So it's my pleasure to introduce Fabrizio Andrea Traf, from Milan, who will speak on a peer-decreterium for good reduction of curves.

Okay, so thanks a lot to all the organizers in both continents for the invitation, especially for changing the day. So yesterday I was teaching, tomorrow I'll be teaching, so today was really the only day on Tuesdays I could come.

So thanks a lot for this opportunity. So what I'm going to talk about is joint work, before I forget, with Andrea Miovita and Minio Tim. Okay, so let's fix first of all some notation.

So k is a complete discrete pressure field.

I will denote by OK its ring of integers. And I will let k to be the residue field, which I assume to be perfect,

of characteristic p, which we very soon will be positive. And actually I will also let k0 to be the fraction field of the lead vectors inside k.

I will also fix an algebraic closure, so the original field of k0, which we forgot. So this is an algebraic closure and gk will be the Galo group.

And we fix once and for all xk to be a smooth, proper, geometrically connected curve over k.

And now the question I want to address in this talk is,

can we decide whether xk has good reduction from informations on xk?

And the first step, the first simplification is just using the semi-stable reduction theorem and allowing possibly some field extension of k.

The first simplification is to assume that xk has already a stable model.

Okay, and then the question becomes, can we decide the reduction type? Can we say if xk, so the special fiber, is smooth or not?

If informations, just the generic fiber. So stable is like in the Riemann force? In the Riemann force, yes. So every irreducible rational component meets the other components in three points.

Not semi-stable, but stable, yes. So can we decide this? And well, the first guess is for the Jacobian of xk, one has a good answer for this question. So we know now when an Avidean variety over k has good reduction.

And so let me call the theorem. So for the Jacobian of xk, we know the answer.

Because, so here's the theorem. So the Jacobian of xk has, so there's an Avidean variety. This has good reduction, if and only if, well, there are two cases.

So let me consider H1 et al of xk bar with L-adic coefficients, L prime. So it's the dual of the first et al homology of the generic fiber of xk bar.

And then we know, first of all, if L is different from p, then that condition that is in smooth reduction is equivalent to require that as a representation of gk, it's unramified.

And from p, there are two characteristics. While if L is equal to p, this can never be an unramified representation

otherwise the p-rank of the Avidean variety would be twice the dimension. And we know it cannot. But the next approximation for L equal p to be unramified is a contented notion of crystalline representation.

So the first is for the AP curves, the well-known neuron of Shafranovich criteria. And then in general, a theorem of Sarate. And for the second one, let me mention,

well, if Jacobean excused reduction, then the fact that this is crystalline is work of H1 from 10. And the converse, that if this representation is crystalline, then the Jacobean introduction is due to work of many people. So there is Farid Mukran, Adrian Jovita, Robert Colliman, and Christophe Berry.

So for that we know. But unfortunately, this is not good enough. So here you don't need the stable model. You just do any Avidean variety. Yes, absolutely.

The Avidean variety in the neural model, and then you can see whether the special fiber of the neural model is again an Avidean variety or not in terms of this. OK, but this is not good enough for us,

because we know that if xk is smooth, which is what we want to know, then the Jacobean of x is an Avidean scheme,

which implies that the Jacobean of xk has smooth reduction. And then we know to characterize in terms of that quandary group.

But of course, that implication is wrong. And for example, you just take xk to be a genus 2 curve, such that xk is two elliptic curves,

two genus 1 curves, meeting at a point. Then the Jacobean of the special fiber is again an Avidean variety. It's just the product of the two elliptic curves,

the two p0, the two genus 1 curves. And that implies this. It implies this. But of course, it is not smooth. And in fact, asking that the Jacobean of xk has smooth reduction amounts to prove, well,

in that case, you know that the special fiber of the neural model, the connected component, is an extension of the p0 of the normalization

by a torus whose rank is exactly the dimension of the homology of the dual graph. So requiring that we have good reduction means that the homology is trivial. That means that the dual graph gamma of xk

is a tree, which is exactly what happens in this case, where the dual graph just consists of two irreducible components and one singular point, which is a very nice example of a tree. So to get a criterion for whether the special fiber of x

is smooth or not, we need some finer information than the one contained just in the first homology or first time homology. OK? And so this has been considered already

by Takayuki Oda, who considered exactly this problem. Oh, yes. Got to catch it.

