 So it's time. So it's my great pleasure to introduce Fabry Ciordiata from Milan, who speaks on the PID criterion 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 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 the adrenovita and the menium beam. Okay, so let's fix first of all some notation. So K is a complete discrete collusion 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 very soon will be positive. And actually I will also let K is zero to be the function field of the big vectors inside K. I will also fix an algebraic closure, solution field of K to C is zero, if you forgot. So this is an algebraic closure and GK will be the Gallo 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 of information on XK? And the first step, the first simplification is to assume 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. And then the question becomes, can we decide the reduction type? Can we say if XK, so the special fiber, is smooth or not? From information, just the generic fiber. So stable is like in the Liemann form? In the Liemann form, yes. Can we reduce some of the rational components, meet the other components in three points? Not semi-stable, non-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 ambient variety over K has good reduction. And so let me recall the theorem. So for the Jacobian of XK, we know the answer, right? Because, so here's the theorem. So the Jacobian of XK has, so there's no real 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-alic coefficients, and try. So it's the dual of the first et al of 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 isn't good reduction, is equivalent to require that as a representation of GK, it's unromified from P, then it's supernaturistic. While if L is equal to P, this can never be an unromified representation, otherwise the P rank of the ambient variety would be twice the dimension, and we know it cannot. But the next approximation for L equal P to be unromified is the contents notion of crystalline representation. So the first is, for the AP curves, the well-known Neural Ogre Schaffer-Levych criteria, and then in general a theorem of silent tape. And for the second one, let me mention, well, if Jacobian of XK is reduction, then the fact that this is crystalline is work of Schaffer-Levych from 10, then the converse, that if this representation is crystalline, then the recommended reduction is due to work of many people. So there is Freud-Muckrand, Adion Yorvita, Robert Coleman, and Christophe Berry. Okay? Okay, so for that we know. But unfortunately, this is not good enough. You don't need the stable model, you just do any ambient variety model. Yes, absolutely. The ambient variety, the Neural model, and then you can see whether the special fiber of Neural model is again a ambient variety or not in terms of this. Okay, 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 Jacobian of X is an obedient scheme, which implies that the Jacobian of XK has smooth reduction, and then we know to characterize in terms of that quality group. But of course, that implication is wrong. For example, you just take XK to be a Genus II curve, such that XK is two elliptic curves, two Genus I curves, meeting any point. Then the Jacobian of the special fiber is again an obedient variety. It's just the product of the two elliptic curves, the two PX0, the two Genus II curves, and that implies this, it implies this, but of course, it is not smooth. And in fact, asking that the Jacobian 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 PX0 of the normalization by a torus whose rank is exactly the dimension of the homology of the dual graph. So requiring that we are good in action means that the homology is trivial, that means that the dual graph gamma of XK is a tree, which exactly happens in this case, where the dual graph just consists of two reducible components and one single point, which is a very nice example of a tree. Okay? So to get a criterion for the, whether the special fiber of X is smooth or not, we need to, some finer information than the one contained just in the first homology or first ethyl homology. Okay? And so this has been considered already by Takayuki Oda, who consider exactly this problem. Oh, yes. I'd love 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 b, and here is his result. So let me fix also a smooth section and let me denote pi1 of L brackets n and n, the lower central series of n, which is a short annotation for pi1, so the L-addic fundamental group of XK bar with section VK bar. So just to fix notation, pL of 1 is just pi1 of L and for example, if we take pi1 of L modulo, pi1 of L2, so this is exactly the bininization of the fundamental group, so it's as a more to the GK representation that already appeared. Okay? I'm mispronouncing, sorry. And the theorem by Takayuki Oda is the following, that my curve has good reduction in the sense that the special fiber XK is smooth. If and only if, if we take those representations, so for all instead it's enough for one, 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 and run fine. No, but you need the base point, you know. Sorry? Sorry? No, I'm still mad, no one offers. So these are runnified for every N and in fact it's enough for every N in between one and four. So what are runnified means that the inertia group acts trivially. Okay? And so let me make some comments, I hope that you can hear it, hear them in Japan if I 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 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 punctured disk and you replace pi 1 of L with the usual topological fundamental group. And in fact the way Oda proves this theorem is by a reduction to the case of the complex numbers. Okay? Because it takes your perm X over OK it deforms it to a germ of a two-dimensional surface which is given by the power series using one variable over the bit vectors. Okay? And then by inverting P it reduces the complex numbers. It shows that the dual graph of the various things that appears are the same. So where is construction? The dual graph of the generating ring-round surface is the same as the dual graph gamma that they've been considering. And then using these computations over the complex numbers it reduces this case theorem in a static context. Okay? I want to say this because we adopt the same strategy for the PIX. Okay? So we go from PIX to complex numbers. Okay? And our main result is that essentially in this situation a certain monotony operator in the PIX 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 automorphism indeed. Yes. So all that takes care of this problem of what happens if you change the base point as well, at least in its figure. All that considers the problem, yes. So it doesn't depend on the point. You mean you can check a priori? Yes. Yes. Yes. We don't. Our setting really fix a point and we prove everything from one point. But in this case, it probably doesn't depend. 