Okay, thank you for your introduction. Thank you for your introduction. This one is more fast, LEGO-based, skill-complete, and critical, because it actually has a lot of hatches. We've done with a little bit, but I think it's a good idea. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Is there a way because you cannot assume what a level 3 can or cannot do? So much reason for our technology to use this formula model. In a formula model, that is a kind of an engine. You want this, no exception under this behavior. And this is the model, this word comes up. The data that we're not going to go against to enforce only behavior is still kind of a choose. say you have a genie, a lot of copies of gulliverties, then you should stop them to check and then we can do the right thing first. If something goes wrong, it doesn't really matter if there is any error in the genie check piece, the right things that you can do are possibly all correct. And the most advanced country's idea is the copy batch card. In a single institution's engine, the institution of a competition is viewed as traversing all the gap gates, all gates in a topological order. And in the batch schedules model, the gulliver first generates a lot of gates for the condition. Then the two parties jointly use the quantitative protocol to decide which of the gates should be checked. Then they also jointly use the quantitative to pass the outside gates into these buckets, and then do some wide-solid salary technique to combine all this stuff together, then do the evaluation. There are lots of work, employees' idea to do the single execution combination. There are also other works that send this idea to execute a single function many times in parallel. And many of our work, all these work, is we develop a method to stop in one of the exorbitant hatches, and we can achieve to one single good gate in every bucket can ensure security. There are other works required to correct this majority. And it also improves the benefit of the second rate from one loss to one total to half, and we can improve this rate to one, some different setting. And the last way we have proved that for a database protocol, no matter how the circuit is, we need at least two gates per bucket, or two gates per two gates. This is a little bit of a limit. Before that, people always think that we have an over and over and advantage. In my thought, I will only focus on the first video. We had read my paper to see the multi-solid, but that will do with us. Let's first talk about wide-solidary. At some point of the protocol, the inventor, I've been writing some capital gates in a bucket. He will get a wide label, WAS, where S is the semantic value of this wide label. And he has to transmit this X to the other label of the whole bucket. How can he do this? If everyone is honest and we have set up such that all the wide labels are at the XOR of their label and the public, not public, but global data, then we can cover who knows everything to see the differences of their labels to the better, then the better can complete the correct of the wide label and not the wide of the bucket. But the things that we are facing are active, but there was three. The challenge is, what can we ensure the quraniness of this difference? If we have an XOR homomorphic half-gain, we have such a solution, we can hash the Z-Label on YA and YB as their label. And as a galveston, this hash of the wide labels chooses the better. Because these hashes are homomorphic, the better can easily complete the hash of the difference and think about trying some difference. Does this solve the problem? Actually no. Because the better can now verify the wide label of the X against the hash of the Z-Label, then he can determine whether X is there or not. So the government knows about this. So our second attempt is to introduce a communication field. We don't hash the Z-Lables. Instead we hash a random label, the P, A, and Y, A, and P, B, and Y, B. And with all the hashes, the random these P, A, and P, B. And we also have two hashes of global data. In this case, we let the government to send all these hashes to the developer before the soldering sprays. Then if two hashes are to be soldered, we let the developer to send the difference of this from here. And the developer can verify the X against the hashes, know that the hash is still homomorphic. Then we can let the developer to send the differences of the washes. And we even can also verify the claiminess. And this technique can also be used to soldering washes with different data. Previously I talked about the washes of the same data. Here we're talking about different data. We have data 1, data A, and data Q, and data B. Now now we don't have a global data. Instead, we have a data hash and data Q hash. We let the scalpers send the hashes to the developer and have two differences to finish the soldering. The eventually we can verify the claiminess of this difference and obtain a use proper difference to obtain the white label on the white page. And now you may wonder why we want to take all this equity to soldering washes with different data. And the main benefit is that this allow advocate to use a local version of different data instead of a global data. This can allow that we fully open a data by revealing his local data. And this data has a whole variable. And this leads to a check in with data. And this is very important for you to kind of choose larger components, which might require you to fully open a common to check. Now, as all these results are based on such an idea, such a morphcache, we have to look at this interactive hash or hashes to realize this idea. I also like to look at the idea of a hash. In IHBergo, we have a standard and a receiver and a functionality. The standard by the message to the idea of functionality, that the parameters in the receiver are hash of the message. And this message can be derived from this message from some public, not publicly known algorithm, as well as the level of the hash is shorter than the level of the message itself. And this has to have some properties. First, it has to be exon-morphic. It's to say the exon of the hashes is the hash of the exon, the hashes of the community. Second, it has to be binding or it's not hasher at all. Then it's to say two different messages that are supposed to have different packages with a very high probability. And this has to be hiding. Let me say it's a hash is still has to be shorter than the message. Such that the receiver cannot learn the entire message. The receiver may list some information for another entire message. And now let's come to a realization. In our protocol, the message is a simple vector of this L. And we have to fit this vector into a new domain coding. You send this to M. Let's fit this encoding into an OT protocol. You still may not only watch some of this vector, but it's an OT protocol. The watch symbol covers up this hash of the message. And let's look at the scripted intuition of this design. If there's a difference in the original message and two different messages, then it might be some, there'll be a lot of difference in the encoding. This is due to the property of the original encoding. Then with very high probability, the receiver will see some of these differences. That's due to that the different message has different hashes. And if the receiver doesn't watch these things, we have such a different grid. But the receiver cannot watch too much, too many implications. Otherwise, this is a technical and linear original encoding. If we watch too much, it will learn the entire message. And even if we don't want to watch too much, we can still set up the parameters just that the binary property can ensure that it will be very small to know. Now, you may wonder that we have to use open up the OT and we might be very sensitive to the message. But that's not actually the case. There is only one message for calling a number of, calling a number of messages. Actually, without loss of generation, let's look at how we hash random messages. The sender first pick a number of Cs and Cs, then these PRCs use 10 symbols, TKC, YT, N, then he use the first L symbol a vector fitting to a systematic written encoding to extend to a length of N. Then he takes the first L symbols as a message, it's a random message. Then he places these Cs into our OT and the receiver watches some of these things. Just like the standard side, the receiver also is PRG with 10 of these Cs, two symbols with a length and number of symbols. To obtain, to obtain a hash, the receiver first preserve the symbols at the first L position, then the sender computes an extra correction between the last L symbols from random people on the PRG and the last L symbols from the encoding, and he has an extra correction that's placed in this extra correction to the receiver. The receiver then applies this correction to the last L symbols from PRG, then he gets the whole hash of the message. This double of the OT, where only is once, the whole time, and the rest of the parts will only send an extra hash for communication for a random message. Like the main protocol security, in our protocol as I've mentioned, we only have to guarantee one correct gate per bucket to ensure security, not like other protocols indeed are majority correctness. In our protocol, a bucket is so called good as long as it's contained at least one correctly generated capital page. Before we proceed, I think, just the concept of better labels. In Showa, there are better labels. The government knows the labels, the government knows which label could be better on the event, because the better doesn't know. So how can the inventor know whether a label has an online place, and a divination, and if they see a label in the bag, because there are hatches on labels, on ones. It can come to the hash of their label, and a white label from the hash on ones, so he doesn't know which has a zero or one, but he can get a fixed set. Then a white label is called value, if the directory can match the hash. Otherwise, the value is called invalid, because if a white label cannot match any of the hatches, it must be a correct label, and come from an incorrect date, and this white label will be discanted for Showa. Now let's see why we can get this duty, if there is only one gate to migrate. Because we have already got all the garbage labels, then we must fix it. We'll show you one of two cases. The first case is that we've got a lot of white labels, then they are the same. Most of them will already have a correct label, a correct date. We have a correct label, and all the labels are the same, so all the labels are correct, then we are sure, okay? So I don't know the case, we don't have one single value, we have two different values, they have a set of white labels, and some of them are this, and some of them are that. Well, because all the garbage labels are discanted, this label remains either their label or one label, so they are excellent data. Now the question is that can we extract some cheaters' input from this data? If we have two, then, okay, I have my input, I have everything, chase a type of cheater, I have everything. So the question is, can we do this? Yes. Those cheaters' input, we have WAX, that means input, and we have patches, and the only thing we need to know is that you guys need to know straight X is a combination grid. So if we can enforce the cheater to generate all the