 Okay, thanks again, you know, I wanted to make a comment about how this conference is going and one of the really sort of great Aspects that is occurring is all of these sort of issues that are being raised in the Q&A and in chat during the talks I think that's really great. You know, I spend a lot of time thinking about how to foster online Mathematics communities and interactions and somehow this community here is doing a really great job at the back and forth and the engagement And that's and and and that's terrific. It sort of warms my heart to kind of see that kind of thing happening happening online. Okay, so Let's Let's talk a little bit about what we did last time on Tuesday so I talked about the stable homo to the groups and why they're important and One of the main ideas that I that I emphasized last time was what you see right at the top of the screen this conclusion That the stable homotopy groups are more than groups. They're more than a ring. They're They have this higher structure that's indispensable that you that you simply have to pay attention to if you really care about about the details. Okay, and We talked about motivic homotopy and we talked about some constructions of motivic homotopy Elements, okay, and then we dove into the set up of the atom spectral sequence Okay, and so there was a sort of formal way of producing a filtration in the spectral sequence, but what really kind of mattered in the end Was what you see now at the top of the screen that first of all There's an e2 page that's in that you can write down in terms of x groups over the steam right algebra Okay, which is entirely algebraic object Hey, and that there's this three-step program that first you have to compute the x groups the algebraic x group x groups That's the e2 page then you have to analyze the atoms differentials in the spectral sequence And then finally you have to interpret the the final answer by solving hidden extensions. Okay, so we talked We already started talking Last time about how you go about computing x groups, okay And I showed you some charts of what these sorts of things end up looking like Okay, and we started talking about the cobar complex and this is where I'd like to pick things up from From last time. Okay, so first of all I've got this cobar complex here that I've written down. It's a polynomial algebra in the Zetas, okay I remember it's a hop algebra. So it's got a product and a co-product and there's a co-product formula here over here on the right Okay, one of the things that we're going to need About this cobar complex is we're going to need to study the primitive elements So a primitive element means that the co-product is of the simplest possible form your co-product of an element Is that element tensor one plus one tensor that out? Okay, and it turns out that zeta one two to the end Those are precisely the primitive elements. Okay, so then the cobar complex looks like this There's a little f2 in degree zero. Okay, and then there's a copy of a bar That's the augmentation ideal of oops, you know what? I'm missing a lot of stars here, right? a bar Star is the augmentation ideal of the dual steamer algebra Okay, and then there's the tensor powers of a bar in higher and higher degrees and there's a differential Right, this is a differential graded algebra. There's a differential here Okay, and this first differential is is just given in terms of the co-product Okay, the higher differentials are also in terms of the co-product They're kind of like an alternating sum type formula in terms of the co-product But we're not going to analyze these into detail, but we will take a brief look at this first differential here Okay, all right, so But we're working here with the reduced right sorry with the with the augmentation ideal, right? Which means that these one tensor Terms disappear. Okay, and so the the the boundary map the differential on a primitive is zero Okay, and so these zeta one two to the ends these are cycles Okay in the homology of this cobar complex, right? And again, that's precisely because they're primitives Okay, that's exactly what this formula is saying. Okay, and the typical name typically we take this so there's a here's a You know, here's a here's a cycle in the homology and we call it h n That's what the h n is probably heard at at some point or not You've been to a talk or study something you hear about these h n elements and this h 1 h 2 h 3 h 4 And this is what they correspond to okay the zeta 1 2 to the ends, okay So that's fine for h 1. Okay, you can get h 1 pretty naively just by looking at the cobar complex Okay, and then you want to keep going right and so the next thing to do is to think about h 2 right of this Of this complex here, right? You want to look at cycles modular boundaries, you know In this in in this degree here, right? But that turns out to be not so easy to determine by hand Okay, this is not something you can simply do by brute force directly with the cobar complex Well, I guess you can but it's a tremendous amount of work okay And one of the problems with the cobar complex is that the dimensions grow exponentially right because you're taking These tensor powers right and so the dimensions are growing very rapidly the cobar complex has great formal properties Okay, but it's gets very hard to compute with Once you get into higher degrees because it's it's really kind of like the wrong model Okay there for machine computation. There are smaller minimal resolutions that you can work with right The cobar complex is not one of them. Okay All right, so I want to um, we're going to spend some time playing around with the cobar complex today because I think that You know picking up these elements and handling them yourself and getting a feel for how things work really gives you some Some valuable experience and some insight into how things work in higher in higher dimensions as well, okay Um, and but before I do that I want to dive into this this topic of massy products Okay, so I've been referring to this higher structure higher structure higher structure for um, you know Throughout these talks and I want to actually make specific what kind of higher structure that I'm particularly interested in Okay, there's lots of different ways of expressing higher structure I use massy products and total brackets because I find them to be useful for the types of computations that I want to carry out Okay, all right. So let's start with a general setup. Okay. Let's start with a differential graded algebra Okay, and this differential graded algebra. So let's make it over f2. Okay, it doesn't have to be over f2 But there are minus signs there are that you have to deal with and the and the formulas involving the degrees get complicated So rather than having to write plus or minus all the time I'm just going to have work over f2 or plus 1 equals minus 1 and I don't have to worry about those minus signs Okay, that's that's the only it's the only reason I'm working over f2 here is just so I don't have to deal with minus okay, so You have a differential graded algebra. You take the homology of that of that object and what you'll get is another algebra Okay, the Leibniz rule is essentially what guarantees that there's a moat that the ring that there's a well-defined ring structure On the homology. Okay, but it's not just an algebra. It doesn't just it's not just a ring It has this higher structure in terms of massy products. Okay I'm going to break down for you what this really means. Okay, so let's start with three elements a0 a1 and a2 In the in a and I guess we want to assume that they're cycles Okay, so the different d of each of them is zero Okay, and let's assume that their products and I put bars over these things right because now I'm looking at their at the at the Homology classes that they represent right a the a little a's are in capital a the a bars are in the homology, right? So in the homology these products are zero, okay So a0 a1 is zero and also a1 a2 is zero in the homology not necessarily in a but in the homology Okay, so here's what that means right the fact that that that the product is zero in homology means that a0 a1 Is hit by a differential. It's exactly what it means to be zero in homology. Okay So you choose an a01 that hits a zero times a1 and you choose an a12 that hits a1 times a2 Okay, you can you can do that basically by definition of these conditions Okay, and then you define the massy product of these three homology classes to be the set of all expressions of the form a01 a2 plus a0 a12 Okay, so this looks like some sort of some sort of complicated formula. Where does this come from right? So over here on the right I've sort of drawn a schematic for how I you know in practice on scratch paper when I'm competing massy products This is how I draw them, right? I write the three elements. This is a massy product that I want to compute Underneath a0 and a1. I write the the element that hits the product Underneath a1 and a2. I write the element that hits that product, right? And I've got this little diagram here and then I kind of cross multiply and add Okay, so that's kind of how I keep track of things the nice thing about that picture is that it generalizes To