 Okay, so yeah, one one's like I'll turn it over to you here but I just wanted to say just a couple things before we get started as one welcome to the seminar, just a quick note on just the fact that attendance is open to everyone so despite graduate student being in the title there is sort of a wide variety of backgrounds here so I just want to draw you know just to be respectful with your friends and colleagues. So if you have any questions, if you have a question for us the easiest way is to just go ahead and politely interrupt at some point. If you have a question you can also type it in chat. I can try to keep an eye on that and really it's speaker. If you have sort of a longer question we'll leave a little bit of time at the end. If anybody wants to ask anything. Right, so I guess with that let me let me go ahead and turn it over to Sarah Griffith from Brown University and she'll be telling us about the tropical study room. Yeah, so I'll, I'll be talking about this algebraic object that is really important to a subset field of algebraic geometry called tropical geometry. It's equivalent to another gadget called the Max plus semi ring, which I'll mention later and sometimes it is also called the tropical semi ring. So a note on the name so the subject was sort of founded by a Brazilian mathematician. He named it. Apparently in, in honor of him. I don't know what his thoughts are on the name I think there's a, there's a general feeling that the name tropical geometry is a little distaste. Yeah, but I don't have a better name for it. Min plus semi ring is therefore both like, let's me avoid the term and it's it's more accurate because there are two possible rings that may use. So tropical geometry really leans on this on this min plus semi ring, and it's not really obvious how it comes about. So if you hear this thing about like, oh if you take like a log with this particular divisor and you take a limit like ends up encoding the the min plus or max plus semi ring structure. Personally, I never found that very convincing. So what I'm going to do is like not so much teach you like tropical geometry proper, which is kind of like theory laden and a little and hawk. But I'll just try and convince you that the min plus semi ring naturally encodes interesting things. So I'll give you for applications of increasing complexity. So the first three have to be like pretty broadly accessible the fourth one will be more ag ish. Okay, so let's go. So the first problem that we can use the min plus semi ring to address is the problem of finding minimum weight paths on graphs. So by a graph here. Well you can work with more generality than this but what I mean is you've got a set of vertices you've got a set of edges you have at most one edge between the two vertexes. And you have non zero weights on each of the paths. So if you are interested in given two vertices specified ahead of time, finding the shortest possible path from one to the other. So linear algebra, as it is useful everywhere is useful in graph theory. So you may be familiar with something called the adjacency matrix for the graph. So for this graph right here. The vertices are labeled v1 v2 v3 v4 and each of those corresponds to column of this matrix called the adjacency matrix of the graph. So the v i corresponds to the ice column. And we put a one in if the vertex, this column corresponds to is adjacent to the vertex corresponding to the corresponding to the vj or j is the number of rows you go down. So for example this corresponds to v1, and then we have a zero here because we don't consider v1 to be adjacent to itself we have a one here because we consider v1 to be adjacent to v2, and so on. So this you can prove my induction has a really nice property, which is if you take the adjacency matrix, and you raise it to the nth power. And the way they entry will be exactly the number of paths from v i to vj. Now, there's kind of some some data associated with our graph that isn't represented in the adjacency matrix, which is, we insisted that our graph edges would have non zero weights. What we can do is we can look at the weight matrix. So this isn't completely filled in yet. But this corresponds in a very similar way to pairs of vertices. So, for example, this still corresponds to v1. We still have a one here, because the path from v1 to be sorry the edge between v1 and v2 has weight one. On the other hand the edge between v1 and v3 has weight two, and that ends up getting represented as a two here. Now what are these little question marks doing here. Well v1 is not adjacent to v4. So we want to use some kind of like linear algebra construction to use the weight matrix to calculate possible possible weights of paths. It's not obvious what we should put in these entries. So one of the first questions is, can we fill in these question mark spots and then use some kind of linear algebra or variation thereof to find these minimum weight paths. Sorry, I just had a quick question. So when you say powers compute the path between vm vj, like if I raise it to say the third power and I look at the ij entry. Like as in if I raise it to different powers won't the ij entry be different. And so, Yeah, yeah, sorry, I may have omitted a word. The the nth power computes the number of paths of length and specifically. Okay, thank you. Yeah, my apologies. Thank you for