5 Weird Texting Habits of Narcissus Narcissus texts differently than other people They have specific texting habits and that can help you to identify them So I am going to give you 5 Weird Texting Habits of Narcissus 1 Love Bombing The Narcissist will love bomb you through text When you first start texting them They will send you a lot of messages They will reply very quickly They will send you long paragraphs To make it seem like they are really interested in you And it may be quite overwhelming They will express themselves in a determined and confident way They will tell you you are perfect And that they can be perfect for you Until it leaves no doubt in your mind that they are pursuing you And that they want to be with you They will try to make themselves perfect for you They will try to be everything that you want and need They will try to make you feel like you have met the right person for you 2 Self-centredness They will tell you about everything they are doing They will talk about their problems But they won't ask about you They won't ask about how you are doing Or how is your day being You just end up helping them with their problems And you may feel like you are doing the right thing But eventually You start to realise that something is not right Because it's just all about them They are not making an effort to share your experience They are not trying to put themselves in your shoes 3 3 Future faking They will use a detailed vision of the future To facilitate bonding and connection In order to get what they want in the present But with no intention of following it up with action They are not emotionally capable of making it a reality But they will tell you whatever they think you want to hear To lure you in so that they can establish their control over you But at some point you will realise that it is all a lie And then they will ghost you Ghosting The narcissist will ghost you They will stop responding to your texts They will dismiss you as unimportant Because they know you are onto them They know you are starting to figure them out And they know that you are expecting a return on your investment So they will disappear Because they were never serious about building anything with you anyway They just wanted your attention and admiration They needed supply But if you question them on it They will deny it They will lie to you They will gas like you They will blame you After they were the ones who lured you into it They made you believe that you were perfect for each other They pulled you into their fantasy world And then they detached themselves from it Grandiosity They will try to impress you They will show an unrealistic sense of superiority They will brag and boast about their accomplishments And how they're better than other people They will say that everyone else is stupid They will send selfies to you In an attempt to entice you To get you attached to them But they have no intention of attaching themselves to you So when all is said and done You end up all alone in this prison that they have created for you You're left alone with your own thoughts You ruminate about everything they said You remember all of the things they promised you And then you look around and realise that you've been duped You've been taken for a ride Because you gave out all of your time, energy, love and passion But in the end you got nothing Nothing but a mirage An illusion An unrealistic hope and wish that can never be achieved Thank you for watching I hope this video resonates with you Please like, comment, share and subscribe If you would like to donate My PayPal link is in the video description Coach the Inquiries You can email me at coach.naqsurvivor.uk Thank you for watching and I'll talk to you soon


The guest, my biggest father, was a little boy, so he said he was going to retire and we should have something in common.


This is Witchspace news for Friday the 19th of April 2019 I'm CMDR Buur In this weeks news Frontier developments show off the next content update for Elite Dangerous We get our first look at the interstellar initiatives replacing community goals and the Elite community meet is this weekend If you enjoy our content then please remember to hit like and subscribe and click the little bell icon to get a notification next time we upload a new video. You can also find us on Twitter, Patreon, Discord and Instagram, links to all those are in the description below Frontier took to the live streams last night to show off the content update that's being launched on Tuesday of next week. The update is packed with features primarily aimed at newer players but it also contains some significant quality of life improvements that will affect all commanders of particular note in the demonstration with a new advanced docking computer that is able to launch as well as land your ship at a space station, spaceport or megaship. The supercruise assist module that acts as a kind of cruise control in supercruise note it's not an autopilot and it also puts your ship in a very cool little orbit once you reach your planet of choice but there's also some new interface icons to better indicate what activities are available at a given location or what threat level it might pose The ship will also now have the option to auto throttle down when exiting hyperspace preventing you from crashing into inconveniently placed stars when reading discord chat or answering the doorbell. There's also more in depth trade tools and a complete pilot's handbook in the cockpit that details how to do just about everything in the game. As I've said before anything that brings new players into the game and then more importantly helps keep them there can only be a good thing for all players and for the future longevity of this glorious thing that we all love so much. As part of the live stream last night Frontier also played a really nicely produced promotional video for what they're calling interstellar initiatives. These initiatives are a replacement for the venerable community goal system and provide a more curated developer supported experience that will in some cases at least change parts of the elite dangerous galaxy forever. The first initiative will see players voting on where they'd like to see a kind of ferry service megaship heading in order to gather guardian relics and then helping in the creation of that megaship. The first interstellar initiative called Bridging the Gap will start on April the 29th and is expected to last for about a month. It's fantastic to see new life being injected into the community goal idea and I'm excited to see groups of players all gathered around a common goal again this time with genuine galaxy changing consequences also megaship bus service. How often does it run? Will we see other galaxy spanning ferry services? So many questions. The full promo video featuring the legend of Will Flanagan is linked below. It's quite excellent and well worth a look. A quick reminder that the elite community meet this year's first major elite dangerous community gathering is happening in Manchester in the UK tomorrow. Unfortunately, we can't get there this year. If you are going, please take all the pictures that you can and grab all the video that you can so that the rest of us can at least share part of the experience vicariously with you. And finally, we're very proud and excited to announce that we've recently been accepted into the Frontier Development's affiliation program. If you click below into our video descriptions, you'll now find links to buy games and in-game extras from Frontier that will help support this channel going forward. That's it for now. Thank you so much for making it this far. We'll be back later this week with more videos. Until then, oh, seven commanders, follow the Greens on the way out and do keep clear of the toast rack. We very much look forward to seeing you next time.


So yesterday, we discussed the ADHDM equations, which describe the modulate space of framed instantons. So I didn't introduce the terminology yesterday. So instantons are the solutions of the equation, the self-dual part of the curvature is equal to 0. And framed means that we identify the gauge equivalent solutions of these equations with the requirement that this gauge transformations approach 1 as x goes to infinity. So the resulting space, which I denoted by m plus k, I should probably introduce the second label for the rank. So we'll be talking about the u-alansantons. So this space has an SUN symmetry, which comes from the possibility of performing a constant gauge formation. So I claimed that there is a certain compactness theorem, which one can prove in this case, which is useful in analyzing the partition function. So let me state this theorem. So in addition to the SUN symmetry, we also have the action of the group u2. So this is the group of rotations. In fact, let me work with this spin cover. So these are the rotations of R4. And the equations, the original equations are invariant under all symmetries of R4. In fact, they're invariant under the conformal symmetries. But we introduced certain deformation when we discussed the compactification of the modular space, which was adding to the so-called real map equation, a constant. So this deformation breaks the rotational symmetry from a spin forward down to u2. This deformation uses a specific complex structure on R4. So in order to single out one equation out of three, so the other two were combined into the homomorphic equation, I have to distinguish a specific pair of homomorphic coordinates. And so this is the result of the symmetry. So it basically acts on b1 and b2 as a doublet. i is invariant. And j transforms as a determinant representation. So if t is in u2, you act. So t is a 2 by 2 matrix. You act with this matrix on b1, b2. And you multiply j by the determinant of this matrix. So the commutator of u1, b2 will be multiplied by this determinant. And that's why this equation is compatible with the symmetry. Now, so this is a big group. And in applications, we only use its maximal torus. So let me denote it as the torus of rotations. So if you look at the fixed point, set of