 So first I just like to say, you know, thanks to the organizers for putting on this fun conference. It was, I thought it was really neat how, you know, as everything started closing down, you know, people react in lots of different ways and a lot of mathematicians said, okay, now we can take this opportunity to put our seminars online and to make online conferences and just reach out to each other. And that's been, you know, I've appreciated the way that the mathematics community has done that for each other. And I think it's really cool. So especially I'd like to thank the organizers of this, of this conference for doing that. And, you know, the speakers for having given some really cool talks that I've all enjoyed so far. And then thanks to everyone else for coming to my talk. So let's see, without further ado, I'm going to talk to you about so an application in condensed matter physics of something called boredism, which, which I'll explain. And then the main calculation, the calculational tool that you can use to compute these things, the atom spectral sequence. So, okay, this talk will be in three parts. So first is from, you know, from SPTs to boredism. So SPTs are the condensed matter physics application. So I'll say very briefly what that is, and spend a little more time talking about boredism, which is a mathematical tool that allows us to understand this physics question. So this is related to the homology of boredism that was in the previous talk, but they're not exactly the same. So then second, there's a number of different ways to compute boredism groups. I'm going to tell you one of them called the atom spectral sequence and walk through an example computation in with the idea that, you know, spectral sequences have this reputation of being a little bit abstruse or terrifying. And in, in many cases, in this specific application, it's actually a little nicer. So I'll indicate what parts, you know, you can teach to a computer and what parts require a little bit of cleverness and show you what the computation looks like. And then in the last part, I'll tell you about a slight generalization of the physics application and some work in progress of my own on using or computing boards and groups and then using them to say things about physics. So then let's see, yes. So boredism is, well, let's talk about the first thing in the, in the three things in my title, boredism. So you may have also heard this called co-boredism and I'm not going to distinguish between the two. So just to begin with, consider all closed end manifolds up to defumorphism. And that set is a commutative monoid under disjoint union. But it's way too big, like, like actually calculating this above dimension two is just, you know, dumb, impossible. So let's take a quotient of all manifolds which bound something of one dimension higher. And in order to get something interesting, you have to say that the thing in one dimension higher has to be compact. So we're saying we want man of closed end manifolds up to an equivalence relation, where if the two, if M and N jointly bound a compact N plus one manifold, then they, then we want to consider them the same. So the board and boredism means boundary. So fun fact, what you get out is an abelian group, not just a commutative monoid. It's called the boredism group and it's denoted omega N. So there's there's many variants of this. So what one that I particularly like is that you can say, well, let's consider boredism of manifold with G structure. So a manifold has transition functions valued in GL and R. And you can so once you choose an atlas that is. And so given a group with a map to GL and R, you could say, well, I want data of a lift across that map. So transition functions valued in G. So for example, if you take the connected component of the identity in GL and then the, so the functions with positive, the matrices, sorry, with positive determinant, then a G structure for that group G is an orientation because you're saying all, all transition functions preserve the orientation. So now, okay, oh, there's a question. I think I saw that someone raised their hand. Is there Hi, I did raise my hand. Sorry, do you just a quick question. Do you mean the derivatives of the transition functions? Um, I don't think I mean, the transition functions are defiomorphisms. So you pro see you want to say the differentials of those transition functions. Oh, sorry, you're right. Yes. Yes, that's right. So you could say it that way, or you could say that the transition functions for the tangent bundle. And I think those should be, those should both work. Yeah, thank you for clarifying. Any more questions so far? Okay. So when I say M equip these with compatible G structures, I'd like a G structure on M, a G structure on W, and the induced G structure on the boundary of W to be equivalent to the one on it. So then you get oriented board ism spin board ism lots, lots of interesting variations. You could also consider board ism of manifolds together with a map to some fixed space X. So this means that so M has a map to X, W has a map to X and the restriction of W, but sorry, the restriction of the map on W is this is the map that you started with on M. So as X varies, this defines something called a generalized homology theory. What that means is that this behaves a lot like ordinary homology in the sense of Meyer v. Torres, the long exact sequence of a pair, et cetera, et cetera. But it doesn't have the same value on a point. And so it, you know, it's not exactly the same and generally a little harder to calculate. So the fact that it's a generalized homology theory is I mean, this is a good sign in terms of algebraic topology because it means we have, we have tools for dealing with them. And indeed, computing boards and groups is a historical classic problem in algebraic topology. So many important concepts such as the Tom space and I think even transversality were invented in order to tackle this problem. So that's a foundational work of Renee Tom and Lev Pontragon. And then further important work was done by many authors, including wall for the oriented case, Anderson Brown Peterson and the spin case Quillen and the almost stable almost complex case, and then many more. So just to give you a flavor of what this, what this looks like, the unrated boards and groups begin. So in dimension zero, we have Z mod two, then dimension one, everything bounds circle bounds a disc. In dimension two, we have a Z mod two generated by RP two. And so it's an interesting theorem that the real project of plane is not the boundary of a compact three manifold in dimension three, everything bounds in dimension four, you have two generators, RP four and RP two cross RP two. And then in dimension five, you, you get the first thing that can't be produced by project of spaces. And so you can pick as your generator this particular space, the wound manifold. So for another example, oriented borders and groups begin Z and dimension zero generated by a point, nothing in dimensions one, two and three. And then dimension four, you get another copy of integers generated by CP two. And then dimension five, you get Z mod two. And again, it's generated by the same, the same thing. So this is just a taste of how these things look. So okay, great. Why? So there are a bunch of applications of this. And here's the one that I care about today. So this is a theorem from 2016 of Dan Fried and Mike Hopkins, which so there's a lot of words on this page that may not have appeared before. So what I will I will explain what some of these words mean. So first of all, there is an abelian group of these things. And I'll say something more about those words. And the point is, if you care about those things, which I will, I will say a reason to care about them later, then you can compute them in terms of borders and groups, the torsion part and degree and in the free, the free some in the degree and plus one. So this is not a naturalizedomorphism, but it can be upgraded into one with, if you say it's slightly different. Okay, so the words. So quantum field theory is hard. And in most cases, there is no, like most quantum field theories that physicists care about, admit no rigorous mathematical formulation. But there are special examples called topological field theories that