 So thank you all for coming, and I'd like to thank the organizers for giving me a chance to speak here. So this is joint work with Joav Len who spoke yesterday, and in the sense my talk is a continuation of his talk, but I don't want to assume that you're there for his talk yesterday. So I just want to begin with a few basic definitions. So tropical geometry is a kind of combinatorial version of algebraic geometry or more accurately a polyhedral version. And I don't want to get too much into the details, I just want to talk about a few examples today. So the analogs of algebraic curves in tropical geometry are graphs. So with or without a metric. So let's first talk about ordinary graphs. So a graph can have loops, it can have multiple edges between a pair of vertices. So combinatorials would call this multigraph. And the genus of a graph, it's simply its first betting number, so it's edges minus vertices plus one. So if it's a planar graph, it's the number of faces. And on a divisor on a graph is just an element of the free of viewing group generated by the vertices of the graph. And so the degree of the divisor is the sum of the coefficients. And so one of the starting points for tropical algebraic geometry was the realization that came in the first in the combinatorics community and then was taken up by algebraic geometers that divisor theory on graphs behaves in a way which is remarkably similar to divisor theory on a Riemann surface. So there is a Riemann rock theorem and so on, but let's first talk about what linear equivalence is. So linear equivalence for divisors on a graph is generated by what's called chip firing operation. So a chip firing operation is kind of an elementary principle divisor. So what you do is you fix the vertex on your graph. So this I'm fixing the red one here. And then what I'm going to do is I'm going to move a what's called a chip. So just kind of a positive integer from this vertex to all neighboring vertices. And if there are multiple edges between a pair of vertices, then I move that many chips. So the divisor obtained by firing this vertex V is going to be one chip at this neighboring vertex here, two chips at this vertex here because there are two edges connecting. And then however many chips I give away, I put the negative of that number at the red vertex. So clearly the degree is equal to zero. And so the subgroup of the divisor group spanned by these chip firing devices is what's called the group of principle divisors. And then you do the same thing as normally in algebraic geometry. The Picard group is the group of divisors module of the principle divisors. Then degree is well-defined on the Picard group. So the Jacobian group is the degree zero Picard group. And there is the standard exact sequence. OK. And so a theorem which is really quite kind of impossible to attribute because it came together as kind of grew as an understanding. Not wasn't just someone proving this is that the number of that first of all, the Jacobian group of a graph is a finite Abelian group. And the number of elements in the group is the number of spanning trees of your graph. So for example, this graph you can count has five spanning trees. And in fact, the Jacobian is a cyclic group of four to five. And so, for example, the Jacobian of a graph is trivial, if and only if the graph is a tree itself, that is to say if and only if it is genus zero. So that's the tropical version of the statement that the Jacobian of a Riemann surface is trivial, if and only if it's a P1, if it has genus zero. OK. So now let's talk a little bit about metric graphs. So in a metric graph, what I'm going to do is I'm going to take a certain underlying combinatorial graph, but now I'm going to put edge lengths along all of its edges. And when I look at divisors on a metric graph, I'm now allowed to put, I treat the graph as a metric space in an obvious way, and I'm allowed to put points on the interiors of edges. So there's now going to be infinitely many points on the graph. And so a divisor on such a graph is simply the an element of the free abelian group generated by its points. So what is the chip firing operation on a metric graph? Well, the idea is the same. You fix a point so it can be a vertex, but it can also be a point in the interior of an edge. And then in every direction from that point, you move a chip and you move them all a certain equal distance, which you can also select. So, for example, if you put a vertex, if this is your chosen vertex, you can find some small distance epsilon and then just move one chip a distance of epsilon in this direction, this direction and this direction. And then you get some divisor, which also has degree zero. And so you take the subgroup of the divisor group, which is spanned by divisors of this kind. So that's going to be the group of principal divisors. And then you do the same thing. You define the precar group as the set of divisors module of principal divisors. You might want to look at the divisors of degree zero. So that gives the Jacobian group. And what's the difference now? Well, so the way to think about