fixed points, under the maximal torus and the maximal torus of the group of the conjugation formations, so those fixed points were classified yesterday. And they were simply isolated. There was a finite number of those points. T rotation cross t gauge fixed points and tuples of partitions whose total size is equal to k. So each partition is a young diagram. So this is lambda alpha for one of the values of alpha. And the size of the partition is the number of squares. So each square, remember, each square was a vector in the k dimensional vector space. And so if you have a vector, if you have a square with coordinates i and j, j goes here, i goes here, then this vector, so alpha ij, this is the vector b1 to the power i minus 1, b2 to the power j minus 1, i applied to the one-dimensional vector, the eigenvector of the transformation from the global gauge rotations corresponding to the eigenvalue e to the i alpha, e to the alpha. So this is the situation when you have the full maximum torus at your disposal. But it may happen that the problem you are solving forces you to consider special specific values of the equivalent parameters. So if these parameters, epsilon 1, epsilon 2, are not generic, you may not have the full torus at your disposal. So remember, OK, so this is my two-dimensional torus. And epsilon 1, epsilon 2 give me a direction, a specific slope for the one-parametric group, which is generated by this transformation. So if I start iterating it, put some coefficient s. I will generate one-parametric subgroup. So if the ratio epsilon 1, epsilon 2 is not rational, then the subgroup will actually fill the torus densely. And that means that you have the whole torus at your disposal. And so the set of fixed points is precisely that. But if this ratio is rational, then you actually have only one-dimensional symmetry group. So then this spans s1 with u1 embedded into u1 cross u1 at some rational slope. And so since this group is strictly smaller than the maximum torus, the set of fixed points will be strictly larger and need not be isolated. So these points are isolated. So let's say this is my modular space. So this is the picture of these fixed points. So these are just examples of these fixed points for the case of general epsilons. But if the epsilons are non-generic, then what happens is that these points become continuous submanifolds. So this is what happens for non-generic epsilons. In fact, if you further reduce the symmetry group, namely you start playing with Coulomb parameters, the parameters which the generators of the global symmetry transformations, your set of fixed points becomes even larger and actually can go off to infinity. So this is what happens if A alphas are non-generic. And that's kind of bad for the partition function. That means that you actually have integrated out something which is massless, can run away to infinity. And so that means that that signals actual singularity of the partition function and requires further analysis. But I claim that if the differences between the Coulomb parameters, so the different eigenvalues, are not of the form epsilon 1 times a positive integer plus epsilon 2 times a positive integer, then this never happens. The fixed point set is always. So let me prove it. Yes. Yes. So there is a case when you specify A's to the value that it front-case number of data. Say what? So you specify A's parameter to front-case number of data or are these choices? No, no, no, not in this situation. Not in this case, but in general, the rest of it. So this is, so if you, when we'll consider a theory with matter, we'll have other types of parameters which are the masses. And so those correspond to, those can be used to truncate the size of young diagrams. So those correspond to the possibility of adding new equations to the set of equations I've written before. And so when you add more equations, you restrict the modular space. So what happens then is that your modular space becomes strictly smaller. So you go to vertices. Vertices? Vortices. It could be, sometimes it could be interpreted as vortices, sometimes it could be interpreted as something else. But that's what you do. So you have extra equations from extra matter. We'll do that momentarily because we will need to consider theories which correspond to asymptotically super conformal theories. And for those theories, the modular space which we're integrating over will be of virtual dimension 0. So ultimately, you actually want to count points, some points. But right now, these quantities appear in the denominator of the localization formula. And so they lead to the poles if something is non-generic. What Maxim is asking is in theories with matter, you have numerators in the localization contribution. And so those can vanish, which simplify life sometimes. OK, so the reason I want to present this compactness theorem is because it's a kind of baby version of a more complicated story, which leads to the non-perturbative Dyson-Schwinger equations. So let me assume from the very beginning that the parameters epsilon 1, epsilon 2, so these are infinitesimal rotations of spacetime that they are integers. So I can rescale them. If the ratio is rational, I can rescale them so that they're integers. And so what we are looking at, we are looking at the solutions of the following equations. This is the infinitesimal version of the equations which I wrote yesterday, hopefully without the typos which I had yesterday. OK, we'll see that in a moment. So what this equation says is that if I do infinitesimal rotation of v1 and v2, I can undo that by the infinitesimal UK symmetry. And now, remember, there was a statement that instead of solving this real moment by equation with zeta and then dividing by the group UK, you can actually forget about this equation. You impose a stability condition, which is phrased in holomorphic terms. And then you divide by the group of general linear transformations. So that allows you to think of phi as a general just k by k matrix. So then i is acted on by the, so let me write this like this. This is phi a. So a is the diagonal matrix with eigenvalues a1 an. i is the map from n to k. So phi is this compensating transformation which acts from the space k to itself. And finally, epsilon 1 plus epsilon 2 j minus j a is equal to, sorry, minus aj is equal to j phi. So this sum epsilon 1 plus epsilon 2 is the linearization of, it's a logarithm of this determinant. So what this equation says is that the infinitesimal rotation and the global gauge transformation can be undone by the local GLK transformation. So this defines the fixed points, the set of fixed points, which we want to study. And so what we said yesterday was that from this equation, well, we know that i cannot be zero because if i were zero, then this condition would not be satisfied. So it means that at least one component of phi is non-zero. So it means that at least one of the eigenvalues of phi has to be equal to the eigenvalue of a. And so the spectrum of phi has the form. So this part comes from acting on the eigenvector of phi with b1 and b2. So this, like in supersymmetry, by looking at this equation, if psi is the eigenvector of phi with eigenvalue lambda, then b1 times psi will again be an eigenvector using this equation. So I'm using that the commutator of phi with b1 is equal to epsilon b1. So I can pull it like here. And so what I get is lambda plus epsilon 1 times b1 psi. So it means that if this vector is not killed by b1, then it is an eigenvector of phi with the new eigenvalue, which is lambda plus epsilon 1. And so by acting with b1 and b2, you generate these shifts. And from this it follows that j has to vanish because for j, the eigenvalues go in a different direction. You see, j wants to subtract from some combination of epsilon 1 plus epsilon 2. So this argument works when the signs of epsilon 1 and epsilon 2 are the same, so if they are of equal sign. So for p and q both, let's say positive or negative. I mean, by scaling I can always assume, let's say, that q is positive. And I'm assuming that both p and q are not equal to 0. So that's important for compactness theorem, otherwise it will not work. So this is an important assumption. So if this is the case, then by these arguments you conclude that j is 0. If the signs on p and q are opposite, then you cannot use this argument because, well, epsilon 1 plus epsilon 2, for example, could even vanish. And so it wouldn't matter. So that will not be enough to conclude that j is equal to 0. We'll need to argue differently. But let's first dispose of this case. Actually, it's more complicated than others. So the idea is the following. Well, first of all, if a's are generic, we can actually split the space k into the subspaces such that phi restricted on the subspace k alpha is simply. And under my assumption that epsilon 1, epsilon 2 are integers, positive integers, well, this is some non-negative integer. So the spectrum of phi consists possibly of several arithmetic progressions, which start at different complex numbers a alpha. And so we can separate the space into the orthogonal summands corresponding to different values of alpha. So it's actually enough to understand it for a fixed value of alpha because they don't talk to each other. And therefore, so what I want to do, I want to find the bound on the norm squared to find a bound on the norm of all these matrices. Since we just proved that j equals to 0, this makes life easier. And in fact, the norm of i, you can compute from my equation, but we'll not do that. What we will do is the following trick. So now I work at fixed alpha. And so by shifting a, I can assume that a alpha is equal to 0. It doesn't really matter. So my equations now have the following form. This is 0. I forget about this. This is p and this is q. So I want to find some bound on this norm. So we now can split the space k alpha into, in turn, some of the subspaces on which phi has eigenvalue k alpha n is the kernel of phi minus n. So n is non-negative integer. So the question is, well, so this integer can be arbitrarily large, not too large, because in the dimension