 introduce some rather abstract objects and today I want to try and get some very computational and concrete methods for getting at these things in very explicit ways. So first of all I want to review some of the very nice constructions that I first learned from Kassan Wickelgren, Wickelgren about Euler classes and local indices, very explicit formulas for these Euler classes. So you remember we had a vector bundle on some say smooth scheme X and with zero section as zero then we have the Tom class and we have some co-mology theory E. Let's work over field K and we have some Tom class in theory Tom space so R here is the rank and of course the Tom the Tom space is just V modulo V minus to zero section so this is really the same thing as the co-mology with supports along the zero section on V and I forgot there's some kind of depending on the type of theory there's some kind of orientation information if the theory is oriented then this funny squiggle isn't there so the squiggle is equal to the empty set in the oriented case it could be the inverse of the determinant of V in the SL oriented case and it could be minus V in the general case. So that's the Tom class that lives there and the Euler class is just the pullback by the zero section of the Tom class okay and this lives in the theory on X again with whatever the orientation is okay now suppose instead of this section S zero we have just some arbitrary section S well then we get the same thing you can always do a A1 homotopy between S zero and S by just multiplying S by T and letting T go to zero so that says this is also the Euler class is also equal to the pullback of the Tom class in here but see when you pull back the Tom class it pulls also pulls the support aspect so if you let Z the S inverse of the zero section then you actually get a sort of a Euler class with supports on Z but this this will depend on the section S because if you move the section S it'll move the zero locus Z so you can't expect that to be independent but in any case so we put it in the notation and this lives in the co homology on X with what are their orientation with supports on Z okay so let's we'll be interested in the special case where yeah the rank of the bundle is equal to the dimension of X so let's just call it D and in that case a general section would have only zero dimensional zero locus so let's assume that let's assume that the dimension of Z is zero but we won't assume that the intersection the pullback of the zero section is transverse so let's just say that Z is some finite set of points and there might be some kind of multiplicities along these points and these Euler class with support is kind of telling you about what the multiplicities are so since it breaks up into pieces this this this guy here this co homology with supports also breaks up into pieces so it says this Euler class with supports on Z is just a sum of pieces Euler class with supports on each of the Z eyes okay and so for simplicity we can just let us assume we can localize around each of these points let's just assume Z is Z a single point just for the purpose of discussion and we want a formula for this local Euler class okay so well the first case let me say I won't say too much about this but the first case is when E is just a typical homology and then this 2dd part so D is equal to R so 2dd part is just the child group of co-dimension D cycles on X which is just the zero cycles because X has dimension D and then if you put the supports condition in you have a purity theorem on Z this is just child zero of the point Z which is just the integers times Z and what you're getting then this local Euler class is just equal to the some multiple of Z and what multiple is it is just equal to the intersection multiplicity of the section S of X with the zero section at the point Z times this single point Z okay so that's sort of the classical case let's do something a little more complicated second case is we take e to be k theory so this is the spectrum representing usual k theory and then the e2d we have a bought periodicity the 2dd doesn't do anything this is just k naught of X if we look at it with supports well I'm gonna do something a little perhaps unsuspected here it's k naught of X is really k naught of perfect complexes on X and k naught with supports is k naught of perfect complexes on X whose co-homology is supported at Z and on the other hand then you have some kind of resolution theorem which tells you that this is the same thing as k naught of the category of modules over the local ring of finite length in other words supported at Z okay and then you have a Davies-Age theorem which says well if I have a module of finite length I can find the filtration so this is this is like the resolution theorem and I have the Davies-Age theorem which says this is the same thing as k naught of just the residue field kz k naught of Z and of course this is just the integers by rank times the class of the residue alright so the reason I went through this is we want to calculate what is this e z we asked well remember this is I take s star of the Tom class of V where I'm taking the Tom class in k theory and remember how we got the Tom class in k theory the Tom class in k theory is the kazoo complex so I have this kazoo complex for the vector bundle V with respect to its canonical section and so what is that that's you call that's a p star v dual going to o v by the canonical section what's called t dual about the zeros of this is just the zero section so that's the structure sheet of the zero section and then you have the wedge two of that etc all the way up to the top wedge okay so when you take s star of that well now the point is that let's say we're working near work near z and let's take a local basis for for v so basis of sections you want to ed and then we can write our section s as some s i e i and the fact that the zero locus is