 I want to start off by saying that um yeah my work has been uh although I don't personally, I haven't personally interacted with Derp very much uh my work has been quite heavily motivated and influenced by like his work um and yeah my my PhD advisor is Karen who was advised by Derp so sort of like my academic grandfather you could say um and yeah obviously one way in which uh influence is going to be displayed in this talk um so I'm going to sort of a high level I'm going to talk about um some tree-like equations so these are these are going to be generalizations of well the generalizations of certain dysentery equations um and then I'm going to describe how uh so core diagrams come into play when solving these all right so uh I mean I'm sure most people here know the conch primer hop algebra um so it's combinatorial hop algebra introduced by Derp uh in the context of renormalization um it's the free commutative algebra freely generated over uh so in this case field F by the set of rooted trees uh the product is concatenation of forests and the co-product looks like this um yeah you're basically uh living a tree at the vertices of an anti-chain um and so what's sort of the relevant property um here uh is the universal property of conch primer um and this is one of the reasons why conch primer is uh important in certain contexts um so it has this universal property uh and it is the only hop algebra up to items more than with this property so what is this property um for those who may not be familiar with it um so if we have a co-algebra a we first have to define something called the hock shroud one co-cycle this comes from uh hock shroud co-homology um and it's a linear map so a linear map uh from the co-algebra to itself um where it satisfies this equation um which basically says what happens if you um take the composition of the met of the uh the co-product and the linear map l um then you basically get two terms you have the identity tensor l composed with the co-product plus this extra term of l uh tensor one um well this is the this is the identity map this is the identity uh yeah so that's what a one co-cycle is um and the conch primer the add a root operator so this is the operator that takes a bunch of trees and you basically uh add add a root and then stick each tree uh as a child of that root um so this operator is a one co-cycle you can check that this equation is satisfied and then this is the universal property uh proved by uh conch primer um we have a commutative algebra a over a field f and we have some linear map from a to itself um then there exists what what the uh universal property says is that there exists a unique algebra homomorphism from uh conch primer to this commutative algebra a such that uh if you compose the homomorphism with b plus that is the same as composing l with the homomorphism so basically this homomorphism is turning um b plus into l essentially um uh and if you assume a bit more about a you get some stronger properties on the homomorphism so with a is a bi-algebra so both an algebra and a co-algebra um in a compatible way and l is the one co-cycle that's where the one co-cycle comes in then uh this homomorphism real well is a bi-algebra homomorphism and if a is also a hopped algebra then the homomorphism becomes a hopped algebra homomorphism so i mean these are both just saying that it's compatible with that structure um um yeah so that's a really nice property um so and then slightly differently uh where are the tree like equations going to come from um or how are we going to motivate them at least um we're going to motivate them by these sub-algebras um so these are um some pop sub-algebras um of kong primer that foici looked at um and they're generated by a family recursive equations of the following form um so we have t of x uh is equal to x times then we're applying b plus to five t of x so phi is some formal power series um and it has a uh basically a non-zero constant term setting it equal to one um so this gives some some like uh formal power series solution with coefficients uh in kong primer um and so if we write uh p sub n for the the nth coefficient of you know the solution to this equation then um foici characterized when the sub-algebra generated by those coefficients by the t ns um sub-algebra uh of kong primer is hop so when it when it's a hop sub-algebra um and the theorem is this so it's a hop sub-algebra if and only if uh the formal power series phi uh has this simple form uh yeah so it's basically just specified by two constants a and b um yeah and it's basically just this simple like when your polynomial raised to this fractional power um so we're going to basically apply the universal property to get uh these tree-like equations that we want to look at um which are motivated by these uh what you call like tree equations uh here so in this case you could you could apply um you could think about applying this to sort of any uh algebra or co-algebra um or bi-algebra structure we're going to apply it to the polynomial algebra uh in a linear map from polynomial algebra to itself and that gives some algebra homomorphism be a little the universal property from kong primer to the polynomial algebra um so there's polynomial algebras just the algebra polynomials um and so yeah you get this algebra homomorphism and then we apply this to the tree equation so we apply it to both sides and that gives this two variable equation um yeah where l is now so b plus has become l this linear map and so one way you can think about this is that um in this case the the algebra homomorphism corresponds to the Feynman rules um that means Feynman graphs associated with Feynman integral via the tree of sub divergences um and this is going to ultimately you can sort of sometimes think of this as a kind of Dyson Schrodinger equation or like a generalization of Dyson Schrodinger equations so here we're just working with any any arbitrary linear map that's not necessarily going to give