So this is essentially an overview

of the most basic way of constructing invariants. The idea in this whole program is to give liftings of classical integral valued invariants like intersection numbers, Euler characteristics, counting curves, that type of thing to quadratic forms.

So let me begin by just doing an abstract view of Euler characteristics, which most of you I'm sure are familiar with, so categorical Euler characteristics.

So this is in the setting of, you have a symmetric monoidal category, and then you have an object X, which has a dual. So that means a dual is a triple consisting of the dual of the object together with two maps,

delta, I'll call them delta and epsilon, where delta is a map from the unit to X times X dual, and epsilon is a map in the other direction that satisfy some well-known properties, which I'll omit.

And for a typical example, just in case you haven't seen this before, I'll just do the very simplest example. We take C to just be K vector spaces, K of field, and then X, if it's dualizable, it means it's a finite dimensional K vector space.

X dual is the usual dual. Of course, tensor is the usual tensor product over K. And then we have this, the usual isomorphism, X tensor, X dual is then the endomorphisms

of the vector space X. So you have the delta is just the map sending, well, what's one, one is the field K, it has a vector space and delta is the map, delta sends one in K to the identity on X.

And then the epsilon, which is often called evaluation map for this reason, it's evaluation map, right? If you have a linear function F and a vector V, this goes to F of V, F of V. Okay, so in this situation, then, so this is just a little aside.

If you have a dual, an object which is dualizable, if it has a dual, then you have the Euler characteristic, the categorical Euler characteristic in the endomorphism ring of the unit, which is just the following composition. You take the diagonal map delta,

you switch factors for some funny reason, and then you take the evaluation map epsilon. So that's the thing. So in our example, the Euler characteristic in vector spaces of a finite dimensional vector space

is just the dimension, but really in the endomorphism ring, the endomorphism ring of K is K. So it's this, the dimension times the identity on K. All right, good.

So let me make a broad and false statement that all, not quite all, all familiar Euler characteristics arise this way. There's a, maybe someone will tell me,

the one example that comes to mind that I don't see immediately of this form is the universal Euler characteristic on varieties over K, K naught bar. So I don't know if it arises that way, maybe it does. Somebody can probably tell me later. Okay, so, you know, for example,

the usual suspects are, well, if we go from say a homological complex in a billion groups, then it's actually will be dualizable means it's perfect.

We assume it's perfect and it's dualizable. So equivalent to a finite complex of free abelian groups, and then a finite, finally generated free abelian groups. And then the Euler characteristic is just the usual alternating sum of the ranks

as Z modules of the homology. And the alternating sum that's sort of interesting, the minus one comes from the tau in the tensor category of complexes of abelian groups. So that's where the minus sign comes in. And then if you have C, a CW complex,

which is, it should be then finite, that's the dualizable condition or equivalent to something finite, then you have the Euler characteristic is just the topological Euler characteristic, which is the ranks of the homology.

And of course, this is of course just induced by sending CW complexes to the derived category of abelian groups by taking the singular chain complex, for example. And then we can take X, a finite type K scheme,

and then we have the eta or their characteristic of X, which is the same thing where you just take the say dimension or say QL, L prime to the characteristic of K of X over K bar. Sorry, which I can tell nobody can read this,

but you know what I'm saying. And- I have a question, but it's just an organization question. Can you make the pages scrolling instead of passing from one another? Is it possible with your- Sure, I can do that. That better?

Sure, how's that? Yeah, that's better, thanks. Perfect. Perfect. Okay, so yeah. And of course, in these two cases, we also have, let me just put on the side, we also have the same thing with compact supports, just replacing homology with cohomology

with compact supports and the same that they call cohomology. So we'd like to get a refinement of this. One thing you could try and do is you could take motives. So if you take geometric motives over a field, so an object here would be dualizable. For example, if you have X smooth scheme over K,

and maybe you have to invert the characteristic due to problems with resolutions and similarities, then you get the motive of X in here, and this will be dualizable. But if you take the Euler characteristic of the motive, well, this lives in N of Z of zero,

which is again Z. So you don't get anything new here, you just get these things back again. But the interesting story is now we take, so here's the, this is all lead into the, we take our symmetric monoidal category C

to be the motivic homotopy category over some nice base S. We usually make S to be a spec of some perfect field, but for a while we can just forget that. And then we have these objects.

If we take X to be a smooth variety over S, smooth scheme over S for finite type, let's even assume a quasi-projective for my own personal reasons. Then you have this motive in here.

So what is this? I'm gonna introduce for functor formalism in a bit. So one way to view this is it's a infinite, it's a suspension spectrum, T suspension spectrum, but another way without writing that, you take the structure of morphism, PX,

you take this lower sharp thing of one on X. So I'll explain what this is in a minute. And then you can even do this for just a quasi-projective S scheme Z, and you have its motive with compact support. This will be the structure morphism with exceptional push forward applied to one on Z.

