 I think a little story about Tom, but shall we do it in behind his back? I don't know He's not here, right Okay, anyway, so Let me just start with the with the setup and then we'll turn back to that so X4 Always a smooth oriented and And compact for manifold closed for manifold and In trying to understand the the smooth structure we as to say the function to it and I call it the genus function G which maps from the second homology to and By associating to a homology class alpha the minimum of Of the genera of surfaces which represent alpha so these are just Smooth embeddings of of X of sigma Such that I lower star of sigma of the fundamental class is exactly alpha and this is closed and Oriented and We hope that this will somehow detect Something about the smooth structure of of the four manifold so before proceeding as you luckily arrived so Let me just join Gordana saying that you know, I met Tom in my mathematical childhood That was already the adolescent years from for Tom with all the benefits and complications And I always remember him sort of sitting in the back of the lecture hall Which somehow changed by now gravity or something have to pull him to the front and You know in the back chatting all the time Discussing it and then you you I Said all the time, you know and then somehow you had the feeling that he doesn't pay any attention And then towards the end or at the end of the lecture He just asked the right question and for me it was very helpful and you know, it helped me To sort of go on and and gave me program for the next sometimes today Sometimes two weeks or even two months to think about and to sort of further develop So the only thing I can wish for you guys who are younger than me, which is unfortunately a Lot of large portion of the audience and it's getting larger and larger for some unknown reason You know, it's very helpful to have a senior colleague Who just asked the right questions and sort of helps you through when you feel stuck so I'm really grateful and thankful for Tom doing that and So much about his questions. So let's Talk a little bit about his answers. So here is here is one So let's go back to to this minimal genus function and we would like to understand it for some manifolds So the first manifold which might come to your mind is s4 Well, the situation is pretty simple there. This is the zero group. So there is nothing to study and the next example is CP2 where there is this famous theorem of Tom and Peter from 94 the advent of cyber written theory, which is the Tom conjecture saying that if So how shall I phrase it? So Suppose that H2 CP2 Which is isomorphic to Z generated by the class age and D is non-zero Then this genus function of CP2 on D times the generator is just D minus 1 Times D minus 2 Over 2 so that's the famous Tom conjecture I was always puzzled that asymmetry like they saw the Tom conjecture and what did Peter get? So they're not really fair. Anyway, so this this Completely describes the the genus function here and there is a generalization of that Which we call the symplactic Tom conjecture or generalized symplactic Tom conjecture and This was done by Peter and Zoltan Like this Anyway, so there is this conjecture. Let me just remind you very quickly. So if we have a symplactic manifold X omega and C is a symplactic sub manifold Sigma is embedded such that The homology class represented by Sigma is the same as the one with the symplactic sub manifold then The genus of Sigma cannot be less than the genus of the of the symplactic Sub manifold, so that's sort of a sharpening that and it leads already to To the resolution of this of the understanding of the genus function in some cases Let me just tell you maybe two examples so the first one is Theorem of Danny Ruberman from 95 Which describes the the function for CP to blown up once so if X is CP to blown up once So the second homology is two-dimensional and Random homology class can be described by a times the generator here and b times the generator there Then first of all we can assume. Okay One remark. This is a diffele morphism invariant So meaning that if you apply any kind of diffele morphism to X then the image of the class will have the same G value as the original one and we can exploit that symmetry every now and then and This is what we will do here So first of all we can assume that a and b are both positive because there are diffele morphisms induced by the plus one sphere and the minus one sphere in these two components and if a is bigger than b then the genus function of this manifold on A b is given by the following rather simple formula It's a minus one a minus two over two minus b minus no b times b minus one Over two and you can sort of check this easily because what you do is under this condition and Both are positive You take CP to You take b lines passing through the same point and then the rest of a minus b line somehow randomly Generically and then you blow up this point and you will get a complex surface representing that particular Homology class so appealing to that statement This gives you a minimal genus and then at a simple arithmetic which I hopefully got right gives you this answer Is it important while in this