have a rig, a rigorous mathematical formalization due to Michael Latia and Graham Siegel. So this is, it's stated in terms of co-boardism, but it's a little, it's a little fancier. There's a boardism category and you work with that. So we're not going to need to really engage with topological field theory in this talk. So I'm not going to say too much more about the definition, but it is something I find cool and would love to discuss if, you know, if anyone wants to hear more about it at some point. So, okay, reflection positive. This is, I mean, it's a technical condition and it is quite technical, but it amounts to some sort of equivalence data under orientation reversal. And so if you, if you are studying topological field theory, you don't have to have this. But if your example came from physics in any reasonable way, it will almost, it will probably be reflection positive. You know, I can't turn that into a theorem because what does physics motivated precisely mean? But this is a, this is, this is sort of not a surprising constraint to put on your, on your theories. Invertible though is more surprising. So topological field theories are easier than quantum field theories, but that does not mean they are easy. So in general, classifying them is very difficult and invertible topological field theories. So I'm going to call them IFTs to save syllables. These are very, very simple examples. They are almost but not quite trivial. And the fact that there are non-trivial invertible field theories is, I mean, this, this is a kind of interesting thing. This is the, maybe some of the first examples you can study when you're trying to understand some fact about topological field theory or even mathematical physics. So invertible means, you know, small or nice or special. So n-dimensional means we're looking at manifolds of dimension n and then G structure is what I said earlier. So I think that's all of the words here. Oh, deformation classes, which means that there, there's, you can, this is the statement about saying, oh, well, I want to consider families of these. And if two things, if one can be continuously transformed into the other, then I want to say that they're the same. I'm not going to say much more about that because those details are not needed in this talk. Are there any questions about this slide or previous slides? Arun, I think it's worth mentioning that some fancy four-dimensional TQFDs like Donaldson theory are neither invertible nor reflection positive. Donaldson theory is not reflection positive. Oh, no. Okay. That's, wait, that's interesting. So then, as Ryan mentioned, this is, so, so certainly one thing which I, which is important is most TFTs are not invertible. So Donaldson theory, I didn't realize Donaldson theory is not reflection positive. That's, that's interesting because it's definitely physics motivated. Okay. So I should, right. So when you do the, when you do the topological twist, you ruin unitarity. So oh, okay. Violate spin statistics, right. Okay. I see. So, so I think the reason this is going on is that there's two ways that people write down TFTs. This is like the Schwartz versus Witten type thing, right? And so if your TFT comes via some sort of topological twist of a supersymmetric QFT, then so is that, should I expect that in those kinds of examples, you'll also ruin reflection positivity? Yeah. Okay. Oh, interesting. Okay. Great. Well, I'm happy to have learned that today. That's not something I knew before because I don't know much about those, those theories. Brun, there's also a quick question from Jack. It's great. That is a topological field theory. The same thing as a TQFT. Yes. Sorry. So sometimes people refer to topological quantum field theories. So TFT, TQFT, topological field theory, topological quantum field theory, you should think of as the same thing. In addition to saving space on the slide, I was recommended by someone that the axioms of TQFT could encode classical field theories as well as quantum field theories in some settings. And so a priori, quantum is maybe not the right word, but people use it interchangeably. So the, sorry, to summarize, yes, they're the same. More questions. Okay. Please keep asking questions as you have them. So maybe this theorem says that boredism is related to some pastiche of words. Great. And now I'm going to tell you why that one application of that pastiche of words in physics. So condensed matter physicists study many things, but among them there are these things called topological phases of matter. And so, and they want to classify them. So what exactly is a topological phase of matter? And that's not something that I can precisely tell you, both in terms of what I know and what in terms of what the, what physicists know as a general formalism. But we have plenty of examples that motivated the theoretical considerations. And these tend to take the form of, you know, take some alloy of a couple metals and cool it to a certain temperature and maybe you like apply a magnetic field or something. And then you get this very weird physics behavior that the people who study these things were not used to. So one concrete example that can occur is you get these things which behave like particles, but they're not bosons and they're not fermions. And so that seems physically impossible. And so the resolution of this riddle is that they're not really particles. They are, you know, there are patterns of entanglement of electrons in these materials which look like particles but are not. And it's convenient when you're studying these to pretend that they're particles and that you're working with some new physics system. But this is, I mean, this is already very strange and very strange is a great place to say, well, let's, you know, crack open this puzzle and understand what's going on. So that's where that's what a lot of condensed matter theorists are interested in. So let's try and classify them. Oh, that's very difficult, it turns out. So we're going to restrict to a very nice, very simple subclass called symmetry protected topological phases. So what this means is that there's a specific way of combining two different phases into another phase. And a phase is symmetry protected topological or SPT. If you can do, if you can combine it with something else to get the trivial. So you can think of these as invertible. And all of this is fuzzy in the sense that we don't have great mathematical formalisms for this. But this, this should feel, this is similar to the notion of an invertible topological field. So just to briefly say, how are these things modeled in terms of physics? How do people work with them? There's a few different ways. But a, the most common I would say what it is you're doing this on some manifold, you triangulate your manifold. And then using only the combinatorial data of the triangulation, you try and write down a Hamiltonian system and then do this sort of quantum mechanics like stuff. So, right. So the physicists studied these and concluded that once you fix a dimension and what kinds of symmetry you care about, you get an abelian group of these. And so there's been in the past decade or so, there's been a lot of research on computing. What, what are these abelian groups? And so these computations proceed via physics methods that sometimes, you know, admit clear mathematical formulations and sometimes don't. But in general, this, we still don't like making sense of these rigorously, still a hard open question. So, you know, you can't just take the space of Hamiltonians and try and compute like pi zero of it or something. That's, you know, we're nowhere and you're being able to do that. So we're going to try and go about this a roundabout way using top invertible field theories. So there's this general Anzatz coming from physics ideas that there, there should be a way to take a low energy effective field theory of an SPT phase, which is some physics thing that we don't know quite how to mathematically define yet. And what should pop out is an IFT in the fully mathematical sense. And this moreover, you know, while we're dreaming, this should be an equivalence between the classification of SPTs and the classification of IFTs, which is great, because we know how to classify IFTs. And there was that thing I said about reflection