So Takayuki Oda answered completely this question I'm addressing in the first case that we have been

considering. So when n is different from p, and here is his result. So let me fix also a smooth section, the curve.

And let me denote pi 1 of l brackets n with n. The lower central series along l,

which is a short annotation for pi 1. So the l-adic fundamental group of xk-bar with section bk-bar. So just to fix notation, p l of 1 is just pi 1 of l.

And for example, if you take pi 1 of l modulo pi 1 of l2. So this is exactly the abelianization

of the fundamental group. So it's isomorphic to the gk representation that already appeared. Misspelling, sorry.

OK, then the theorem by Takayuki Oda is the following, that my curve has good reduction in the sense that the special fiber x small k is smooth.

If and only if, if you take those representations, so for all instead it's enough for 1, prime l different from p, that's important.

The gk sets given by taking the quotients of my fundamental group by higher and higher terms

of my lower central series are un-runified. No, but you need the best one to. Sorry? Oh, yeah, OK. You have the smooth surfaces, OK. Sorry? No, excuse me. I have no other question. So these are un-runified for every n.

And in fact, it's enough for every n in between 1 and 4. So un-runified means that the inertia group acts trivially. OK?

And so let me make some comments. I hope that you can hear it. You can hear them in Japan. I don't want to write them. So there are several comments. The first comment is that in fact, Takayuki Oda proved the same result in the case where you have a family of limit surfaces

over a punctured disk, such that at the disk, they have stable reduction. And then the result is the same, where you have to replace gk with the fundamental group of the punctured disk. And you replace pi 1 of L with a usual topological fundamental group. And in fact, the way Oda proves this theorem

is by a reduction to the case of the complex numbers. OK? Because it takes your current x over OK. It deforms it to a germ of a two-dimensional surface, which is given by the powers using one variable over the vectors.

And then by inverting p, it reduces the complex numbers. It shows that the dual graph of the various things that appear are the same. So based construction, the dual graph of the generating real-world surface is the same as the dual graph gamma that we've been considering. And then using these computations

over the complex numbers, it reduces this theorem also in the reality context. OK? I want to say this because we adopt the same strategy for the p-addicts. OK? So we go from p-addicts to complex numbers. OK? And our main result is that essentially, in this situation,

a certain monodromy operator in the p-addicts setting or in the complex setting are the same, which we found very interesting. Is it clear a priori that the choice of the base point does not change the condition? Yeah. Because it computes it up to inner optomorphist, indeed.

Yes. So Ola takes care of this problem, of what happens if you change the base point as well, at least in his paper. Ola considers the problem, yes. So it doesn't depend on the point. You mean you can check a priori? Yes.

Yes. We don't. So in our setting, we fix a point and we prove everything from one point. But in this case, it doesn't matter. But actually, we didn't think about what happens if you change the point or if you don't have, for example, a point, what happens.

But if small k is algebraically closed, you have a problem. But what if not? OK. Secondly, yes? So it means that you prove it does not depend on the choice of the smooth point. Yeah, the condition does not depend. But I think he also proves that if you consider things

up to inner optomorphism, the description that you get, explicit description of the action does not depend on the point. That's what he wanted to say. Yes? Just if you start with a curve of such,

Jacobian has good reduction. And the curve not and is at least of genus 2. Then there are cyclic et al. There is a cover of the curve of order l prime 2p,

such that the Jacobian has no more good reduction. There is a defect which appears already for suitable cyclic cover.