just concentrated at z means that the this set of elements s one up to sd a sequence of elements is a regular sequence in the local ring o x at z and when you have a regular sequence the kazoo complex for that regular sequence remains exact and that's what happens when you pull this back so when you take s star of the kazoo complex or the monocle and use this basis this is just the kazoo complex for o z o x z to the d with respect to these elements s one of the sd so that's what you get it's the same thing as what I wrote here except it becomes a little more concrete it's just that we have o x so we just call o x o just o to the d going to all the old mod the sections we call that thing old bar and this is just the map you know where the bases on the e i goes to s i and then you have the various exterior powers on it all the way up to the top exterior power and so that remains exact okay but since it's exact so here we're over here now where it's a perfect complex with cohomology supported at z but since this is this is a resolution of all bar that says by the resolution theorem we're really getting just a class of all bar in k naught of modules of finite length over the local ring and then let's the day visage would say well I have something of finite length and then I take filtration so that the relative pieces are actually vector spaces over the residue field but to make it a little easier let's just assume that the residue field is contained in the local ring that's not really much of a restriction because if I want to make the computation I can always increase the base field to do that necessary or pass to a completion in any case when you do that when you apply day visage what are you getting then you're getting just the well on one hand you're getting the length of all bar as an all module that's the multiplicity times the class of the residue field and this is just equal to the dimension of all bar over the residue field as a subfield times the class of KZ okay so I'm sure you all know all about this it's just going over in some gory detail so this is the that's the formula for the local Euler class in that case now what I'm really interested in is the case of the Hermitian K theory so there again we have the member of the Tom class of V was the same casual complex together with the quadratic form and the quadratic form was I took the term in degree I so I think of this is V star V shifted by I it's in homological degree I and then gets paired with the term in degree D minus I of the dual sorry then just the wedge product goes into the determinant the top wedge will be determinant of the pullback of the dual so that's just the pullback of the inverse of the determinant of V but in degree D so induces the map of this Q which you think in just the casual complex answer itself into this determinant V shifted by D okay so that's the that's the casual complex with its quadratic form and then this local gadget here is just s star of this pair and where does this live well this is supposed to live let me just write the formula supposed to live in a not to be with supports on Z on X so what is this well these are quadratic form so let me call that GW growth in the bit of the perfect complexes whose comal co homology are supported on Z let's say over the local ring oh but the values are not in oh or in hey or whatever they're in this element this complex determinant inverse of the V shifted by D so that's where you end it's so this is the analog of K naught instead of K naught you have GW of the same kind of object and the value of the quadratic form is this inverse determinant in degree homological degree D so all right yeah that's a that's a question for you are you gonna Q&A I can't see the answer yes that's right how's that that better we'll be able to see it for a while okay so that's the that's where it's living and then we have the resolution theorem which tells you that this is the same thing as GW of modules of finite length over all with values in the inverse determinant now what I'm not telling you here is to get the GW to get the GW you have to say what the duality is and here there's a little technical point that the duality is given by X so we're not going to talk about that in any details so let me just hide that under the rug and then you have the Dave visage part which tells you that this is the same thing as GW of the residue field with coefficients in the determinant inverse of V but the fact that this X group you remember in sort of local growth and de-duality the X group corresponds in some sense to the top wedge of the ideal sheaf defining Z so you have to tensor with that or the maximal ideal of my local ring modulo it's square that turns out to be where it lives so there's a sort of purity twisting going on here so this goes to something so we have a bunch of rather complicated gadgets and this goes to something in here this is a fairly explicit gadget I mean these are just quadratic forms on kz vector spaces with values in this line this is this is a this is a particular one-dimensional kz vector space just to make things canonical and we should be able to write down a formula so there is a formula for this this is the formula just do a classical grand really building on some work from the 70s from Shaya and Schwarz so that's what I want to tell you about okay so we're trying to try and compute this thing okay so I'm not gonna show you how you'll go from here to here I'm just gonna show you what the answer is okay so I'm gonna leave this up here and so remember we had V what do we have we have V and we made a choice of a basis this is our local basis and we had our section s which in terms of this local basis we can write as si yeah okay and then we also have our local ring O which contains this maximal ideal and we'll make another choice we'll choose some generators for the maximum ideal we're talking about a smooth