this is that the Jacobian of a graph is a compact group over z. So that's a finite group. The Jacobian of a metric graph is a compact group over r. So that is to say it's a real force. And in fact, it can be identified with h1 coefficients in r mod h1 coefficients in z. So it's a real torus of dimension g. And in fact, it has much more structure than I'm able to describe today. It's what's called a tropical principally polarized abelian variety. It has something called an integral structure. It has a dual torus to which it is isomorphic in some sense. So there are quite a few details that unfortunately I'll have to state. So now let's talk about our Kirchhoff's theorem for graphs. So Kirchhoff's theorem for an ordinary graph says that the number of elements in the Jacobian is the number of spanning trees, which is to say it's the number of subsets of the edge set with a property that the complement is a tree. Now it's easy to see that if you have a graph of genus g, then the number of edges you need to remove to obtain a tree is exactly g. So here we're summing over certain g element subsets of the edge set of the graph. Now, the Jacobian is a torus. And the fact that it's a tropical PPV in particular means that it has a certain product on it. So it's a Riemannian manifold. So you can ask what is the volume of this Riemannian manifold? And so the answer is given by this paper on Baker-Couperbergen-Schokere, which is that here's what you do. You just take this sum where you sum over all g element subsets of your graph, such that the complement is a tree. But instead of taking a one for each such subset, what you do is you just take the product of the edge length that you're removing. So for every g edges that form such that removing them forms give you a spanning tree, you take a product of those g edges and then you add up all of these volumes that you get. So it would seem for dimension reasons that the volume of the Jacobian should be the sum. But there's kind of a quirk here that when you translate the metric geometry of the graph into Riemannian geometry on your Jacobian, the dimensions all double. So in some sense, the unit of length on a graph is meters squared and not meters. So, for example, if you have a graph, which is a circle of length L, then the volume of that circle viewed as its own Jacobian is actually going to be square root L and not L. That's just kind of a slightly unusual circumstance. OK, so now I'm telling you that the volume squared of the Jacobian is a sum of certain sets, right? And so now think about it this way. If I tell you that a certain polygon has a perimeter A plus B plus C, you may be tempted to ask, well, maybe this polygon is a triangle and A, B and C are the lengths of its sides. So the question that naturally arises is, is there some kind of way of representing the Jacobian of a metric graph, cutting it up into pieces such that this decomposition of its volume as a sum represents just simply this decomposition of the Jacobian into pieces. And so this is what's called the A, B, K, S decomposition of the tropical Jacobian. So here's how it works. So what we're going to do is consider the tropical apple Jacobi map. So you're mapping from the g-th symmetric power of a curve to the g-th symmetric to the pic g. And it's exactly the analog of the algebraic one where you're just taking g points and you look at the line bundle of the generate, which I'm just identifying with a corresponding divisor. Now, in the algebraic setting, this map has degree one, right? It's onto and if you want to construct the pre-image map, so the locus of indeterminacy is precisely the theta divisor, so it's the set of line bundles which have a non-trivial section. OK, so what we want to do is we want to understand the structure of the tropical apple Jacobi map. So here's how the structure works. So if you have a metric graph, then the symmetric product carries a natural cellular decomposition in the following way. So if you have a set of points, a set of g-points on your tropical curve, you just look at the edges on which those points lie, right? And so each point is going to lie in a certain cell. Some of them might lie in several cells, which are indexed by g-tuples of edges of your graph. So if you have a graph and you're looking at SIM 3 of it, well, you can just say, OK, well, let me pick any three edges on the graph. I could pick the same edge twice. And then I'm just going to allow points to move along those specific edges. And that gives me a cell in the symmetric product. OK, and so now the main idea is that the cells in the symmetric product come in two flavors. So I'm going to say that C of f is a break cell. If the edges, along which I am moving the points of my divisor, have the property that their complement is a spanning tree of my original graph. So that's going to be called a break cell. And all other cells I'm really not going to care about. OK, and so what the ABKSD composition looks like before I state the theorem. Let me just show you the picture. What ABKSD proved is that the apple Jacobium apple looks kind of like this. So this