Thank you very much for coming to the second lecture of this mini-course. Here is the plan for today. First, I'll explain the theory of Tamkin's metrization and the Kondasevich-Soborman essential skeleton. Second, I'll introduce skeletal curves, which is a key notion in the theory. Third, I'll explain where do skeletal curves come from in practice, natural sources of skeletal curves. And fourth, I will introduce naive counts of skeletal curves. And finally, I will give a proof of the symmetry theorem by skeletal curves as an application of the theory of skeletal curves. We will see other applications of skeletal curves in the next lectures. Okay, so let's start with the first part, Tamkin's metrization and the Kondasevich-Soborman essential skeleton. The idea is the following. By a college, non-alchymidian analytic spaces have very complicated underlying topological spaces. For example, the analytic p1 is an infinite tree, containing infinitely many vertices and infinitely many branches. And the Baikovic analytic elliptic curve is infinitely many trees attached to a circle. It is impossible to visualize Baikovic analytic spaces in higher dimensions. But they contain very nice piecewise linear subsets called skeletons. In general, skeletons are not unique. They depend on the choice of formal models. But if we are given a volume from omega on the analytic space, then we can define a unique skeleton, sk of omega, associated to the volume from omega. Thus, for Kalabiya variety, where we have a unique volume form up to scaling, we have a canonical skeleton called the essential skeleton. For example, the circle inside this elliptic curve is the essential skeleton of the elliptic curve. Here is the history of essential skeleton. In 2000, Kondasevich and Soboman constructed an essential skeleton inside non-archimedian analytic Kalabiya space, X over C double parenthesis T of maximum degeneration. Here, C double parenthesis T denotes the field of formal Laurent series. Their method is the following. First, they define a weight function psi on divisorial points, X div inside X using semi-stable models of X. And then they define the essential skeleton sk of X inside X to be the closure of the minimum locus of psi. After that, in year 2012, Mustafa Nikas extended the weight function psi to the whole analytic space, so it is no longer necessary to take a closure for defining the essential skeleton. Then in 2017 and 2018, Brown-Mason and Mori-Mason-Stevensson, they extended the weight function and the essential skeleton to pairs. And in 2014, not really in chronological order, Michael Temkin made a vast generalization. He bypasses completely the use of semi-stable models. In this way, he is able to extend the theory of weight function and essential skeleton to any non-archimedian base field, not necessarily of characteristic zero or discrete valuation. And moreover, his theory works in the relative situation for any analytic space X over another analytic space S. His method is the following. First, he provides the shift of scalar differentials on X, omega X with the maximal seminorm called scalar seminorm, which is the maximal seminorm making the differential D from the shift of functions on X to the shift of differentials on X a non-expensive map. He calls this maximal seminorm scalar seminorm. Then this gives rise to a seminorm on the canonical bundle kx by taking top exterior power of the shift of scalar differentials. Now, if we have a volume from omega, and if we apply this seminorm to omega, we obtain a real-valued function. And the Temkin proved that this real-valued function is equal up to a constant to the Kondasevich-Sobomar-Mustatonicus weight function in the situations where the weight functions are well defined. And the essential skeleton in Temkin's language is just the maximal locus of this scalar seminorm of this volume from omega. This is roughly his method. And since we will need Temkin's formulation to establish some properties of essential skeletons for some of our proofs, here let me give more details of Temkin's construction. So first let us define some seminorm at the level of rings. Definition given a seminorm of the ring B and a homomorphism of rings, phi from A to B, we equip omega B over A, the module of relative scalar differentials with the scalar seminorm by the following formula. For any element x in this module of relative scalar differentials, we define its scalar seminorm to be follows. First, we write x as sum of C i d B i, where C i and B i lie in B. C i and B i are elements of B. And we take the maximum over i of the norm of C i times the norm of B i. And then we take inf of this maximum over all possible ways of writing x as sum of C i d B i. So this gives the definition of scalar seminorm at the level of rings. And Temkin proves that gives a canonical characterization of this scalar seminorm defined by the explicit formula. He proved that this scalar seminorm is the maximum seminorm that makes the differential D from B to the module of relative scalar differentials a non-expensive a-homomorphism. Now let's consider the global geometric situation. Given f, a morphism of k analytic spaces where k is any non-archimedian base field, we apply this above definition at the level of rings, we obtain immediately a pre-shift of scalar seminorms on avinoid domains. Then via shiftification, we obtain so-called scalar seminorm on this shift of relative scalar differentials. And similarly, we have a canonical characterization as in the lemma above. Temkin shows that this norm defined from shiftification is simply the maximum seminorm on this shift of relative scalar differentials making the map D from the shift of functions to the shift of differentials a non-expensive map. And now if we take top exterior power and arbitrary tensor product, we obtain the scalar seminorm on pulmonary volume forms. There is a small technical point is that in fact, we have to consider so-called geometric scalar seminorm after passing to algebraic closure in order to get better properties. Now here is a theorem of Temkin. For any pulmonary volume form omega, we can take its scalar seminorm and we obtain a real-valued function on x. Temkin's theorem says that this scalar seminorm of omega is an upper semi-continuous function. The theorem says that this real-valued function seminorm of omega is an upper semi-continuous function. Now we make the following definition. We define the skeleton of x associated to any pulmonary volume form omega to be simply the maximum locus of the scalar seminorm of omega. It's possibly empty if the maximum doesn't exist. And we denote this skeleton associated to omega by sk of omega, considered as a subset of x. So this definition of skeleton depends on the choice of some pulmonary volume form. And now let's introduce the definition of essential skeleton, which is just a union of all such skeletons over all possible volume form. So here it is. Definition essential skeleton. Let k be a non-alchimedian field of characteristic 0 and let x be any smooth k-barite. We define the essential skeleton of x denoted as sk of x to be simply the union of skeletons associated to all omega, where over all log pulmonary volume form omega. And by definition, a log pulmonary volume form is just a section of this line bundle, which is some arbitrary tensor product of the logarithmic canonical bundle. And here we take any SNC compactification x in y, and so y is any SNC compactification of x, and d is the complement of x. And one can show that this space of sections is independent of the SNC compactification we choose. So we can just choose any SNC compactification, consider all pulmonary volume forms as sections of any tensor powers of the logarithmic canonical bundle, take the associated skeleton, and then take a union, and this is by definition the essential skeleton of x. Since we have taken union over all volume forms, it's just canonically associated to x. Let's introduce a notation for later use. When a compactification x in y is fixed, it's usually quite natural to consider the closure of the essential skeleton of x inside the identification of y. And we denote this closure by sk bar x, and sometimes we call it the closed essential skeleton. So that makes sense if we have a compactification fixed. Let's give some examples of essential skeletons. First example, we take x to be the algebraic torus. In this case, the essential skeleton of x is homomorphic to Rn, and it lives in the identification of the algebraic torus. One can show that the essential skeleton of x is in fact a birational invariant. With respect to volume forms, of course. So if u is a log-Kalabiya variety containing azarisky open torus tm, m being the co-character lattice as in the previous talk, then the essential skeleton of u is just equal to the essential skeleton of the torus, and it's homomorphic to mr, the lattice m tensored with r. So it's just Rn. So for our log-Kalabiya, the essential skeleton is very simple, just Euclidean space. Second example, we take x to be p1 minus some closed points. In this case, the essential skeleton of x is equal to the convex hull of these points. So recall that the analytic p1 is an infinite tree with infinitely many vertices and infinitely many branches. And we take out some closed points from this tree. The closed points, they are points on the boundary of this disc. Then the claim is that the essential skeleton of the punctured p1 is equal to the convex hull of these points. So here we take out four points, four closed points, and then the essential skeleton is the convex hull of these four points, which is this red subtree inside this infinite tree. Example three, we take x to be an elliptic curve with bed reduction, whose electrification is infinitely many trees attached to a circle. In this case, the essential skeleton is just the circle inside this analytic space. Then we have a two-dimensional analog of this example three, where we take x to be a k3 surface with maximal degeneration. And in this case, the essential skeleton is homomorphic to s2, two-dimensional sphere inside the n-identification of x. The final example we want to give is the following. We take x to be m0n, the modular space of p1 with n marked points. Then we show that the essential skeleton of x is homomorphic to chop 0n, the modular space of rational tropical curves with n legs. So we show this by considering the classical DeLene-Bernford compactification, m0n bar of x, consisting of stable n-pointed rational curves. And then we show that it gives rise to a minimal compactification. And we further deduce that the essential skeleton is just the usual, the skeleton associated to the compactification. And that skeleton was previously studied in the work of Abramovic, Caporoso, and Pain. So that's, for the moment, that's what I want to explain for the theory of Tamkin's matrization and the essential skeleton. Now let's turn to the next section. We will introduce the notion of skeletal curves, which is a key notion in the theory. The idea is the following. Let's consider an analytic curve C in the log-calabial variety U-analytic. We have our log-calabial variety U-analytic, and the verification of our log-calabial variety. And inside we have this blue essential skeleton, piecewise linear subset embedded in this analytic space, this blue essential skeleton. And we consider some analytic, this red analytic curve C inside our calabial. If the dimension of U is greater or equal to 2, then by dimensional reason, the curve C never meets the essential skeleton. Because the points in the essential skeleton are valuations on the generic point of the variety U. And the points in the curve, the curve C is a one-dimensional subspace. The points of the curve C, they are at most of dimension one, while the points in this essential skeleton, the points in this essential skeleton is of top dimension. So this curve C has no chance to meet this essential skeleton, just because of dimension reason. But we can let the curve C touch the essential skeleton of U, if we allow the