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K algebra homomorphisms can be used to understand the geometry of the vanishing set of some KL divergence, which relates to learning behavior. Jet schemes are a local construction in algebraic geometry which provide a geometric interpretation of the data of a point, and PSI is a K algebra homomorphism which maps a function to its value at a point P. Jet schemes are used to represent the pre-image of a point in a vector bundle, and the origin is special in that it is the only point where both the gradient is zero and F is 0. This video provides an explanation of how to use K algebra homomorphisms to understand the geometry of the vanishing set of some KL divergence.

K algebra homomorphisms can be used to understand the geometry of the vanishing set of some KL divergence, which relates to learning behavior. This geometry is seen in the arc scheme, which is the limit of jet schemes. A K algebra homomorphism from Simvastar to Kt mod t squared can be determined by the sequence of scalars p and q, which is a tangent vector. Tangent vectors are derivations of functions which send a function f to a pair consisting of its value at a point and an expression. This data is a tangent vector, and germs are equivalence classes of functions which agree on a neighborhood of a point. Derivations of the ring of germs are one definition of a tangent vector.

Jet schemes are a local construction in algebraic geometry which provide a geometric interpretation of the data of a point. A zero jet is a K algebra homomorphism from the quotient of SIM V Star by the ideal generated by f, determined by where the variables xi are sent to. Two jets are K algebra homomorphisms from something into KT cubed and KT squared respectively, which can be composed with canonical K algebra morphisms to send two jets to one jets and one jets to zero jets. A one jet is a tangent vector that knows where it is and the map extracts the evaluation at the point to get the point itself.

PSI is a K algebra homomorphism which maps a function to its value at a point P. It takes in a vector of variables xi and subtracts a vector of constants pi to get qi t plus RIT squared. Taking derivatives of this function with respect to the variables results in terms of order 0, 1, 2, 3 and higher. The zeroth order term is F evaluated at P, the first order term is F's gradient at P dot q t, and the second order term is a product of Qi plus RI t to the power of the indices i1 through i n. To get a T squared, either zero powers of T from Source One and two from Source Two, or one from each, or two from Source One and zero from Source Two must be used. When I is all zeros, only the FP term contributes.

Given a polynomial, a two jet of a variety can be calculated by taking the triple pqr, where p is on the curve, q is the gradient of the polynomial at p, and r is a dot product of the gradient with itself. This can be demonstrated using the example of a nodal curve, y2 - x2 + x + 1, with two jet equations p2 - p1 + 1 = 0 and q1(-3p1^2 - 2p1 + q2)p2 = 0, cut out of R4. This lemma provides a way to interpret the Hessian Matrix as a pairing of bilinear forms on the tangent space at P.

Jet schemes are used to represent the pre-image of a point in a vector bundle. At the origin, the dimension of the fiber is two-dimensional, while away from the origin it is one-dimensional. J2 is the second jet scheme of a nodal curve, which is a union of lines. A point on this line is represented by P and Q, and the inverse of this point is all the PQRs such that P and Q are given, and R satisfies a quadratic equation. The solution to this equation is a one-dimensional space, and the fiber of J2 has a dimension one less than the space of R's. The origin is special in that it is the only point where both the gradient is zero and F is 0.

Jet schemes are a way of understanding the geometry of the vanishing set of some KL divergence, which is an analytic function. This geometry relates to learning behavior, and can be seen in the arc scheme, which is the limit of the jet schemes. K algebra homomorphisms can be used to understand this structure, which relates to generalization error. The speaker is unpacking this in the case of one jets and two jets, and is aiming to explain how the structure relates to learning behavior.

A K algebra homomorphism from Simvastar to Kt mod t squared can be determined by the sequence of scalars p and q, which is determined by the K algebra homomorphism itself. This data can be thought of as a tangent vector, as it is given by the formula PSI PQ, which sends xi to pi plus qi t for all i. This bijection between K algebra homomorphisms and tangent vectors is proven, demonstrating that it is reasonable to think of the data as a tangent vector.

Tangent vectors are derivations of functions, sending a function f to a pair consisting of its value at a point and this expression. This data is a tangent vector, and germs are equivalence classes of functions which agree on a neighborhood of a point. Derivations of the ring of germs are one definition of a tangent vector, and for smooth functions, this is the usual definition in differential geometry.

A zero jet to a variety Z is a K algebra homomorphism from the quotient of SIM V Star by the ideal generated by f. This homomorphism is determined by where the variables xi are sent to, which are denoted as Pi. F must be sent to 0 for the homomorphism to be well-defined. The zero jet scheme of Z is the space of all images resulting from this homomorphism.