And this should be, of course, Z not X. And certainly if S is a field of characteristic zero, these are dualizable objects. And if it's a perfect field, and I think due to you, if you invert the characteristic,

then they're also dualizable. So let's say usually dualizable. So in this case, these are usually mostly dualizable. One case where there's absolutely no problem is if you assume that X is smooth and projective over S,

then it's dualizable without any real restriction, it doesn't have to be a field anymore. Okay, in any case, then you get these categorical Euler characters which I'll write in this fashion. And this lives, now we take, well, if I take this over S, this will live in the morphism ring

of the unit object in S. And the same thing for the thing with compact supports. Also in there. But note, in case S equals spec K, K perfect field,

then we have the theorem of Morel. Which tells us what this endomorphism ring, well, I guess maybe characteristic nine equals two, I think he claims this in characteristic two also, but let me just sweep that under the rug. If you take the endomorphisms of one here,

this is isomorphic to the Grothendieck bit ring of non-degenerate symmetric bilinear forms over K, where you, of course, group complete with respect to orthogonal direct sum. You do the Grothendieck bit ring. Okay, so that's the, you get these interesting gadgets here,

living in here in case you have S spec of a field. And in general, you still get interesting gadgets in S and then you can try and study them. Okay, so the basic theme, I'll tell you something about the various properties of these gadgets and say a little bit more

about the four plus functor formalism that I've already used here. And then I'll explain a bit about, a little bit more about how you get these things to have duals, and then how you can try and compute

these Euler characteristics. So the computation we'll use eventually will come the end to this really nice theorem of the Deglise, Jin and Kahn. It's a notific version of the classical Gauss-Bonnet formula. And in order to get to that,

we have to talk about all those things. And in order to make computations with it, we also need to specialize to theories which have orientation. So I'll try and fit that all in. That's the more or less the program. Okay, so maybe I, passed the next page in spite of scrolling.

Okay, so here's the next topic is duality. Nope, let's see, some properties, stop, turn my page. Some properties, let's just do the case of a field,

compact support one. Okay, so first of all, if you take X, which is smooth over K and projective or proper, then, well, I guess this is true over any S,

then the guy with compact supports is the same as the original one, or without compact supports. That's actually easy to prove. This is also true for general S.

Okay, but one interesting thing, which I don't know how to prove and over general S, if the dimension of X is odd, then this Euler characteristic is a multiple of the hyperbolic form.

So H is the form H of X, Y equals X, Y, or if you like, X squared minus Y squared characteristic, not two. Okay, so that's the sort of form that everybody understands and sort of trivial, you get rid of this H

if you pass from the Grothendieck bit ring to the bit ring. So it says that this is just a trivial gadget in the odd dimensional case. So the first computation that I know of was due to Mark Hoyle, who showed using a Lefschet's trace formula

that the Euler characteristic of PN, think over an arbitrary base, is the following. It's equal to, I'll explain the notation in a minute. So we've already one plus N over two times H, this is N is even,

and just N plus one over two times H if N is odd. So I've already explained what H is, this form, if you have U, a unit in the field, this form U is the form X, the one dimensional form, X goes to UX squared. So there's a quadratic form,

or as a bilinear form, it's XY goes to U times X times Y. Okay, so that's the first computation. And I think at the beginning, that was really the only thing that anyone could compute. It was even a question. So this thing about thing being odd, that came much later. It was even a question of what is the Euler characteristic

of a Saveri-Brouwer variety. So I'll just mention that. If you take X over K, then the Saveri-Brouwer variety dimension, and then the Euler characteristic of X

is the same as the Euler characteristic of the projective space. So not so interesting, but actually this follows from these two facts we mentioned before, because if N is odd, then we know it's hyperbolic, which is the odd case for projective space.

And if N is even, then you can split the Saveri-Brouwer variety by an odd dimensional field extension. And that is fairly, the Grobendy bid ring is fairly insensitive to that. So you can basically reduce that. Okay.

Let's see. Oh yeah. Let's see. And why is this true? Well, that's pretty easy. That's just because the Euler characteristic of X over K

is this categorical Euler characteristic of the motive. But this is the same as the categorical Euler characteristic of the dual of this motive. That's just a formality in any symmetric monoidal category.

And as we'll see in a minute, this is the same for dual is the motive with compact supports. So we'll see that in a minute. This is later. So,

now this Euler characteristic of compact supports is actually interesting. From the following point of view, I had mentioned briefly this universal Euler characteristic K naught bar. Hey, I'm not going to say anything more about this. If you know what this is, then fine. If not, then just ignore me for a minute. And this extends to,

or descends to a ring homomorphism from K naught bar to the Grobendy bid ring. So the so-called motivic measure. So you can use computations here, which are often very difficult,

but some interesting ones, a lot of interesting ones have been made to make computations here. All right. So let's go on. I want to talk about four plus functors. Depends how you count them.

So what's the situation. If we fix our base base scheme S, then we can send X and S scheme to the motivic stable homotopy category over X.

And this defines a functor from say, nice, I'll say finite type or quasi-projective, whichever you like, contravariant functor to triangulated categories. Of course, this has lip things to infinity categories, but we won't need that.