higher products. We're not going to talk about them But there are four fold and five fold and end fold massy products And you just have to make sort of bigger triangles bigger lattices of of things hitting other things in order to compute those Those higher massy massy products. Okay. All right So one way to think about what this this massy product is is measuring, right? Is it look at this triple product a0 a1 a2, right? This triple product a0 a1 to a2 is zero for two different reasons Right, it is zero because the first two elements multiply to zero And then when you multiply with a2 bar, you still get zero, right? It's also zero because the last two elements multiply to zero and then when you Premultiply by a0 you still get zero. Okay, so there's two different reasons Okay, and what and when those two different reasons are expert are the two terms in this formula, right? Here's the reason when the first product is using the first product is zero Here's the reason where the last product is zero, right and you're adding them or taking their difference And and that's the massy product, okay? And this is a common theme in homotopy theory, right? When you when when something is zero for two different reasons typically you assemble them You put them together or you look at their difference and you get a higher order invariant, right? Okay, so one of the difficulties in dealing with massy products and one of the the things that say students when they're learning about this Not exactly well-defined operations. Okay First of all, they're only defined when these two conditions Happen in the first place. Okay, but even worse They're they're not well defined in the sense that you don't get a single element in homology You get a subset. I wrote subset here for a reason you get a subset in the homology Okay, and the reason you get a subset is that there are some choices here You have to choose an a01 and you have to choose an a12 Okay, and if you make different choices depending on the specific properties of the computation at hand Um, if you make different choices of a01 and a12 you can end up with different elements in homology Okay, and so what we say is there's some indeterminacy in these total in these massy products or total brackets Okay, sometimes depending on on the specifics. Okay, we're not going to see any of that in these talks But it's just an important thing to keep track of and inevitably like all of the most sort of Or yeah The most famous or important mistakes in say the computation of stable homotopic groups of spheres Typically arise because people were it were not careful about these types of indeterminacies This is exactly that this is the most the slipperiest point in this whole subject is dealing with these indeterminacies and remembering to keep track of them Okay, so that's sort of a warning, but we're not going to really explore it Okay, what are we are going to explore is an extended example? Okay? I'd like to talk about this massy product h0 comma h1 comma h0 in in the in the homology of the Of the uh of the steward algebra. Okay, so there's a question the question is if a dga is formal Are the massy products zero? Yes, that's exactly right. Okay, so a dga being formal means that it's that it's That it's quasi isomorphic to a dga in which the differentials are all zero. Okay, and if the differentials are all zero Then the only way that a a product like this could be zero in homology is if they if is if they're representatives A zero and a one product were already zero, right? So this product is already zero and then a zero one is zero and a one two is zero Okay, and so then you end up with zero in in your massy product. Okay, so all mass product Let me rephrase that all massy products contain zero in a formal dga, right? But again because of the indeterminacy there they could actually be larger they could not just necessarily equal zero Okay, another question. Is there a geometric interpretation of these products? For example the dirham dg algebra of forms. Okay, so yes, there absolutely are geometric interpretations of these massy products In say like the cohomology of a space. Okay, so one good example of those that type of geometric interpretation Is the boromian rings, right? The boromian rings are these three Three rings right in r3 three loops in r3 They are pair-wise not linked Right, so there's they're not linked pair-wise, but the entire configuration of all three rings holds together. Okay um and and so uh, so When you think in terms of the cohomology of the complement of those three rings The fact that they're the fact that the rings are pair-wise unlinked is talk is is Interpreted in terms of cup product There's two each of those rings corresponds to a cohomology class And the cup product of those cohomology classes vanish That it corresponds to is if and only if those links are those those those loops are unlinked Okay, so pair-wise the product the cup products are zero But the fact that the boromian ring configuration does not fall apart Is precisely saying that the massy product of those three cohomology classes is non zero It's precisely that condition. Okay. Um, so that and that's a really really nice interpretation of Um of mass products, but it's kind of unrelated to what we're going to be talking about here. Okay A question when do these um question when do these indeterminity indeterminacies not exist? Okay, so the indeterminacies always exist in a formal sense It's just that sometimes the indeterminacy can be zero. Okay, so what's what in a little more detail the indeterminacy Is like a subgroup. Okay, and so it's you what you really have here What what these um with these massy products really are are cosets with respect to some subgroup Okay, and the indeterminacy is the name of the subgroup that you have a coset with respect to Okay, and it's possible that that subgroup is is the zero group In which case we say there's no indeterminacy, but what we really mean We really mean that the indeterminacy is zero. Okay, so there's a little bit of abuse of of language there There's another yeah, okay. There's another question in the chat. Yeah, yeah But could you explain a bit the ekman hilton duality between massy products and toda brackets? Okay, so i'm not sure what that question means But let me say um, let me give you my answer to the compare and contrast Massy products and toda brackets. Okay, so massy products. Uh, well, first of all, they're really the same thing Okay, so if you put these things on in the correct categorical framework You're working here with these dga's you're working in some derived in the derived category of a right and these called these Homology classes are maps or you know or derived maps or something like that. Okay Then a toda bracket and a massy product end up being exactly the same thing. Okay Um, but I tend to think of I use the term massy products when I'm thinking in this very concrete algebraic type of situation That I have in front of me here. Okay, and I use the term toda bracket when I'm working in a more homotopical context Okay, so when I'm talking about the stable homotopy ring stable homotopy Groups have toda brackets when the pair-wise products are zero you have toda brackets But it's not kind of an algebraic linear case. It's some sort of like somewhat non-linear You know situation with a homotopy category and I use toda brackets. Okay I'm not going to dive into the formal definition of toda brackets You can make them work in any in any of triangulated category I think even maybe maybe even less than that. Um, you can talk about toda brackets in a triangulated category Uh, and then the massy products are sort of like a special case. Okay So my use of massy product and toda bracket is very much cultural, right? It depends on which aspects I want to draw attention to when I want to draw attention to the fact that these things are algebraic and computable I'll use massy product when I want to draw attention to the fact that they're that they're expressing Homotopical information. I'll use toda bracket. Okay But as to say again in these talks, we're not really going to dive into the the toda brackets much Okay, these are great questions. I'm glad people are bringing these up. These are, you know, the I mean these massy products are really Like at the heart of this whole subject, right? Like what when I think about, you know, what, you know What skills I learned on the way to being able to carry out these computations, you know, to a very large degree it is the It is my ability to work with massy products, right? My familiarity with massy products and toda brackets that has allowed me to kind of carry out out these things Recently a student referred to my database of massy products that I carry around in my head, right? And I have all this information and when I see an element I can recognize it in those terms And it's a really really sort of useful for getting at getting getting it all kinds of stuff. Okay So let's talk about an example. Okay, h0 h1 h0. Okay, the product h0 h1 is zero Okay, and so this this these pairwise products are zero Okay, which means we can at least write down this massy product. Okay in the cosmology of the steward algebra Okay, let's