positive, but these examples are expected to be reflection positive. Because, you know, as Ryan mentioned, the issue is something called taking a topological twist, and that does not come up in this case. So assuming this Anzatz, we have a math map, which, okay, well, you know what happens when you assume, but assuming this Anzatz, we have a way of computing these using math. So great, next step, turn the Anzatz into a rigorous theorem. Nope, that's very, very hard. So that's not the next step. So that is, I mean, that is something that would be great to have. But right now, I don't think we haven't, we understand these enough, even at the level of physics, to state precise conjectures. So we're going to have to, if we want to test this Anzatz, we better do something else. So one thing we can do is we can compute a bunch of boredism groups and use them to say, here's what the Anzatz predicts in these examples. And you can compare with what the physicists produced by a bunch of other methods. And well, if they agree, that is a good sign that your Anzatz is doing the right thing, especially in non-trivial cases. So this has taken up in the same 2016 paper of Dan Fried and Mike Hopkins, and then in a subsequent paper in 2017 by Jonathan Campbell. And the conclusion is that the answers agree. So that's nice. Oops, sorry. So this tells us that this approach is, you know, this is a good sign that this is something that we could use to understand physics better. And indeed, in the last couple of years, physicists have started using this to this courtism method to compute groups of SPTs in other cases. So next, I'm going to go on to showing you how such a computation might look. But before I do that, are there any more questions that I can address? So coming up next, the atom spectral sequence. But so there are several steps. And some of them are easier. And some of them require either ingenuity or trying everything you get until you run out of ideas. And I'm going to show you what those steps look like on a specific example. And that's G-Boardism for this particular group. Great. That's maybe not everyone's first example of a leak group. So what this is, is pin plus is a version of the spin group, which itself is a double cover of SON. And pin plus is like, I want spin, but I also want to do it on unoriented manifolds. And then we're taking a semi-direct product with the integers mod n or the cyclic group of order n. So here this acts on this where pi zero is the mod two and it acts by inversion through pi zero. So this is, I mean, I'll tell you later some reasons for caring about this particular one. So putting a G structure on a manifold for this is sort of like having a spin structure and a cover whose structure group is the dihedral group. But it's not necessarily, it's not exactly that, but that's the kind of data you have. So if you're doing geometry with this thing or physics, you know, you have analogs of things like the Dirac operator, but maybe they don't behave as nicely as you'd like. And then you have this dihedral group floating around. So you could turn this into, you know, you could translate the borders and groups here into some particular kind of SPT. And it's, you know, that would be already an interesting question, but sorry, that would be an interesting application. But I care for a slightly different reason, which has to do with crystalline phases and which I'll talk about in the last part of the talk. So the, I picked this example though, because it is neither too simple nor too complicated, I think. So hopefully we'll illustrate the concepts nicely. And so that combination of things, you know, the Goldilocks zone in terms of, in terms of how elaborate it is and the fact that it is an example that people care about, or at least that I care about, is why I picked it. So for this next section, there's, a lot of this stuff has been known since the 70s, or maybe even earlier. But there's this paper from a couple years ago by Agnes Bodry and Jonathan Campbell called A Guide for Computing Stable Homotopy Groups. And this is a really good expository reference for all the stuff that I'm about to show you. It fills in all the details that I'm not gonna, that I'm not gonna dredge up. It goes into, you know, it doesn't make assumptions on what you know beyond some basic facts about algebraic topology. And it works through several examples. So when I have been learning this stuff, this paper has been a really great way to, you know, to fix some of my confusions and to just teach me how to actually work on this stuff. And so if you're interested in these things, that's the premier reference I would recommend as a place to, to learn more. So here is the signature of the atom spectral sequence. There's, there's a bunch of things and I'll, I'll explain in the next couple slides what they, what they actually mean. But briefly, if you haven't seen spectral sequences before, that's okay. The idea is that the left hand side, the E2 page, is an approximation of the right hand side. And ideally, it's, the left hand side should be easier to calculate. So let's just do a sanity check. X is some space and these are its homotopy groups. We'll worry about the two, the two and a half later. And here we have some thing and we have cohomology. So in, for most, for many spaces that you might care about, cohomology is easier to calculate than homotopy. And certainly this is true for manifolds or things that are built out of manifolds or most spaces that what the algebraic topologist comes across. So therefore we might, we might expect this is indeed easier to calculate assuming X is nice. The law of conservation of effort says this can't be too much easier to calculate because this, it's, this is saying cohomology tells you homotopy. Okay, well, there's a lot of stuff going on here, but that's the hope. So something interesting has to happen in this text. So, but it is true that this is easier to calculate than this and directly and this provides information on the right hand side. But the approximation is not perfect. There are things called differentials and hidden extensions, which get in the way. And I will talk about those a little bit and why you don't have to worry about them in, or how to not worry about them. So the first thing is the steamrun algebra, which is denoted mathcal A. So this is by definition, it is the algebra of stable mod two cohomology operations. So what that means is what are all the things you can do to cohomology naturally. So given a mod two cohomology class produce another one in a way that's compatible with pullback and with suspension. So this is a complicated algebra. It's graded by the degree. So if you raise degree by K, you're in degree K of the algebra. It's not commutative. And it's also infinitely generated. So this is, this is large. But, and so what this is saying is that if you, if you understand the steamrun algebra, then you understand, like only acting on cohomology, you understand a lot about your space, like it's on the topic. So a sufficient but slightly too large list of generators is things called steamrun squares. So square K raises degree by K and we do this for all natural numbers. Defining them is a little bit prickly. So I'm not going to, not going to get into that. But they, you can work with some axiomatic properties they satisfy, which usually suffices. So the upshot is elements of the steamrun algebra act on cohomology. And in particular, the co, the mod two cohomology is always a graded module over this. And this is natural. So pullback maps and cohomology are linear with regards to this module structure. The module action does not, it play, it, it does not make this into an A algebra. So cut product does not play perfectly with this, though it does play interact pretty well. Just not quite good enough. So that's, that's the first thing. Now what's X'd. I should have called the slide. Thank you, X'd, but I forgot to. So X'd is a way of giving an algebra and two modules. You can get out an abelian group, which is built out of, out of extensions of length s up to some notion of equivalence. So you may have seen X'd one as measuring ways in which you can write, say, two abelian groups and say, okay, well, I'd like to put them together in a way that might be more interesting than a direct