So in other words, we have this guy, and if we have a map F from X to Y, then we get a pullback map. That's the contravariant functoriality from SH of Y to SH of X. And this functor has an adjoint. So we have a adjoint pair,

F lower star, F lower star goes in the other direction. And also in case F is smooth, there's another functor, which is left adjoint to the F upper star.

So again, a covariant functor. And this is the one I use to describe the motive of X by taking X smooth over S, and then just taking the push forward of the unit with respect to this lower sharp gadget for the structure. Mark, there's a question about compatibility of characteristic, characteristic and base extension.

So I guess it's a compatibility with pullbacks. So we- We're not available with any symmetric monoidal functor. So pullbacks are symmetric monoidal. So yes. Okay, thanks. Okay, so those are those guys.

That's if F is smooth, we have this left adjoint. And in general, we have the exceptional adjoint pair F lower shriek, F upper shriek. And again, if F is smooth, well, you can compute the thing. So, well, we'll have to get to that in a second. But this leads to the following gadget.

If you have V to X, a vector bundle, then you have this suspension operator, which is, so here's the projection P, and then you have the zero section S, which I'm sorry, probably can't read that, it's too small, but you call it S, the zero section.

So what do you do? You take S lower star and then P lower sharp, P is smooth. So you can do that. And you also have its adjoint, which is the same as its inverse,

which is S upper shriek, P upper star. So this is actually an invertible gadget. And this extends to this suspension map from the K-theory space of X to endomorphisms

of these triangulated symmetric monoidal category, or the triangulated category SH of X. Okay, so, and now I just wanted to say, if F is smooth, then we have all these functors,

we have this lower sharp and this upper shriek and the upper star and the upper shriek, and they're all related, then F lower shriek is the same as F lower sharp composed with suspension with respect to minus the relative tangent bundle for F. So it's the kernel of the map and the tangent bundles.

And F upper shriek is similarly, you suspend with respect to this guy, composed with F upper star. So they're all related in the smooth case. Okay, and so let's see. Oh yeah, and there's another bunch of lists of things

I was supposed to tell you. One other thing is you have the one, you can phrase Morel-Voyabotsky, Morel-Voyabotsky purity theorem in this language. It says, if you have a closed immersion of smooth gadgets over smooth schemes over S,

then that gives you an isomorphism of you can take the structure push forward with respect to the structure morphism composed with the inclusion map. And that's the same thing as pushing forward

with respect to the original guy, but first pre-composing with the suspension operator with respect to the normal bundle of Z in X. So that's what I meant. Or maybe I should, maybe you write that the other way. The normal bundle of Z in X, I'll just write it this way.

Confusion. All right, and that's statement from Morel-Voyabotsky. Purity, and you also have, if you send X goes to this MX, which is, I already described, it's TS or sharp one on X,

defines a functor M from smooth things over S to homotopy category. So if you have a map of schemes that induces a map on the motives and Z going to the gadget with compact support,

which is this guy here, extends to a functor. This uses a little more work and see, well, where does it go? You take schemes of finite, save your favorite finite type category,

but you restrict to the subcategory of projected morphisms and then it's contravariantly functorial. So if I have time, I'll explain how you do that, but let's just say, so it means if you have F from Z to W proper,

then that induces F star from the motive compact support of W to the motive compact support of Z. It's not so hard to construct that, but let me just avoid it. All right, so that brings us to duality. How do we get duality?

SH, well, let's take X to be smooth, so I'm going to refer to this to Mark Hoiwa's paper. This was of course done in various forms by a lot of other people, starting with Boyavodsky,

then Hu, then a lot of other people. Hu did a lot of work on this, but I think the nicest and the most complete form is in Hoiwa, that's the nicest one I know. So we take X to be smooth and proper over S

and then the statement, which is I've mentioned that this motive of X is dualizable and it's dual is the motive of the compact support. So how is this? We have to define these maps delta should go from the unit to,

so you write the tensor operation as wedge due to topological bias. So how do you do this? Well, you first have the map, well, this is a proper map, so we can take P X upper star, go into the motive with compact support on X, right?

We have a pullback map for proper morphisms, okay? And now what is this thing? This guy here, this is this P X lower shriek of one on X, but P is also smooth,

so this is P X lower sharp of suspension by minus the tangent bundle of X over S, let me just avoid the over S on one on X, all right? And this thing here is the same thing that the smash product here comes from just product on X. So what you've done here is you,

if you like in the first variable, you're just taking the P X lower sharp, the second variable you're taking the P X lower shriek, but that's the same thing as just twisting by the minus the tangent bundle. So this smash product is really equal to P X cross X lower sharp

of you suspend by minus P two star of the tangent bundle on X, one on X cross X. And then from that description, you see that the usual diagonal map from X into the product gives you a map here. So I'll just call that, that's the usual map induced by the map

on the usual motives, but twisted by this minus the tangent bundle. So that's Delta, right? So we're supposed to get a map that way. And what's the Epsilon? The Epsilon works for any smooth variety, smooth S scheme.