So what we want to do is to try and to understand this theorem in the case when l is equal to p. And in that case, we are in trouble

because this is not for n bigger or equal to 3. It's not in a billion group anymore. So it's not a GQ module. So you cannot apply contents functions. So the first thing I want to do is to show you how you can replace those objects with things to which you can apply contents theorem.

And then I want to outline the strategy for proving the result. Concerning the question that I asked, so you have a morphism from GK to the morphism group of this pi 1l.

But then you can map it to the outer morphism group. And this is indeed independent up to isomorphism of the choice of this section. Exactly, yes. Now, is it enough to know that the one going to the outer is unromified or is it necessary? No, it's enough. So in Oda, Oda really computes things

in the outer morph, in the outer of this. So outer morph is modular inner outer morphism. And he has his criterion there. But he also has a result proving that if you have good reduction, then the action is unromified on this. So it's enough, the weaker condition is enough. OK.

But in our case, we don't work up to the inner of morphism. So that's why I place the result in this one.

Very efficient in this.

To replace for this quotient, so that indeed we

get QP vector space with a GK action. And the replacing is given by the following. So define n in n. So we take the group algebra defined by this group.

This has, sorry? P. Well, in general. Then in a moment, we have l equal to p. And I will drop this condition.

Oh, sorry. P of pi n. Yes. Yes. OK. If you don't like it. Yeah. That's true. Yes. All right. Sorry. So let's take n equal to p. So that was if you take QL.

Yes. So I take this group algebra defined by this group. This has an augmentation to QP, sending all the elements to one. And you take the quotient modular nth power of this augmentation idea. And this I denote as a straight EN with etaL on top.

And then the remark is that if you take E etaL to be

the limit, so it's the unipotent completion of this group algebra of E and etaL, this is a non-commutative but co-commutative of algebra,

where the quantification sends an element, a class, an element of p1p through gp sub g.

And in fact, by the theory of unipotent completion, the near-potent completion of our group, so that sits inside E etaL and coincides, in fact,

with the elements, let me call, x in E etaL such that x is a group-like element. So it satisfies this condition.

And what we want to do is to replace the analysis of the gk action. And this map actually commutes with the gk action. So we want to replace the analysis of the gk action on this to see whether it's un-ronified or not, with the analysis instead of the gk action on this thing

here. Because now each E and etaL is a gk model. So it's a qp vector space indeed with an action of gk. OK? Sorry, so you're seeking?

In the case of E etaL, you don't need to, I mean, you are worried about the topology of pi1?

I don't think so. You don't think so? I don't think so. I don't think so. You need some anchor. Are you? Yeah, now I don't, I should think about it. OK, thank you. I don't remember right now.

OK. And so another remark is that, in fact,

if you take the dual of that E etaL, defines a pro-algebraic group whose representations

in five-dimensional qp vector spaces

classify unipotent eta sheets on xk bar.

OK? And the reason we want to dualize is that that algebra is non-commutative but co-commutative. So if you dualize, you get something which is commutative

but where the hope algebra is non-commutative. So just a small question. So I think one thing associated with this model is another thing, namely the kind of Lie algebra of this group, pi 1p over pi 1pn.

So because it's, is it the case that this is kind of a unipotent algebraic group over... Yes. Well, it's open, but then you can take the Lie algebra of this. So is there a way to describe, in terms of the Lie algebra, what are you taking?

So this is a five-dimensional... Maybe you can take something smaller, but I really want to tell you that there is a deep relation between

unipotent fundamental groups, which have essentially this property, and those qk vector spaces with the gk action that I've defined. OK. OK, this part is exactly the limit of... OK? OK, now I understand.

