Strongly minimal groups interpretable in o-minimal expansions of fields
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Hasson, Assaf
We prove that if D=(G,+,\dots) is a strongly minimal non-locally modular group interpretable in an o-minimal expansion of a field and dim(G)=2 then D interprets an algebraically closed field K and D (as a structure) an algebraic group over K with all the induced K-structure. I will discuss some key aspects of the proof that may be of interest on their own right. Joint work with Y. Peterzile and P. Eleftheriou.
 I've mentioned that this is, I've attended all four neostability meetings and I remember distinctly speaking in three and I mean this is the third one and each time I change the title of my talk so I'm just keeping up the tradition. So I'm going to talk about joint work with Kobe and Pantelis and it has to do with Zilber's trichotomy, the restricted Zilber's trichotomy of Zilber's and I think that it was in 2006 that Kobe suggested the following conjecture, strongly minimal structure, interpretable in an O minimal field say is either locally modular or interprets an algebraically closed field. So I've been working on variants of this conjecture for many years and there are a couple of comments that should be mentioned almost immediately so one comment is, so comments. The first comment is that by results of Kobe and Sergei if a field, if a stable field is interpretable in an O minimal structure then it is pure. So at least this part of Zilber's conjecture is out of the way if you restrict yourself only to structures interpretable in O minimal fields or O minimal structures. Yeah well whatever I mean it doesn't really matter. Sorry well if the structure is locally modular it won't interpret the field. Now the second remark which is more relevant to the approach to the problem is that by all results of Kobe and Charlie if an ACF is interpretable in an O minimal structure it is two-dimensional. So this means that the conjecture really splits in two there's the two-dimensional case where you expect to get a positive result where you expect to actually have a local a non-locally modular structure which interprets a field and two other dimensions where you expect not to have strongly minimal non-locally modular structure is interpretable. So it's in a way a vacuous instance of the conjecture the result is a non-existent would be a non-existent result. Now so there's a theorem of Kobe myself and the and ALF which says that the conjecture is true in the one-dimensional case. So if you have a strongly minimal structure that's interpretable in on the one-dimensional O minimal structure it's locally modular. Okay now the four-dimension for dimension d greater than 2 this is open even for for ACF 0. So note that any structure interpretable in ACF 0 is interpretable in the reals right so the conjecture so the analogous conjecture for locally for strongly minimal structures interpretable in ACF 0 is a special case of this conjecture and the case of dimension greater than 2 is totally open even in ACF 0. I've been thinking about this for a long time and that and even good points to start thinking about this. There are some ideas but they are not very not developed. Now how do you go about proving instances of such conjectures so so we restrict to the two-dimensional case and how do you go about proving such things then usually I mean if you go to the heart of essentially every proof of instances of Zilberst's trichotomy then it's done in two steps. The first step is you construct a group you show that such a look it is not known so I said that ACF 0 is interpretable is interpretable in in the reals. So if you have a strongly minimal structures interpretable in ACF 0 it's interpretable in the reals so this is a special instance so the conjecture for strongly minimal structures interpretable in ACF 0 is a special instance of this conjecture. This is a generalization of the conjecture on ACF 0. Oh minimal dimension. Well of course it's a risky dimension I mean molly rank whatever. So the way you usually prove such theorems is you first prove that there is a group interpretable in the structure and then you prove it and then you prove that there's a field interpretable using the group structure. But in practice it quite often goes the other way around because you have no idea how to produce the group because you don't understand well enough the situation so you first study what's going on when you have a group and then you produce a field and from that you hope to gain enough insight into the situation that will allow you to construct a group. So that's what we do in our theorem so the theorem that we prove is the following let g let's write it this way. B two-dimensional group definable in an O minimal expansion of a field or n. Let D be g plus and some other structure be strongly minimal and not locally minimal. Yeah all let's say interpretable in 10. Then there exists D interpretable algebraically closed field k and an algebraic group h over k over k in k such that D is isomorphic to h with all its k-induced structure. So what we're really saying is that g is an algebraic group with all the induced structure coming from the field k. So this is in a way the best possible result you could hope. Is that enough? Okay. Pure algebraic group when algebraic group is interpreted as an algebraic group with all its geometric or algebraic structure. Yeah. Yeah precisely. Sorry. D this structure. So I have