curve C to be defined over a big non-alchimedian field extension, k prime of k. And here is the surprise. And soon as some k point of the curve C touches the essential skeleton of U, then the whole skeleton of the curve C must lie in the essential skeleton of U. So we observed that in general by dimensional reason, there is no chance for a curve C to touch this green essential skeleton. But if we allow the curve C to be defined over a big enough non-alchimedian field extension, then as soon as some k point of C touches this essential skeleton SKU, then the whole skeleton of C will lie in the skeleton, the essential skeleton of U. Now let us give the precise statement. We fix some log-calabia variety U, the orange U over k, some volume form omega on U, U in Y, here Y, some SNC compactification, and let D be the divisors at infinity. We denote by D essential inside D, the union of essential divisors. By essential divisor, we mean divisors where the volume form omega has a pole. So here in the picture, these dark blue curves denote essential divisors while this light blue, light blue curve is a non-essential divisor. And now we will consider some curve C in Y, this red curve that touches some points of the boundary divisor. So as we said, we must, if we want the curve to touch the skeleton, we must pass to a big enough field, base field extension. So let k and k prime be a non-archimedean field extension, and we choose C, a rational nodal curve over k. We consider F, a k prime analytic map from the base change, from the base change of C to the base change of Y, such that the pre-image of F, the pre-image by F of the divisor D is equal to the pre-image by F of the essential part. In other words, the curve C meets only essential divisors at infinity. And furthermore, we ask that the pre-image of the essential divisors is some linear combination of k points, P i in C, such a curve, which mainly lies in the interior U, and when it hits the boundary divisor, it hits only the essential part at some k rational points with some multiplicities. So F is a k prime analytic map between the base changes, and we consider the composition of F with this natural projection map given from the base change. So we have made a base change, and we have the natural projection of base change, and we consider the composition which we denote by F y. Now the claim is that if F y of x lies in the essential skeleton of U for some k point x, then F y of the essential skeleton of the base change of the punctured curve C naught is just C minus the marked points. So then F y of the skeleton of the punctured curve will lie totally in the essential skeleton of U. In other words, the whole skeleton of the curve lies in this essential skeleton of U. And recall from the example that we mentioned above, the essential skeleton of such a punctured curve is just equal to the convex hole of all the marked points in the analytic space. So this is a precise statement, and we call such F skeletal curves. Here is an example of a skeletal curve. We take U to be the algebraic torus, and we have seen from the examples above that the essential skeleton of the algebraic torus is just R n, this blue plane, essential skeleton homomorphic to R n, and we take our curve C to be just a p1. So it's an infinite tree, and we choose four marked points in p1, 1, 2, 3, 4, four marked points in p1. Then the essential skeleton of the punctured curve C naught, C minus the four marked points, is just the convex hole of these four marked points, which is this red sub tree inside this infinite tree. And now we consider a map from this p1 to the algebraic torus. So as we said, in general, this map, the image of this p1 has no chance to meet this blue essential skeleton, just because of dimensional reason. But if we pass to a big enough base field extension, then it might happen. And the theorem says that if some k point of the curve C hits the blue essential skeleton, then the whole skeleton, this red sub tree, the whole skeleton of the curve will lie in the essential skeleton of U. The major advantage of skeletal curves is that they have canonical tropicalization. Since the map fy maps the skeleton of the curve into the essential skeleton of U, we can just restrict this map fy to the essential to the skeleton of the curve, and then we get some tropical object from some finite tree, some tree, which we denote by gamma, to this polyhedral object. And this restriction is independent of any choice of retraction map from the identification of U to the essential skeleton of U. So in general, for general curve, this image of the skeleton of the curve will does not lie in the essential skeleton of U. Therefore, to get anything tropical, we must further compose with a retraction from the identification of U to the essential skeleton of U. But this retraction is not canonical. For example, different minimal compactification U in y gives different retraction maps. So then for general curve, different retraction maps gives different tropicalizations. But for skeletal curves, the compactification does not matter. We always have a canonical tropicalization. And we call this restriction the spine associated to the skeletal curve. So in the example above, the spine, the associated spine is simply the map from this red sub tree to the blue to the blue plane, this red curve. And this is canonical independent of any choice of retraction. Now let me explain the idea of the proof of the skeletal curve theory. Let's first recall the statement. We have some narachymedian field extension k prime of k and a rational curve nodal rational curve C over k. And we consider a k prime analytic map of the base change of C to the base change of y such that the curve hits the boundary divides meets only essential boundary divisors at some k points. And we can see the composition of F with the projection map of from the base change. The claim is that if f y of x, if f y sends some k point of the curve to the essential skeleton of u, then f y sends the skeleton of the base change of the punctured curve, which is just C minus all the marked points to the essential skeleton of u. In other words, the whole skeleton of the curve lies in the essential skeleton of u. Here is the idea of the proof. So for the proof, we put the map f above into a family. And we consider the skeleton of the family and also the skeleton of the base. We want to relate various skeletons together. In order to put the map into a family, very naturally we consider a home scheme consisting of all maps from the curve to y analytic. And then we consider the subspace of the home scheme h consisting of all maps f from C to y analytic of the same curve class and the same intersection pattern with D as the given one. We have the following diagram. So h is some space of maps. Over h we have the universal curve, which is just a product. Since it's just a space of maps, the domain curve doesn't change. So it's just a product, c times h. We have two projections, p c to c, p h to h. Then we have the universal map from the universal curve to y, which we denote by e. And we consider also the map phi from the universal curve to c times y. Whose first factor is projection to c and the second factor is given by the universal map. By the deformation theory of curves, we can show that the map phi is a tile over some dense Zariski open subset of the target. It's generically a tile. Furthermore, using deformation theory of curves by computing the tangent spaces of h, we show that the volume from omega on u in y, the volume from omega on u gives rise to a volume from omega h on h. So it induces a natural volume from omega h. Then we do an explicit computation. One can see that the pullback of omega, omega is here, the pullback of omega by e and the pullback of omega h, omega h is on h by the projection map p h. They agree on p h horizontal tangent spaces of the universal curve. So they may not completely agree, but they agree on horizontal tangent spaces. This implies that for any one form alpha on the punctured curve c, if we pull back alpha by the projection map p c, and we wedge the pullback of omega by e, this is equal to the pullback of alpha by p c, and a wedge the pullback of omega h by p h. It's just because they agree on the horizontal tangent space, the two forms. So if we wedge anything vertical, we get equality. We denote this by equality star. And the second, for any k rational point, k point x in c, which is not a marked point p i, we consider the evaluation map at x, e v x, which is a map from this space of maps h to u, just evaluating at x to u and l. Then since such x gives a horizontal section gives a horizontal section of this projection p h, we see from since these two forms agree on the horizontal tangent spaces, and if we pull back using the horizontal section, we see immediately that the pullback of omega by e v x is just equal to the volume from omega h. And that implies that the pre-image of the essential skeleton of u by e v x is just equal to the essential skeleton of h associated to the volume from omega h. This is because again, by the deformation theory of curves, we can one can see that the evaluation map e v x is generically a tile, and that implies that pullback of skeleton is equal to skeleton of pullback. So here pre-image of skeleton by the etanus of e v x, pre-image of skeleton is just a skeleton of pullback. So this omega h is pullback, and we denote this equality by double star. Now let's pick one fiber of our family. So we choose any point f in h, h is the space of maps, we choose any point f, we denote by c f the fiber of the universal curve at f. So it recalls that the universal curve is just a product. So the fiber at f is just some base change of c. And f h is the space of maps, f is a point in the space of maps. So f gives a map from the fiber c f to y analytic, which is just the restriction of the universal map e from the universal curve to from the universal curve to y analytic. There should be no c here. And it's natural to denote this map f because it's really given by f. Assume, now assume that f x lies in the essential skeleton of u for some k rational point x. Since f x is just evaluation of f at x, so this equality double star implies that f lies in the skeleton associated to the volume from omega h. Because we assume f x lies in the skeleton of u and f x is just evaluation of x at f.  