of k is fixed. So we cannot have more than k, little k values for this n. But the danger is to have the matrices be with matrix elements which are arbitrarily high, arbitrarily large. So what we want to show is that even though these are non-isolated fixed points, they actually sit inside some big ball in this modular space. So this is what we want to find. We want to find the bound on the size of this ball. So let's compute. Let me define the following quantity, delta sub n. It's a trace over the subspace k alpha n of b1, b1 dagger, plus b2, b2 dagger, plus i, i dagger. So you see, b1, when acting on the space k alpha n, sends it to the space k alpha n plus p. And b2, this is exactly the repetition of this argument. And b dagger acts in the opposite direction. So the norm which I wanted to compute is just a sum of these partial norms. So now we need to find some identity for these norms. And what we can do, we can, using the real moment map equation. So here, for this normal analysis, I'm using the unitary picture. So I'm only devising by the unitary to summations UK and using the real moment equation. So from this equation, I'm using the j is equal to 0. So what you can see is that this guy is almost everything that you see on the left-hand side. What you see on the left-hand side is bb dagger plus i, i dagger minus b dagger b. So we can rewrite this as 1 over zeta trace k alpha n zeta times 1 plus b1 dagger b1 plus b2 dagger b2. This is just the algebraic manipulation. And now using this property, well, you see that now b1 takes this space and pushes it further into the space with eigenvalue n plus p. b2 does it with the map set to the space n plus q. So you can estimate, use the fact that the trace is cyclic. So this is the dimension of the space k of n from this unit. And then you can estimate it by delta of n plus p plus delta n plus q. So you get some kind of sequence of telescopic inequalities. And eventually, so you use this inequalities, you effectively increase the value of n until you get to the point when this space k alpha n is simply 0. So k alpha n is 0 for n much bigger, let's say, than k. Just bigger than k, yes. So from that, you can get an estimate. And the conservative estimate is rather high, but it doesn't matter. It's finite. Could you repeat again how you found b1 dagger b1? OK, so this is the definition. So this is the real moment equation. So now I want to, I'm claiming the following. So the trace in the space k alpha n of b1 dagger b1, I want to rewrite as a trace of the space k alpha n plus p of b1, b1 dagger. So normally under the trace, you can exchange, but you have to be careful in which spaces the operators act. So b1 acts from k alpha n to k alpha n plus p. And so now, since in this formula, k alpha n plus p appears outside, that's the space at which I'm taking the trace. And so this is less than or equal to the same quantity where I'm adding something which is non-negative, namely b2 b2 dagger plus ii dagger. So this is a very conservative estimate. I mean, this is small, and I'm increasing it by, which is, so if I divide by zeta, this is delta n plus p. So you can introduce generalized Fibonacci numbers for this problem to package this. So usually Fibonacci numbers are solving the equation of that kind, I think. So the generalized Fibonacci, so you start with f1 equals 1, f less than 0 equals to 0, and then you impose this condition. So just like for ordinary Fibonacci numbers, you can write a formula using the eigenvalues of the two-by-two matrix. So here you solve the Fermi equation lambda to the p plus lambda to the q is equal to 1. So it has a max of p and q solutions. And so in terms of, so you can write the formula for f of n as a sum over all solutions of this equation, lambda to the n, and some prefactors, initial conditions. The point is that it shows that these numbers, they grow, but at most exponentially. So there is some constant by which you can bound them. And from this inequalities, so the inequalities of that type, delta n is less than k plus delta n plus p plus delta n plus q. Again, I'm making conservative estimates. And increasing the right-hand side, it follows that delta n is less than equal to the sum k n plus prime fn plus prime plus 1, which means that this can be bound by k exponential. Say it again? Should we see the minus delta? Yes, but that's probably the same. You can multiply by, oh, OK, maybe I won't. Right, so it follows that the sum over delta n has some finite bound. OK, so that's the way this argument works. So a little bit of work, but not too much. And for now when p and q have opposite signs, then the following trick works. If pq is less than 0. Well, first of all, you need to prove that j equals to 0, because now you cannot use the eigenvalues. I claim it nevertheless. So how do I prove that? Well, let me assume now that n is one-dimensional. So this is what we did here. We split. So the problem splits into n-Abelian problems when the Coulomb parameters are generic. And so let me do it now in the Abelian case. So the equation B1, the commutator of B1 B2 plus ij equals to 0, let me take the trace of this equation. So it will imply that ji is equal to 0. So now, you see, normally I would have to put a trace here. But trace of the 1 by 1 matrix is equal to the matrix itself. So this is because i was acting from n to k. So this is the n to n operator, which is one-dimensional. So it's a number. So j annihilates the image of i. Now, if I multiply this equation by B1 or B2, in fact, if I multiply it by any power of a linear combination of B1 and B2 and take a trace, so this is 0 because of the moment of my equation. But now, if you look carefully, you can rewrite this as the commutator of alpha B1 plus beta B2 gamma B1 plus delta B2, provided that alpha delta minus beta gamma is equal to 1. Now you see, you have the power of the matrix times the commutator of this matrix with something, and then you take a trace. So that means that that part of the trace is 0. And so you are left with j multiplying alpha B1 plus beta B2 with power n for any alpha, beta, and n. So if you knew that B1 and B2 were commuting, that would imply that j vanishes polynomial of B1 B2 applied to i, which is by the stability assumption is equal to k, which means that j vanishes on k, which means that j vanishes. But B1 and B2 may not commute, but that's OK, because now, if you try to insert a commutator of B1 B2 into some word involving B's i and j here, you can use the moment of my equation to replace it by as a product of two words. And so by induction, if you prove that j annihilates all words in letters B1 and B2 acting on i up to the, let's say, depth k, then by this argument, if you have a longer word, and then inside you have a commutator of B1 and B2 that vanishes. That means that inside this, when you close any word with B's by j, inside B's have actually commute. And that means that this argument works, j vanishes. And then it means that actually they do commute, because if j is 0, then B1 commutator of B2 is equal to 0. OK, so that's nice, but that's still not the full story. But actually, the full story is almost over. I claim that if this is the case, if epsilon 1, epsilon 2 are in rational relation with opposite signs, then actually the corresponding fixed point is invariant under the full two-dimensional torus. So if pq is less than 1, actually it follows that any epsilon 1, epsilon 2 fix this configuration. And this is how we prove it. Look at the following operator. So suppose p less than 0, q is greater than 0. So I will raise B1 to the power q, B2 to the power minus p. If p is negative, this is positive, so it's an allowed operator. Well, from these fixed point equations, it follows that phi come used with n, because the weight of this combination is actually 0. So B1 has weight p, and B2 has weight q. So p times q minus p times q is equal to 0. And so n, so curly n. So curly n is a nilpotent operator which commutes with u1, which is generated by this epsilon 1, epsilon 2. Therefore, so if I decompose k as a sum of the eigenspaces, so these are the eigenspaces of phi. So u1 is generated by phi, in the sense. Each space kn is invariant under the action of this nilpotent operator. And then there is a theorem by Jacobson and Morozov, which says that a nilpotent operator can be included into the SL2 subalgebra. So it follows that there exist operators h and n star such that the commutator of nh with n is n, commutator with h with n star is minus n star, and n with n star is twice h. And so this h becomes the second grating, so that kn and this further split into the knm. And that means that now you have the u1 cross u1 symmetry, so h and phi. And then we are back in business as yesterday. So we have the full two-dimensional torus of symmetries. So that proves that as long as both epsilon 1 and epsilon 2 are non-zero, so this is what I used, the set of fixed points will be compact. OK, so now let me say something new. So so far, we discussed pure n equals 2 d equals 4 theory. Let's add some matter. It is convenient to, in some sense, so if you want to add matter while preserving the asymptotic freedom of theory, and so if we only add matter in the fundamental or by fundamental presentation if you have a product of unitary groups as a gauge group, then the choices are actually limited. So you only can work with quivers, quiver theories. And the condition that the theory is asymptotically free or asymptotically conformal means that these quivers have to be of the ad type, either a fine linking diagrams of ad type or finite linking diagram of ad type. And that means in turn that the matter content of the theory can be obtained by certain orbital truncation from the n equals 4 theory. So it's all can be gotten by Orbe faults of n equals 4 superannual theory with maybe much bigger gauge group, possibly with some massive deformation. So it could be n equals 2 star theory. And so let me simply explain what happens to the adHM data when we pass from n equals 2 theory to n equals 4 theory, because everything else will be straightforward orbital truncation. Now, unfortunately, unlike