So first of all, I'll remind you about graph polynomials from the first lecture. So the data of a Feynman graph was a connected graph G, and it had some external momenta which were called qi, and each qi is in typically real space that I mentioned d, but we assume the momenta are non-zero, and we had masses, certain number of non-zero masses sitting on internal edges of the graph. The edges of the graph will always be labeled in this talk, and this dimension of spacetime is an even positive integer. So over here, I'll draw an example that will be considered throughout today's talks. So we'll study this in detail. So G will be the phonon graph, so it has two momenta, q1, q2 coming in, edges labeled 1, 2, 3, 4, and the thickened edge means that it carries a mass, so we have m1 non-zero mass, but the mass is m2, m3, and m4 are zero, we just ignore them. This external momentum is considered to be zero, so I don't write it, and we had momentum conservation q1, q2 equals zero. So to such a graph, in the first lecture I explained how to associate graph polynomials. There were three of them. There was the Kirchhoff polynomial, which was the sum of all spanning trees, and I called them spanning one trees, and then the product of the edges not in that tree r for e. Then the, that's also called the first semantic, this is the second semantic polynomial, is the sum of spanning two trees, and you take the product of, same thing again, but this time you multiply by the total momentum coming into one of the trees, t1 or t2, it makes a difference. Total momentum entering t1, and then the thing we're most interested in is polynomial I called xi, which depends on the masses and the external momentum, and it is defined to be phi gq plus the sum of the squares of the masses times psi g. So in our, in our favorite example, we have psi g equals and phi gq.  Okay, so we were interested in the Feynman integral, I gqm, as a function of masses and momenta, and it had some, some gamma factors that I'm just going to ignore. So it's an integral over a certain domain sigma, one over the first semantic to the d over 2, and then this ratio, and here we have ng minus hd d over 2 omega g. So let ng be the number of edges of the graph, and hd is the number of loops, the first Betty number. Yeah, and so as explained in the first lecture, the degree, psi g is homogeneous of degree equal to the loop number, and the degree of phi and hence xi in the alpha variables, the string of parameters is hd plus 1. That's very important. Okay, so what was sigma? Sigma was the domain in projective spacer dimension ng minus 1, and it was the coordinate simplex. So sigma, it's the region with projective coordinates in the Anglo-Saxon notation, where alpha i are real and positive.  So the goal then is to, as I explained in the last lecture, is we want to understand this as a period of some cohomology. And that will take the entire talk today and a bit of the beginning of the talk next week. So we want to write i g qm as a period or family of periods of cohomology. So we want to think of it as a pairing between a Durham class, a differential form, and some cyclic integration. And so what? What is this letter? You want to say that. Sorry, yeah, so the capital xi is like this, and I hate writing that because it doesn't look like anything. So on the board, I prefer to write a line through it. Otherwise it just looks like three random lines. I know my Greek teachers as well would be a poor. A small question about your graph. Is it allowed to have one edge connecting one vertices to itself? Yes, that's called a tadpole. Yes, and do you assume your graph will be primitive or not for the moment? Primitive, in what sense, primitive? I mean the octave. No, general. No condition whatsoever. No condition whatsoever, and in fact, I won't consider primitive until much later. All the native things are going to work suddenly for the even. Yeah, so it works for the even because that's the case we're most interested in. Because as we all know, we live in 11 dimensions, which is, yeah, okay.  So obviously when D is odd, you get a square root here. You have to do something, there's no problem. I think one can work with that. But I haven't thought about it. I won't. No, for what you do, you maybe need D equals three. So dimension, or you pick a dimension? Yeah. So I think I believe that people who sometimes do three dimensions work in four and do a productive expansion from four dimensions down to three. But I'm sure that everything works for odd, but it's a different kettle of fish. Anyway, so the issue here is the same issue as I explained on the example of z to two last time, is that we want to interpret the domain of integration as a homology class. So we look at its boundary which is contained in the place where all the alpha is vanish. See, these are called the coordinate hyperplanes. And the problem is, which is exactly the issue with the z to two integral last time, is that the boundary of, or I mean the domain of integration itself, not necessarily its boundary, we can just scrap that, meets the singularities of the integrand of the differential form. And this poses problems. So