product. So X'd is just saying, well, let's do that for things of larger, like longer sequences. So defining, defining the group structure showing, showing that this is in fact an abelian group is a little bit annoying. So I'm not going to worry about that. But that's concretely what this is. And so alternatively, you can say, you can define it very quickly by saying that this is the right-to-right function of HOM if you are, if you like homological algebra, but you don't, you know, those are different definitions that different people will prefer. So there are two, there are two indices here. And the second one is just given by degrading on the Stienron algebra. So you say, I want to compute these extensions, but I want to take this thing, the beginning of the extension, and shift it up in degree. So I always think of this as an actually moving upwards. And so later when there are some pictures, you will see that, like why I do that. So this notation is reminiscent of the suspension of a space, and that's not a coincidence. Finally, this thing says that you only recover the two completion of the homotopy groups. And so that under reasonable circumstances, including anything that, you know, any of these tau and spectrally, sorry, any, any choices of X that you might consider for computing spin boredism, then what this tells you is the free part of the homotopy groups and the two torsion part. And that's it. There's no information on odd primary torsion. And this sounds sad, but at least in the case when this is telling you boredism groups, which, you know, I'll have to say something about what kind of X tells you this, then there are other ways to compute odd primary torsion, which are usually easier than the two torsion. So I'm not going to worry about that much in this talk. But it is, I mean, there's, there's lots of fun ways to do that. And again, they're usually not as much, they're usually not as painful. So, okay, before I go on, let's, can I check for questions again? Are there any questions? Yeah, so are those other methods that are easier, not like the mod p atom spectral sequence? So that's one thing you can do. But another, so I'm going to talk a lot about spin boredism today. And it in low dimensions. And so what you can do in that case is if you're, if you're over an odd prime, that spin spin boredism looks the same as oriented boards. So m spin is equivalent to MSO once you once you invert to. And so therefore, you can just use, for example, via tier Hertzberg spectral sequence for oriented boredom, which is a lot sparser than for spin boredom. And so if you care about things up to dimension five or six, because those are the things that physicists care more about usually, then you that's often good enough. And you don't have to take out the odd primary atom spectral sequence. And then I mean, there's, there's other ways of doing it. But that's, that's one. And then there's also the fact that, you know, the atom spectral sequence at odd primes is, I think, often easier itself. Though not in this case. Um, did that answer your questions? Are there more questions? That answers my question. Okay, thanks for asking. Okay, so I think I haven't, I haven't heard any more questions yet. But again, please, please keep asking questions. I'm always happy to hear them. So this, you know, what algebra I told you is infinitely generated. So computing its x is hard. Um, okay, well, that's not, that's not an implication, but it is true. And we should so we'd like to make this problem simpler. So let's make an assumption that we care about spin borders. So where is in the manifold of the spin structure and a map to X, where X is a space, maybe a spectrum, if you know what that is. So I'll go about justifying that assumption later. So hard work of Anderson Brown Peterson from the seventies, or maybe late sixties, showed that spin borders is determined by things that we understand. So if there's something called connective real K theory, which is not a little KL, and then there's some other stuff, which is usually related to KL theory. And below degree eight, in fact, spin boards and isomorphic to KL theory. And so this below degree eight is particularly good, because as I said, physicists tend to care about things in low degrees, maybe not string theorists, but that's, you know, there are other ways of dealing with this above degree. And now something really convenient happens, which is that the cohomology, mod two cohomology of KL, this spectrum, the representing object for KL theory, is this quotient of the Stienron algebra by a subalgebra. And I will define the subalgebra in just a sec. But the upshot is that you can invoke a change of rings theorem that someone proved about X and massively simplify your problem. So the the input to the added spectral sequence looks the same. The only thing I've changed is that there's an A1 here. You take you take the X, so extensions by two of your your cohomology, but now it's by as modules over a much smaller algebra, it's generated by two generators, and it's finite dimensional even because they satisfy some relations. So this is tractable enough that you can like work things out on a piece of paper and draw pictures and generally like actually do stuff. So this this is a very, very fortunate thing for for these examples, because otherwise we'd all have to work a lot harder. So, okay, it looks like I put a pretty strong assumption on there. So now, so now I got to write these borders and questions as spin boredom or something. And that's not always going to work. So in it turns out, depending on the kind of question you ask, this can be easy or hard or impossible. And so the running example conveniently enough satisfies this assumption, and I'll say what kinds of things do. So one large class of examples is something called a twisted spin structure. So what you do is you pick a vector bundle over some space. And because this is a virtual conference, maybe you should pick a virtual vector bundle, which is a formal difference with two vector bundles. So if you if that is a new concept, don't worry too much about it, because we're not going to deeply engage with it. So now given another vector bundle over another space, a twisted spin structure is a map here map from m to x, so that you can pull back v and then a spin structure on these two together. So, okay, the point is a lot of important examples can be described in this way. So spin, a spin C structure is a twisted spin structure where x is CP infinity and v is the tautological bundle. Pin plus and pin minus structures, which are unoriented versions of spin structures, admit descriptions in this way, where this is rp infinity, and then many more. So if your problem is spin like or feel sort of like spin, or if you're a physicist, if you're looking at fermionic phases, then there's a really good chance that you can that you can invoke your assumption by writing this as a twisted spin structure. So I'll get to the example in just a moment, but the fact that is that helps put these two together. So going from twisted spin structures to the assumption that it's spin spin boredism of something is that the boredism groups of manifolds with this structure is the spin boredism of something. And that something is, well, xv is called the tom spectrum of v. And then the minus dim v is saying, well, let's shift that so that it's cobalmology groups are in the correct degrees. I'm not going to define tom spectrum. If you've heard of a tom space, and if you've heard, and or if you've heard of a spectrum in homotopy theory, then this is a spectrum built out of tom spaces. If you've not, then the point is that tom spectra are they are nice spaces from the lens of algebraic topology in that it's possible to understand both their co homology and sometimes with a lot of work, their homotopy groups. So the fact that they have something to say about boredism is, again, it's quite fortunate. So for our running example, PIM plus semi-direct CN, this is equivalent to spin boredism over the classifying space of the dihedral group. Sorry. First, let me say what this is. So d2n is, you know, it's the finite group of rotations from flexions of an n-gon. So bd2n is pick pick a as any example of a space which has d2n as its fundamental group and whose universal cover