 Morning everybody, like Cornelia, I wish very much that I was in Bourg with you this morning. I'd hoped I would be, but life is still too complicated. So bon anniversaire, Thomas. I can't believe you're so old because you're almost the same age as me. So you should be a young guy. So I'm gonna talk today about subgroups of products of surface groups, which is something I've thought about for a long time. And in fact, the main construction I want to talk about is something that I thought about quite a while ago, but it's only recently that I really convinced myself that it's interesting enough to talk about and write about. And so I want to try and convince you that a very elementary looking construction is an interesting new way of constructing finitely presented groups, and in particular, finitely presented subgroups of direct products. And I'm gonna concentrate mostly on free and surface groups. So that's the main theme. So it's a new method for constructing families of finitely presented groups. And I'm gonna emphasize the case where the groups constructed are subgroups of direct products of surface groups or free groups. And one of the interesting features that actually is behind a lot of the applications is that when you see these groups, they have remarkably small generating sets. So these subgroups are gonna, the size of the generating sets is gonna be small compared to the number of factors M. Yep. So just some vocabulary in case I forget to say it. So if I write D of G without explaining it, that's always just the minimal, the rank of the group, the minimal cardinality of a generating set of the group to save space when I write G to the M, that just means the direct product of M copies of G. And so this is the term I tend to lapse into and forget that not everybody says it every day. So a subgroup of a direct product is called a sub direct product if it maps on to each factor. So that's a standard object in algebra. So the idea would be that if you didn't map on to each factor, you just replace that factor by the projection of S. So generally when you're looking at subgroups of direct products, it's you reduce to the case where the subgroups are direct, a sub direct product. And you say that a full sub direct product is full if it intersects each of the factors. And again, that's just sort of an obvious reduction because if you didn't intersect a factor, you just project away from that factor. And of course, in Thomas' honor, I should say something about Cayla groups and just to remind you, a group is a Cayla group if it's the fundamental group of a compact Cayla manifold, for example, a smooth complex projective variety. And in the background of some of this, which I'll just say a little bit about at the end, there's this wonderful mystery still of which groups are Cayla groups, right? This is a question I've found fascinating for a long time and it still seems deeply mysterious to me. Okay, so just, so I feel slightly embarrassed because I'm about to tell you something that everybody in the room knows, but just again to fix vocabulary. Why do we think about finitely presented groups? Well, because it's just a natural condition. So a group is finitely presented if and only if it's the fundamental group of a reasonable compact space or equivalently if and only if it acts freely and co-compactly on a simply connected CW complex. So that's sort of the first reasonable topological finiteness condition, because as you increase the niceness of the space X on which the group acts, you get higher finiteness conditions and they're all non-trivial. That is to say they place non-trivial constraints on the group. So just to remind you, so groups of type FK, if it has a classifying space that is the fundamental group of space with a contractable universal cover and that classifying space has finite K skeletons. That's type FK. And then if you take just the cellular chain complex of the universal cover and act on it, you'll see that a group of type FK is a type FPK. That just means there's a projective resolution of the trivial module Z by Z gamma modules that are generated up to dimension K. So that's just a homological version of the homotopic finiteness condition FK. And then this can be an awkward thing to work with. And often when people are proving theorems about things not being of type FK, FPK, what they really use is this condition here, which I call weak FPK, which as you'll see comes up naturally in some geometric situations. So this is a less standard definition. I'd like it to become more standard is that a group gammas of type weak FPK over some ring, which I'm just going to stick to Z. If the homology of the group with coefficients in Z is finitely generated up to dimension K, I'm not just the homology of the group, but the homology of all of the subgroups of finite index as well. Okay, so this is just, I say this is the way that one normally, this is how one normally show that things are not of type FPK. So if you type FK, you have type FPK. If you're finely presented, you can reverse this first arrow. And then FPK implies weak FPK, but you can't reverse this arrow in more circumstances. So those are the finiteness properties that I want to think about. Now, finiteness, just if, finite presentation can be a hard thing to determine. So for easy groups, it's easy. For finite groups or finitely generated obedient groups, but you already have to, you need to have some structure and think a little bit to see why finitely generated in the potent groups of type FK for every K, type F infinity. And then you really have to think quite hard to see that naturally occurring geometrically interesting groups like SL, SLNZ or mapping class groups are out of FN, you really have to think to see why those are finally presented. And it's not, that's not just a matter of getting used to things. It really is genuinely hard to show that groups are finally presented. So I like these little examples that Beer and Strabel looked at. Take already for some two-step solvable groups, metabelian groups is not obvious. If you take Z and invert six and then act on it by another copy of Z, if you act as multiplication by six, you get this group, if you act as multiplication by two thirds, then you get this group. Those are both finitely generated, in fact, two generated groups, but the first is finally presented and the second is not. Okay, now there's lots of people in the room who'll know how to prove that, but it's