compare the z to two example from last week. So now to explain this issue, let me do an example. So let's look at this graph. And for reasons that will become clear later on, I'm going to label the edges two, three, and four. And so the graph polynomial here is alpha two, alpha three, plus alpha two, alpha four, plus alpha three, alpha four. So let's call x subscript psi g to be the vanishing locus of this graph polynomial in p, p two. This is the zero locus where the graph polynomial vanishes. And we call li the ith coordinate hyperplanes where alpha i vanishes. And so we have the following picture. The graph hyper surface looks something like this. Well, because it will appear as a quotient of this graph when I contract the edge one. And so later on it will, this example will serve a purpose later on. That's why. The coordinate hyperplanes look like this.  This is x psi g. And the domain of integration is this simplex here. So in this situation, I'm assuming say that ng equals hd over two so that I'm not considering the polynomial psi. But that's, it's just a toy example. And exactly as in the last lecture, there are bad points here which are not normal crossing. In fact, there are three bad points. So the bad low bad points where the domain of integration meets the graph hyper surface are the intersection sigma psi psi g and they are exactly the three vertices of the simplex. So as I explained last time, the solution is to blow up these points to make the situation normal crossing. And the picture we get after blow up is this and then the three bad points become copies of p one. That, sorry about that, I should have drawn it like this. And the white lines no longer meet graph hyper surface in the new picture. We get a new domain of integration which is no longer a triangle but a hexagon. So I'm going to call this sigma tilde g, which I'll define in general. And these are exceptional divisors epsilon 24, epsilon 34, epsilon 23. So what do I want to do today? We need to understand the general geometric situation. So first of all, we want to determine the bad loci in general. Then we want to blow them up. And then the third part will be to study the structure of the resulting space. I did in the first lecture. I'll say it again, the alphas are called Schringer parameters. Oh yeah, so the alphas are called Schringer parameters. And you want to do this in a canonical way in order to get a well defined motif? Exactly, I want to do it in a canonical way uniformly for all Feynman graphs. We'll get to this point, but of course these are families of hyper surfaces. Depending on masses and momenta. I'm just going to work fiber by fiber because I don't want to say a lot about families. But it poses no problem whatsoever. And one could be very precise about how it works. But I'll just pretend that q and m are fixed and generic. And one could be very precise about what generic means. So first of all, a lot of properties of graph polynomials. And the first one is called contraction deletion. So this is going to be half an hour of just graph theory, or maybe more. So if e is an internal edge of our graph, then define g minus e, which is the deletion of edge e. So you delete the edge e, but you retain its end points. So for example, if we had a graph like this, and this is the edge e, then deleting the edge e gives you a new graph, which now has two connected components. So this is g, this is g minus e. So for most of the rest of the lecture, so I may be a little bit sloppy about this, I'm going to consider subgraphs defined by sets of edges. So an edge subgraph. So that means you take a subset of edges and you include all the end points as vertices. So sometimes I may write this, I may forget to write gammas contained in the edges, but we think of it as a subgraph of g. So in this case, we define the quotient or the contraction. You always take the vertices at the end of all the edges, and no more. Exactly, you take the minimal subgraph that contains those edges that's a graph. So what is the contraction? What would you do is you delete all the edges in gamma, and then you identify all the vertices lying in each connected component. So if you have gamma has a bunch of connected components, each connected components gets contracted down to a single vertex, gets squashed down to a point. No masses for now, they're going to come later. Actually, thank you, that's my next comment. So the quotient, yeah, you're right, thank you. The quotient, first of all the momenta, the quotient g gamma will inherit the external kinematics. And in general, it will inherit the massive edges. If they were massive before, they will remain massive. It actually depends on the situation. I'll say more about that later. So here's an example, let's take g and contract the subgraph 3, 4. This is just this graph. And g, contract the edge 1, is the other example, which is why I considered it. What would you do if one of the edges you contract has a mass? No, it pays no role. That corresponding edge does not appear in the quotient. So for now I'm just talking about the quotient. So that mass does not appear, does not an edge of the quotient, so it's gone. But later on I'll explain exactly what happens to the masses momenta, and it's quite subtle, but it'll come. It's because the internal momenta goes to infinity, so the mass is... I hadn't thought of that, but that's probably a good explanation.  