is SIMG of your curve. This is PICG. So PICG is a torus. So locally, you can just think of it as Euclidean space. It's just this nice manifold. SIMG is what's called a polyhedral complex. So it's something that's put together from from these blocks. And then the break cells evenly cover the in SIMG evenly cover PICG. Whereas the other cells that are not break cells are kind of sticking out and are contracted by this map. So you can think of the other cells as being kind of the tropical version of all of those line bundles, that all of those g tuples of points, which have a non trivial section. So this is the this is kind of the the locus that you're contracting. And the break cells evenly cover PICG. OK, so in algebraic geometry, if you have a map of degree one, which is, you know, the blow up of something that, of course, that map does not have a section, right? Because if you try to construct the inverse map, there's an indeterminacy locus and tropical geometry that actually doesn't take place. Whenever you have kind of a map of degree one, you can just take due to simple continuity reasons, you can take a section of this map. And so this is the this is the exact statement of the results of the ABKS, which, in fact, were actually proved earlier by Mikhail Kanan-Jakov in 2008, is that the tropical Able-Jakovian map has a unique continuous section. So if you have an element of PICG on on a graph, every element has a unique preferred representative as an actual sum of points, which is certainly not the case in algebraic geometry. And so what they also proved is that this decomposition of, so this map has a section and the image of the section is precisely the union of the break cells. And so the volume formula has a geometric interpretation in the sense that you can try to find the volume of the Jacobian, which is, of course, the volume of PICG because they're the same thing. So just one of them is a translate of the other by summing together the volumes of these cells. And so what they proved is that if you take a break cell here, then the volume of its image here is just the product of the edge length divided by a factor, which happens to be the volume of the entire Jacobian. Whereas if you have a cell which is not a break cell, then it gets contracted by the map, so the volume of its image is zero. And so then you can find the volume of the Jacobian by just adding together the volumes of all the break cells. And so what you get is this weighted sum. And then it comes with a coefficient, which happens to be the reciprocal of the volume that you're trying to find. And so then if you multiply out by this volume, you get volume squared, isn't this sum? Okay, so these are the results of ABKS. And so what I wanna talk about today is a version of all these results for the tropical print array. Okay, so first of all, let's talk about double covers. So you all have mentioned them in his talk yesterday. So a double cover of a graphs, for me would just be the naive notion you just treat the graph as a topological space and you look at a covering space of degree two. So any such cover is yellow, of course. And if I'm looking at double covers of metric graphs, I'm also going to require that my map be a local isometry. So all of the edge lengths in this graph are induced from edge lengths in this graph. Okay, and so it's an easy exercise to see is that if the target graph has genus G, then the source graph has genus two G minus one. So the numbers work out exactly as they do in the algebraic setting. Okay, so before we talk about metric graphs, let's just try to talk about ordinary graphs. If you have a double cover of finite graphs, then these two graphs have Jacobian groups, which are finite groups. And there's a natural way to define the push forward or the norm map from the Jacobians, which is essentially just take any point and map it downstairs. So take any divisor and map it downstairs. Okay, and so a theorem that Joab and I discovered last year, but which turns out to have been known in other language by combinatorialists, is that, well, we want to say in the algebraic setting, the prime variety has a, the kernel of the push forward map when you have a tall level cover of curves has two connected components. And then the connected component of the identity is the prime variety. Now these are finite groups. So it doesn't really make sense to talk about connected component of the identity. So instead, what you have to do is introduce a certain parity on this kernel. And then the prim group is just the even subgroup of the kernel corresponding to this parity. And then you could say, okay, well, I've defined a certain prim group for a double cover of finite graphs. So what is the order of this prim group? Okay, and so this is the formula that we found that as I wanna emphasize was actually known before. By the way, if you ever prove a, if you ever write a paper with a combinatorics result that you think is new, when you post that paper, make sure to cross-list it on math.co. That will save you some possible embarrassment later. Anyway, so literally the next day