to distinguish somehow between the group and the structure that in which it lives. No. No it will not be. Yeah no. Automatically it will be. Yeah. Automatically it will be the same. It's a it does yeah. It is definable over D. Yeah I guess it is. Yeah it must be. The isomorphism is definable over D. Indeed. Yeah indeed. So those structures are really by interpretable. The g in the field. What about parameters? Just to be sure. The right hand side. This h. The right interpretability if I'm not mistaken is with out parameters. Out parameters. But what we consider D is as interpretable in n. Well we can add parameters in n. So nk is a pure field. So what is the problem? Well that this is that in this sense. Yeah well h may be expanded by constants. I don't know. I mean we don't we never traced. We were not interested in constants and we were very liberal with uses of constants and in the proof. So well I mean it's a legitimate question. Asking whether you what parameters are needed to interpret everything. But sort of it was difficult enough as it was. And the conclusion is very strong because it says precisely what structure on this kind of part. Yeah I agree. Now note that for arbitrary strongly minimal sets we cannot hope for such a good resort because the examples of Boris of those non-algebraic risky geometries which are interpretable in n. Structures. So you cannot hope that if you do not assume the group structure to begin with then what you end up with is purely algebraic. So this is a phenomenon that happens only because we have the group. Okay. So I want to dedicate the rest of the talk for describing parts of the proof. So how do you approach this problem? So idea of the proof. Well there are quite a few of them. So I would focus on the first part of the proof really. So what do we have? How do you approach such a problem? Then ultimately if you go to essentially every proof where the Zilberst-Drake-Athom is proved and it boils down to defining tangency in your structure up to algebraicity. Defining an equivalence relation or an algebraic relation which causes tangency on plane curves. So that's what we're going to do. And what tools do we have? So the idea is, as always, and this goes back to Zilber and Janjera Vinović, that if you have, if your curves were, so we work in G square and when I say plane curves I mean the definable subsets of G square that are of rank one. Okay. So plane curve for me are just one dimension, one morally rank one subsets of G square that are definable in D. So if your curves were algebraic and you had two tangent curves, so of course I'm drawing a real geometric picture. The picture, complex picture is does not look like that. But essentially the idea is that if you had here a tangency, a point of tangency, then by Rousses theorem or the argument principle or something like that from complex analysis, this is a double intersection point. So if you perturb a little bit your curves, then here you will get, if you get, if you have, if you perturb the curve a little bit, then here in a small neighborhood of your point you will get two intersection points because this was a double intersection point and here you will not lose intersection points in a small neighborhood of the original intersection, of the other intersection points, essentially because again of Rousses theorem or something like that. But of course we don't have, so this is our aim. Our aim is to somehow show that if you have two curves which are tangent, then if you count the number of actual geometric intersection points without multiplicity, the intersection points that you actually see in the structure, then you will get a smaller number of intersection points. That's essentially the only tool that we have to work in this level, to work within this level of generality. Now of course our curves are not assumed to begin with to be analytic, so we don't have Rousses theorem, we don't have the argument principle, we don't have anything like that. Okay, so the problem breaks into two really. First we have to show that if we have a tangency point, regardless of what that means, if we have a tangency point then if we perturb our curves we gain intersection points. This is first, a first part of the problem and the second part of the problem is to show that if, that again if we perturb the one of the curves, then we do not lose intersection points next to, in small neighborhoods of other intersection points. So we somehow have to show that if we have tangency, then perturbing increases the number of intersection points next to the intersection point, but does not lose intersection points in other places. Now if, again, if you change the argument a little bit for the for the analytic case, then what you can say is that you can use instead of Fousche's theorem or the argument principle you can use Fousche? Sorry, sorry. This theorem says that essentially the number of, that if you perturb a little bit an analytic function then the number of solutions to an, if you look at F of X equals zero and then you take a small w, w close to zero and you look at the solutions to X of X equal w, then the number of solutions counted with multiplicity does not change in small neighborhood. It's essentially right the degrees invariant under, the topological degrees invariant under homotopy. This is essentially the essence of the theorem. So if you, if you replace, if you don't want to work with those analytic theorems and you want to work with a