the pure instant on case where there was this nice reciprocity theorem mapping adHM data to gauge fields and back and forth, in the n equals 4 case, I'm not aware of such theorem, so what I'm going to present will be really construction by analogy, not systematic derivation. So n equals 4 gauge theory has, in addition to the gauge field, it has six scalars, which we can organize into the complex scalar sigma and its conjugate as n equals 2 case. And the other four, I will twist, let's say in the Waffa-Wieden way, and so they will become a scalar and the cell dual 2 form. And so n remains the gauge field. So this is done in order to give a certain mathematical structure to the localization equations, which we will be analyzing. Again, I'm working on flat space on R4. So in fact, I'm not changing my theory in any way. You can actually just use some flavor indices to label your fields. But I will single out one of the supercharges out of 16 supercharges of this theory. And from the point of view of that supercharge, it's convenient to write the theory in this way. So the equations in which we will be localizing our theory deformation of the instanton equations, I think it starts like this. So you add the commutator of the cell dual 2 form with the scalar, and then there is another term where with some gymnastics you can produce a bilinear map from cell dual 2 forms into cell dual 2 forms. And then there is an equation that da. So this is just cross-project. Yeah, it's some kind of cross-project. It's the, so if you think about the cell dual 2 forms as a triplet of SU2, there is a, well, it's a triplet of SU2 times the adjoint representation of the Lie algebra. So you have the, that's not three-dimensional, so three times the Lie algebra, right? There is a Lie algebra structure there. There is also a structure, there is also a structure constant. Anyway, but, and then gosh, so what is the, so I guess, so this, here I want to write an equation of this type, I think. So this equation is valued in, again, in cell dual 2 forms, valued in adjoint bundle. And this equation is valued in one forms. Yes, well, on R4 it's kind of tricky because it's not compact, so you have to be careful. Presumably with some vanishing conditions at infinity, these equations imply that B and C are actually equal to 0. But the point is that now that you have more fields and more equations, and you produce the supersymmetric, I mean, the path integral over these fields and their superpartners, even though physically it will reduce to the same, to the integral of the same modulus space of instantons because on solutions B and C vanish. So these equations imply that B and C equals to 0. But since you use more fermions to impose these equations, the actual measure of integration over the modulus space of instantons will be not 1, as in n equals to 2 case, but it will be essentially the Euler class of the tangent bundle or cotangent bundle. So you upset the sigma and sigma bar to 0? Sigma and sigma bar are the, so they are in the equivalent multiplet. So if you like, there is also the equation of this form. This part is not changed. I'm not, this is, so, I mean, I'm working in the paradigm of fields equations and symmetries. And so this is the part of the symmetry multiplet. These are the equations, and these are my fields. So what's nice about these equations is that they actually kind of match the set of fields. Previously, we had the fields were one forms, but the equations were self-dual two forms, and the symmetry were scalars. So there was some mismatch. You were comparing one forms to, so in n equals 2 case, you're comparing one forms to self-dual two forms and scalars. But here, we have one forms here, one forms here, and we have self-dual two forms here, self-dual two forms here, and we have a scalar here and a scalar in the symmetry. And so this is an example of what digraph Moore called a balanced topological theory. So we can mimic, right, sorry, one more thing. So these equations, you can actually derive from some kind of super potential.  So if I vary this functional W with respect to B, I'll get the first equation. And if I vary it with respect to A, and I didn't make a mistake, maybe I didn't make a mistake, I will not get the second equation, but the second equation means that the gradient of, so maybe I made a mistake in the first equation. So the gradient, so the variation of W with respect to A is equal to the gauge of summation generated by C. So maybe I don't need this part. And the variation of W with respect to B plus is the gauge of summation of B generated by C. So these equations are kind of generalized gradient flow equations. So that's what we will mimic on the ADHM construction. So I claim that in order to produce the same measure, which I just erased, the Euler class on the module space of instantons, the ADHM data will have to be supplemented by two more matrices. And the equations will be now, so we'll have more equations. So the first equation, so I start with the ADHM equation as a complex map, but then I add to it the conjugate of the commutator of B3 and B4. So somehow you should see part of it in this structure. You see, if I choose complex structure on R4, then B plus can split as B2 0 plus B11 omega plus B02, which is conjugate. And so B2 0 is like B3, and B11 omega plus IC is like B4 in this notation. And then I will have equations. B1 commutator would be 3 plus B4 commutator would be 2 dagger is equal to 0. B1 B4 plus B2 B3 dagger is equal to 0. And finally there, or not finally. So then there is equation B1 B1 dagger plus B2 B2 dagger. And so I can size with the sum of all four commutators plus i i dagger minus j dagger j is equal to zeta. And finally B3 i minus plus j B4 dagger is equal to 0. And B4 i minus j B3 dagger is equal to 0. So you can compare these equations to the equations which follow from the superpotential. So the superpotential will not respect the symmetry of rotation of four matrices. It will be something like B commutator B1 B2 plus ij. And the equations will be that the gradient of w respect to the field commission conjugate is equal to the gauge variation of this field x generated by B4. So for example, if x is B3, the derivative of w respect to B3 is the first part of the addition equation. And then the gauge transformation of B3 is a commutator of B3 with B4. And so you take commission conjugate. So that's the first equation. And if you take the variation respect to j, you'll get B3 times i. So B3 times i. And this is minus j B4 is the gauge variation of j. So that's the logic here. Again, so here is a vanishing theorem which is easy to prove that these equations imply that B3 and B4 are equal to 0 when zeta is positive. But now if you set up the calculation, the integration of 1 over the modulus space of solutions of these equations. And now we have more symmetry. So the symmetry is u4 is SU4. Sorry, it's not quite SU4. It's the maximal torus of SU4. So you can multiply matrices B a by any phases provided that the product of these phases is equal to 1. So the first equation, you multiply the commutator of B1, B2 by T1, T2. And this commutator of B3 and B4 is multiplied by T3, T4, which is inverse of T1, T2 by this condition. So it means that compared to the n equals 2 case where we had two epsilon parameters, the parameters of rotation, here we have three parameters. So we have epsilon 1, epsilon 2, epsilon 3, and epsilon 4, which is negative epsilon 1 plus epsilon 2 plus epsilon 3. The physical meaning of these parameters is, so epsilon 1, epsilon 2 correspond to the rotations of spacetime. And epsilon 3 is the mass of the adjoint hypermultiply. So this is the mass of adjoint. So now if you do the localization computation for the integral of the module-specific solutions to these equations, which means literally what? It means that for every variable here, so you have six types of complex variables, or eight Hermitian matrices and four rectangular matrices, you introduce the fermionic variable of the same type. And then for each equation, you introduce a pair of boson and fermion, the Lagrange multiplier for the equation and its fermionic partner. And then for the symmetry, so the symmetry is UK, you also introduce a small multiplet like sigma, sigma bar and eta. And then you write the standard homological field theory Lagrangian, which enforces these equations and takes care of the symmetry, which would deform using this global symmetry and the global framing symmetry, which still acts on i and j. Then the result would be the partition function, which is again the sum over all n-tuples of partitions. And then you'll get the product of the eigenvalues of phi acting on the tangent space to the modular space of instantons at the corresponding n-tuples of partitions. Let me call them lambda. So these lambdas have the differences of the Coulomb parameters plus integral linear combinations of epsilon 1, epsilon 2. And now you'll have in the numerator the same eigenvalues shifted by epsilon 3. And even though it's not obvious, this expression is actually symmetric under exchanges of epsilon 3 and epsilon 4. So these equations were completely symmetric. But there was no preferred, I mean there was no, well I can exchange b3 with b4 and b4 with negative b3. The equations remain the same. And so there should be symmetry which exchanges these parameters. And that means that when I call epsilon 3 a mass of the adjoint hypermultiple, I could equally well call epsilon 4 the mass of the adjoint hypermultiple. The notion of mass is slightly ambiguous in the mega background. So you say that in this case the proof is not known. But if you can construct the explicit form of the gauge field, et cetera, from given this solution. No, this is what I don't know. I mean, oh sorry. Of course, the equations, since the equations imply that b3 and b4 equals to 0, the equations reduce the old adjoint equations. So of course, at the level of, at the set theoretical level, of course, you have the correspondence. So you get, it's the old instant which you get from the old b1, b2, and g. But these equations have the meaning