variety x over k really over kz and we'll choose we'll choose parameters t1 up to td okay so now of course the fact that the s that z is the zero locus of s just says that all the si's are in M so it says you can write si as a sum a i j e j so these a i j's that's essentially the partial derivatives of s with respect to the t's that's all that is all right and then great so now we have this element epsilon which is just the determinant of this matrix but I'm gonna view it in old bar remember old bar was old modulo the s's and it's not unreasonable the a i j's are not gonna be the ideal generated by the s's unless something funny happens in fact it's not because we divided by the t's all right so we've got that and now what do we do with that we choose so there's a choice here a kz linear map trace from old bar kz which is normalized by saying that this trace of this epsilon is one it'll turn out that the eventual the class of the quadratic form we get will not depend on the choice of this trace okay so once we've done that what do we do then we have a bilinear form kz the shea storage bilinear form where x y is just equal to the trace of x times y and we have the quadratic form q of x is just the trace of x square just what you get from the bilinear form okay so we made some choices we made choices of the e's and the t's and this is a quadratic form in kz so now I want to get a quadratic form not in kz but in this funny line and that will get rid of the choices so I just so the class of this the answer is the class of this local spoiler characteristic is exactly this quadratic form multiplied by well it should be in the inverse of the determinant so I take e1 wedge up to ed maybe it's ed up to e1 I think there's a sign issue here so that generates the determinant so it's inverse or if you like the wedge of the dual basis generates the inverse of the determinant and similarly if you take the images of the ti's there in m you look at them in m mod m squared and take this then this lives in a z tensor the inverse well it has values it's a map the quadratic form mapping o bar to a z tensor the inverse determinant of v tensor the inverse determinant of m mod square and you it's not hard to check that that's independent of the choices and that's the answer so that's a very nice result of Cass and Luekelgren so let me just make this even more concrete let's just take d equal to one and forget the bases just get the form so let's say s is just a unit times t to the n so then all bar has basis one t t to the n minus one call that the one up to the end and then okay we write s it's u times the use a unit this is u times t minus to the n minus one times t so this is our a it's a 11 if you like and then so the epsilon is just equal to ut to the n minus one bar in old bar in other words modulo which is all modulo t to the n all right and now if we take we can take the trace we make our choice of our trace by just saying trace of vi is equal to zero if i is not equal to n and let's see it should send this this epsilon is of course just u times the n so it should be one over you if i equals n okay and now we can just write down what our form is the bilinear form let's let's do that then we get a matrix this thing of x y is just equal to the trace of x y and the matrix in our basis is just one over you and this is so anti diagonal matrix okay now of course in quadratic formland one over you is up to a square that's up to a square is the same as you so this is isomorphic if you think about it you have you can pair the one with the one and the other one with the other one etc so you think about it a minute if n is even this is just n over two copies of the hyperbolic form just by pairing the corresponding off diagonal elements and if n is odd you have the middle element which gives you a one over you which will call a u just for a simplicity of notation and then you have oh sorry minus then you have the remaining off diagonal elements which contribute hyperbolic form and there's an even nicer way to write this this is just the one dimensional form u this is the form x goes to u x square that's the rotation times this form n epsilon so this is the number n viewed as a quadratic form and what does i mean by that i mean n epsilon is the sum of ones but it's the sum of minus ones minus one to the i i equals one the n minus one so because the hyperbolic form is a one plus a minus one you can see that this gives you exactly the same answer and now if we take a general d but assume well another case let's assume that our s is a sum u i t to the n i times e i then well i mean you can see what the computation is this quadratic form q s will just be the product the one dimensional form product of the u i times the product of the n i so epsilon that's an easy computation it follows from this and finally the the other simple case if it is if s is just sort of the non-degenerate case it's just gotten by a scalar matrix so here the let's assume that the a i j is symmetric and non-degenerate then this chaise-storch form is just equal to the determinant okay so and you can i mean in principle you if you have the the section explicitly you can find the form explicitly so it's it's nice and explicit all right so um let's see it's time for some further applications so i want to see how this works in a geometric situation and this is what we call the riemann provitz formula which is also can be viewed as a calculation of Euler characteristics using Morse theory all right so um let me just mention the theorem suppose we have field k x smooth objective over k c of some dimension d c smooth objective curve over k everything let's say irreducible and f from x to c um just some subjective morphism okay and let's suppose so this will have critical points in other words a general point of c let's assume that subjective and separable so it means that a general point of c the inverse image