more topological theorem, then what goes on here is really that those functions, if think of those as, as analytic functions, ignore the fact that these are curves really, but locally you can think of them as analytic functions, then analytic functions are open and the difference between two analytic functions is also an analytic function, so it is also open. So if you have a point of intersection here and you perturb it a little bit, then openness assures that you don't lose intersection points. So at least solving this problem amounts to showing essentially that locally our curves are graphs of open functions. If we can show that locally our function, sorry, our curves are graph of open functions, then at least the first part, the first problem of the problems of keeping track of intersection, of intersection points after perturbation does not, is more or less solved. Okay, then there's the problem which is, I have the paper roughly of showing that if you have a tangency, then you actually gain intersection points, but that's a different problem. Okay, so I want to talk about how to approach the problem of showing that those are really open maps, that what we have are open maps. So, so first goal show that generic curves, that curves, that plane curves are locally graphs of continuous open maps. Well, so if we were in a compact setting, then to show that, then to show this, it would be enough to show that plane curves are closed. Right? So that's the first step. So the first step, not necessarily just that it has O minimal dimension too. Oh, as a set, okay. As a set, as a set, the universe of these two-dimensional, in the sense of the O of the underlying O. In G square, yeah. Yeah, so we assume throughout that you can, you can always, this is a group interpretable in the O minimal, definable in the O minimal structure. So you can embed it as a closed subgroup of n to the k for some k. And the topology is the affine topology. No, so plane curves are molly rank one, molly rank one, D definable subsets of G square. So plane curves, molly rank one, D definable subsets of G square. So, any questions? So this continuity openness that's in the O minimal topology? Yeah. Yeah. So the first theorem is if S is a plane curve, then S, the frontier of S, which is the closure of S in the O minimal topology, those points in the closure which are not in S is finite. And moreover, in, well, this is the canonical base of S. So this is contained in an algebraic, if you want the, yeah, so those points are algebraic in the sense of D over S. S knows about the, its frontier points. Yeah. Of the, canonical parameter. Yeah. Yeah. Sorry. A code for S. Yeah, if you want. Okay. Now here I want to say that there's an issue. Here already this is really not obvious because consider the following example. Take your group to be the additive group, but now what you do is take C and split it into three stripes and take any definable bijection in the O minimal structure which just switches those two bits and push all the structure, push forward all the structure you have in D through this bijection. Now this will just not happen. So this will be continuous with respect to the O minimal topology. Right? Because we have here two large sets where things are obviously not continuous with respect to the O minimal topology. So we have to be working with the group topology. So we've embedded G into an affine space where the topology, the induced topology is the group topology but if we don't assume this then we have to assume that we're working with the group topology and here there's a, here's a question. You want to prove this theorem, you want to prove this conjecture without assuming that there is a group. So you have to find a topology on G, on your strongly minimal set, call it whatever you want, that will make this true. How do we find such a topology? So my guess is, and it's not completely outrageous, my guess is that if you have a strongly minimal structure that is not locally modular, interpretable in an O minimal field, then there exists a unique topology on, definable topology on that strongly minimal structure which makes this statement true. Coming from the O minimal structure and you have a definable basis, I mean if you want to manifold topology. And I think that in general this is an interesting question, I mean if you have a strongly minimal set in a general topological setting and you want to exploit this topological setting, you cannot assume that you're just working with the affine topology, you somehow have to find a manifold topology on that structure to make things work and how to find it is I think an interesting problem. So how to prove this, so idea, so suppose that you have your curve and this is B, your frontier point, then what you do is, you take a generic curve, you fix some large family of plane curves, non-local modularity provides you with a large definable family of plane curve, you take a curve, a generic curve passing through that point. Now assume that this curves comes from a family of more than degree one, so that it has a unique generic type. So there's a unique generic intersection number of curves from that family with our target curve S. Right, now the idea is that if you move your curve a little bit, then if all is well, if here the intersection is transverse of those two curves, if the intersection is transversely and you move a little bit your curve here, then here you will not lose intersection points because of