themselves. Even though the solutions are simple, but the equations are more important than the solutions. Somehow, I would like to have a way of connecting the equations to the equations without assuming that the equations imply this thing. Because also, if you, instead of r4 you work on some other manifold, the vanishing theorem does not hold. So then you have, anyway. So from this, by looking at these equations and observing that these equations have a symmetry, let's say, SU2 symmetry which rotates b1 into b2, you can perform some orbit faults which will replace this r4 spacetime by the ALE space. That's one option. There is another SU2 symmetry. So the symmetry which these equations actually have is the product of SU2 cross u1 cross SU2. So this SU2 acts on b1, b2 as a doublet. This SU2 acts on b3 and b4 as a doublet. And this u1 scales simultaneously b1, b2, and scales in opposite way b3 and b4. So if you have a discrete subgroup of this SU2, or if you have a discrete subgroup of this SU2, and so there are AD classifications of such subgroups, you can impose additional equivalence condition under the subgroups and produce other equations which will produce the ADHM data for quiver gauge theories. So from this, you get for free, essentially. You get, so these are quiver gauge theories on r4, or if you also impose the orbital projection by gamma 1, it will be quiver gauge theories on r4 mod gamma 1. But now you can say, you can observe one thing, that here you have some kind of symmetry between 1, 2 space and 3, 4 space. And these two procedures look symmetric, but the equations are not quite symmetric because you have ij, which couples to b1, b2, but nothing couples to b3, b4. And so you can try to fix that symmetry by introducing one more set of fields, more data. So we introduce new space and tilde. And maps i tilde, which maps n tilde to k, j tilde, which maps k to n tilde. And modify these equations. So here we add additional terms into the real moment map. And then now symmetrically, so b3 wants to annihilate i. So b1 wants to annihilate i tilde. So now life looks truly symmetric. So what these equations describe, they describe the instantons, which live on the intersecting worlds. So in the brain world picture, in the brain world picture, you have n brains, which span r4 with complex coordinates z1 and z2. And then there is n tilde brains, which live on c2 with coordinates z3 and z4. And these two complex planes intersect at one point. And this is the point where there is some interaction between this data. So these equations imply that separately b1, b2 plus ij equals to 0. So this follows from these equations. And separately b3, b4 plus i tilde, j tilde is equal to 0. So this is the adhm equation for instantons. So you get un instantons on c2, 1, 2. And you get un tilde instantons on c2, 3, 4. But you don't know in advance what is the instanton charge of the solutions. So you may have the solution where you have charge k1 on this plane, k1, 2, and charge k3, 4. On that plane, complex plane, so on that four dimensional space. And the sum of these charges is greater than or equal to k. So some of these instantons are actually shared by both planes. And so in what sense are they shared? Well, this is algebraic condition. So what we need to do, we need to look at the space generated by applying to the image of i polynomials in b1, b2. So this is the space k1, 2. And similarly, with b3, b4, i tilde, and tilde. And so these two spaces, the sum should be equal to k. So this is the stability condition. This is what follows from this equation when z is positive. But these two spaces may intersect. And so the dimension of intersection, this is the number of instantons trapped at the intersection. So I call the intersection the cross. And this modulate space is the modulate space of crossed instantons.  Now the game which you can now play, so there's a partition function for this cross instanton configuration. You can first integrate out instantons living on one plane. And that will produce an observable in the theory on the second plane. And then the compactness theorem, which again holds, will imply that there's a function of certain equivalent parameters, which from the point of view of the observer living on the horizontal plane are just some spectral, zero spectral parameters. The correlation functions of those observables have no poles. And from that, follow Dyson-Schminger equations. So compactness theorem implies regularity of correlation functions, implies word identities of safety, actually. So from these Dyson-Schminger equations, one derives B, P, Z, and K, Z equations of two-dimensional conformal field theories. So let's just say that I'm done, finished. Thank you. No more questions. Sorry, sorry. The fact that in some instant the charges shared by 1, 2, and 3, 4 seems to mean that they kind of make a bounce data or interact with the other. But still you're solving the actual solution you're constructing seems to be the self-dual solution, not the more complicated solution. So talk with each other, but the profile is not looking good. Well, you see the point is that, well, they talk to each other, in particular, through the real moment map equation. So in order to produce the instant on-gauge field via usual edichem construction, you need this equation and the real map equation, which involves only B1 and B2 and INJ, not B3 before. So I don't know how to produce something like a gauge field using solutions of these equations. Because now once you added second brain, second set of brains, it is not true that B3 and B4 is 0. So this is no longer 0. So qualitatively, there are solutions for which these two parts are very far away, separated. So then approximately, you can construct instantons of some charge k1, 2 leaving here and instantons leaving here. But what happens differential geometrically when they all come together, nobody knows. This here is only the algebraic geometric description which works, only in terms of the holomorphic data, not in terms of the differential geometric data. So the fixed points on this modular space consist now of two types of young diagrams. So some young diagrams grow in directions epsilon 1, epsilon 2. And other young diagrams grow in directions epsilon 3 and epsilon 4. And so generically, they have separated in complex plane. But sometimes they touch. And so when they touch is when the possible poles may appear, but then they come back in a theorem that the poles actually cancels. So this configuration I call the butterfly. When two young diagrams start touching, these are like wings of a butterfly. And to the observer who lives on the horizontal plane, things look in a slightly mysterious way, because it appears that the instanton can actually disappear. So you start with the configuration of instanton charge k. And then one box of young diagram can disappear, gets chipped off. It doesn't disappear in the combined system, because it just goes off into the other direction. And so this is what happens when these two diagrams kiss each other. So they kiss. And then the box disappears in one diagram and gets attached to another diagram. And so the poles cancel between these two contributions. Maybe I should describe the solutions of these equations in the case of k equals 1. Maybe it's helpful. So if k goes to 1 and n equals to 1 and n until this equal to 1, so everything is one dimensional. Then all matrices are just numbers. All commutators vanish. So you have very simple equations to deal with. Oh yes, I forgot one more equation. j tilde i minus i tilde dagger j equals dagger equals to 0. So one proves very quickly that j actually vanishes. And so the equations which you have are, you only need to look at these equations. So b3 i equals to 0, b4 i equals to 0, b1 i tilde is equal to 0, b2 i tilde is equal to 0. And these are all numbers. So bA is just a complex number. i is a complex number. i tilde is a complex number. And then you have a real moment equation which says that the absolute value squared of i plus absolute value squared of i tilde is equal to zeta. So this is a three-dimensional sphere. But now we have several branches. So if at least one of the numbers b3 or b4 is non-zero, then i equals to 0. And so you have a branch which is asomorphic to c2. So it's parameterized by b3 and b4. On this branch, i equals to 0. Then the norm of i tilde is fixed. And up to location of formation, i tilde can be set to be just square root of zeta. So that's one branch. So inside this branch, there is a special point when b3 and b4 vanish. If they vanish, then i may be non-zero. Similarly, there is a branch when b1 and b2 are non-zero. So then i tilde is 0, and i is equal to square root of zeta. And then there is a special point, 0 when i tilde may become non-zero. And then there is a branch when all b are equal to 0. So these equations are solved trivially. And here you have S3 mod u1. So this is Cp1 parameterized by the ratio i to i tilde. And so what this module space is, you have one instanton which can sit either on the vertical plane, so this is that branch, or it can sit somewhere away from the origin, or it can sit somewhere on the horizontal plane, or when it reaches this point, so this is the position of this instanton, reaches the intersection point, then it actually starts thinking whether it wants to stay or it wants to go, and has the whole two-dimensional sphere worth of choices. And so that's the extra component of this modular space. And so now in the classification of the fixed points, they remember that we were supposed to have n-tuples of young diagrams, so n plus n-tuple, n plus n tilde, tuple of young diagrams of total size 1. And so we have two choices, either one box empty or empty one box, and so these two choices are these two points. So these are these two fixed points in this case. One of them is visible to the observer in the horizontal plane, and another one is not visible. So when instanton has gone to a different branch, to us it seems that we lost instanton charge. Once everything is a bit in string theory, nothing is lost. Go string theory. Thank you.


