correspondences. So you can define the composition of correspondences as we do normally by taking tensor products and all that. And then there is a CRM. And this is the non-trivial part of the second part. So the CRM tells you that if you take two Frobenius correspondences, and you compose them, if you take two real numbers whose product is not rational, then you get the corresponding Frobenius correspondence. So the composition of Psi lambda with Psi lambda prime is Psi lambda lambda prime. However, when the product is rational, it's more subtle. And one can understand intuitively very easily why it has to be like this. The point is that if you take an irrational number, this irrational number is the corresponding correspondence is a little bit stable. So it's one of the angles of the line or the line? So it is a line. What matters is where the line, the slope of the line is lambda. This is given. And then whenever you have a yellow set, you slide it up. So this is a map. It's a map of the myrings. All right. So what happens then is the following is that when the two numbers are irrational, so they have some robustness, but their product is rational, then you will not get lambda lambda prime. You have to make this lambda lambda prime, which is rational. You have to make it stable. And for that, you have to compose it with the tangential deformation of the identity, which is a certain correspondence, which is quite simple. It's just adding an epsilon. In the definition. And as a corollary of this, now we can compute the intersection number of two divisors of that form. So of course, we shall be interested in divisors which are integrals of Frobenius correspondences in the square. And because of the composition law, you can now define the intersection number of two such correspondences by computing their convolution, their composition, if you want, and then intersecting it with the diagonal. And the one uses the previous thing. And not also that it's not enough to have the Frobenius correspondences is when you do the veil stuff. You need also the correspondence, which is point cross the variety, and variety cross the point. So is this NU going to be the? Yeah, of course. The NU will exactly give you the same. And as I said, in the veil thing, you need to have point cross and we have it here. Why? Because look at the picture. Look at this picture. In this picture, I was taking slopes. But I can take the vertical slope and I can take the horizontal slope. It's like a slope of the Riemann vertices for curves. For curves. Of course, I mean, I don't want to say it. Because then, if you say this, you are out of the game. But of course, what we have been missing for all this time was the geometric setup. And of course, what we are missing still, and this is a big, big, big missing thing, is the Riemann-Roch formula. So as you know well, as you know very well, I mean, Mathuc Tate, and then Grotendick wrote notes in which, in fact, Mathuc wrote in his memoir that in this paper, they missed a boat. Grotendick got it. So as you know, if you add the Riemann-Roch, then OK. But I think it's a very good motivation to try to develop a Riemann-Roch formula of suitable type in this framework. And I want to finish with one thing. I want to finish with the following fact. My thesis teacher, Jacques Dix-Mier, studied my Contrandu note with Katia Konsani very much in detail. We celebrated his 90th birthday on the 26th of May. And last Saturday, he gave me a telephone call in which he pointed out to me a paper which I had written in 1967 that should be related to our Contrandu note. So I looked at this paper. And his paper actually gave, very quickly, the meaning of this story in non-commutative geometry. So as being, you see, with Katia, we had written papers in which we were using a word which many people had used. We called it the absolute point, what we are looking. Because of the sort of motives. But we had no idea. Now let me explain to you why non-commutative geometry. This is the right name and why it gives you this. And the reason is the following. You see, in non-commutative geometry, when you talk about a topological space, it's encoded by a cistargebras. And the idea of Gelfand, many people, you know, is that you will extend ordinary topological space. It's not by topos, but by taking cistargebras. Now what happens is that when you work with fallations, you find out that you shouldn't be that strict. Namely, two different cistargebras can correspond to the same space if they are moritaicuvalent. So what really matters is the moritaicuvalent class of an algebra. And then what you find, these are theorems of people working in cistargebras, is that in each moritaicuvalent class of cistargebras, there is a unique representative which is stable. And what does it mean to be stable? It means that you