the assumption on the intersection that it is transverse and here you will gain a new intersection point. So the number of intersection points of a generic curve passing through that point will not be the number of intersection points of a generic curve of the family with this target curve. Okay, is that clear? This is a frontier point, right, so it's not, so this is not an intersection point of this curve with S. Now because it's in the frontier, you can perturb it as little as you want and it will gain a new intersection point in S. This intersection point did not occur in the original curve we started with. Now if you can assure that the curve you started with intersects S transversely everywhere else, then the perturbation will not lose intersection points because only tangency can allow you to lose intersection points and therefore the generic curve passing through here will have more intersection points with S than a generic curve passing through here. Now we can assure that if B is generic in G square, we don't care about its genericity in S. If B is generic in G square, then this always happens that all the intersection points are generic, are transverse. But the problem is who assures you that the B is generic in G square? So the idea is because we have a group we can move around our curve and put B in such a situation where it is generic in G square and then the intersection will behave well. Now this is essentially the only point where we are using the group topology in this argument. So it's a very, very delicate point and I find it very, very curious that there's an extremely delicate point where if you use the group then everything works, if you don't use the group then obviously this is wrong. Not the group topology, the group operation, the group topology you use? Well you need to use the group topology in order to know that the translation is continuous, a homomorphism and therefore it maps from tear points to frontier points and stuff like that. So this is a curious point, this is something to look deeper into and understand what's going on because if you want to prove the theorem without assuming a group then you will have to find something like that and this is where things could go wrong. By the way in this goal are you showing that algebraic closure is definable closure? Why would it be? It's locally a graph, locally in the sense, what do you mean locally? Locally in the sense of the ambient or minimal topology. Oh, so it's not implying that. No, no, no, not at all. Not even. What do you mean? Just look topology graph around the point. Yeah. It's, we prove something stronger but I don't want to go into the details. So as I said if we were in a compact situation then this would imply that also the openness result but if we're not in a compact situation then things become delicate because what the situation, think of the following example. So consider G plus, sorry, C plus and F where F of x, y is x and 0 if y is equal to 0 and x, y, 1 over y if y is not equal to 0. Then this, the graph of this function, the graph of this function is closed in C square but it's very non-continuous and I'd be very glad if anyone could give me an easy proof showing that this is not strongly minimal. I mean of course by our theorem we know it's not strongly minimal but proving that this exact specific example is not strongly minimal is not so trivial I think and what happens here is that we have infinitely many poles in this function and the next theorem that we prove is that this set S cannot have infinitely many poles. It cannot have infinitely many points where if you tend to a point in G then the curve goes to something that has no limit in it. This follows immediately, I mean the fact that this is the only obstacle for S being open follows immediately from this theorem. So the next theorem is S has finitely many poles so since we assume that our group is embedded in NK with the ambient topology then this just means that there's a neighborhood of the point that S is unbounded. The pole is just a point where and this is an extremely hard argument, sophisticated geometric argument and I think that when we were working on it we were definitely thinking that there must be something simpler but even for that example it is not that easy to find an argument showing why this is not strongly minimal. So maybe if someone can analyze examples like that and understand them we can come up with a nicer argument showing that this. So a pole, so A, so definition assuming G subset N to the K with the ambient topology A in G is a pole of S if for all open U containing A U times G intersected with S is unbounded. So it's really what you would think a pole should be. So now it's an immediate corollary of this argument that if you only have finitely many poles then S is in fact an open relation which means that it maps open sets to open sets as in the same sense as F is an open function. So this is really the essence of the problem of proving that S is open and this gives us if you want a starting point for what comes next because as an immediate corollary of that we can get for example that if you have a function that is smooth at the identity then it's determinant Jacobian has constant sign in a small neighborhood of the identity and this is what this is the engine that allows us to understand how the differential, the Jacobians of these local functions if you want look like and this is really the machine that allows us to conclude how things really look. But this is a whole different part of the paper so I guess it's a good place to stop.