after we posted it, the criner from, I think he's from University of Minnesota pointed out that this result is not. Okay, so what we want to do is we want some kind of version of what it means to count spanning trees, except that now instead of having a single graph and we're trying to find the order of its Jacobian, we're trying to find the order of the prim groups of the kernel of this F of a double cover. Okay, and so here's what you need to do. You need to count the number of odd genus one decompositions of your graph G. These decompositions are stratified by rank. And so each, a odd genus one decomposition of rank K comes with a coefficient of four to the K minus one. So let me explain what these odd genus one decompositions are. Okay, so first of all, how am I going to describe a double cover? Well, so this, if you just kind of give this as an extended exercise in your algebraic topology class, I'm sure the students would come up with something like this. So if you have a double cover of graphs, what you first do is you choose a spanning tree for your target graph. So this will be in thick black. And then it's pre-image is a pair of trees. It's a spanning course for the source graph. Okay, and so then all of the complimentary edges come in two flavors. There is, there are edges with the property that their pre-images are somehow trivial in the sense that, for example, this black edge here, its pre-images connect the same vertices on the same tree as they do down here. And then there are the blue edges whose pre-images are flipped. So for example, this vertex is connected to this vertex via this blue edge on this graph here. This vertex is connected to the corresponding vertex on the other tree via the blue edge, right? So essentially to specify your graph, you just have to specify your spanning tree in thick black and then say which of the remaining edges are you going to color blue and which of them are going to be thin black edges? Okay, and so there's a natural way of identifying the set of choices you could make this way with H upper one of your graph with coefficients in Z mod two. Okay, so now here's what I'm going to do. I'm going to say, okay, now I have a graph of genus G and genus one decomposition of this graph is a way of removing G minus one edges from the graph. And so here's how this is going to work. So if you have your graph here, this graph has genus three, let's try to think what can happen if I remove two edges on the graph. And if you play around with it a little bit, you'll see that either one of two things happen. If you remove two edges from the graph, either you have one or possibly several graphs of genus one. So either your graph falls apart into a bunch of cycles or it doesn't or there's going to be a component of genus two and maybe a component of genus zero. And so a genus one decomposition of the graph is a way of removing G minus one edges from the graph such that what remains is a collection of cycles. And the rank of this decomposition is simply the number of cycles that you get this way. So removing this edge here and removing this loop is a genus one rank one decomposition. Removing this edge, this bridge and this edge here, genus one rank two, this is not a genus one decomposition. Okay, so that's what a genus one decomposition is. And I'm going to say that a genus one decomposition is odd and only after we came up with this terminology we realized that that shortens to O God. So an O God is a genus one decomposition with the following properties. So remember that all of we have a curve with some marked edges in it and that curve has a double cover, right? So now imagine removing a pair of edges from this graph and let's also remove their pre-images here. And so now your double cover is restricted to this collection of cycles that remains. Now it's a topological exercise that if you have a cycle and you have a degree two cover of it well either that cover is trivial or it's not, right? So the A because there are two representations of I of Z into Z mod two. And so I'm going to say that a genus one decomposition is odd if the restriction of this cover to each of these genus one components that you get is a non-trivial cover. And when is it a non-trivial cover? Well, the answer is that if there's precisely an odd number of edges in each cycle that remains. So for example, here, there's an odd number of edges in this cycle. Here there's an odd number of edges here and odd number of blue, sorry, blue edges here that we're counting. This is not odd because there's an even number of blue edges along this cycle. And so if you look at the covers that you get from these three decompositions, the way you can tell is that if your number, if you have an O God, if you have an odd number of edges in your cycle, then the pre-image of the corresponding component, each of these components is going to be connected. Whereas if you have a genus one decomposition that's not odd, then this cycle here, for example, is going to have two pre-images. So this is the trivial cover. Okay, so an odd genus one decomposition is a way of removing two