will test or the algebra by the algebra of compact operators in Hilbert space. There is only one such. You take the separable Hilbert space, OK? You get the same cistargebras, all right? You have to know that there is only one Hilbert space, OK? And only one algebra, k. So the single point now, you see, it's not right to say a topogical space is represented by c of x. No. A topogical space is represented by c of x times k, OK? So in particular, a single point is represented by this algebra of compact operators, by k, OK? All right? Because it's up to moritaicuvalent. And so up to moritaicuvalent, the right algebra representing a point is k, OK? So the single point is represented by this star algebra k. And then, what do you find? That's an easy exercise, is that when you look at the semi-group of endomorphisms of this cistargebras, modulo-inner automorphisms, which you always do in all commutative spaces, you find exactly the semi-group n cross. And how does it act? It acts on the compact operators by Frobenius type, endomorphisms, which are essentially doing amplification. Namely, you have an operator, and you sort of repeat it, you put it inside of itself k times. Now, you can organize this in such a way that it's an action of n cross on the compact operators by endomorphisms. It's compatible with a product, that's all the properties. What does it mean? It means that in all commutative geometry, we have Frobenius endomorphisms for any algebra because they are tensile by k, OK? And we have the structure of the point. And so as I said, you have a natural action. And what is the result of Dix-Mir, 1967, which put me into this? How can it be translated in our terms? You see, because now k, now k becomes an involuted algebra in this topos, because it has this action of n cross, OK? I am done, OK? So now, because it defines a shift. So we can take the stock of the shift at any point of the topos, OK? What we get is not a cistargebras because it's not complete, it's an active limit of cistargebras. We can complete it. We get what is called an infinite separable matroid cistargebras. And by the theorem of Dix-Mir, what you find is that all the infinite separable matroid cistargebras appear in that way and once and only once. And Dix-Mir had noticed, reading our note, that the classification was the same, OK? So I think this is a good way to end. Of course, we didn't discover this topos not at all this way. We discovered it by the link between the arithmetic site and the epicyclic site, which has to do with cyclicomology and so on. So, I mean, you know, the route was totally different. But it's very satisfactory that, you know, we seem to be closing really on what is the point in the absolute points, somehow. OK, thank you. Matroids are not matroids. No, they are not the matroids of Gelfand, yes. They are cistargebras, yes. So what happened to the earlier formula? So you see, it looks like a thesis where this kind of attitude, right? You have attitude, so to speak, right? Presentation des défunctions, c'est pour l'editif. Et qu'est-ce que c'est pour l'editif, si vous vous interprétiez? Oh, c'est l'earlier formula, parce que quand vous faites le logarithmique dérivé, bien sûr, vous avez un summe, vous savez, c'est ce qu'il fait, bien sûr. C'est pourquoi vous avez un summe de places. Ma question, si vous essayez d'appliquer à toutes les défunctions, c'est de la façon dont vous ressentez, par exemple, oui. Ah, c'est ce que je n'ai pas essayé. C'est ce que je n'ai pas essayé. Vous voyez, pour les autres fields, les gens ont généralisé la classe adulte et tout ça, mais je ne suis pas sûr que c'est de la même manière. Parce que ce qui était important dans ce que j'ai expliqué, c'est que nous avions de l'adélique. Nous avions vraiment de l'algebraic géométrique. Et de aller au l'algebraic géométrique pour les autres fields, l'un de nous devra être extrêmement careful. Extremement, vous savez, il ne sera pas juste, je veux dire, qu'on pourrait, vous savez, commencer par Adèle et tout ça, mais je n'aime pas ça. Je veux dire, les pièces que nous avons prises, maintenant, c'est d'être dans l'algebraic géométrique. C'est l'algebraic géométrique, qui est spécial, c'est un caractéristique. Et ça, c'est un travail pour le développer, mais c'est là. C'est vraiment l'algebraic géométrique et c'est l'algebraic géométrique. Et, vous savez, pour moi, c'est une immense joie pour le fait que l'algebraic géométrique a inventé ce algebraic géométrique. Je trouve cette idée marvellante et je trouve que, ok, ça a été utilisé dans les mathématiques, mais sa puissance est immense. Juste pour vous dire, c'est l'algebraic géométrique et la puissance mathématique. Parce que, c'est une question d'expression. Exactement. Oui, oui, oui. Donc, nous allons prêter que nous pouvons marcher jusqu'à ce qu'il y ait, vous savez.