 I've mentioned that this is, I've attended all four  neostability meetings and I remember distinctly speaking in three and I mean  this is the third one and each time I change the title of my talk so I'm just  keeping up the tradition. So I'm going to talk about joint work with Kobe and  Pantelis and it has to do with Zilber's trichotomy, the restricted Zilber's  trichotomy of Zilber's and I think that it was in 2006 that Kobe suggested  the following conjecture, strongly minimal structure, interpretable in an O  minimal field say is either locally modular or interprets an algebraically  closed field. So I've been working on variants of this conjecture for many  years and there are a couple of comments that should be mentioned almost  immediately so one comment is, so comments. The first comment is that by  results of Kobe and Sergei if a field, if a stable field is interpretable in an  O minimal structure then it is pure. So at least this part of Zilber's conjecture  is out of the way if you restrict yourself only to structures  interpretable in O minimal fields or O minimal structures. Yeah well whatever I  mean it doesn't really matter. Sorry well if the structure is locally modular it  won't interpret the field. Now the second remark which is more relevant to  the approach to the problem is that by all results of Kobe and Charlie if an  ACF is interpretable in an O minimal structure it is two-dimensional. So this  means that the conjecture really splits in two there's the two-dimensional case  where you expect to get a positive result where you expect to actually have  a local a non-locally modular structure which interprets a field and two other  dimensions where you expect not to have strongly minimal non-locally modular  structure is interpretable. So it's in a way a vacuous instance of the conjecture  the result is a non-existent would be a non-existent result. Now so there's a  theorem of Kobe myself and the and ALF which says that the conjecture is true  in the one-dimensional case. So if you have a strongly minimal structure that's  interpretable in on the one-dimensional O minimal structure it's locally modular.  Okay now the four-dimension for dimension d greater than 2 this is open even for  for ACF 0. So note that any structure interpretable in ACF 0 is interpretable  in the reals right so the conjecture so the analogous conjecture for  locally for strongly minimal structures interpretable in ACF 0 is a special case  of this conjecture and the case of dimension greater than 2 is totally  open even in ACF 0. I've been thinking about this for a long time and  that and even good points to start thinking about this. There are some  ideas but they are not very not developed. Now how do you go about proving  instances of such conjectures so so we restrict to the two-dimensional case and  how do you go about proving such things then usually I mean if you go to the  heart of essentially every proof of instances of Zilberst's trichotomy then  it's done in two steps. The first step is you construct a group you show that such  a look it is not known so I said that ACF 0 is interpretable is interpretable in  in the reals. So if you have a strongly minimal structures interpretable in ACF  0 it's interpretable in the reals so this is a special instance so the  conjecture for strongly minimal structures interpretable in ACF 0 is a  special instance of this conjecture. This is a generalization of the conjecture on  ACF 0. Oh minimal dimension. Well of course it's a risky dimension I mean  molly rank whatever. So the way you usually prove such  theorems is you first prove that there is a group interpretable in the structure  and then you prove it and then you prove that there's a field interpretable using  the group structure. But in practice it quite often goes the other way around  because you have no idea how to produce the group because you don't understand  well enough the situation so you first study what's going on when you have a  group and then you produce a field and from that you hope to gain enough insight  into the situation that will allow you to construct a group. So that's what we do  in our theorem so the theorem that we prove is the following let g let's write  it this way. B two-dimensional group definable in an O minimal expansion of a  field or n. Let D be g plus and some other structure be strongly minimal and  not locally minimal. Yeah all let's say interpretable in 10.  Then there exists  D interpretable algebraically closed field k and an algebraic group h over k  over k in k such that D is isomorphic to h with all its k-induced structure.  So what we're really saying is that g is an algebraic group with all the  induced structure coming from the field k. So this is in a way the best  possible result you could hope. Is that enough? Okay. Pure algebraic group when  algebraic group is interpreted as an algebraic group with all its  geometric or algebraic structure. Yeah. Yeah precisely. Sorry. D this structure.  So I have to distinguish somehow between the group and the structure that in  which it lives. No. No it will not be. Yeah no. Automatically it will be. Yeah.  Automatically it will be the same. It's a it does yeah. It is definable over D.  Yeah I guess it is. Yeah it must be. The isomorphism is definable over D.  