combination bits with this data, then if the divider can get the data, it can get the combination bits, even if the combination bits get the X, then we are good. Okay, how can we do this? We do this from a dynamic proof, we get the governor who uses data to generate the states he or be used in iHash. What kind of this? From a dynamic proof from iHash and then having it states, and then when that doesn't happen, also hash it to states, and send them to the divider or the receiver of iHash. Then the receiver also obtains the states, if you say hash of the state from the sender directly, then he obtains states from this from the out of the OT in iHash, then the states the generator has, but we use the generator with the combination bits, PAPB, blah blah blah, then the states the receiver has, we use the generator with the hash of the combination bits, PAPB, blah blah. Then the generator needs to prove that, okay, iHash's new hashes, these are hashes, I really come from states, and the states really come from a data, but if the receiver helps you verify that the hashes, the sender needs matches with the states that are obtained from the from the iHash, that the, oh we got the OT from iHash, even all the way from states, they really believe that we can see that, okay, the generator really uses data between the states, between the the combination bits, and we can see that if the receiver obtains this data, then the receiver will be able to spread out the combination bits, and with the right label, the generator will be able to spread all the inputs of the cheater, and all these proceed, they are expensive, really expensive, but they only take one single student, no matter how the, how the size of the competition, they have one single student during the setup setting, during the single, during the setup phase, and there's another challenge about in practice, because we all know that 8.9 is a very powerful method and it runs 10 times faster than 1096, but it has an elimination, it only works on 192 bits inputs, however, if we want to achieve 192 bits security, our one label that reminds us a vector of symbols, it's less than 432 bits, so we cannot directly fit them into the ASMI, our solution is to compress it into a vector, and today it fits stronger labels than fit into ASI, and how can we compress them? For a one label of for L symbols, it has omega symbols of the coding, of the public with the name coding, I would leave in the IH, though we don't know, but we don't know where our disease is, and there are only L and omega symbols on the main, then we have to use our randomly, random 4 grand matrix T to multiply with this vector, the negative result, the result is kappa zero, and it lengths to 180 bits, surprisingly, with our answers, the entropy in the original method is 180 bits, but the entropy, in which the entropy in the result of this multiplication is smaller than 127.99, you can end up 5 or 10 bytes here, it says as you have lost almost nothing from this compress, and this idea is very similar to the leftover hash lemma, but if you apply this leftover hash lemma directly in the backbox way, then the result is presently, is presently bad, we have we have sort of like 80 more bits than Japanese lemma, and this is our result of our work in some state applications, basically this WMK protocol, which runs the CWF, is slightly faster than us, in most of the settings, but has its own elimination, why is that this WMK protocol is very kind of slow, maybe 100 times slower than our work when accessing the input wires, the other elimination is that this WMK protocol cannot handle reactive competition, such as O-DRAP or PORGRAP, which I opened the last day, PORGRAP something like that, and our protocol is always almost faster than the best previous decos that protocol settings on DES, I think, when, this DES is a very short input with very relatively huge competition, which we cannot attribute to our advancing input wire processing, besides our value is slightly larger, and this becomes a bottleneck, and for some special applications, such as AES, there's another alternative we can do in decos that protocol, wire soldering is much, much more intensive than that one goes, and if we can use large components, for example, addition and mass, or supplies, as the country's academic unit, then we can skip the soldering for the other internal wires, which can be very frequency, and this typically needs to be fully open of a couple components, and our technique of soldering different wires with different dilemmas can standardize all its needs, and so our protocol is already quite efficient if we will survive as a country to pack an academic unit instead of any case, we still get a double ohang speed up 100% and save half of the bandwidth, and our conditioning is available until half, and I have to mention that we have much more scalable ohang, extreme scalable go-up for work based on this work, thank you, I'm really getting questions, thank you. When you have these different deltas, are you losing the bricks or property, or through wire soldering, can you have something like lexor? Yeah, we can still have three extra, basically, I can do, say we have, if you have different layers, you will have all the wires on the bucket, so you still have three extra, because extra is down on battery level, but I think it's on data level, so you still have the bricks or what's the extra? Extra cost, basically, you have extra extra information for wire up. Well, question? So what's the security properties of your hash functions? This basically is a hash function, it only works on two parties, and you let me, I'm also going to be interactive, and it's not like normal hash, it doesn't have a hiding, it doesn't hide all of these in your message, it only hides a random position, so it only works on written message, but you use extension with your hash with just... No, I don't, we don't need a shashash, I'm very busy, we just use that. Random one, it stands for hiding, don't have a computational hiding. No question? No question, let's thank the speaker again. This is the end of this session, thanks all the speakers for the last presentation, and thanks all of you for being here today. Enjoy your coffee break.


So welcome to Paris Piquin Tokyo Seminar. So it's my great pleasure to introduce the speaker, the first speaker on this Zoom session today. So the speaker is Artur Sejou Lubla, also he will tell us the prismatic during the series. So please start. Okay, well, hello everyone. And well, thank you very much to the organizers for giving me the opportunity to speak and for setting up everything in this very particular context. So my talk of today is about prismatic to donate theory. And so everything I will discuss is a joint work with UNS Anschutz. And for all this talk, I will fix a prime number P once and for all. So the goal of our work was to prove some classification results for feasible groups of various kinds of rings. And the main tool we used to do this is a recent theory of prisms on prismatic community, which has been developed by Bat and Sholzu. So the plan of my talk will be the following. I will first spend some time, maybe 20 minutes, half an hour to discuss a little bit prisms on prismatic community. So recall the basic definitions and constructions. And then I will start speaking about the joint work with Uranus. So first of all, I need to tell you about which kind of rings we want to classify feasible groups. And these strings are called quasi-syntomic rings. So this is what I will do next. I will explain the definition of these strings. Then I need to tell you by which kind of objects we classify feasible groups over such rings and recall them fit our prismatic Jordanic crystals. So I will explain the definition and some basic properties of the subjects. And then finally, I will explain our main results and say a few words about the proofs and some corollaries of these results. Okay, so let me start with prisms and prismatic community. And I should make it clear like everything in this part is due to bat and shoulder. So the theory of prismatic community relies on two important basic definitions, the notion of a delta ring and the notion of a prison. So let me start with delta rings. So I will assume always that my rings lives over Z localized away from P. And then the delta ring will be a commutative ring A together with a map of sets, just of sets, delta, which goes from A to A and which has the following properties. So first of all, it maps zero and one to zero. And then you want to prescribe how delta behaves with respect to multiplication and addition. And this is given by these two formulas which may look a bit strange the first time you see them. So delta of a product, if X, Y for all X, Y in A is X to the P delta Y plus Y to the P delta X plus P times the product of delta X by delta Y. And then you have a similar formula for the addition. So delta of the sum X plus Y is just delta of X plus delta of Y and then you need to correct it by this term here. And so observe that if you use a binomial formula to expand this X plus Y to the P, then all the terms except X, P and Y to the P which are cancelled here are actually divisible by P. So this expression makes sense in any ring. You don't need to assume that the ring is P or should free or anything like this. Okay, so this is what the delta ring is. So a ring A together with a map of sets satisfying these three properties. So if one has a delta ring A delta, so you can do the following operation. So you can define a map phi which goes from A to A and which sends X to the P plus P times delta X. And then you check an exercise that the identity is satisfied by delta which I had before over there. Actually exactly what you need to check that this map phi, just a map of sets from A to A is actually a ring morphism. And moreover, just by definition, it's a lift of four binos.  X goes to X to the P. And conversely, assume you start with a ring, commutative ring A together with a ring morphism phi which lifts for bnus modulo P. Then if you assume that the ring you have is P torsion free, you can actually just like divide phi X minus X to the P by P and define this way delta structure on your ring. So in other words, in first approximation, you can sing to delta rings as rings together with a lift with a ring morphism which lifts for bnus, right? But the two notions are not exactly the same thing when you have P torsion in your ring. Okay, another point of view which I just want to mention about delta structures is the following. So if you have a ring A, then giving yourself a delta structure on A is in fact the same thing as specifying a ring morphism from A to the ring of length to bit vectors over A which will be a section of the natural projection on the first component. And the recipe for this is assume you start with a delta structure, delta on your ring A. Then you just look at the map from A to W2 of A which sends X to X comma delta X. And once again, you can check that the axioms about delta which tells you that you have a delta ring is exactly what you need to verify that this is a ring morphism. Okay, and this point of view is useful because it allows you to prove that this category of delta rings contrary to the category of rings with a lift of robinus has all limits and collins that you can just compute on underlying rings. And so in particular, this forgetful factor from delta rings to rings, we have both the left and the right adjoint. And the right adjoint is given by the bit vectors function. Okay, so that's first remark. And then another remark which is more an exercise that you can do. I said that, I mean, there are examples of delta rings with P torsion, but it can never happen that you have a delta ring in which P to the N is zero for some A. Okay, this is something you can check just using the definition of the delta. Okay, very good. So we will see some examples later. Oh, wait a second. You know, it's a trivial remark that zero is a delta ring, so P to the N. Yeah, okay, I assume that zero is different from one. But okay, yeah, if you want. No, but if you want all limits and collimates, then maybe. Okay, yeah. Then I should have said that, there is no non-zero delta ring in which P to the N can be zero. Thank you. Okay. Okay, so as I said, we'll see examples soon, but now I'll come to the next definition, namely the definition of a prism. So what is a prism?  