not a tall obvious, right? You need to, that's, this is thinking about these sort of examples as where Beer and Strabel started BNS theory. It's not a tall obvious that even a concretely given group like this, or SLNZ even, it's not obvious to show that a group is finally presented. So it gets hard even for fairly small groups that have a fairly small description and it quickly becomes impossible. That is to say, if you move into the world of subgroups of direct products of hyperbolic groups, say of free groups, then you can't tell which finitely generated subgroups are finally presented. So already if you take 12, there's no algorithm that will take a set of 12 elements but direct product to two free groups of rank two and tell you whether or not the subgroup they generate is finally presented. Okay, so once you think of finite presentation as a difficult thing to prove, unless it's obvious, right? So unless the group you're given is given to you by a finite presentation or else is given to you as the fundamental group of some compact space or all the space. Okay, so that's all propaganda. Just to warm up to say, I'm now gonna describe to you a whole new families of groups which one knows are finally presented. And I wanted to do that propaganda because the way I'm gonna say it, I'm just gonna say that we're dealing with these families of finally presented groups. But in the background, there's quite a bit of theory which I'll go over quickly as to why these groups are finally presented. Okay, so here's a very simple picture that I want you... So rather than do the general construction, I'm gonna do a concrete example. Okay, so I'm gonna describe to you a new families of finally presented subgroups of FM, that is direct product of M copies of the free group. And I've just fixed the rank of the free group to BR. I was doing it with rank two, but I thought that's misleading. It's really, there's nothing special about two, we'll just do it for R. So I'm gonna keep A1 up to AR. There was letters as the generators of my free group F. Right, now there's no free groups here. This is just zeros and ones. All right, so here's a matrix. And I just wanted to stare at this matrix and when I explain to you what it is, you'll see it's nice and easy. So I've just written the numbers one up to M. I've taken M to be 18. It's because that's the widest thing that tech would let me fit on the page. And what I've written in the columns is just the binary expansion of the numbers at the top. Okay, so one, two, three, four, five, and so on up to 18. So I've just written the binary expansions of those numbers as the columns of this matrix. And then, so the rows will give me an element of the field of two elements, a vector space of dimension M over the field of two elements. And I want to treat that as a multi-index and I'm gonna take each generator of my free group and I'm gonna raise it to the index given by the rows of this matrix. So for example, if I take generator AI and I raise it to the first row, then I get this element of free group to the 18th power. The second row gives me that and so on down to the fourth row. Okay, I'm going over this slowly just because it's really easy and I'm not hiding it in here. Can everybody in the room nod? So I know everybody's stared at this, just move your head a little bit for me. Okay, everybody's got what's going on here because now this is a really easy and I don't want it to be mysterious. Okay, I've just written out the numbers from one to M and I've used them to define these elements, just use them as indices to define these elements of a direct product of free group. So I've got rank R and in this example, I've got four rows, so I'll have four R elements. Okay, and so that's what I've got. And so the subgroups I want to talk about, I just take, first of all, I just take the subgroup of direct product of M, copies of the free group, generated by those elements I had on the previous page. Okay, so I've done that for each generator. I've got these four elements in the 18 case and I just got to take the subgroup they generate. I'm also going to consider a slightly bigger subgroup, which is that subgroup together with the diagonal elements. So for each generated free group, I just take the diagonal elements of the direct product. And those are the first examples of what I'm going to call binary subgroups and a direct product of free groups. And just to remind you the standard terminology, if you're like me, you always worry about whether this starts with zero or one. The lower central series is just you take repeated commutators of a group. Okay, so here's the theorem about these groups, which is what I want to talk about. So take a direct product of M copies of the free group of Frank R, R, at least two. And the first important point about these groups is they require very few generators, right? So let me just go back here. So I'm going from one to M, I took M to be 18 and writing out the binary expansions, which means the length of that binary expansion is only like log to the base two of M, okay? So I've only got logarithmically many rows in that matrix. And for each generator of the free group, I was at these elements of the bottom, I'm doing that for each I in my free group. So I've got R for the number of generators of the free group and then the number of rows in that matrix is just one plus the floor of one plus log M, log to the base two. And then for this blue subgroup B1, I just added these extra diagonal elements so add an extra R elements. So in each case, if you think of the free group being a fixed rank, then the number of generators is just logarithmic and the number of direct factors. Okay, so that in that sense is rather a small subgroup, but it's quite big in the sense that it's a full sub direct product. So it's easy to see that they're sub direct products. That is they map on to each direct factor, each of the free groups direct factors by construction. It also intersects each factor in quite a large subgroup. In fact, it contains some term of the lower central series of the direct product. So in particular, it contains that term of the lower central series of each free group. For the first one, for the red subgroup, it actually contains the M minus first, M is the number of factors, term of the lower central series. For the blue subgroup, the bigger one, that contains an earlier term of the lower central series, actually M minus one over two plus