So we get a very nice characterization of f now just from our hypothesis. And recall our goal is to show that f of the skeleton of the punctured c0 the punctured fiber c0 f lies in the essential skeleton of u. So in order to show that let's compute this pre image, pre image by phi of the product of the skeleton of c0 times the skeleton of u. We want to show this, we compute this product, we will use the map phi. And by definition the skeleton essential skeleton of c0 is just a union of skeletons associated to all possible log volume forms on c0. Here taking union over log volume forms or puluri log puluri forms they are the same. But first equality is by definition of essential skeleton. Next using Temkin's theory of matrization, Temkin's matrization theory one can show that skeleton of product is equal to product of skeleton. So here we have product of skeleton and it's equal to skeleton of product with respect to the wedge of the volume forms. And the next recall that phi is generically etal by deformation theory. This implies that pullback of skeleton is equal to skeleton of pullback. So here we have pre image of skeleton by some etal map and this is equal to skeleton of the pullback of the form by this map. Now recall by definition phi has two factors. First factor is the projection to c. Second factor is the universal map e. So by definition of phi, by definition of phi, this is just equal to the skeleton of pullback of alpha by pc wedge pullback of omega by e. And now we apply our explicit computation, this equality of forms on horizontal vector spaces. We apply our explicit computation, we deduce that this is equal to skeleton of this wedge product. So we replace this wedge, this pullback of omega by e by this pullback of omega h by ph. And then to summarize this by definition again is just the essential skeleton of c0 of the punctured curve times the essential skeleton of omega h. And we observe that by Tamkin's matrization theory, a point z lies in the essential in the skeleton of a product x times y if and only if z projects to a point y in the skeleton of y of big y and the z lies in the skeleton of the fiber. x y. So a point lies in skeletal product if and only if it projects to skeleton of the base and moreover it lies in skeleton of the fiber. Therefore, since f lies in the skeleton associated to the form omega h, so we think h as the base here, therefore for any x in the skeleton of the fiber, the skeleton of the fiber, the punctured curve at f, this computation, the equality above between this one and this one shows that x just lives in the pre-image of the product of skeleton because by what we just said, f already lives, so we look at this line, f already lives in the skeleton of the base. Now if we choose any point in the skeleton of the fiber, then this point actually lies in the skeleton of this total space and that is just equal to the pre-image of this product of skeleton. So this shows that x lies in the pre-image by phi of this product of skeleton and we deduce that just to recall the definition of phi, we deduce that f of the skeleton of the punctured curve at f lies in the essential skeleton of u. In other words, the skeleton of the curve maps to the essential skeleton to the essential skeleton of u. So proof complete. Remark, by adding extra k points to our curve c as marked points, the above argument has a stronger and perhaps more surprising result. We can show that the convex hull of all k rational points inside the fiber cf maps to the closed skeleton, sku, the closed essential skeleton, which is just the closure of the essential skeleton in this fixed compactification by analytic. So not only the skeleton of the curve maps to the skeleton of the target localabia, but the convex hull of all k points will lie there. So that is all I want to say for the proof of the skeletal curve theorem. If you did not follow every line of the proof, no worries. And now we will move to the next topic. So the question is, the skeletal curves seem so nice. They have canonical tropicalization and we will be using them for many purposes. So the natural question is, where do they come from in practice? And in the next section, we will talk about the natural sources of skeletal curves. Let's first make five minutes break before moving on to the next section. So the skeletal curves, they seem so nice, but where do they come from in practice? And that's what I will explain in the next part of this lecture. So let's explain where do skeletal curves come from. Recall from the Frobenius structure conjecture that we are interested in counting rational curves in Y with prescribed intersections with the boundary D. So we have Y, some SNC compactification of our log-Arabial U. And we are interested in counting this kind of red curves whose intersection numbers with the boundary divisors are fixed. Or if we can also phrase it in terms of the interior, in other words, we are interested in puncturing the rational curves in U with prescribed asymptotics at the punctures. Anyway, U is what we ultimately care about. So let's fix some notations for convenience. We have a tuple, both p, consisting of pj, where pj are integer points in the skeleton. So in my last lecture, I give an explicit formula for this SKUZ, which is just zero disjoint union with the positive integer multiples of essential divisorial valuation. And now, in this lecture, I explained the theory of essential skeleton, and they are just integer points inside the essential skeleton. So we fix this tuple in order to prescribe intersections of our red curve with the boundary D. Some pj can be zero, and we call such j internal marked points. For example, we can have an internal marked point, p4, internal because they maps to the interior. And for nonzero pj, we call such j boundary because these marked points are supposed to go to the boundary, and we write pj as in this explicit form, multiples of some divisorial valuation, mj times nuj, and the divisorial valuation is just some divisor at infinity. So we can always assume that nuj is given by some component of our boundary D after making some blow up. Now let's consider the modular space, the modular stack, mu, both p beta, consisting of n-pointed rational stable maps from some nodal rational curve c with marked points pj to y of class beta, such that each boundary marked point pj meets the interior of the divisor dj with tangency order mj and no other intersections with D. So exactly the sort of modular stack we consider in the Frobenius structure conjecture. And if we pick an internal marked point pi, then we can evaluate at this internal marked point and we obtain something in u. And we can also take the domain and take the stabilisation of the domain, we obtain a point in the Dlingman for the stack of n-pointed rational curves. So recall that the domain of a stable map may not be stable, thus we need to take a further stabilisation in order to get a stable curve. And we put them together, we have the natural map phi i, very analogous to the map phi we considered in the proof of the skeletal curve theorem. And now we have the theorem source of skeletal curves, which says that phi i over the skeleton inside the target has finite fibres. And moreover, the fibres, they consist of skeletal curves, which just means that the pre-image of phi i by phi i of this skeleton inside the product consists of skeletal curves. So that's the way we produce skeletal curves in practice. And just a small point, here we consider closure of the skeleton, so it's a bit stronger than we just consider skeleton. And that is important in the theory because we also want to consider degenerate domains. And in the proof of associativity, for example, it's, and also just in the classical theory of Gromov-Witton invariance, it's useful sometimes to degenerate stable maps and to break them apart. So that's why we also consider the closure of skeleton, which will contain these degenerate curves. So that's the way we produce skeletal curves in practice. And the proof is the following. For finiteness, we again use the deformation theory. We can show that for any fixed modulus of domain, the fiber of phi i at mu is finite at al over some zarisky dense open subset of the log-calabi yaw. And skeletalness follows from the skeletal curves here. So here finiteness allows us to count curves naively without using virtual fundamental classes. And let me explain now how do we count them naively using this finiteness result. So let me explain now naive counts of skeletal curves. The above theorem source of skeletal curves suggests a simple definition of naive counts associated to spines in the essential skeleton of yaw, which we explain now. And the study of properties of such counts is the main technical foundation of our theory. So recall we have our natural map phi i going from the modular space of stable maps to the modular space of stable curves by taking domain modulus and to our log-calabi yaw by taking evaluation of some internal market point. Now any stable map inside the pre-image by phi i of the skeleton in the target is skeletal by the above theorem. So we have a canonically defined spine, which is just we take restriction of f to the skeleton of our curve and this maps to the skeleton of yaw by the skeletal curve theorem. So here we take the closure of the skeleton, it doesn't change much, just more convenient to work with, because otherwise it's just infinite curves in like in Rn. If we take closure, it's just more convenient for notation. We can say where infinite point goes. So that's a very minor point. And conversely, given any abstract spine h from some graph, some tree to the skeleton of yaw and some curve class beta, we want to count all skeletal curves of class beta giving rise to this spine h. So this is our goal now. We want to define the count nh beta, which is supposed to be the number of skeletal curves with the spine h and the curve class beta. So first the question, what is an abstract spine in the essential skeleton of yaw? First, observe that the essential skeleton of yaw has an intrinsic conical piecewise integral linear structure. The idea is the following. So if we take any S&C compactification of yaw, we obtain a simplicial cone complex structure on the essential skeleton. And now such structures given by two different S&C compactifications, they are just related by some piecewise integral linear map. So therefore, this essential skeleton has some intrinsic piecewise integral linear structure. And thus it makes sense to define a spine in the essential skeleton to be a piecewise integral affine map h from some nodometric tree to the essential skeleton. Here is a picture. So now we consider such nodometric tree gamma. This is our essential skeleton and we consider a spine inside. And we denote by vj the set of one valent vertices of gamma. Let us first consider the case of extended spine. In other words, let's assume that all the vj's are infinite vertices. And we denote by pj the weight of vectors at every vj. In other words, just the derivatives. So these purple vectors are pj. And we denote the whole all the pj, we put them as a tuple both p. So here we have five one valent vertices v1, v2, v3, v4, v5. And v5 shoots up vertically, which means that the leg v5 is mapped to a point. The map h can be constant on the whole leg. And in this case, this p5 is zero the derivative. So recall that we said that the essential skeleton of m0n is homomorphic to the modular space of tropical curves, rational tropical curves with n legs. And in fact, this holds also after taking closure. So the closed essential skeleton of m0n is actually a homomorphic to the modular space of stable extended nodal rational tropical curves with n legs. That's gamma. The northern. Sorry. Yeah. Oh, I have a question on this. So the trap bar is just a naive closure of the tropicalization. The trap bar is a compactification of the modular space of tropical curves. So you're all internal legs of infinite lengths. The legs, yes. I allow some edges to have infinite lengths. Yeah. Because I don't know. Like the, I think the Jonathan Weiss and the melody Chen, they have, they define the trap bar, like which is, yeah, I don't know. Is it, is it this coinciding with the Jonathan Weiss and the melody Chen's definition of the trap bar? So this trap bar 0n, I think it was first considered in the paper by Abramovic's Caproso and the pain. Okay. Powered tropicalization of modular space of stable curves, probably. And with the show, we show that here the essential skeleton is just the essential, is just the skeleton given by the classical Dulin Man for the compactification. And then we apply a result in the paper of Abramovic, Caproso and the pain, which identifies the skeleton associated to the Dulin Man for the compactification with this modular space of extended tropical curves. Okay. Thanks. Yeah. So they are really natural objects when we consider compactification. Yeah. So we have our nodometric tree. And it's just a point in this modular space of tropical curves. So by this homomorphism, we obtain a point in the skeleton of M0n. And recall, we have our natural map phi i from the modular space of analytical curves to the modular space of domain times our local abial. Inside, we have a product of skeletons. And then we have a point gamma in the skeleton of the first factor. And we also have the point Hvi. So in this picture, it's just this point H of V5. We also have this point in the skeleton of U. So the pair together gives a point in the target. And now we just take the preimage by phi i of this point in the target. And by the skeletal curve theorem, the preimage is just a finite set and consists of only skeletal curves. But now we have a finite