 Thanks everybody for making it out today. Today we have Clifford Smith from the University of North Carolina at Greensboro, and today we'll be talking about the non-crossing bond post sets. So take us away. And in case this does make it to the internet, my name is Smite, but that's okay. Oh, I'm so sorry. I was not sure. No, no, no, no. It's, if you're from Ireland, I think you're supposed to say Smith. We just like think ourselves are, I guess my grandfather thought he was fancy. Or something and was going to call himself Smith. So thank you for letting me know. So, yeah, we're going to, I'm going to talk to you today about the non-crossing bond post set. This is joint work with Joshua Hallam, myself and Matt Farmer. It was begun when Josh was a postdoc at Wake Forest University nearby where UNCG and Matt was his master's student now, Matt is a PhD student of mine at UNCG. And Joshua is tenure track at Loyola Maramount University in California. All right, so we're going to start our story with four partially ordered sets, three of them lattices. The first lattice that we're going to talk about is the partition lattice. That's capital Pi N. So you collect all the set partitions of the standard and elements set one through N. And you put a partial order on this by setting Pi 1 to be a less than or equal to Pi 2 if every part in Pi 1 is a subset of part in Pi, a part of Pi 2. So another way to think of this is you obtain Pi 2 from joining parts in Pi 1 until you reach Pi 2. So this partial order turns out to be a lattice if you can form a meat of two partitions. The smallest partition that is less than or equal to Pi 1 and Pi 2 and that turns out to be just a kind of shattering of Pi 1 and Pi 2. You take apart from Pi 1 and apart from Pi 2 and if they're non-empty you take their intersection and these are the parts of the meat. So joining is just the meat of the elements that lie above both Pi 1 and Pi 2. Here we see an example of Pi 4. There's the partition where all the elements are in singleton classes. There's the next level of the lattice where I've taken two elements and joined them together to form a part one two or a part one three. In this example you can see that I've joined one, two and three to make one, two, three. And so we have partitions with two parts. There's two different types, one with three elements and one element set and then there's some other ones with two element subsets and then at the top is the, the maximal element of the lattice, the partition of just one part one, two, three, four. The partition lattice is ranked. The rank is the n minus one number minus the number of parts of Pi and this is because the covering relations are obtained by taking exactly two parts and one partition and joining them. And that gives you the element that covers an element that covers that partition. And so, you know there's an n minus one steps of these joins you're you're at the top of the lattice and all the chains between any two elements all the maximal chains between any two elements are the same length. You just join two parts together at a time until you reach the top element. So here we see the same lattice Pi four with the levels, the ranks so you'll notice that it's exactly n minus n minus one minus the number of components. So we're going to be talking about the characteristic polynomial of these postsets and lattices so there's the Mobius function which is defined to be one for the zero element. And then you define the Mobius function of every other elements so that all the intervals. If you look at any x, and you look at all the wise in the interval between zero and x, the Mobius values for those elements should sum to zero. Another way to put it is that new effects is negative the sum of the Mobius values of all the elements strictly less than x. So if he happens to be a rank post that as many of our postsets will be, you can form the characteristic polynomial, where you sum up over all the elements the Mobius value of the elements, weighted with the power of T, and the power of T rather than being the rank of the element is the rank of the post that minus the rank of the element. So, for example, here is our Pi four, the Mobius value of the zero element is always given one. Here's an interval, the Mobius values have to sum to zero. So this one has to be a negative one. Let's see. This one has to be negative one. This one has to be negative one here is another element. It's above these two so there's an interval. The Mobius values have to sum up to one so this Mobius value is a one. If you look at this one, it's got three elements below it, and this one, and so on this interval, the Mobius values have to sum to zero so you get one minus one minus one minus one and then this one has to be two, and so on. And the characteristic polynomial is just to collect the sum of the Mobius values for each rank. It's run a little bit backwards so T cubed gets the Mobius values at this rank, T squared gets the Mobius values summed at this rank so that's a negative six. So Mobius values summed at this rank, and so that's two plus two plus two plus three, and that gets you 11 and then there's one Mobius value at the top and that's the constant coefficient and that's a Mobius value of negative six. So the non crossing partition lattice is a restriction of the partition lattice where we forbid parts to cross. So that means you can't have members of one part a and a prime interleave with members of