one. Now, so I started off with this propaganda that it's hard to see if groups are finally presented. These groups are all finally presented, okay? So I haven't come anywhere near giving you a finite presentation, but in fact, they are all finally presented that I shall explain, but they have some, what I think of as exotic finiteness properties. So the first one, the smaller one, it's finally presented, but it's not of type WFP3. In other words, it has a subgroup of finite index whose third homology is not finitely generated. In fact, the third homology itself is not finitely generated in this case. For the bigger blue one, for M equals four, in fact, it's the whole of the direct product of four copies of the free group. For M bigger or equal to five, it's finally presented and it is of type FB3, but it's not of type FB4. In fact, it's fourth homology will not be finitely generated. And crucially, I'll explain why I'm interested in this eighth point a little later on. If you fix C, a null potency class, then there's some polynomial depending on the class you're looking at, so that M is the number of factors, remember? So when M gets reasonably big, bigger than this polynomial evaluated at log of M, which doesn't take very long, of course, right? Now, polynomial and log is soon smaller than M. Then when that's the case, then the bigger of these subgroups, B1, does not contain or does not even virtually contain the C-th term of the lower central series, okay? So if you ask what's the earliest term of the lower central series that this subgroup contains, as the number of factors gets bigger, for a short time, you'll contain the commutator subgroup and then the next term of the lower central series, but the first term you contain is going to infinity quite quickly as the number of factors goes to infinity, okay? So those are what I think are the interesting properties of these subgroups. So what I'm gonna do, having started with a very concrete example is explain the ideas as to how one knows where these properties comes from, and then I'll explain why these are just an example of just an example of a more general construction and then finally I'll say something about what these got to do with Taylor groups. Sorry, but I just did that with, I did that with free groups to make it as clear as possible. I could do the same thing with surface groups. So if I take the standard generators for surface group, so then I think of the surface group of genus G as being an image of the free group of rank 2G, and then I just take those subgroups I had in the direct product of free groups and I push them down to surface groups, okay? So I just push them forwards to surface groups and I have this fancy B naught and this fancy B1 and they have exactly the same properties as the ones I had in the free group, okay? So here's here are new families of sub direct products of surface groups that have these interesting properties, small, small generating sets, they're finite presented, they contain these terms in the lower central series and they have these varying finiteness properties. Okay, now where is all this coming from? So this is a story I was talking about a lot about 10 or 15 years ago, about 10 years ago, I guess. So some of you will have heard me say this in the past but let me remind you of it. So I said earlier that trying to understand that the finite generated subgroups of direct products of free groups is a hopeless task, right? They're just too complicated. But there's this remarkable theorem of Bamsleig and Roseblade from 1984 that started off a whole search to understand the finite presented subgroups, direct products of free groups. And what they proved is that the finite presented subgroups of a direct product of two free groups are only the obvious ones that are either free or they're virtually three times free, okay? So a full sub-direct product is only finite presented if it's a finite index. Now, when you've got more factors, life gets more complicated and this again, of course, has turned into a big part of geometric group theory over the last 20 years. So starting with the examples of Stallings and Beery and then the Bessvena Brady groups and finiteness properties of subgroups of rags. But this is just going back to the classical Stallings and Beery construction. So if you take a, so say you take a direct product of M free groups and map onto Z so that each factor maps onto or just maps non-trivially, then the kernel will be finitely presented, but it won't, and it's in fact of type FM minus one, but it's not of type FM. In fact, it's M thermology will not be finitely generated. So when I talk about subgroups of Stallings, Beery type, what I mean is sub direct products that virtually contain the commutator subgroup. So up to messing about with subgroups of finite index, they're kernels of maps to a billion groups. Okay, now for a long time, that was those were the only finely presented subgroups of direct products of free groups and surface groups that we knew. But then I want to remind you of how we came to understand more. So first of all, part of this, so this is a long story series of papers I did with Howie Miller and short or subsets of them. In the end, the right class of groups to look at are not free groups or free groups and surface groups. The right class are the limit groups. What's Leo calls limit groups that is fully residually free groups. Just to remind you a group is fully residually free if you take any finite subsets, then you can map your group onto a free group so that this your favorite finite subsets injecting. And free groups of course are fully residually free and importantly, surface groups are fully residually free. And then along series of papers culminating in this one in 2009, what Howie Miller short and I proved is if you take a, oh, sorry, that should be a product. If you take a product of non-Abelian limit groups, then you can sort of characterize what the finite presented subgroups are. So you restrict to full subdirect products that is you intersect each factor and you project onto each factor. And then what we showed is if this thing is finally presented, then in fact the projection onto each pair of factors must be a finite index. So that's a big generalization of the bounds like Roseblade theorem. For some, again, up to messing about the subgroups of finite index, every finite presented subgroup has to contain some term of the lower central series of the direct