construction? It is but then you realize that this manifold has an orientation reversing diffele morphism Which gives you x back and so the same formula works for the case when b is bigger than a And if a is equal to b then a little bit of thinking helps you to see that The complex curve is a collection of these joint spheres and you just tube them together So all these classes can be represented by spheres. So in particular if a b the homology class the homological square is zero then the class can be represented by a sphere Okay, so this really used the this this symmetry built in the manifold and the very similar argument applies when you take s2 cross s2 so the same Or similar formula for As to cross s2 Here again the two generators are denoted by a and b and the minimal genus function is given by This simple expression if a b is not equal to zero and when one of them is zero then again you can represent it by a sphere Okay, so these are sort of very simple cases and Mainly because you have a Symmetry and you can reverse the orientation of the manifold. So what happens if we take a slightly more complicated case So let's consider x to be CP2 blown up twice So this trick now will not work anymore and indeed we don't know the answer for the entire function Let me just tell you one statement. We found with Zoltan in this case. So suppose that alpha is a homology class and Now the second homology is three-dimensional and I denote by a the component here and b1 and b2 on the two CP2 bar components and As always I can assume that they are all non-negative and so here is what what you can do so if Alpha square is is non-negative Then you can apply this these defilmorphisms generated by Embedded plus one minus one plus two minus two spheres and by a defilmorphism You can assume that a is bigger than the sum of the two elements, so Let me say it again the original in the original class this Inequality might not be satisfied but applying defilmorphisms right Rightly chosen and not very complicated. You can assume that which then puts you in the same position as here So you just take a copies of the projective line b1 going through this point p b2 going through The this point q and then the left over just chosen Generically and so you get a complex curve You compute the genus after you blow up the two points and then you get a representative of that class satisfying this condition Which is a complex curve, so we again appeal to the tom conjecture the sort of resolution of the tom conjecture And we get the minimal genus the formula is not that nice it's it's rather an Algorithm together with a formula and then you get a complete understanding of this case and a little remark if Alpha square is equal to zero then that just means that a square is equal to b1 square plus b2 square So that might ring some bells, you know, these are the Pythagorean triples And then you run this algorithm and it turns out that the one satisfying this equation Will have b1 or b2 equal to zero. So these are very simple classes and for those g is equal to zero So said differently in this manifold Similarly to the once blew up and the product if you have a homology class with homological square zero Then that can be represented by a sphere Okay, so this is the first manifold in this sort of growing complexity or growing Manifolds growing in size which we know has an exotic pair So let me just very quickly remind you how to construct a Smooth for manifold which is homeomorphic to this axe but not the film or fiq to it and let's examine for a Short-time the genus function of that manifold. So so here is a construction of an exotic CP to blown up twice so this is due to Ahmedov and Park and I don't know maybe 2008 something like that 2009 and the construction goes as follows So let's take the zero surgery on the trefoil not Which gives us a fiber three manifold it fibers over the circle with fibers the torus So if I multiply it with s1 I get an oriented for manifold as always and this fibers over t2 With t2 fibers. So this is a torus vibration over over the torus We can equip this manifold with a symplectic structure symplectic This needs a little bit of a thinking because typically torus fiber Surface bundles over surfaces are symplectic Except when the fiber is of genus one you have to make sure that the fiber is homologically trivial a non trivial But in this case you have a section So the section will show that the fiber is not trivial and indeed we take a section So now I will draw and I hope it will be okay so this box symbolizes this for manifold and I take a section which is a Torus of self intersection zero and I take another torus which represents twice the fiber So it needs a little bit of a braiding and then I take the two intersection points and blow up one and smooth the other so blow Up and smooth If we go through this procedure what we get is Genus two surface of self intersection zero inside this manifold blown up once and the next step is to take this this gadget and fiber sum with Sigma two cross t2 and then there is a little yoga of computing the signature and The Euler characteristic of the resulting for manifold. This is very easy But unfortunately, it will not be simply