you much that's all that interesting but the universal property sort of points the way to what would be interesting to look at which is uh one co-cycles arising arising from co-algebra structures on the the polynomial algebra um f of y yeah and just a comment you could also you know not look at the polynomial algebra you could look at some other algebra and co-algebra structures on that algebra and then consider one co-cycles for those um it may be of interest but we're going to focus on the polynomial uh really bi-algebra here um or co-algebra at the very least um and there are actually two classic uh graded co-algebras on polynomials so it's the binomial co-algebra and the divided power co-algebra um and they're isomorphic um but they which one we look at does matter here because we're keeping the algebra the same but varying the co-algebra um so we'll get different results um so for the binomial co-algebra the co-product is the following um yeah you get this binomial coefficient you're breaking y to the n into y to the k tensor y to the n minus k um now if it's uh what we get then is that uh we get this limit uh defining or you know describing the one co-cycles for the binomial co-algebra um oh this should be f of y uh uh so what this says is that there is uh so for each such one co-cycle there's a power series f of z defining it uh and the action of l on y to the n um is given by so you substitute in a differential operator uh into f this power series it's formal power series and then that is applied to t to the n and then we're integrating all of that from zero to y yeah so that gives the the one co-cycle for the binomial co-algebra um and then the divided power co-algebra so what does that look like it looks basically the same or we're just dropping the binomial coefficient from the co-product and then the corresponding one co-cycles look also very similar they're also defined by some formal power series f and in this case you replace the differential operator here by a different a slightly different sort of operator that sort of behaves similarly so if you look at its action on or how it acts on uh y to the n instead of uh dropping the exponent by one and then multiplying it by n you just drop the exponent by one so all this operator does is drop exponents by one basically um or if you have constants uh they go to zero um so we're substituting that into f just like we did up here and then instead of integrating we're just multiplying by y um yeah so that's what that's what the one cosine was applying for the divided power co-algebra um yeah so we have these two co-algebras and we can look at uh which we like equations for each of them so now we have uh the equation we were looking at before and l is now uh it could be a one co-cycle for the binomial co-algebra or a one co-cycle for the divided power co-algebra and we're sort of interested in solving both of these um and we can think of these as uh dysentering equations in particular the form or corresponds to a dysentering equation um for a class of Feynman graphs generated by we're recursively inserting at one place in a single primitive graph um if we keep the Feynman integral unspecified then we get exactly this we're unspecified just means like you consider an arbitrary like Laurent uh Laurent expansion um Laurent series expansion of that of whatever the Feynman integral is um then you get basically this this first equation from the binomial co-algebra the divided power co-algebra doesn't really correspond to a dysentering equation at least as far as I know in the same way um but yeah uh it's sort of motivated by by the same background um so what I was kind of interested in what we were kind of interested in um along with Karen um and what others have looked at as well in the past um is solving these uh via some like or as some like weighted generating functions indexed by some nice combinatorial objects um this has already been done for this first equation um with the binomial one co-cycle but not for the second one no one's no one's as far as I know until the second one before um and so how was it done for this first one and how did we do it for the second one also is uh well the weighted generating functions are going to be indexed by core diagram um so let me briefly explain some of the background here um core diagram is perfect matching this is at one up to two n so that just means you're pairing you're pairing these elements up um so we might represent it by this this sort of diagram where the chords are just these little lines uh yeah and we have to look at um we wanted to find some parameters in order to specify the solution to these tree like equations um and those parameters are going to be defined in terms of the core diagram and its directed intersection graph um so the directed intersection graph uh has the chords as vertices the direction intersection graph of this core diagram has the chords as vertices and two chords are adjacent if and only if they cross so like literally you look at the diagram and the two lines crossed and the chords cross yeah yeah so that's an intersection graph associated with each core diagram um and it's directed in the sense that uh so one chord that comes earlier in the diagram so it's it's a source uh first end point comes earlier than the first endpoint of the other chord then then we put a directed edge from this i'm saying this chord to the next chord um so from earlier chords to later chords we directed that in that way so we're going to be interested in certain special chords that are in a sense terminal so they're terminal in the sense that there are no outgoing edges in the directed intersection graph um but they have no outgoing incident edges um so in particular what that means for the diagram is a terminal chord uh has no uh chords that cross it to the right so for example this chord diagram right here has three terminal chords uh basically these last three chords because they have no chords crossing