product. In fact, it'll contain the M minus first term. Yeah. And if you're not a finite index or unless you're really one of the most obvious subgroups you've got to subgroup the super infinite index, then one of your homology groups will not be finitely generated. So there's these repeating themes about terms the lower central series and the failure of homological finiteness properties. And so this essentially gives you, this is part of a story that allows you to characterize all finite presented residually free groups. So there's that theorem again. But I want to focus for a minute on this thing that's in red that associated to any subfinite presented subgroup for direct product of limit groups in particular surface groups. You always have this invariant, the conal potency class, the last term of the lower central series of the direct product or some finite index that your subgroup contains. So the Stallings-Berry type, the ones that came earlier, the kernels of maps to obedient groups, they'll be of conal potency class two. And then a challenge, an obvious challenge arises here. So you have the obvious subgroups of direct products, you have the Stallings-Berry type, how do you construct finite presented subgroups of direct products of freer surface groups that have these higher conal potency classes that they don't contain the commutator subgroup that the first term of the lower central series that they contain is the next one. The one after that all the way up to the M minus first. And that turns out to, now that was a bit of a struggle. It wasn't obvious how to construct examples that weren't either obvious examples or of the Stallings-Berry type. Okay, now I'm gonna come back to that in a second, but first I want to concentrate on this first bullet point in this general theorem about sub direct products of free and surface groups and limit groups. So here I said, here the directions, if you're finite presented, then you virtually project on the pairs of factors. But you might ask about the converse, was it got a subgroup of a direct product of surface groups and that subgroup, if you map it to each pair of factors that's virtually onto, does that force the subgroup to be finite presented? Well, not only is that true, but this turns out to be true in remarkable generality. I understand this theorem, I did prove it, but I still find it deeply mysterious because it sounds like something that should have been known in the 19th century and proved in a first course on group theory, but it doesn't seem to have a really easy proof. And so the VSP theorem, virtual subjection to pairs, is so these aren't limit groups now, these are just any finite presented groups. So it's the VSP theorem. So if you take a direct product of finite presented groups and you have a subgroup with the VSP condition, that is you map to each pair of factors and that map is the image is a finite index, then that forces the group to be finite presented. And moreover, it will contain some term, virtually contain some term of the lower central series and it will be closed in the profinyte topology, which is another useful fact in different contexts. And so just to emphasize, I haven't missed out any hypotheses here, this is just about subgroups of direct products of finite presented groups. And moreover, and this is gonna be important in the context of our binary subgroups, this actually can be turned into an algorithm that if you give me finite presentations with these groups, G1 up to GM, and you give me a finite subset, and I take the subgroup generated by that finite subset, if it's got the virtual subjection to pairs properly, that is you check to see if you virtually subject each pair of factors, which can be a hard thing to do, but if it does, then you can, the algorithm will construct an explicit finite presentation for you. So in the case of our binary subgroups, this can be used in theory, well, I'm hoping this can be coded up, that actually you can construct finite presentations of these binary subgroups. So that's the theorem which is behind why these binary subgroups are finally presented. Okay, is this virtual subjection to pairs theorem, which I say, if you haven't seen me talk about in this in the past, it doesn't follow just quickly from the definitions. It uses this one, two, three theorem about which fiber products arising from short exact sequences are finally presented. But nevertheless, with it in hand, it's a nice, very simply stated criterion. Now, having seen the significance of projections to pairs of factors, there's an obvious question. Well, what can you say about subgroups matter on to triples of factors or four tuples of factors and so on. And so that has been explored in these papers by myself, Harry Miller and short, but then there's a very nice paper by Desi Koshlukova. And then by my student in his thesis, Beno Kukuk looked at this problem. And here's a summary theorem that involves results from all of these theorems that I like this nice if and only a statement. So if you take a product of limit groups, non-Abelian limit groups, so what's of most interest to us today is surface groups and take a sub direct product, then the following conditions are equivalent. Your subgroup will project or virtually project on to reach capable of factors, if and only if the subgroup is finally presented and of type WFPK, weekly of type FPK, okay? So in other words, you virtually subject on to reach capable of factors, if and only if the obvious homological condition works that your homology and those of your subgroups of finite index are finally generated up to dimension K. Now, that is often useful when it's used with this lemma and this lemma is dead easy, but it's very useful. If you've got a subgroup direct product surface groups and you project on to capable of factors, then you will contain some term of the lower central series where the class of the term is just M minus one number of factors minus one divided by K minus one. So the more factors you project on to, the earlier the term of the lower central series you contain and if I show you the first case, then you'll know how to prove this. So first case, the first interesting case is M equals three and K equals two. So if you project on to each pair of factors in the product of three things, then the claim should be that's two divided by one. So you should contain gamma two, you should contain the commutator subgroup of each factor. And all you have