clearing that up. Let's see. So let's try just like naively trying to compute some paths from v1 to v4 and see if it suggests some kind of like algebraic operation that we can relate to the matrix. Okay, so one thing is, the strategy we're going to pursue is going to involve taking powers of this weight matrix and reading off some kind of some kind of entry that we've generated by that method. Now, what we might expect is that, say, the nth power gives you the minimum weight path of length n between two vertices. That's what you can read off the entries, because as was just clarified, that's what you get with the adjacency matrix. We've already managed to find a path that lands at the endpoint we'd like. It would be convenient if we could just consider a path of length and to be say a path from the vertex we're trying to go from to the vertex we're trying to go to, and then a bunch of like stand still moves of weight zero. So we're going to include not moving not moving as a valid step that increases the path length by one. Okay, so let's look at different path lengths and see how we can potentially get from view one to view four. So if we have paths of length one. Well, we could stand still so I didn't bother drawing that in. Or we can take what the weight one path to v2, or the way to path to v3. As paths to v4. These are like infinitely bad, because they don't end at v4. We'll assign these right here, the value infinity. If you look at the paths of length two, we can go to v2 and stand still we can go to v3 and stand still, we can go to v2, and then to v4. And in that case the total weight of the path is one plus four, these first two were still infinitely bad. So we can go to v3 and then before, and the total weight of that path is two plus one seems to be a stray infinity here I'm not sure what that's doing. Probably an example I deleted. On the other hand, if we look at like paths of length three. We've got all the all the prior ones, essentially interspersed with do nothing moves. For example, this one can be regarded as the path of length to append to do a do nothing move. So that's one plus four plus zero. And so on, is that clear to everyone. I'm going to take those yes please feel free to interrupt me or like ask for a backtrack or something I don't mind at all. So, once we've got all the possible weights of paths to v4 of a particular length, which since we allowed stand still moves will include paths of shorter length, essentially. What we need to do is we need to look at the possible path lengths that we determined, and we want to take the minimum among all of them. Okay, so, in order to formalize and make rigorous this kind of naive calculation. We define the so called men plus semi ring. So, this, as a set consists of the real numbers and also infinity. And it has two operations. So I'll call this, I guess, oh plus and I'll call this oh times. Times is great because it's not a tensor product, but whatever. Okay, so what are the operations. So, the plus operation, the O plus operation is going to take the minimum of two numbers. So that's the comparison step we did at the end where we got a bunch of paths of different lengths and we found the least one. The times operation is going to add two numbers together. So we're going to add in like the usual sense of real numbers. Now, why is, why is, why are these defined these way this way well they they enjoy the usual distributive and associative properties that you associate with a ring. So the big difficulty there is that there's no, there's no subtraction. There's no such as life that's why it's called a semi ring. So there's a canonically isomorphic object called the max plus semi ring, where instead of O plus taking the men of two numbers. It takes the max. So it's the same thing. But you know some sometimes one convention is used sometimes the other. So, our earlier sort of naive calculations together with this definition suggests that we let the weight matrix have remaining entries, infinity, in order to indicate that we think of the distance between non adjacent vertices as essentially infinite. Okay. Now, if you take w to the end I wrote three here for some reason specifically. That's why. So it's pretty clear that the minimum possible path length from view one to view for. Let's see. Will be of length at most two. Now if you take w three, which is certainly long enough, which is certainly a high enough power to capture the paths of length to then the IJ entry will in fact to be the minimum path length from via to VJ. Because if you look at the site. There's a third power with respect to these operations. So to multiply two matrices together, you do completely ordinary matrix multiplication, but wherever you would multiply. You do times and wherever you would add, you do O plus. So if you get out will encode minimum path lengths between vertices of length at most whatever power of the matrix you're taking. I have a question, is that the weekend. Yeah, so when I say, let's see, the minimum. Yeah, that's just a typo, the minimum weight. Wait, sorry. I may have been misspeaking this whole time. Okay, so is, is the meaning of that clear. Wonderful. Thank you. Okay, so here's sort of the heuristic this suggests, whenever we have an optimization problem. We can encode it and max or equivalently men plus. When we can do two things we can