So recent studies have shown that email is still one of the most effective ways to get through to prospects and clients but how come your emails are not getting returned. We're going to talk about a few tips and tricks on making sure that your emails get returned in a more timely manner. Hey everyone, thanks for watching. We're talking about emails in the meetings industry and how to get your email to meeting planners returned a little bit faster either with a response to the question that you've asked or just acknowledgement of your products and services. Now, some of you may be having deja vu because this is round two of email tips and perhaps you're thinking where is round one. So check out the first video I did about email tips a few months back, but this is round two. So a lot of different things have come up and we're going to tackle some of the different strategies, new strategies that you can use to ensure that your emails are getting returned from your meeting planner prospects. Tip number one, make sure you have a clear subject line. I so often receive emails that the subject line is simply the word hello. It's not going to compel me to respond. I have 400 hellos coming into my inbox every day. So if there's a clear subject line then I know exactly what information you're looking for. And I'm able to address it quickly, especially on those high stress days when the emails I can get out to you need to be quick in nature, but also help you move your project forward. So help me out. Help me move your project forward by giving me a clear subject line and a clear call to action in that line. Tip number two, please give me some frame of reference with the name of the program or project that we're working on together. There are quite a few meeting planners in the industry who are juggling multiple priorities in a single day, either multiple programs or multiple projects reports, etc. And the information you're providing is one piece in a very, very large puzzle that they are working on. So again, just help them out. Give them some frame of reference. What is the name of this program? What are the dates that this program are running over? Anything that can help potentially jog a meeting planner's aging memory into remembering why you and I are working on this project. Tip number three, have a clear signature line. Yes, the signature line is the place at the bottom where you have your name, perhaps where you work and perhaps other contact information details like an address or a phone number. You may think this is a given. However, there's a number of meeting professionals and meeting partners out there whose signature line is still a bit vague, if at all present at all, which again makes it really difficult to figure out what program are we working on together. I can't even remember sometimes what hotel you're working at. And please, please, I get it. I need to work a little bit better on my memory skills. Trust me. I'm going through that training as well. But help me out. If your signature line can have a little bit of some clues as to where you work, that would really help me with my frame of reference as to what program we're working on together and what information you are looking for from me. And of course, my favorite tip of all time is the good old call to action or CTA. It is a lot easier for meeting planners to respond to your request when they know what it is that you are looking for. So make your call to action clear. You can do that with a question mark right in the email behind the question that you're asking. That way we know exactly what it is that we need to address. And again, the quicker we can address those questions, then you can move your projects along.  But when the emails are vague and there's no CTA attached, we are left guessing what it is we're supposed to respond to. And then often the email just kind of drifts away and gets deleted. So please give me a clear call to action. I will answer that call to action. Finally, do your research, especially if you're doing mass communication emails to a number of clients or a number of prospects about what's happening at your hotel. Doing your research and finding out potentially that planners history with the hotel or planners history, even with you as a salesperson and customize the message a little bit better. I know it's easy. I know it's easy to draft an email and hit send that goes out to three or 400 meeting planners and prospects. But when a meeting planner opens a mass communication and knows that it's not directed towards them because there are some blatant errors and how the communication is being worded. That's already going to leave us with a bitter taste in our mouths. So make sure that you do your research, but then you're tailoring that communication to be very personal in nature. Now I've included some examples of this in the blog post, but you have to go over to Leanne Calderwood.com to look at the blog post and see some of the examples of what it is I'm talking about. And I know when you see it, you're going to realize, oh yeah, you know what, I've either seen these emails or I sent these emails out. And now I see why it might not be the most appropriate action for every single planner to do a mass send out. So take a look at the examples at the accompanying blog post, which I've posted below here this video as well. So hopefully these email tips have helped you today. Email isn't the only way to connect with meeting planners. We love connecting in a whole bunch of different ways, be it social media, voicemail or networking in person. And there are best tips and tricks to connecting with us in those formats as well. The best way to find all of those is to download the how to getting meeting planners attention download over at Leanne Calderwood.com. It's free download, free free to use and I encourage you to take a look and see how best to connect with us. Have a great day and I really hope to see you soon. Bye for now.


Hey guys, it's Nate, AKA The Foot Account, and welcome back to the channel. It might be a huge Monday of FIFA 23 content today because we are expecting the SBC that everybody wants to see, the Max 87 rated hero upgrade pack. I think there's a very, very high chance of that SBC dropping today. We'll talk about that, and of course, what it is gonna do to the market because a lot of people will have to sell cards in order to do this SBC, so I'm expecting a lot of prices to go down today, but I'm also expecting some player prices to go up. On the SBC fodder side of the market, and of course, there's gonna be plenty of coins to be made today with hero cards themselves and so many other areas of the market.  Also, we should get upgrade packs today. Hopefully they have 80 plus packs return, and I wanna talk about how the market's gonna react even before the content drop today. Will we see our normal Monday market rise with the footchamps rewards being given out? We're gonna talk all that and more in today's video, so if you're excited for this icon pack, hit the thumbs up and subscribe if you're new. If you're like, Nate, what in the world is this icon pack you're talking about? Let's just quickly go over the basics. On Thursday of last week, EA added this pack to the code, 87 overall Max Hero, and every day since then, people have been waiting for this pack to drop. In my opinion, this pack is the most, I guess, has the most anticipation of any pack that we have had dropped in FIFA 23 yet, especially in terms of an SBC pack, and the reason why we feel so strongly about this being today is just one week ago, we had the Max 86 icon upgrade dropped on Monday. It was two squads, it was pretty good value for a Gamble SBC, 140,000 coins, and a lot of people went out and did it because it was the first icon upgrade of the year, and honestly, it wasn't terrible value. Now, I didn't do this one, but I know a lot of people did, and what we're looking for this week is basically just a more high version of an SBC like that, of course, including the Hero cards. Now, Max 87 rated, so what heroes are gonna be packable from this today? You've got Yaga Turei, you've got Marquisio, those are probably the top two cards that you can pack from the SBC today. Govu, Smolarec, Cordoba, Robby Keen, Harry Kool, the list here is pretty nice. I mean, there's only like, I think 10 or so heroes that are not packable, like you can't get Volar, you can't get Ococha, you can't get Lucio, Machirano, you know, for Lan, there are some really top tier heroes that won't be packable from this SBC pack, but a lot of really solid ones. As you can see by the price, right? 1.2 mil for Yaya is the most expensive card, the cheapest one, Lars Ricken, 22K. You do have a fair chunk of heroes that are under 100,000 coins, but you have a lot of heroes that are between 100K and 300K as well. So, when we think about a price on this SBC, I'm thinking it's gonna be similar to the Max 86 icon upgrade, but I think EA know that hero cards are just more hyped in this year's FIFA than icons are. Let's be completely honest, right? EA added a bunch of new heroes in this past year, compared to FIFA 21, I'm sorry, FIFA 22, wow, went back two years there, but they've added a bunch of new heroes. These guys have really nice cards, and I just think that with the way heroes work for chemistry, heroes are more hyped than icons this year, right? That's a pretty well-known fact, and you can gauge it based off of prices on the market as well. I think that EA might make this SBC a little bit more expensive than the 86 icon was. I'm thinking somewhere around the 175,000 coin range for that SBC today, and of course that makes you think, Nate, what kind of fodder and what kind of SBC cards are gonna be going up on the market today? Because obviously, people are gonna have to go buy cards to do this, right? Yes, I do believe just like last week, you're gonna see SBC prices rising on this game, SBC fodder, as we call it, right? The cheapest players by rating. I'll put the link to this page in the description. This is a very, very helpful page if you don't know where this is at, I'm flipping, but this, again, as we take a look at the fodder market, cards prices are slightly up, right? If you guys bought SBC fodder during the weekend, 85s are rising above 9,000 coins, 86s are in like 14K, above 14K range, 84s are 2.2, and 87s are about 18K. I do think that if the SBC is gonna be a little more expensive today, one way they could do that is make it an 84 rated squad and an 87 rated squad. I also think 85, 87 would be a potential requirement for the SBC today, but hopefully it's less than 200K. If it's more than 200K, I'm gonna be completely honest. People will still go and do it. I mean, I'm gonna be very tempted to do this SBC today regardless of the cost because the gamble of getting one of these cards, and