is contracted. Excuse me. So it turns out all such spaces are homotopy equivalent and everything that we're going to do today doesn't care which one you pick. So given a representation of d2n, you can produce a vector bundle over the classifying space by saying, okay, well, I want the monogamy on my representation, or sorry, I want my monogamy on the fiber to be given by, so on a loop which is given by an element of this, I'd like the monogamy to be how that loop acts, how that element of d2n acts on the representation. So let's do that for the standard representation in dimension two of the dihedral group as rotations and reflections. And then we're going to have to take the virtual bundle, which, okay, whatever that means, it's something that can be dealt with. And so we get this notion of twisted spin borders. And so that means that if we understand the tom spectrum associated to this data, then we can start computing the e2 page, so the input data of the atom spectral sequence, which, okay, this still seems elaborate, but many parts of what I just told you simplify. And it's actually not as bad as it seems. Before I go on to the tom isomorphism, actually, more questions. So first of all, if you want to understand a tom spectrum, so we want to understand the monocochromology. And there's a theorem which just says that the monocochromology is the same. And okay, there's a degree shift, but we shifted down in the same degrees. So this actually will, that degree shift will end up going away in our example. So typically, so this is an isomorphism of abelian groups, but not of rings. And typically, we write this as x goes to the tom class times x. And there's reason that this is all for a good reason, which when you look into the topology of how this is actually implemented, there is that's, you know, this notation is the correct notation to use, but that's out of scope for right now. So I want to explicitly mention that this is not just some cute theorem that we happen to have in our pocket here. This is an important, important theorem in the algebraic topology of manifolds. If you want to set up generals, like it can be used to prove along with some other stuff, punk or a duality. If you want to understand punk or a duality or orientation for generalized cohomology theories, you use this. If you want to use, if you want to say how intersection theory relates to the cup product, you use this. And there are a bunch of other things. So this is, this is a good theorem. I know, okay, all theorems are good, but like this is a particularly good one. So we know the cohomology as an appealing group. Now we need to determine the steam rod algebra action. And, okay, we really only need square one and square two. But the general story is not much harder, at least in theory. So the idea is that the vector bundle V has characteristic classes called Schieffel-Whitney classes, and they tell you the action. Now in practice, if you have some particular example, determining the Schieffel-Whitney classes of your vector bundle, you know, it's not, it might be completely trivial. It might be a little bit nontrivial, but it usually, you know, there's sort of ways that you can check and routine ways to check that allow you to obtain this information. So thus far, the cohomology of the tom spectrum is, you know, not, not too, not too scary. The Schieffel-Whitney classes are a little bit interesting, but there's nothing, you know, usually nothing spicy going on here. And so therefore you get the steam rod action. So the concise way of saying it is the i-th steam rod square on the tom class is the tom class multiplied by the Schieffel-Whitney class. So then you say, okay, well the carton formula tells you how to compute square i of u times some other class x in terms of this, the action on u and in terms of the action on, on, on your cohomology class x, which came from the base space. So in short, if you know the Schieffel-Whitney classes and the cohomology on the, on the base, then you know the cohomology, the steam rod action on the cohomology, the tom spectrum. And so this is the kind of thing where, you know, I'm sorry I said it very concisely, but after playing with a couple of examples, it becomes, you know, like it's pretty routine. So thus far we haven't needed to be creative yet. Just, it's the kind of thing that after you play with a couple examples, you start to get the feel. And now, you know, every topology talk should have some pictures in it. So we're going to draw these, these modules. We're going to draw this algebra and it's, so it's, the pictures really helped me at least to get a grasp for how to work with these things. So the rules are you draw a dot for every some end as an, as an abelian group and you draw them in, in height corresponding to their grading. So increasing grading is going to go upward. I think this is called the optimistic convention for drawing these things. So you want to keep track of the, the action of square one and square two, the generators of A1. So first, first, you know, so let's go back to square one. We draw a straight line from A to B. And for square two, you draw a curvy line. Oops, this should, this B should be C because I want to say square two is indicated by a curly line. Sorry about that. So here's a picture. So A1 is a module over itself. And it has eight eight generators as an abelian group. So here's, here's the identity, which is a cohomology transformation. Here's square one. Here's square two. This one is, okay, do square two and then do square one. And so forth. This would be do square one and then square two. And so the point is there are relations. So for example, square one, square two, square one is the same thing as square two, square two. So this is, I mean, this, I like this picture, but it's, like I said, it helped me really get, go from, okay, here is this nebulous thing in terms of, defined in terms of other nebulous things to, okay, well, this is kind of cute. So just as another example, here's the cohomology of infinite dimensional real projective space. And so there's one, there's one generator in each degree, sort of like how the module cohomology of RPN is like that, but it stops above degree N. This one, well, there's no degree N, it just keeps going. To actually determine the steenrod module structure, it actually follows because this is really generated by a single element in degree one, you can use the basic axioms of the steenrod algebra to generate, to compute these, these square ones and these square twos without having to think too hard. It just, you just multiply some, some polynomials. And that, that's totally something you can teach a computer to do. So maybe this gives, gives an impression of like what these things might look like. So the next example is going to be a little bit fancier. It's our running example. So what's going on here? So the cohomology of BD2N has N plus one generators in degree N. So this starts small, but then it keeps getting, you know, it gets kind of big. So I've only shown, so what I've done is I've taken everything, every element in degrees seven and below, and then that generates several, you know, several summands of the entire cohomology. And then I just keep those. So this includes everything in degrees seven and below, and then anything you can get to by starting with them and using A1 acting. So this one will keep going on forever. So what's there to say about this? Okay, so these are the Stiefel-Whitney classes in terms of the generators. There's ways to calculate this. In this example, you can look it up in Peter Teichner's thesis. In other examples, you might have to use some group theory representation theory, but it's usually not too, too bad. So let's look, looking at the structure here, we have a bunch of copies of A1 maybe shifted up by a degree. So these things are nice. We'll see later that they're easy. And then we have this infinite summand, which is not a cyclic module and not even finally generated, but that's okay. Many things are nice because we only have to worry about degrees up to like, say, five or six. I've colored each summand a different color of the rainbow because we're going to be, we're going to be in, when we do more stuff with this, you'll see what, how they contribute to the, to the X, because, well, I'll