And this algebraic group, I'll just denote this way, this algebraic group. The reason that I want to sort of shift our interest from proving things about unipotent fundamental groups,

because in this case, for example, you have a formalist, so you would have a direct way to construct it, so much easier, to, instead, something of this type, which in context where it's not clear that you have a Tanaka formalist, like in the crystalline context and so on,

instead those objects are well-defined. So we want to sort of shift our attention from those objects to their dual, those objects. OK? OK, so that's one remark.

And the second remark... Yeah, it comes later. And now the theorem that we proved is that, one, that for every n in N, the finite dimensional QP vector space

with the gk action is semi-stable, in the sense of content, in the sense of content. And second, that xk is smooth, if and only if, for every n in N,

and again, it's enough to prove it for n between 1 and 4, En et al. is crystalline, or equivalently, same condition, and the monotony on this semi-stable of En et al. is trivial.

So that's what we want to prove, that the special fiber is smooth, if and only if the monotony of such an object is 0.

So we have made a first step translation of our problem.

So the second remark that I want to make, actually let me give a definition. So I want to show you that those En et al.

that I've defined now very directly, can be described as follows. So you take a universal pointed, unipotent, et al. shift on xk bar, of length less than or equal to n, that I'm going to define now.

You take the fiber of that along the section bk bar, that gives us our En et al. So let me define, first of all, what this universal object is. So define unipotent, et al. pointed over xk bar to be the category of pairs.

So one, V is a unipotent et al. shift on xk bar.

Unipotent means that it's an iterated extension, the trivial object, which would be Qp in our case. And two, if you put a et al. shift on xk bar,

you can take the inverse image under vk bar of V, and small v, so that's just a five dimensional Qp back to space, and it takes small v to be an element in it. And morphism of pairs of that type,

or just morphisms in the category of et al. shifts with the property that the marked section on the fiber is sent to the marked section. And an object En et al. En et al.

in unipotent et al. dot xk bar of length less than or equal to n, meaning that it can be written as an iterated extension of Qp shifts which come from the base,

so they are trivial, is universal if for every v and v in unipotent et al. dot xk bar of length less than or equal to n

there exists a unique morphism from En to V,

sending small v, En et al. to V. So there are two notions of length, so there is a length in the sense of the number of copies of Qp, but you can also say that you just are interested in trivial. Exactly, that's what I said, maybe not clearly.

So here, length means that it's an iterated extension of shifts which are trivial, meaning that they come from the base, so they are trivial representations of the fundamental group of xk bar. OK, so they are of higher dimensions, OK. Yes, yes. So it's less restrictive than the other notion.

Anyway, the unit is on k, the universal is for k bar or for k bar? k bar, k bar, k bar, that's the point.

So you see there is no gk action on those objects. By universality there is a gk action on the universal object if it exists. But then it is also universal for the sinks? On xk? Well, I don't know.

I prefer to work over xk bar. I think it is because, I mean, I'm using the xk bar, I consider gk action, and by uniqueness it should also be compatible to all. So I think it is, but for us xk bar is good enough.

And now, in our case, it's not that difficult to prove that it exists as a universal object, because, well, what does it mean to give a unipotent shift? It just means to give a unipotent representation of the fundamental group.

So instead of giving the shift, I give you the representation. And the representation is what? It's exactly e and eta. e and eta, so it's associated through the representation q1p modulo i n,

which gives you, as a representation of the geometric fundamental group, using left multiplication.

So that gives an action of the fundamental group, and that gives correspondingly an eta shift. And you can prove that it is indeed unipotent in that sense. And what will be our element e and eta? So what is the fiber dk bar star of this?

Well, it's just this where you forget the action, right? Which I denote as e and eta, which was this without the action. And what is the element e and eta, which is universal?

It's just the element 1, which definitely lies in here. And indeed, you can also prove directly the universality, because if you look from the point of view of representations,

if I tell you where 1 should go to, then by using left multiplication by pi 1, you get a unique map from all these objects to this. So in that case, it's very explicit.