another part B and B prime. So you can't have parts that have elements that do this with one another. And so a partition is called a non crossing if no two parts of it's none two of its parts cross. And so we're going to denote that with NCM. And it's a sub lattice of pie and it's closed undertaking meets being non crossing is that is so it's definitely going to be a sub lattice. And it's going to be ranked as it's lattice. Sorry, it's ranked. And here we see the characteristic polynomial of it, and there's not a whole lot of difference between NC for and pie for there's really only one partition that you have to throw out you have to throw out 1324. So the partition with parts 1324 because they interleave all they cross one another, all the other partitions are okay. So that changes Mobius values a little bit, because it's such a small lattice this element of course disappears, and it's Mobius value of one disappears with it so this negative six gets reduced to a negative five. So in distinction, the coefficients and instead of being one minus six 11 minus six there one minus six 10 minus five. The bond lattice of a graph is another lattice it's based on the partitions, the partition lattice, but you only take certain partitions, you take those set partitions. These parts induce connected components in your graph. All right, so it's all the set partitions but you only take parts that induce connected components in your graph. Another way to look at it is as a set of sub graphs of G sub graphs that span and whose components are induced. There's just a one to one correspondence between these. And if you look at it in terms of graphs. One bond is less than or equal to another if the graphs are contained. And so the meat of two bonds is just the intersection of the two graphs. So this is a ranked post set, and you can see its rank function there. Just to remind you here is the characteristic polynomial for pi for, and I'm sorry this should be an NC. This should be right so if it's pie for and if it's if the graph is K for there's no restriction right. Any part. That's a subset of KN is going to induce a connected component because all the edges are there. Sorry, that's the point I wanted to make with that slide. If you want to pick any partition one for is one part to three is another part. Those parts will induce connected components because all the edges are there. The story gets interesting if you have a graph that's not the complete graph for instance this twisted for cycle. The partition one for two three has to be thrown out look at this part one for it doesn't induce a connected sub graph so this thing gets tossed. These get tossed for the same reason one for does not induce a connected sub graph and neither does two three. But that turns out to be all of them. And so some movies values disappear and the movies values adjust to give you a characteristic polynomial that you see here. And here's another example where a lot more things get thrown out. So that's the graph underlying it and very little survives of the original five for. So what we did or what Josh's idea was to kind of complete the picture you've got the partition lattice. You've got the non crossing partition lattice behind it. Then you can introduce graphs and you've got the bond lattice of a graph. How about the non crossing bond. I guess it turns out to not be a lattice but it's the non crossing bond post set so it's the it's the missing fourth corner of the quadrilateral. So, let's start again with a graph on the vertices one through and again the non the members of the non crossing bond post set are those non crossing set partitions. Whose induce connected components and we furthermore require that these components don't cross one another. So if you look at it from the graph perspective you're looking at sub graphs, such that the components of HR induced and don't cross. The non crossing post set is not necessarily the non crossing bond post set is not necessarily a lattice and might not even be ranked but the non crossing bond post set on K for again is the same thing as the non cross the bond post set. Or for restrictions but here's an example where things are different. Here's pie four and the characteristic polynomial. And you'll see that 1324. While it's a element of the bond post set. It's actually a crossing bond, it's components cross one another. Now, we are fixing a cyclic ordering of the vertices so all this crossing is with respect to a cyclic ordering of the vertices so you see now the graphs are are arranged like this so my four is there. And NC four just has that one thing thrown out and the two blow it thrown out and the various function and characteristic polynomial adjust. So, the non crossing bond post set. That's not be a lattice. Here's an example. If you just look at the graph that just consists of the edge one for so that's one part that contains one for and the rest are single tens. And you look at the other bond three five that just consists of the edge three five and all other points are signal tens. These are bonds in the non crossing post set. And one for you join three five should be the unique minimal sub graph that contains both one four and three five. So here is a minimal bond that contains one four and three five, but you can go around the other edge of the cycle and get another minimal bond containing one four and three five. And they're not. They're not comparable. So, the join of these two