the fiber over c will be smooth and then there'll be some places where the fibers is not smooth let's assume that um f has isolated in other words finitely many critical points the critical points in other words are just where the differential is zero so let's say those are points x1 xn and a technical assumption we don't really need this but uh just to make the formula simpler let's assume that the field extensions are all separate so for example if k were a perfect field that would not be so then we have a formula for this quadratic Euler characteristic minus one to the dimension of x times the Euler characteristic in the quadratic sense you know this lives in gw okay there's the following so you have the sum over the critical points of what well you have these um ah i forgot an assumption we assume further assume awesome sorry we have an isomorphism row of the um utilizing sheaf we're just the one forms on c isomorphic to the square of some line sorry that's an important assumption i forgot that in okay so then when you do that this d notice that um df is a section of not just it's not a section of omega on x it's a section of omega tensor f star of omega c inverse so this is where the Euler classes would normally live but we're going to put them inside of this guy here by a row tensor l to the minus two or f star of l to the minus two so i've kind of run out of room with my explanations so remember we're thinking of df is living in this vector a section of this vector one and so if we do that we get the sum i equals one to n of the we take the trace of this quadratic form from k i extend onto k of the local Euler class for this df living in omega x tensor f star of the minus two so that's the that's the contribution for the critical points and then we have sort of the general fiber also contribute something so minus the degree over k but now this is all numerical information we take the usual first turn class of my l pull back and then times these are the usual first usual turn classes so in the chow chow ring if you like the omega x over k times the pullback of the first turn class of the omega c over k the i minus one is number times the hyperbolic one sorry for the formula giving a little messy so so why is this um so that's the so this is uh if you take the rank so the rank applying the rank gives you the classical Riemann-Hurwitz formula and if you assume that k sits inside of R and you take the associated signature this is just Morse theory this gives you the Morse theory computation for the Euler characteristic of the real points but this is sort of a quadratic lifting of both of those computations okay and let me make um perhaps one little remark from this formula um let's suppose that the dimension the dimension of x is odd well everything is hyperbolic here this part is already hyperbolic but we already know that a um Euler characteristic of a smooth projective odd dimensional variety over k is hyperbolic so that gives you a relation it says there's this funny relation on the on this term it says that the sum of these traces of these local shortness here is also hyperbolic so um I think it's I don't know what to do with it but I think it's an amusing and perhaps surprising relation and if you make you can even make this quite specific uh if you maybe a little exercise for yourself is to take the case where x is just a hyperbolic curve mapping to p1 ramified at some set of points say 2g plus 2 uh say a subs quo reduced closed subscheme of p1 of degree 2g plus 2 over the base field by taking the square root of the defining equation and then you get a funny relation on the um corresponding differential so if you take that case it says if you take p2g plus 2 in a x so degree this that degree with um which is a separable polynomial so all the roots are different in the algebraic closure then if you take um x tilde to be the zero subscheme then you have a tracemath from uh the Groten-Dieg-Wittring that this will be let's assume that this is uh their base field again is uh separable so this is a well separable polynomial so it's a separable extension of the total quotient field this might be several pieces and then it says that if you take the trace this tracemath apply to the one-dimensional quadratic form two over the derivative of p evaluated here this is hyperbolic probably easy to tell but it's just a special case of this formula which then generalizes a higher dimension okay so i want to turn to um so that's a little bit about Euler classes i want to turn to um Euler characteristics for computation okay so i'm going to remember recall this theorem um with raxit that says if we have x over k smooth objective then this Euler characteristic has an explicit form it's equal to this um sum of the hp x omega q's shifted by q minus p together with this um cup product form composed with the trace map so x dimension d this will go from h d x all right um so the thing is this is um mostly hyperbolic so we already know that if the dimension is um odd then everything is hyperbolic but you can see that because if the dimension is odd then you're always pairing an hp q with an h d minus p d minus q and those are never the same group so you're always pairing one group with its dual and that's automatically hyperbolic and so if you throw away those parts you see the only thing this is this is equal to some hyperbolic part plus the form just in the middle dimension so let's assume that d is equal to 2m then this is just equal to the cup product form on h m x omega m into k which is just saying x goes to the trace of x squared so that's a very explicit just one piece now since you have such an explicit form what you can do with this is this this enables um sort of uh twisting constructions so let me just say a word about this let's just take the case sorry mark just real quick kirsten says that you have too many m's in your formulas oh too many m's