Jet schemes are a local construction in algebraic geometry which provide a geometric interpretation of the data of a point. For m=0, the jet scheme is isomorphic to the point itself, and for m=1, it is the tangent bundle to the point, consisting of a point in the variety and a tangent vector. For m>1, the jet schemes are less familiar objects, but still provide a local interpretation of the data of the point.

Two jets are K algebra homomorphisms from something into KT cubed and KT squared respectively. Composing these with the canonical K algebra morphisms sends two jets to one jets and one jets to zero jets. A one jet is a tangent vector that knows where it is, so the map extracts the evaluation at the point to get the point itself.

PSI is a K algebra homomorphism that maps a function to its value at a point P. It is a bijection of two Jets to affine space with triples (P, Q, R). Applying PSI to an F in Sim V Star will yield a scalar multiplied by the partial derivatives of F at P. The result is an equation that holds in Sim V Star.

PSI is a function that takes in a vector of variables xi and subtracts a vector of constants pi to get qi t plus RIT squared. Taking derivatives of this function with respect to the variables results in terms of order 0, 1, 2, 3 and higher. The zeroth order term is F evaluated at P, the first order term is F's gradient at P dot q t, and the second order term is a product of Qi plus RI t to the power of the indices i1 through i n. Higher order terms are not needed.

In the formula, there are two sources of terms with T, Source One and Source Two. To get a T squared, either zero powers of T from Source One and two from Source Two, or one from each, or two from Source One and zero from Source Two must be used. Tuples of non-negative integers that add up to less than or equal to two are necessary to get a contribution. When I is all zeros, only the FP term contributes. Tuples where one entry is non-zero and is a one add up to one and contribute the first line of the formula.

This lemma proves that given a symmetric matrix H, a bilinear form can be defined by summing the product of two vectors, Q and H, according to H. This bilinear form is then used to calculate the t-squared terms of a product of t with a dot product of the gradient of F at P and Q. The second term in the bracket is the pairing of Q with itself according to the pairing where H is the Hessian of F at P. This lemma provides a way to interpret the Hessian Matrix as a pairing of bilinear forms on the tangent space at P.

This summary explains how to calculate the space of two jets of a variety given by the vanishing of a single polynomial. A two jet is a triple pqr, where p is on the curve, q is the gradient of the polynomial at p, and r is a dot product of the gradient with itself. This is demonstrated by an example of a nodal curve, where the equation of the curve is y2 - x2 + x + 1, and the two jet equations are p2 - p1 + 1 = 0 and q1(-3p1^2 - 2p1 + q2)p2 = 0. This is cut out of R4, with coordinates p1, p2, q1 and q2.

The geometry of the pair of equations can be understood by mapping J1 to J0. At the origin, the gradient is zero and F is 0. For all other points on the curve, the pre-image is a line of points PQ in the jet scheme such that Q dot some non-zero vector is equal to zero. The origin is special in that it is the only point where both the gradient is zero and F is 0.

Jet schemes are used to represent the pre-image of a point in a vector bundle. At the origin, the dimension of the fiber is two-dimensional, while away from the origin it is one-dimensional. The jet scheme of a regular variety is trivial, meaning the pre-image of the origin is the same as the original vector bundle. J2 is the second jet scheme, which sends p, q and r to P, Q and R.

J2 is the second jet scheme of a nodal curve, which is a union of lines. A point on this line is represented by P and Q, and the inverse of this point is all the PQRs such that P and Q are given, and R satisfies a quadratic equation. The solution to this equation is a one-dimensional space, and the fiber of J2 has a dimension one less than the space of R's.

A one-dimensional affine space can be given a vector space structure, creating a line that lies over each point. Taking a point at zero, Q, the pre-image of that point is all the things of the form zero Q r, with the equation being the Hessian multiplied by Q. This pairing will be equal to zero when Q2 minus q1 is equal to zero, creating a three-dimensional variety of J2Z over both the lines where q1 is equal to Q2 and Q2 is equal to minus q1.