to observe is, so you want all the basic commutators. You wanna have to, you want to say that your subgroup S contains each element of form X comma Y one one. And all you have to observe is because I project on to the first two, on to the sit side, the first and the last factor, there'll be something that goes X, some garbage and then one, right? That's just saying I project on to G one times G three. Well, knowing that I project on G one, G two, three times I get that, knowing I project on to G one times G two tells me I can get Y comma one with some garbage. And the point is when you take brackets, garbage bracket one is one, okay? And so therefore you get that basic commutator. And that simple-minded calculation, once you've seen that, you can just see how that would work with projections to cage pools, okay? So that gives you all basic commutators up to the weight you want, that gets you the term of the lower central series you want. Okay, so that's all the theory, right? So there's quite a bit of work, but it was done a few years ago. And I want to just exploit that work very explicitly to construct new families of groups. So let's go back to our example, just to remind you, I just written out the binary expansions of one up to M and I use the rows of that matrix to define these elements of the direct product and I get these subgroups B naught and then B one, where I throw in the diagonals and they have all this properties. So the rank is just by construction, I only use that many generators. And if you look at the rank, the image and the optimization, you'll see you need at least that many generators, you know what the rank is. This thing about containing this term of the lower central series, all you have to do is check that this B naught subjects pairs. That's the key property, that this subjects pairs. So why is that? Let's just go back and look at this example. So I claiming that if you take, so you take say that, let's say you take the fifth and 12th factor of your direct product, you project your subgroup onto it, what, why am I getting the whole thing? Well, all you have to observe is that this binary number and this binary number are different. So somewhere I will have a one zero, which means that when I raise my generator to the power multi-index one zero, that will give me the element a comma one in the direct product. And then you have to also realize that somewhere you'll have a one one, but it's really very easy. You just think about by construction, this binary encoding forces you to get the generators of each of each, the a comma one and one comma a for each of the generators a of the free group and in each pair of factors. So honestly, that's really, if that hasn't struck you already, when you're halfway through your next cup of coffee, you'll just think, oh yeah, that's obvious. It's a little less obvious, but it's a little less obvious why I actually defined this binary array. That isn't the first way I thought of it. I just over time realized that was the concise way to do it. Why does the blue group throwing in the diagonal subject triples? Again, by the time you finish your next cup of coffee, you'll convince yourself of that, but you might need to do a little bit to see it, but it's equally elementary. Once you've convinced yourself that B naught projects pairs, it's easy to see a B1 subjects triples. The harder thing is to think why are these were the right things to choose. You'll also be able to convince yourself that B naught does not itself project onto triples of factors. Okay, so actually you can see that already in this array. If I take the first three factors, one, two, three, then only the first two rows of my array have non-zero entries in them. So that's all that matters, right? So when I do this raising the elements of the free group to those indices, I'll only get two R elements. And I can't use two R elements to generate a direct product of three copies of the free group. And so for certain triples of coordinates, B naught will not map onto triples. And so because of this theorem, because of the if and only if nature of this theorem, here telling you which things have which finiteness properties according to their projection properties, that's what tells me this theorem, this point is gone. What tells me that 0.5 here is true. B naught is finally presented because it maps onto pairs of factors, but it's not of type WFP3 because it doesn't map onto triples. And similarly, you can see that B1 does not map onto four triples. And that's what tells me this 0.7 here that you get groups of type FB3, but not FB4. This final property about the co-numpotency class going to infinity with the number of factors, I hope I've said enough to convince you. I said it's actually proved to be a bit of a struggle to construct subgroups of direct products of free groups and surface groups where the numpotency class wasn't too, where there weren't of Stallings-Biery type. And this gives you a very soft way of seeing that you can force the co-numpotency class to go to infinity with the number of factors. And while that is, it's just a calculation with the WIT formula, that's a bit technical and I'm not gonna go into it today, but it's not a difficult thing because you project to a product of free numpotent groups and you do a calculation with the WIT formula to estimate the number of generators you need. I might just mention that some background to... So this construction does solve a problem that was asked by Minasian in 2015. So Ashart wrote a nice article on subgroups of right-angled-arting groups and acquire a few generators. What he proved is if you take an n generator, residually free group, then it was previously known you can always canonically embed it, in fact, in a direct product of limit groups, then you might ask how many limit groups? Can you limit, can you restrict that according to how many generators your group has? And Ashart proved, yes, there's a sort of universal upper bound with such should be an E, sorry. It's exponential of n cubed limit groups is what he proves you need as an upper bound, but he asked, can you do it with only polynomially many? And the answer is no, that this construction of these binary subgroups show that you can't do better than a single exponential. So now that these groups, they only need log m generators, but you can argue because they contain a direct product of m free groups, their free product rank is m, they cannot be embedded