run processes together with linear benefits or costs. So that's encoded in the operation of times. And almost extreme of those combinations of processes will matter. So that's O plus. So to relate that to what we just did. So we can combine the two processes of moving from V1 to V2 and from, I guess this was V2 to V4. And we can combine these two processes of moving these paths. And then we can compare which of the two produces the most extreme outcome and we end up with the minimum weight path. So less abstract example. You can put water pipes or transit methods end to end. And in that case, the one with the least capacity will end up controlling the total capacity of the combined ones. Or you can put them side by side. And in that case, the two capacities just like add to each other. So that's the kind of like optimization problem that we might want to look at. Okay. So problem two is completely different. So this is a question about finding approximate solutions to numerical questions. So everything on this page will become clear when I do an example. So suppose we have a bunch of real functions, FIJ in our variables. And we have an equation that consists of taking a product of a bunch of these FIJs and adding them together and seeing when that's zero. So suppose we want to understand the solutions to that equation. They are potentially horribly complicated. But it could be that their magnitudes are not so complicated. So, in that case, what we do is we write each FIJ as a product. So the product will consist of two parts. So there will be a part that we don't really care about that much, which is a part of roughly constant magnitude. And then there will be a kind of magnitude control term, consisting of 10 to roughly the logarithm of the FIJ. So once we have this, we can rewrite the equation using these. So I wrote C1 here to be like a product of a bunch of CIJs. And then this is multiplied by 10 times while you add the exponents that you came up with together. So now your equation looks like this. Okay. Well, if we're going to have a solution to this. So that two of these terms must be canceling each other out. So what that means is that these magnitude control parts of each of these have to be about the same. So, in other words, we have to have that the max of these sums that are occurring in the exponents occurs twice. Does that make sense to everyone? Okay, I'm seeing some notes. So that's encouraged. Okay. So let's take a let's take a deliberately horrible example. So let W be a virus trust function. So these you may recall are the. They're the continuous but nowhere differentiable ones, I think. Anyway, the point is that they're discussed in. And they're bounded. And we can like adjust the bounds a little bit something like that, whatever. So we'll say that zero is less than or equal to the magnitude of the virus trust function, which is less than or equal to two. And now we consider possible solutions to zero equals 50 plus virus trust of X times 10 to the X squared minus 100 plus 10 times virus trust of X times 10 to the X plus three. Okay, so I mean this is, this is, this is disgusting. Completely impractical, except that surprise this equation is engineered so that each of these terms has a really simple magnitude part. So, we can look at that through the manipulations down here. Okay, how do they end up with oh yeah yeah yeah. So we're sort of factoring as much magnitude as possible out of this part right here we want to put it in the prior form it's almost there but not quite. And when we do that. We have hours of 10 that are living inside come out and live with this 10 right here. And our equation is now five plus wire stress of X over 10. So very consistent magnitude times 10 to the X square plus one minus one plus virus trust of X over 10 again very consistent magnitude times 10 to the X plus. Okay, 32, but whatever. Let's see. I think, I think there was supposed to be a zero here and the rest of the slide is correct. Okay, there we go. Okay. So the solutions have to be near the sets where these two exponents match up. So in other words, near where x squared plus one is equal to x plus 32. So the possible solutions can only only occur near x equals one half times one minus five root five, or x equals one half times one plus five root five. Guarantee of existence here. This is just a necessary condition. Okay. So a quick question from the chat is how do you. Sorry, I guess there are two questions. Magnitudes here today mean absolute values. Let's see. I don't think I quite caught that was the question like does the does the magnitude consider absolute values or something. Yeah, or by the magnitude is that does that mean the absolute taking the absolute value. Yeah, yeah, absolutely. So we're going to apply like, essentially this logic with non ordered fields. And yeah, so that'll battle that won't arise until part four as far as as far as this application part. If you if you actually wanted to show existence I think you would need to like do a more careful analysis involving something like the intermediate value theorem. And then you would probably want to care about sign. But just for producing this like necessary condition that tells you that sort of bounds the region. Well, okay, I guess it doesn't boundary, whatever it kind of tells you where to look. You don't really need conditions