Indeed. Yeah indeed. So those structures are really by interpretable. The g in  the field. What about parameters? Just to be sure. The right hand side.  This h. The right interpretability if I'm not mistaken is with out parameters. Out parameters. But what we consider D is as interpretable in n.  Well we can add parameters in n. So nk is a pure field. So what is the problem?  Well that this is that in this sense. Yeah well h may be expanded by constants.  I don't know. I mean we don't we never traced. We were not interested in  constants and we were very liberal with uses of constants and in the proof. So  well I mean it's a legitimate question. Asking whether you what parameters are  needed to interpret everything. But sort of it was difficult enough as it was.  And the conclusion is very strong because it says precisely what structure on this  kind of part. Yeah I agree. Now note that for arbitrary strongly minimal sets we  cannot hope for such a good resort because the examples of Boris of those  non-algebraic risky geometries which are interpretable in  n. Structures. So you cannot hope that if you do not assume the group structure  to begin with then what you end up with is purely algebraic. So this is a  phenomenon that happens only because we have the group. Okay. So I want to dedicate  the rest of the talk for describing parts of the proof. So how do you approach  this problem? So idea of the proof. Well there are quite a few of them. So I would  focus on the first part of the proof really. So what do we  have? How do you approach such a problem? Then ultimately if you go to  essentially every proof where the Zilberst-Drake-Athom is proved and it  boils down to defining tangency in your structure up to algebraicity.  Defining an equivalence relation or an algebraic relation which  causes tangency on plane curves. So that's what we're going to do. And what tools do we have?  So the idea is, as always, and this goes back to Zilber and Janjera Vinović,  that if you have, if your curves were, so we work in G square and when I say  plane curves I mean the definable subsets of G square that are of rank one.  Okay. So plane curve for me are just one dimension, one morally rank one subsets  of G square that are definable in D. So if your curves were algebraic and you had  two tangent curves, so of course I'm drawing a real geometric picture. The  picture, complex picture is does not look like that. But essentially the idea is  that if you had here a tangency, a point of tangency, then by Rousses theorem or  the argument principle or something like that from complex analysis, this is a  double intersection point. So if you perturb a little bit your curves, then  here you will get, if you get, if you have, if you perturb the curve a little bit,  then here in a small neighborhood of your point you will get two intersection  points because this was a double intersection point and here you will  not lose intersection points in a small neighborhood of the original  intersection, of the other intersection points, essentially because again of  Rousses theorem or something like that. But of course we don't have, so this is  our aim. Our aim is to somehow show that if you have two curves which are  tangent, then if you count the number of actual geometric intersection points  without multiplicity, the intersection points that you actually see in the  structure, then you will get a smaller number of intersection points. That's  essentially the only tool that we have to work in this level, to work within this  level of generality. Now of course our curves are not assumed to begin with to  be analytic, so we don't have Rousses theorem, we don't have the argument  principle, we don't have anything like that. Okay, so the problem breaks into two  really. First we have to show that if we have a tangency point, regardless of what  that means, if we have a tangency point then if we perturb our curves we gain  intersection points. This is first, a first part of the problem and the second  part of the problem is to show that if, that again if we perturb the one of the  curves, then we do not lose intersection points next to, in small  neighborhoods of other intersection points. So we somehow have to show that if  we have tangency, then perturbing increases the number of intersection  points next to the intersection point, but does not lose intersection points in  other places. Now if, again, if you change the argument a little bit for the  for the analytic case, then what you can say is that you can use instead of  Fousche's theorem or the argument principle you can use  Fousche?  Sorry, sorry. This theorem says that essentially the number of, that if you  perturb a little bit an analytic function then the number of solutions to  an, if you look at F of X equals zero and then you take a small w, w close to  zero and you look at the solutions to X of X equal w, then the number of solutions  counted with multiplicity does not change in small neighborhood. It's  essentially right the degrees invariant under, the topological degrees  invariant under homotopy. This is essentially the essence of the theorem.  