in fewer than n limit groups. So in fact, if you've got an n generator, residually free group, then you get this lower bound and you need exponential of n limit groups in order to embed it in direct product. So that wasn't the main motivation here, but it is nice that it settles that question. Okay, but it's time to think about why this might be more interesting than just one or two examples. So I've concentrated in order to make you see, just to emphasize how elementary the construction is on these two subgroups, b naught and b one. But the key properties as we saw in that brief excursion through what we know about subgroups of surface limit groups is it's all about whether or not your subgroup projects onto catapults of factors. So in fact, if you give me your favorite integer m, that's gonna be the number of direct factors in this product. If I choose any m distinct numbers and write their binary expansions, I will get a different subgroup, which will again be finitely presented. And according to how I choose these binary numbers, it will project to catapults or where k varies according to what numbers I choose. So this is an extremely flexible construction where you can get infinite families of groups with varying cone opotency class and varying homology binitis properties. And I really believe that this is a new and very sort of rich flexible source of finitely presented groups. Because you've got this machine in the background, the sort of one, two, three theorem and the virtual subjection to pairs theorem, telling you that these are finitely presented groups and you have an algorithm for generating finite presentations. So it's really a very flexible way of producing finitely presented groups and particular finitely presented subgroups with direct products of free and surface groups. But now, okay, you start in a free group, but just as we push that forward to a surface group, you can push it forward to other groups. So if you give me any m to pull of r generator groups, then I can take these binary subgroups and direct product free groups and push it forwards to the direct product of the groups you gave me. Also, you might pull things back. So if you've got one of these binary subgroups of a direct product of some groups GI and you have maps from HI, you can pull back and this whole engine of the VSP theorem and these projection theorems will tell you that the pull back subgroup will also be finitely presented. So here's the key lemma that pullbacks preserve virtual projection to cage pulls in the coronal potency class, but they'll typically increase the rank or push forwards. They do not increase the coronal potency class and they do not increase the rank. And so in particular, the pullback from the free group to any residual free group in particular, the surface group take these binaries. So you can take these binary subgroups of direct product free groups and push it forward to a product of limit groups as I did earlier, surface groups. We could also pull it back. They take, think of the surface of genus G as the boundary of the handle body of genus G. So take the standard map from a genus G surface to a genus G free group and pull back these binary subgroups. That will also give you interesting new classes of finite presented subgroups of direct product of surface groups. And crucially, we've got the coronal potency class going to infinity. So these are not of Staling's Beery type. They're really new classes of finite presented subgroups of direct products of surface groups. Now, before I talk a bit more about those explicit examples of sub direct products of surface groups, let me just mention one other application which is in a totally different direction which is something more algebraic was interesting a couple of years ago. For a long time, I was fascinated by this old work of Philip Hall from 1936. And what Philip Hall proved, he was looking at products of direct powers of finite simple or finite perfect groups. So if you've probably all heard at some point in your life you take a direct product of 19 copies of A5, the alternating group A5, that only needs two generators and direct product of 20 copies needs three generators. And then you have to go up to something like a, by the time you get up to a thousand you need four generators and so on. It grows very slowly, the number of generators you need for direct powers of finite simple groups. And those ideas of Philip Hall called the Eulerian functions of groups, they were extended by Jim Weigold and other people to cover infinite groups. I just want to mention some recent progress in this and how these new binary subgroups help. So let me start with a very simple example, Z to the M, Z to the M is not Z cross Z, it's dot dot dot cross, sorry about that. So of course you need M generators and the number of relators you need is quadratic in N. You can see that because the rank of the second homology is a lower bound and that actually is the number of relators you need. So when H1 and H2 will give you lower bounds on the number of generators and relators you need for direct products. But the remarkable thing is as soon as those homological lower bounds vanish everything just really collapses completely. So this bound that the number of generators that's what Philip Hall and Weigold and others were doing. So take a, but the number of relators I think hasn't really been looked at until I wrote this paper a couple of years ago. Take a finite presenter group and take a direct power. So take M copies of the group. If the homology first and second homology vanish then as these people proved you only need logarithmically many generators. But the thing that's new is you only need logarithm cubed relators. Now I'm really curious to know whether this logarithm cubed is the truth or not or whether you can just get away with logarithmically many without the cube. Now I'd like to advertise this that the number of relators you need for direct powers is remarkably small if the homology vanishes and quite how small I'm not sure. But this new construction of binary subgroups enables you to do a bit more, right? So I've just said, if you take direct powers of things with no homology in dimensions one and two then you need logarithm in many generators and only logarithm cubed relators. But now these binary subgroups let you go from, those old arguments are very much about direct powers and not arbitrary direct products. But now with these binary