okay so we're talking about jet schemes so I introduced uh I still haven't really defined jet schemes but I started with somewhat abstract discussion which was meant to put jet schemes in the context of SLT at least at a higher level and the idea is that we're trying to understand the geometry of the vanishing set of some KL Divergence that's an analytic function maybe we can find an equivalent polynomial and the geometry of that Vanishing set of that polynomial uh relates to learning behavior in ways we're trying to understand one of those ways is to try and understand what information is in the resolution and then there's some conjectures of Nash and results and sort of a whole story there but at some level some of the information we care about in the resolution which the theorems in watanabe's book and elsewhere tell us that that structure relates to say generalization error some of that structure can be seen also in what's called The Arc scheme so the arc scheme is the limit of the jet schemes xn for all n okay so that's a very high level reason to care about jet schemes what I want to try and do over the remaining lectures in this series is to try to steadily pin that down into something that is more like learning behavior in some sense you might recognize for example the dynamical system of a network being trained or or rather perhaps more realistically sampling from the Bayesian posterior as a process so yeah that's the AIM yeah this was kind of some sort of abstract categorical way of of realizing what the Jets to a scheme are it's not important really of this one over your head but the the point here was to get to in some abstract way justifying that we care about K algebra homomorphisms like this this is an engine so I'm still going to continue with this as a sort of abstract definition I then unpack that a little bit in the case of One jets and two Jets or rather just one Jets um maybe not here I think the next set of boards explain this right uh so I think I finished on this calculation yes so what I was doing in this pair of boards was to explain how a k algebra look at the red board open so simvastar V is our space simvastar is polynomial functions on the space that's a k algebra K is our field it's always going to be an algebraically closed field for Simplicity uh well I guess since we're talking about um real functions I I shouldn't say that perhaps uh yeah I take that back so K is any field we're looking at K algebra home morphisms from simvastat a a is this

quotient ring ktmon t squared which uh just means linear combinations of 1 and T with the relation that t squared is zero okay so that's the algebra and then I was Computing for you what classifying K algebra homomorphisms from simvastar to a and the way I did that was to look at the Taylor series expansion of f so f is a function a polynomial function on V viewed as an element of sim V Star I can take the Taylor series expansion and if I apply to it some K algebra will always in PSI and I've written here PSI sub PQ now that's because given any PSI I can evaluate PSI PQ so given how should I say this given any PSI from simvastar to a by the Universal Property of Civ V Star which I also discussed last time such a k algebra homomorphism is determined by what it does to the generators and those are the X I's those are the coordinate functions on V okay so evaluate PSI on x i that's an element of a an element of a is a linear combination of one and T called the coefficients pi and Q I so those are scalars okay so any PSI determines these the sequence these two sequences of scalars Pi and q i and I'll collect them together and call them p and Q okay so that data p and Q is determined by PSI and determines PSI by the Universal Property so I'll just write PSI PQ for the PSI that sends x i to pi plus q i t for all I okay so if I fix up PSI and I evaluated on an F what I get is what we computed here now I think we must have run out of time because that's kind of an awkward place to stop so maybe I'll finish that thought and and then continue on so the point of doing that calculation is to convince you that it's reasonable to think of size as tangent vectors so at the upshot is that or first of all every K algebra homomorphism simply star to KT mod t squared is of the form for some p and Q and V unique p and Q okay so that's what K algebra morphisms between those two rings look like they're in by ejection with p's and q's pairs so there's a bijection which I wrote earlier and you could say the map is this way in the sense that given PQ this formula defines a k algebra homomorphism and that map is a bijection okay so I've proven that I've proven there's this bijection uh okay now I need to convince you that it's reasonable to think about this data as a tangent vector and the formula hopefully makes that clear right so PSI what is a tangent vector well there's many ways of defining tangent vectors but usually you would say a tangent Vector is a a function assigning to

a germ of a function at a point a scalar maybe that's not the way you think about tangent vectors maybe you think about tangent vectors as equivalence classes of Curves through a point that's fine um but this is an equivalent point of view so this is the point of view which says that a tangent Vector is a derivation of functions and we see here the latter point of view so PSI sends a function f to well its value a pair consisting of its value at a point that's not a derivation but this part here is a derivation right sending F to this expression is the tangent Vector Q and so under the normal prescription if you were to assign to q a tangent Vector at P it would be exactly the function which assigns to F this second term I underline okay so basically by definition of a tangent Vector this data this PSI is a tangent Vector together with this extra information the evaluation at P which says where is the tangent vector okay so um this is the argument that one Jets to V equal tangent vectors to V V is just our fine space and the left hand side was defined to be these K algebraomorphisms right so I've now given a geometric interpretation of these K algebra home morphisms any questions so this is kind of wrapping up last lecture I guess sorry uh germ yeah so germ just means an equivalence class of functions where the equivalence relation is that they agree on a neighborhood of a point so if I have a point p U V so a germ you would take a pair U and a function f which is a defined on U and it would be equivalent to the pair v g if and only if there exists an open set w contained in U intersect V and containing p with f restricted to W equals g restricted w so the equivalence classes of that relation are called germs and if you take the set of all germs that's a ring and derivations of that ring so assignments of scalars to elements of that ring of germs which satisfy the leibniz rule with respect to multiplication that's one definition of what a tangent Vector is if you take the functions f and g if you take them to be smooth functions so germs of smooth functions then you get the usual definition of a tangent Vector in differential geometry cool thanks yeah okay [Music] um well that's fine uh we've talked about one Jets but only to affine space which is a bit boring so now I want to talk uh so yeah I'm going to say something about um jets to more interesting varieties ultimately we're interested in Vanishing sets of functions like K and H back on the first board right