crossing them to the right but everything else does have a chord crossing to the right so they are not terminal so those are going to be important for defining the solution to these three line equations and a diagram uh is so one one particular type of diagram that's especially relevant are one terminal diagram so they're diagrams that have exactly one terminal chord um and if there's only one terminal chord or there is always one terminal chord and that is the last chord so the chord whose endpoint comes at the very end um whose second endpoint whose uh sync comes at the very end one of the thing that we need which is uh so there's a there's one standard way to order to give a total order on the chords of a chord diagram the most standard way is just to look at the first end points of each chord so the source of each chord um and that's just ordered by that there's another there's a few other ways to to define a total order on chord diagrams at least a few other ways um and one that will be important here is called the intersection order um and how this order works is you label the root chord so that's the very first chord this very first chord one and then you remove it and when you remove it you'll get a bunch of nested connected components um so nested connected chord diagrams so they're connected in the sense that the the intersection graph is connected um and then what you do is you label the first connected components recursively and then you label the second one and so on and the labels determine the order um yeah in general that order is not going to uh correspond to this like standard order by the first end points of each chord they're actually going to diverge quite a bit in general um yeah so okay what does the solution look like to these three like equations we will focus on the the divided power one um so first we have to mention um something that we'll need to define the solution to the divided power uh feel like equation um so one thing that one might notice is that there are exactly two chord diagrams whose intersection graph uh if you forget the directions is an induced cycle so that's a cycle with no uh well no chords um so a cycle that uh yeah is also a whole um and those two induced cycle chord diagrams are what we're calling the top cycle and bottom cycle so this one right here and this one right here these are the only two ways to make uh an induced cycle of any particular size um as a chord diagram so okay then we can define the solution um well mostly define it at least um so we look at the the divided power uh tree like equation that we had before um it is uniquely solved by the following power series formal power series in x and y so its formal power series is indexed by top cycle free chord diagrams so these are chord diagrams that don't have any top cycles as sub diagrams so um this chord diagram doesn't appear it's forbidden to appear as a sub diagram of these chord diagrams um so you can think of it as sort of like halfway to tree to to tree to diagrams that are just trees uh if we forbid if we forbid both top and bottom cycles then we just have trees um but it's sort of halfway there um so those are those are what's indexing the formal power series solution um and then we have uh so the x variable is sort of counted by as is counting the size um of the chord diagram and the y variable is uh counting the uh position of the first terminal chord these are these terminal chords there's going to be a first one um in the intersection order it's its uh index in that in the intersection order um that's going to be this t1 of c so it's uh y is sort of counting the uh the index of the first terminal chord um minus one and then uh someone unusually we're sort of applying the divided power one goes to this um so yeah um so really for each chord diagram you're getting uh a y to the i for all i up to the index of the first terminal chord minus one is how that works or no just the index of the first terminal chord um yeah and then uh to this we have a few different weights um so one of the weights phi sub c is just some weight that's determined by the coefficients of the the uh full power series phi that defines tree like equation um those i'm not actually going to define but they're relatively simple and they're just determined by c and then sort of more importantly is this f fc um these this is a weight uh that's determined by the co so these are the these are the coefficients f i of that formal power series f that defined uh the divided power one core cycle so this l div um so this these f i's are the coefficients of that formal power series f defining the one core cycle and this weight is defined by so basically we're indexing we're taking a product of these coefficients indexed by differences in the consecutive adjacent uh terminal chords so the indices of these terminal chords in the intersection order we're looking at those and then we're taking the differences of consecutive indices of these terminal chords um those differences are going to index these terms in this product of the uh f i coefficients um and then we also get uh an f zero raised to the uh size of c minus k um yeah so basically you can just think of this as just some weights that's determined by uh the one core cycle and the positions of the terminal chords in the intersection order um yeah so that's the solution relatively nice um and okay so now what we're sort of interested in is thinking about how we can analyze this solution and what sort of uh combinatorial questions uh does this motivate and one of them is one of the obvious ones is counting uh these top cycle free core diagrams that haven't been counted before um and in particular we would like to determine um the number of top cycle free diagrams of size n and uh with the first terminal chord having index k so those are those are precisely the diagrams that index this solution um so for the rest of the talk I'm basically going to describe what that count is um um