Well, A is a delta ring. So usually in the notation, I will just forget the delta. Okay, I'll just say A is a delta ring without mentioning delta. It's implicit. So you have a delta ring, A, and then you have some ideal, I, inside A. And again, this pair has to satisfy some properties. So first of all, you require that I define the Carti divisor on spec A. So it's just like locally principle generated by a non-zero divisor. Then you also require that I is P iidically complete. And for technical reasons, you should mean this in the derived sense. Also, soon we will make some assumption on the ring, which I'm sure that in practice, like the derived and the classical completions agree. So first approximation, you can just think that this is classically P iidically complete. And then the last condition, the most important one. So I has to be a pro-zarisky locally generated by a distinguished element. And what is a distinguished element? By definition, it's just an element of D, which has the properties that delta of D is a unit. Okay, and once again, in practice, I will always be principal. So you should just remember that a prism is a pair A, I, where A is delta ring, and let's say I is principal generated by a non-zero divisor, which is distinguished in the sense that it's image by delta is a unit. And moreover, the ring should be P iidically complete. Okay, I mean, the important condition is the last one. And so now I can give two examples of such prisms.  Okay, so I mean, remember, so you need to check three conditions. So first of all, I had assumed that P, there is no P torsion, so P is non-zero divisor. Then by assumption, I also require that my ring is P complete, in this case, I is just P. And finally, the only thing you have to check is actually that delta of P is always a unit. Okay, and this, I mean, this is also what you need to do this little exercise I gave before that it cannot happen that P to the N is zero. I mean, the way to prove this is just to check that delta P has to be a unit in any delta. Okay, so that's one first class of example. And now here is another interesting example of prisms. So let me give one more definition first. So we'll say that a prism is perfect if it's for benus phi. So remember, whenever you have a prism, sorry, a delta ring, you have these deltas and you can define a phi which leaves for benus by the formula that phi to the X is X to the P plus P times delta X. So what you require is that this ring morphism phi is actually an isomorphism. And then the claim is that actually the category of perfect prisms is the same as the category of integral perfect to it rings. And how do you see that? Well, I mean, you can define a function in both directions which are quasi inverse of each other. Let me tell you how. So in one direction, assume you start with an integral perfect to it ring R. Then you can do this classical Fontaine's construction. You can consider a inf of R. So what you do is first you tilt your perfect ring. So you get a perfect ring of characteristic P, R flat. And then you take its ring of bit vectors. This is what is called a inf of R. And a inf of R comes with a natural map theta which goes from it towards R. And I mean part of the definition of, or at least consequence of the definition of an integral perfect to it ring is that this will be principle generated by a non zero divisor. And you check that in fact, this generator which is usually denoted by Xi in PRD code theory is in fact, distinguished in the previous sense. So namely, oh, so I should have said first that this ring is P torsion free. R flat being perfect. And so it has, it comes with a natural Frobenius and you just take the delta structure attached to this Frobenius leaf. And then my claim was that the generator of this idea is actually distinguished. And well, the way, I mean, you can check this by proving more generally that in such a delta ring and the delta ring of this form, a bit vectors of some perfect ring an element is distinguished it on only if it is primitive of degree of degree one, which means that when you write the expansions of some of Taichman last times powers of P, then the coefficient of P has to be a unit. And this you can check for. Okay, so this is one, one functor. And then if you want to go in the other direction, it's even more simple. You just mod out, you have a prism AI, which is assumed to be perfect. And then you just look at a mode. How do you recover delta? How do we get delta? Yeah. Well, okay. I said that if the ring is P torsion free, which is a case for this ring, then having a delta structure is the same as having a Frobenius ring. And you have a Frobenius on the ring of a bit vector of R. So you just take the delta structure attached to this, to this Frobenius. Okay, and so this proposition is, I mean, this example is the reason why I guess Baton Schultz described prisms as some kind of deep perfection of the category of perfected rings. Because you see that inside the category of all prisms, I have this subcategory for my perfect prisms. And this are exactly the same thing as perfected rings. And I is like choosing I, my ideal is B, like it's the same as choosing some un-tilt of the tilt of my perfected ring. Okay. Okay, so once you have done this, namely introduce delta rings and prisms, you can define the prismatic side. And there are several versions of it. So for us, the one which is really relevant is the absolute version. So let's start with R. There's a question. Could you unmute the person who asked the question? Yeah, I did not hear anything. Takashi, could you unmute the person who asked the question? Yeah, so I'm trying to do that, but it doesn't work. Yeah, maybe there's a question on the microphone. Oh, there's a problem on the microphone. I, yeah, I can't do. Okay, so maybe he can ask his question by chat, okay? Okay, okay. Okay, so I define now the absolute prismatic side. So let me fix a ring R, which is assumed to be periodically complete. Then the absolute prismatic side of the ring R, which will be denoted by R prism. Okay, so this symbol is supposed to be prism, even if it appears as a delta here. So as a category, it's just the opposite of the category of all bounded prisms, B, J, together with a ring map from R to B mod J. Okay, so here there is one adjective which I did not define yet, bounded. So it's just again, a technical condition about which you can forget in first approximation. It's telling you that you have your prism, B, J. And what you require is that B mod J, here, as a bounded P infinity torsion.  This is a condition you put for technical reasons which have to do with derived versus classical completions. Okay, but I mean, basically an object of the side is a prism with a map from R to its reduction, B mod J. And then you put a topology on this category. So you define covers to be morphism of prisms. P J goes to B prime, J prime. So morphism of prism is an obvious notion. So it's just a morphism of delta rings compatible with the delta structure which sends J into J prime. And you say it would be a cover if when you just look at the underlying ring map from B to B prime, it's P J completely facefully flat. So this means that if I take the derived answer product of B prime with B modulo P J over B, I mean, first of all, it has to be concentrated in the zero and it is then facefully flat over B modulo P J in the usual sense. So it's just a slightly weaker notion than the notion of facefully flat ring morphism. Someone would just require a condition modulo P J for the reason that again, everything is assumed to be complete and you want a notion which is stable under completion. So instead of looking at facefully flat morphism, you just look at P J completely facefully flat morphism. Oh, there's a question. Okay. Okay. So first for the prison several technicals. So in the morphism of prison, you require that J goes to J prime, but I suppose it should follow the J generate J prime. Is it correct? Yeah, this is true. Yeah. And also you need to have finite disjoint unions of, I mean, when you have the topology you also want the risky covering of this N like N opens. Here you just have covering by one thing. Of course you need to add like the disjoint union of N level or the copper. Well, if it is a product of situations, then it's taking all of them will be a covering by N thing. Okay. So yeah, I agree. Okay. So it should just be generated by these covers. Okay. Yeah. Okay. No, no other question. Okay. So this is the definition of the prismatic side. And then you have two natural pre-shifts on this side. So one is denoted by O-prism and the other one is denoted by O-prism bar. And this is the front door which define on the prismatic side which send a prism BJ on the prismatic side to B. So this is for O-prism. And O-prism bar will send a BJ to B mod J. And something that shows the check is that these two factors are actually shifts on this prismatic side. And they both have a name. So this shift prism will be just called the prismatic structure shift. And the other one, O-prism bar is called the reduced prismatic structure shift. Okay.  