so Jets to algebraic varieties okay I'm just going to stick with one equation but everything I'm going to say generalizes to multiple equations so we're going to suppose we have a function on our space V I'm just going to identify this with the polynomial ring in the excise I recall that the notation I introduced earlier is that V has a basis E1 through e n and x i is the Joule basis and Z is going to be the vanishing locus of f so think about there's V and here's Z there's maybe a p and maybe there's the gradient okay so what's an inject or an mjit 2z so an mjit to Z as compared to V right I've already defined what an m jet to V is it's a k algebraomorphism from SIM V Star to KT mod t to the M plus 1. so an m j to Z a sub variety is by definition the K algebra morphism from The Ring of functions a ring of polynomial functions on Z which is the quotient of sim V Star by the ideal generated by f now I'm just presenting these as definitions uh the motivation for this definition comes from unpacking the content in this uh I'm handing you this definition and then we'll unpack it and see what its geometric content is but uh yeah there's this in some sense no no motivation Beyond just observing that it has some information that is geometrically geometrically relevant well give more of a motivation but we won't do that here okay and the amp jet scheme of Z is the space of all the images again this is not the place to introduce schemes I just want to have that idea in your head that this is what an mjet is and if you took the set of all mjets that set has more structure it's a topological space it has a notion of regular functions on it it defines what's called a scheme and we'll compute an example so I'll compute the two jet scheme of of a singular curve hopefully at the end of today's talk and you know it will just be defined by some equations and you won't need to know what a scheme is okay so um let's give some examples let me take m equals zero so if m equals zero and a zero jet to Z is just a k algebra homomorphism from this quotient to KT modular t to the 1 which is just k so what is a k algebra morphism from this quotient ring to K well that's determined by where I send the variables right so let p i be defined to be the image of the variable x i in that quotient ring well the PSI to be well defined it has to send um F to zero what does PSI do to F if it sends all the variables to Pi well it just sends if you think about it it's just looking inside F and replacing each x i by p i

which what which is what it means to evaluate F at p okay so we see here that zero Jets 2z which are these size are exactly the data of so if p means the point with these coordinates zero Jets to Z that is size if you extract the data of the pi that is the image of the variables it's exactly the data of a point where f is zero but that's the same as a point of Z okay so zero Jets to Z are the same as points of Z and the nth jet scheme are denote by ZM or I take it back I'm not sure what notation I was using earlier maybe it was ZM but I'm going to write j m z the nth jet scheme have said okay so this means that j0 of Z is just Z or rather it's canonically isomorphic to Z what about if m is one well that's what we just discussed so then PSI is okay algebra on morphism from this quotient KT modulo t squared if I take pi plus q i t and set it equal to psi of x i as we just discussed then the equation zero equals the image under PSI of f is exactly looking at the previous board the first equation that means that FP plus the gradient of f dot Q T is zero but this is in KT modulo t squared in which 1 and T are linearly independent so that says two equations it says zero is equal to f p and it says that 0 is equal to a gradient of F at P dot Q well this just says that P is in z and this says that Q is in the tangent space to Z at p because the things that are orthogonal to the gradient are exactly the things that are tangents all right so these two equations say that uh p q are in the tangent space the tangent bundle to Z for the tangent bundle to Z is a space where the points appears consisting of a point in Z and a tangent Vector at that point Q okay so what this says is that J1 which is the space of all One jets is the tangent bundled to Z One jets to Z are tangent vectors to Z all right so this is meant to convince you that at least for m equals zero and M equals one the notion of an m jet to an algebraic sub-variety has a clear geometric content it just gives you points respectively and tangent vectors once we get to two Jets and Beyond there they're less familiar objects any questions is it is it bad to say that if we are only concerned about local property um near a point we can always restrict cultivation to similar to what you have done here which is like I think we don't need to care about the global scheme structure we just need to care about applying patches yeah that's right yeah as in most of algebraic geometry things can be done locally so the jet scheme is a local construction