So let's see, so here we have MC of X smash M of X. And the same thing I said before works here, except those factors get switched. So this is just the suspension of minus P one star with a tangent bundle of X, one, okay.

And so now we can use Morel-Boivodsky purity and I should have said something there's, well, there's also through the adjoint property, we have the adjoint property gives us a map

from the identity on the SH of X cross X to Delta lower star Delta upper star, just using the adjoint property of Delta. Delta is the diagonal map from X into X cross X.

And we combine that with the Morel-Boivodsky purity isomorphism. And you see, you get a map of this gadget here to the following thing. It's just a PX lower sharp, but this is viewing X by the diagonal of what? We have to twist this guy here,

of course, pulled back to X and it just becomes the tangent bundle again by the normal bundle of the diagonal inside the product. Y to one on X. Okay, but as we all know, the normal bundle of the diagonal inside the product of X with itself

is just equal to the tangent bundle. So we're ended up with no twisting. So this thing here is just equal to the mode of X. That's X is smooth, we have a tangent bundle, right? And then we have the usual push forward map by the covariant punctuality of this mode of operation to the mode of S, which is just the one on S.

So that's the map epsilon X. So those are the two structures and Hoi-Wad's theorem is, it says that MCX together with this delta X and epsilon X is a dual.

So A means up to unique isomorphism is a dual of MX. All right. Oh, and moreover, maybe I should just say, and once you have a dual, you can also take duals of maps. It gives a duality and maps between dualizable objects

and the it's map PX star, which was defined here is actually just the dual of PX lower star. Okay, so let's see.

That brings us to, I think I'm doing okay for time. That brings us to the next topic orientations. Okay, so if we have an object in my stable homotopy category besides,

let's just, I think I'll just work. Well, can I work over S? Let's see how it goes. Work over S. Let's assume that this represents a cohomology theory, which means it should be a commutative ring object. So it has a commutative multiplication in the category

and it has a unit object, a unit morphism from the one. And then if you have such a gadget, I said it's a cohomology theory. What's the cohomology? It's a bi-graded cohomology on some object is just a hom in the category

into, well, you have this two variable family of suspensions. I'm not gonna say too much about that. You probably all know about it, but this is just two variables where this corresponds to taking smash product with S, the usual sphere to the A minus B, smash with the GM to the B,

assuming that A is bigger than or equal to B and B is bigger than or equal to zero. And then you've inverted enough things in here to make this make sense for all A and B. Good. So that's a quick and dirty description of what this cohomology theory is represented by a commutative ring object in the category.

So what's an orientation for E? Is the assignment of Tom classes or vector bundles.

In other words, if you have B to X vector bundle

R, then you should have an element Tom class in. So I'm assuming here X let's just restrict the X being smooth over S. You have an element in the E cohomology

of this Tom space of B, where the Tom space of B is, well, I guess I didn't really define this, right? This is the P X tree with suspension of one on X. What I mean more concretely, it's just you take B modulo B minus the zero section,

view it as an object over S. The same thing. Okay, that satisfies certain properties, satisfying blah, blah, blah, which I'm not gonna tell you. But the upshot is, is this leads to a Tom isomorphism called theta V,

which goes from the E cohomology of your X to the E cohomology with a shift on the Tom space of V.

And what you do is you take an element here, X, and then you can pull it back to V and then take the cup product with this thing here. There's a cup product from the cohomology of V and cohomology of the Tom space of V to the cohomology of the Tom space of V in here.

So that's an isomorphism. And the conditions that the Tom classes have to satisfy I say that in case that the vector bundle is trivial, it's just the suspension isomorphism. All right, so that's an orientation

and we have other flavors of this thing. So what's an SL orientation, or even more general an SLC orientation is the same kind of thing,

except you require a bit more rigidity is the assignment for each V, you have a V as above, plus an isomorphism of the determinant of V with the trivial bundle, or in the SLC orientation case,

you relax this a little bit. This is an isomorphism of determinant of V with the square of some invertible sheaf. Then you get a Tom class.

And where does this live? Yeah, it lives in the same place. Okay, and induces Tom isomorphisms in the same way.

Okay, so that's a brief description. Now, what good are these? Well, these are good for defining Giesen maps or push forward maps,

just by identifying in some sense the, so how am I doing here? So I think I'm gonna get to the end. Okay, so how do you do this? Well, you take, if you have F from Y to X,

let's say smooth over S, but this is proper. Then we already have this F upper star from PX or shriek one on X. That's the motive of compact support. PY lower shriek one on Y.