cycles, G minus, sorry, is a way of removing G minus one edges such that the restricted cover is non-trivial on each connected component. Okay, and so we can count the genus one decompositions for this cover that we have here. So it turns out that there are eight of them that are odd of genus of rank one and four that are odd of rank two. So C one is going to be eight, C two is four. And so I predict that the order of the perm group is eight plus four times four, so that's 24. And so you can go back and try to count the spanning trees. This one we counted has five spanning trees. This one, if you want to count, it has 240 spanning trees. And so you take the ratio of these and you divide by two because remember the perm group is only half of the kernel. And so you get the predicted result, the number of elements in the perm group is 24. Okay, so it's so far so simple. And like I said, this is a result that was already known in not in the modern language, but it's already understood about 40 years ago. And so now what I want to do is I want to say, okay, so let's look at the tropical prim variety. So now if I have a double cover of metric graphs, then you can try to construct the prim variety of this in the same way that you do an algebraic geometry, which is that you look at the push forward map, the norm map on the Jacobians, you look at its kernel and you try to understand what that kernel looks like. And so there are two papers about this by Jensen and Lennon-Olerge that, so together they proved the following that essentially the same holds true as, the same statements hold true as in the algebraic setting, which is that the kernel of this map has two connected components. The connected component of the identity is a tropical principally polarized ability variety called the tropical prim variety. And in fact, the principal polarization on it is exactly half the principal polarization that's induced from the ambient Jacobian. And everything tropicalizes nicely, so I'm not really gonna have time to talk about the relationship to algebraic geometry, but essentially if you have a family of curves, so a curve over a non-archimedean field, then you can look at what's called the Berkovich analytification of your curve, look at the Berkovich analytification of the Jacobian. And then if you look at the skeleton of all that, then what you get is the tropical prim variety. So there is a direct relationship to algebraic geometry, which unfortunately I don't have time to explore. Okay, and so what I want to do now in my remaining few minutes is to just talk about a version of the ABKSD composition for the tropical prim variety. So first of all, is there a formula for the volume of the tropical prim variety? And so the answer is yes, and it's very similar to the ABKS formula. So the order of the discrete prim group is a sum over all o-gods of your graph, of your target graph, where each o-god is weighted by a factor which is four to the power of rank minus one. And so what we proved is that if you want to find the volume of the tropical prim variety of a double cover of metric graphs, then you just take the same formula except that you weigh it by a term which is volume F, which is just the product of the edge length that you're removing every time you have an o-god. And so again, there is this strange situation that the units don't really match up in some sense that what you get is a volume squared, not the volume. Okay, and so again, the question is, so I have proven a formula for the volume of the tropical prim variety. Is there some kind of way of taking your tropical prim variety and cutting it up into pieces so that the volumes of the individual pieces correspond to the terms in this sum? So the answer is yes, but the way to do this is more complicated than what you do with the Jacobian, as you would expect. And the idea is that instead of looking at the ABL Jacobian map, I'm gonna look at what's called the ABL prim map. So I'm going to look at the G minus first symmetric power of the top curve. And then I'm going to map it to the prim variety by just taking some divisor and mapping it to D minus the action of the involution on this divisor. And so to be extra careful here, you don't exactly map to the prim. You map to the connected component of the identity of the prim, of the kernel of the norm map, which has the same parity as G minus one, right? So for example, if you take a point minus the involution of that point that always lands you in the odd component of the of the prim, not the unit component. Okay, so there was a four-chlor result, which is that in the algebraic setting, the ABL prim map has degree two to the G minus one. And so we're very thankful to Sebastian Castellana-Martin who took the time to write up an 11 page appendix to our paper where he proves this and a number of other useful statements about the prim variety. So I encourage you to read our paper if only to look at his appendix because I think it's marvelously well-written. Okay, and so our main result, which I'm completely out of time now is that the tropical prim map is what's called a