so if you want to do the amp jet scheme of Zed is locally pieced together from the jet schemes of open subsets of Zed which you can assume to be air fine so I'm presenting here the affine story but in some sense that is the whole story is that that's how it fits together sorry sorry yeah what is missing from uh this local story so what kind of information it could be well it's a bit similar to the blood I know it's like topological right well yeah and sometimes nothing is missing the the jet scheme is those local pieces together with how they glue together but the information of how they glue together of course uh is real information so um it's not that you know everything by just knowing the the F line opens right okay all right so I want to talk about two jits let me see and yeah maybe it's a rather I want to talk about two Jets to algebraic sub varieties which uh first of all I just want to recapitulate that calculation with the Taylor series but keeping more terms so that we understand what two jets are and I think that's best done on the next set of boards so we'll just move over oh whoops okay I forgot that we we had Matt's talk here last time so uh yeah we should proceed further okay um so two jits so note the canonical subjective K algebra morphisms from KT modular T cubed KT modular t squared okay and these just send T to T to T yeah I think I'll call these Pi 2 3 and pi one two I I guess the swimming is a bit distracting that's the fault of the location that's sorry here thanks okay so the point what do you mean sorry what do you mean by sending T to T and K in the last map yeah yeah right good question though yeah there's no tea there yeah thanks um okay so these K algebra morphisms mean that we have natural maps from three Jets or rather two Jets to One jets to zero Jets right because well remember that a a two jet is a k algebra homomorphism from something into this guy so then I can compose with this and get a gay algebra home morphism into this guy which is a one jet so two Jets determine One jets one Jet's determine zero Jets uh and we've already seen this this in action so what I'm writing think about how you would extract a point from a two jet sorry from a one jet and so I just said that one jets are tangent vectors but sort of based tangent vectors right a sort of a tangent Vector that knows where it is so if you take where it is that's a map from tangent vectors to points and that's exactly what this map here does so so we take PSI PQ it just extracts well the evaluation at

p right that's the map that sends a function to its value at p and that's the function the K algebra homomorphism that corresponds to the point P maybe that's mixing metaphors a little bit so it's it's either you can think about it as the K algebra homomorphism but since PSI PQ to evaluation at P or you can think of it on the other side of the bijection as PQ goes to to p okay so M Jets determine n jets for every n less than or equal to m well so what well that what that means is that in understanding two Jets I just have to understand the extra information right so a two jet is in particular the information of a one Jet and some extra stuff so I don't need to interpret sort of two-thirds of the data okay um so let's see lit PSI I'll start with a two jet just to F on space and then I'll introduce the equation so let's just suppose we're given a k algebra more morphism PSI what's the data in there well again we just evaluated on generators but now we get three scalars per generator right so we're going to set po plus q i t plus RIT squared to be the image of the if generator right one TT squared or a k basis of KT mod T cubed all right so that's the data that determines PSI um and that's basically all there is to say about it right by the Universal Property to tell you a size exactly to tell you these three scalars for every eye so that's it that's a bijection of size K algebra amorphisms or two Jets to affine space with triples all right so velvet P be that point and Q similarly and r as well that's a bijection by the Universal Property of the symmetric algebra okay but we still want to know what this PSI corresponding to pqr actually does right so if you give me up PSI and it has p q and R as its data what does it do to an F so that's the calculation I want to Now do so given PSI as above and an F in Sim V Star well again let me just take the Taylor series expansion of f but now the calculation will be a little more interesting because before I applied PSI and then threw away terms of order two and higher right because t squared was 0 in the codomain of PSI but now PSI is a two jet so in the codomain of PSI t squared is non-zero so I will keep more terms times all this stuff okay so apply so that's an equation that holds in simply start now apply PSI to it which is a k algebra homomorphism so that's a scalar that's a scalar remember right I evaluated that partial derivative of F at p these are actual polynomials so I'll get PSI of that but that's PSI of this times

sub PSI of Etc and the powers can go outside the PSI as well so I'll skip all that intermediate stuff and just get straight to psi if is I'm just going to write I underline for these indices here and I'll write High underline factorial for that maybe it won't be that lazy okay so I get PSI of X1 minus P1 to the i1 dot dot dot PSI of x n minus p n to the i n okay so what is PSI of x i minus pi well that's PSI of x i minus p i PSI of x i is pi plus q i t plus RIT squared right so that is q i t plus RIT squared which is t q i plus RIT sub PSI of f um over I 1 over I underline factorial uh so each of these is going to get replaced by by that for a suitable eye so I'll get t to the power of all those I's and that's i1 Plus I2 through i n so I'll get t to the modulus of I meaning i1 plus dot dot i n and then I'll get q i it'll be the product from I equals 1 to n Qi plus RI t J rather to the i j now remember all of this is modulo T cubed so any terms of order three and above I don't need to worry about okay so let's think about how we get quadratic linear quadratic and cubic terms out of this expression let's start with order zero terms well yeah can you see what the order zero term maybe I've made a mistake here ah yeah um I guess we don't actually need to do the calculation for the zeroth order term and first order term although we maybe I'll sketch how you can see them um just to point out why we only need to compute the order two term that was kind of the point of drawing this diagram at the top of this first board here right so if I take PSI which lands in KT modular T Square and I'm post-compose with the map to KT modular t squared so that takes Alpha plus beta t plus gamma t squared and just sends T to T but now t squared is zero so that's Alpha plus beta t all right so I know what the first two terms here are because so they must be F of P and the gradient of F at P dot q t plus something right I just don't know what the something is but I know that the first two terms must be this hello is that you've been I can't hear you no I can hear it today okay somebody asking a question I couldn't hear you all right uh Okay so I hope that's clear so I I mean we can we can see it from the formula it's just no no reason to go into it so if if I want to get the zeroth order term then I need I to be I underline to be the zero Vector which means I'm not taking any derivatives and then here I also get zero Powers so I just get F evaluated at P which is the zeroth derivative of f evaluated at P if