yeah yeah there we go hey yeah thanks thanks kirsten so this uh too many m's just thank you i see that so this enables twisting so let's just take d equals 2 first non trivial case and then we're just looking at a quadratic form on h1 x omega x going by this k x goes to trace of x square now if we we're working over a field k but if we go up to k bar we know that h1 x k bar is just the base extension here and we have the same form right we have the exact same form here and so the the h1 x sits inside here as a k subspace so explicitly what it says is you can take the form so for q you can take this let me just call this form q you can take q for x over k bar and then restrict it to this guy here this will land in k and this is just equal to your original form q so now usually what happens is you might have a natural basis for the thing over the algebraic closure where you can easily compute the form and then you just have to see i have to find now a k basis of this k subspace and that will tell me how to compute my original q so for example one way you do this so example if uh pg of x is equal to zero then this h1 x k bar or even over k is just equal to k bar times let's say and the q is equal to zero q is a different q so let's say h h1 of x o x equal to zero then this thing is just k times the group of x over k bar and you the this form here q x k bar is just equal to the intersection form on pic extended k bar whoops extended k bar linearly so that's often something you can compute explicitly and then you have this question of thin binding this k vector space in here and making a basis so that's the that's the twisting construction and i just want to remark is you can't do that in the golden digvid ring the golden digvid ring doesn't have descent but if you have an explicit form then you can do this twisting construction okay so i won't running out of time let me not do an example of that but instead let me turn to another explicit case explicit case of hyper surfaces so uh one simple case is a diagonal hyper surfaces so in other words generalized for my hyper surfaces so that's you have a hyper surface given by a sum of like this ti to some power e so this is sitting inside of projective space of dimension n plus one so this is a hyper surface of uh degree e dimension n and let's assume n is even all right so um in that case the remand let me just tell you what the answer is the remand herwitz formula you can apply it to the following thing you can take the map you have a rational map of x to p one which is just sending x zero to x n plus one to the last two coordinates of course that's not defined when the last two coordinates is zero that defines a hyper surface z it's uh co-dimension one sub-variety of x z which is just another diagonal hyper surface in a projective space of um smaller dimension so if you blow this up then you have a morphism here f you can apply the remand herwitz formula to this guy plus uh induction for the z plus uh a blow up formula you have a formula for the blow up let me be interested in time let me just say that or the Euler characteristic and that leads the following formula that um in this diagonal case the Euler characteristic um it's quite simple it's one dimensional form e plus some multiple depending only on n and e of the hyperbolic form this is again if e is now if e is odd the odd degree case is simpler and the even case you have an extra term times again the product of the ai plus a multiple of the hyperbolic form this is for e so for example if if e is two if you have a quadric then uh this so in other words then you're just taking a quadratic form every quadratic form can be uh t can be diagonalized so this is some quadratic form q and this product of the ai is just the discriminant of q and so you get the um right the Euler characteristic of the quadric x is just equal to two plus minus two times the discriminant and then plus i guess n over two times the i so quite explicit and you get this well maybe a little disappointing you just get this rather elementary invariant of the quadric in the Euler characteristic all right so um that's nice but not every hyper surface is diagonal so what can you do for a general hyper surface well there's there's a way to do that also let's uh make the general hyper surface so you let's take x inside of t n plus one over k uh smooth and defined by some degree e polynomial does the question mark uh what is a and e it's a number so a and e yeah thanks a and e is a number um i mean you can write down what it is it's um the number that it is let me put it this way you can find it's not hard to find a formula for the Euler the topological Euler characteristic of a hyper surface of degree e in p n plus one let's say you know that then the rank of this quadratic Euler characteristic is that number on the other hand the rank of the h is two so you have that equation to solve in other words the this a and e in the odd case is the topological Euler characteristic of x minus one then divided by two and in the even case it's the topological Euler characteristic minus two divided by two yeah thanks for the question okay so um back to the general case uh let's say we take our um hyper surface defined by um degree e polynomial and we assume it's smooth and let's assume that e is prime to the characteristic of k then we have the Jacobian ring so this is uh Jacobian ring um this is defined by taking modding out by the partial derivatives of f and under this assumption that e is prime to the characteristic the Euler formula tells you that the original f is in this idea and what that means is of course f being smooth means that f together with all of its partial derivative have no zeros in the projective space which means if you look in the affine space this thing here has zero locus um this ideal has zero locus just zero in the affine space which means that the Jacobian ring is a finite