subgroups you can just take direct products. So take a direct product of M groups each of which needs at most our relations then the direct product will only need logarithmically many generators. And you always contain some term of local series. So that's the, just focus on this corollary down the bottom here by pushing forward these binary subgroups you can show that you take a direct product of perfect groups each of which needs at most our generators then the direct product only needs a logarithmically many generators. However, I cannot extend the argument about needing few relators, okay? So I really don't know what happens. I'd like to have a complimentary theorem that says if you take direct product of groups with H1 and H2 zero then the number of relators only grows like some polylogarithm. But I don't know if that's true or not that the estimate for direct powers very much is a homological argument that very much uses the fact as a direct power. So I've no idea what the universal upper bound on the number of relators you need for direct product of super perfect groups is. And I think that's an interesting question. Okay, but in the last five minutes, type five minutes. Yeah, in the last five minutes, I'd just like to go back to, I like to say that something a little more about these subgroups of direct products of surface groups that are not of Stalin's theory type. So the challenge was to construct finite presented subgroups of direct products of surface groups with higher conal potency class. I not the ones that come from maps to a billion groups. Now, these binary subgroups are not the first. In fact, there's a more delicate construction that gives more precise results. It's in my paper with Howie Miller and short. Well, we proved it. If you tell me what conal potency class you want then you can construct a subgroup of direct product of free groups or surface groups that will intersect that will contain exactly the right term with a lower sample series. Now that's quite a delicate construction that uses the Magnus embedding of the free group into the group of units of the ring of power series and two non-commuting variables. This old embedding theme of Magnus is a very useful theorem in many contexts. And you can then do calculations in these power series rings to control null potency in particular. And so you can construct explicit subgroups of direct products of free groups that have contained a particular term in the lower central series. And they're not that hard to describe once you know what you're doing. So for example, there's two parameters involved like four and three here. In a direct product of four copies of the free group of rank two, you put in the basic commutators and then you put in these elements, the diagonals and then AA squared, A cubed, A to the fourth. You can guess what you do if you've got more factors you just extend these elements in the obvious way. And those will give you subgroups of direct products of free groups that will contain the appropriate term that your favorite term of the lower central series and no earlier one. And so these particular examples have already been used in another context. For example, Claudio, Yosirisneric, and Roman Tessa used them to study the Dane functions of residual free groups to get arbitrary polynomial lower bounds on Dane functions. And you can get interesting subgroups of direct products of free groups by pushing forward or pulling back these subgroups but also these new binary subgroups. So I want to just finish by saying something about that. So I hope Tom knows how much I appreciate his beautiful work on Kaila groups. I've really been fascinated for a long time by Kaila groups and particularly what Misha did about using harmonic maps to understand cuts in Kaila groups and then this fabulous paper of Misha and Tomah about cuts in Kaila groups. That really got me interested in the subject. And what appears there is that when you have maps from Kaila groups to many other things, for example, to free groups, they'll factor using holomorphic maps to surfaces. And this makes it clear that the sub direct products of surfaces play an important role in the understanding of surface groups. I also, I don't really need to mention this but it's just such a beautiful thing that this paper of Tomah and PIP that a Kaila group is cubelable if and only if it's virtually a product of surface and free obedient groups. And that's really a fabulous theorem. And what I'm saying, it's not nearly near this level of sophistication but it's very much in the spirit of trying to understand the role of sub direct products of surfaces in Kaila groups. And in their original paper, Tomah and Misha asked these questions. So which subgroups of direct product groups, sorry, which subgroups of direct products of surface groups are Kaila? When is an algebraic variety, a complex algebraic variety inside of a product of women's surfaces? Which sub direct products of surface groups do you get by looking at the fundamental groups of those varieties? More generally, if you've got a Kaila manifold mapping to a product of women's surfaces by a holomorphic map, what are the images in pie one of such maps? So these are sort of the three fundamental questions connecting sub direct products of surface groups to the study of Kaila groups. And then the first sort of non-trivial examples of sub direct products of surface groups that are Kaila are in this important paper of Dimka, Papadimra and Sukiou in 2009, where they show that the generic fiber of, if you take a product of surface groups and map it to an elliptic curve by taking ramified covers of degree two in each factor, at the level of pie one, that's a stalling's berry. The kernel will be a stalling's berry group. You're mapping to do a free obedient group of rank two when you're taking the curl on that map. So one can phrase what they did is saying that there exist subgroups of stalling's berry type, which are Kaila groups. And so then you naturally want to know, well, which subgroups of stalling's berry type are Kaila groups? And that's the main topic of Cloudeo, Yosiris and Rick's thesis where I'll prove what I think are beautiful results about this situation when you have a product of women's surfaces mapping to an elliptic curve, which of those kernels will be Kaila groups? And Cloudeo really answered that question completely. And then subsequently in a work, which I know he thanks Toma for inspiration on he