on the signs in order to narrow it down that far. And there, there's a second question about how you know, how many factors of 10 to factor out here. Yeah, good question. So, the actual answer is that we just want these, these parts in here to have like roughly similar magnitude. Yeah, so there's no there's no like canonical answer for that. See anything else. Some folks are answering the questions in chat. Okay, great. Okay, so the third problem. So this is proof. I warned you guys on Twitter, this is going to be contrived. So, you know, hang in there. So the third problem is finding redundancy while optimizing over time. So, in this scenario, you're designing a protocol for powering down a nuclear reactor in very bad circumstances. By the way, this is hypothetical. So the power, the overall power on the plant P is going to go to zero over time. I've drawn this, I've drawn this nice diagram. And I'll go through and sort of explain, like, what's up with it and roughly what it's trying to encode. So, there are six systems by which I'm referring to these like gray boxes right here. I'm sorry, I didn't. I feel like I didn't respect color blindness in this palette very well so my apologies for that. Let's see. So, these are the six systems right here. So, each of them receives power from three sources. So we've got the main power. And then we've got something called the fizzy modulator here on the left, and we've got something called the spinning modulator here on the right. So, these are these are real nuclear reactor technical terms although it may not seem like it. No, they're not that was a lie. So kind of the idea here is that the power coming out from the fizzy modulator to these various systems, and the power coming up from the spinning modulator to these various systems is a function of two things which is first off, power as a whole, and second, how much of the main power you choose to shunt to those modulators. Okay, this is this is let's let's put this slightly more formally. So the thing you control is how hard the modulators work that is how much power they're outputting to various systems as a function of the main power. And then each of these receives power based on both the main power and the functions you determined that tell you how hard the modulators are working. So each system's level of activity is a function of the inputs. I've drawn up above here so these are the equations so for example. This, this system right here has the equation one over p times s, which means as the power goes to zero. So this, this term will say that the system works harder, and this term will say that depending on how hard the spinning modulator is working, the system will the work of the system will vary according to that, according to a factor. Okay, so each system's level of activity as a function of the inputs is shown above. And the goal is to have as many systems with little activity as p goes to zero as possible. So, you know this is, this is just like basic and clear reactor science I assume you all knew this coming in. Okay, actually anecdote. So I used to work at, I used to work at Java juice, and I was sitting in the back, reading one of the newspapers results and looking through the classified ads and there was an ad that said, nuclear power operator wanted no experience necessary. And I was really curious about that so I called the number and it turned out you were signing up to work on a nuclear sub. So, there you go. And the equations for the various systems up above here. So these color coded bars are supposed to indicate sort of like how much power it's getting from each of these and that's reflected in the equations but you don't worry about it. So, here's how, here's how you optimize having as few systems active as possible. The power goes to zero. So, you write f equals p to the x s equals p to the y. And what you're trying to do is you want as many of these to be minimized. Yeah, you want as many of these, which in some cases I've combined the terms to be minimized as p goes to zero as possible. So how did we get these, how did we get these functions. Well, we know that f is equal to p to the x, where x is like some some variable we can choose that determines how f will behave in terms of p, that's the control you have. You set a number x, and then that controls f in terms of p. Well, if we substitute that in up here, then we get the equations down here. So you want as many of these to be minimized as p goes to zero as possible. Okay, so if you look at the exponents of each of these. What we're asking is that as many of 2x plus 2x plus y 2y plus 2x y minus one and three need to be minimal as possible. So, if we look at the men plus arithmetic encoding of the question that I just indicated we were trying to solve. We're trying to get multiple monomials to matter in the following equation right here, when I say multiple monomials to matter these pluses are taking the men of each of these. And so you want multiple men's to be the same. Is that. Is everyone sort of on board there. Do you just treat these sort of like formal sort of in like a formal polynomial ring where you actually think of a real real value. Yeah, so this is this is all real valued stuff. Yeah, so these x's and y's are variables that you will input actual numbers into. And then you will apply these operations that are like