And with hindsight I realised that I should have defined things slightly differently, but it's okay. So define another notion of contraction. It's g double slash gamma, so this is not a standard notation, but I think it's helpful. So it's g slash gamma if gamma is a forest, so it has no loops. But it's the empty graph, or zero graph, whatever you prefer, if the subgraph you contract contained a loop, at least one loop. And then the contraction deletion identity is the following. We take any internal edge in the graph, and then the first graph polynomial satisfies, you delete the edge e and multiply by the corresponding Schringer parameter, and the constant term is the contraction in this sort of brutal sense. And the second semantic polynomial satisfies a similar identity. So here it's clear, when we delete an edge you have the same external vertices, so they keep their external legs, so there's no ambiguity about what this means. What can happen of course is that when you delete an edge the graph becomes, is no longer connected, and in this case this graph polynomial will vanish, and that's what you want. The equations are correct precisely because of that fact. Okay, so the only thing that we're going to need from this is the following identity. So the first graph polynomial restricted to the locus alpha equals naught is just psi double slash e. And from the definition of psi, the restriction to alpha e equals naught is psi g double slash e qm. So this tells us something about the geometry. It tells us what happens to this graph hyper surface when we intersect with a coordinate hyperplane. There's a beginning of a recursive structure because you get the corresponding hyper surfaces of quotient graphs where you contract an edge. So that will be important. So now I'm going to state some what I call factorization theorems. These are absolutely crucial for everything that I'm going to say. The entire crux of the story is hidden in two or three identities that I'm going to write down now. And everything relies on that. So G be connected again, connected Feynman graph. And gamma, a subgraph defined by a subset of edges with several connected components, gamma up to gamma n. So the number of loops in gamma is just the sum of the number of loops of each connected component. And the first result, so this is not new because it's certainly in a paper of mine with Dirk Kreimer. And I'm sure these identities are in principle been known for a long time but not necessarily been written in this form. So we can write the first semantic polynomial, ultraviolet. Don't ask me what it means. Actually, this formula is in a paper by Bloch and Orrin Kreimer. The following one is in a joint paper with Kreimer. And I'll explain in a minute why these in principle been known for a very long time. Phi uv gamma g. So here we have two approximate factorizations of the graph polynomials. We need to explain what that means, what r is. Do you want to put also the whole uv on the first r? I've deliberately not done that for a subtle reason. I'm glad you spotted that. Where r dot in either case, gamma g, is of degree strictly larger than h gamma. So h gamma is the degree of this part. This has degree h gamma in both cases. But the r has a remainder time. It has higher order in the edge variables r for e for e in the subgraph gamma. So let's do an example. So in this example we have, let's just look at the first graph polynomial r. So here we can write the first graph polynomial. You can check that it can be written alpha 3 plus alpha 4 times alpha 1 plus alpha 2 plus alpha 3 alpha 4. Where the subgraph is 3, 4 and a quotient graph 1, 2. The external momentum play no role for the first graph polynomial. And it's also equal to, so it factorizes in two different ways, alpha 4 plus alpha 1 alpha 3 plus alpha 2 alpha 3. And here the subgraph is, and the quotient is a tadpole. So here we have the subgraph here has variables alpha 3 and alpha 4. And to leading order in these variables, this is a degree 1. But the remainder is quadratic, it's a degree 2. So if we think of alpha 3 and alpha 4 going to 0, this goes to 0 at order of epsilon squared. And this goes to 0 at order of epsilon. So in the limit as these subgraph variables go to 0, the remainder terms will disappear. And we'll just get the product on the left. That's very important. So a couple of remarks. So in a very special case, so in a very special case where gamma is subdivergent, then these factorizations are sufficient to derive all the main, well, most of the main results certainly of renormalization theory. So you can prove BPHZ theorem, and you can derive the Canon-Somansic equation just from the existence of these factorizations in a very special case, for this very special family of subgraphs. So this was done in a paper with myself and Dirk a couple of years ago. So that's why I say physics knows about these factorizations and has done for a long time