elements does not exist in the non crossing bond post set. So, early on we were able to characterize when the non crossing bond post that is a lattice, and in some sense the obstruction, the potential obstruction that I just showed you on the last slide is the only obstruction. If you have very simple bonds just an edge E and just an edge F and they cross. If the joint of those two bombs always exists for crossing edges, enough, then NC, the non crossing bond post that is a lattice. And another way to look at that is that for all crossing E and F, there's got to be a join and it turns out the join is the unique minimal induced connected sub graph containing E and F. So this join of E and F the unique minimal component and EF is what we're going to call it. Sorry. Right, so if G has two components that cross you can take an edge from one and an edge from other that will cross and this unique minimal induced connected sub graph on exist there will be no way to get a connected sub graph going from one component to another to both contain E and F. So we're just going to assume that G has no crossing components from now on so that, at least our non crossing bond post that will have a maximal element the graph G itself. We're able to characterize when the non crossing bonds that post that was ranked when it has no crossing components, the non crossing bond post that is going to be ranked. If and only if well unsurprisingly if you take one bond, and it's covered by another. You get from one to the other by joining two components of age. There's a proof here and I think I'll just skip it. Here's an example where the non crossing bond post it is not ranked. So, this is actually a cover relation, H is covered by G, but you can clearly see you did not join two components to get from H to G. And the reason is, if you just try joining components. Just, you just can't do it. Let's say I tried to join two six with three five. Well there's no edges connecting them. So, if I try to make a part out of two six three five it's not a connected component. So I can't do that. If I try to join one and if I tried to join one and three five, I can do that because there's an edge connecting one five and that'll give me a connected component but then that edge is going to cross this component. So you can just go through all the, all the possibilities and this, there's just no way to go directly by joining two components you have to join them all at once. So you have to jump ranks. So, I want to talk about an interesting combinatorial formula for the characteristic polynomial of the bond lattice, because we're going to be trying to repurpose that combinatorial formula to get a combinatorial formula for the characteristic polynomial of the non crossing bond post site. So we're going to start with the combinatorial formula for the bond lattice first. So we're going to put a total order on the edges in G. And we'll say a non broken circuit with respect to this order is a spanning forest such that if he is not an edge in the forest and, well, when you add the edge you'll create a unique cycle, a path in the forest together with. Similarly, if you join an edge that's that's within a component of the forest you'll create a new cycle. What you don't want is you don't want this e to be the minimal element of that cycle with respect to that partially in order. I guess you could say a broken cycle is you take a cycle and you remove its smallest element. So that's what you would consider a broken cycle so a non broken circuit is something that is maximal and doesn't have a broken doesn't have a broken cycle in it. So for instance it's going to be a forest right if it has a cycle in it, then it has a broken cycle in it just remove the smallest weight edge. So let's let NBC K count the number of K edge non broken circuits. So the idea is every time you add an edge. If you create a cycle. There will be an edge less than that on the cycle. So let's do some counts. We're going to put 12 less than 13 less than 24 less than 34 as our total order. How many K edge forests are there, there's just one one forest with no edges. How many forests are there with one edge. There's four of them. And I don't have to worry about creating any broken cycles because there's no cycles to create. And the two edge non broken circuits there's six of them there's the two edge forests. Now we can get into trouble if we have a non broken circuit of three edges. There's one spanning forest of three edges that we have to throw out we have to throw out this one, because here's an edge that I could add, and it's going to be the minimal element on its cycle, because one to just happens to be the minimal and these turn out to be exactly the coefficients of the characteristic polynomial of the partition lattice in this reverse order, where the number of edges and the degree of T add to three. So the relationship that this comes from is the reason this comes about is Whitney's theorem of the characteristic polynomial so the chromatic polynomial of the graph is the polynomial that such that when you plug in T you get the number of property colorings of G. So, if Whitney's theorem was that if G is any graph on and vertices and you have any total ordering on the edge set whatsoever. Then the number of non broken circuits of K edges is invariant with respect to the order you can pick any order and you'll still get the same number of non broken circuits on K edges. And the characteristic