Something you can deduce from this is that you could also consider the factor I prism which sends just BJ to J itself. And then you, this is also a shift on this side. Okay. So you have these two shifts, I mean these three shifts on this prismatic side. And this is the ones we will use later on. But before doing that, I want to mention that, okay. I defined the absolute version of the prismatic side. You could also do the following. So let me fix to start with a bounded prism, A comma I. And then I require that my ring R is leaving over AI. So R is a P-complete A mod I algebra. And then you can define the variant of the absolute prismatic side where everything leaves over A. So it will be denoted by R over A prism. And it will be the category of all prism BJ which leave over AI. So together with a map of prisms from AI to BJ. And as before, you want that B mod J receives a ring morphism from R which now is required to be a morphism of A mod I algebra. So everything leaves over my bounded prism AI which I fixed at the beginning. And the topology is as before. And this is the version that a bachelor use and what they do. I mean, one of the main objectives of the paper is to compare what you get using this notion with other classical, more classical piadic homology series. And so to do that, I mean, once you have defined this site, you can define prismatic homology. So I keep the same notation as before. So AI is my fixed prism and R is leaving over A mod I. And then the prismatic homology of R over A which will be denoted by prism of R over A is simply the homology of my shift or prism on this relative prismatic site R over A prism. And well, actually to be more precise, you only make this definition when R is assumed to be a formally smooth over A mod I. You could do it always, but it's not well behaved. And the way bachelor define prismatic homology for general A mod I algebra is using left-hand extension from the smooth case in the same way as you would define the cotangent complex from the shift of degree one differential in general. But at least when everything is smooth, then the definition is just the homology of the shift on this side. And as I said, what bachelor's would do is they compare this new homology theory with other piadic homology series. And I just want to mention two comparison results I prove, there are many of them. Namely the hot state and the crystalline comparison series. So for this, okay, so I recall notation, AI is fixed prism, which is assumed to be bounded. And R is formally smooth over A mod I. Okay, so the first result is the hot state comparison. So it tells you that, I mean, let me start with the right hand side. So here you have your prismatic homology complex, which I defined before prism R mod A. And you take its direct answer product with A mod I over A. So it would be the same as considering the homology on the prismatic side of my reduced prismatic super shift or prism bar. And I consider a homology of this in some degree I. Then the claim is that this is canonically isomorphic as an R module with the module of degree I differential forms on R over A mod I. And here, I mean, it's implicitly assumed to be periodically completed. Up to some small twist, which is denoted by this symbol with an I. It's a boy kissing kind of twist. So I recall the notation below. If you have some A mod I module, M, you will denote by M twisted by I, the tons of product of M with I mod I square I times over A mod I, okay? So this is a rather surprising result when you sing a little bit about it, because I mean, you have made this definition of prismatic homology just using data rings and prisms. And you see that naturally when you compute it, I mean, when you compute the reduced prismatic homology, the homology groups of this complex, then you see differential forms showing up. Okay. And as a remark, I said before that- Do you have a question? Yes. So you didn't define what is be completely smooth. I imagine that completions of smooth things or direct limits, fitted limits of those are be completely smooth, but what is the exact definition? So I would say that I have a math A to B. I would say it's be completely smooth. If I take the derived answer product of B with A mod P over A, and I require this to be sitting in your zero and being smooth over A mod P. In the classical finite type sense. Yes. Okay. Other question? It's okay. Go ahead. Okay, that's a remark that I said before. If you want to define prismatic homology in general, you don't do it just by computing homology of the structure shift on the prismatic side. You do this process of left-gen extension. But once you have done this, I don't want to explain it in detail, but you can check that this hot state comparison result will actually generalize as follows. Namely, if you have some A mod I algebra, R P complete, but not necessarily smooth. Then it's reduced prismatic homology. So this complex over there, the base change to A mod I of prismatic homology. It actually comes equipped with a natural filtration, which is increasing, and which has the properties of the graded pieces of this filtration, which is called the conjugate filtration, are given just by wedge powers of the cotangent complex of R over A mod I. And suitably shifted and break is interested and periodically completed. Okay. So I mean, this is something you can directly deduce from the definition, from the previous hot state comparison plus the definition of both sides in general. Now I wanted to point this out because one, I mean, we will see the cotangent complex appearing later, later again. And basically the moral of what you can remember from this statement is that hot state comparison gives you a way to like have some control on prismatic homology or at least its reduction modulo I in terms of the cotangent complex. So if you have some information on the cotangent complex, you can usually deduce some interesting properties of prismatic homology. Okay, and then the next statement is the so-called crystalline comparison. So this is a case where you assume that in your fixed prism AI, I is generated by P. Okay, then in particular, because R lives over A mod I, it means that P is zero in your ring R. And then you could ask the question, how does this prismatic homology relate to another interesting prism, homology theory in characteristic P, namely crystalline homology? And the answer is that actually they are almost the same. So if you compute crystalline homology of R over A, then, well, it is the same as prismatic homology except that here on the right hand side, you have to twist by Frobenius. So you take pull back along the Frobenius of A. Okay, so in particular, if you know what prismatic homology looks like, you recover crystalline homology. You can not necessarily go the other way because A is not assumed to be perfect. So phi of A is not necessarily an isomorphism. But at least if you know prismatic homology, you recover crystalline homology. And this is compatible with the Frobenius structure on both sides. This is also quite surprising because this way you get a definition of crystalline homology without using divided powers in anything like this. And the key technical statement to check this is a funny exercise that if you have a P-torsion-free delta ring, and if you have some elements in this delta ring so that it's first divided power is in your ring, then actually all the other divided powers are also in the ring. This is something you can check using the existence of the delta structure on the ring. This is one of the key inputs to in the proof of this crystalline comparison. Okay. Ah, there's a question. Yes. Just a clarification in the theorem, you wrote phi A upper star, the kind of pullback, but since you are always working in the derived, the complete things, is there some completion there or is this algebraic? Maybe I'm confused, but. Wait. I'm not sure now. I don't think you need to complete, but. What you say is that the delta. Yeah. No, go ahead. No, as far as I understood, you characterize your things like, when the relation with forms is only after modding by I and also maybe P, to P complete. So it seems that everything is something's derived complete relative to your ideals. And so you, and when you take phi A upper star, maybe it destroys this. I'm not. Yeah. Okay. I'm not sure. I should check again. Okay. Okay. Good. So, okay, that's all I wanted to say about prismatic homogene general. No, I turn to prismatic do-don theory itself. So as I said that I need to explain over which rings we want to classify feasible groups and by which kind of objects we want to classify. So I start with the rings. So there will be again definitions. So we, the rings we will consider are called quasi-syntomic. So ring R is said to be quasi-syntomic if it satisfies the following conditions. So first of all, it's P complete. So this, this we always assume everywhere and with bounded P infinity torsion. So I recall that this just means that there exists some integer n so that everything killed by a power of P is already killed by P to zero. Okay. And then the really important condition in the definition is that you want the cotangent complex of R over ZP to have P complete tormplitude in degree minus one zero. So this means that you take this cotangent complex and you take its derived answer product with N for any R mod P module N. And then you want that this object lives in degrees minus one zero and the complex of R mod P module. Okay. So this is the absolute notion somehow. And then you can also define what the quasi-syntomic morphism is. So it will be a morphism of P complete with bounded P infinity torsion rings. R goes to R prime, which, okay. First you want that R prime is P completely flat over R. So I already explained what that means. And then you want that the relative cotangent complex of R prime over R as P complete tormplitude in minus one zero. Okay. And you can also define what the quasi-syntomic cover is. This would be useful later. It's the same definition, but instead of requiring that the map is P completely flat, you want it to be P completely phase fully flat. Okay. But so the important condition is really the condition on the cotangent complex. And this definition is due to, I mean, it appeared in the paper of Batmouro and Chelsea on topological orchid homology. And the idea is that it should extend, well, in the world of periodically complete rings. So there's some trouble. We lost that speaker. Ah, yes. Coming. Hello. Ah. Okay, does it work? Sorry. I think it... Yeah, no, we convert our connection. What? Okay. Is it okay? Yes. Sorry. So is it good now? Yeah, it's good now. Okay. Sorry. I think the connection, okay. That was, I don't know. Okay. So I was just saying that this definition is an extension of the classical notion of symptomic, well, of LCI ring and symptomic morphism. But you don't make any Nussarian or finite type assumption in this definition. So, yeah, before giving examples, one more piece of notation. So I will denote the category of all quasi-syntomic rings by QC. And then you can look at the opposite category and consider the topology which is defined using the quasi-syntomic covers in the above sense. So, as I said, like maps which are quasi-syntomic and which are P-completely phase-free. Oh, sorry, P-completely phase-free. Okay. And then as a notation, if R is an object of this side, I would just denote by R with small letters QC, the sub-site which is formed by all rings which are quasi-syntomic over R. And again, undone with this quasi-syntomic topology. Okay. So I give examples of such rings now. The first example is just to justify the claim before that. This generalizes the classical notion of LCI ring. So the claim is that any P-complete and necessary ring which is locally complete intersection is quasi-syntomic. And, well, this is checked using, I mean, actually Avramov gave a characterization of such rings LCI in terms of the cotangent complex. But here you just need the easy direction of Avramov CRM to prove that any such ring is quasi-syntomic. Okay. So that's one first class of example, but you also have like huge rings in this category of quasi-syntomic rings. So, namely, I claim that any integral perfected ring is quasi-syntomic. And the reason for this is, okay, well, remember you have to check, okay, first of all, it's P-complete by definition. Then you can also check that for perfected ring, if there could be some P-torsion, but like the P-torsion is just the same as the P-torsion. So the first two conditions are checked. And then you need to check this condition in the definition about the cotangent complex. And for this, well, you observe that this map from ZP, I mean, the canonical map from ZP to R, it actually factors through, you