But this thing is the same thing as PX or sharp minus the tangent one on X. This thing is PY or sharp minus tangent one on Y. So you see if you have,

so this induce maps on the hom into your E. So if E is oriented, he has an orientation, then the F upper star induces the F upper star upper star, which we'll call F lower star. We'll go from EAB of X to EA minus 2D,

E minus D of Y, where D here is the relative dimension of F. And if E is only SL or SLC oriented,

then you get the same kind of thing. You get F lower star, but now you'd have to take, well, let's look at the SLC oriented case,

more general case. It gives you a push forward map. If you twist, I have to say what this thing is. If you can twist your theory by the relative dualizing sheets, the top wedge of the differentials, you have some Y middle L on Y,

does the same thing in 2D. E minus D on Y, twisted by S, insert L. Where here, the theory on some Z say, twisted by some invertible sheet M,

means you take the theory on the Tom space of M for M and invertible sheet. So you have to trace this out, but that turns out to work. And now you have the general case,

which is I think first appeared in the work of De Guise, Jin and Kahn, where you just say, I mean, you just use this. You just denote the theory to just say,

you're just doing this here. You define EAB of let's say X with respect to a vector bundle of rank R to be EA plus two R, E plus R of the Tom space of D. And you extend it to virtual bundles

in a fairly formal way. And so you get in this notation, in this sense, it's more or less notation, except you have to make sense of it. You get F lower star going from EAB of X minus the tangent bundle of X going to EA plus,

sorry, the tangent bundle of F. Oh yeah, sorry, just a remark. The F lower star should go from EAB Y to EA minus D. Oh yeah, I got everything else. So let me just change this here. That's it. Sorry, sorry, that was. No problem. I knew I was gonna do that.

Thank you very much. And this should be then, yeah, this should be Y. Sorry, oh dear, oh dear. Confusing everybody here. Thank you, sorry about that.

Okay, so you're right. So something, or if you like, maybe I should just stick with it.

Okay, so that's the general case. All right, so let's see. Kind of running out of time, so let me just come to the interesting examples in a second. So I have to say something about Euler classes.

Well, I mean, now more or less it's formal. If you take V, a vector bundle, with zero section S, rank of V equals R, then you have the Euler class

in your cohomology theory E of V is just S upper star S lower star of one on X. And where this lives depends on, in general, it will live in this. That's the general case,

but this will specialize to this in the SL oriented case. And we'll specialize to perhaps the more familiar guy here. Whoops, sorry about that. In the oriented case.

Okay, but the formula is the same. So, and then finally we have, we want to compute, we want to count things. So we need the degree maps. And what's that?

Well, we can consider this if you have now X, it's just the dual of the pullback. If you have X over S, smooth and projective, then if you like, you have the degree map

going from your theory, let's say a relative dimension D, and it goes from E to D, D of X minus the tangent bundle of X. This is just equal to, this is my structure map PX is just PX lower star.

Okay, right. And of course, again, this specializes in the SL oriented case, this specializes to, again, this determinant inverse of this, but this is of course, just the dualizing sheet going into the same thing.

And in the, this is the SL oriented case. And in the oriented case, this is just without any decoration. Okay, so this last one is the usual degree map that you see from things like chow groups or K theory. Good, so in particular,

we have even in the most general setting, we have the Euler class of the tangent bundle of X. This lives in H to D, D of X

minus the tangent bundle of X. So then we can apply the degree map to it, to get this degree of the Euler class in, well, not H, let us see. That was a misnomer.

E, sorry. And this lives in E zero, zero with my base scheme S. Okay, so now, I think I'm doing good. So now we have all the ingredients to describe, state the motivic Gauss-Monet theorem.

This is due to Deglise, Jin and Kahn. So it says that we take X over S smooth and proper.

Then coming back to our Euler characteristic, this motivic O over S is just equal to the degree of the Euler class of the tangent bundle in which theory in the unit theory.

So this is living in one, zero, zero of one, well, of S, but this is just N.

So it's living in the right place. They all live in the right place. And since I have a few minutes, I mean, the proof is actually. So you still have 10 minutes because we started the five minutes late. Ah, great, okay, so plenty of time. Thanks very much.

So let me give a quick, in spite of that, I'll only give a sketch of the proof, the main idea. It's just you have to reorganize the definition. Well, the first thing is a trivial reduction. It's of course, the definition of this Euler characteristic is you have this composition of these maps.

You take, what do you do? First, you take delta, then you compose with this twist map, and then you compose with epsilon. But this is a map from one S to one S. But now if you apply, of course, if you're, you know,

you have a ring and you multiply by an element in the ring, it's the same as the, does the same thing on maps of the ring to itself. So if you apply hom into one S, which is now I'm thinking of this as my E, then that says that this Euler characteristic here,

is just gotten by applying these maps to the unit. In, right, okay, so in other words, let me just say it, it's equal to delta X star composed with tau star composed with epsilon X star

applied to the identity on the one on S, which I'll just call one. Okay, so what is that? Let's see, well, let's remind ourselves. Well, what's delta? Remember delta was this guy here.

This goes into MX times its dual, which is the MC. And how was this constructed? We took the dual of the structure map going into here,

and then we applied the diagonal map. Okay, then we take tau, that switches things. Of course, this thing here, remember, this was equal to PX over sharp

of minus tangent one over X of one, but this is just the Tom space. This is just the Tom space of minus the tangent one. Okay, and we still have a similar delta here. And since delta doesn't, lives in the diagonal,

doesn't care about the tau. So now what's this thing here? Remember this guy here was the Tom space, if you like, of minus P two star of the tangent bundle X. And so this guy here is the Tom space minus P one star of the tangent bundle X.