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 you have, if you have a small... ah, if small k is algebraically closed, you have a point. Yes. That... what if not? Okay. Second... 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 it also proves that if you consider things up to inner automorphism, the description that you get, explicit description of the action does not depend on the point. That's what I wanted to say. Yes? Just, if you start with a curve such Jacobian as good reduction and the curve not is at least of genus 2, then there are cyclic etal cover of the curve of order L prime to P such that the Jacobian has no more good reduction. Okay. There is a defect which appears already for suitable cyclic cover. Okay. Okay. Yeah. Okay, so what we wanted to do is to try and 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 equal to 3. It's not in a billion group anymore. So it's not a GK module so you cannot apply Fontan's punctures. 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 Fontan's theorem. And then I want to outline the strategy for proving the result. Concerning the question that I asked, so you have a from GK to the automorphism of this I1F but then you can map it to the outer automorphism. And this is indeed independent of the choice of the section. Now is it enough to know that the one going to the outer is unlamified or is it necessary? No, it's not. So in order order really computes things in the outer and out of this. So automorphism is modular or inner automorphism. It has this criterion there but it also has a result proving that if you have good reduction then the action is unlamified on this. But in our case we don't work up to the inner automorphism. So that's why I praised the result in this one. So we want to replace the result so that we get Qp vector space with a GK action. And the replacing is given by the following define N in N so we take the Qp algebra the group algebra defined by this group this has Well, in general at the moment we have a equal to p and I will drop this condition sorry, it's Qp of pi N yes, yes if you don't like it yes, that's true yes right, sorry so let's take a equal to p otherwise I could take Qn yes, so I take this group algebra defined by this group this has an augmentation to Qp sending all elements to one and you take the quotient of this augmentation idea and this I denote as a straight En with ethyl on top and then the remark is that if you take ethyl to be the limit so it's the unipotent completion of this group algebra of En ethyl this is a non-commutative commutative of algebra where the commutification sends an element a class, an element of in Q1p through G tensor G and in fact by the theory of bunch of completion the nipotent completion of our group so that sits inside p ethyl and coincides in fact with the elements x in ethyl 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 Gk action so we want to replace the analysis of the Gk action on this on the wrong side or not with analysis instead of the Gk action on this, then here because now each En ethyl is a Gk model so it's a Qp vector space in the action of Gk okay sorry in ethyl you don't need to how are you are you opposed to your pylon I don't think so I don't think so no I don't think so you're in the wrong car you're in the same car now I don't I should think about it okay thank you okay okay and so one another remark is that in fact if you take the dual of that ethyl defines a broad ray group whose representations in five-dimensional Qp vector spaces classify unipotent ethyl sheets on Xk bar okay 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 whole algebra is non-commutative okay so that's a small question so I think one can associate the smaller thing namely the kind of Lie algebra of this group pi1p over pi1p and so because it's is it the case that this is kind of a unipotent ethyl algebra group over 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 of what are you taking so this is a five-dimensional yeah maybe you can take something smaller but I really want to tell you that there is a deep relation between unipotent fundamental groups which are essentially this property and those qk vector space with the gk action that I have defined okay okay this part will go with exactly the limit of okay okay now I understand good and this algebraic group I'll just denote this algebraic group the reason that I want to shift our interest from approving things about unipotent fundamental groups because in this case for example you have a tanacan formula 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 tanacan formula like in the crystalline context and so on instead those objects are well defined okay so we want to sort of shift our attention from those objects to their dual those objects okay okay so that's one remark and the second remark yeah comes later and now the theorem that the truth is that one that for every n in n the final dimensional qp vector space with a gk action is semi stable in the sense of full tan in the sense of full tan and second that xk is smooth if for every n in n and again it's enough to prove it for n between one and four e and et al is crystalline or equivalently same condition and the monogamy on the semi stable of e and et al is trivial okay so we that's what we want to prove that the special fiber is smooth preferably if the monogamy of such an object is zero so we have made the first translation of our problem so the second remark that I want to make actually