Patriot Prime Reviews is a channel for adult collectors and may not be suitable for children under 13 years of age viewer discretion is advised Hey, what's going on guys Patriot Prime here once again with another Transformers review, but before I get started I need you guys to do me a favor if you're watching this video right here and are not a subscriber of Patriot Prime Reviews Please hit that subscribe button right now. It won't cost you a thing Will help me and my channel out tremendously. Also, make sure and visit my sponsor toyhacks.com Toyhacks provides reproduction and upgrade decal sets for transformer toys from generation one to the latest modern figures Make sure and stop by the toyhacks armory where they can equip your figures with new weaponry in multiple colors and toy stages provides backdrops for figure displays and photography Each purchase with toyhacks earns you RoboSense that can be used for future purchases. You can check your balance anytime in your cart Toyhacks is a company run by collectors for collectors So check out toyhacks.com and make your collection stand out from the rest and tell them Patriot Prime sent you now on to the review The featured button this video is Transformers generation selects Sandstorm this is one bot I have been anxiously awaiting he kept getting delayed over and over again But I finally have him now in my collection now taking a quick look at the box This is your normal generation selects box plain brown cardboard with a spray painted Autobot logo here transformers war for Cybertron and That's pretty much it and then you've got a decal saying what's inside and inside the box We have sandstorm in all of his glory and I have been looking forward to this guy because I am very familiar With his color scheme So now without further ado, let's get sandstorm opened up out of this packaging and check him out and welcome to Patriot Prime Reviews Now once you get sandstorm all opened up and out of his packaging You'll see he does come with a sheet of instructions that as usual are very well illustrated and very easy to follow Though sandstorm here is the seventh use of the Earthrise seeker mold So you should know how to transform this guy by now. He also comes with his two null rays That are the exact same as all the other seekers and we'll just plug those in to his shoulders and And now we have sandstorm all armed and ready for battle now sandstorm here is a repaint of Earthrise ramjet in a desert color scheme and what's really cool in jet mode He has a desert camo color scheme and I really like that as I said in the intro I'm very familiar with that look now taking a closer look at sandstorm I really like his paint applications This color scheme really works for the guy and totally makes him look like a different figure Of course, he is a different character.  I do like the paint applications for the eyes I mean that is spot-on. He's got black here for the chest thrusters a Brown waist brown biceps down here on the knees more white and red black down here on the shins and the toes There's also some brown back there for the heels He also has this translucent orangish yellow Cockpit that I really like I'm not a fan of the translucent plastic, but sometimes it really really works now Sandstorm here has the exact same articulation as the other seekers the heads on a ball joint Can you look up look down do a complete 360 the arms can do a complete 360 they can go up They can go down there is a bicep bend and a bicep rotation though mine is really really tight There is no wrist rotation due to transformation the same with the waist So it's kind of sad that these guys lack waist articulation But oh well legs can go in and out They can go up and down though They do hit right here on the hip skirt unless you bring that up. There's also a knee bend and Hip rotation no knee rotation and there is an ankle tilt man The joints on mine are tight and I absolutely love it I've had so many loose figures lately. So there we have sandstorm in robot mode now Let's check him out in jet mode Jet mode is where sandstorm shines and is actually the reason I picked this figure up I absolutely love the looks of this jet. He's got the flying tagger mouth right there on both sides of the nose cone Which looks spectacular not to mention He's got the black nose cone that the seekers need to have the orange cockpit looks great, and he's got this awesome Desert camouflage all over the jet. I wish the desert camouflage pattern carried over to Robot mode after seeing jet mode robot mode. It looks a little plain I love that. He's got the brown decepticon logos on each wing He's got some light tan paint applications there on the tail fins I mean overall this jet mode looks Fantastic and I couldn't be happier. I love this. So now for a quick comparison here is generation select sandstorm With earth rise ramjet and man what a difference a paint job makes Plus, I love the fact. He's got that big black nose cone that these guys really really needed So there you go guys one final look at generation select sandstorm in jet mode And now for some quick size comparisons here is transformers generation selects earth rise sandstorm With generation one sandstorm Generations sandstorm, and I think this is the first time I've moved my figure from that shelf since I bought him And earth rise ramjet now take note. My ramjet here has the non f weapons Which really add more to his g1 look compared to the original seeker weapons that he came with Transformers generation selects earth rise sandstorm is a great addition to your seeker armada My only complaint with this figure is I wish he had the desert camo look in robot mode I would love to have had that there on his chest. Maybe his arms But it is what it is. He's still a great figure and comes highly recommended. So there you go guys transformers generation selects sandstorm So does a transformers generation selects earth rise sandstorm belong in your collection This is one for the seeker collectors. This is not a mainline character Which is great for a generation selects toy.  