dimensional a algebra and more more over you have n plus two variables and you have n plus two equations it's also a complete intersection the idea complete intersection okay so this uh the general theory of such rings tells you that it has a one dimensional socket in other words you look at palm of k which is in other words you take k modulo the variables mapping into this ring there's a one dimensional image of that and so it has a socket one dimensional and it's generated by this shear storage element so in other words you write the partial of fi with respect to that's partial of f with respect to ti you write it as a sum the aij times tj when you let epsilon be the determinant of this aij the bar so this is in jf then the socket is equal to k times epsilon so we have a quadratic form we have the shear storage quadratic form from jf or let's say and going with the quadratic form or the pairing maybe i just went down the okay so um let's see i need the trace form but i'll say what that is in a second you have to choose a trace form i choose the trace form since jf is graded see these are all all these partial derivatives are all degree e minus one so jf is a graded ring so i define it this way i take x yj and if these are homogeneous then this is equal to zero ah so what's the i see what's the degree of this thing this is in degree to count let's see the derivatives have degree e minus one that means these coefficients have degree e minus two and there are n plus two of them so this is in degree e minus two times n plus two and so for homogeneous elements x and y the pair will be zero if degree of x plus degree of y is not equal to this magic number and it's equal to lambda if x times y is equal to lambda that's absolute in other words the trace form is just the one which says epsilon to zero and all the other graded components in jf to zero okay so what does this have to do with anything well on the other hand we have this really beautiful theorem of clemens and griffiths arising from griffith's study of rational integrals back in the seven days which says that there's an isomorphism from hq x omega p isomorphic to the certain graded piece in this jacobian ring and it's in this degree q plus one times e minus n minus two so this is a generalization of the classical theorem on residues which tells you that the you look at the case of sorry this is not quite right I have to put a little prim here and this is for p plus q equals the dimension n so I'll tell you what the prim means in a minute it usually doesn't mean anything but yes so if q is zero and p is n this is just this is just the canonical this is just the global sections of the canonical sheaf and we know what the canonical sheaf is by the adjunction formula so this is just the generalization of that to all the cosmology so and what's the primitive guy hq x omega p primitive is equal to um zero if p plus q is not equal to m it's equal to the h the whole thing if p plus q equals n and p is not equal to q in other words they're both different from this half and what is it in the interesting case well if p equals q equals half the dimension so this is m which is n over two then we have the class of a linear section so this x is in a as a hyper surface in a projective space and we have the class of a linear section of dimension let's see equal of co-dimension n over two this lives in h m x omega m and we have our quadratic form the cup product followed by the trace map so this is equal to the perpendicular with respect to l okay so you're losing exactly one class it turns out that the p plus q is not equal to n everything in the co-homology of x in degrees different from n comes from the projective space so it's not so interesting all right so that's the clemen griffith's theorem and now we have two things we have let's see we have of course and this is perpendicular with respect to the intersection form which defines our quadratic Euler characteristic then this j f is telling us should be telling us something about the remaining part of the quadratic Euler characteristic right we know what the quadratic Euler characteristic is sorry we know what this q on l is it's just equal to the degree of l squared which is equal to my e so we know what that part is and this is telling us what the the remaining part where p plus q is not equal to n this just contributes a hyperbolic part in the Euler in the quadratic Euler characteristic so we might expect that this quadratic form here this j is storage quadratic form on these this part of the jacobian ring tells you exactly the Euler characteristic and that's almost right so what the theorem is is essentially due to clemen's griffith's although not quite but more or less is that the j is source form on this direct sum over q of this jf q plus one where is it q plus one times e minus n minus two is equal to well let's see you have to multiply by minus e and then of course we have to add in this e due to the l and then plus the hyperbolic form and i think it comes out to be um let's see what is this n times the hyperbolic in over two yes once in over two times the hyperbolic form is equal to this okay so you have a really very explicit way of calculating these things and you can even use this to check the previous answer on the diagonal hyper surfaces and the answer is yes they agree okay so i think um i'll stop here i wanted to uh say a few more examples and also talk about the theory of characteristic classes for um for bundles in uh bitcoolmology but i absolutely don't have time for that so thanks very much okay yeah thanks mark bill thank you for a wonderful talk i particularly enjoyed all the examples um and i want to ask if you think there are other types of examples one should try to work out oh yeah there are lots of examples so i mean one example just yeah i mean it's hard to tell exactly what kind of what kind of examples you might be