how we get that count is we're going to we're going to find a bijection so I'm not really going to describe that describe how to prove it but the way you would do this is you find a bijection to some other combinatorial object that has already been counted or some of the combinatorial objects that have already been counted um that's the most simple the most straightforward way at least in the sort of most combinatorially illuminating way um and what are those other objects uh they're going to be triangulations um so triangulations in this case are plane graphs in which every bounded face is a triangle so the un unbounded face um is not necessarily a triangle so a triangle is uh just three vertices all of which are mutually adjacent um and yet we're forcing every bounded face to be a triangle but um the unbounded face may not be a triangle it might have more than three vertices and we're actually specifically going to look at rooted triangulation so we're rooting at a boundary edge so this is an edge on the boundary face on the bounded uh the unbounded face the exterior face um and we'll call vertices that are not on the unbounded face interior vertices so on unbounded faces those vertices are going to be interior vertices um um and the want the vertices on the unbounded face will be exterior vertices so this is just what one of these triangulations look like all the bounded faces are triangles we're rooting at this edge up here um and uh yeah the unbounded face is not a triangle because that's more than three vertices uh yeah I mean it could be a triangle of course but in this case it is not a triangle um so then what do we get uh well there's an old result uh from the 60s by uh William Brown um that counts that counted the number uh he counted the number of rooted triangulations with a given number of interior vertices so in this case n and m plus three exterior vertices there always have to be at least three exterior vertices uh and the number of these of number of these is given by this uh fraction of uh factorials so that's a relatively nice you know explicit expression um and it turns out this is exactly the object uh these are these exactly the objects thrown by Jackson with top cycle free core diagram so we have the following theorem um one can find a bijection between uh connected top cycle free diagrams oh this is one thing that I forgot to mention actually um the uh the solutions here uh oh did I skip over it I did uh this is actually uh these are these I should have said these are connected top cycle free core diagrams so they have to be uh the intersection graph of these diagrams has to be connected um yeah so there exists a bijection between uh connected top cycle free diagrams with n chords and the first terminal chord having index k in the intersection order t1 is k um and rooted triangulations with n minus k interior vertices and k plus 1 exterior vertices so in some sense this is almost a more natural object to work with um because instead of you know dealing with this the intersection order first terminal chord we're just dealing with uh interior vertices and exterior vertices of a triangulation um yeah and that was it thank you thank you Lucas all right I see a question in the q and a um where Marcus asks if there's any hope to calculate the values for the top cycle free diagrams of size n but then where you have information on the other terminal chords or whatever the analog of yeah of the other terminal chords yeah um yeah you would certainly want that um to know more I think it's yeah I think it's probably possible um perhaps also by looking looking further at the uh the related parameters in the triangulations um and getting another bijection or just looking directly at the diagrams and explicitly counting them um but I haven't looked at the number of other terminal chords is going to correspond to in the uh in the triangulations um that is a good question I'm not entirely sure um yeah you that there'd be some sort of recursive thing um yeah I'm not entirely sure I haven't thought too much about that all right there are other questions I had a quick question just in general the divided power out of hop out of up versus the binomial one I mean if I think about it naively right this is exponential connection I mean one x is primitive and the other it's group like so can you somehow transform transport this relation between the hop fighter brass to your to the different results for the two cases or do you have to do independent calculations for both um I think generally a few independent calculations um there may be something there it would be really nice if there was some way to uh yeah to transform that that connection between the like the co-algebras of the hopped algebras um to yeah these like core diagram solutions uh I don't know of such a way but that would be nice if there was thank you well and as people are thinking of remaining questions maybe just as a comment the side you didn't emphasize is that you have a more conceptual way of proving the original core diagram result that my original proof of was ugly quite frankly so there's value on both sides even if the connection between them is not completely clear any other questions yes I have a quick one still um can you hear me yeah yeah uh so um so with this universal property then can you just go out um to uh hop algebra and core diagrams this way too or not uh is there a hop algebra and core diagrams that's yeah that's that's another way to phrase the question yeah um that would be nice uh I thought a bit about this but I don't know if such a hop algebra um it would be nice if there was something like that and that was relevant here um yeah I guess the the follow-up question would be if if a hop algebra of core diagrams could be set up then what would be the common total objects indexing the formal solution to the datas and swing equations in that hop part of the question all right seeing no further questions let's thank Lucas again