And the other remark that I was doing is that due to the universality, even if it's not clear that this object has a gk action, it comes indeed equipped with a gk action, which is, if you do the computation, being used by conjugation on the geometric fundamental group,

given by this section, which is the gk action that we are considering on e and eta. So we write this universality gives gk action on e and eta,

and on e and eta. So before outlining the strategy for proving our result, let me revisit the notion of semi-stability,

in a way which is more useful, more genetic for us. And for that, let me fix a uniformizer in OK.

And let me define OCRIS to be the chaotic computation of a divided power envelope, O with respect to e and the kernel of that map.

So that will be generated by some eigenstate polynomial. Now, this O has a log structure, which is defined by z. And now, I define a log to be the chaotic completion

of the log divided power envelope of what? So I'll write just the map. So on the one hand, you can do the usual construction of 10,

taking the inverse limit of OK bar with respect to Frobenius. You can take the vectors that makes a map to OK bar hat.

This is the usual map on the front end. And this is the map that I've described in this way. Here, you have a log structure defined by z. And here, you take a log structure defined by taking the time period lift of the system in F plus OK bar,

given by choosing pn roots of pi. So that has a well-defined map with two log structures. You take the log divided power envelope. So let me just write what the answer is. And this is just the usual divided power envelope of this

with respect to the kernel of the map to OK bar, which is from 10k chris. And then you have to take the chaotic completion of an element where you divide the two elements giving on the left and on the right the log structure, which is usually called U.

And it is for us, as I said, this time period lift of a choice of pn roots of pi divided by the z minus 1. OK?

And as usual, I let d log equal to a log where n goes to t. This is a ring which appeared already in the work of Kato. And the-

What is t here? Sorry? Sorry? What is t? Call the definition of t. t is full times t.

The log of the multiplicative period of gem. So writing it this way, you see that b log has the derivation

that I don't want to describe in detail with respect to z. OK? Of which decrease is given by the horizontal sections.

And Kato, for example, proved that if you take- So there is a monotony operator which is given by the residue of this connection. And this is stable from 10 is exactly the elements of b log on which this monotony operator facts importantly. OK? And what I want to tell you is that-

So there is a notion of admissibility worked out by Christoph Brei which works instead of from 10 is simply stable, this b log.

And Brei proves that two notions are equivalent. So from 10 is stability is equivalent to- So this is Brei, b log admissibility.

OK? So this b log is, in some sense, the log-crystalline cohomology of OK bar like the- Yes. We have to speak to that. Yes. Yes, absolutely. Yes. So that's why it's more useful, in a sense, from the point of view of geometry

than b semistable. So it's b log admissibility. And while here there's a minor nonsense that I don't want to get into because, contrary to this semistable, whose invariance in the GK is k0, so from 10's functor spits out k0 natural spaces,

the invariance of b log and the GK are more complicated. But since all the models that we are getting are et al, you can prove that there is a map which is Frobenius squared from the invariance to OCRIS.

Oh, sorry, I should do this. Sorry about that. It's OK, it's OK. Yeah, I realize. So instead you can really work with this ring. So what I want to say is that instead of getting k0 vector spaces,

you get modulus over OCRIS. So essentially over that O with P inverted. Right? So on a disk, in a sense. Moreover, if you take this semistable of our representation,

so b is GK representation, then this semistable of content can be recovered from d log of brai, which is defined by tensor with d log and taking invariance,

maybe doing this, modulo z. So it's the contribution at z equal to 0. And the monotony of this semistable of v

is the residue of the connection which you get from this. OK?

OK, so now we are ready to outline this strategy of the proof. And then we want to make two remarks. Can you compute the d log of d and d log?

Sorry? Can you compute the d log of d and d log? Yes. Yes, that's what we are doing. That's what I'm saying now.

Did you define d log? So d log of b is just d tensor qb with b log

and then you take GK invariance. And maybe, as I said, you want to get something manageable, so you apply Frobenius squared to get to something, or crease with b inverted.