connected. In fact, it will have very significant b1 The b1 will be four. So on the result do for Well, I call it lottinger surgery To get the manifold which will denote by AP Ahmed of Park So lottinger surgery is a little bit like Dane surgery, but now one dimension up So we take a torus with trivial normal bundle We take it out and we glue it back by a diffele morphism Which can be characterized by a single curve like t2 cross d2 is a four manifold Which can be decomposed by having one zero handle to one handles and one three two handle And you only have to specify where the two handle goes So the the theorem is what they proved The first is that you can do you can choose these four tori in a way that what you get at the end is simply connected And why is it exotic? Well, that's a rather simple observation That the genus function of this manifold When you restrict it to those alphas Where so alpha is in the second homology This is definitely non-zero Since you have these genus to surface in here and in there both are symplactic It's clear here. It's very visible here You have to vary a little bit why the this double is a symplactic, but it's not very complicated So you have a genus to surface which is symplactic so it minimizes genus in its homology class So it's a the function is definitely not constant zero So it differs the the formatting for differs from from the from CP to blown up twice So this is kind of a lesson to learn, you know at the current state of techniques to show that they are not the film or fake It's pretty easy We have all those theorems and they just give it for free essentially to show that it's simply connected Well, that's a challenge, you know You have to use the one-compane theorem which you taught to your students many many times Told them to be careful about the base point and then you are not careful about the base point. So, you know Okay, so So Here I give you a sort of complete description of squares positive squares or non-negative squares What happened with what happens with negative squares? Well, we know much less Cyber written theory does not really help directly in understanding the minimal genus of negative squared homology classes We have to look for other tools. There are other tools But they give you only partial results like bounds on the genus Such tool for example if alpha happens to be a characteristic class So the complement is spin then you try to sort of glue to another spin manifold and apply The the 10 8th conjecture or the 10 8th theorem of Ruta and get some understanding of the possible minimal genus You cannot expect a sharp result because we don't expect the 10 8th conjecture to be sharp So these are bounds which tell you some restrictive cases alpha should be characteristic so you can exclude them to be Spheres for example the same strategy works in in the AP manifold Of course because it only uses homological argument Here you know a little more because the cyber written invariance of the AP manifold are non-zero So you have an adjunction formula not only for classes with non-negative square But also for negative ones coming from the same paper of Peter and Zoltan So we know we know a little more But we don't know the exact shape of this function and it would be very interesting to see Because this manifold does not come by itself So we have these four lottinger surgeries, which we can do in a way that the result is simplactic but also we can sort of tweak one of those to get an infinite family of of exotic structures where the Exoticness is reflected by the value of the cyber written function not by the Configuration of basic classes the basic classes will be the same But the value is different the function takes on them and we have presently no idea how this value Influences the genus function, so it would be nice to see and well I'm waiting for the right question from Tom to proceed with that So here is a little conjecture we have So we spent a lot of time trying to find a sphere in this AP and we didn't So what do you do if you don't find something you think that it doesn't exist? so conjecture AP does not contain Homologically essential In bad its fear I have no Good argument why this should be true and the only foundation of of stating this is because we didn't find any Speaking of which, you know, it's always a nice Question to try to understand spheres in four manifolds and let me quote another old question so spheres in the K3 surface So a very simple-minded question is what is the most negative number? We can present as the self intersection of a sphere of a sphere in K3. So here is the the current Best result this goes back to finnishian and mihalkian 97 That there is a sphere In the K3 surface such that the self-intersection is minus 86 and We tried to be that we couldn't let me give you a construction how to Find a very negative sphere in the K3 surface. So first of all, what is the K3 surface? So here is one construction you take s2 cross s2 and you take four copies of the vertical and four copies of the horizontal horizontal S2 and you take the double-branched cover. It's an even class. So you can consider the double-branched cover This is a singular curve So the result will