just go ahead and do that in the next slide. But first questions. Yeah, I have one question. Great. I was a bit confused. So you is the Tom class, right? Yes. So isn't you like in the, the degree of that class is like the dimension of the manifold? Yes. So that's generally correct. And so we've done a trick here. I'm going to go back to the slide where I say it where, in this trick, so can people see when I highlight? Yeah. Okay, great. Good. I haven't been doing this for, for half an hour, for no reason. So this minus dim V that, that I didn't really talk about is saying, okay, instead of shifting up by degree, we're just going to move everything down to sort of the Tom classes in degree zero. And so this amounts to desuspending your space, your space, and you can't actually invert the operation of suspension on spaces, which is why this is a spectrum and not a space. So spectra is the category of, you can define in several ways, but one of them is you could take topological spaces and invert the suspension function. So basically because of this sort of parlor trick of putting every, putting the Tom class in degree zero, you don't get the shift. And so that's why in this, the Tom class U is in degree zero, exactly like you asked. Does that, I realize that's a bit, you know, pain no attention, but to the behind the curtain kind of answer, but is that hopefully that is a helpful answer? Definitely. Okay, thanks. Are there more questions at this point? Okay. So just to summarize, we've determined the cohomology is an A1 module. Oh, and I should say something about how I, how I did this picture. So once you know, once, so given the Stiefel-Whitney classes and the cohomology, the, this formula I mentioned, oops, right here, is pretty, it is pretty formulated. So you can teach a computer to do this. And indeed that's what I did. So I, it multiplies some polynomials together and you tell it which polynomials to multiply. And so then I sat there and spent like 15 minutes just actually tracing out the action. So everything that has happened so far is pretty routine. And maybe the hardest part is figuring out the characteristic classes, but there's, you know, there are tools for doing it. So even though this maybe looks a bit complicated, it really, you know, after playing with a couple of examples, it becomes easier and even kind of like a fun puzzle. So on to Xt. So fortunately, Xt commutes with direct sums. So we can just work with each one of these rainbow summands at a time. Now, how do we actually do that? Well, the first example is we had all these copies of A1, but shifted up into different degrees. That's what this sigma k does is it says move it upwards a bunch. So in that case, Xt of A1 is an A1 module, is pretty simple. You're just going to get one sum in in degree s equals zero t equals k and everything else is going to vanish. So those guys are nice. How about that orange sum? Well, there are a couple of different ways of computing Xt. And the first one is to cheap it out and look it out, look it up in someone's paper. So the paper of Bodrian Campbell that I mentioned earlier has a bunch of these that they, you know, along their way to making computations and examples, they compute the Xt of several import, like commonly occurring summands. So that's one way to do it. And Jonathan Campbell's other paper, the one I mentioned in the beginning, also has a couple of examples that he and Agnes don't do in this paper. So great. But what if you don't want to do that? The most powerful tool that I know of for in terms of effort to effectiveness is that a short exact sequence of A1 modules induces a long exact sequence of Xt. And this is very much like the long exact sequence in cohomology and is true for roughly the same reasons. So you can use this to say, well, I understand these simple modules, well, sorry, I understand these easier modules, and I put them together into something more complicated. And here's how that affects Xt. So that is another technique that I find myself using a lot. Finally, there are ways to compute it directly by maybe playing with extensions or writing down a nice projective resolution or something. And this really is a last resort in that it's, depending on the example, it might not be complicated or it might be terrible. So I would try these things in this order. But the upshot is this also is fairly routine. In fact, put Chatham and Dexter Chua have a program online which can compute some of these. And so often, you know, this part, once you know the cohomology, it's not exactly press a button, but it's not too much harder. So here's the Xt of that running example. And I'm going to explain this picture in just a bit. Sorry, I'm going to explain it right now. So first of all, the caveat I should say is that the X axis is T minus S, and the Y axis is S. And this is the standard convention when doing the atom spectral sequence. It makes things a little easier to process. So if you remember, the red, yellow, green, teal, blue, and purple summands were all copies of A1. So we just get something in degree S equals zero. And they don't really, you know, there's not much more going on. The orange summand was fancy. So we get this thing. And then there's going to be more stuff out here that we're not going to worry about, because we're going to care only about these low dimensional groups. So how did I get this? Well, the answer is I looked it up, but you can compute it by describing that module as an extension, and then using the long exact sequence trick. So what are these lines? So the Xt that I just said, this Xt created a Boolean group admits an action of X over A1 of F2 against F2. And so there are two elements of interest there. One is called H0, which raises degree by S degree by one. And one is called H1, which moves them diagonal. So H0 will correspond to, once we, in the end-boardism problem, corresponds to multiplying by two. And H1 in the end-boardism problem corresponds to multiplying by something called eta, which if you're doing, so in-boardism, this is, of course, this is the circle with its non-trivial spin structure. Or, okay, I guess either, but if it's spin-boardism, it better be the spin structure. So this extra data will be useful, as I will tell you in just a sec. So we need to compute differentials. So in this t minus s comma s grading, a differential degree R goes one over to the left, and then R degrees upwards. So we can have d2s here, it looks like, in these degrees, or we could have a d3. And I'm not saying, oh, it must be the blue summand or it must be the purple summand. It's just, this is just how the figure came up. So we need to deal with these differentials. And this is usually why people are, have a healthy respect for the anti-spectral sequence. But in this case, there are some, there are some important tricks we can use that make it not so bad. So the first technique, and you know, if you were playing Pokemon, then if you tried this, it would say it's super effective, is that differentials must be linear with regards to these h0 and h1 actions, meaning they must commute. So let's look at this thing. I have an h1 action here. The h1 sends this to zero. So the, so the, therefore, this differential maps to this differential. And so let's suppose this differential were nonzero, then this differential's also got to be nonzero because it hits this non, this thing, which is nonzero. And well, you know, you can't have something zero hit something nonzero. So that's out of luck. So in other words, this differential has to vanish because otherwise it would turn into zero matched to nonzero. And the same argument goes here, except with h0. So h0, these things die, and so you can't, you can't hit this because then you'd also have to hit this. So both of the d2's here must vanish. We can't see anything about this one, but we will soon. So you can't always use h0 and h1 linearity to kill all differentials in these problems, even in low dimensions like this one. One, excuse me, one common trick is that the addon spectral sequence is functorial, like everything in algebraic topology except for the splitting in the universal coefficient theorem. So you can map to or from something where you already know the differential dimensions. This requires some creativity. It's not like, you