So the strategy. So step one, as in ODA, we choose a deformation x tilde,

to the form of a spectrum, but you can also algebraize it, it doesn't matter, of such that the singularities,

so here the singularities since we have still reduction, are of the form x, equations of the form xy minus p, and here you want to replace p with z. R of the form.

So the completion of this at a singular point is of this type. That you can do. You have a stable curve, you have the deformation theory of the Linn-Manford, you can prove that you can do that.

No, but in general, the notion of stable curve that they allow is such that you can have xy is equal to the power of your new uniformizer. So I want to have a regular thing, yes, instead. He did the best change of the field. Yes. How do you make, but the stable model is unique,

so if it is not regular, you cannot... So if the original, in the original stable model, you cannot assume it's, I mean, you want to treat okay the students, it's not necessarily regular, it could be with the singularity that they said,

xy is equal to the power of the uniformizer. And then the deformation also cannot be, then the deformation will also carry this problem. I don't know if you allow,

if you can get rid of it somehow, but... In the early stable, not in the stable. There is some, of course, the terminology depends on the author, but clearly, in the sense of the modernized spaces of stable curves and so on, like you have, if you were, so in this sense that the modernized,

I mean, I'm up to the, yeah, okay, like in the Limnapfer, as we said, then you have to use this notion that you allow the... Yeah. So I have to think about this. Okay, thanks.

So step two. So step two is to prove that, and this is the key point, that this is b log admissible, but more. We know what this d log of it is,

and d log of e and eta is obtained from the fiber. So sorry, here we had a section b,

let me also deform the section b tilde, is obtained from the fiber along b tilde

of a universal, the round object of x tilde relative to o of length less than e.

So it means that if you play the same game as before you take unipotent, the round, unipotent modules with the connections, unipotent since they are extensions of the trivial shift with the usual derivation,

then there is a universal object, and actually it's fiber along b tilde, which is over o, if your base change to o crease is the same as what you get from v log. Okay, step three.

It doesn't depend on the deformation. It doesn't depend on the deformation, no. But we need a global deformation to perform this step two. Sorry, can you show us step two? So step two is... Can you just show us step two? Yes, sorry.

So each e and eta is b log admissible in the sense of drawing. And it's b log, which is an o crease human-spot module, is obtained from the fiber along the tilde, so by pulling back, a universal, the round object over x tilde.

Yeah, thank you. Thank you. Okay. So step three, base change to the complex numbers

exactly as in order, by sending such a way that these factors via o crease, so for example I send it to p times z prime. And then our x tilde will give us a family of...

Well, and here I can sort of get to a disk, so we shouldn't go too far, exchange to this, and get from x tilde a family of Riemann surfaces

s to the disk. It's a complex disk.

We need a property

that the dual graph of our Riemann surface s

at zero is equal to the dual graph of x t. Okay? Now, you base change d log with its derivation,

with its connection, you get, we said, a universal fiber along the tilde of a universal round object that's also a module with a connection, a logarithmic connection, logarithmic at zero. Okay? And what we need to prove is that,

let's step four, prove that the monotony on d log e n et al tensor over ocris

with my disk, or monotony on this base change to disk, is the battery, the battery realization,

has battery realization given by the action of the fundamental group of the disk minus the zero acting

C, fundamental group of a fiber.

And now we can use ODA, this complex analytic computation.

So can I have two minutes more? We started. OK, so let me make two comments. So for step four, we use the interpretation. This is the interpretation.

So take for the fiber long material. Because this is the interpretation which commutes with base change. So if you change this thing here to the complex numbers and then to the disk, you exactly

get something that has bad serialization. So it's some composite mesomorphism at the topological level, which is described by this action of inertia, topological inertia on that portion. And you also need something like out in approximation to get the convergence. Yes, yes. OK, you're not changing a lot.