be a singular four manifold and either you resolve it or you make the following trick First you get rid of the singularities by blowing them up. So let's blow up all these intersection points There are 16 of them. So we get s2 cross s2 blown up 16 times And take the proper transforms of these eight curves and take the double-branched cover along them Well, what you get you have to take my word for it. That's the K3 surface and I also would like you to I would like to invite you to see a lot of spheres in that in that manifold namely every branch curve After the blow up it will become a minus four sphere and when you take the double-branched cover branch along this curve It will become a minus two sphere the the the Exceptional divisors of the blow-ups these are minus one spheres and they are branched in two points where they intersect the The branch locus so they will also become minus two spheres and I claim that these four horizontal four vertical guys Will give you four copies of minus two spheres which intersect each other like this. So there are four of these crosses 20 homology classes all together. We take one more class here This will also become a minus two sphere. So they are joined by this Last one. So all together you have, you know, there are the crosses all together. We have 21 minus two spheres and If you smooth the intersections by the appropriate orientations, this will give you a minus 82 sphere Well, not as good as the as this theorem, but it's pretty close and Let me just point out that what do we really need use here Well, we construct a very specific vibration over the K3 surface So if we go to these left sheds vibration story and we analyze the monodromy the total monodromy of the of the elliptic vibration is AB to the 12th and what we did here We just portioned it partitioned it into four equal pieces. So we just took This decomposition. So I'm saying that 12 is equal to three times four. So it's pretty It's it checks. Okay. Well, and now comes the trick because 12 is also AB to the four to the cube and This gives you a very similar curve configuration with 22 curves and this will give you minus 86 so you can improve by by playing around with a little bit with the with the factorization and this led us to the following construction so So this is a proposition again this I Worked that with Zoltan last year and so it just says that if we take the elliptic surface E n then it has an embedded sphere of Self-intersection and then there is this silly formula minus 44.2 times n plus four over five five minus r where r is the mod five residue of n So R is is a is an integer between zero and four and Determined by n. So this just recovers minus 86 in the for the K3 and gives you other examples of of this this Self-intersection and this led us to the following question conjecture that for every axe smooth for manifold as before with Non-trivial cyber-witten in variants and as always B2 plus bigger than one So sorry, there is a C. There is a universal constant C such that for every axe and that embed its fear we have That the self-intersection of S is at least C times the second betty number of X and So at this stage after trying to Prove or disprove or do something with that question the best C. We found was minus five so We don't really know again. There is this problem with negative self-intersection surfaces very little Tools we can use again for characteristic classes We can do the this fiber summing because the complement becomes spin If it's some even multiple of a or two-power multiple of a characteristic class We can take branched covers and again apply the 10 8 theorem or use the 11 8 theorem for gases But this does not apply for every homology class But whenever it does it sort of verify this this question Okay so So here's a next manifold which is a little more Challenging, so let's consider two CP two so we sort of out of the realm of Holomorphic geometry or or symplectics, so it's like CP to connected some CP to Yes, so There is a constant that for every x and whenever you have an s then it should satisfy here So here again the homologies are represented by two Integers and that is the the obvious bound so the genus of a b To CP to is definitely less than what you get from CP to like the the most simple minded a minus 1 a minus 2 over 2 Once again, I assumed that a and b are both Positive I was thinking to call the two components and mr. But then I decided it silly, so I don't do it I tried Yeah, so So this you know if you have to bet you would say that This will never hold I mean that's that's too simple minded and it turns out that it's not completely true so once again using this branched cover and Spin ten eight conjecture ten eight theorem trig. There is a paper of Jim Brian Which shows that the this bound gives the right answer for the class three three and six six So sometimes for very low numbers somehow the difference between the ten eight and the eleven eight doesn't really come into effect and you might be lucky to get something like that and We know that this cannot always hold so for example, we can take the class three zero and This can be represented by a fish tail in the first component and two lines in the