especially since they're new and there's a lot of great cards in here, the gamble of it is pretty enticing. I can't lie, and I think that's why there's so much higher for this SBC today, but that's kind of what I'm expecting for the price, and again, if it's anywhere under 200,000 coins, I do think you're gonna see SBC fodder cards going up in price, and I mean, as I think about my club right now, I actually need to stock a couple fodder cards in my club, and I know that I'm not doing this to try to make coins in these cards because I think if you were gonna make the optimal amount of profit on SBC cards for today, I think you would have already bought, right? Because some of these card prices have gone up a little bit, but if you don't have any fodder in your club or any cards to do this SBC, I would say it's not a terrible time to go out on the market and just go in club stock, buy one, buy yourself enough players to get the SBC done today or to at least cover your basis, basically, to do the SBC for a little bit cheaper because once this SBC does come out, it's gonna be a mad dash of people going onto the market and trying to get players to add to their club that they can then go and use to get this SBC done. So that'd be my recommendation right now is I'd go on the market if you don't have anything and stock your club because it's gonna get pretty crazy today, honestly, 14 and a half K for a Mendy. Like I'm gonna need this because I am gonna do this Hero Upgrade Pack SBC. I'm not skipping out like I did in the Icon Upgrade just because that's how hype it is for me and I think for most people in the game. So I do expect fodder to rise. Some people might ask like Nate, are these cards over-invested? I mean, technically, there is a lot of investing that's gone on with this side of the market. I totally understand that, but I just think that with the amount of demand this type of SBC will have, of course the ball is always in EA's court because if they make the SBC today too expensive, then they could make some people not do it if it's way too expensive, but I feel like way too expensive for an SBC of this caliber would have to be 300 K plus. They'd have to really make it way too expensive for SBC fodder to drop today because I just feel like this pack has so much hype. So I feel like there's a very, very slim chance of seeing SBC fodder prices go down that much today. I think more so, it's just gonna be so much demand to get this done. And I think you'll see those prices go up. So there's not too much of a risk there and if prices do go down a little bit, I definitely would not panic sell because again, we have a big promo outcome this Friday with the new out of position change promo. So I just think that holding your fodder, even in the case that it would stay the same price or go down a little today, it'd definitely be something you wanna hold on to throughout the week as it has potential to rise up with more content to be dropped, messy player that months still out, stuff like that. So let's talk about where the market's gonna go today other than just SBC fodder. Well, you might be thinking about hero cards like Nate. If hero cards are gonna be packable, isn't Yagatore going to crash in price? Isn't Robby Keen, Marquicio and Al Awarian, aren't these guys gonna crash in price? Well, the answer is yes, slightly, but also I think that a lot of these guys are gonna end up going back up. This is, I think what we're gonna see is the same exact scenario that we saw last week with icons and what I wanna show you is last week, a very popular base icon that was packable from the Max 86 SBC, right? I think you can see it right here. Hold it was 872K, kind of dropped down into the panic before the SBC was released. Last Monday, hold it went from 700 and actually it shows 790K. Hold it was down at like 730, I think. There was like an hour or two of panic selling right after that icon SBC was released and then all of the top tier icons that were packable from that just went rocketing back up in price. Hold it went from like 730 all the way up to 900,000 coins and so many other of your top tier hero cards and even some of the ones that aren't that top tier, I think about Baby Blanc who is not that expensive of an icon card but last week he was in the low 300s and then rocketed up to the high 300s. He was like 370, 380 for a couple days last week before now dropping back down a little bit more. I think that's how the hero market's gonna react today. I think guys like Yaya Tore especially, Yaya Tore was 1.3 mil on the weekend, right? Before this leak happened, for this SBC to drop and now he's down at 1.169. This is a card today that might not drop any lower than like, I don't even, he might not even go down a total of like 70 or 80,000 coins. Like seriously, I think that people are gonna be so on top of the market today that yes this SBC will have a lot of hype but I also think that prices are gonna rebound very fast today. So watch during that first hour after the SBC drops some of your most popular and honestly the hardest to pack hero cards that will be in this SBC. Markizio, the Yaya Tore that we just pointed out some of those guys, maybe Joe Cole, Cordoba, Smolarec, Govu, those guys are gonna be tough to pack. Not very many of people are gonna get it from a max 87 hero SBC today. So people are gonna sell these and panic to go do the SBC and fear that their price are gonna go down but then what actually is gonna happen is people don't pack the Markizio or the Yaya Tore that they think they're going to and then of course they have to go back out to the market and buy those players back up for their team and there's just a lot of hype for heroes in general when an SBC like this comes out. So that's what I would really, really watch out for today. So if you wanna make coins, watch out for some of these hero cards that are absolutely gonna drop in price today because of panic selling around the content time after the SBC is released but then of course watch them as they rebound back up later as people go back out to the market and buy the cards that they were not able to pack. So that's what I would say for today on the market. Be very careful with the lower tier ones though.  He was up to 110, now he's 85,000 coins. I still think there's some hype there. Peter Crouch is like 40K, Tim Cahill's 42,000 coins. Somebody who's really low tier hero cards might even get really low and then bounce back a bit just because there's still semi-popular cards on this game. Ole of course with the nice links to Erling Holland both being from Norway. So really watch out for the hero cards today but in general if you think about it last week we had this happen too. 150,000 coins ish to get this SBC done. A lot of people had to sell cards from their teams to go out and get this SBC done. So it doesn't just impact the hero cards in the market. It kind of impacts everything else as well. I do expect to see a pretty sizable just market dip today at content on rule breakers cards. I mean these rule breakers that are impacts from rule breakers team too, they might even get supplied a bit, right? Last week we had two different packs. A premium mixed players pack and a small prime electrum that were the rewards out of the two SBCs inside of the icon upgrade. If that sort of thing happens today, if it's a two or three segment SBC I think you'll see some supply hit the market as well. That could drop some of the prices on these. Rule breakers team two cards and then I think you'll just see some panic selling on the rest of the market. Now if you're trying to make coins specifically that that's your only goal today on the game. Watch Road to the Knockouts because these guys have games this week. Tomorrow is a UCL Tuesday with UEFA Marquis matchups. Games and upgrades are on the line for a lot of these players. If there are price drops today on any of these cards especially if they are cards that are very close to getting an upgrade like Obama Yang. A guy who won his first round game two weeks ago when the Road to the Knockout promo was first out. Keep an eye specifically on those types of players for big drops. Wanda Saka would be another one and then potentially having their prices rise later on this week. So that's a great part of the market to be watching today. Once to watches as well. I mean, think about it. These guys are getting upgraded on Wednesday, right? Lewandowski, they got the upgrade.  I think right now it looks like he's down to 370. Nah, that's an undercut. Just kidding. You know, some of these guys that are getting plus ones or plus twos on Wednesday because they hit that threshold for the upgrade, right? Renato Sanchez is gonna get a plus three because PSG won. So that sort of thing. If you see drop offs on these cards I think that's a great watch as well. And even your out of packs team of the wheat cards and rule breakers team won. Think about Calvin Phillips or PK. Cards that have risen really, really well out of packs. If you see a big drop on any of those I think it'll create a great opportunity to get into some trading aspects of the market today. Because again, like I said, a lot of what you would see today on the market is some panic. I think you would see some panic of people that have to go sell a card or two from their team to go and get the SBC done. But then of course they go, they pack somebody. It's not who they wanted to get. They go back out on the market and buy those cards right back for their team. So watch heroes, watch out of packs card today. Even some of your meta golds. I remember last week I traded with literally everything. Golds, icons, heroes, out of packs, informs. There was a ton of movements on the market last week with the icon. I can only imagine it's gonna be even crazier today with a hero SBC. So definitely watch out for all of that stuff. And again, I do think you would see a bit of a market bounce back too. Because so many people have been waiting for this SBC to drop for all the panic and stuff that's gonna ensue afterwards. I really do think that you will see some nice price rises. Even though, you know, it's not Mondays aren't days, especially after 6 p.m. on Mondays are not days that you really usually see a huge market rise on. It's usually in the morning on Mondays. But I think with everything that all the anticipation that's been building up to this SBC that we should see today, I think there will be some nice rises on the market later today. So definitely watch out for that. Now, that's a lot about one big SBC. Of course it is very hype. But what else is happening today on Monday? Well, Mondays always bring upgrade packs, right? Now, we had the 80 plus upgrade pack. I think that was during road to the knockouts. Last week during week one of rule breakers we had the 78 plus pack dropped again. Now, could