let me give my definition so I want to show you that those o and e and et al that I've defined now very directly can be described as follows so you take a universal pointed unipotent et al sheaf on xk bar of length less 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 e and et al okay so let me define first of all what this universal object is okay so define unipotent et al pointed over xk bar to be the category of pairs so one v is a unipotent et al sheaf on xk bar unipotent means that it's an iterated extension of the trivial object which would be qp in our case okay and two but if you have such a unipotent et al sheaf on xk bar you can take the inverse image under bk bar of v and small v so that's just a five-dimensional qp vector space okay and it takes small v to be an element in it and morphism of pairs of that type are just morphisms in the category of et al sheafs with the property that the marked section on the fiber is sent to the marked section okay an object et al et al being unipotent et al dot xk bar of length less or equal to n meaning that it can be written as an iterated extension of qp sheafs 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 or equal to n there exists a unique morphism from et al to v sending actually it's en et al sending small v sorry yeah 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 want I said maybe not clearly so here length means that it's an iterated extension of sheafs which are trivial meaning that it comes from the base so they are trivial representations of the fundamental group okay so the higher dimension okay yes so it's less restrictive than the other notion but anyway it's an iterated on k the universal is for k bar for 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 but then it is also universal for the things on xk well I don't I don't know I prefer to work over xk bar I think it is because in xk bar it comes to the gk action and by uniqueness it should also be compatible so I think it is for us the xk bar is good enough and now in our case it is very difficult to prove that it exists as a universal object because what does it mean to give a unipotent entire sheaf it just means to give a unipotent representation of the fundamental group so instead of giving the sheaf a unipotent representation and the representation is what is exactly e and eta e and eta so it is associated to the representation qp q1p n which give you as a representation of the fundamental group using left multiplication so that is an action of the fundamental group and that is correspondingly an eta sheaf and you can prove it is 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 is just this where you forget the action which I have denoted e and eta which was this without the action and what is the element e and eta which is universal is just the element 1 which definitely lies in here and indeed I will 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 translation by left multiplication by pi1 you get a unique map from all this object to this so in that case it is very rare remark that I was doing is that due to the universality even if it is 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 on the geometric fundamental group given by the section which is the gk action we are considering on e and eta okay so we have this universality it gives gk action on e and eta and on e and eta okay so before outlining the strategy for proving our result let me revisit the notion of semi-stabilized in a way which is more genetic for us and for that let me fix a uniformizer in ok and let me fix let me define a quiz to be the reality completion of the the bio-power envelope with respect to p and the kernel of that map so that we will be generated by a semi-stabilized polynomial now this O has a log structure which is defined by z and now I define a log to be the periodic completion of the log the bio-power envelope of what so I'll write just the map so on the one hand you can do the usual construction of log 10 taking the inverse limit of ok-by-model ok-by-model aspect of Frobenius you can take the limit vectors that meets a map to ok-bar head ok this is the usual map on the front end and this is the map that I described in this way here you have a log structure defined by z and here you take a log structure defined by taking the circular lift of the system in ok-bar even by choosing p-inputs of pi ok so that has a 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 10 to increase and then you have to take the fiali 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 circular lift of a choice of p-inputs of pi divided by the z minus one ok and as usual a log equal to a log where I am going to take this is a ring which appeared many in the work of Cato and what is t here? sorry what is t? t is full time t no, the log of the multiplicative period of GM so writing in this way you see that b log has a derivation of what is described in detail with respect to z of which decrease is given by the horizontal sections and Cato for example proved that if you take so there is a monotony operator which is given by the residue of this connection and this table of fountain is exactly the elements of b log on which this monotony operator happens importantly and what I want to tell you is that so there is a notion of admissibility worked out by Christoph Breuil which works instead of fountain and Breuil proves that the two notions are equivalent so contents and instability is equivalent to, so this is Breuil so b log admissibility so this b log is something called the log crystalline chronology of Pocré-Bart like the yes, we will speak to that yes, absolutely so that's why it's more useful in a sense from the point of view of geometry yes, so it's b log admissibility and while here there's a minor noisense that I all want to get into because contrary to this table whose invariance in the GK is K0 so from then Spantos speaks out K0 matrices the invariance of b log and the GK are more complicated but since all the modules that we are getting are et al, you can prove that the map which is Frobenius square from the invariance to Ocrease oh sorry I shouldn't do this sorry about that so instead you can rework with this ring so what I want