set. And not all curves inside this finite set are good. So we further restrict to a subset F i H beta consisting of stable maps whose spine is equal to H. So this subset, this set phi i inverse, it just says that our curve has the correct domain and the internal mark point i pi maps to the correct place. That's all. It doesn't say anything about the spine. That's why we consider a subset with the right spine. And then the count n i H beta that we want, that was our goal, we want to define. We just let it be the length of this subset considered as a zero-dimensional analytic space because probably we have some new potents or multiplicities. If we pass to a big enough infinite, if we pass to a big enough algebraic closure, if we pass to an algebraic closure, then it's enough to take just the cardinality of this set. So we define the count n i H beta to be this length. And n i H beta just means the number of skeletal curves associated to the spine H curve class beta and by evaluating at the ice marked point. So intuitively, this number counts these purple rational curves, close the rational curves with the given spine, given red spine. And more generally, we consider also non-extended spines. Sometimes we call it truncated spines. In other words, we allow some one-way lent vertices, vj to be finite. So the idea is to use toric tail condition to define the counts associated to truncated spines as in the first lecture. Here is the picture. We have skeleton of our log-cala-bio. And we consider a truncated spine. So here, the vertices v1, v3, v4, they are finite vertices. And v2 and v5, they remain infinite vertices. And in order to count the skeletal curves associated to such spines, we recall that we have a torus inside u with co-character lattice m. And this implies that the essential skeleton of u is equal to the essential skeleton of the torus and is homomorphic just to m tensor with r, rn. And now we can extend the truncated spine, this truncated spine h together with curve classes. And we obtain an extended spine h hat and an extended curve class beta hat. So I wrote since regarding curve classes in blue just to mean that you can ignore it if you are not familiar with the theory. They are more auxiliary. So we can just focus on the spine. So we apply the constructions above. We apply the constructions above to this extended spine h hat and extended curve class beta hat. We obtain a finite set, fi h hat beta hat as above consisting of closed curves with spine h hat. And now we consider subset, a further subset satisfying the toric tail condition. We ask each punctured tail disc to lie inside our torus. So then we are ready finally to define our count associated to such a truncated spine to be simply the length of this subset considered as a zero dimensional analytic space. So intuitively this number just counts this kind of open curves with given spine. And by open we mean curves with boundaries. So that's the definition of our naive counts. And we have the following theorem concerning this number, this counting number. So assume the spine h is in the general position. More precisely we assume h is transferred to walls inside the skeleton of u. So I will introduce the notion of walls in the next lecture. Here let's just imagine that h is in some general position. And then the count hi ni h beta, meaning the number of skeletal curves associated to the spine h and the curve class beta by evaluating at i mark point is independent of the choice of the internal mark point i and the nor of the choice of the torus inside. So remark the independence on i used to be called the symmetry zero and had a tricky proof by a deformation in variance. Now we have a much more conceptual proof via skeletal curves and let me sketch below. So that shows another application of skeletal curves. We can get a conceptual understanding of this independence of the choice of the mark point where we evaluate. So if there are any questions you can ask. Otherwise I'll just go to the proof of the symmetry theorem. Yeah, so let me explain the symmetry theorem via skeletal curves. So the symmetry theorem is just the independence of our count on the choice of the place, the point where we evaluate. I mean if you think why this is true it's not really obvious because we evaluate at an internal marked point i and we want to show that it doesn't depend on the choice of this internal marked point. So maybe we want to move if we have two different places where we evaluate maybe we want to move from one place to another. But the trouble is that when we move from one place to one place to another at some point we will across some walls and the spine is no longer transverse. So this kind of deformation in variance no longer holds if we move across walls. In general, we will have some wall crossing formula if we go through move across some walls. And here the way we want to show via skeletal curves is that we can actually move through the walls if it is a skeletal curve. So let me give more details. So the idea is to move from one place to another in the skeletal curve setting and in that setting we can go through the walls without some complicated wall crossing formula. Yeah, so let's recall the setting from the proof of the skeletal curve theorem. We have a home scheme parameterizing maps from domain curve C to the target Y and we consider a subspace consisting of maps with a given intersection pattern with the boundary and also some given class data. And we also had this natural maps. We have universal curve to projections. Universal curve is just a product and we have the natural map phi. First factor is just projection to C. Second factor is the universal map. And on Y we have volume from omega and on H by deformation theory we produced a volume from omega H. For any point F in H we denote CF the fiber of the universal curve at H and we denote the map, induced map again by F because that's what F means. So recall from the proof of the skeletal curve theorem, F being skeletal is equivalent to F lies in the skeleton of H associated to the volume form and phi X lies in the skeleton of the target if and only if F lies in the skeleton of H and X lies in the skeleton of the fiber. So that is what we have shown. The main point in the proof of the skeletal curve theorem. If you're confused about the rest, just yes. No, you can't formula what F lies in the skeleton. F is a map. F is a map but F is also a point of the space of map. Ah yes, sure. Sorry. Yeah, so F is a map but it's also a point in the space of maps. Yeah, so here we really showed that F as a map is skeletal if and only if F as a point lies in the skeleton. So that's what theorem says. Yeah, so now we assume F to be skeletal. In other words, we assume the point associated to the map lies in the skeleton. So we have a canonically associated spine which is just given by restriction H restricted to the skeleton of the fiber which is the same as the skeleton of the curve. So it maps to the skeleton of U. So that's all what we have done in the proof of the skeletal curve theorem. And now let delta denote the graph of the spine H. And here we make a claim. Assume that the spine H is in general position. In other words, assume it is transverse to walls. Then the skeleton of the fiber, Cf inside the pre-image by phi of delta is a connected component. So recall from this equivalence or recall from just from the fact that the whole skeleton of the curve lies in the skeleton of U. The skeleton of this fiber just lies in the pre-image. And we claim that this subset is a connected component. So I drew a picture for your understanding. Recall that our natural map phi goes from the universal curve C times H to C times Y. And we have the graph of the spine delta inside this target C times Y. And we have phi going from C times H. So this is C times H. H is the base, the space of maps, and every fiber and this total space is product C times H. Every fiber is curve C. So if we take pre-image by phi of delta, by the finiteness of phi, we obtain some graph inside the product C times H. We obtain some graph. So this fiber, the skeleton of this fiber Cf, it lives inside the pre-image because this goes to the skeleton as we have a skeletal curve. But we also have some other pieces. And the claim says that this fiber is actually a connected component. They do not, it doesn't touch with other fibers. So it's not difficult to see that, to show the claim. First, by the equivalence star, we see that this fiber, the skeleton of fiber is equal to the fiber of the pre-image to the fiber of the pre-image over F. So this implies that since it's a fiber, and the fiber is always closed. So this implies that the inclusion is closed. And we are left to prove that the inclusion is open. And we suppose the contrary, we pick some, so suppose the contrary, we pick a germ of a path, like this green germ, starting from the fiber, the skeleton, this skeleton of a fiber, we pick a germ of path zero epsilon, going to the pre-image by phi of delta, starting from this fiber and then goes out. And we can, so since the image of by phi of this alpha lies in this product, we can write it, write alpha as two components, QTFT. Maybe I should say that since, so since we have shown that the pre-image of phi, so alpha is a germ of path in the pre-image by phi, but we have shown that the pre-image of phi is just the product of skeleton of C and the skeleton of H. So we can write alpha as two components, QTFT. QTFT is some point on the curve and FT is some points in the modular space of maps. And we denote, since everything is a skeletal here, we denote by HT the span of FT. And now observe that the condition that alpha lies in the pre-image by phi of delta or phi alpha lies in delta and the delta being the graph of H0, this just implies that HT of QT is equal to H0 of QT. So we have QT fixed, a fixed point on our curve and it implies that for this small deformation of our map H, the image of this point doesn't move. And then by the continuity of tropicalization from FT to HT and the rigidity of transverse spines, this I will give more details in the next lecture, we deduce immediately that this equality must imply that HT is constant. In other words, there is no way to perturb HT, no way to perturb the spine while keeping this equality. So intuitively it's very simple, we have a spine and we have a fixed point QT and we fix the image of that point. Then if this spine is transverse towards, we cannot move this spine, it's just fixed at that point. In other words, this HT is constant. And if HT is constant, it means that QT FT lives in the pre-image of this fixed point, Q0 H0 Q0 for NET. And that is a contradiction to the quasi-finiteness of the map phi. So I said that by deformation theory, generically over the target, phi is finite at all. So in particular it's quasi-finite. But here we just produced infinitely many, we just produced a germ in the pre-image by phi of some point. And that's a contradiction, so that completes the proof of the claim. Yeah, and the claim produces us this nice connected component. And so let's just, I just explained the proof of the claim, but let us recapitulate what is the statement of the claim. So we have our natural map phi from C times H to C times Yn. And we have a skeletal curve, we have a skeletal curve F from C to Yn. And we assume that the associated spine is transverse. Then the claim says that the skeleton of the fiber CF inside the pre-image by phi of the graph of H is a connected component. And now observe the following. First, observe that the first factor of phi decides exactly where we evaluate for the second factor. The second factor of phi is a universal map. And the first factor the first factor of phi is the projection to C. So the first factor determines where we are evaluating for the second map. And furthermore, observe that if we take some of degree of our map phi restricted to this skeleton fiber, and here the degree makes sense exactly by the claim. Because we know that the map phi is finite at tau, generically over the