interested in but just to test your computational powers um that there's this i think it's called this uh quine cubic in the surface case we look at this equation nice cubic surface and then uh what's the Euler you can use this they can i mean it's something you can do sitting in front of the tv you have nothing better to do than the Euler if this is the defining equation for x then calculate you get a 25 we know the rank has to be um what does it have to be has to be nine because that's the Euler characteristic of a cubic surface and this five is a little surprising but it's not so surprising because it means that this thing is actually singular mod five but it's not singular mod three so even in the characteristic so the theorem doesn't work in characteristic three but since you can specialize this it says that this form really does calculate the Euler characteristic also in characteristic so can write down your favorite examples can we compute the Euler characteristic of k three surfaces well i mean you could do it for quartic surfaces in p three certainly by this method you could um write down any equation and it's probably not so hard to find the answer for any specific equation it just um i would think if you you could might even be able to get an answer or say the general cubic equation but the formula would probably look rather complicated and mysterious too many too many coefficients but um yeah i what about relating it to a degree six curve over which the k three is branched off um yeah because then it would require a modification of this method so this is really based on the this Clemens Griffiths theorem telling you how to compute the um hodge cohomology in terms of the jacobian ring and if you take a branched cover well it's in a it's in a um weighted projective space naturally so you have to um modify that but i'm pretty sure that that should work there should be a way um to make that work it's not really special for projective space you need the basic theorems you need some bot vanishing theorem the difference in weighted projective space is um you have to be careful about the singularities in the weighted projective space so that's some technical problem but if you stay away from the singularities i'm sure that that would work um uh following up on that you for your riman herwitz you have this multiplicativity of Euler characteristics that um uh you know it's great it doesn't hold generally but in the case of a branching over a curve you show that um the whole the Euler characteristic of the whole total space is the base Euler characteristic of the base times the Euler characteristic of the general fiber minus these correction terms yeah and in general we don't have this multiplicativity that you could ask um you know maybe in the case of uh k3 branched over a degree six some nice multiplicativity for the p2 and then uh uh i see uh huh um i mean if you take if you well if you take it over if you take a cover branched over a curve then you can try and do it by the riman herwitz formula also just by taking a pencil in p2 and um counting you you get um if you take a general pencil then you'll i mean that might be a way to do it you get a general pencil and you'll have the um the singular will be ordinary double points so it should be fairly explicit to calculate that i mean it's a finite map it's not a pencil of curves um well i was thinking about the k3 surface the case yes finite map to p2 and but it's branched on a degree six curve right it's i guess it's the two to one case is that which k3 you're talking about yes yes yeah so then if you take um you take a general point in p2 and project to p1 from that point then that will if you blow up your k3 at the two points lying over the general point then you have a morphism of the k3 to p1 and the singular fibers of that will be when your line is tangent to the degree six curve great so if you take a general point you can arrange that so it's just tangent in an ordinary it's you know an ordinary tangent and that will just give you an ordinary double point a quadratic singularity on the k on the fiber over the k3 and um you know it'll just be a single point so there'll all be rational points so you don't have any trace to compute so it should be fairly explicit you need a you you'd get something in terms of this cover i mean i guess you can figure out how many you know how many double points are there how many how many tangencies are there for degree six curve you have that extension of the base field on the degree from the degree six curve the the sink the ramification points for that projection to p1 that's some fairly large extension and presumably the the trace the trace form for that extension then suitably modified by the value of your local local indices at the double points upstairs on the surface will give you the answer but how explicit that is i don't know i i think i see what you're saying that sounds really fun i i have a um more basic question too it had if could you go up to where you had the um the the form on the jacobian ring this one um no farther up before the yeah right so see how you have degree um uh x plus degree y is not equal to the the degree of epsilon in there is it a fact that um the uh the um the that in the jacobian ring all the terms of this degree are a multiple of epsilon so are you saying that the way jacobians rings work out it's one-dimensional that's i mean that it's also the degree not this is thought the sockle is in this degree but the sockle yeah i didn't say this this is also equal to the jacobian ring yeah thanks of this degree we should have seen that great thanks thanks thanks for pointing that out yeah that's exactly right the sockle is equal to that graded component that degree component in the in the jacobian ring yeah thanks good point okay any other questions i don't think so then i want to thank mark again thanks for a wonderful talk