can factor it through the CETA map. So I should have written that this second map here on the right is Fontaine CETA map. And whenever you have such a composite, you get a triangle for the cotangent complex. And now you observe that, well, what is a inf of R once you mod out P? It's just R-flat, which is a perfect ring of characteristic P. And whenever you have a perfect ring of characteristic P, its cotangent complex over FP is just zero. Basically here is just that if you take any X, like it's always of the form Y to the P for some Y because the ring is perfect. And then DX is just like P, Y, P minus one, DY. And if P is zero, this is just zero. So this way you can check that the cotangent complex of something perfect over FP is zero. But this just tells you that mod P, I mean the P completions of my cotangent complex of R over ZP and the cotangent complex of R over a inch of R agree, right? Because in this triangle the other term will vanish after P completion. But so then once you know this, you can just, you are reduced to describe this cotangent complex of R over a inch of R. But then this map is a subjective and the kernel theta is by properties of perfected rings is principal and generated by a non-zero divisor. So this means that then the cotangent complex is just the same as R, but shifted leaving in common logical degree minus one. Okay, so this way you check that in fact, in this case the cotangent complex even has term amplitude in degrees minus one, minus one, it's even better. And then from these two class of examples you can construct other examples if you want. So you can take a smooth algebra about perfected ring and take its P completion or you can take a perfected, sorry, yeah, take an integral perfected ring and just mod out by a finite regular sequence. So if it's again, if it is again bounded P infinity torsion, then this would give you another example of quasi-sintermittent rings. And here I list some examples. So you can take the theta algebra in one variable over OCP. You can take OCP mod P or you can take this characteristic P perfect ring, FP T one over P infinity and mod out T minus one. So these are all examples of quasi-sintermittent rings. Okay, so I think that's basically all I wanted to say about quasi-sintermittent rings, but I just wanted to try to convince you that many interesting examples of rings are actually quasi-sintermittent. Okay, and one good point of this but more short to definition is that it's purely a definition in terms of the cotangent complex. And we have seen before, this was a hot state comparison theorem that if you have some control on the cotangent complex, because of this hot state comparison, you can usually deduce things about prismatic correlation. Okay, now I turn to third part. So filtered prismatic do-donate crystals. This will be the objects we will use to describe our feasible groups. And as you can guess, the definition will use the prismatic side. So to state it, I first need one observation. Let's take R to be quasi-sintermittent. And then the claim is that, okay, you have a natural morphism of topos which goes from the category of sheeps on the prismatic side, the absolute one, to one of the category of sheeps on the small quasi-sintermitt side of my ring R. And if you wonder how this, what does this morphism of topos come from? Well, it's defined as a composition.  So it will be P-complete. And the claim is that this defines a co-continuous function from the prismatic side of R towards the big quasi-sintermitt side of her ring R. And then you just restrict. Okay, but I'm, only difficulty is checking this co-continuity of this function. But if you are familiar with crystalline homology, it's like very similar to what you do when you, you go from the crystalline side to the ethyl or the risky side, except that here we work with something a bit more general. We work with the co-continuous. Okay, and now what I will do is I will take my prismatic structure shift or prism on this ideal prismatic shift, I prism, I just push everything using this morphism B. So V was my notation for this morphism of topos. I just push everything down to the co-sintermitt side. And then I claim that, okay, actually you have a natural surjection from oprys to O. And I would give a name to the kernel of this morphism of shifts. So O here is just the structure shift on the co-sintermitt side. This kernel is denoted like this. So it's what is called the first piece of the Niagara filtration on this prismatic shift oprys. Well, the reason for this notation is that, I mean, there is this notion of Niagara filtration of a prism, which is defined for any, you could define for any positive integer I, here I only need the first piece of the Niagara filtration. I would just take as a definition that it is the kernel of this surjection, which I did not explain. And then a property of this is that, well, because any data as a Frobenius, so this shift oprys will come with a Frobenius morphism Phi. And one can check that this kernel has the properties at Phi of the kernel. So Phi of the first piece of the Niagara filtration of oprys actually lies in I-pris times oprys. So morally on this first piece of the Niagara filtration, the Frobenius is divisible by, well, let's say the locally I is, the ideal is generated by a non-zero distinguished element. And the idea is that on this first piece of the Niagara filtration, Phi is divisible by this distinguished element. This question from- Yeah, give a question. I cannot hear. Can you put the microphone on? You should activate his, I guess you can unmute him. Ah, yeah, now his microphone must be on. Internet, I don't- Oh yeah, now you can speak. So why do you write I-pris times oprys? I think I-pris is an ideal in oprys. Ah, yes, sorry, sorry, yeah, it's a typo. Yes, thank you, sorry. Yeah, it's just because later it will appear. Yeah, sorry. Okay. Okay, so then the definition of a filtered prismatic eudonic crystal is, so let me again fix a quatericentamic ring R. And then a filtered prismatic eudonic crystal is by definition of collection, a triple M, fill M on Phi M. So M is finite located free oprys module. First of all, then fill M inside it will be some oprys sub-module. And finally, Phi of M is morphism, ring morphism from M to M, which is assumed to be a Phi linear. Okay, well Phi is a Frobenius of oprys. Okay, and then you ask three conditions on this triple. So first of all, you want that this Frobenius Phi M sends a fill M to I-pris times M, okay? But then there is an obvious oprys sub-module of M, which also have this property that Phi M of it is leaves inside I-pris times M. Namely, consider the first piece of the Nagat filtration of oprys times M. Then because of the previous slide, we know that this is contained in I-pris. And so this sub-module, first piece of the Nagat filtration times M, also has a property that Phi of it is contained in I-pris times M. And then the second axiom you have is you want that this sub-module is in fact contained in fill M. And once you have done this, then you know that M-module of fill M will be a module of our op-pris module or the first piece of the Nagat filtration. But remember that op-pris module or the first piece of the Nagat filtration was by definition is just the structure shift of the quiescentomic site. Oh. So M-mod fill M will be an O-module and you ask that it is finite locally free. And then the third condition is that the image of the filtration by the Frobenius is big enough in the sense that it will generate I-pris times M as an op-pris module. So if you are familiar with Christian Doudonné theory, again, it's very reminiscent of the usual notion of a Doudonné or filtered Doudonné crystal. And the third condition is just in this setting would just be a reformulation of the condition that the filtered crystal is admissible in the sense of Quotainy. So that's somewhat just the obvious extrapolation of this classical notion to the setting of the prismatic site. Okay, and then notation is if R is my quiescentomic ring, I will denote by Df of R the category of all filtered prismatic Doudonné crystals over R. And the morphism R. So obvious ones, I mean, they should be op-pris linear and they should be compatible with Frobenius and with the filtration. Okay, and so now I can come to the statements of the two main results we proved. So I will fix again once and for all, now quiescentomic ring R. And let me take G to be a feasible group over R. Then I can define M prism of G to be X1 of G by op-pris. And here when I write this curly X1 is supposed to be like the local X groups in the category of Abelian sheeps on the quiescentomic site. So you check that your feasible group G, the reduction to the case of finite locally free group schemes defines an Abelian sheeps on the quiescentomic site. And op-pris is also an Abelian sheeps on this site. So you can take X1 in this category. And then you also define fill M prism of G to be the same thing except as you replace op-pris by the first piece of the Niagara filtration. And then the first main result is that, so if G is as before, then this triple M prism G fill M prism G and Frobenius of M prism of G, which is just by definition the Frobenius coming from the Frobenius on the prismatic shift here. Then the claim is that this is actually an objective of this category Df of R. So this is a filtered prismatic geodontic crystal of R. And in what follows, I will denote it by M prism G underlying. Okay, and as a remark, if P is zero in my ring R, so in the characteristic P situation, you can check using the crystalline comparison serum, which was discussed in the first part, that in fact, this is just the same thing as the usual function you will find, people have studied in crystalline geodontic theory. Which you find, for example, in the book of Bertolo, Brin, and Messing. So this is nothing new in characteristic P. Okay, and then our second main result is that, well, this filtered prismatic geodontic factor underline M prism, which associates to G, M prism of G underline. It actually realizes an anti-equivalence between this category of Bt of R of visible groups over R, and the category Df of R, which I defined before. And as a bonus, I mean, as a byproduct of the proof, you actually obtain that the prismatic geodontic factor, G goes to M prism of G, is already fully faithful. But if you really want to get an equivalence of categories, so if you want to be able to describe the essential image, you need to add the filtration in the picture. Okay, so now I make a few remarks about these two results and about the way we prove these two results. So first remark is that, as I said before, in characteristic P, you just recover the usual factor from geodontic theory. So in particular from serum two, this classification result, well, you can deduce that the crystalline geodontic factor is an equivalence for all quasi-syntomic rings in characteristic P. And, well, this was actually already known, of course, in some cases. So if you look at LCI rings, which are also excellent, then fully faithfulness was proved in the end of the 19th by To-Yong and Messing. And more recently, Pecolo has proved that this factor is actually an equivalence when the rings is, okay, no serum LCI, and more of a F finite. So this means that the Frobenius is finite. And this particular implies excellent. So in this case, the result was already known. Okay, then I wanted to explain now