I take the first derivative then uh you know you can see you get to choose exactly one of these indices to have the value one and the other zero and so you get the first derivative and the dot product comes from from this part here but I'll focus on the second order term since that's the new information Okay so uh let's say t squared stuff plus lower order terms can you turn it down please a little Russell and if you're finished can you go back in the other room you finished your food okay so uh let's see which terms will actually contribute and I'm just trying to understand my notes Here so there's two places we can get T's from right we can get them from T to the power the modulus of I or we can get them from uh those brackets so maybe the thing to do is to worry about when we get yeah okay about it that way so here's our two sources of teas in this formula right let me call them Source One and Source two now if we're going to get a t squared then either we get uh zero powers of T from Source One and Two from Source two or we get one from each or we get two from Source One and zero from Source two yeah there are no other possibilities so let's think about what it would look like to get uh one from let's think about which way I want to write the terms okay uh why is this obvious um so we can throw away cubic terms right it's a working modular T cubed so we can throw away any terms that yeah so we should think about the index I I guess right so this index I here we only care about indices that are um well we get contributions if that's equal to 2 right because we can take the cues here and multiply them with T's here but only sums that have indices with two or less contribute so what tuples of non-negative integers can you make that add up to less than or equal to two well you can have zero all zeros where you can have a one somewhere that will add up to one and that's the only way to have the mat up to one uh but uh yeah I just don't think that this factoring out the T has really helped is it um okay so the this term here came from this was having a single i1 non-zero right okay so the when I is all zeros it just contributes the FP term and there's no there's no other bits contributing the tuples I underline where one entry is non-zero and is a one that is the sum the tuples where the modulus is one those are the things that contribute something like this first line here right right because T times q j plus rjt is exactly PSI of x j minus P J so these these here are exactly the cases where

is equal to one uh the next line these two together are the cases where I is equal to two right it's just whether or not I have I equal to J or I different to J and then you can pay attention to the factorial and see that the coefficients are correct so I'm not double counting or whatever okay so these are the terms I need to pay attention to and then I get to throw away terms that are cubic in t so for example if I expand out that product of t squared with t and t it's going to be zero or if I take that product it's going to be zero Etc so I'll just form that throwing away and I'm left with t squared Okay so let me take the product I get from multiplying this t with this Factor here the other Factor where I multiply the T with the QJ that's the dot product term on the left hand board at the bottom the gradient of F at p with q right so I've already accounted for that I'm just getting the t squared terms so what I get here is the sum J equals 1 to n of r j at p that's that's the term we just talked about what's going on with my it's all okay so the next term at the moment to fix my pen the next term comes from the next line how do I get a t squared well I have to choose Qi and Q J when I expand the brackets right so that's some I less than J q i q j and then in the third line uh I have to choose qi plus lower order terms in t so to conclude this is t squared r dot gradient of if at p Plus and this is an interesting sum and the whole point of doing this calculation is to discover this expression here sum over i j q i q j times the second derivatives of f all right any questions about this calculation now I'm going to interpret it but this was the hot pot okay so definition given a symmetric Matrix h I'm going to define a bilinear form by saying take two vectors pair it according to H by just doing this which is of course sum over i j q i h i j q j Roblox just crashed on me okay so this is a standard way of producing a bilinear form from a symmetric Matrix it doesn't have to be a symmetric Matrix to write down this formula of course but if we want a if we want a symmetric form it needs to be a symmetric Matrix so the second term in that bracket is the pairing of Q with itself according to this pairing where H is the Hessian it's a definition uh The Matrix of second order derivatives of F at p is the Hessian of F at p so we've now proven so you can interpret that Matrix as a pairing a bilinear form on the tangent space at P right Q are tangent vectors so the Lemma we've proven by the