go through how you might actually explicitly compute this massy product Okay, and we're going to use the cobar complex. Okay, so h0 is another name for zeta 1 right zeta 1 We've talked about this is a primitive element right and h0 is the name of the cycle that zeta 1 represents Okay, and h1 is represented by zeta 1 squared. Okay now Rapidly, you're going to see that the notation is going to start devolving into imprecision right technically You know, I can't I'm going to talk about zeta 1 as if it is an element as it is a cycle an element in x Well, technically zeta 1 it represents h0 and you know, and I don't need do I need to put bars or brackets around zeta 1 Well, I'm going to sort of drop a lot of that extraneous notation because it just gets too cumbersome Okay, so we've got zeta 1 and zeta 1 squared Okay, and I'm just rewrote the cobar differential that I had written on a previous slide just so we can see it in front of us Okay, all right. So now I'm going to write down a few I've written down a few of the specific Co-product formulas that we're going to need. Okay the reduced co-product on zeta 2 is of this form zeta 1 squared bar zeta 1 Okay, and here i'm writing tensor and here i'm writing a bar and that's just sort of tradition right traditionally in the cobar complex We use these bar symbols instead of tensors, but it means the same thing Okay, so that's the reduced co-product on zeta 2. Okay, and look what that's saying that's saying look zeta 1 squared That's h 1 zeta 1 is h 0. So this cobar this this this co-product right here is telling you immediately That h 1 times h 0 is 0 notice the order here, right? It's h 1 times h 0 is 0 Okay, so that's this already sort of relevant to what we want to do. Okay now here's another Co-product on zeta 1 cubed Okay, if you look at zeta 1 cubed and you look at its reduced co-product Here's what you get you get zeta 1 squared bar zeta 1 plus zeta 1 bar zeta 1 squared Okay, and why is that I wrote a wrote over here in green a little bit more about how you would actually do this right So zeta 1 cubed is zeta 1 times zeta 1 squared Okay, and the co-product is a ring map. It turns out so you can you can break it down this way Here's the co-product of zeta 1 Here's the co-product of zeta 1 squared. Remember these guys are both primitive, right? So we know they are co-products Okay, and then you've got four terms here right two terms here two terms here You've got four terms altogether, but Um, but two of them are the one tenths of the element and the element tenths are one Which we got a rig which we get rid of right we're looking at the reduced co-product and the mixed terms are these two Okay, so that's where this formula came from right? It's the sort of like the two cross terms right in at least in American Schools we often talk about foil right first outer inner last right and so this is the The the the outer and the inner are here right and the first and the last are kind of like, you know reduced Okay All right. So what does this say this says that well, we've got h 1 times h 0 plus h 0 times h 1 Right. This is telling us that h 1 plus h h 1 h 0 plus h 0 h 1 is 0 Okay, which is also useful to know it and together with this formula That's telling you that h 0 h 1 is 0 right because these things aren't necessarily commutative in I mean extra of the standard algebra is commutative because because some because of because of some additional Additional additional outside fact that I haven't talked about we already know that the co-modular the steward algebra is commutative For other reasons that I don't want to get into okay, but um, but here we're explicitly showing that both products in both orders are zero All right. So those are the co bar Those are the co bar differentials that we're going to need okay So now let's set up the diagram that I talked about earlier right here are the three elements Who's massive brush I want to compute there's h 0 there's h 1 and there's h 0 Okay, and here's the element that kills zeta 1 bar zeta 1 squared Right, it's the sum of these two right the sum of these two kills zeta 1 zeta 1 zeta 1 bar zeta 1 squared on the other hand It's zeta 2 that kills zeta 1 squared zeta 1 right and you can see the non commutativity of the co bar complex Is really kind of key here. It's really showing up here right in when you when you change the order It's a sort of it's a different thing right okay, so The cross you cross multiply right and so the massy product is supposed to be zeta 1 zeta 2 zeta 1 bar zeta 2 Right plus this guy right which is zeta 2 zeta 1 Plus zeta 1 cube bar zeta 1 right so that's what I've got here right so this is what the massy product is supposed to be Okay, but there's a problem. We look at that thing and that doesn't we can't recognize that as the cycle What is that thing right that doesn't look like any kind of expression that we recognize these zeta 2's in there I mean and the zeta ones are supposed to be h zeros But the zeta 2's aren't permanent cycles and the zeta 1 cube is not a permanent cycle And so what does this really mean right so what we have to do is we have to rewrite this thing Find something that it is comologous to Or homologous to I guess something that it's homologous to that's sort of in a better more useful form Okay, and so but so this is the this is the expression that we're supposed to be working with right Okay, and so what is that and we got to go a little bit further to understand that here Okay, so in order to understand this thing. I'm going to look at one more Co-product okay on zeta 1 zeta 2 okay, and when you look at the definitions and you work it out You get these four terms you get zeta 1 bar zeta 2 zeta 2 bar zeta 1 zeta 1 cube bar zeta 1 That's the same as these three but then you get one more term Which is zeta 1 squared bar zeta 1 squared Okay, and where this where this comes from the way to compute this just like before to compute You compute the unreduced co-product on zeta 1 zeta 2 using the ring structure You get six terms here two of which are the you know are reduced when you look at When you look at You get rid of when you look at the reduced co-product and four of which survive Okay, and those are the four here. This is zeta 1 squared zeta 1 squared For example comes from this term 1 tensor zeta 1 Times zeta 1 squared tensor zeta 1 right those two together give you that fourth term for example Okay, so what this is saying Is that this expression these three terms that we computed as the massy product Are homologous to this fourth term right because their sum is a boundary right and so this massy product is then Also detected by zeta 1 squared zeta 1 squared bar zeta 1 squared and that we can recognize as h1 times h1 or h1 squared Okay, so there it is There's a non trivial massy product in you know in in something right that you've seen now Okay, and as I mentioned earlier, it's really key that the so the non commutativity of the co bar complex Is really kind of key here right that zeta 1 bar zeta 1 squared is different than zeta 1 squared bar zeta 1 okay And shan tilson raises a great question. Is it obvious that it is an equality instead of an equals? Okay What sorry is it is it obvious inequality rather than an inclusion right and the answer is That it is that it is not obvious that shan's instinct is absolutely correct that you don't stop here You don't say ah h1 squared i'm all done You always have you have to train yourself to always ask Is there indeterminacy? Do I need to worry about that? Okay, and then it turns out by inspection In this degree that there is no indeterminacy. Okay, but it's it's a totally valid question Okay, so you're gonna have to go on you have to go check that separately Okay, um, so the another question. Why do some of the terms disappear when you pass to homology? Okay, so I think what you're talking about maybe is why we had these three terms here And then we reduced it down to just one term here, right? And the point is that these we're working when we're working in homology, right? We're working modulo boundaries, right? So these this this sum of four elements is a boundary Right and so therefore these three elements are equivalent to this one element It's not so much that uh from the sixth term expression. Oh, okay So here why do two of them um go away two of them go away because we want the reduced co-product The reduced co-product means that you should not pay attention to the to the one tensor x and the x tensor one terms Okay, and the reason that you ignore those two terms If you go back and you look at exactly why the co-bar the way in which the co the homology of the cobar complex Is x and you're very careful with the details on why you know on how the cobar complex is a Is a free a resolution of f2 and and how you to hom into f2 and then take homology to get x And you you go through that sort of carefully through that homological algebra story. You see that you need the reduced You you need the reduced cobar complex. Okay Okay, so that's why you're dropping those those those terms there. Okay, uh question Can you convince a computer to go through this sort of computation or is the homologous replacement a limiting step? No, the homologous replacement