harmonic morphism of polyhedral complexes of degree two to the G minus one. And so perhaps instead of trying to explain what this is, I'm just going to show you a picture. So remember in the four Jacobians, the picture of what the ABL Jacobian maps look like is this. So you have cells corresponding to break divisors and you have all the other cells and these cells evenly cover the car group and all of the other cells are contracted. So in our case, the picture is a little bit more complicated but it's also kind of reasonable to understand is that there are cells corresponding to ogods and those cells cover your prim except that now you might have several cells that are ogods in the symmetric product over any particular cell of the prim variety. But if you count the number of cells that you have over any particular point, you're going to always get two to the G minus one. So there's a way of defining degree for these cells that are not contracted and the degrees are always powers of two in the same way in the way that this degree is locally constant in some sense. So if you have a cell, if you have two cells of degree two to the K meaning along a co-dimension one cell, then if there's another one here then the total degree here is the total degree here. And so the sum of all of the degrees and the pre-image is two to the G minus one. So well, unfortunately, there's no short way of saying this but just say I have completely out of time that if you have a element of the prim group then it has exactly two to the G minus one representatives of the form D minus Yoda of D where you count those representatives with certain specified weights. And so this is the tropical version of the statement that the ABLA prim map has degree two to the G minus one. So I think I'm going to stop here. So thank you very much. Thank you very much. So is there any question, Mark? I have a question which is maybe a bit philosophical. So do you expect that there is a way to connect the two results so that it's not a coincidence that the both degrees are the same but that in principle one could be used to prove the other. So that would of course be the ideal way to prove this result. You would say that first you prove that this degree, this map has degree two to the G minus one in the algebraic setting then degree is preserved by tropicalization in some sense. And therefore this result also holds in the tropical setting. We tried to do this and the machinery has not yet been developed for it. So we think that this can be done. In some sense, this is above my pay grade. So I think somebody else, not me, but yeah. So this is certainly something that's plausible. I've talked to a few people experts and they said that yeah, this is something that sounds feasible but our proof is entirely combinatorial. You essentially just count how many pre-images you have and then as you move around you make sure that everything glues together nicely. So yes, there should be some general theorem from which this follows but the techniques haven't been developed yet. Thank you. Other questions? I have only one curiosity. So do you think that it would make sense to consider also ramified double covers? Like not different? Yeah, so first of all, let me just say a few things. Before you consider ramified double covers you want to consider all un-ramified double covers. And in tropical geometry, un-ramified is a more general thing than being a free cover. So there's a collection of covers where you can have what you naively would think would be called ramification where an edge might have one pre-image or a vertex might have one pre-image instead of two but some of those covers are un-ramified in the tropical setting because essentially they are the tropicalizations of a tall double covers of curves. And so this theorem is actually true. I even erased a comment because I thought I would extra this. This theorem is also true for these un-ramified double covers except that the kernel that is now has a single connected component and the principle polarization. There is a principle polarization on it but it's not the restriction. It's not half of the restriction of the ambient one. And so we believe that all of these results also hold for the prim variety of un-ramified double covers of curves which just haven't worked that out yet. And then if you have ramification, yeah, the natural question is, let's say you have a cover of tropical curves which is ramified at exactly two points. Is that still a tropical prim variety? We don't know. I mean, I didn't make a list but there are exactly three papers about the tropical prim variety at this point. These two and the ones that I'm talking about. So there's plenty of low hanging fruit and a lot of work that remains to be done. But yeah, so looking at that's certainly a nice, I don't know, master's thesis program. Look at ramified covers and see if you can prove something about their tropical prim variety. Okay, thank you very much. Other questions? Marks, observations. Okay, so if not, we thank again Professor Zaharov and we go further on. So I ask the next speaker to prepare.