was able to answer this question about in the Code I Mentioned One case, which Code I Mentioned One subvarieties of a product of women's surfaces will give you which subgroups of direct products of surface groups do you get? Can you show a Kaila by exhibiting Code I Mentioned One algebraic varieties? And he was also able to answer this question, question three in the three factor case. That's all beautiful work, which I want to mention, but I also want to emphasize that this is all strictly restricted to the co-obedient case, where these are all subgroups of stalling's berry type. And the question that I'm interested in now, and this is something I'm talking to Cloudeo about, which of these, so we have now examples, the original BHMS examples in these binary subgroups, do they give you any, so these have higher conal potency class, right? So they're not of stalling's berry type. So they're not kernels or virtually kernels of maps to obedient groups, but they contain some other, some later term of the lower central series. Can those things ever be Kaila? Would be the first question, right? So are all Kaila subgroups of direct products of surface groups, either a finite index or a stalling's berry type? Now I'm not saying that's a conjecture or anything, I'm just, we have very few examples, okay? So we're just starting to play with this, but can it ever be the case that these subgroups with higher conal potency class are Kaila, okay? So let's look at the ones we know. And so in particular, one of the things that Claudia and I can prove is that if you take these BHMS, the original non-stalling's berry type subgroups and you pull them back from the free group to a product of surface groups, those ones are not Kaila, okay? Now that isn't, that isn't, by modern standards, you think there's a trivial argument like just saying it's got some subgroup of finite index with an odd betty number or something like that. And that actually uses this deeper work of Claudio on this work mentioned here that he answers this question about holomorphic maps to products of surfaces. He really answers that question in the three factor case and we use what Claudio did in that context to prove that these first examples of sub direct products of surface groups of higher conal potency class, those are not Kaila. And then for various of these binary subgroups I've talked about, we also have results showing that the push forward and pull backs of those are also not Kaila, but I still think there's a lot of work to do. And that is the end of what I could say except to wish Bon Anniversaire encore à Thomas. And thank you to the organizers for organizing this very interesting meeting and for inviting me. Other questions? Is V1 of type F3? Sorry? V1, you say it's type HP3? Yes. Is it of type F3? Yes, it is, yeah, it is, yeah. Yeah, it's of type F3, but not type F4. So did you try to build the code important for Kaila groups, did you try to propose a different result? I tried, but I haven't, yeah, I haven't succeeded. Yeah. I find positive results on Kaila groups hard, right? I think it's an interesting question because it seems to me it's really quite mysterious, right? Because, yeah, so what are the tools? What are you gonna use? Well, presumably you have to think about which Nopotin groups are Kaila groups and then try and pull it back somehow. But even making elementary observations seems hard. So I don't have a good candidate. I also have a question. So you say you could not estimate the number of relations in this binary- Yeah. So group? No, it's like, yeah, I couldn't. So this is something I gave a talk at Oprah Wolfach in the computational group theory meeting last week. So I was trying to interest some computational group theorists in doing this. You can in principle generate, well, in principle, and I think in practice generate finite presentations of these binary subgroups. But the problem is that it won't give very efficient presentations. So even if you go back to, so this theorem here about getting this small number of relations, if you give a computer, if you give a computer, say just a finite presentation of a finite simple group and then you ask it to compute a presentation of a large direct power, it won't do this very efficient thing unless you really massage it. So this proof that you only need the small number of relations, it uses, it's a homological argument that uses the vanishing of H2 ensures theory of central extensions. And now in theory, that could actually be programmed to give, that could really be programmed. But it really very much uses the fact it's a direct power. And I'll just say the basic idea of the trick is suppose you just got G cross G. Well, and you want to present that direct product. Well, you can just take a, think of the right-hand one acting on the left-hand one. You can think of it as a semi-direct product, it's just that the action happens to be trivial. So you can tell the generators of the second factor to act by conjugation on the first factor say A and B, the second factor. Act on the first factor is conjugation by A and conjugation by B in the first factor. Now, so that's a semi-direct product, but you're acting by inner automorphisms, that's a direct product in disguise. And the trick is that if you've got the relations in the second factor, then because you're acting as conjugation in the first factor, that will already force the generators, the relations that you want to be central before you've imposed them, just that action will force them to be central. And so what you'll get without imposing actual relations is a central extension of G acted on by G. And then if G has trivial H2, that central extension by arguing with universal central extensions, you can force to be G. So that's the basic trick, but it really requires that those two Gs are the same. And so it's only gonna work, that trick's only gonna work for direct powers. And it could be that for direct products, you can't get away with this small number of relations, but I don't know. There might be some other trick I'm just missing, but in particular for these binary subgroups, I don't know how many relations they need. I just don't know how many relations they need. And the obvious bound will grow quadratically with the number of factors M. Other questions? Maybe there are questions online? Not seeing anyone, but as usual, just unmute yourselves. Well, if there is no questions again, then thanks Martin for the look.