target. So the degree makes sense. But if we restrict to a subset, the degree may no longer make sense. And here it still makes sense because we restrict this finite at tau map to some connected component. And then the degree is still well defined because the map remains to be finite at tau over some thickening of this connected component, some neighborhood. So the degree makes sense. And we take the degree and we take a sum of such degrees over all skeletal curves whose associated span is equal to h. And that is exactly the following count, nw hw beta, where we count the number of skeletal curves associated to the span hw, which is just the span h. But we add an internal marked point at w, meaning that we add some internal leg at w, which is contracted the leg. And then we consider the count of skeletal curves associated to this augmented span and the curve class beta by evaluating at the added marked point, w. And the left-hand side is equal to the right-hand side by the definition of this count. So now we can conclude the symmetry theorem for transverse span, the count. Now we see that the count nh beta is independent of the choice of the internal marked point. Because here we see that the left hand doesn't depend on the choice of w. And the right-hand side is the count of skeletal curves where we evaluate at this point, w, and w is allowed to move everywhere. So the count is invariant when we move w anywhere along the span. And this shows the symmetry theorem. Furthermore, we can show that adding or removing internal marked points does not affect the counts at all. So this is an illustration of how we use skeletal curves for establishing important properties of our counts. And we will see further examples of that in later parts of the lectures. So here for the symmetry property, symmetry theorem, actually we can have different proofs without passing through skeletal curves. But for other properties, we must use skeletal curves. And here it's nice to see that using skeletal curves, we really have the freedom of moving the point w everywhere. If the curve is not skeletal, there's no way to cross a wall while keeping the invariance. As a proof of the symmetry theorem, we don't move across the wall if we don't use the skeletal curve. So that's what I want to explain today. And for the next lecture, I will talk about deformation invariance, which is, and also many other properties of the counts that finally leads to the proof of the associativity of the mirror algebra. And for deformation invariance, as we said, usually it only holds outside walls. When we cross a wall, we are supposed to have wall crossing formula. We no longer expect deformation invariance. But for skeletal curves, actually, there is some trickier deformation invariance that somehow similar to this situation about moving around this marked point across walls. For skeletal curves, we can actually move across walls a little bit, as long as it's sufficiently transverse, but not really transverse. For non skeletal curves, it must be transverse in order to have deformation invariance. But for skeletal curves, we can relax a little bit the transversality condition. And that's actually important in the proof of associativity and also in the proof of wall crossing formula. Because in associativity, I mean the definition of structure constants, if you remember from the last lecture, the place we evaluate, we ask the point to go to Q. And the Q is, although it's a very generic point at the level of analytic geometry, it's a very special point at the level of tropical geometry. So all the spines that appear in the definition of structure constants, as in the previous lecture, they are all very special. They are not transverse at all. So in order, and of course, we can make them transverse if we don't ask the marked point to go to Q, but to go to some place, to go to some point sufficiently close to Q. But then we will have the choice of asking it to go to either the left of the wall or the right of the wall. Or if there are many more walls, then we have even many more choices of chambers. But in general, we have the choice of asking it to go to the left or go to the right. And it's not clear at all whether the structure constants for the marked point going to the left is equal to the structure constants for the marked point going to the right. And this going from left to right across in the wall, we have to use the theory of skeletal curves again. So I will explain more about that in the next lecture, the next amount. Thank you very much for your attention. Okay, thank you very much. And maybe you have time, maybe don't have time for the questions. Actually, I have a very simple question. You have this variety H, which has the same dimension as Y. It also has logarithmic volume form here. Yes. Yeah. But is it? Yeah, so it means that you can start to reproduce from some local abbey or another local abbey or in a sense. Yes. Yes. And does this H contain the torus? Again, if you assume that Y contains the torus. This H. Yeah. H is a cover of the torus. Probably itself is not a torus. Ah, it could be more. H is really the modular space. Yeah, I see. And also, we don't really have a good compactification of H. Ah, so it's not local abbey or more. It's a chromified cover, yeah. It's not clear whether it's local abbey or not, because we only consider the essential skeleton of H associated to this particular volume form. Yes. Yes. Yeah. Maybe there are other volume forms. Yeah. Oh, maybe this volume form has zeros. Yeah. It can have zeros or yes. Yeah. Okay. Okay. So thank you.


I got to get some water. I wasn't thinking, and I started my video, and I don't have any water. So don't go anywhere. If you're coming in, just stick around. I'll be right back. I'm going to close this window, because the wind just started ticking in. Anybody here? Oh, good. I didn't miss anybody. And I have you guys on a wobbly table. Anybody who might come in later, just came on to talk. I've got my ears on. I have these ones, and then I have a pastel one. I want to get the small world one. I was going to buy it for myself for my birthday, and I forgot. Hello, Roberta. How are you? Everybody tell me how you're doing when you check in. Hello, Lisa. Better put on my glasses so I can see who's here. I swear, I wish so much that I could play my music when I'm chatting with you guys, because it just affects my whole mood, and how I talk, and what's going on. I want to hear from everybody. I want to hear what's happening, how your week went, how you're planning for the weekend, if it's making a big deal. If it's, hi, Susan. I was hoping you'd see I was on, take a break. Hi, Cheryl. If it even makes a difference. Here, Susan, are you wearing your shirt? We could be twinsies. Literally was just sitting here feeling kind of lonely, feeling kind of, you know, I'm having all these mixed emotions. They're opening up some of our beaches, but not the parking lots. It's like so many mixed messages. Hi, Jimmy. You know what, friend? You are not alone. I did too. I have been, and I'm up way more than, that's what started this mood. I'm up more than I care to be, and I thought maybe somebody else would like to talk about it. Hi, Tony. Hello, Susan. Oh, I didn't do a live on my birthday. Hi, Patty, honey pie. What was I doing on my birthday? On my birthday? I wasn't in the greatest of moods on my birthday, so I didn't really feel like chit-chatting.  But I did get a whole bunch of flowers. I got to show you guys this. Tell you something funny. My friend Cindy sent me a new cup today. I got a new cup. Look at, you know what, the color doesn't come across well, but it's a beautiful baby pink. And I love, love, love it. She's like, I know you don't need it, but she, every time we FaceTime, she's always got her cup. I'm going to show them the flowers, Patty. And I'm like, I love your cures, and I want to have matching cups when we FaceTime. So she sent me one for my birthday. It's really pretty. It's 30 ounces. I love it. This guy just needs a break from time to time, right?  First of all, hi to everybody else. Grow back in a minute and say hi to everybody. So I got flowers on my birthday, and I read the card, beautiful flowers, and it was from Oliver.  Nana loves flowers. Nana loved that Oliver thought of me, because he's with this other grandmother this week. Jimmy, I hope you're able to stick around so that when I get over this story, we can talk. So then Steve and I later, obviously much later, we ordered a takeout barbecue for dinner, which I have to tell you guys was really good. And when Steve went to answer the door to get the food, there was another vase of flowers just sitting on our porch. So he brought it up. Same note from Oliver. It was the same order, only different. I'm going to show you the flowers in a minute. So I immediately called Casey, Oliver's dad, and I'm like, I love my flowers. Thank you, boys. I love you. Casey, check your statement. Something's going on, and they sent, somebody sent another thing of flowers, because none of them were marked as far as where they came from. I'll be gosh darned yesterday when I was at the park when I came home. There was a third vase of flowers on my porch. And that one, on the third one, they had text me to see if anybody was home. So I text back and I said, excuse me, but where did these come from? And they said Vaughn's on tap-oh. And then I obviously didn't get any more, but we don't know what happened. We don't know if, like, the order was sent to three different places and they all fulfilled their end of the bargain. But Casey wasn't charged more. So let me show you guys my beautiful flowers. So I'm going to put, these are all the cards, the Oliver word for word, what he wanted on the cards. I'm going to put them in an envelope and put them in the Oliver drawer. This was the first vase to come. Absolutely beautiful. This was the second one to come. It was, if this was the only one that Casey would have ordered, it would have been a pit of waste of money. But of course, I would have loved them anyways. And then this was the third one to come. Nice and big. And I love flowers, you guys. I just love them. So that's about Flower Story. Pretty guys back over here. So we had a good laugh over that. We thought that was pretty funny. And one more note on Oliver, since he is with his other grandmother this week on yesterday, they had to put down one of her dogs. So he has had a tough week. He was very, very, very close to this dog, just like he would be with Kitty and how he was with Clara. So other than that, my birthday, again, it was very, how can anybody have a fun birthday during this time, right? It's just a drag.   