we see a player pick SBC today? Personally, I'm hoping that EA don't drop the player pick. I mean, it would be nice. But at this stage, like the rule breakers promo has just been like kind of L content the whole entire time. I'm not expecting EA to go crazy today with that side of the content. I mean, we wouldn't complain about a 75 plus player pick because even if they do that, that's gonna just create more hype content. I mean, and we're not gonna complain if they do that but it almost just feels like, I guess what I'm trying to say is it doesn't feel like it should happen because nothing inside of this promo has been super insane or a huge surprise thus far. I mean, completely honest. So that's why I don't think we're gonna get player picks today, but the 80 plus upgrade pack or the 78 plus pack again would be very, very possible. Which again, those have not been terrible value whatsoever. As people maybe have a chance to pack some rule breakers cards. Maybe it's Rivery you're after. Maybe it's Tiago. Maybe it's Havertz or something. One of the more hype players that are in this promo but you know, ooh, 370 for Tiago. He's dropping a little bit. Nice. I do think they will get some sort of upgrade pack today. That's almost a given, right? It's a Monday. So that happens. So the combination of the hero pack and the upgrade packs today are definitely the two most hyped things in the day. Now also, also today, what I would look out for, you know, last Monday with the hero pack, they also dropped Essien and Trezegey. We have the two other, I think it's Cassius and Vidditch have been leaked as icon player SBCs. Mondays maybe the new day where they drop icon player SBCs. They might not do anymore today. I do think this could happen in, you know, the next couple of weeks though is that we continue to see these drops. So just be careful. Maybe another icon player SBC today. I wanna throw something out there though. I think EA is gonna change it up this year. Maybe for the first time ever we'll have specific hero player SBCs as well. You know, maybe since they're basing hero content, you know, a lot more into the game with, you know, the hero upgrade packs and adding more heroes to the game. I just think that would be a nice touch, especially for as they're gonna upgrade heroes throughout the year, maybe not give us like a yaya SBC right away, but you know, a medium to lower tier, maybe like a Papan SBC that you could do for like a really cheap price, or, you know, I don't know, maybe like a Govu or somebody like a newer hero like Rudy Valer, maybe you could add in an SBC for him. Maybe we're too early for something like that as well since these cards are not that expensive, but I just have a feeling that that could be something that EA would do at some point this year is drop foot hero player SBCs, like we have icon SBCs. I feel like they could just drop a hero SBCs in that section as well. And I think that'd be some welcomed content. People would enjoy that because heroes again are kind of more hyped this year than icons are in general. So other than the upgrade packs today though, I'm not sure if we're gonna get to anything super crazy. There's really nothing leaked right now on the market. I mean, or on Twitter, there's nothing leaked besides the hero pack expectations. And that's really about it. So hopefully we do get some cool content today. Now, really quickly elsewhere on the market and on the game, we did have some new cards dropped. Yesterday we had a dynamic duo between the Syria team, teammates of Posh and Marco Ornadovic. Ornadovic looks really good and so does Posh. I think, again, this dynamic duo concept is a W. It doesn't give us the most insane cards in the world, but it gives you two really dope cards for just stocking your club with nice position changes. And I think they're both lengthy actually. Yeah, Ornadovic is lengthy by default. And then I believe Posh is as well. But let me tell you guys this. I mean, if you have some extra 83s in the club, both of these are just an 83 rated squad in a team week. Easy, easy, easy to do. These are great dynamic duo SBCs to get done, especially if you think about maybe starting to prepare your club for like an icon swaps further down the line. You know, if you do this, get it done. You automatically have some nice and tradable, decent players in your club that you can use and have some fun with when serious specific objectives come out. And it's kind of more like the card collecting aspect of the game that we talk about. Of course, a lot of you guys that are even watching this video have extra interest in, right? Whether it's making coins or putting some good value cards in the club. I would just say that, you know, don't skip these on these all the time. Now I've been somebody who has not done any dynamic duels yet, but I think this might be the first one that I do, especially with upgraded packs this week. I'm gonna have a couple of 83s to rinse into these two SBCs. I've got a couple untradable team of the week cards in my club. They keep releasing that 80 plus team the week every week. I do it once, put one of those team week cards in here. You know what I'm saying? So I just think that these are something that to not skip out on because, you know, especially once we get those objectives upcoming, which I don't think are gonna be too far away. I'm probably like hopefully a month away from icon swaps, maybe a little bit longer than that. But even a swap program in general, we shouldn't be super duper far away from that. And just stocking these types of cards in the club is really nice for getting objectives done and starting your club off right in that direction. So I like that content. We have the Domintin Sanchez, the Merino and the Roldan. Domintin Sanchez for me is just another Prem Centerback. I mean, I am a Spurs fan. I'm gonna have to give this card a go. He does have really good work rates. He has got the unique body type as well. Domintin Sanchez was a very OP meta gold card a couple of fevers ago, very hyped for sure. Lengthy by default. He is two-star, two-star, which is not that great, but his stats look decent except for composure and reactions.  And then we have Merino here, which is we're basically calling him Robert Tone 2.0. He can be lengthy with a multiple of different chem styles, high and average body type, high-high work rates, three-star, three-star. And, of course, the link to the ever-popular Striker, which I somehow am forgetting his name right now. I haven't been to Sorloff. How could I forget about Sorloff, right? He's got a link to Sorloff, which I think is a little bit of extra hype there. So, will we see a nice rise into Monday on a guy like Merino? You guys might remember last week we saw the huge rise on Robert Tone. And his price went absolutely flying. Now, I don't know if we're gonna see that with Mikhail Merino today. I picked up a couple of these just under 60,000 coins. I really don't know if this guy's gonna fly today in price. I'm not expecting it. I think people are seeing what happened last week and they're expecting that to happen this week, but with a big hero upgrade SBC on the horizon. Last week, that max 86 icon, we didn't really expect it on Monday as much as we expect the hero today, so that anticipation, I think, will hurt some people's motivation. Wow, 58K with the shadow, I'm gonna get that. Okay, I'll take that, not a problem. I just think that might hurt some team buying today on Monday. I do think the market can still have a decent rise today, which we just got one heck of a deal on a Merino there. I'm gonna list this up right away, take my couple thousand coins of profit, quick flips. I still think that we have a little bit of a market rise in store for us today. I think we can absolutely see some of your meta golds that have just been low all weekend, especially if there's some team of the week potential. You know, I think about a guy like Usman Dembele who's already started to rise up a little bit in price after his performance yesterday. His gold card is starting to go up. Holland, of course, had a couple goals over the weekend. Some people are maybe thinking that he could be a team of the week. I think that some of these guys could potentially go up on the market today and some of your other meta golds in general, right? I feel like I always check Benzema.  That's, I think, almost the cheapest that I've seen him. 56, 57K, 53K on a snipe right there. So just keep an eye out on your gold meta market today and some of your out-of-packs cards and even some of the rule breakers, I think, could have slight rises today. But I think with that hero pack being in mind for a lot of people, that might hurt the rise on some of those cards. If you're wanting to get somebody like Neymar for your team, Neymar is kinda low right now, 1.17. These are cards that I would look at. If you even want to invest in team of the week for a quick flip and buy in the panic today, even though I'm not a huge fan of this team of the week in terms of investing, I do think that some of these guys might get cheap today with the hero pack coming out with that panic in general. So keep an eye on some of those as well if that's an interest for you on this market. But really, I just think today is gonna bring us a pretty crazy day all centered around that hero pack and everybody trying to get this pack and trying to gamble, right? And then after that, it's just gonna be some crazy market movements of a lot of people going back out and buying cards for their teams or maybe trying out heroes that they had seen a lot of people pack pull. Like, that's another aspect of when you have a gamble pack like we could potentially have today dropped on the game, a lot of people will just see there'll be a lot of hype around heroes and that just kind of pushes their prices up in general, which is why, again, I think for me personally today, I'm gonna be looking at a lot of hero cards and road to the knockouts because if I see a guy like Abami and drop back down to like 530K where I almost bought him at yesterday on Sunday and now he's like 600,000 coins is kind of crazy. But if this guy drops down like that much for a hero SBC today, then that's the kind of card that I want to be involved in. A very rare card that can really easily bounce back and is very hype on this game for whatever reason that it is. And it looks like we just sold that Merino for a cheeky couple thousand coins of profit. Boom, trading live on YouTube video. You absolutely love to see it. So big day ahead today on Monday.  Stock some cards in the club if you need to and get ready to make some coins. It's gonna be a fun one.  Twitch.tv slash foot account and link down below in the description. I'll see you guys there today. It's been Nathan with Accountant. Catch you guys later. Peace out.