calculation on the previous board is that there's a bijection between triples pqr and two Jets reason to perform this calculation is that now we can say what the geometric content of a two jet is because we can evaluate it at F and see what it means for it to vanish and that tells us what two jets are the sub variety are so what we've done so far is we've proven there's a bijection between triples and two Jets D cubed where PSI pqr of an F is as I said the beginning we just have the two the one jet part and then the new information is this part okay it's not meant to be clear what this term in Brackets means some are it's a DOT product of the gradient and then there's this term which is the pairing of Q with itself with respect to the Hess in at p okay so that's a two jet this formula tells you what two Jets are up to understanding what determine the brackets means so next I want to Define just write down for you the equations that would determine the space of two Jets of a variety given by The Vanishing of a single polynomial and then I want to compute what all this means for um a nodal curve but are there any questions about the last few boards has anyone seen this pairing before okay um all right so let's do the let's do the example we think about way to do it so I'm going to do the same curve we discussed in the context of blowing up so the example is f is y squared minus x squared X plus one gradient of f is this V is just two dimensional I'm going to write X for X1 and Y for X2 well so uh this is what j0 looks like that is the the curve itself okay so what about the space of watch it's well remember one Jets were the same as tangent vectors that is pairs p and Q where FP is zero that is p is on the Curve and the Q dot with the gradient of F at p is zero so let's write out what that means in equations so p and Q are just both pairs of scalers right so AFP equals zero means P satisfies the equation of f thought about as an equation in P1 and P2 so we'll say P2 squared minus P1 squared P1 plus 1 is equal to zero that dot product there means q1 times minus 3 p 1 squared minus two P one plus Q2 to P2 is equal to zero and that's it so the jet scheme is cut out of if K was R I guess it I wrote it as R might as well stick with that um so this is in R to the four with coordinates P1 P2 q1 and Q2 so I've just told you a pair of equations so to cut out the first jet scheme as a sub variety of R4 okay so that's it that's all the jet schemes it's just this pair of equations

all right well let's think about what the geometry of this pair of equations are the easiest way to understand it is to think about it in terms of its map to j0 uh maybe I'll draw it on the next okay so here's J1 here's j0 so that's j0 I think I just drew so let's think about J1 and there's this map here right which is just p q goes to p but in J1 we shouldn't sort of be prejudiced about what's more important P or q but they come together right so just think about J1 Z as being something cut out of R4 by some equations and we can interpret that if we like as uh pairs of pairs we call the first pair p and the second pair q but J once I know exists on its own before that interpretation okay so what does it look like well let's think about of a different there's different kinds of points right so we need to divide into the case where the gradient is zero and the gradient is non-zero well that's here right the only place where the gradient is zero and FP is zero right so where is the gradient zero well if you do the calculation you'll see that well it's at y equals zero and where either X is equal to zero I guess we can just do it um so where's the gradient zero well it's y zero and x minus three x minus 2 is equal to zero so that means either X is zero or X is equal to one minus two thirds but minus two thirds zero is not on the Curve right so that means that the only point where both the gradient is zero and F is 0 is at the origin and everything else fits into this category over here all right so what does that mean uh remembering that this equation this equation here says Q dot gradient is zero so let's pick pick a point here and think about what the pre-image is let me call this pi what is pi call that P regular what is that well that's the set of all pairs PQ in the jet scheme such that the first point is p rig right now the way I drew it the gradient at P reg is non-zero so this is all the points of the form P rig Q where Q dot some non-zero vector is equal to zero how many of those are there well that's a linear equation in Q right so that's just the vanishing of a linear form on the cues so it's got one less dimension then the space of queues which is two dimensional so that's just a line right so this here is just a one-dimensional vector space okay so there's a lines worth of stuff sitting over that regular point and that's true for all the points of the Curve apart from the origin which is somehow special okay is that clear so there's over every Point downstairs

a line the blue stuff taken together makes up the pre-image of Pi at every Point apart from the origin but the origin is special so if I take the origin and take Pi inverse of 0 well that's all the points that look like zero comma Q what's the equation on Q well Q dot product with 0 is 0 but that's no equation so that's just all cues okay so that's two dimensional it's isomorphic to V itself so up here at the origin I have something two-dimensional okay so the way we say it is that we have one dimensional fibers fibers meaning these pre-images away from the origin but at the origin we have a two-dimensional fiber so that's what j1z looks like um if you fiber it over j0 you get this uh well it's a vector bundle every fiber is a vector space but the dimension is not constant okay so I'm going to go further and do the second jet scheme and that's where I'll finish for today but any questions about this map first wait dan the vector bundles have to have the same dimension I mean I guess I mean so it's it's relative spec of a of a of a Sheaf but it's not a locally free Sheaf so yeah I guess I'm Miss speaking I shouldn't say Vector bundle that's right I mean away from the origin of this yeah that's right that's good enough right yeah yeah so maybe it's worth in connection with that comment noticing that if you took away the origin so if we take j1z remove the pre-image of the origin and we take this remove the origin so that maps to that well that's uh actually a trivial Vector bundle so that's just that's just the base space so j0z remove the origin cross ah right those are just all the blue lines uh now the reason for that is that if and that's a general fact is that if you take the regular part of Z and take its so this is actually isomorphic to the the jet scheme of the regular part I'm sorry the jet scheme of a regular variety is just trivial you'll just get something like this you'll just get the original thing crossed with something um so that's why the jet scheme of something that doesn't have singularities is not interesting okay so let me sort of stack on top of J1 now J2 and maybe I'll call this row so remember rho sends p q r p q now why are we doing J2 well we don't quite understand that um that uh I guess I haven't actually properly explained what J2 is have I um so maybe let's do that quickly so if I go back to this formula uh where is it here right so the formula in the Lemma PSI p q r yeah maybe I should write so um hence two Jets of Z which is the vanishing locus of f