For myself, I need to say that, you know, I'm conflicted about this weekend because I feel like they're giving so many different messages on what you can and cannot do. I just need to be realized. I got it. I have a roof over my head. I have food. I'm taking care of. I have no worries. And, you know, am I making any sense? So I didn't get any birthday presents, except for this one from my buddy, Cindy. Steve was in the doghouse for a few hours. So let me go back and see who's here and say hello to Cheryl. Jimmy, I hope you're here. Susan says hello. Patty Roberta. Anna says, hi, everyone. Hi, Anita. I wish there was a way we could send you a message when I'm thinking about you that is not in the comments. Oh, Anna, that's kind. Thank you. I'm on Instagram. I can always be messaged on Instagram under the same name. Not today, but I should. Not today, but you should. What? Oh, be wearing your shirt. I can't see anything. Aren't they beautiful, you guys? Let's see. Anna says, I think it's better to make sure you're feeling and doing your best all around, not just on the scale. Oliver will get a kick out of this. Susan says, the day you were born is always important regardless of quarantine or not. You need our gift. Oh, thank you, Susan. Thank you so much. That is very sweet. I know. And you know what? The only, I can't believe I'm saying this, good thing about him being with his grandma when it happened was when we had to put Clara down, he didn't. Okay, perfect, Jimmy. He didn't get to say goodbye to her and he brings it up often and I feel horrible about it. So, at least he was able to say the dog's name was Snicket. And it was, it was her time. And so, but other than that, the weekend is upon us. And, Jimmy, let's talk about the scale. Every morning when I wake up, I'm just so, because I wake up and I'm like, you know how you take that minute to think about your life. What do I do yesterday? What am I doing today? Do I have to go to work? Do I have to go to rep from school? That's right, run quarantine, right? And I have just put on so much of my weight. It's slowly, it's like, okay, when I look, this is the odd thing. When I look at my tracker, because I do track everything because I tracked what I had for dinner for my birthday. I tracked the cake. I track everything. And when I look at my tracker and when I look at when I, when I weigh in and everything, I'm pretty much for the most part maintaining within three pounds. But my body is just bigger than three pounds and I don't understand it and I know measurements and I know plateauing and I know all the stuff from doing this for three years. And today is just one of those days where, you know, I have three pairs of shorts that I can wear. Right, Patty? I have to go buy a pair of shorts. I mean, I don't have to. I don't want to because I feel like I'm caving. And, but then having, you know, some special foods for my birthday, I think kind of sent me over to the brain where I'm thinking, yeah, Patty, it's so weird. I'm thinking, OK, well, I have my yummy birthday food. I still have cake left. We're on a holiday weekend. And this is not something I normally do. I know it's very common. Hi, me, my Beth. I know it's super, super common for us in our community to do this. But it's not something that I do. And I am having the mentality of, I'm just going to go this holiday weekend when it's not even a, I mean, it's a holiday for people who it applies to. But we're not, you know what I mean, we're in quarantine. It's not like we're going to, it's not like I have the excuse that I'm going to go to a big party or barbecue or something. And just saying, shine it.  I can't believe that I'm at this point and I'm feeling lonely in this. And that's why I came on, because I know I cannot. Hi, Rosie. As Jimmy pointed out, we can't be, Jimmy, you and I are not the only two people in this whole community that are struggling with this. And I don't know. Patty, yes, Sag here. And that is totally how I'm feeling. I was feeling like state of marshmallow man this morning. I got dressed to take the dog out. And I said my regular shorts, a couple of them do have elastic and a button. But my muffin talk, hey, what you doing? I just, I look in the mirror and I don't like what I'm seeing. And I'm, I'm, hi, Margot. It's okay to struggle this. I need you, Margot, we need a leader in here. Sometimes it's kind of give yourself to take a break. And I completely agree with that. And I agree with that, but I don't know. I'm having, I don't, I need help. I need to talk this out with you guys. I need to see if anybody else is feeling this way. If I continue to, it's a small cake. It's not even a big cake. Steve and I continue to eat it till it's gone. And I tracked that. And there was something else that I was, oh, I'm not going to say what, because again, I don't want to trigger anybody. Yesterday, I went walking, I went to the park and I've walked every day except for my birthday. And after I walked, I went to a fast food place and I had something. Now let me tell you a couple of things. We don't eat fast food. We are not a fast food family.  It's just not, we don't eat fast food because we're so fabulous. We don't eat fast food. We just don't. It's just not our thing. Just like we don't drink soda. It was not something prevalent in my family growing up, nor when I was raising my children. And then Steve's either. He sometimes, like if I'm, when I was working, if he needed something, he would run over and grab something, but we just don't eat fast food. And I'll be damned if I wasn't sitting in my car or eating something. Totally out of character for me. Marvis is your best friend. We're feeling this way. What would you say to them? I know. I don't know. And that's what I'm struggling with. I don't know what I would say to her. And I think she's struggling right now too. And I find myself coming up short of what to say to her. Because in normal times, I think that we would have a better definition of why what's going on is going on as far as our bodies.  And I don't even know if that makes sense. Oh, Lou says, totally relate. Oh my gosh, I was at CVS earlier in Fudge Brownie and I jumped into my basket and I ate him before I got home. That never happens to me with the Reese's. And no, of course I haven't had a piece of cake today. No, I have not had a piece of cake. But it is tracked. It is tracked. I don't want to trigger, I try. But it's so foreign to us to not have a normal routine and it keeps going on and on. Another thing I'm struggling with today is, and I'm not, this isn't going to be all about me, whatever anybody needs to talk about, please bring it up. 10 points. That's how much a Reese's is too. I think that's the average. So many mixed emotions about this weekend. They're opening a couple of our beaches, but not all the beaches. I think, you know, some folks like my husband aren't taking it that seriously. Broken record, I know, but with this being a holiday weekend. Okay, this is what I'm going to say. Maybe one of you guys, I know one of you guys can help me with this. I feel guilty for making him stay home because I feel like that's what we need to do. We just need to stay home. Our quick trip to Costco, quick trip to Sprouts, that kind of stuff. Well, I know he's going to want to go golfing this weekend. And if he doesn't go, he's going to pay out and then I'm going to feel guilty. And I don't know what to do about that. But back to the body stuff is just so weird. Margo, I'm so glad you're here. I'm glad all you guys are here. Beth says, I totally understand Anita. I shook things up myself this past week, got back to bases and tracked every single thing and was real honest with myself so far, so good. But that's the thing. I have tracked every single thing, tracked everything. I haven't gone over my weeklies. I've used weeklies, of course, because it's my birthday, but I haven't gone over them. You guys know me. I always have my water. This week I walked and I gained this morning. I gained significantly. Oh wait. Oh, I think I just lied. I think I thought I gained, but maybe I maintained. Hi, MD. Thanks, Lou. I love it too. I have this one, which is just like little diamonds. It costs Steve about $40,000. Okay, no. And then I have one that's pastel. Actually, the story of the pastel one is a good one. I was carnage. We were at Carnation Cafe. Let me look up and see what my weight was this morning. Okay, you know what? I lied. I maintained this week. I was exactly the same as I was last week, but I'm still too high. I'm getting way too close to a decade that I have no place being near. No place being near this decade. And I'm two pounds away from it. But let's jump in with a funny story. Well, not funny, but something to talk about. So I saw this necklace in this little store called... What is it called? I always mess it up. Fortunosity or something like that.  And I ran next door and I bought this necklace for myself. Okay, let's see. Margo says, I think we're all struggling with the fact that we can't control what's happening. So I find myself obsessing over the things I think I can control, though it may not actually be true like my body. I'm disappointed in myself, Margo. And you're right. I wouldn't say to my friend, I'm disappointed in you, but I am.   I know, but Lou, I'm maintaining is so high, but you're right. Let's take the positive. Hi, Sue's. How is it going? Okay, you drive. You don't talk. No talking. My shirts, my arms. It's like, I don't know also too if it's an age thing. Okay. When I eat ribs or cake, how can it possibly all go to my arms? Why isn't it an equal opportunity offender as far as weight goes? Things are changed up. That's the thing is I'm eating a lot of stuff. Normally I am a boring, same stuff eater. I have my lay wraps. I have my boar's head. I have my overnight oats, and I have been changing things up. Thank you. I still can't pronounce it. Fortuosity. Fortuosity. I know I can see it, but I can never say it. Gosh, Patty, my arms are so... Okay, so here's the thing. I got some birthday money, and I'm going to go online and buy shirts with three-quarter length sleeves. Now, I have friends who don't care. They're like, it is. I care. Okay. They bother me. Now, I haven't seen your arms. I don't know about your arms. I don't look at other people and their arms don't bother me. My arms bother me. On your way to see your mama. She lives alone, so I'm going to chat with her. Oh, yay. That's fabulous. Oh, it does work, and I wish I had somebody to sit with today. That's why I bombaid you guys because I was just... I keep looking at the clock, and I can't believe it's only two o'clock. It's like, dude, tick-tock. It's got to be five o'clock somewhere. Steve's got to be home. I can't do... I wouldn't even do sleeveless. You know what? You know what, Patty? I won't even sleep in sleeveless. I won't let Steve or Oliver see me sleeveless. Beth says, I had two pair of jeans. I could no longer wear. I had to stop eating all my salty food. I'm pretty sure that was one of the big problems. But see, and there you go. You know what's going on. Now, after my birthday meal and the cake and stuff, I'll know what's going on, but am I going to add insult to injury? I just don't know.  We've had this discussion before. We will have this discussion again. I do not consider myself a failure as a Weight Watcher because I am up right now at all.  But for some reason, this life situation right now is not so helpful. Oh, I don't... So those of you who came in later, look, I got a new water bottle from my buddy Cindy, the one who moved to Texas. This is the stress and all that stuff, but it's like... Also, I did go to my Zoom meeting this morning. Isn't it pretty, Lou? It's like this really pretty. It doesn't come across on the screen as pretty as it is. It's a perfect baby pink. Soft baby pink and I just love it. She's so funny. I FaceTimed her when I was opening the box because she put about three rolls of tape on there. She goes, I know you don't need another cup. She says, but every time we're FaceTiming, ballerina pink. Yeah, that's it. That's the ticket. She says, you always comment on my cups. So now we can be twins when we FaceTiming. That'll be fun. So, and then that was another thing. When I FaceTimed her to thank her, she was in her truck. I'm like, where are you going? She's like, we're going camping. I'm like, I totally shouldn't have called you because now I feel even worse because I want to get out. But I can't. So just deal with it, Anita. I'm a little scared because my birthday is Monday on Memorial Day. If I case, I don't see you happy birthday friend. And with the restaurant sort of closed, the restaurant sort of closed, sort of open. I'm worried my old favorites are not going to be opened dining. Okay, wait, I'm not reading that right. I'm a little scared because my birthday is Monday on Memorial Day. And with the restaurant sort of closed, sort of opened, I'm worried my old favorites are not going to be opened dining. I don't know why sometimes, like when I go back and I read stuff that I don't get the first time, I don't know why I'm like such a knucklehead someday.  