know, necessarily beating your head against the problem for years until it gives away, gives it away. But like, this is the kind of thing that's hard. Like it's when you first read these arguments, or sorry, when I first read these arguments, I was like, how did someone come up with this? And then later, you know, later you start to pick up the tricks. So this part, this part requires some ingenuity, or in the absence of ingenuity, trying everything to think of until one of them works. Another good way to do this is to use topology. So this is a conference about topology. And so let's, let's, and this, we've been sort of mired in algebra for a little while. So what we're going to do is use, use the fact that these are groups of manifold, borders and groups of manifolds with some structure. So I'm going to, I should move a little bit faster in the next couple of minutes. So there's something called a Smith homomorphism, which says, okay, in degree five, I obtained a manifold with this structure, and then that produces a pin plus manifold in degree four. And so this is something like take a Poincare dual of a certain characteristic class of your D2N bundle. And then check that that's surjective. And it's already known that this is Z mod 16. And so therefore, a surjective map on a Z mod 16 says that this must also be Z mod 16, which means that these, so each sum in is, is something of order two. So two, four, eight, 16. Oops, we can't kill this with a differential. So this, so this differential is also zero. Great. So there's a lot more on Smith homomorphisms in a paper of let's see, Ryan Thorngren, Zohar Komarkovsky, and Haslund. I got the authors in the wrong order, because that's not alphabetical. But they discuss some generalities on the Smith homomorphism. And then Ryan is here today, which is pretty cool. So next up, there could, there's something called hidden extensions. And what, and this is just another way of seeing these three are all Z two summands. Could there be, what, what, is this really Z two plus Z two plus Z two? Or is it Z two plus Z four? Do these come by? So it turns out if they're, if you map to them by H zero, sorry, an action by H zero lifts to multiplication by two in the final answer. So you get multiplication by eight from a non-zero element to a non-zero element. So this has to be a group of orders 16 rather, sorry, this has to be Z mod 16 rather than Z two plus Z two plus Z two plus Z two. However, the converse is not true. We have to rule out a multiplication by two here. So functoriality is still useful. And using arguments from, you know, geometric topology where you actually play with example manifolds are also useful. In this case, we're just going to use algebra. So it turns out that eta, which is the, the non-trivial spin circle is two torsion in, in boredism. And so if this is multiplication by two and this is multiplication by eta, and this is non-zero, then we've just said that two eta times something is non-zero, which is a problem when two eta is zero. So that solves this problem. Often, extension problems can actually be quite difficult. So this is one of the things where you, that really requires, you know, ingenuity and perseverance. I mean, this example, it didn't, but in other examples, it might. So this is, this is the part that really requires, you know, that makes this more interesting. So here's the example or sorry, here's the answer. So we have a Z two, we have a Z two, we have a Z two, we have a Z two. In this dimension, we have three copies of Z two and the hidden extension was trivialized. So we get Z two plus Z two plus Z two. In this dimension, like I said, we got Z mod 16 from these three extensions. And this dimension, we have Z two plus Z two. So that concludes the computation. I'll, I'm running a little behind schedule, but I'm going to very quickly say what, you know, this application that I find interesting. But just to summarize this computation, you know, there's a bunch of steps and they all involve these big, these kind of big moving pieces in algebraic topology. But almost all of the steps are pretty routine. And there are, and, you know, work well when it comes to play with a couple of examples to get the grasp. The things that require cleverness or persistence, you know, also, you tend to pick up the tricks once you play with these a little more. But it's sort of like they're small enough that this, you know, I've really felt like this is something where I can make computations and actually understand how to answer these questions. So again, they can be kind of like fun puzzles where you get to play with these pictures. So I'm going to try and spend just three minutes on this quickly. So there's a generalization in condensed matter physics from SPTs or topological phases to things called crystalline phases. And so the idea is that instead you care about symmetries, which, you know, act on space. So let's say you actually have a crystal and you want to do things which are translation invariant in that crystal structure. So given some way of modeling phases, such as the lattice Hamiltonians that I briefly mentioned, you might say, okay, well, I want to look for phases that are invariant under this. And sure enough, there are interesting examples discovered by physicists, both people working with actual materials and theoretical physicists. And so this led to this notion of crystalline symmetry protected phases. So now there's a group acting on space. So this is nice, except that the argument that, oh, yeah, just take the low energy physics and get it and get an invertible field theory. At best, it's not so straightforward. And at worst, it just might not apply. So physicists are classifying these things. And we'd like to understand that mathematically. So we'd better do something different. And we means Frieden Hopkins, who last year put out a paper, which made the following kind of argument. So physics suggests, oops, typo number two, that topological phase on a space X for a fixed symmetry type, it's sort of like a generalized homology theory. Or maybe it is a generalized homology theory, once we figure out how to make this mathematically rigorous. And this, this is itself, you take borders and you do something to it. So crystalline phases for a group acting on space X should therefore be given by the corresponding G equivariate generalized homology theory. So this is within the realm of equivariate homotopy theory, which that's that's a word. But in many cases of interest, you can completely reduce it to a non-everian question about borders and groups. So great, the techniques that we have just spent a while talking about still apply. So this will be, I think this is my, this is my last slide. So let's say you want to implement this for a particular physics example, which is called fermionic phases, the symmetry group is the dihedral group acting by rotations and reflections. So if you then take Frieden Hopkins proposal and put it in this language, you get exactly the running, in the borders language for this case, you get exactly the running example we've done. And five physicists at the end of last year put out a paper which computes in dimension three, what this group is by different methods of physics. And so they get the same answer that we do. And so last bullet point, this is a suggestion that we're on the right track and that the physics methods and the math methods agree, which is evidence for the nuance of Frieden Hopkins. So we work in progress that I'm hopefully going to get out on the archive sometime soon. I check this with a whole bunch of other symmetry groups, some of which I can compare to preexisting calculations and some of which no one's have not already been looked at in physics. And most of the predictions agree, but there's some interesting disagreements that I'm looking forward to delving into. So that's, that's all that I have to say today. Thanks. Thanks so much for being here and thanks for your time. And let me know what questions you have. All right, awesome. We can go ahead and thank Aaron one more time for, for an excellent talk. You guys are free to unmute yourselves and clap or clap if the participants can also type clap for applause and Jen. I will go ahead and open it up for questions. Does anybody have questions for? Hello. I