second But one of them comes with a positive and the other with the negative orientation So somehow in homology they cancel and so the local picture is just The the mirror image of each other So you just tube them together and you get a genius zero representative of that class But in general, it's a it's a pretty challenging problem to actually give estimates for for the the genus so in a project with a Marco more and gone We found the following So he is a little proposition So it says that if we have the genus function on Two classes again, I assume that a and b are both are now negative This is always less than we take the usual Kronheim and Rovka formula for both and We can subtract a term Which looks like this under the assumption that a and b have the same parity and and Well, you know this only becomes interesting if a is much larger than b If if a is a little bigger So I always assume that a is bigger than b if it's only a little bigger than somehow this doesn't Give any sharper bound, but if it is much bigger for example, we take the classes Twice a an odd number two In this case the bound of of gym applies, it's twice a characteristic class It gives an expression in terms of p. It's a quadratic Expression with some leading term this formula gives another quadratic expression with another leading term and or bound Sort of lowers the upper bound by a linear term So we don't really touch the essence the the difference in the in the leading Coefficients, but it shows that there is a room to to improve Okay, so Well, I started with an example of s4 or this genius bound story and and then we just slipped through that and I said that you Know none of these apply So we have to sort of improve or approach a little bit and here is an idea to include as for For access before consider The four manifold M Which sometimes we denote by x circle, which is just x minus the forehandle the interior of the forehandle So this becomes now a four manifold with non-trivial boundary the boundaries s3 and then we can associate with the slice genus function which goes from the set of knots in s3 to the natural numbers and then it sends a knot to the minimum of You know the the slice genus in X in in M And again see infinity and embedding So we can somehow refine this This search for surfaces when we have a boundary then we allow to Surface to have non-trivial boundary. It should be properly embedded and we just fix how it should intersect the boundary and So what can we say about this function? Well, there are a few very simple statements So for example, it's very easy to see that for s2 cross s2 This is just the constant zero function because we represent we I just draw a curvy diagram for For s2 cross s2 once you gave me a knot k. So here is your knot k and then It needs a little bit of a thinking that what you get is actually s2 cross s2 And you attach a two-handle along the knot k. So that will be my slice disk and the same applies For CP2 Connected some CP2 So for these many folds, this is pretty simple. It's more complicated if you ask the same question for CP2. So for For CP2 This function is Unbounded this is a little note. We prepare with our array and Allison Miller and and Marco Marengon What happens with that really heavy for many for like a k3 surface for k3 Well there were that this question was sort of floating around for a while now and here is the latest I know about so there is this theorem of Marco and the student of mine That if the unknotting number of the knot is at most 21 then this This value is zero So every knot is slice if the unknotting number is less than 21 So from practical purposes under any other circumstances. I would say that these are all the knots I mean you don't know any not which is a little like But this 20 not is 21 is very suspicious if you remember the configuration. I was drawing here. It had exactly 21 Nodes and that was an improvement which has 22 So somehow it's very much attached to the second homology of k3 And so this is exactly what their argument uses So maybe in order to go from 21 to 31 or something a genuinely new idea is needed So still I wouldn't write this problem off Okay So I'm still not at s4. So let's Let's do that So let me any question The homology class. No, they just floating around they can be so sliced us and Then there are all variants like you fix the homology class or you fix it to be each slice where you require this F to be homologically in a Inessential and then you can do much more and I will leave it for the competent personnel to enlighten us about that But what happens if if you know you don't have any homology. So suppose That M is potentially exotic D4 But that's the real question in the in the story whether these exist or not. So let me give you two definitions So the first is that M is small If M embeds into D4 and large otherwise So the terminology is borrowed from the story of exotic our force similarly So like our force Similarly put into two categories and let me just denote by SM the set of slice links in M So so far I always at Stack stock to two nulls But now assume that we are taking all the links and a link is sliced If every component bounds a disk and this disk got disjoint from