is not a limiting step. Okay You absolutely can train a computer to do these types of computations What you need what you do what the way you train a computer to do this Is you put an ordering on all of these different terms and all these monomials And so so that there's sort of a preferred form for every homology class And then when the computer gets something that isn't in reduced form it it it always Puts it into reduced Preferred form first and then it compares it. Okay, and this is um, you may have heard the expression a curtis table Okay A curtis table is kind of is sort of a way of organizing this type of kind of choosing a canonical Well, it's not a canonical but choosing a preferred representative of each homology class is very much related to that kind of kind of thing Okay, so um the the issue with a computer Computer can do this one just fine the issue with a computer in the cobar complex is in higher degrees Where it just grows exponentially and you just get out of the linear algebra becomes unmanageable rapidly Okay, so we did that one by hand. Okay, that was not a huge amount of work But it was some work right and that's just this tiny little thing about like, you know about about pi two Right h1 squared is about pi two, right? So it's not something we want to do a lot of right if we can avoid it okay, so what I want to talk about is a better more efficient way of carrying out these types of computations of x groups, okay, and the uh the kind of like the the Philosophically speaking it's like the best the most efficient way for humans to think about x compute x groups is with the may spectral sequence Okay, so the idea of the may spectral sequence is to filter the cobar complex That's c star c star is the cobar complex filter the cocom cobar complex by primitives remember by uh by this sort of thing Okay, and um and then when you do that, okay You so you choose basically you're choosing some filtration on the cobar complex Okay, and then e1 is the associated graded of that filtration, right? And then there will be a spectral sequence converging to what you want Okay, so it turns out that the the e1 page is really nice. It's just a polynomial ring on certain elements h ij okay And h ij corresponds to zeta i to the j Okay, so we talked about the h and h1 h2 h3 h4 before those course in this notation Those are h10 h11 h12 h13 and so forth Okay, but there's also ones involving zeta 2 to powers Okay, these guys themselves are not primitives But the the idea is that they are primitives modulo primitives, right? And so what you're what you're really doing here Well, with sort of anytime you filter right is you're working with the cobar complex But you're sort of working modulo higher order terms, right? And that helps you kind of keep track of the details you ignore the distracting um technical terms and then and and then you and then you work in a much simpler algebraic Setting okay, and this is um and this is this is a highly effective process Okay, it's been done by hand to 70 steps Okay, ten ten gora did this in the classical context um decades ago out to 70 stems and i've done it by hand In the c-motivic context out to 70 steps Okay, but you can't do it forever like it gets pretty it does eventually get pretty difficult Okay, but you there's no way you'd ever be able to compute the cobar complex to 70 stems by hand You know if you I don't even think you can get to seven stems by hand um Seven stems by hand in the cobar complex. Okay, the x group is spits out is bi graded So this is a triply graded spectral sequence. That's exactly right x itself is bi graded There's this topological degree and there's also the atoms filtration Okay, uh, so that's showing up when you think about the cobar complex the way that's showing up say like in the cobar complex Where's a little diagram here in the cobar complex? There's internal degrees internal to the steward algebra the steward algebra is already graded, right? Square one square two square four. They have different degrees, right? And then there's also the homological degree of the cobar complex Okay, and you have to reindex for an atoms chart, but it's this there's the two dimensions There's the two gradings right if of x coming from the homological homological degree and the internal grading Okay, but now so so the answer is bi graded which means this thing is tri graded That's right. There's also a may filtration on these guys as well And that has to do with sort of how primitive these things are Okay, so the the things that are primitives themselves the zeta ones are in the lowest filtration things that are primitive Modular primitives are in the second filtration and and so on and so forth So these hijs have a may filtration and i'm not telling you what those degrees are All right, so How do you learn how to compute with the may spectral sequence? Well, the way to learn the may spectral sequence is to study a simple case Okay involving a of one Okay, so a of one is a small example that behaves that has It's sort of like the first kind of non trivial example that behaves like the steam run algebra It's got the first complication involving the steam run algebra But only one complication and so it makes it much more manageable Okay, a of one star is defined to be f2 adjoined zeta one and zeta two only so no zeta three and higher Those are thrown out modulo zeta one to the fourth and modulo zeta two squared Okay, so there are eight monomials here, right? This is an eight dimensional Algebra over f2 right eight dimensional vector space over f2 right so this is not so big okay So we want what I want to study is x over a of one Okay x x of a of one x of this Much smaller ring right then then the full steam run algebra Okay, that's that's our goal Okay, and if we learn how to do that then we'll have learned a lot about how the may spectral sequence works in general Okay, so when you look at this thing, okay, let's note that zeta one and zeta one squared are primitive Okay, zeta one of the fourth is zero and the higher powers of zeta is zeta one is zero So they're not relevant Okay, zeta two is not primitive, but it's primitive. See here's it's here's its reduced co-product here Right that that middle term is what makes it not primitive But it is primitive modulo primitives right if you mod out by the primitives zeta one and zeta one squared Then the co the reduced co-product becomes zero right and so that makes it sort of in the second layer in the second stage of the filtration okay All right, and so the upshot of all of this is that the e one page for a of one is polynomial again But it's only got three generators. Okay. So this is h zero or h one zero in my earlier notation This is h one or h one one in my earlier notation And this is v one or h two zero in my earlier notation. I'm going to call it v one because well You know, you hear p chromatic homotopy theory or you know, and he'll put it some periodicity here BP hear people talk about the v ends all the time, right? Well, this v one it that's that's that's that v right that That's really the name that it deserves it connects to a lot of other related mathematics, but it's also h two zero Okay, so there's the e one page. I drew over here on the left side of the screen I drew a picture of the e one page Okay, so it's a free polynomial in three elements and when I go through and I check the degrees H zero is here's the identity at the origin. Okay, h zero is right here in degree zero one H one is right here in degree one one and then v one is right here in degree two one Okay, and then it's polynomial, right? So I have all these multiples of h zero I have all these multiples of h one. I have all their products of h zeros and h ones in here Right. So the black stuff those are all the products of h zero and h one Okay, and then I have v one, right? But then I have to multiply v one by all the multiples of h zero and all the multiples of h one And so because the picture gets messy quickly I drew these two little green arrows to indicate a big sort of infinite triangle of h zero and h one multiples on v one And then here's v one squared that again has h zero and h one multiples and a v one cubed and a v one to the fourth And a v one to the fifth and so forth. Okay, so that's the e one page with all the degrees put into place Okay Now the d one differential is relatively easy to compute because you can look it up in terms of the co-product Okay, the co reduce co-product of zeta two is zeta one squared bar zeta one And that's saying that d one of v one, right v one corresponds to zeta two is equal to h zero h one Or rather it equals h one times h zero It's really but but we we we already know if things are commutative and so the order doesn't really matter Okay, so the d one of v one equals h zero h one. Let's look at that on the chart That's this red arrow here the differential on v one hits this product h zero h one. Okay Okay, so that's uh, so so there's a differential in there Okay, and then of course all these multiples of v one hit all these multiples of h zero h one, right? Just by uh by linearity, right? Okay, and then v one squared Is a d one cycle because of the Leibniz rule, right? d one on v one squared involves