polynomial has as its coefficients, these non broken circuit counts in reverse order. Okay, so with a little bit of argument, we can use this to get a non broken circuit formula for the characteristic polynomial of the partition lattice. First of all, let's remember what the characteristic polynomial is you look over all the bond lattice, excuse me, you look at over all bonds. You take the mu of the bond and then the appropriate T polynomial with it. I'm grouping them according to the degree of T that they get. Right here I'm substituting in the actual ranks of the bond lattices it's n minus the number of connected components of G. Moving on to the next page. It's a not hard argument to show that the characteristic polynomial the chromatic polynomial and the characteristic polynomial of the bond post that are related in this way. And then it's just a straight substitution, and you get a formula for the coefficients of the characteristic polynomial of the bond post the bond lattice. So there we go. A formula for the coefficients of the characteristic polynomial of the bond lattice in terms of non broken circuits. And there they are the one minus four or six minus three that we saw from that previous example are on the previous page. For the same graph. If you look at the non crossing bond post that the characteristic polynomial for that is this one T cubed minus four T squared plus five T minus two. What you saw in many examples was the phenomenon that the absolute value of the coefficient for the non crossing bond post that was just less than or equal to the absolute value of the same coefficient for the bond post that characteristic polynomial. I was thinking that maybe since there was a combinatorial count for the coefficients of the bond post that polynomial maybe there would also be a combinatorial count for the coefficients of the characteristic polynomial of the non crossing bond post that. And in fact that's what we found. We slightly modified, or slightly generalized the notion of a non broken circuit to a non crossing non broken circuit. So the non crossing non broken circuits with K edges are the non broken circuits with respect to the total order on the edges that also have non crossing edges that also have no crossing edges. And it turned out in many cases that these were precisely the counts that you needed to get you the polynomial, the coefficients of the characteristic polynomial for the non crossing bond post set. So, for example, six goes down to five look at that T coefficient that corresponds to non crossing that corresponds to broken circuits non broken circuits of two edges. And there are six two edge forests that are non broken circuits. And one of them is crossing and has crossing edges so we got to throw that out. So the six goes down to a five. Exactly what it should be. So for the constant coefficient, the three is supposed to go down to the two. Now what was the three. It was the three edge non broken non broken circuits. So, not including this one. This is non broken. This is not a non broken circuit because of this edge. There's one more that we have to throw out this one because it's got crossing edges in it. So, the three goes down to a two, as it should. And we get the polynomial. So, this is what we found the characteristic polynomial for the non crossing bond post set can in some cases be given as a formula where the coefficients are common editorial counts. of objects specifically non broken circuits that are also non crossing. But only in certain circumstances. Or the ones that we found there's there's examples where we couldn't make this work, but we were able to make this work if the non crossing bond post that was the lattice and had what we called a NC NBC ordering or non crossing non broken set ordering. Or if G was a perfectly labeled graph. I'll say a little bit about what both of those things mean. So in particular, you know, unlike the situation with Whitney's theorem, the number of non broken circuits with K edges did not depend on the total ordering you put on the edges. Unfortunately, when you're looking at non crossing non broken circuits, the counts do depend on the ordering. So, which is part of the reason that you can't have a universal statement here. But if you happen to have an NC NBC ordering. So what is that you look at the joint of E and F any to crossing edges. The minimal connected induced sub graph containing edges in F, if you require that that always have an edge G less than E and F in your ordering. And in fact, you do get these combinatorial counts. And in case that seems obscure. I will. I will try to make it sound a little better in that if you're given a graph we have a very efficient algorithm that will either find the ordering that will make this a true theorem, or will prove that such an ordering can't exist. So at least, you know, it's not like it's a collection of graphs that we have no idea about that satisfy this theorem we know if you're given a graph we can tell you if it'll satisfy this theorem. The other situation in which we have this theorem is if he is perfectly labeled. So that's when you can put an order in on the edges of the, on the vertices of the graph and all the back neighborhoods of a vertex format cleats. So, there's more results in the paper that we wrote that got published in electronic journal combinatorics. And a number of families of graphs and tabulated some properties that the non crossing bond