to say is that instead of getting K0 vector spaces you get modules over Ocrease so essentially over that all will be inverted right, so on a disk in a sense and moreover the if you take a this semi-stable of our representation so b is GK representation then this semi-stable of Fontaine can be recovered from the log of Braille which is defined by tensor with b log and taking invariance like this module z so it's the contribution of z equal to 0 and the monotony of this semi-stable of b is the residue of the connection which you get of this of b log of b at z equal to 0 okay so now we are ready to define the strategy of the proof and then we want to make two remarks can you compute the log of eta? yes yes, that's what we are doing that's what I'm saying now can you define the log so b log of b is just b tensor qp with b log and then you take GK invariance and maybe as I said you want to get something manageable so you apply it for v squared back to something so the strategy so step one as in order we choose a deformation x theta to the form of spectrum you can also algebraize it of such that the singularities so here the singularity since we have still reduction r of the form equations of the form xy minus p and here you want to replace p with g r of the form so the completion of this at a single point is of this type that you can do you have a stable curve and 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 uniformize so I want to have a regular thing instead he did the best charge of the field yes how do you make but the stable model is unique so if it is not regular you cannot in the original stable model you cannot assume if you want to treat all cases it's not necessarily regular it could be the singularity that they said xy is equal to the power of the uniformizer ok and then the deformation also cannot be then the deformation will also carry this problem I don't know if you allow you can get rid of it somehow it's really stable there is some terminology depends on the author but clearly in the sense of the modular spaces of stable curves and so on like you have if you were so in this sense that modular I mean I'm up to the like in the limbo as we said then you have to use this notion that you yeah so I have to think about this so step 2 so step 2 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 D and et al is obtained from the fiber so sorry here we had a section D let me also deform the section d theta is obtained from the fiber along d theta of a universal VRAM object x theta relative of length so it means that if you play the same game as before you take unipotent VRAM unipotent modules with the connections unipotent since there are extensions of the trivial shift with a usual derivation then there is a universal object and actually it's fiber along d theta which shifts over all if your base changes to okris it's the same as what you get from D log step 3 it doesn't depend on the deformation it doesn't depend on the deformation but we need a global deformation to perform this step 2 sorry can you show us the step 2 so step 2 is can you just show us the step 2 yes so each en et al is D log admissible in the sense of drawing and it's D log which is a concrete human smart module is obtained from the fiber along d theta so by pulling back a universal VRAM object over x theta thank you so step 3 base change to the complex numbers exactly as in order by sending saturated these factors via okris so for example I send it to p times z prime ok and then our x theta will give us a family well and here I can get to a disk we shouldn't go too far let's change to this and get from x theta a family of Riemann surfaces s to the disk it's a complex disk to meet the probability the dual graph of our Riemann surface s at 0 is equal to the dual graph ok now you base change D log with its derivation with its connection you get we said a universal the fiber along d theta of a universal VRAM object that's also a module with a connection a logarithmic connection a logarithmic at 0 ok and what we need to prove let's step 4 prove that the monotony in et al tensor over okris with my disk or monotony on this base change to disk is the the materialization has materialization given by the action of the fundamental group of the disk minus the 0 acting on c fundamental group of fiber ok and now we can use order this complexity computation so we can have 2 minutes more we started ok so let me make 2 comments so for step 4 we use this interpretation 3 for the fiber along d theta because this is the interpretation which commutes with base change ok so if you chase this thing here to the complex numbers and then to the disk you exactly get something that has bad materialization so with some comparison as a morphism at 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 the convergence yes yes yes self changes a lot this whole part is really in order so that's why also I don't record 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 all this paper where it does the ladder case and why do you write z going to p z no just because I want to factor by a okris because these modules deliver okris so you have this p d envelope which you also have divisions by powers of p that you want to get rid of but the power series over z prime and if you want to oversee then you can the choosing parameter z prime or p z prime doesn't seem to change anything yes ok sorry that's ok sorry yes absolutely ok so the key point is step 2 ok so for step 2 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 x zero relative to okris x zero is x modulo p not modulo pi modulo p so these are unipotent extensions of the trivial isocrystal so I also refer to p so universal objects objects e and crease and e and crease exist we need a relative comparison between this e and crease and e and etal etal was the universal etal sheet on x k bar here you have a universal crystal because how do you get a comparison between the fibers long b k bar and the fibers long b tilde of this universal object so here since I have the formation I use interpret things as modules over between 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 I have chosen compatibility and that will give you on the one hand via the