a two right, but two is zero here. We're working over f two And so v one squared is a cycle at least for d one Okay, so let's take that picture And let's look at what happens in e two. Okay, so remember this differential v one killed h zero h one and all the h zero One h one multiples, right? And so that hollows out the inside of this triangle, right? The inside of this triangle is all now wiped out, right? All that stuff was set to zero Okay, and the green triangle is has disappeared because none of it survived, right? All of the green elements all supported differentials hitting these black elements So the green triangle is completely gone. Okay, it's it's it hasn't survived to e two Okay, and then there's v one squared and it's got h zero multiples and it's got h one multiples But again, the triangle is hollow, right? Because v one cubed v one cubed came in and hit H zero h one times v one squared Okay, so you've got this this hollow triangle pattern repeating On one v one squared v one to the fourth v one to the sixth and so forth Okay, so written in term in more precise terms. You've got a generators h zero h one and v one squared Right, but you've got to kill off h zero h one Okay, so that's what the e two page looks like Okay, and now you have to analyze the d two differential Okay, and when you look when you inspect the degrees what you discover is that possibly there's a differential on v one squared That could hit h one cubed. Okay, so that's this question here does v does d two of v one squared equal h one cubed It's possible for degree reasons. That's such a thing occurs Okay, but just because it's possible doesn't mean that it does occur and you have to figure out how to compute it So one thing you could do is you could inspect the cobar complex, right? And since we're studying h one cube we can look for z zeta ones Squared bars zeta one squared bars zeta one squared. Okay, but that's pretty hard Okay, things are getting big enough now or fight playing around the cobar complex is more difficult Okay, so what typically happens here? Is that you do something a little bit indirect and ad hoc? Okay, and because it's ad hoc that means that you have to do some work, right? This this is you know that Um to to kind of figure out what's what's going on right and sometimes you have to be creative and and you know In inspect things and try stuff out. Okay, so what I want to use Is this massy product that we calculated earlier that h one squared is the massy product h zero comma h one comma h zero Okay, we did that in x over a not an x over a of one But the exact same thing works symbol for symbol works exactly the same and you get the same A massy product in x over a of one. Okay, so you have that Okay, so let's look at this string of equality start with h one cubed Okay, well h one cubed is h one squared times h one So there it is there's h one squared, but I should I rewrote as a massy product times h one Okay, so far so good, but no nothing interesting has happened now The next equality is where the action happens. Okay, there is a formal relationship between massy products, right? When you have a comma b comma c and then multiplied by d That always is equal to a times the massy product of b comma c comma d As long as both massy products are defined Okay, that always works out happen. That's a formal thing you can do in any dga You can you can prove that those those these two expressions are formally equal Okay, so and I I would call this a massy product shuffle, right? So you shuffle this formula Okay, and you get h zero times the massy product h one comma h zero comma h one This is a different massy product than the one we had over here Okay, and very often something special happens when you shift from one massy product to another You can see structure that was otherwise not so not so obvious Okay, now we haven't computed h one h zero h one Okay, we could go back to the cobar complex and compute it Okay, and x over a this one is non-zero. It works out. Okay Um, but what about in x over a of one? Well at x over a of one This massy product ends up living right there, right where the red dot is That's the degree in which that massy product h one h zero h one lives And we can we can see from inspection. There isn't anything there, right? There are no possible non-zero elements. The only element there is zero Okay, and so this massy product has to be zero Okay, so you get h zero times zero and you get zero Okay, so what we just did Is we we did something kind of clever and not obvious, right? We we wrote down an expression for h one cube using massy products We shuffled that and we proved that it had to be zero Okay, and therefore H one cube so h one cube has to be zero it must be hit by a differential Okay, and then this is the only possible differential and we do get that differential Okay, so this is a really kind of slick argument, right? And this is the sort of stuff that we do all the time in higher and higher in more and more complicated ways In higher and higher steps, right? This kind of shuffling around and inducing relations, you know, based on shuffles And then which then imply differentials is the name of the game. Okay question Do you have a favorite reference for an exhaustive list of such shuffle and juggle formulas? No I do not have a favorite reference for For that. I don't really know of a good reference for that kind of thing Um Some of these things are written in ravenel's Complex co borders in book like maybe in one of the appendices or something like maybe it's not an appendix anyway Are some of these things are written in in ravenel's book Some of them are written in the down in the various sort of like large Like, you know manuscripts that I've produced on these on on this topic of computing stable homotopy groups But that's not really exhaustive either Okay, so I think there's actually a reason why Um, there aren't any kind of like, you know, basically what happens is when someone needs a particular result They kind of write down a proof write down a lemma and then they go and use that result All right And there's a reason that no one has written an exhaustive list of such shuffling juggling formulas The reason is that we don't really know what all of these shuffle juggle formulas are Okay, we know examples of them And we can try to apply we can try to extend the the low dimensional examples we know into higher examples and But we don't really know what all the relations are This is kind of an open question And I think there's probably some room for some sort of like You know operad like, you know structure here to describe what all the possible Relations on these massy products actually are It's not something that someone has ever done But I think, you know getting at a infinity, you know a infinity structures are kind of related To to what i'm talking about, but they really aren't i'm thinking first of something much more algebraic than just sort of saying a infinity But um, but this is sort of an interesting question. What are all the what are all the formulas? So we don't really know Okay, what do the axes in these pictures represent? Okay, so that's a good question. What are the axes represent? So the the vertical axis represents the homological degree Okay, the sort of like the degree of the cobar complex if you're in the 19th stage of the cobar complex Then you'll be up here at height 19 in the picture okay the um The rather than tell you what the horizontal degree means I'm going to tell you instead what the sum of the degrees means the sum of the degrees corresponds to the internal degree of the steinward algebra Okay, so h zero here is in is in total degree one right zero plus one is one Right and that corresponds to square one in the steinward algebra This guy is in degree total degree two one plus one equals two And that corresponds to square two which is degree two Okay out here for x over a not an x over a one but for x over a out here You have an element h two and degree three comma one total degree four and that's corresponding to square four total degree four Okay, so the vertical axis is the homological degree the sum the x plus y That's the internal degree. Okay, but the x degree the reason we draw these things this way is that the x degree Then corresponds to the stem when you use these things to compute stable homotopy groups Then a column corresponds to a stable homotopy group Okay, so I always index things this way so that we can see the stable homotopy groups easily easily Okay Andy Baker reminds me that peter mays matric massy products paper does Contain a lot of those formulas. He's absolutely right. That is a great reference for for For the types of formulas that we're talking about here Even that's not exhaustive, but it is a good place to To look at it. However, you will need to set aside some times You're going to look at that paper You will need to set aside some time in order to familiarize yourself with the technical notation For what's going on in that paper, but sometimes that's just necessary Okay, thanks very much anything. Thank you for reminding me of that. He's he's absolutely right Okay All right, so what we did is computed a may differential d two on v one squared