tropical but they're like, they're not like just formal operations. Let's see. Okay. Any further questions. I'm particularly curious as to whether I got the role of x and y across. Okay, we can come back if it was unclear, please, please interrupt me. Okay, so let's draw the points in the XY plan where multiple of these monomials end up being the minimum. Well, if we do that. We get this. So these are the coordinate axes. And this is exactly the set of values of x and y for which multiple of these monomials end up mattering to the tropical outcome. If you've done some tropical geometry before you will recognize this as the tropical curves associated with the with tropical polynomial ring up here. Okay, well, if we look at this, we see that we need to examine the following possibilities. So this is, I'm not quite sure why I wrote this this way but this is what x and y are sort of by definition. And the ones that Mac that minimize the, the most number of things have to be located at three for negative one zero, negative two zero, or negative one three. So these line segments right here are what you get when two of the monomials end up having the same men, and these vertices right here are what you get when three of them end up having the same men. Okay, so we can also read off from this. There's a request would you mind revisiting what how x and y are defined. Yes, absolutely. Let's see. Yeah, so the control that you have over the system is what the output of the fizzy modulator and the spinning modulator is in terms of the overall power of the system. And what that means is that you have the ability to choose a number x and a number y. The contribution of the fizzy modulator to the behavior of each of these systems will be determined by p to the x. And likewise for the spinning modulator. You'll set s equals p to the y, and each of these equations up here. And we'll determine how hard the system is working. So x and y are real numbers that are that are under your control. Let's see. Okay. Another thing we can read off from this is that you can't minimize activity for more than three systems at once. Which I think is is like very non obvious from the presentation of problem or even from the tropical polynomial. Okay, so this is more AG right here. So this kind of relates AG stuff to the to the previous stuff I've talked about. And I'll try to be like explicit enough that the AG terminology doesn't just like overwhelm you if you're not like familiar with it. Problem four is this is this is not an actual technical term I just made it up for this is dilation stability in very affine varieties. This is an actual term, I have no idea why it's very strange. So if you have a close sub variety why of an algebraic tourists TN, you call that a very affine variety. Let me indicate what an algebraic tourists is. It's related to but definitely different from the product of s ones that you think might be used to if you are not like AG brain poisoned. It's the product of affine space minus the origin with itself and times. So it's, I guess, you know in the cases where you where you can say this it's homotopy equivalent to the tourists that you used to. But yeah, in AG that's what we call the course. Okay, so suppose we have a very affine variety. I should mention, if you, if you don't know what a variety is this, this might be a little rough, but you can think of it as a complex manifold, possibly with singularities, and you'll be fine. So if that's if that's more approachable. Just think of that. Okay, so one question is, what does why look like near the origin. So this is maybe not exactly the question will end up answering but it at least leads us to like think about the question will end up answering. So the usual AG answer is that there's a procedure called blowing up that you can do at the origin and AN. And then you can take the closure of your very affine variety within the blow up. And if you look at the exceptional, the exceptional fiber, you get a sub right of PN that tells you what directions, your why is coming in at. So, you don't quite need to understand the details there, but it's a procedure that exists. It gives you is like a bunch of tuples of numbers over your possible arbitrary field of definition. So, you know, algebra is considered that geometry but like, you know, normal people may not be as happy. So you might ask, can we understand this question numerically. Well, let's say for the sake of simplicity that why is actually like a hyper surface inside the tourists. And by that I mean that why is defined by the vanishing set of one equation. One function f inside of the coordinate ring of the tourists which looks like this. So, that means you've got a field and you would join a bunch of variables and their inverses and you take one polynomial in that ring. Okay, so we're going to experiment with this suggested concept by seeing what the limit as the approaches zero of f of z looks like as we approach zero from different directions. Okay, so this is not a well defined notion yet. So I'll figure out what that means, we would like to do something similar to the reactor problem. So in the reactor problem, we had one parameter in that case it was the power, and we set each of these variables to be defined in terms of that parameter. So the parameter is sort of like was, it was a regulator that we could