comparison on the left on the right hand side you get this debug and on the left hand side you get the fiber long b tilde of e and crease which is what appears here ok so here is what we developed this comparison as a morphism in the semi-stable case when you have a deformation with adrenovita and for the existence instead of these universal objects it's a combinatorical result with proceeding by induction on n guarantees once you know how the x i 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 combinatorical result that we worked out so here actually there are indeed 3 2 theorems a uniqueness of those universal objects as I just said and then 2 is this comparison result at the 2 levels ok so it particularly tells you that this is admissible in that it's setting and this is the associated iso-crystal in the sense of faultings for example I'll stop here, sorry for being there over time ok so maybe we start from Tokyo we get there are some questions from Tokyo and Beijing and that does such a period what I said so so in the end I have a period of a lot of result of all of that so I want to inform that this period result from comparing the period of fundamental group with the era of fundamental group so comparisons between 2 different crimes that's what they're saying, P and L yeah so for example on topology so you can compare them yes so I'm going to make a similar argument if possible this week I don't know the only way I know it is as I told you here so pass into the complex numbers not to the eladic please yeah so I don't know if we can compare them are there some other questions from Tokyo? I think I think that in any cases there are these 7 is too also for the open gap but in in the eladic case I think that that that so because there is no much of the outer greater representation so there should be some difficulty so how do you think of me? no so I agree that in the eladic setting we cannot prove this good reduction criterion in terms of automorphism, modeling and morphism why all the complex numbers you can but the difficult direction is to prove that so if you know that the curve has good reduction then you know that the oldest ends 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 then the difficult part proves that if those modules are not crystalline then the curve is not smooth but all this result tells you something stronger that the monodrominal science is not simply non-trivial on this en but even up to in an automorphism so you can use this result in that sense but I agree that in the eladic setting we do not have at least for the time being the full strength of all the theorem it has the characterization in all of the fundamental group yes thank you what ok ok that's all from Tokyo thank you so is there any question in begin ok I wonder if the bound for N is optimal in our 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 2 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 sensual series you get something trivial while if you go to the module 4 it does not so this theorem is sharp in that sense yes and since our theorem is it's also the same thing it is sharp for the outer for the outer so it's possible that for the non-out for the left it's possible to check its computations ok so then in the first you have questions 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 SA so how you formally how you conclude so formally you conclude the following to this disk of this module is the fiber of a universal the wrong object on your family of real surfaces that is associated by this bent comparison to a universal representation of your fundamental group of a fiber ok 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 which is given by the fiber of this ok 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 connected connection the residue ok but here are the complex numbers so this fundamental group is what it's the topological it's a topological fundamental group yes and you can really get at the end to our yes because you are not interested on the right hand side but on the monomony on the monomony yes ok so I have also just one question so there are some theorems for fundamental groups for instance by Shiho or by Vologotsky so how it compares to your work ok so here really I should say a few words about previous work ok so there is certainly work of Vologotsky for example but he works instead of going from the right hand side and proving that this is the associated crystal he works on the opposite side so he starts on the round side and then he applies some relative Fontana 5 theory but then that means that his results which are exactly this work whenever the unification index is within a crystalline context so it's less than p-1 and whenever this n is less or equal to p-1 over 2 ok why instead of 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 Olson in the case of proof deduction when ok is wk but I think he has partial results but we know it's straight about the dimension so no dimension 1 which here we used to have this global deformation and there is also work by a former student of Falkings Majid Ajahn who did the case of a fine curve 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 3 there ok so some combinatorics how this gets easier but also in this case it worked he did it in the case I think again of good reduction so he proved the 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 yes and take any Galois curve of the joint fiber which is a finite p-hole yes then potentially the Jacobian of the curve has also good reduction the Jacobian if we start say with a curve of genus 2 which is generating two elliptic curves which are super singular can we expect the same thing you see you want to apply this criterion to check if we start with a curve like this we take a Galois curve of the joint fiber which is a p-hole finite p-hole we look at the three stable reduction is the georgraph 3 I don't know do you know no it cannot be a local statement because there are Galois curves near the ordinary double point which introduce non-trivial graphs I don't know you think about the semi-stable reduction of the curve upstairs I don't think the question I don't know no question other questions if not let's thank the speaker