Okay um and And and joey bovey feist tower points out that ravenville screen book does have some of these formulas as well Okay All right, so now you can go from e two to e three Okay, and unfortunately The way these things fit on the screen. I can't really show you both of them so easily at the same time Okay, so what happens is that v one squared hits that h one cube and notice that there's all these One multiples of v one squared and all those h one multiples of v one squared are hitting all these higher h one multiples Okay, so this whole blue diagonal line doesn't survive and this whole black diagonal line gets killed Okay, but the vertical line Does survive right this guy does not support a differential Okay, and so that's what we're going to see down here. Okay the h one cubed and higher were killed The blue diagonal line did not survive, but this h not tower does survive Okay, and then there's v one to the fourth and the pattern repeats And you'll have an h not v one to the sixth and a v one to the eighth and so forth Okay, let me point out just as another exercise, right? this h not v one squared is an indecomposable in in in uh e three in terms of the ring structure But in terms of the higher structure It has a decomposition as the bracket h zero comma h one comma h one squared Okay, and I wrote down here the little bit of the little details that you need in order to compute this bracket In the e two page h zero h one it was already zero in the e two page And so it's this is just d of zero equaling zero But h one cubed is hit by v one squared right and then you cross multiply right to get h not v one squared Okay, so that's a little massive product here in in x over a of one just as another illustration of whether whether that's useful or not Well, that's another matter Okay, all right, so there it is and then for degree reasons there are no more possible You can look at all the elements and you can just see that there aren't any possible And so you must have e three equals e infinity Okay, so the um so this story really like what on the one hand this this a of one is a very sort of easy Is it much easier example than the cosmology of the full steenward algebra? but on the other hand it also really get captures the flavor of what's Of what's going on Okay, there's really a lot of truth that if you wrap your head around what all the details of what's going on here Then you're in a good position to study the full steenward algebra Okay All right, so that was sort of like our our exercises for the day Okay, and let's get back to some a little talking about things at a slightly higher level. Okay, so Let's talk about Motivic x groups, right? If we want to compute with the k-motivic atom spectral sequence We need to compute x over the motivic steenward algebra. Okay, and mod I'll use the notation mp All right for the motivic homology of a point, right? So you have x over a mp mp converges to the Motivic homotopy groups and what you need to know in order to get this spectral sequence started You need to know the motivic homology or co homology duly of a point Okay, and you also need to know the k-motivic dual steenward algebra. Okay, so here's what happens typically Okay, and by typically I mean for for a large class of fields, right? and there there are Exceptions when like the prime p is equal to the characteristic of the field k then things get squirrely Right, you know, but but sort of in general typically what happens is that the that the motivic homology of a point Is you take the milliner k theory mod 2. Okay, so I don't want to get too much into milliner k theory Um, but let's so let's just say here that this is an this this milliner k theory mod 2 This is sort of a these are arithmetic invariance of your field. Okay And then you just adjoining An element tau kind of a free polynomial variable tau So you have milliner k theory, but then you also have a tau. Okay, that's what the homology of a point typically looks like okay, and then the uh, the uh, the uh, the The dual steenward algebra typically looks like this big monster here Okay, and let me talk through these form this formula a little bit to um wrap rather heads around it. Uh, typically at Typically at Well, okay, so the formulas I've written down here are for p equals 2 the formulas at odd prime So there's a question about odd primes, you know, what's going on at odd primes So at odd primes the formulas are known typically, but the formulas are a bit different Especially this the dual steenward algebra, right the dual steenward algebra has a different as a somewhat different appearance at odd primes But it is typically known Okay, I just want to specialize to p equals 2 to not write We're not going to really do anything with the odd primes and I want to write down the the complicated formulas and not use them Okay, so let's look at what's going on here. Okay Classically, we had the zetas polynomial and the zetas here. Motivically, we have tau's tau i's And also xi i's. Okay, so two families of polynomial generators But there is A a relation Okay, there's a relation that tau i squared equals tau, right? And that tau comes from the coefficients, right tau xi plus one plus ro Ro also comes from the coefficients. Remember ro is our name for the class of minus one in the milner k theory Okay, ro and then tau zero xi plus one. That's two these guys and then finally another term plus ro tau i plus one Okay, so this formula is kind of a nightmare and we need to break it down and think about it carefully if we're really going to understand it Okay, what is a reference for the p equals odd case? Well, these things go all the way back to vavadsky if you look at Oh, what's it? Uh I I forget the name of the article, but these for sure these things go back to vavadsky These formulas have been written down in a bunch of different places Um A couple off the top of my head a couple of plays powell wrote an exposition of this that you already wrote down the formulas Um borghazy wrote down an exposition of these things right wrote down the formulas There's lots of we we I I can give you something precise afterwards But they're sort of they're sort of all over the literature by now I mean there are mistakes in various places. So you got to make sure you want to get one. That's right, but But but basically the to be actually the odd case is a little bit easier where it behaves a little bit more Like the classical one. Okay. It's this relation is really kind of exotic thing that happens at two okay So when you look at these formulas What you realize is that the the the situation really breaks down into two Cases, okay The first case is where your field contains a square root of minus one if your field contains a square root of minus one Then um, then then this class row in milner k theory mod two Is zero right if minus one is a square then it's zero mod two in milner k theory Okay, and so row is zero and you have a much simpler formula. It's just the tau i is tau times c i plus one Okay, that's a much simpler formula Okay, and then what happens to the x groups over k is that you just need to compute the c motivic x groups Okay, and right, uh In the comments put in the reference to vivatsky's Where he wrote vivatsky's paper where he wrote down the the dual steward algebra. Okay, um Uh, so you need to do the c motivic x and then simply tensor with milner k theory Okay, and when I say c c is my name for any Algebraically closed field of characteristic zero right it doesn't really matter that c it matters that it's algebraically closed and it's characteristic zero Okay, so so that's what you uh, so so you can so in other words You don't really need this case really boils down to the c motific case and then there's this arithmetic that shows up also Okay, on the other hand There's the case where minus one is not does not have a square root Okay, and then the class row is non zero and then x is more complicated So this is the more complicated case and we'll need to filter by powers of row and deal with a row box and spectral sequence And that's something that we'll talk about tomorrow Okay So let me say a little bit about the c motivic Atom spec So what this is telling us here is that we should really need to spend some time With x c and the c motivic atom spectral sequence because the behavior of the c motivic atom spectral sequence Is going to kind of dominate this entire case where we're minus square root of minus one Where minus one is the square root there will also be arithmetic phenomena occurring for a general k in this case But the first order phenomena are already seen by c. Okay, so we can't skip that step We have to think about c more carefully. Okay, so What i've done here is i've copied down the formulas for the homology of a point The homology of a point and for the dual steward algebra in the case When k equals c and on the right i wrote the analogous computations the classical one so we can see what happens, right? Here we have f2 a joint tau versus just an f2 Okay, and um, we have this Over here, we had a polynomial algebra and here we have two families Of of generators, but then we have these relations that intertwine them. Okay So when you look at these formulas, here's