You know what, take a little of the stress off. Susan says, I need to start lifting weights when life isn't chaos. See, I don't want to lift weights. That's the thing. I guess I could, I don't think Susan, your arms are much smaller than mine. So it's, it's totally, there's no comparison. Besides that my arms right now, I do not think other than surgery there's anything I could do about it. I'm 58 years old now. But I don't want to be, I don't want to lift weights. I don't want to sweat. One thing I do want to do though is I do want to get back in the pool. See, not Susan. I can't, I'm not committing to getting on it this weekend. I have to be honest. I'm still kind of, let's commit to getting on track this weekend. We can all, you guys can if you want. But I cannot commit to something that I know I'm not going to do. And this weekend I know I'm not going to do it. I am in a frame of mind and I love your positivity. I love my honesty. I would never ever, see there you go. Anna's going to be your cannibality partner Susan. I knew somebody would meet your challenge. I just don't want to, I would never say I was going to do something if I knew I wasn't. And I just don't have it in me right now. Thank you so much, Fawn. I appreciate that. Everybody missed it. I wasn't on. I haven't been on. I haven't made any videos. But yeah, and that's, you know, that's the thing too. Swimming, I think that's what has me about this weekend a little squirrely. Swimming is my thing. Water is my thing. And I have no access to it. Unless, unless I just get in my car and drive over to Julie's, jump the fence and swim in her pool. No, I'm just, I'm not, I don't know. I'm just, here's the thing about me though is I know what I'm saying. Oh, trash truck. But like I haven't eaten lunch yet today. And I was thinking, oh, I would really like to have several different options in my head. And then I sat and realized that I could have something that I have here for less points and be just as happy. So I'm just a mess. I am a mess. Also, sadly, let's see. I know we'll all feel better. Sometimes you just have to, and I will when I'm ready.  Sometimes we don't even pay attention. Kind of an autopilot. Thank you, Louise. Sorry about that, guys. He's going to back out and be out of here in a second. I was also thinking that I would like to have some, some Mexican. So here's the thing.  And I can have Mexican, but I want real Mexican. You know what? It's all frame of mind. You know how we get like this around our birthday. See, my birthday starts on Mother's Day and then my birthday the Memorial weekend. Your, your whole head, your heart, everything just goes into it's party time. So maybe I will just relax for the weekend, even though I never fully relaxed when it comes to my eating. Or else I would be back at my full weight gain. And Monday morning, Tuesday morning hit the bricks running. Thank you, Tamsen. They're, they're great. I gave Steve the rest of the chicken. I just remember this morning. Then Louise, that's awesome that you found something that works for you to do. I've been getting into my little painting. I'm going to go back over and do that when I'm done here with you guys. I just wanted to come in and talk. I needed to talk to somebody today. Does Steve ride in the golf cart? Yeah. Yeah. He, he, he rides in the cart. Susan, so what's your plan?  It was good, Lou. And you know what's even better is when I make it with instead of salsa, pizza, pizza sauce. Cause I do have pizza sauce. I could just find it that day. And I make it like a pizza with the mozzarella and I put my pineapple in it. Do you have a message about the pineapple by the way? Oh no, you know, Steve's with his friends. I wouldn't go with him. That's, that's like if, I don't know, if I was doing something with my girlfriends and he wanted to come along that would just wouldn't work. Everybody emotionally eats.  Is that what I'm doing? Am I emotionally eating? Because I've never been an emotional eater before. And so maybe that's why I am just like, because it's a new feeling to me. Usually I keep everything at bay beautifully. I can tell you. And I say this every single time. There is most if not 95% of things in my life that I am not good at. I'm just not. I'm not a good housekeeper. You know, I just, I'm, but I'm a great way watcher and right now I'm not being that great watcher, great way watcher that I've put a lot of time and energy into being. It was really good and you guys could use, what else could you use? I think maybe you could use like a waffle iron to do that too. You know what? And spending so much time alone isn't helping either. It's, let's see, oh by the way, did you construct a strawberry cheesecake? Wasn't it so good? Pineapple, yes. Oh, maybe I'll make some of that tonight because I still have the stuff. Thank you for that idea. I'm getting a lot of feedback on my Cheerios, yogurt and pineapple too. People are like digging it in their life. We love it for dessert. And I'm like, dude, I'm eating it for breakfast. Dottie, Dottie's been around forever. Is she caught up on her website? Like is she, because restaurants come when they go. Is she still, is she still a go-to? You can be a great way to watch her and still be struggling. Great watchers struggle like everyone else. Thank you for that. I don't want to be heavy again, you guys. I don't want to be heavy again. Okay, Nina, then watch what you eat this weekend. I've been watching what I've been eating for three years. There's people here watching this video right now that can vouch for me. That even under the most terrific circumstances as far as eating something wonderful, I don't.  And I just, I don't know. Cheryl, that's the problem. I don't foresee going back any time soon. I cannot see wearing a mask at the park. I think it's going to be quite some time before I get there. And you know what, normally it's okay because I can go out and go to home goods and go to the market and just pill around town and talk to people because I have no problem talking to people, obviously. But I can't do that right now and it's becoming a problem. Yeah, it was a great side, Lou. Let's see, Louie said, I made those chocolate gooey cookies you posted a while back. Oh, they're good, aren't they? They were the best warm right out of the oven. The next day, they're not so swell. Patty, I don't think you're still in here. But if you are, Patty made the best brownies the other day. You don't know what I'm talking about. W.W. Maintaining Patty at Instagram. She made some sheet brownies that were just delish. She had the same problem I did when she pulled them out of the oven. Ooh, doggies. But she said the next day they weren't so great. Tamsen joined W.W. and Digital this morning. Oh, fabulous. Great, what took you so long? I'm teasing you, of course. That's great news. No, it's just not something I care to do, Susan. It's like, it's him and his gig and his friends. And then I appreciate that. And I don't want a fringe on that. Do I like tea? I don't. I don't like tea. Yoga with Adrienne on YouTube has a video of the homemade tea recipe. It's so good. You know what? Let me stop right here and tell you guys that I apologize. Like you guys are like giving the ideas and I'm like, no, no, no, no, and I'm sorry and don't personally please. That's what I'm struggling with up here is exactly what I'm bouncing back at you. Okay, Patty is still here. The bottle washers are shut down in Utah. I don't know about here. I don't know.  You know, I see in the news that these beaches are open. Those beaches are open. And you sit here and you look at the reporter on the beach, right?  Let's just go take a walk on the beach, right? We're not thinking that collectively as a group. I'm thinking that. But then when you sit and think about it, the whole wide world is going to be down there. There will be no walking on the beach. There will be no social distancing. There will be none of that. Because that's going to be in Ventura County. The beach that's open is in my county. But the folks in the counties on the other side of me are all going to swarm down there. I know. I will get it to click back in. And I need to just stop looking in the mirror and relax a little bit. I need a friend in person is what I need. I wish, you know what, if my, I'll tell you, and we talked about this this morning, if my friend Cindy hadn't moved for those of you who don't know, my buddy Cindy lived in the same complex as me. And we did everything together and then she moved to Texas. If she hadn't moved, we totally would have quarantined together. We wouldn't have gone back and forth to each other's house. And all this being alone, I think is just starting to take away at me. I find for myself, three PN is the hardest part of the day to keep focused on lifting thoughts. Yeah, Tamsen, you know what? I'm with you there, sister, because you know, we wake up and it's a new day and we've got this and as the day goes on and nothing's going on. I have turned off the news. I'm not talking about the schooling. I've got all my cousins who are all my children's age with children all over his age are all typing on their face, but they're freaking out. And I don't know why they're freaking out when there's, as of right now, no decisions have been made. I've been in touch with Oliver's teacher. She's like, we don't even know what's going on. So it's just like everybody feels like everybody's making more trouble for themselves than they have to. And I'm not going to do it. I'm fine. It's completely understandable. This is like a moving target. You can't even identify the problem yet. It will come to light in its own time and then you'll feel better. Thank you, Anna. Thank you for that. And I do appreciate you guys offering up ideas like going with Steve or doing yoga or drinking tea. I totally appreciate your thoughtfulness and putting it out there and trying to give me ideas. When I get like this, ask my mom, oh boy. Oh, my water's empty. But you're right. It'll click back in. I'm just having an off couple of days. I started with my birthday when you know who. What did I do with it? My husband pulled out a piece of paper out of the printer, grabbed a Sharpie and wrote happy birthday to me and left it here on the table. That started my mood off not so sweet. Body of car, dang it. You're at the store every day. I agree. People need to calm that down. Lou, I don't even watch the news when they're... I'm not watching it and I'm not reading it. If I see the kids talking about the schools and what are we going to do? The thing that disappoints me about today's generation and worrying about their kids, it's lovely they're worried about their kids. But it seems like they have no faith in our education system. You know, the teachers, the people, they're going to work hard to do the best they can for our children. I choose to believe that. And it's like they're ready for a fight. You know, it's like they want to get in there and throw down. Throw down. They don't like this. And then why don't they just wait and see what happens and then throw down or not even throw down at all. Let's give these fine educators, these people who work so hard, so hard. Why don't we give them a chance to try to figure out what's going to happen? People do need... People need to calm down Lou. Put on some tunes, do my painting, maybe talk to myself a little more even though I was just never ending. Talk to my little doll. Oh, she's over there sleeping. And just remember that whatever comes on, I can get off. But it's, you know, doing it all over again. But I guess that's life, right? Nothing... What does my mom always say? Nothing is guaranteed in life. That's not what she says though. But she's right. Susan's completely alone in New York. No one. I have been alone in this concrete jungle. Yes.  And I'm so sorry and please stay safe. You're never doing it all over again. How am I never doing it all over again, Margot? I'm so glad you're here. So glad you're here. Because when I listen to your videos and stuff, I can't get... You can't get interested in anything. Patty, see, I can't either.   