So if you, if you replace, if you don't want to work with those analytic  theorems and you want to work with a more topological theorem, then what goes on  here is really that those functions, if think of those as, as analytic functions,  ignore the fact that these are curves really, but locally you can think of  them as analytic functions, then analytic functions are open and the  difference between two analytic functions is also an analytic function, so it is  also open. So if you have a point of intersection here and you perturb it a  little bit, then openness assures that you don't lose intersection points. So at  least solving this problem amounts to showing essentially that locally our  curves are graphs of open functions. If we can show that locally our function,  sorry, our curves are graph of open functions, then at least the first  part, the first problem of the problems of keeping track of  intersection, of intersection points after perturbation does not, is more or  less solved. Okay, then there's the problem which is, I have the paper roughly  of showing that if you have a tangency, then you actually gain intersection  points, but that's a different problem. Okay, so I want to talk about how to  approach the problem of showing that those are really open maps, that what we  have are open maps. So, so first goal  show that generic curves, that curves, that plane curves are locally graphs of  continuous open maps. Well, so if we were in a compact setting, then to show that,  then to show this, it would be enough to show that plane curves are closed.  Right? So that's the first step. So the first step, not necessarily just that it  has O minimal dimension too. Oh, as a set, okay. As a set, as a set, the universe of  these two-dimensional, in the sense of the O of the underlying O.  In G square, yeah. Yeah, so we assume throughout that you can, you can always,  this is a group interpretable in the O minimal,  definable in the O minimal structure. So you can embed it as a closed  subgroup of n to the k for some k. And the topology is the affine topology.  No, so plane curves are molly rank one, molly rank one,  D definable subsets of G square. So plane curves, molly rank one, D definable subsets  of G square. So, any questions?  So this continuity openness that's in the O minimal topology?  Yeah. Yeah. So the first theorem is if S is a  plane curve, then S, the frontier of S, which is the closure of S in the O minimal topology,  those points in the closure which are not in S is finite.  And moreover, in, well, this is the canonical base of S.  So this is contained in an algebraic, if you want the, yeah, so those points are algebraic  in the sense of D over S. S knows about the, its frontier points.  Yeah. Of the, canonical parameter. Yeah. Yeah. Sorry.  A code for S. Yeah, if you want.  Okay. Now here I want to say that there's an issue.  Here already this is really not obvious because consider the following example.  Take your group to be the additive group, but now what you do is take C and split it  into three stripes and take any definable bijection in the O minimal structure which  just switches those two bits and push all the structure, push forward all the structure  you have in D through this bijection. Now this will just not happen.  So this will be continuous with respect to the O minimal topology.  Right? Because we have here two large sets where things are obviously not continuous  with respect to the O minimal topology. So we have to be working with the group topology.  So we've embedded G into an affine space where the topology, the induced topology is the group  topology but if we don't assume this then we have to assume that we're working with  the group topology and here there's a, here's a question.  You want to prove this theorem, you want to prove this conjecture without assuming that  there is a group. So you have to find a topology on G, on your strongly minimal set, call it  whatever you want, that will make this true. How do we find such a topology?  So my guess is, and it's not completely outrageous, my guess is that if you have a strongly minimal  structure that is not locally modular, interpretable in an O minimal field, then there exists a  unique topology on, definable topology on that strongly minimal structure which makes this  statement true.  Coming from the O minimal structure and you have a definable basis, I mean if you want  to manifold topology. And I think that in general this is an interesting question, I  mean if you have a strongly minimal set in a general topological setting and you want  to exploit this topological setting, you cannot assume that you're just working with the affine  topology, you somehow have to find a manifold topology on that structure to make things  work and how to find it is I think an interesting problem.  So how to prove this, so idea, so suppose that you have your curve and this is B, your  frontier point, then what you do is, you take a generic curve, you fix some large family  of plane curves, non-local modularity provides you with a large definable family of plane  curve, you take a curve, a generic curve passing through that point.  