You're like I said, if you're a seeker fan, you're not going to be disappointed Now guys, if you enjoyed the video, don't forget to like don't forget to subscribe And don't forget to hit that bell icon to get notified when I upload new reviews Also, if you're in any position to help out the channel I offer channel memberships here on youtube and I also have a patreon page Once again guys, this is patreon signing out


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Hey everybody, how you doing? This is Gus G. I'm here at NAMM 2014 over at KORG USA booth supporting Blackstar amps. Well, I just finished my first solo record. It's called I'm the Fire coming out March 18th in the US. It's got a lot of guests on it. So yeah, looking forward to hitting the road with that later this spring. It's a combination of a lot of different vocalists, different bass players, drummers. I've got Michael Starr on it, from Steel Panther, Matt Slavent from Campbell Mass. Jeff Scott Soto is singing on it. I did a track with DeVar the day, Billy Sheehan's playing on it, David Ellison. A lot of good people, a lot of great musicians and some of my heroes. You know, I'm just kick-starting the solo career thing so I didn't know really what to expect or how this would go down. But I had a great band with me, Jeff Friedel and Marty O'Brien and Matt on vocals as well. It was a great night. It was a great party. I'm glad that Blackstar gave me the opportunity to put the showcase here in Hollywood for a lot of friends and press people and industry people. I think it was a fun night. Actually, I'm starting touring in March in my home country back in Greece. I'm doing a co-headline tour with only John Roth. And then moving on, possibly, we're putting schedule together. Either we're going to start in Europe or the US, I don't know yet, but all the dates are going to be announced very soon. Well, the website is GusGeeOfficial.com, same thing on Twitter, GusGeeOfficial or Instagram. And on Facebook as well, which is the opposite, official GusGee. GusGee and I play Blackstar Amps.


TIGER, Financial News Network T.F.N.N. Headline, News Update Hi everyone, Miles the Chapman for this two o'clock TIGER Financial News Network Market Update. That was a 413. Yesterday it was down at 300 points at one moment, it closed down to 200 and look at this nice big green candle. Most important thing is that the 9 is still over the 14 period moving average in the daily chart. That's what I've been looking at as residual strength. This is actually a little bit more than residual strength, but we need to see how it closes and what happens tomorrow. Key support now is at $33,530, I'd say, just off the cuff. Looking at the S&P, S&P has also had a very nice candle today. It's trading at this point. It's trading up 1.57 percent. That was only at 1.2 percent. S&P is up 63, at 41.19. And both in this particular instance, the 9 is still positive. It's held the inside track support level. What's really important about this is that it's helping the weekly chart. If you just go sideways for a while, it's going to garner enough strength to move that weekly chart into the 42.30s. But at this particular point, we had 41.20, it's a bit of a struggle to see if we can take out the 41.69 high that was made just about a week and a half ago. What we're looking at in the QQQ, that's different. Basically, when sideways from the 321.63 high, double topped the 321.42. About eight days later, pulls back sharp. It looked ugly yesterday and the day before. All of a sudden, this is fantastic with magic and other stocks like what have we got there? We've got a Microsoft as well doing very nice. And that's helping the weekly chart. So key support is now the 315 area. If at any point it starts to trade in the 322s next week, that's going to be a big deal. That's going to be very positive. Looking at the high WM, it's being very weak. It's coming back a little bit here. It's at 173. Gold is really the issue. Gold is trading up about unchanged at 1997. Gold, if it breaks above 222, that's going to be good. But if it slides under 1973, that's so good. So have a great day. Stay tuned for great programming. Check out my opening call newsletter. And I'll be doing a webinar on Wednesday the 3rd. Check it out on the front page of TFNM. Have a wonderful day.


USS Ross, one of three Arleigh Burke-class guided missile destroyers for deployed to Naval Station Rhoda, has returned for scheduled maintenance as part of a four-month patrol. As the ship pulled into the familiar location, some very familiar faces were anxiously waiting. We're very excited. The daddy's coming home. It's been nuts because we PCS from Washington, D.C. ourselves so this has been lots and lots of work so we're very happy to have him back. USS Ross, USS Porter and USS Donald Cook will continue to conduct patrols to support a strong U.S. forward presence. From Naval Station Rhoda, I'm Corporal Tony Nardiolo.