Yes. This whole part is really in ODA. So that's why also I don't recall here how exactly it is. So this deformation and also this last step of going from this formal thing to the disk is really in ODA's paper, where it does the analytic case.

And why do you write Z going to PZ? No, it's because I want to factor by OCRIS. Because these modules, they live over OCRIS. So you have this PD envelope, where you also have divisions by powers of P that you want to get rid of. But the power series over Z prime and PZ, if you want to go with C, then you

can, the choosing parameters, Z prime and PZ prime, it doesn't seem to change anything. It doesn't change anything. Yeah, OK. Sorry. That's OK. That's OK. Yes. Yes. That's a problem. Yes. Absolutely. Thanks. OK. So the key point is step two.

OK? And so for step two, we need to prove, first of all, a theorem, which is our second theorem,

is that in the category, so you can take unipotent crystal pointed over X0 relative to OCRIS. X0 is X modulo P, not modulo pi, modulo P.

So these are unipotent extensions of the trivial isocrystal. So I also import P. OK? So universal objects, objects, ENCRIS, and ENCRIS exist.

We need a relative comparison between this ENCRIS

and EN et al. EN et al was the universal et al sheet on XK bar. Here you have a universal crystal. Because how do you get a comparison between the fibers along BK bar and the fibers along B tilde

of this universal object? So here, since I have the information, I always interpret things as modulus over X tilde with the connection. You do it by comparing the universal objects. Once the comparison is done at a relative level, you just base change this comparison

via the section which you have chosen competitively. And that will give you, on the one hand, by the comparison on the right hand side, you get this d log, and on the left hand side, you get the fiber along B tilde of ENCRIS,

which is what appears here. OK? So here is what we developed. This comparison has a morphism. This is a stable case when you have a deformation with Adrian-Yorita. And for the existence instead of these universal objects,

it's a combinatorial results with proceeding by induction on N, guarantees once you know how the xi of these groups of these elements behave for i equal to 0, 1, and 2, it tells you how actually to construct the N plus 1 step.

It's a combinatorial result that we worked out. So here, actually, there are indeed two theorems. One is about the existence and uniqueness of those universal objects, as I just said. And theorem two is this comparison result at the two levels. So it particularly tells you that this is admissible in the rapid setting,

and this is the associated isocrystal in the sense of findings, for example. I'll stop here. Sorry for being there. We're fine. Thank you. OK. So maybe we start from Tokyo. We get some questions from Tokyo, and then Bijin,

and then Piers. This is your theory. We need to find our friend. So in Canada, you have a periodic algorithm without the rule of law. So I'm wondering if you can give us

a way of comparing the periodic fundamental group with the periodic fundamental group. So comparisons between two different crimes, that's what you're saying, P and L? Yeah, yeah. So for example, if you're on topology,

so you can compare them more to your character. Yes? Yeah? Yeah. So I wonder if a similar argument is possible. This week, I don't know. The only way I know it is as I told you here. So passing to the complex numbers, not to the ladic

phase. Yeah. So I don't know the way to compare them. OK, so another question from Tokyo?

I think that in ladic cases, there is a, this theorem is also for the open curve. Yes, yes. It was a- But in any case, I think that, so because there is not much of the outer,

outer critical representation. So there should be some difficulty. So how do you think of it? No, so I agree that in the periodic setting, we cannot prove this good reduction criterion in terms of automorphism model in automorphism.

Why? Over the complex numbers, you can. OK? Yeah. But the difficult direction is to prove that, so if you know that the curve has good reduction, then you know that all these En's, which I've told you,

are crystalline. That's the easy part. And then you get the easy part. So smooth implies crystalline, you always get. So the difficult part is to prove that if those modules En are not crystalline, then the curve is not smooth. And all this result tells you something stronger

that the monodromy of science is not simply non-trivial on this En, but even up to inner automorphism. So you can use this result in that sense. But I agree that in the periodic setting, we do not have, at least for the time being,

the full strength of the order theorem against the characterization in all of the fundamental group. Yes? Thank you. One. OK. OK, that's all from torture.