Okay, so that was a delta, that's tau, and now what about epsilon? So remember how we did this? We applied Morel-Boivodsky purity to the unit of the adjunction.

So what did this go to? Remember this went to this thing, which was the Tom space of, it was really minus the tangent bundle plus the normal bundle, but this cancels out. And the canceling out is the Tom,

the canonical Tom isomorphism for X going just to the motive of X. You think about it for a minute. This thing here is, you know, this is, well, okay. This is really the, well, I think I skipped a step here.

Sorry, let me go back a little bit. Well, think about this thing. You think, let's think of the Tom space. This is, you can think of this as you take the Tom space, let's just cheat a little bit here. This is like taking the Tom space of the normal bundle of X where I view the normal bundle of X as a bundle

on the Tom space of minus the tangent bundle of X. So this is a bit of a stretch, but that's a nice way to think about it. Now, if you think about it that way, what have we got? Well, here we have this space, the Tom space of minus the tangent bundle,

and we're taking delta X into here. This is really just the zero section reduced by the zero section of the normal bundle viewed as a bundle over the Tom space of minus the tangent bundle of X. That's my S inserted here.

And now this thing by the canonical Tom isomorphism is isomorphic to MX. So I'm still describing epsilon. And now to get epsilon then at the end, I apply the push forward here back to one on S.

So let's see, it's a little messy here, but this whole gadget here is the epsilon. All right, so let's try it. What is here? This is the guy we want to calculate. What is delta X star? I still have a few minutes.

Mile star epsilon star. One. Well, that's the same thing here, delta X star. Well, the same thing. Oh, I didn't want to do that.

Come back. That was close. This is the same thing as, well, if I follow it this way here. So this is the same thing as I take PX star of one. So that's one on X.

In other words, it's one X. Then I take theta of that. That's the Tom class. So what is this? This Tom class is just S zero lowered star of one on X. And then I take S zero upper star of it.

That's the Euler class, right? With respect to the Euler class of the tangent bond on X. And then, oh, then I take PX dual of it, which is just the degree. So there you are.

That's the sketch of the proof. Okay, so- There's a question about the, so the classical Gauss-Bonnet formula relates curvature to Euler characteristic and the questioner wants to know the intuitive interpretation of the algebraic analog. Well, the curvature we view as representing the Euler class

in the topological sense. So that's the connection. So the curvature, right? You integrate the curvature. The curvature is, you can view as a cohomology class representing the Euler class of the tangent bundle if you put a metric on it by the Tia.

Who does this? What's it called? Help me out, Chris. I think that's, yeah, that's it. The degree is the integral of the- Yeah, and then the degree is the pushboard. That's the integral of the Euler class as represented by the curvature. So it's a kind of quadratic curvature.

Right, exactly. Okay, so that's the theorem. Now, let me give, here's a application. It's a theorem to myself and Arpan Rekse. Which says the following.

Let's say, now let's take X over K. So I think we only proved it in the projected case, but probably it works in the proper case. Then the Euler characteristic of X over K is the following explicit quadratic form.

You take, to phrase it most nicely, you should work in the derived category. So I take these Hodge cohomology groups,

I view this guy in degree I minus J, so I apply the shift, and I take the direct sum over all I and J. Where it's going just between zero and the, so let's say D equals the dimension of X.

Okay, then I need a quadratic form. So the quadratic form Q is as follows, where Q is gotten by taking H I X omega J tensor with H D minus I X omega D minus J.

Unfortunately, the degrees cancel out. I take the cup product, this goes into HD on X omega D. And then I have a trace map here. So of course, this is the familiar omega X over K.

I have the trace map to K. So that gives me the quadratic form. And let me say one word about the proof. So almost out of time, we apply the Motivik-Gauss-Bonnet formula

to the case E equals KO, Hermitian K theory. So the point is the unit map from one on K to KO. So this is living in GWK induces an isomorphism

from GWK, that's the endomorphisms of the unit by Morel's theorem to KO of K degree zero zero. So you can compute. So here's where our Euler characteristic lives. And you can compute it here by applying U to it.

And then you use the work of Kalmanis and Hornbostel, which give sort of explicit formulas for all these push forwards.

Okay, so let me just conclude by saying, as far as I can tell, recent work of Bachmann and Bickelgren have extended this to work over an arbitrary base, where you just assume that two is invariant.

Okay, thanks very much. Okay, thanks a lot, Mark. So, I don't know how everyone applauds, probably. I'll take it as given. So, I'll start with a question that was asking before. So, it's about the application

of this formalism of other characteristic formulas who braided monoidal categories, which are some weakening of symmetric monoidal categories. Oh, and so the question is, is this case relevant to the motivic case? No idea. I don't know. Is there an Euler characteristic in the braided?