If you guys aren't watching Margot Poros at her YouTube station, you need to click on her name right now and go over, subscribe and watch and listen to her and hear her stories. She just wrote a book. She's fabulous. I love her two pieces. It's like coming back to school after summer. I love to listen to some things, but don't have to learn the alphabet. I guess that's true. That's true. But that's exactly how I feel. There's some folks... It's true with everybody, right? There's some folks we just can't listen to for whatever reason and then there's folks that we can't wait to listen to. And you're one of those. Susan's worried about her job. Anna, I think it's just under her name Margot Poros. Margot, let me look it up. There's a lot of educators in here, Susan. I'm sure all you guys are worried about your jobs. She can answer it quicker than I can.  I think it's time you come to... Yeah, so just... Anna, just click on her name right here. Thank you, Lou to the rescue. Lou, I need to tell you something when we message. Something I've been meaning to tell you and I keep forgetting. So when we're done here, we message me and tell me to tell you what I mean to tell you because I keep forgetting to tell you here, thank you. I want to tell you is that when we have somebody here that has a channel, I want to shout them out. But I don't want to shout out other folks when they're not here because then I forget other folks that aren't here and when I play it back, I feel horrible. Did that make sense? That's all I wanted because you're so awesome at finding links for everybody. And Mima, I forgot to meet you guys. Mima has a channel, Mima, Living Healthy. Please click on her name and go to her channel. She's another soft, easy listening channel. Susan is cooking healthy food, staying on track because that's all that I can control. Reading, doing puzzles, paid by number. Yeah, I don't know, just, oh, but it's true. And you know what? To be honest, I haven't even been watching weight-watching content at all because I'm just not interested in it. Isn't that interesting? Isn't that interesting that I'm not interested? I'm watching, the first thing I watch every morning when I wake up is Stephen Colbert and then I watch Fallon and I watch Kimmel. Anything that's going to make me laugh, David Spade and then I watch my Disney folks. Some of them I come and go with some of them but I'm not watching so much of the, and I'll tell you why. When I do watch it, Leslie Jordan. Who's Leslie Jordan? See, I'm often afraid to say my feelings because then I feel like somebody's going to take it as a dig when it's absolutely not a dig at anybody. I had somebody say to me once that I didn't support other YouTubers and there couldn't be anything further from the truth. If you look at my collaborations, I think that kind of speaks for itself. But a lot of the other YouTubers, they bounce from plan to plan and that's their prerogative, that's why WW has three plans. Some throw in their worries over fat, some worry about the carbs. And I am just, you know me, I'm at the KISS, I keep it as simple as possible and that's what's always worked for me. You know Leslie. Yes, okay Leslie Jordan on Instagram. Okay, I didn't know he had YouTube. Okay, thank you, Patty. I didn't know he did YouTube. But yeah, that's what I, right now is this particular time. So since I don't meal prep, I don't watch the meal prep even though I do tune into Christy because she's another calming force and I love her, he is hilarious. So that be that, you know what I mean? Oh, she's amazing Anna. Oh my God, I love Stephen Colbert. You got to be off color to get a good laugh. Oh my gosh, we lost power. Oh no, just the season in south, oh no, we have some really pretty weather here right now. I just wish I had all you really pretty friends here with me right now. And I'll stop whining now. I didn't come on here to whine, I came on here to talk. I needed to talk.  Send you something. John Krasinski and his SGN have been deteriorating emotional feel good during all this. I don't think I know him. John Krasinski and his SGN. I'll have to look that up. I have no idea who or what SGN means. I don't know how to answer that. I don't, I clearly after whining about it for the last 45 minutes, I don't feel like I have anything to add the conversation about weight loss right now. It's just, does that even make sense? Sandy's on a roll, you know. I'm so, so gosh darn happy for her. And I feel like I need something different to talk about. That's what I feel like. Thank you, Susan. People, you'd be amazed people to have they take things. Oh my gosh. Thank you, Margo. Well, that's good to know. Maybe I'll pop in then. And the zoom ice. God, I need to stop complaining. Let's stop complaining now. I'm looking at this pile of wonderful music in front of me. I told you guys that I played some music for my dad over the phone. Didn't I? This one just came all of a sudden. Tampson. Oh no, I hope you're safe. You know what? I'll be in a decent mood, figure out what Steve and I can have for dinner because Oliver is coming home this weekend. So this will be our last couple days by ourselves. Even though I'm telling you, man. He was on the chopping block for a couple days there. He shares stories of the world doing kind things right now. He played Jim on the office and his Jack Ryan. Totally lost, but I'll check it out. I'll check it out. And who was it? Me, my sense of color. I'm totally into off color humor. I like all humor, but I'm not. I'm off color. I do not craft. No, I do not craft. I never did craft. And when I tried, you know how all your friends are doing some, you know what I liked back when I was a little girl? I loved macrame because it was simple. But I just, my hands, it's hard with my hands even though right now I'm having a good bout. I'm just not a crafty yell. I don't have the patience. I have the patience of a nap. That's why I'm not a big movie watcher. Aren't they all on the chat? Oh my gosh. This birthday. Hi Barbara. Barbara baby. Click on Barbara's channel to everybody please. Who was it that just, Liz just bought one of Barbara's paintings. Yay for that. I was so excited to see that because I know Liz would appreciate your work. She's another one I love. Love Liz. She's struggling right now too. No, I'm just not, I'm just not, I don't enjoy doing things with my hands. The best way to get right to the subject.  Isn't that interesting? And then grow up to have problems with my hands. I like, I like my photography. My camera's broken. Photography's my thing. Liz is wonderful. She's just wonderful. She's, she's funny. We talk a lot and she just, she gets my jokes. She gets my jokes. Thank you, Lou. You guys for listening, I greatly appreciated it. I just needed to talk. I was sitting, like I said, I was sitting here watching the clock. Go minute by minute. Just needed to talk. I actually stood outside and talked to my neighbor for a while. A gentleman with a very thick Spanish accent. I did, it was kind of tough, but not bummed Barbara. Yeah. I'm up even though I maintained. I'm not liking the way my body feels. I'm not liking what I see in the mirror. I'm just, you know, you know how my ebbs and flows. I wish I could be you, Barbara. I wish I could just, because I can just see you every morning getting up, being your beautiful self and having your coffee or your tea and taking a shower and doing your beautiful hair and your beautiful makeup and, but Barbara, that's the problem I'm having today is when I get like this, I do for the most part, look for the positive. You guys bring it to me. I see it. And you know that about me. And today I don't feel like it. I'm no stinkin positive. And that's why I do because you guys are awesome. I should eat though. I haven't eaten anything and I'm getting a little hungry. We had a car chase today and the news I had just fall.  I know, Barbara, and you guys are always so awesome about it. I'm not down. I'm disc, I'm not content. I'm not content. I'm squirrely. And I don't like the way my body looks and feels and only I can do anything about that. And I'm, I do appreciate you guys though and I'm sorry if, I don't drink tea. I'm not, I'm not like everybody else which just kind of puts me in a weird spot too. I don't know what I don't want. I'm absolutely drinking my water, Laura. I'm always and no matter what, I'm drinking my water. Oh, thank you, Susan. I'm always, I'm always posting pictures of something and everything and I feel like it's a big, oh, I forgot to show you guys my earrings that Denise sent me from Dish with D. She sent these to me. They're me. They are so me. I would have picked these out myself, aren't they pretty? I'm going to send her a pair back. We've got a thing going back and forth. Thank you, you guys. It is, maybe I'll go sit out there and do my, my paint by numbers. Put on some, now I need to decide if I want to listen to some Boston Nova or some classic rock or some classical. What's going to really, I think maybe I need to look like just blast some, some like, some Jethro Toll or some Zeppelin or something like that. Thanks, you guys. I'll let her know or next time you see your Taylor, you saw them. I love them. We're both the, got our things for earrings and now flowers. So yeah, I think that's what I'll do. I'll just put on, it's tomorrow weekend so we can party and like it's 19, no wrap here, no wrap, no polka, no country, but maybe some Sergio Mendez.   Hi Diane, welcome.  Tamsen, again, with the theme of this, with this particular video, I can't stand either of those bands.  I've got it all laying here in front of me. I've been on a huge Oscar Peterson. Uh, does anybody here listen to Oscar Peterson? There will be an extra point for anybody who listens to Oscar Peterson or at least knows who he is. I'm going to sit here and I'm going to wait until one person says Oscar Peterson. I don't know who Zach Brown is. I don't, if he's country then he's two country. Who you did? Did you like him? Is he not the most amazing piano player you've ever heard? Margot C. You and me. Oh. He's just, was just amazing. And then I watched a video this morning where he was doing a duet with another piano player. I know his name. I can't think of it.  He's just amazing. If you guys, you know that stuff on YouTube too you can check out. See how I brighten up music. Music's my jam. Music and photography. Extra point for Lou. We've been doing it then. Gene Harris. I found a lot of my Gene Harris and my Diana Carl. So I got it all, I got it all down there.  And that's okay. I love it. But he's so funny since I introduced him to it. He's got his favorites. That's for sure. What did she say? Extra point for Lou. It was really amazing Margot. I'll tell you what, let me tell you this and then it'll lead you to it.  Shoot, it's buried over there. Hold on. I'll message you to you. It's a song by a different composer but on YouTube when he plays it and then all the songs that come on underneath that or I shouldn't say songs because it's not really songs, composers. Compositions. They're just amazing. I'll message it to you. Let's see. Oh yeah, Linda's absolutely. That goes without saying. And her with Ann Savoy and her duets album, Linda Ronstadt's duets album is phenomenal, you guys. All right, guys.  I sincerely appreciate you guys coming in here and listening to me whine, offering up all the help for me to just deny it. She lost it. You guys. I'm going to go message her. Thanks, you guys. I'm going to go reach out to my friend Liz. I'm going to have a bite to eat and then I wish I could think of that one song I want. Quiet, quiet. Whoever wants to stay, let me hold on a minute. Let me see what it is. Because it's on my YouTube, on my iPad, and I watch it. Oh, it's quiet. Quiet stars on a quiet night or something like that. All right, I'm going to go eat. I love you all. You're fantastic. Next time I see you, I promise I will be better because I won't come back until I am. And if anybody needs to message me, come on over to Instagram. Do you know, Marco, what I'm talking about? Listen to Oscar because it's a Sabine song, but Oscar Peterson kills it, man. Okay, I'll catch you kids later. Hugs to you all. There you go. Lou, Lou, I'm going to have to start paying you, lady. Bye, guys.