that this category DF of R may seem a bit abstract, but there is an interesting class of quasi-syntomic rings for which you can make it more explicit. Namely, you will say that the ring is quasi-regular semi-perfectory. If it is quasi-syntomic, of course, first, and then you also want that there exists a perfected ring which maps subjectively onto R. It has to be big enough, in this case. Examples of such in the list of examples I get before, where you can look at any perfected ring is obviously, because we know it's quasi-syntomic and the second condition is trivial. Any perfected ring will be quasi-regular semi-perfectory, but also like a quotient of such a ring by a finite regular sequence, which has bounded P infinity torsion. So these are examples of such rings. Well, for such rings, it's like for perfected rings in the example we saw. In fact, the quotient and complex after P completion is just sitting in the minus one. And for this reason, you can check, using watch-take comparison, that for such a ring, the prismatic side has a nice feature that it admits a final object in this case, which you can just describe as, if you want the absolute prismatic homology comes with a natural idea. And then also you can check in this case that if you take the first piece of the Nagat iteration on this prism to be just the inverse image of I by Frobenius, then the quotient is isomorphic to I itself. So two examples of computation of this initial object or the final object. The first case is, if I is perfect to E, then this final object is just given by the ring A-inf of R together with the kernel of the theta map. So, okay. In this case, the prism is perfect. So Frobenius is an isomorphism. So there's always some choice that you can work with the map theta or with its pre-composition with Frobenius minus one, and then consider theta tilde. It's just a matter of conjunction. Okay, that's the first example. And then another example, if you take a ring, which is called a regular somaiparfectoid, and with P0 in this ring, then this initial prism is just the same as a crease of R together with the idea generated by P. So you can compute it in several situations. And then we make a definition. It's very similar to the one we had before. A filtered prismatic Jordanian module over R will be a collection M, field M, and phi M. But no, I just have, instead of having finite locality free modules over the prismatic shape, I just have a finite locality free module over this ring, prism R, a certain sub-module of it, field M, and a file in R map, phi M. And you just ask the exact same actions as before, but no, you only work with modules over this ring, prism R. So I want to repeat them because I don't have much time left. But okay, so you can make this definition. And then the claim is that if my ring is quasi regular somaiparfectoid, I can in fact just evaluate all the object I have on this, so I should have written final, on this final object. And the claim is that this gives an equivalence between the category of filtered prismatic dunes, crystals over R, and the category of filtered prismatic dunes modules over R. So in other words, in this case, you can make the category more explicit. You can just work with modules instead of working with this object, this category DF of R. And moreover, if the ring is perfect to it, then you can do even better. Then you can even forget the filtration. So just look at the forgetful functor for this category of filtered prismatic dunes modules over R to simply what you can call a prismatic dunes module. But in this case, it already has a name. People call them minuscule break is in FARG modules. And then the claim is, in this situation, it's this forgetful functor is in fact an equivalence. So in other words, as a special case of the CRM2, you see that you recover the fact that for perfect to it ring, peaceful groups over such a ring are classified by minuscule break is in FARG modules. But this is a bit cheating because in fact, we need as an input for the proof of CRM2, we need a special case of this. Namely, we need that peaceful group over a valuation ring, which is perfect to it and has algebraically closed fraction field is the same thing as a minuscule break is in FARG module. And actually it's not difficult to do. You can reduce the case of all perfected rings from this special case using VT sum target. But so we need this as an input. Okay, so can I just take five minutes to finish? How should I stop now? Okay, go ahead. Okay, so I try to finish quickly. So I also want to mention that as I said, in general, you really need the filtration if you want to describe the essential image of this prismatic do-donate function. But I just said before that in this remarks that for perfected ring, the filtration is actually unique. So it's not needed to state the classification theorem in this case. This is not true in general, but this also works for P complete regular rings. And as an example of this, just take the ring of integers in some discreetly-valued extension K of QP with perfect residue field. So for example, take a ring like ZP. And then also this case was, of course, already known before. So then you can prove that these were groups over this such a ring are classified by so-called minuscule-bray-kissing models. And this was, this has been done by Bray and Kissing, at least for all P, and then extended by Kim, Lau and Liu to do all P. But we can also recover this, namely by first proving that for such a ring, you can forget the filtration. And then checking that you can just evaluate when there is a natural prism attached to such a ring once you choose a uniformizer. It's not a final object, but still you can check by some reduction to the perfect situation that in this situation, evaluation on this object is again unequivalent. And so, and then we check that the factor we have is actually the same as the one which has been studied by Bray and Kissing, all these people. But now the good point is that some of you directly learned in the correct category. So the proof works uniformly for all P. You don't need to make a special argument when P is two. Okay. And then just two words about the proof. So for theorem one, it's not surprising. So you just follow the strategy of Bertot Le Brin and Messing in their book. So you have this definition of this triple. You want to check that it is a filtered prismatic dosonary crystal. And the idea is, well, you have to understand how what is X one looks like. And in fact, for any group object, in a topos you have, I could have some device to make computation about this X groups, at least in low degrees. So this is explained in the book of Bertot Le Brin and Messing. And you are reduced to compute some prismatic commodity groups. So first step, you use a theorem of Reno to reduce to the case of P-Zero groups which comes from some Abelian scheme. And then, via this Bertot Le Brin Messing partial resolution of any group object in the topos, what you just need to do is understand precisely prismatic commodity of Abelian schemes. And a key tool for this is provided by the Hodge state comparison theorem for prismatic commodity. So, but the ideas are really similar to the one of Bertot Le Brin Messing. And then, theorem two, the proof is more difficult but I just want to point that the key idea is to use quite a symptomic descent. So here, I introduced just before the notion of quasi-regular semi-perfected ring and the very nice feature of this site, this quasi-syntomic site, which was observed by Byte-Maurot and Cholso is that this quasi-regular semi-perfected rings they actually form a basis of this quasi-syntomic topology. Or by any quasi-syntomic ring by extracting enough piece roots you can always make it quasi-regular semi-perfectory. And this means that once you have defined this factor, so once you have proofs here on one, then you can, to prove that it's unequivalent, you can somehow reduce to the quasi-regular semi-perfectory situation. And then everything is more concrete, as I said, instead of having crystals, you just have modules over this ring prism arc. Okay, and then the hard work starts, but I wanted to say this because it shows that even if you want to, if you are only interested in getting classification results over necessary rings, then the proof is such that actually you really need to work with this big category, QC, because the argument is by descent from this very big quasi-regular semi-perfectory. Okay. Okay, and then I will skip this point. I just want to finish by saying that there are some natural questions which are left open. First of all, it would be interesting to see what happens for more general baserings. We only prove results about quasi-syntomic rings, but for example, in the work of Tsing, you can find some results for very general, periodically complete rings. But I have no idea how to do it for more general rings. And then another thing we don't have is something like deformation theory in this setting. So for this prismatic duodonné functor, so for the usual crystalline duodonné functor, you have this gotonic missing deformation theory, which is very powerful. But here, we don't have any analog of this. And then finally, final remark is more anecdotal, but it's about the case of perfect rings and characteristic P. I just want to point that in this case, well, because if you live over an object of the prismatic side of such a ring R, it will be P torsion-free in particular. So you can always map this prismatic structure shift on this side to the same thing where you invert P and take Q to be the quotient of this injective map. And then you check that, well, in this situation, the prismatic duodonné module of any piece of a group G is in fact the same thing as home from G to the push forward to the quotient-symptomic side of this shift on the prismatic side. I mean, this is just obtained by looking at the associated long exact sequence. And then the natural question is whether the, I mean, Fontaine was the first to give a general definition of the duodonné module over such a perfect ring of characteristic P. And the definition looks a bit similar, except that instead of this mysterious push forward here, you have this shift of VIT covectors. And so it would be also interesting, and we did not do it to know how to relate this object here to Fontaine's original definition of the duodonné counter with VIT covectors, but without choosing the general crystalline comparison, which of course implies this comparison, but just directly using the 2D finish. Okay, so I'll stop here. Thank you very much. So thank you very much for the interesting lecture. So are there any questions? Yes, Luke. Would you unmute him? Yeah, I'm trying, but it doesn't work. I don't see one. Luke, could you unmute yourself? Luke, can you put the microphone on? Now you can hear me. Yes, yes. Okay, so I think you skipped one of the last slides where you were writing something about classification of finite, locality-free community group schemes. You discussed PDVC group, I presume that you, but presumably you can also classify truncated BTs and maybe more general. Yeah, so that is... Oh, I can't hear you well. Okay, so it was just that I wanted to see this slide again. Yeah, okay. But then is it in terms of a module or whether a complex, maybe? Okay, so we only do it for over perfected rings because, I mean, the idea is the same as the one you find in the paper of Kissing where actually you use the fact that you can just identify the category of finite, locality-free group schemes with the category of like two terms complex of peaceful groups with an isogenic between them. And the only difficulty in general for general quasi-syntomic rings is I'm not sure by which kind of objects you would classify finite, locality-free group schemes. So for perfected rings, as I said, for peaceful groups, you can forget the filtration.  