So, I guess it's the question. I don't know anything about it. I don't know if there's a nice theory of dualizable objects. And I suppose, I mean, there must be traces in those cases, right? I assume. So, maybe the

person asked the question could respond and let me know what the story is. But it could be interesting if you could sort of refine SH to have this braided structure. Maybe that would explain something. There is an interesting situation going on where looking for something

that makes GW more complicated, because there are interesting formulas for these Euler characteristics. I mentioned the universal Euler characteristic, K naught bar, and there's interesting work of Gatcha and Shenda, I think, who define an interesting Euler

characteristic with values, you know, it's from the Hodge polynomial, right? So, it has a variable in there, and it would be nice, it has the property that when you, which I should have

mentioned for these Euler characteristics, one nice thing about these Euler characteristics is when you take the rank for the Euler characteristics of X over K, you get the usual topological Euler characteristic, and if K embeds into the reals, then the induced signature function on the Euler characteristic gives you the Euler characteristic of the real points.

And these Gatcha, Shenda, Hodge polynomials, I believe, maybe I got that wrong, but there's polynomials with a variable in them, and when you evaluate at one you get the topological Euler characteristic at minus one, you get the real, the Euler characteristic of the real points, and so it would be very interesting to be able to refine these motivic constructions to

end up with something living in, something where you have a variable and evaluating at minus one corresponds to taking a signature of some quadratic form and evaluating at one corresponds to the rank. The present just doesn't exist, and maybe in this framework of graded symmetric monomial categories such a thing exists, I don't know.

Okay, so I guess that answers the question. So, I have another question that I read. So, if a vector bundles E admits a sub-bundle of odd rank, we have two times the Euler class is zero,

do we have an analog for our motivic Euler classes? With etra, maybe? It's a sub-bundle. Yeah, sure. Because, yeah, there's a theorem that says, well, there's an analog, it's not zero, what it is is the following. So, you can pass, like, let's look at KO, and let's see, I'm pretty sure this is true,

let's look at the simplest situation, where one nice theory, which still gives you the quadratic Euler characteristics, is by taking X, you look at the cohomology of X with coefficients in the Milner-Witt K-groups. So, this thing specializes to the cohomology of X

in the Milner K-groups. So, and this thing, for example, you have this block formula, this is just chow n, and here the Euler characteristic V of rank R just becomes the

top-churn class of V. But here you have this lifting of V, and it will specialize to this, but will also have quadratic information. And here the theorem is that if the rank of V is odd,

then you have this element in here, eta, which sort of gives you the kernel of this map, then eta times this Euler characteristic of V is zero. This is like, eta is like multiplying by

two. And now when you invert eta, then this gives you a class in the bit ring cohomology. If you take this guy here and invert eta, you get the sheaf of bit rings. And what that says is this guy here, if you take the bit ring valued Euler class, this is equal to zero

if the rank of V is odd. And now the Euler class in short exact sequences, I guess, is multiplicative, so that will tell you that you have the same theorem. That after inverting eta, and I'm not sure if this is true in any theory after inverting eta, it could very well be. I'm blanking on that, but I know it's at least true in this theory,

that if you invert eta from this very nice theory here, then that same theorem is true. Okay, thanks. And there's a comment. How about a converse of this property?

Which property? If we have this vanishing, maybe in the cohomology of the bit sheaf, then what can we say? Ah, well that's a very, I mean, that's the Euler classes in here have been extensively studied to ask questions about splitting of vector bundles. So this goes back in here,

looking at the Chern classes, there's a long study by Murti and many others that if you work over an algebraically closed field and you have a vector bundle on an affine, smooth affine x of dimension d and the vector bundle has rank d, then it splits off a trivial

summand if and only if the top Chern class is zero. And then this was extended by many people but I think the most definitive version is now due to Krishna where works over an algebraically closed field over an affine x but you don't assume that it's smooth or anything

and you get the same result where you have to say what the child group is. He places this with a certain Euler class group. And the Euler class group, I think is closely in general, is closely related to this. Now you ask what happens over non-algebraically closed field and this question has been answered and I don't know exact state of affairs in what generality

it's true but this type of group here was used by Asock and Fazel to answer a similar type of splitting question over non-algebraically closed field. We have to look at their work to see exactly what the state of that is. Okay, thanks. So we have other questions. So

the one question is that the construction of refine Euler characteristic in H-theory depends on resolution of singularities and the question is how are we doing that in characteristic p?

We're only talking about smooth projective varieties. I don't know, refined in the sense of the quadratic we find. Is that what the question is about? I'm not sure if this is another question. In this case, you really don't need resolution

of singularities because we're taking x to be smooth and projected. So we don't need a nice theory of hodgecomology on smooth varieties to do this. The way this goes is by working entirely with permission k-theory and then you this result of

Kalmay's Hornbostel is saying when you, I didn't tell you what the orient, see this ko theory, I ran out of time. So the ko theory is slc oriented and you can write down explicitly what the tod classes are. For k-theory the tod class,