how I want you to think about this. Okay, look this guy here This dual steward algebra is a deformation of the classical stammer steward algebra with parameter tau Okay, so what does that mean? That means that if I invert tau, right? Look at the generic fiber if I invert tau. Well, look what happens when you invert tau when you invert tau ci plus one is just Expressible in terms of tau i squared ci plus one is not a generator. You don't even need it Okay, and so when you invert tau you just get a polynomial a free polynomial algebra in the tau i's Okay, and it's really the same as the classical steward algebra Okay, so if you invert tau you just recover the classical steward algebra Okay, on the other hand if you set tau equal to zero, right? That's the special fiber You set tau equal to zero then you have an exterior algebra on the towels and a polynomial on the ci's Okay, and that turns out this thing looks actually quite similar to the odd primary classical steward algebra dual steward algebra Okay, and that's sort of an intriguing link. Okay, so this is how you should think about it Well, this thing is almost polynomial except for these towels that don't make it quite polynomial when you invert the tau Then it does become polynomial Okay, so what i'm saying here about this deformation is very concrete very hand. Yes the copra and and the co-products coincide So when i say that that i when i invert tau and i get the classical steward algebra Yes, the co-products that includes the co-products coinciding. Okay Okay, so i'm so what i'm saying here is something very concrete and specific and right up on the screen You can just look at the algebras and you can see the deformation Okay, but this is a shadow of a much deeper phenomenon. Okay, so the deeper phenomenon Is that cellular it turns out That cellular c-motivic homotopy theory is a deformation of classical homotopy theory Okay, the special fiber of this deformation is the um Well, yes the generic fiber is classical homotopy theory That's what i mean by deformation of classical homotopy theory The special fiber is an algebraic category of bp star bp co-modules Okay, so bp star bp co-modules is some complicated name that you may or may not understand Understand, but let me just say it's a derived calibre category. It's just a straight up algebraic thing Okay, and so what's happening here is that there's these like there's these sort of there's these surprising connections between cellular c-motivic homotopy theory classical homotopy theory and this algebraic category of bp star bp co-modules and this Deformations framework. This is really what drives this recent progress that we've made in computing the computation of the stable classical stable homotopy groups Okay, so let me and there i've listed the names of a few people who have worked on various parts of this sort of deformation story Let me say that something here. This is a great place to end that there is um That there is clearly something really important going on in this deformation perspective Okay, and we know about a few examples and we have kind of used those examples to carry out some computations But there surely is a nice general theory of deformations of homotopy Theories and probably I mean deformations of infinity categories or some derived algebraic geometry type thing Okay happening here and that these are special cases Of it. Okay, so pister govsky has done some work. He has produced a family of such types of different deformations But even pister govsky's framework is not general enough like there are some examples of deformations Um or things that ought to be deformations that I would like to sort of study from that perspective and in order to compute them That don't really fit into any of the kind of like the known frameworks that have already been been developed Okay, so someone and there I think there are a lot of people at this conference who know a lot more about abstract homotopy theory and infinity categories than I do Which is not saying much because I don't really know that much about that stuff anymore Um, uh, I think this is a nice project for someone to pick this up and really kind of put these examples on a frame On on a on a solid framework Okay, all right. Um, and then a question. Well, let me let I think I think we should wrap it up here because we're out of time So I'll stop on that note next time. We'll um, we'll pick up with some more about the c-motivic Uh, the c-motivic steamer and algebra and c-motivic computations Okay, so let let's first first us thank Dan for for the talk Thank you. So everyone is clapping of course. So and and and yeah, I just wanted to mention that you you you you have noticed that there is some lag in the connection So it's just to remind that uh, dan is actually in in the u.s And it's great that we can make some connections in us and europe So the lag is just the trace of that. So it's it's good to see that we can make conferences like that without Trying to accommodate the climate and things like that. Okay So now it's time for for for questions. I think so very the first question by by shan. Maybe we can still ask Yeah, I will I I can stick it around. I'm not in a hurry. So um, so I will go ahead and answer questions as many as many as come in Okay, so um, shan asked what is an example that doesn't fit into the current available frameworks. So Um, I would like to see our motivic homotopy theory as a deformation of Maybe c-motivic homotopy theory or something like that um, I Seems to me that that's t2 equivalent homotopy theory is a deformation of classical homotopy theory as well um And I'll talk a little bit more that about that tomorrow if I have time but um, you know, what happens what happens here in the c-motivic case is that basically, you know The point is that when you invert tau when you take the c-motivic situation and you invert tau then um, you get um You get classical. Okay. So in the r-motivic case If you you could try to invert tau But the problem is the tau itself doesn't really exist tau is sort of more like a periodicity than an actual element And so inverting tau is a little more com- complicated But barons and shaw have have studied sort of tau periodic Armotivic homotopy and they show how it's connected to c2-equivariant homotopy So that's another so so c2-equivariant homotopy should also be a deformation of armotivic homotopy theory and so forth So on and so forth. Okay. Um So, uh Um, so the you know, it's these you know, it's these examples, right? And then I would like to there there are indications that I would like to You know, you would like to it's kind of like, you know somehow You know, we we are now seeing c-motivic homotopy theory as you start with classical homotopy theory and you like Motivisize it somehow, right? You add this extra grading and you make it more interesting And I would like to take c-motivic homotopy theory and Motivisize it again, right add another grading and making it more interesting. Okay So there there's all all kinds of examples that there there are There is computational evidence that there's all kinds of interesting examples for these sorts of deformations Then of course you probably want to look at the modular stack of all possible deformations And the global sections of those things tell you something universal and really powerful and you know, and and and who knows Right. And now I'm just sort of like shooting off buzzwords. I don't really know what I what I mean by that Okay, um sources to to get started on this idea about deformations. First of all, let me say again Like I think this is like this is like a really important direction. Like, you know, like that That that we're going to see a lot more results coming out of in in the coming decade And so I really like I think this is definitely something that that that I hope people will look at so I'm good source to start with Um, understanding these deformation perspective. There are there there's two places to start. Um, first of all There is a paper of pister golf ski about synthetic spectra Okay, that's one good place and there is an article by george myself krausa and rica About it's the the title of the of the article is motivic modular forms Okay, um, but it it all it sets up this kind of a deformation Framework and we use it to construct motivic modular forms. Um, and people are my secretaries in the background are Um Are our posting links to those references pister golf ski's reference got posted and Someone can look up the motivic modular forms one also Okay, uh, let's see here mark levine asks How does the fact that classical homotopy theory is a deformation of bp star bp modules give new information on classical homotopy theory? Right co modules, right? Okay. So here's what happens. Let me maybe I have a slide later, right? So we'll look at this. We'll jump ahead here here. Here's a diagram. Okay So this diagram at the top of the screen kind of describes the relationship Here's c motivic homotopy theory and then, you know, you have a functors to classical inverting tau setting tau equal to zero You end up in this category of bp star bp co modules Okay, so this is an algebraic category which which means that machines can do all of