each other So in particular the linking numbers are all zero Okay, so here are two statements we proved in this case so, so this is a work I did with a Alberto Cavallo and so we claim the following so if M is small Then The corresponding set of slice links should be the same as for D4 first and if M is large and Not only simply connected but geometrically simply connected So it has a handle decomposition without one handles So a geometry simply connected then S is Of M is strictly bigger than S of D4 So this is the last statement. I would like to finish with but I would like to give two remarks So the first one is that what is really a small Exotic D4 well this would provide a counter example to the Schoenflies conjecture The Schoenflies conjecture states that if you have an S3 embedded in S4 smoothly Then one side should be the regular D4 and this would give a counter example to that So the the first line suggests that don't try to prove your potential Counter example to the Schoenflies conjecture by Sliceness of knots right by by this because you will get this has to be the same So this is a fairly simple argument The next one is sort of you know, I was thinking how to call the the the links living in the difference So these are the Obi-Wan knots, you know, these are our only hopes So if you want to prove exotic nest along these lines, those are the only hopes and So what do we need so first of all we need examples like that And we have to detect that they are not in S of D4 So we do need further refined link invariance Which are probably not coming from gauge theory because they tend to behave the same for Exotic D4's than for D4's So Kovanov Kovanov type Kovanov-Rozansky something like that should be useful in this case and let me make a final Remark about this assumption. So what do we expect? So for a closed manifold, it's an open question whether simple connectivity Implies geometrically simply simple connectivity So if you have a closed for manifold, which is simply connected Can you present it with the without one handles? We don't know What happens if we have a manifold with boundary? Well, if the boundary is not S3 and the manifold is contractible Then there is an argument of Casson saying that they are never geometrically simply connected You always need a one-handle So it's up to us to decide whether a manifold with S3 boundary Behaves like a closed manifold or a manifold with boundary. I could make an argument for both but that doesn't spare us from trying to find such examples and then trying to find the Obi-Wan link and then maybe at some good day. We might find an exotic S4 So until then I think I'm stopping here Yes, so that's a yeah So there is an upscaled version of this the function and then of course for links there are various notions of slice genus and It just says that it's hopeless and Maybe, you know, maybe the Schoenfliss conjecture is true. I don't know but there is this option For the Poincare I would not bet Yeah, so this one so where do we use that? So like I tell you what what what is the guy? What's who is Obi-Wan? Well, you take a handle decomposition and there are no one handles so the Attaching circles of the two handles are exactly the and then the argument is a little longer But this this is what you would like to show that those knots Don't that link is not slice in the regular disc this one No, otherwise, I would tell you right now Unfortunately, this is like so close, but of course, you know You have to you have to be extremely lucky because the lower bounds as the numbers are starting to grow So here is three three six six the the lower bound gets weaker and weaker In a sense of what do you really expect because you can use the ten eight and we expect the eleven eight to be true But even if you assume the eleven eight, we didn't get a single example where this is sharp That's very sad any other ideas No Lower bound no do we for it for a beef in general So none, you know, you can try tau or all kinds of these invariants and and they might exist or Kavanaugh homology and There is nothing I could say it would be nice There's the surface look like so the standard diagram is this right and so what we do is so You X you you consider CP to without the forehand and we would like to put that a knot here Such that the knot will be sort of it will be a connected some of two knots and One of them will be slice on one side and then we have an estimate on the on the On the genius on the other side and this is a slice not in d4 So you can sort of finish it off and I will just tell you the knot and then I will not tell you how they are related to the To the circles then the knot is a nn minus one Connected some With its mirror so the one a negative torus not and the and the same top positive torus not clearly it bonds a disc up here and The you know this plus one can be used to introduce all the negative twists And then so this is a this will be a disc and the other one You have to suffer a little bit to to position this plus one nicely so that you gain a little bit of a genius So that's roughly the the the idea and then a minus one or n and plus one I don't remember well, I didn't tell you what n is so That's an old trick