I mean tom class, the tom class for k-theory for a vector bundle is just the kazool complex you get by taking the canonical section of the vector bundle and you have a similar story for ko theory. You take the kazool complex and just the usual map of wedge i of

a vector space and wedge r minus i of the vector space of the vector space at dimension r into the determinant gives you the quadratic form. And then the problem in this theorem is that that tells you essentially what the orientations here are and then you have to just, and that was used by Kalmay's and Hornbostel together with Groten-Dieck duality theory

to tell you what this push forward is. And then the only real problem here is to say this explicit formula is the one given by the general theory once we have these orientations. And so that's essentially a calculation since you know what the what the ko theory is after you invert

eta for a projective space, I think of odd dimension. So that's how the proof here goes and then the extension by Bickelgren and Bachman is saying okay we do this in a fancier way using using frame correspondences and constructing spectra that way, but I think the idea is roughly

the same. You basically have to get an idea of an explicit formula for this push forward map from a projective space down to the base. But you never use hodge cohomology as a cohomology theory directly. It just comes in because that was what arises in looking at what

the Euler class is in ko theory. Okay so next question, we have still three questions. So a remark about the fact that in the C2 or Z mode 2 maybe equivalent stable case there is

also big grading so the question is that does this motivate an equivalent version of classes and characteristic? Oh I think there's a lot of work done in that. There are people who've constructed equivariant, I mean there's a construction, Hoiwa's work

talks explicitly about SHG. So you take some group and you look at G equivariant motivic stable homotopy theory. So this is in there and then people have for C2 equivariant people have constructed versions of real k theory and related it to Hermitian k theory.

So that's definitely been done. I haven't seen what these constructions of the Euler classes would do in that case although there should really be a connection especially because if you look at the C2 equivariant classical stable homotopy theory that's very closely

related to there's a functor from the real from SH of R to SH C2 where the complex conjugation on a real scheme gives you the C2 action on the complex points. So that's how you map from SH over the reals to C2 equivariant classical stable homotopy theory

and there the signature comes up again because in fact the Grodendieck bit ring of the reals is exactly equal to the Burnside ring of C2 which is the endomorphism ring of the unit in the C2 equivariant theory. So there's a very

close connection there but other than those few trivial comments which I've lifted from other people I don't know anything specific about. Okay so there's also a question about so it says that in case of vector spaces of our field the Euler characteristics always land in

the image of a ring of integers on the endomorphisms of k and so the question is is there a similar phenomena in the motivic case? Ah well let's see there is let's see there's a special ring where these things land well we'll talk about computational methods there is um

yeah let's put it this way um we have the the sort of technical problem is that we don't really know exactly what the endomorphism ring of the unit is over the integers

but and we also don't quite know what k so we can approximate that by ko of the integers and if you invert two we know exactly what that is that's just the growing big bit ring of z you join one half and there's been a lot of work recently about trying to patch over this problem at two so in fact Bachman and Bickelgren mentions explicitly

that if you what one supposes that k over the integers is just the growing big ring of symmetric bilinear forms over the integers then if you have any gadget which is

unramified in whatever sense you like over the integers then its corresponding degree of the Euler class or whatever quadratic invariant you calculate for it should land in the growing big bit ring of the integers so for example if you look at Euler classes of universal bundles on Gerasmanians and if they're of the right degree to take the degree then you'd expect

them to land in the growing big bit ring of the integers which makes life easy because that's a very simple ring it's just gotten by passing to the growing big bit ring of the reals which you detect everything it's just detectable by the rank and the signature so it would make a lot of calculations very easy the current state of affairs is you get this up to

a two torsion element which comes from the one half in the growing big in the z adjoined one half and then Bachman and Bickelgren say okay well then we can just make explicit evaluations at other primes and eliminate this two torsion element by computing elementary invariants like

discriminates so except for this minor flying the ointment i think that would be the analog that it doesn't land in the integers it lands so unramified classes would land in growing big of the integers and if you're for example if you had a hypersurf of some some projected variety over k which comes from something which is smooth and projected over

z adjoined one over n then you'd expect that the Euler characteristic would land in the growing big bit ring of z adjoined one over n something like that okay it's true except at

two we're not quite sure but it's expected i think that's the that's the analogous answer so there's the last question about uh is there a formula expressing viola class of a vector bundle v over x in terms of say viola characteristic of x and and v or maybe

the tom space of v or some other object ah in terms of the in terms of the Euler characteristic of x and the Euler characteristic of the tom space or of v or well the Euler characteristic of v is the same as the Euler characteristic of yeah although if you take the

Euler characteristic compact support that's the Euler characteristic of the tom space i guess so um i hadn't thought about it but it's probably not because the point is that the Euler characteristic is essentially multiplicative so when you're taking the Euler and it also satisfies cut and paste or maya via torus so when you take the Euler characteristic of the tom space

of v let's say v has rank r it just multiplies the Euler characteristic of the base space by this by the Euler characteristic of reduced projected space of a sphere which is minus minus one so which is um minus one right the quadratic form minus one not the number minus one

so it just multiplies by that quadratic form minus one raised to the rank of the vector bundle so you don't really get any information about the Euler characteristic of the vector bundle that's coming by looking at how the zero section sits inside the tom space that's really

what's given that's the other class is done by that and that's not given by the Euler class of the tom space okay so that's the end of the question i guess thanks a lot mark