do have a question, but it's not directly related to this talk. So in I read your recent paper about comparing faces of Z2 dig wrap with the, the generalized semi-ons. Yes. So you mentioned in that, so basically, if I understood your paper correctly, the first part is talking about how to reconstruct a Hamiltonian that looks very similar to the tone code from the data of a TQFT, right? Wait, so you're saying that you're talking about how to start with a TQFT and get the Hamiltonian? Right. Okay. So, all right. Yeah, sorry. Keep going. So let's, let's say we want to do that. And you did mention, like, as a remark that you have to be careful about the, the sort of combinatorial structure that you put on your manifold. Like, whether it's actually a setup structure or a cell complex. Yes. So that's right. So I was wondering, and you did give an example where, you know, things actually don't work out in the case where you have a setup complex. So I was wondering if there's a, like a more general sort of obstruction, this sort of problem. Oh gosh. That is, that is a good question. And unfortunately, I don't have an answer. I'll tell you what I do know. And also just, so just for context for, for everyone else. So this is, so when I mentioned there's this, let's see, there's, I sort of skated over this during this talk, but the question is, how do we actually model these phases and actually say something mathematical about them? And the standard way of doing it is putting, you know, a combinatorial structure on a manifold. And so the, the question is about, well, in some cases, the combinatorial structure matters. So there's an example where if you want to do a CW structure, everything is fine for the Torc code. And that, so that's nice. But in another example called the generalized double semi-on model, that's not okay. And you get, you get nonsense. So in that example, you have to impose a triangulation and even stronger than that, only certain triangulations. And every manifold has one of those. So that's not a problem. But it's like, well, if you have some arbitrary phase, you might want to know, well, what conditions do I have to put on? And unfortunately, I don't know of a general result. I don't even know. I mean, there's plenty of examples where I wouldn't be able to tell you the answer. It's just like, this is the thing that worked. And in general, I wouldn't know what to expect. Triangulation seemed to work in, no, actually, no, I'm not going to say anything because that might be wrong. Unfortunately, I just don't have a good answer for you. But it's a, it's a cool question. What was the question was, how, how do you make Hamiltonians for these? I, if I, sorry, if I understood the question correctly, it was, is there a way of knowing when you have to use a triangulation instead of a CW structure or a restriction like that? Oh, I see. To construct turns in the Hamiltonian, basically. Yeah. Yeah, I, I wish I knew that answer. That is a, that is a good question. Yeah, we always said, just an actually, just an answer to that. It would be very nice to, Just to think about, like, sometimes you need a, a branching structure, right? Like if you want to express, say you want to express cup products on co-cycles. Actually, in a general CW complex, that's pretty hard to do. And in triangulation, it's much easier, but you can, you can do it so long as you locally order your vertices. It's actually quite similar to the talk we heard earlier about discrete Morse functions, but Oh, yeah, that's true. Yeah. Yeah, I'm not sure what to say about that when sometimes somehow you think that, like, a branching structure is kind of like a framing of the manifold. And when you have, if you need to choose a branching structure, it says you're using the tangent bundle in some non-trivial way. I'm not sure. Are there results which say something like, you know, you pick an arbitrary phase in this dimension and you can realize it with it using, as long as you choose a branching structure, is that like some of the stuff that Kapustin and collaborators are doing? That would be awesome. We can do it for group cosmology phases. Okay. Oh, well, that's, that's good. That's it. I mean, I like those. Yeah, those, those are, those are particularly nice. Yeah. But they're, they're very much like, because group cosmology has no reference to manifolds, it's like almost an entirely combinatorial thing to begin with. So you can run that, what you would do for these partisan phases, it's kind of, you know, you have the eta invariant, what are you going to do? Yeah, how do you, so yeah, like, how do you make something that geometric, first of all topological and then like combinatorial? Yeah. So this is, I mean, that's, I, that's something that there's some, there's some work on that, right? Like, there's this, someone had a paper which worked on, did this for the pin minus phases in dimension two and the pin plus phases in dimension four and tried to come up with common flow stuff. Right. Yeah, that's very, there's various approaches. I'd say none of them have completely worked yet. Yeah. So to give an indication of the fact that mathematically understanding topological phases are hard, this is one of the many questions that would have to be answered in order to have a mathematical theory of these lattice Hamiltonians representing topological phases. So this is the kind of thing we're like, you know, there are many things that we just barely know or don't barely understand yet. Can I ask a question about, you said in your, in your upcoming work, you have some places where you, you find some disagreements with, with some classifications in the literature? Yes. Is that because, so I don't know this paper that you, that you mentioned, but I could imagine that they're doing some kind of like group supercomology or something. So they're like, are I restricting themselves to a subgroup of the possible phases and then you are just finding the missing ones, the case. So caveat what I'm about to say I haven't checked yet, but I'm pretty sure it's true. So what I know is that, so these five authors use what's called like extended group supercomology that, that a couple of them have developed. And so what that means is they, this amounts to looking at, okay, Anderson dual something, something, k theory truncated between degrees zero and three. So where ordinary group supercomology is, I think just degrees zero, zero and one. So they get one more piece of information. And they, but they're not seeing everything. So the computations that I'm making are boiled down to k theory truncated in degrees zero and four. So there could be interesting stuff in that in that degree four, which they don't necessarily see. This is in dimension two plus one. Yes. So in that case, I would expect that you're just missing some possible z factors from like this, you're from the gravitational trends simons, whatever, however you want. Right. Right. But they shouldn't always be z, like, I mean, I, there can you know, not always, but one, because they kind of just sit on top. But, okay, but the, so the difference between what they compute in, in a different case, and what I get. So when there is a difference, it's torsion is the issue. So that's, so I don't, you know, I haven't yet figured out exactly what the, how to describe the missing phase, or compare the two. My best guess right now is that it's, it's because of this difference between lopping off everything below above three and lopping off everything above four. But I'm not fully certain. I also don't know, I don't know how to view it from that physics perspective of gravitational trends simons. For me, it's going to be a question of, here is a calculation of some generalized homology group. And here is a calculation of what I did, the Bordeson group. Right. I'm just talking about P one basically, like P one gets you a Oh, I see. in a four dimensional Bordeson that kind of like, when you Anderson dualized gives you a z in three dimensions. Yes. That's why I was kind of thinking that z sort of just like sits on top. It's like a summoned. Right. But maybe you're right. It can be involved. It's after three 30 years, we might have to wrap up. Okay. Yeah, I mean, I'd love to, if people have more questions, I'd love to, you know, feel free to reach out.