are in bijection with well remember the two Jets of Z are the same thing as K alge by definition K algebra omorphisms from simvi Star modulo F to KT modular T cubed but that's a k algebra homomorphism from simv Star to KT modular T Cube that sends F to zero and K algebra homorphisms from Simply star to KT modulo T cubed are of the form pqr so sending F to zero simply means that the stuff on the right hand side here is zero so then we get three equations right we get f p is zero which simply says that P is in Z we get as before the dot product is zero which says Q is a tangent Vector but now we get this final equation which is quadratic in q and this is the great equation that we don't quite understand yet all right so this uh this here is J2 right P triples pqr subject to those equations those are the equations of the second jet scheme of set which I'll now go back and compute in the case of this nodal curve that's where we okay so here's J2 um all right so again we're going to think about uh different easier to think about the fibers than to think about the thing directly so let's draw this is J1 so there's a bunch of lines so J1 is kind of a union of all the blue stuff here there's plane and more lines okay um let me pick a point here so that's a PQ right so this is the this is the P I'm sort of visualizing j0 sitting inside J1 uh by the zero section here but that the point in j1z is this PQ so a point on this blue line so Q is sort of the direction that goes along this line what is uh row inverse of p q well it's all the pqrs uh well such that uh p and Q are the given p and Q and R satisfies that equation I wrote down right so r dot product this is minus a half Q times the Hessian of this with respect to this now the point I chose is a point where the gradient is non-zero right so again the right hand side is something it's some real number it could be zero I don't care but the gradient is non-zero so I've got an equation that looks like r dot product with some fixed thing is equal to some fixed number and that's not linear in r but I can find a solution so let r0 for example just take some multiple of the gradient itself so find some solution than any other solution if you're taking the difference with r0 has a zero dot product with a gradient and that's a linear equation so a set of solutions has Dimension one less than the space of ours the R's have Dimension two so that has a one-dimensional space of solutions so what I get is that this fiber here looks

like my r0 plus some one dimensional space which if you think about it is in bijection to the tangent space that's it okay so that's a one-dimensional affine space if you'd like but I can give it a vector space structure so that's another line that lies over that point so there's a whole line worth of stuff lying over that point and that's an arbitrary point maybe I should draw it a different color red okay so every Point PQ has a line worth of pqrs over it so be like there's a two-dimensional space of stuff lying over p in J2 okay that's not that interesting what's interesting is what's happening above the origin so now let's compute that let's take a point uh let's take a point here which is zero Q for some q and compute what the pre-image of that is well it's all the things of the form zero Q r the equation is the same our DOT product with the gradient but the gradient is zero at P because p is zero so the equation is just Q paired with itself give zero so the r doesn't do anything all right so what I get is that this is well it depends on whether that equation is true or not right so if it's true then I get as many r's as there are possible to take which is a two-dimensional space just a v right so I get V if that's true and I get the empty set otherwise all right okay so what is the set of cues for which that pairing is equal to zero well let's compute the Hessian so the Hessian uh so remember if maybe let's go back this in back over here so the Hessian will be uh this is at zero right we don't care about it anywhere else it doesn't matter anywhere else so with zero I take the in the top left I have the second X derivative which gives me minus three um so that's minus three sorry minus this is the hard part right multiplying into just minus six x minus two at x equals zero these are both zero and here I get two so I hope that's What minus two two okay so then the pairing so that'll give me V if so that pairing will just be minus 2 q1 squared plus Q2 squared is equal to zero well that's Q2 minus q1 okay now finally we have some geometry times Q2 plus q1 I'll think about what that looks like in that blue plane that I drew down below it's just the these things right okay so in J2 uh we have sitting over each of those each point of these axes where q1 is equal to Q2 and Q 2 is equal to minus q1 so sitting over both of those lines I have sort of a copy of V cross the line right so it's a three-dimensional sort of Subs variety of j2z so that's what sort of sits over maybe