Thank you. So is there any question in beginning? OK. I wonder if the bound for En is optimal in your theorem 1. Yes. That is shown already by Oda. I think in the first example that I discussed, the case where you have a genus 2 curve, which

degenerates to two elliptic curves, there it computes things explicitly. And I think it proves that even if it proves that in old. So maybe I don't know if that level is correct. But it proves that module, the first step of the lower central series, you get something trivial,

while if you go to the module 4, it does not. So its theorem is sharp in that sense. Yes. And since our theorem is a case, it's also the same thing. You mean it is sharp for the outer Ota-Morphism? For the outer. For the outer Ota-Morphism. So it's possible that for the non-outer Ota-Morphism,

you need less? It's possible. OK. So I should check his computations. OK. So then in first you have questions. So I have maybe just at the end, in fact, I did not get completely

how you conclude after your step 5. So you get to this complex case. Yes. And then you get information on S8. So how do you formally, how do you conclude? So formally you conclude the following. So the base change to this disk of this module

is the fiber of a universal, the wrong object on your family of Riemann surfaces. OK. That is associated by this better comparison to a universal representation of your philanthropic group of a fiber. And then the result is that when you take the fiber,

the representation, so also the fiber then, are related in the sense that this is the local system associated to the module with connection, the connection, which is given by the fiber of this.

And then you have to prove that the topological inertia is trivial if and only if at the level of the associated module with the local connection, the residue is 0. OK. But here we are at the complex number. So this fundamental group is what? It's the topological fundamental group.

It's the topological fundamental group. Yes. Yes. And you can't really get at the end to our? Yes. Because you're not interested on the right hand side, but on the monotony. On the monotony. Yes. OK. So I have also just one question.

So there are some comparison theorems for fundamental groups. For instance, by Shiho or by Vorigovsky. So how it compares to your work? OK. So here maybe I should say a few words about previous work. So there's certainly work of Vorigovsky, for example.

But he works instead of going from the right hand side and proving that this is the associated admissible crystal, he works on the opposite side. So he starts on the wrong side. And then he applies some relative Fontana 5 theory.

But then that means that his results, which are exactly this, work whenever the modification index is within a crystalline context, so it's less than p minus 1. And whenever this n is less or equal to p minus 1 over 2.

Once I'm working on the opposite side, I think we can sort of go beyond this point. Also, I think there is work that I should mention. So let's see.

Yeah, the work of Martin also, and in some cases, in the case of group reduction, where OK is wk. But I think he has partial results, but we know constraints about the dimensions, so non-dimension one. And here we used to have this global deformation. And there's also work by a former student of Fatim's, Majid Ajan, who did the case about fine curves.

Because in the fine curve case, the combinatorics involved in the case of this proving the existence and uniqueness of these universal objects is much easier because H2 is trivial. So some combinatorics gets easier.

But also in this case, he did it in the case, again, of group reduction. So he proved a comparison between these two whenever you have a fine curve with good reduction. OK. I think there was another question. Yeah. If you start with a curve with good reduction

and take any Galois cover of the generic fiber, which is a finite p-group, then potentially the Jacobian of the cover has also good reduction. The Jacobian. OK.

If we start, say, with a curve of genus 2, which is generated into two elliptic curves, which are supersingular, can we expect the same thing,

you see? Do you want to apply this criteria to check if? We start with a curve like this. We take a Galois cover of the finite fiber,

which is a p-group, finite p-group. We look at the three stable reductions. Is the dual graph a tree? I don't know.

Do you know? No, no. It cannot be a local statement because there are Galois covers near the ordinary double point, which introduce non-trivial graphs.

No more. You think about the semi-stable reduction or the curve upstairs, and then you take the quotient. I don't know. Good question.

Other questions? If not, let's thank the speaker.