So it's not difficult to guess by which kind of objects you will classify finite, locality-free group schemes. So it could just be a module of your ring prism R which is just a inf of R in this case of projected dimension less than one killed by your power of P with a Frobenius and a Vershebu. But in the general situation where you should also take the filtration into account, I'm not sure how to describe the finite, locality-free group schemes. So that's why we only did it for perfected rings. Actually, this was already known in all cases except maybe when P is two by work of love. I had another question. You mentioned work by Tzink. So is there a relation between displays in this theory? Yes, this is another slide I had to skip here. So, okay, let me take R again to be, first one reduces to the case of quasi-regular semi-perfected rings by quasi-syntomic descent. And then, well, you have this natural map from prism R to R, whose kernel is the first piece of the Nagat filtration on this prism. It's a map of delta rings. So it's a map of rings. But I said that the forgetful function from delta rings to rings has a right adjoint, which is the bitvectors function. Yes. So this map induces a map of delta rings from prism R to W of R. And using this observation, you can actually check that you have a function from our category DF of R, filtered prismatic Jordanian modules, to the category of displays of R in the sense of a thing. It's not an equivalence. I mean, the classification of thing is both more general and more restrictive. It's more general in the sense that, I think he only assumes a ring to be periodically complete. But then he has to restrict to formal visible groups, because when P is two, there are like some difficulties, usually. So this is due to the fact that this factor we get from DF of R to displays of R is not an equivalent. But it is when you restrict to any important objects. In the terminology of thing. Thank you. Thank you. Okay. Thank you. Thank you. Yeah. Then of R. Okay. So I remember the concern in this, welcome to the net, there were some after thing, there were in particular some papers of law, where I think he looked at, for example, complete mixed characteristic, notarian local rings with, residue for either perfect or not perfect. Then he gets some using thing. I mean, he gets, he does things with using windows and friends. So they get a very simple linear algebra things, which classify PD visible groups over mixed characteristic regular rings. But sometimes he needs to, to assume it is a result is different and the residue is not perfect. But in any case, he has some, do you get a relation between what you do and is still, but maybe there are several such references. I don't remember exactly, but I remember it's quite simple. Like you only need to give me, you write R as a quotient to formal series over a coin ring, and you just have to give to some finite free modules and some maps, composition equal to the equation or something like this. So is it possible to relate it to your? Yes, as that's what I mentioned here. So as I said, for such a P complete regular rings, this, the filtration is also not relevant. So you can describe everything. I think, as I said, using currency realms, you can get a natural prism attached to this situation. So for example, in the simple case where R is just okay. Yes. You just take this boy kissing prism. So you just take like the bit vectors of the residue field, double bracket U for some formal variable U and E is some Isenstein polynomial, which is like determined by, after you choose a uniformizer, you have a natural map from this frag S towards your ring R and E is a generator of this, of the kernel. And then, yes, then you get a quite simple classification in this case as modules, finite locally free modules over such a ring together with the Frobenius, which has the properties that it, like after linearization, it's co-kernel is, after you invert E, it becomes the linearization becomes an isomorphism, the linearization of Frobenius. So, but for checking that the factor is really the same as the one which is used in law. At least we checked it in this particular case, but this is the same as the function that people usually consider. But in the marginal situation, I'm not sure we checked it. So there was also a technical question about whether the rest of the field is, when the rest of the field is not perfect, I think he needed to work like, like sync with only formal, pretty visible group, but maybe I don't, do you get, do you get easy to, that is specific? Well, I don't have a, I mean, I think when the difficulty, that low on contours is that, I mean, he uses crystalline Jordanian theory. Yes. So somehow you have your ring R, some of us you reduce to R mod P where you can use crystalline Jordanian theory, but then you need to go back from R mod P to R. So you need to use Gotendic-Messing theory. And, but the issue is that this, like this divided power structure on the idea and generated by P is not important when P is two. And Gotendic-Messing theory only works well when the divided power structure is important, right? So I think for this reason you have to do some, usually you have to do some extra work when P is two. But here somehow I think this issue does not appear because in some sense you directly, with this prismatic Jordanian function you directly land in the correct category. Like, again, back to this example where R is okay. In the original work of Boyd or Kissing, actually you, what you first do is you produce a factor from the category of feasible groups over this ring R towards a category of filtered modules over another ring that's like S, which is like the P completion of the PD envelope inside this ring fracas. And this is usually just called curly S. And then you have to do some semi-linear algebra to check that this is indeed the same as the category of minuscule Boyd-Kissing modules. But here we directly get a factor from feasible groups over R towards minuscule Boyd-Kissing modules. So somehow we avoid this difficulty and that's why we don't have any assumption on P. I have another small question. Maybe what is, so you said that in the definition of a prism you have a Cartier divisor. You can, are there examples or it is not globally principled? I know I don't know any example writes, writes not principled actually. I think all the examples are no, the ideas are principled. I see basically one question Can you hear me, Ophar? Yes. Okay, so the question was, does there exist any coincidence between the filtration of prismatic delinear crystal and the filtration on bun G on the Farc-Fontaine curve since Farc proposed that using chromatic filtration one can correspond perfect with BT1 with bun G. One can what? Correspond perfect with BT1 with bun G. I don't know what that means. I'm just reading what is. I think the BT1 is like Berserk detail of level one. I guess so, but I guess what is perfect with BT1 is BT1 over a perfect with bun G. I don't know. I don't know what it means. But maybe BT1 over a perfect with, okay, I don't know. But I don't know any relation with bun G. I mean, of course, Burkis and Farc modules are related to modifications of vector bundles on the Farc-Fontaine curve, but. But bun G is like the bun G bundle, so. Yes. So I don't see. For a group, for a group, they do it with a. I guess, well. Yeah. So Fontaine. G should be a reductive group or something like this. So I don't see any relation. Okay. Well, if I, so it's not a, I'm not, I will need more to see it in more precise form. So, so, yeah. So as far as I understand, okay. Yeah, okay. So by any case, you prove that the same construction of display is canonically correspond to what you do to what you do using these functors to displays, you get exactly things construction, which is defined for more general rings. Right. Yes. Okay. All right. So, yeah. So then this partly answer, because the work of Law and Son is kind of trying to use this display to with some complicated tricks to, which I don't, okay. So it's, and of course you get also the crystalline the other theory from your. Yes. You need to use amorphous in between relating the crystalline. Okay. But you wrote it. I don't remember if you put it in the slides. I just put a remark. I mean, yeah, basically you can check that the usual.  You have the crystalline side. Let's say you just push everything to the, to this quarter centering site and make the comparison there. And then the claim is that the two categories you get are actually the same. So. For, for, for the steel by P. Exactly. Yes. You get a, okay, I have to, this was a. I mean, basically this follows from the crystalline comparison. Okay. Okay, thank you. So I will have to think. Okay. Not the slides would be online at some point. Okay. Thank you. You're welcome. Okay. Goodbye. Thank you very much. Goodbye. Yeah. Have a good day.


Footage of Vladimir Tsima Bursov, a soldier of the Ukrainian army returning from Russian captivity, has been released. Bursov who lost 38 kg during his 20-month captivity, was a soldier of the 56th Infantry Brigade. Vladimir lived in Mariupol with his wife and daughter before the war. Before his capture, he was 191 cm tall, weighed 95 kg. In April 2022, he was captured while fighting for the metallurgical plant and was able to return only after 20 months during the exchange of prisoners. During this period, his weight dropped to 57 kg. Within one month of his release, doctors diagnosed him with a number of diseases, some of them chronic. Photos of the former prisoner taken before the war and after his return from captivity have been distributed on telegram channels.


This television is from Columbia. This television, a view of the isothermal dendritic growth experiment, experiment actually located in Columbia's cargo bay as part of the United States microgravity payload. The IDGE experiment, as it's called, studies the crystals that form metals on Earth, those crystals called dendrites. Dendrites can be formed much more purely in weightlessness and in much more pure form than they can in Earth's gravity. Thus, hopefully providing scientists a better insight into the basic structure of metals. Columbia, for Clark Park. Go ahead, Casey. Focus right now, as you see. And I think you can see what's happening to the particles, not better out here. And you can see sometimes the particle reaches the interface. It just joins the collection, and another collection gets dislodged, goes in a very different direction, so you lose track of the particle altogether. We copy that, Casey. I appreciate your focusing on these different levels. That's going to help the team out. I think it'll make it easier to measure the particle sizes. Oh, sure. This is Mission Control Houston. This view coming from one of Columbia's payload bay cameras, looking down on the Earth as the orbiter flies. 152 nautical miles above the surface of the Earth, passing over the central portion of Africa.