compare the other variables to. Okay, so technical technical term. And so if you have a valued field, which means a field equipped with something called evaluation that sort of axiomatizes properties of taking a logarithm, which goes from the unit group of your field to the additive group of our the image being referred to as gamma. Then we say that that valuation that generalization of the logarithm is called split. If there's a choice of splitting for this shortage X sequence right here. There's a theorem that if K is algebraically closed, then a splitting always exists. It is also a theorem that didn't write down that your choice of splitting actually like doesn't matter for anything in the theory. Two choices of splitting will always give you like a canonically isomorphic collection of results. So why, why would we want a splitting. So the point is to mimic this property of power series where you can calculate evaluation by factoring out a bunch of powers of the relevant variable. So, let me actually write down what that, what that indicates. Let's see. So we write, if we have a function F that we want to take the logarithm of, we write it as F equals two to the W, F tilde with the valuation of F is equal to W. So splitting lets you always, always do that. Okay, so assume that K is valued and has a splitting. So, and for the most part, it's okay and tropical geometry and the valuation is trivial. You kind of recover like more, more torque geometry stuff when you do that. So let's assume otherwise for a moment. So geometrically, if you have this ring right here. K join a bunch of variables in their inverses and K is equipped with evaluation. The picture you want to have is this. You have a n dimensional tourists. So that's each of those variables gives you one factor of affine space minus the origin. And then you imagine that all of this is sitting over a tiny little infinitesimal punctured disk around the origin. So I don't really have. I don't know if I can like explain if you're not sufficiently, you know, unhealthy in an AG way. Why that's like the correct way to think of it. But that's kind of like what the algebra of such a ring is including. Okay, so, in particular, I would suggest that you think of this variable T that arises in the splitting as being actually a lot like the zi. It's just that it's encoding movement along sort of this infinitesimal access. So you can kind of think of your variety as TDN times. Negative epsilon epsilon cut out zero. Sarah, can we go back to where he was introduced. Yes, absolutely. So T is yeah so T is T is the splitting right here. So that means if you have a particular value of your valuation in mind. Then you can write it as the image of some variable, and this will be like compatible with the relevant group operations. So the way to think about that is if for example you have like a discrete valuation so this is like the case of looking at power series at a point. Okay, for complex for holomorphic functions. Well the, the image of the valuation map associated with that will be the integers. And for each of those integers. There is a function Z to that integer. So that the valuation will output it. So that's like a sufficient explanation. I don't know if I can think of this is like a uniformizer. Yeah, I don't know. Let's see. So I mean yeah certainly certainly like in the in the discrete case. Yeah, take a uniformizer. Okay, I guess what I want to ask is so we have this variety. And so you have this wise in the torus, and it is in which is in a and K affine case piece is required to be this valued field. Right. And for example it could be like this, this run series. Yeah, absolutely. Okay, let's see. So, here we're going to remember that that F is the is the function that our wise define defined as the vanishing set of. So we're going to learn how to take what's called the tropicalization of F. So the way you do this is that polynomials in this this ring right here. Have a canonical. This ring is canonically a vector space of this coefficient fields with basis the monomials. So you write it in terms of that basis. So you've got this, the sum right here these are the coefficients associated with particular exponents. And you define the tropicalization of F to be essentially, you take exactly what you wrote down and you replace the operations with O plus and no times. You know it's important that you do this like canonical decomposition first, and then you get something in the tropical semi ring that is your usual tropical semi ring. Hang on a sec union infinity. There we go. That is your usual tropical semi ring, equipped with a bunch of variables. Okay, so for an example of this procedure. So if F is this thing right here. So we're going to rewrite this in terms of no meals and the C sub I. And let's see. Oh yeah that's right. Yeah I forgot to mention, you do something to the coefficients. What you do is you take the valuation of the coefficients. So replace operations and take the valuation of the coefficients. Sorry. Okay, so you write these things in terms of the no meals and then you factor out as many powers of tea as possible. And then that will tell you what the valuation of the coefficient is. So for example, here. T has valuation zero. Here T minus one times T cubed has valuation three. So here you get a coefficient of zero remember coefficient of