the work here, right? They can completely understand what's going on here Okay, and so what you What what so you can you can carry out your spectral sequences in great detail by machine here? Okay, but then you because you have these functors you can pull back You can use naturality and you can pull back Differentials along this line and you can get atoms differentials in the c motivic case from these algebraic differentials Okay, and so you get you get the and these atoms differentials are supposed to have homotopical Are supposed to have topological content, but we're extracting them out of machine data Okay, this doesn't give us all of the atoms differentials, but it gives us most of them It gives us a lot of differentials. It gives us a huge head start and then you can Once you have that huge head start you can do ad hoc arguments to find the remaining atoms differentials And then once you compute these c motivic groups, then you get these classical groups just by inverting tau Okay, so it's it's our ability to use machine data At a sort of a bit the deformation is the theory behind how we can use machine data, right? And so it's it's it's a very it's a nice illustration of sort of like what 21st century mathematics is supposed to be like Right, we have to prove theorems. We have to sort of do some work Right, but the point is that we're we're we're proving theorems that allow us to use machines, right to to find things that we That we want. Okay Let's see have you thought about a criterion for on a motivic spectrum that says that there isn't much difference between So no, I have not I mean, yes We have thought about that kind of thing Um about how about the comal that sort of like the motivic invariant of of a motivic object x would be related To the kind of the corresponding invariant the corresponding topological invariant of the betty realization, right? So that's what shawn tilson's asking is like how do motivic invariants correspond to the classical invariance of their betty realizations so The answer can be Complicated for example, there's this object s mod tau. Okay Um, I just mean the co fiber of tau or it's a two cell complex a motivic two cell complex Okay, um, and it's got a very rich theory. I mean it's algebraic, but it's as it's as rich as bp star bp co modules Which is rich. Okay, well typically but topologically s mod tau is contractable. It's just a point Okay, so you can't see anything about s mod tau classically. Okay, so in general, they're going to be very different Okay, however Many of the common things that we study Right. What happens is that their classical their e star what you're writing is e star star of x They're free Say over e star star like say you might study a spectrum x whose co homologies s mod tau's co homology is not free But you know kgl its co homology is free And so things whose co homology is free over tau tend to behave a lot like their classical counterparts That's probably the best answer I can give there Okay Are there any purely out algebra geometric techniques used in motivic homotopy theory that are useful in the calculations Or is it merely for formal reasons that the deformation of the motive of the world helps in calculations The latter case do you expect that once deformation has been made more precise more abstract deformations could provide additional insights Also, what's my favorite topological space? Okay, so that's actually that last question is actually a question I can answer Okay, um, so let me go back here to the the deformation Okay, so, um Actually, I want to go back here to these to these formulas Okay, here's what I need to do everything that I do Here's what I need I need these two formulas that I put at the top of the screen All I need is the homology of a point and the dual steering algebra And I can get up and running and compute x groups and and carry out an out of spectral sequence That's sort of all I need to get going. Okay, so vavadsky establish these formulas, right? And this is different deep difficult hard algebra geometric work Okay So in that sense we're sort of we're using algebraic geometry to when we whenever we're computing semotopic homotopic groups and therefore classical stable homotopic groups as well however This deformation one of the consequences of this deformation perspective Okay, is that we can actually at least, you know Because we we have these we have this other model for cellular semotopic homotopy theory now Okay, and if we use this more topological model for semotopic homotopy theory We can get at those computations of vavadsky more naively without ever doing any algebraic geometry Okay, so there was this kind of like there's this this this is there There's this obvious sort of philosophical question about like why Is algebraic geometry somehow useful for computing classical stable homotopic groups? Okay, and and and this was a question that puzzled us for you know a decade or more two decades, right? And and and I think we sort of have an answer now, right? And the answer is that it wasn't the algebraic geometry that was being useful Okay, it was the deformation that was being that that that is useful for classical homotopy theory We probably never would have discovered this whole deformation perspective if it hadn't been for morrell and vavadsky coming along And can instructing their algebra geometric categories. Okay, so we absolutely We owe it like irreplaceable debt to those guys to their vision, right for having led us here But now with hindsight, right? We see oh, it's not really so much algebraic geometry that's current It's occurring. It's these deformations. And yes, absolutely other deformations would be interesting Okay, I am currently using another deformation To to learn new things about classical homotopy theory apps That's why I want to look at the moduli stack of all deformations and you know and and so forth because there must be some sort of Universal best thing maybe to work with that tells us every, you know, what we want to know Okay, and finally my favorite topological space is rp infinity, right? It has to be rp infinity Um, and we get into why that's but that's that's another matter, right? All those all those steam rod operations and so forth, right? Okay, um All right, what else so shawn tilson? Explains that I didn't he didn't mean betty realization. He meant relating your work over c to other fields like q um, so I think that the the connection There is that you really what you really want to do is you want to work Simultively first. This is the place where there's the least amount of arithmetic occurring, right? And so you um, so you want to sort of sort out the simulative situation first Okay, and and that's the kind of thing that I'm saying here for example, right? For example in this case when mine minus one is a square Computing x groups is just computing simulative x, right? So just do the simulative case instead of the general case, right? However, when you get into the atom spectral sequences things can change. Let me again jump ahead to a slide I was going to show tomorrow but here Here's a picture of sort of like so the black stuff is simulative xed Okay, and the red arrows are supposed to indicate that there's this extra millner k theory, okay? And they all go off to the left. That's the way the degrees work out Okay, and it turns out that there can be arithmetic Differentials, okay that like there can be an element from like d r Like tau can support an atom differential that hits something in millner k theory Okay, and so that you have to handle. Well, if it's arithmetic, you have to do something, you know algebraic something, you know In something an error. Yeah, you have to do something arithmetic there in order to understand that differential So that's I think the way that you pass from c to q is you first kind of do the c stuff that gives you the basic framework And then you have to handle the additional Complications from the arithmetic as well, and you know glenn wilson's work and is a good illustration of how this sort of thing goes Okay Yeah, so There's a comment by yann collet arpe infinity because oh about back to arpe infinity, okay You know arpe infinity is short. It's a k. It's a kz mod 2 1, right? It has like it supports every possible um steward operation, right? Millner used products of arpe infinity to kind of detect everything about the steward algebra, right? So our arpe infinity is sort of like the fundamental Example, right? It is sort of as complicated as it could be from the perspective of cohomology operations, right? Which is the way I think about spaces and so it's you know as a cell it's an infinite cell complex It's got one cell in each dimension and the attaching maps or things you can study, right? This is sort of a very interesting example from a lot of different perspectives And I'm curious to know who asked that question It was an anonymous commenter, but anyway Okay, so Seems no more questions. So that's already 15 minutes of questions. So that's good so We will end up now and I give you Next meeting tomorrow at 1 p.m. It's over the last talk of mark levin Okay, thank you again dan. Yeah, thanks. Yeah. See you tomorrow. Thanks. Yeah, see you tomorrow