post that had some of the properties we considered where ones that have these joins ones that have this NC NBC ordering ones that are too connected or perfectly labeled. And the non crossing bond post that properties, we examined where the property that it was graded or lattice or had this combinatorial theorem for the coefficients of the characteristic of the characteristic polynomial. We also considered shell ability of the chain complex. So the, as we mentioned, you have this combinatorial interpretation, when you have an NBC ordering. It turns out you'll always have this particular ordering when the graphs are too connected. So, if the graph is too connected, you also have that combinatorial theorem so it's not a non negligible class of graphs for which you have this combinatorial theorem to connected ones will work. There's some deletion contraction results known for the chromatic polynomial and, and hence for the characteristic polynomial of the bond lattice. We were able to produce some deletion contraption results for the non crossing bond post that although they're not, they're not really that pretty, further directions that we were working on at the time that we were looking at this we're trying to guarantee trying to find conditions under which the non crossing bond post that is ranked. It would be nice to know that, at least so that we would know what are the situations in which we can even discuss characteristic polynomials, but a good criteria for this eluded us. One thing we noticed was that, if you have a perfectly labeled graph, the coefficients of the non crossing both on non crossing bond post that appeared to be log concave and every example that we've tried. This is a major recent theorem that the coefficients of the bond lattice are log concave so it might be interesting to at least show that the coefficients of the non crossing bond post that characteristic polynomial are log concave. But we didn't. We didn't do a whole lot on that. It seemed like a hard problem as well. Right. Well, I think that's what I had to say about it. So thanks for your attention. Thank you. Everybody could thank you. Thank you. And are there any questions for Clifford. I don't know if it's a question or thought but this yeah this the result of what is it other proceed to and Oh, right, right. Yeah. And cats is so hard. Yeah, yeah, yeah, exactly. You can tell why we didn't want to spend a lot of time on On the other hand, it's on the other hand, it's perfectly labeled so, you know, maybe, you know, so maybe it's not that hard, but right. Yeah, I mean I was just thinking, you know, there's this closely related conjecture due to, I think the unimodal version was due to Rota and then the log concave version was due to Welsh at the sterling numbers of the second kind are I wonder if you have you looked at those rank sizes. In this case for No, no, we did not actually. That's, that is interesting. Yeah. Well, it's something to add to further. It might be even harder, but. Yeah, yeah, maybe. Yeah. That's always my favorite of the two questions. Yeah, no. Yeah, they're great questions. Yeah, thank you. Any other questions for Clifford. There was stuff in the chat. I don't know, maybe. Just thank you. Okay. Well, thank you. Thank you so much. Thank you. I'd like to invite everyone to thank our speaker again and, you know, thank you for a very wonderful talk. I actually, you know, I wasn't sure if I should ask, because I don't know that this is a well formed question. Oh yeah, sure. Maybe maybe this was a little bit delayed, but there was one slide where you had, you were talking about this condition for the non crossing postsets do not be a lattice or to be a lattice and you know you have this condition. That's right. Yeah. So every, every pair of edges are searching that there has to exist a unique way to find an induced sub graph. Yeah, right. Yeah. Yeah, and they, and they look, they look, they look very interesting actually they you have a crossing edge, you know, you have an edge and then you, the, the MEFs those those joins of, of ENF, they're like a path, there's ENF, and there might be, there might be just one inch. Okay, so there's a path between them, right. There's a path between them and then the very last vertex of the path might connect to one endpoint of E or both. So it's like either like, like it just the path ends and eat, or it's like a little dumbbell at the end, and the same thing with F so it's like a dumbbell graph and. And so you can actually test for this pretty easily looking for such structures is not hard I mean you can, you can write a terribly inefficient algorithm to do this, in other words. There's probably a better way to do it, but yeah. Yeah, I was trying to think of what that might look like and you just, you sort of answered my question. Yeah, no that's exactly what they look like. You know that's why it's. So that's why that one example didn't work. Let's see where was it. Oh there. That's it. So you can see there's, there's kind of like another way to go around. Yeah, I think there's the, there's like a minimal. There's like a minimal in connected sub graph but you can sort of go around the cycle in two ways, and that's that that's the problem. Right. Okay. Thank you. Yeah, sure. So. Thanks again, Cliff. And I think, you know, thanks everyone for coming out for our street seminar this week. And I think, go ahead and end there and I'll see everybody next week. See you. Thanks.