zero is with tropical times so it does not like kill the term. And here you get a coefficient of three and so on. Okay. Now, if we substitute. We do the w I for Z sub I, and w I here is some choice of weights that are in the image of the valuation. Then the tropicalization of F evaluated at the W sub I is calculating the exponent of the term which will dominate as T goes to zero. So like filter through. We're almost done by the way I know we're almost out of time. Things are winding up. Okay. So, for example, if we have F of so here are weights are we choose w one equals one choose w two equals negative two. So, if we evaluate F of T to the one comma T to the negative two, which is this substitution right here. Then we get out one minus T times T to the negative two plus T minus one times T squared plus T minus one. And as T goes to zero the term that will dominate here as well is the one with the most negative exponent. Which is that's right here to the negative two. This will kind of fade away. Okay. So, as it happens if you take the tropicalization of F and you apply it to the weights one negative two. Then you're computing exactly the same thing you're taking the minimum of negative two, which is this exponent here to which is this exponent here and zero which is the exponent associated with this. And you get negative two corresponding to that. Okay. Okay, so suppose that's only one of the terms, only one of the monomials for a particular choice of weight w contributes to top F of w. Then, as T goes to zero F will approach that monomial. So, there's a couple possible cases here. So if alpha this exponent here is is zero. Then F is approaching the coefficients. And so the vanishing set of F will approach the empty set. And F will approach a non zero constant. If alpha is not equal to zero. Then F is approaching this monomial right here. And so the vanishing set of F will degenerate to a bunch of axes and planes and infinity of an. To be clear, we might have like, say, Z one over Z two is what we end up with with a particular choice of weights. Let's see. No, that's not a good example. Okay, I'll come back to it after I after I go over time. So from, from TN's point of view what this is saying is that again the vanishing sets of F is just like disappearing from our field of view as T goes to zero. So, this suggests the definition call the points are and we're multiple monomials contribute the vanishing set of top of F. So if we restate what we just said, the vanishing set of top of F encodes the weightings of variables for which simultaneous dilation according to the weights can yield the generations of why the actually live inside of TN. Okay, that's all I have to, that's all I have to present. So I should have stepped an example for this in here, but it doesn't matter. I went one minute over time anyway so it is actually a brilliant decision on my part. Let's go ahead and thank Sarah for a wonderful talk. Anybody, anybody have any questions. Sarah, can you go back like a page or two. No, this one, this one, this page. So, in your example, there was one plus T. This had no power of T. So shouldn't be when you take the valuations that should this be a negative infinity or infinity. I'm sorry for the X negative to the negative two. Sorry, I believe you're muted. Sorry. Yeah, so this says this has a valuation zero. So you can, because it's because it's a unit essentially in the, you know what I mean in the in the valuation ring associated with it. Yeah, you know, like one every T has valuation negative one T has valuation one so if you multiply the two together you get like something that's pretty similar to one plus T so it has valuation zero. It's weird to see there's zero in the coefficient. Yeah, yeah. Yeah, I think it's best to just try not to think about it. You know, which is arguably true of all of us but So there is there's a question chat. We get off here. The question is that when we pick off the trivial valuation this recovers some work stuff. Any idea or hints on what that might be. Yeah, let's see. So for one thing. If you take a polynomial and you have a If you take a tropical polynomial and anything about this for a sec. Okay, no, no, the correct statement is, if you have, if you have a polynomial and the valuation on your field is trivial. Then, when you take the associated tropical hyper surface. So, for example, like the curves we saw in the case of two variables that is like the one dimensional skeleton things. What you will get out will actually be the skeleton of a fan. You'll get something fan like like with non trivial valuations you can get sort of like polyhedral complexes. Like this. And then with with trivial valuations, you will only get things that that look like if you take a vertex. And you look at like a little neighborhood of that and you extend it out forever. You'll get things that look more like that. So in general the connections between like, between torque varieties and tropical geometry are pretty close. So my suspicion is that if torque stuff were developed today would probably be written in terms of tropical geometry. I want you to sort of like a canonical source in the connections there which is, there's a textbook Diane McLean and chapter six is all about connecting torque varieties and and tropical stuff. Any other questions. Thank you. Thanks for once more.