Now assume that this curves comes from a family of more than degree one, so that it has a  unique generic type. So there's a unique generic intersection number of curves from  that family with our target curve S. Right, now the idea is that if you move your curve  a little bit, then if all is well, if here the intersection is transverse of those two  curves, if the intersection is transversely and you move a little bit your curve here,  then here you will not lose intersection points because of the assumption on the intersection  that it is transverse and here you will gain a new intersection point. So the number of  intersection points of a generic curve passing through that point will not be the number  of intersection points of a generic curve of the family with this target curve. Okay,  is that clear? This is a frontier point, right, so it's not, so this is not an intersection  point of this curve with S. Now because it's in the frontier, you can perturb it as little  as you want and it will gain a new intersection point in S. This intersection point did not  occur in the original curve we started with. Now if you can assure that the curve you started  with intersects S transversely everywhere else, then the perturbation will not lose intersection  points because only tangency can allow you to lose intersection points and therefore  the generic curve passing through here will have more intersection points with S than  a generic curve passing through here. Now we can assure that if B is generic in G square,  we don't care about its genericity in S. If B is generic in G square, then this always  happens that all the intersection points are generic, are transverse. But the problem is  who assures you that the B is generic in G square? So the idea is because we have a group  we can move around our curve and put B in such a situation where it is generic in G  square and then the intersection will behave well. Now this is essentially the only point  where we are using the group topology in this argument. So it's a very, very delicate point  and I find it very, very curious that there's an extremely delicate point where if you use  the group then everything works, if you don't use the group then obviously this is wrong.  Not the group topology, the group operation, the group topology you use?  Well you need to use the group topology in order to know that the translation is continuous,  a homomorphism and therefore it maps from tear points to frontier points and stuff like  that. So this is a curious point, this is something to look deeper into and understand  what's going on because if you want to prove the theorem without assuming a group then  you will have to find something like that and this is where things could go wrong.  By the way in this goal are you showing that algebraic closure is definable closure?  Why would it be?  It's locally a graph, locally in the sense, what do you mean locally?  Locally in the sense of the ambient or minimal topology.  Oh, so it's not implying that.  No, no, no, not at all.  Not even.  What do you mean?  Just look topology graph around the point.  Yeah.  It's, we prove something stronger but I don't want to go into the details.  So as I said if we were in a compact situation then this would imply that also the openness  result but if we're not in a compact situation then things become delicate because what the  situation, think of the following example.  So consider G plus, sorry, C plus and F where F of x, y is x and 0 if y is equal to 0 and  x, y, 1 over y if y is not equal to 0.  Then this, the graph of this function, the graph of this function is closed in C square  but it's very non-continuous and I'd be very glad if anyone could give me an easy proof  showing that this is not strongly minimal.  I mean of course by our theorem we know it's not strongly minimal but proving that this  exact specific example is not strongly minimal is not so trivial I think and what happens  here is that we have infinitely many poles in this function and the next theorem that  we prove is that this set S cannot have infinitely many poles.  It cannot have infinitely many points where if you tend to a point in G then the curve  goes to something that has no limit in it.  This follows immediately, I mean the fact that this is the only obstacle for S being  open follows immediately from this theorem.  So the next theorem is S has finitely many poles so since we assume that our group is  embedded in NK with the ambient topology then this just means that there's a neighborhood  of the point that S is unbounded.  The pole is just a point where and this is an extremely hard argument, sophisticated  geometric argument and I think that when we were working on it we were definitely thinking  that there must be something simpler but even for that example it is not that easy to find  an argument showing why this is not strongly minimal.  So maybe if someone can analyze examples like that and understand them we can come up with  a nicer argument showing that this.  So a pole, so A, so definition assuming G subset N to the K with the ambient topology  A in G is a pole of S if for all open U containing A  U times G intersected with S is unbounded.  So it's really what you would think a pole should be.  So now it's an immediate corollary of this argument that if you only have finitely many  poles then S is in fact an open relation which means that it maps open sets to open sets  as in the same sense as F is an open function.  So this is really the essence of the problem of proving that S is open and this gives us  if you want a starting point for what comes next because as an immediate corollary of that  we can get for example that if you have a function that is smooth at the identity then  it's determinant Jacobian has constant sign in a small neighborhood of the identity and  this is what this is the engine that allows us to understand how the differential, the  Jacobians of these local functions if you want look like and this is really the machine  that allows us to conclude how things really look.  But this is a whole different part of the paper so I guess it's a good place to stop.
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