story you're talking about it's a blue jay relievers name is mike balsinger and he is actually um entering the mix with a potential precedent setting civil lawsuit so balsinger was pitching for the blue jays on august 4th 2017 in houston the astros hit him for four runs four hits three walks and one third of an inning 29 pitches and balsinger told us a today i don't know if i've had a worse outing in my professional career i remember saying it was like they knew what i was throwing they're laying off pitches they weren't laying off before it's like they knew what was coming that was the thought in my head felt like i didn't have a chance astros won 16 7 and it ended up being balsinger's last appearance in the majors he was demoted by toronto the next day on monday today he fired back filing a civil lawsuit suit in los angeles superior court accusing the astros of unfair business practices negligence and intentional interference with contractional and economic relations according to the suit filed by ben meseles of garagos and garagos in los angeles balsinger stated that the astros unlawful and tortuous business practices have had consequences far beyond wins and losses and strikeouts or home runs it could have cost him his career so he is seeking damages in three distinct lanes with the suit including for himself he's asking the astros to forfeit the nearly 31 million in bonuses from the 2017 world series championships with the funds instead going to charities in los angeles focusing on bettering kids life so he also wants to create a fun for retired baseball players who need financial assistance so that makes it actually look pretty good don because he's not looking to gain anything for himself but that's what we're looking at here we're looking at lawsuits being filed i still think you might see some people decide that maybe bet on games now that gambling is legal in certain areas do they have a kick coming um i don't know but that that's that's opening up a whole pandora's box did you


Do I sound weird? Unless you're normally high pitched, you sound fine. Whoa. Whoa. Whoa! How do I look at myself? I can't remember. Okay, okay! What's a little bit of that?   The return.  Alright, what's the... Give me the rundown. I don't know, cause we started and everybody went everywhere.  Currently, this is a house that Ted built that Connor, Schlatt and I are living in. Okay. This is just Sonic rolling away. There he goes. There he goes.  Do you know where Eddie is? Apparently, Eddie was looking for me. Yeah, he showed up when you were fixing your mic. We can go find him. Yeah, yeah, yeah. You're gonna have to teach me how to play Minecraft, cause I haven't played it since the last time. Epic has happened, which was, I think a year ago, almost exactly. I don't know what you're talking about. There hasn't been a first time. Yeah, yeah, you're right. Sorry. The first time. I had this dream. Where is he? Yeah, I have no idea where he is. He showed up extremely excited when you logged in, but I guess he walked away afterwards. Yeah, he wasn't willing to wait for me. You know. Eddie! Hello! Look at you. What's up, Ethan? Look at you. You're verified now.   Look at you.  Yeah. I stepped out of a limo right before streaming. That's crazy. Damn. Eddie, do you want to show me your house? Yeah, but it's really far away. It's like over the mountain back here. Go ahead. Let's go check it out. Why not? Yeah. I gotta change my push-to-talk button because it... Oh. There you go. Yep, that's it. Okay, guys, I'm stuck in the snow. You're good, you're good. You can't jump on that one. Oh, you can't jump on it. Okay, thank you so much. Yeah, you can't jump on some powdered stuff because you'll sink and freeze. That was scary. Thank you so much for saving me. Hey, you're just taking voice-activation to just meet with them. Wow, that felt... That was so awful.  I don't know what's going on, but I'm uncomfortable. Eddie's just thinking about some shit, apparently. I was changing my settings. I just said I was changing my settings. Wait, Eddie, you had your push-to-talk binded to pee? Yeah, it's very simple. It's thumb on the pee. Thumb on the pee. Thumb on the pee? I'm doing it now. You know, Eddie, I'm doing it right now exactly as you're describing it. I'm feeling ligament pain. It's an exercise. Wait, Eddie, where's your house? We got distracted. Wait, so you had your voice thing on pee? Oh, my God. Yep, here it is. It's on an island. And Swagger's is like Red Island. Oh, I see it. I see that orange stuff. Damn. It's like a real island this time, Ethan and not Boomer Island, which was like a patch of dirt we built. What island? I miss Boomer... Well... And we're up. We didn't get to really finish the house, and I don't love the front of it, but it's got a great view. You ready for this? Oh, it's all, you know what they say, Eddie, location, location, location, and you've got it. Look through the glass here. Oh, beautiful. Apparently Matt is giving a live concert at Spawn. I was about to say, I know we just got here, but apparently there's something happening. Let's go. Let's catch that show. This cave is kind of cool. Yeah, it stops there, but I've been calling it the big ol' hole. I like it. Might have to set up in the big ol' hole.  Do that one.  What I love about Matt is if people are there for him, that crowd would be unfazed. Completely unfazed. And then he would probably do it. Go buy my new album, Ouch, on Spotify, Amazon Music, Music Prime, Netflix, HBO. Let's hear some ouch. I am Matt. I'm gonna sing some ouch. Ouch. Ouch. Ouch.   Yeah, I can do that. Yo, my dad, he's not like the rest. He's kind of like a cute girl, just minus the breast.  I've never loved my mom.  I know a crank. Game plays. Hi! Hey. How are you? We're just hanging out. I'll do a little thing. Hey! I'm gonna go and find things.  What is going on? Trevor? Ethan? Hey! What's the whole vibe going on here? I saw Eddie's house and I saw the spawn and that's kind of it. This is our little cruise dock. We got Jacob, Jaden, Ross, Stonepaw. It's pretty spread out actually this time. I feel like we have to figure out a pretty good way to get between everyone. Maybe a bridge system. That's true. Is there a dolphin? Yeah, of course, man. I heard it before. I might try and find some good old materials because I only have bread and seeds right now. So I should find things to make up. Take some more bread. Take some wood. Take some sticks. I'll give you some iron. Look at this. I'm gonna try and build a house. Hey, let me just give you one of these. Hold on. I'll give you a little hug. Hug me back. Figure it out. No, it's fine. I'll just hug you. Hold on. I'm getting a little tired. I'm getting a little tired. I'm feeling a little tired. Hold on. Well, I'm gonna go off on an adventure. Bye-bye. Hey, how's it going? I'm going to build a house over by where Eddie was at. So you can start your stuff. I do. I want to build a house over by where Eddie was at. I was tempted to put a house inside of that little cave. I think you should do that, honestly. Trevor gave me some materials. So I got to make some things. Yeah, he gave me some wood and some sticks and some iron. That's kind of epic. I like this SMP, am I right? Do you have torches? Because I can't see anything.  Oh, true. Yeah, just crank it. I'm home. I'm living here. It's decided. Oh, look at you. Look at you with the echo. This is pretty sick. Thanks. I made it. Look at you.  I made this. You made this whole last thing. I like this tiny one right here.  Yeah. What if I lived in the hole of your Mount Tussie? It's kind of like your mailbox. Oh, yeah. This will be nice. I can't just say that it's your maid quarters. It is kind of like my maid quarters, isn't it? Look at that. There we go. I like the echo. I'm going to hollow this out a little bit.  Wait, were you talking and I couldn't hear you? A little bit.  Or is that just through Minecraft? No, that's just the mod. That's sick. That's so cool. Wait, check this out.  I'll put it further down. There you go. Place it somewhere nice. Don't spend it on one place, guy. Oh, I won't. Let me get some water. You know what I just thought of? Eventually, we could literally dig a hole from right here to the other side of the mountain where the spawn is.  That's actually a good idea. A secret tunnel, if you will. Secret tunnel. Secret tunnel. Check this out. I got something nobody else does yet. And in Craft Minstable? Craft Minstable? It'll be over here if you want to use it. I don't know how, but thank you so much. Water for me. Yup. A little bit of this. A little bit of that. Nice. This is a work in progress. Trying to figure things out, you know? Trying to figure out where we want things to be and whatnot. This is Sneak's house.  He's got one of these. Of course I know how to make one of those, so it's not like I would need that. A little bit of grandiose for my taste, you know? See, this is what humbleness is about. This is how to be humble. You're Minecraft streamers? They don't know how to be humble. They never had to grind. They just, oh man.  We've never met, we've never really talked and I don't know anything about his career. And now I'm going to get cancelled and my career what's left of it is going to go down the drain. What else should I make? Nice. I don't know what's going to go there. I know what's going to go there. This is what she call a chest. There are many like it. But this one is mine. Tommy would make a bed. Yeah. Yeah, as if Tommy would not make a bed. If I hit him right in the face. Yeah. You ready? Oh man. Tommy and he could never. Maybe what I'll do. Look at that. Wonderful. Beautiful. Actually. Nice. Make it sound almond. Is that a person? Hello? Hi. I'm behind you.  Jesus Christ. Fuck. More than the last one.  That's weird as hell. I promise you. Nothing will be worse than me jumping into a ravine and crinios screaming my name five seconds after. Oh man. By the way my house is right over here. If you need anything. Oh thank you so much. I'm just trying to find my way back to spawn. I think it's just zero. I think so. I think it's just like straight this way. No it's this way. Towards the sun. Left of the sun. What good people. This is crazy. I lied to you about spawn it's this way. I was like I was sitting there like it's like zero. Then immediately I like walked around the mountain. I was like oh god no it's not. Oh there it is. Okay okay okay. I should remember what the coordinates are.  I'm going to sneak on down here. Kind of pog. I didn't. I was just seeing what was there you know. What I wish YouTube would do is this. They would give you a warning. Like they private the video and they would say like You guys talking about YouTuber money? Yeah. That's good to see you. Our new mail video down for crack smoking. I saw that. Did it look like it was real crack? Kind. I mean I don't know what crack looks like. Ethan slick. I don't know what crack looks like. Listen I don't know what crack looks like. Remember when it's honest when you guys cooked crack? Listen man well there's no proof of it. There's no proof. Yeah the video is gone. I think it's already been tamed. Come on horse.  I'm going to get this horse. I love my ankle. Nice. Well I can't use it. And I don't have a fishing pole so I can't fish for a saddle either. But trust me horse. If I find one. You'll be mine. How are you? I'm doing good man. Just hanging out? Chilling? Yeah just chilling. Fuck. How did you do that? Do you want to see some? Show me some things. Okay hold on. Watch this. You gotta empty your hands.  Oh no that was on purpose. Shit. That's so cool.  Did you make these? Yeah dude. I found out I can animate them in blender and put them in Minecraft. This is my favorite one.  That's so funny. I also do this. Hold on. Whoa. I like the fart one.  Alright guys thank you guys so much. I'll see you guys next time. I love you. Bye.


The study examined two plant-based emulsifiers, soybean lecithin and p-protein isolate, for their ability to encapsulate an antimicrobial molecule, transcinamaldehyde, Tc. The results showed that both emulsifiers were able to encapsulate Tc successfully, but the p-protein isolate was more effective than soybean lecithin. Additionally, the p-protein isolate produced smaller particles than those made by soybean lecithin, which may have contributed to its higher encapsulation efficiency. Furthermore, the p-protein isolate was found to be more stable under heat and humidity conditions than soybean lecithin, suggesting it could be used as a natural preservative in food products. This article was authored by Fatemay Bagi, Sami Naimai, Emily Dumas, and others.