because it's essentially equivalent to being able to renormalize. But the funny thing is that renormalization doesn't use the full strength of these equations. It only uses the case where the subgraph is subdivergent, and the general case of these identities is not used, but it will be absolutely crucial in the construction of the cosmic Galov group. A third remark is that you can ask, since this factorization property is very closely related to renormalization, you can ask what are the possible polynomials that satisfy these factorization identities? And if you think about it a little bit, you can check that they uniquely determine the graph polynomials. So the factorization formulae plus a very little input, in fact, uniquely determines. So there are remarks on this in my paper with Kremer, but one thing we never got round to was to explore this idea and see how rigid the Feynman rules are if you demand that they be renormalizable in this algebraic geometric sense. So what's the general characterization of R, the remainders? It can be anything. It's just any polynomial that will be homogeneous, but in those variables corresponding to the subgraph, it has higher degree. And would you call it input? Input, well, first of all, for psi, so if, well, it depends what, you can choose what actions you want. But if you give yourself a class of graphs, say these graphs, and you want this formula to hold, and you want this formula to hold, and you specify what psi of an edge is and psi of a tadpole, then those two axioms will uniquely determine the first semantic polynomial. So there is no choice. Physics gives us the unique possible choice for which these equations hold. And so this is a photo experiment that we never got round to, is to try to, well, there's an obvious game to play how unique are the final rules if you impose a certain number of desired conditions. Okay, so here comes an important point that will seem trivial at first, but will be important next week. But the graph polynomial, the quotient graph polynomial is never zero, but it can happen that the second semantic polynomial of the quotient graph is zero. And so I kind of, I did, it was the case here because here I was a little bit sloppy. Here, this quotient graph here really has two internal momenta, q1 and q2. But because q1 plus q2 is zero, this is equivalent, as I explained in the first lecture, to a graph with no incoming momenta. And so this has vanishing second semantic polynomial. So this is an example where the quotient has vanishing second semantic polynomial. Oh, I'm ahead, wow. Okay, so we want to understand this phenomenon now, which is related to infrared sub divergences. So definition, a subgraph gamma, which is connected. So I take a connected subgraph, is momentum spanning. So this is a phrase I made up. It's not standard terminology as far as I'm aware. Momentum spanning, if for every vertex v in the big graph, which carries some momenta, so such that the total incoming momenta, so this is the momenta, the total momenta entering, total momentum entering that vertex. Momentum spanning, if every vertex which carries some momentum, then that vertex is in the graph, is in the subgraph. So it's a subgraph that touches every external leg that carries some momentum. And a general graph, gamma, an arbitrary subgraph not necessarily connected, is momentum spanning. If some connected component, if it has at least one connected component, and hence exactly one connected component of gamma, is momentum spanning. So momentum spanning subgraph is a subgraph that has exactly one component that touches all the external legs, the non-zero external legs, and all other components do not touch any of the external legs. Example, okay, example. So here in this case, a momentum spanning graph would be this subgraph, because it meets the only two momentum carrying vertices here and here. Let me draw them in a different color, here and here. Another momentum spanning graph would be 2, 3, because it has a connected component that meets everything. I don't know what else, what is not momentum spanning? So this subgraph here, 3, 4, is not momentum spanning, because it doesn't see this. But you could also take a trend, yes. Sorry? You could also take a trend. Sure. Yeah, I'm not going to write the full list of all momentum spanning. It'll take time, and I need to refine the condition later. So there will be another condition to do with masses, and then I will give the complete list of those subgraphs a little bit later on. Do you have to see something by the way from the engine? The vertices are the one with non-trivial momentum. So in this case, so this vertex does not carry momentum, because the incoming momentum is zero by momentum conservation. So the linear gamma is momentum spanning precisely when the phi of the quotient is zero. That's my next comment, thank you very much. Yeah, so with a caveat, with a caveat that we have to assume generic momentum, because you could have all q squared equal to zero, for example. You can easily construct examples for which this does not do. But we don't consider that. Yeah, so with sort of generic kinematics, I'll be vague, but actually it's very easy to write down the condition. It's actually that the partial sums should be non-zero. That's the only condition. For generic kinematics, gamma is momentum spanning, if and only if the quotient has zero vanishing seconds of magnetic polynomial. Okay, so now this leads me to the infrared factorization formula. And I'm led to believe that this is new, in fact. I've certainly never seen it anywhere in the literature. I don't understand the formula that you wrote under generic kinematics. What do you mean here? Put the square. This is a sum of vectors in d-dimensional space, and this is the Euclidean norm. For all qi? For all i, for any subset of external legs, you take the sum of the momenta and you take the Euclidean norm. But if you take a subset of all vertices, then the sum of qi is zero? Absolutely right, so it's all i strict. Thank you. I was not planning to talk about this. Yeah, you're right, absolutely right. But here, what would the square be like in the Euclidean space? Yeah. No, no, I'm not. Okay, so I'm using algebraic geometry. So the masses of momentum are variables in complex space, affine space. Everything makes sense. The geometry makes sense over complex numbers. But the square is the Euclidean norm. When you say generic kinematics, you assume that each edge has a non-zero momentum or not? Each edge. Each vertex. No, no, I said at the beginning. So some vertices have zero momentum, and we specify which vertices have zero momentum, like here, it's the bottom one, and the others which are non-zero. Not for all i inside of EG. Oh, sorry, sorry, sorry, yeah. See, this is what happens when I don't use my, when I improvise. For all i contained, I think it was called EXT. Well, I know I had, I think I defined a set of, of vertices which carried kinematics in the first lecture. So that's the notation I used with the external half edges that carried. Yeah, I don't want to get into these details. It's very simple, but it's just fiddly to get there. So to answer T-ball's question, I've forgotten what the question is. Yeah, so what will happen is we'll have a family over some space, and the Euclidean region will be the region where the queues are real or something like this, and in a neighbourhood of that space, everything will behave beautifully, and we can have well-defined analytic functions. And I mean we can extend over a complex space, and then we'll have some terrible discriminant that we don't understand. Yeah, so this is a trivial condition if, yes. If they're real, this condition is trivially implied by the ones I had in the first lecture. Okay, so the infrared factorization theorem. So let gamma, an edge subgraph, which is momentum spanning, with components gamma 0, gamma 1 up to gamma n, where gamma 0 is the connected component which is momentum spanning. And I should add a remark that when you have a momentum spanning subgraph, of course it inherits, it inherits the external kinematics, external legs, external kinematics, because it's precisely connected to every single external leg that carries momentum by definition. So then we have the following factorization formula. Phi gq equals phi gamma nought q. That makes sense because gamma nought inherited the external legs. Then we multiply by the first semantics of the other graphs, and then we take the first semantic of the quotient graph and a remainder term, which I now call the infrared remainder term, where the same story as before, the remainder term has higher degree than the leading piece. So it has degree bigger than h gamma plus 1, because this part here has degree h gamma plus 1. So r has strictly higher degree in the edge variables of the subgraph gamma. So as I briefly mentioned, I don't think this is in the literature anywhere, and to my surprise I think it's a new result. So maybe implicit in some of the very old literature, though not stated explicitly. Is there a similar result of psi sub g? No. So there's an asymmetry in this formula because psi always is sort of, the factorization of psi only involves psi. The factorization of psi involves psi on one side and phi on the other side. And by contrast to the contraction deletion, which involves phi on both sides, and as an exercise you can try to prove the contraction deletion and the factorization and vice versa. So that's quite a nice exercise. And when you do it using this, you realize very quickly that the symmetry in the two phi's here tells you that there must exist a symmetric formula with the phi on the other side. And so they have to both exist. And it's this asymmetry that the phi involves the psi's and the phi's, that means that they're two different formula for phi, but only one for psi. Yes, I was a bit surprised when I saw this. Okay, so back to this example. Let me erase this. So now I can put infrared. So I'll just take a connected subgraph. So let's take the subgraph 2,3, whose quotient is 1,4.  And the remainder term is alpha 1, alpha 2, alpha 3. And let's do another one. We can take the subgraph given by this massive line. Of course masses play no role in this current discussion. And here we get q squared, alpha 1, alpha 2, alpha 3 plus alpha 2. And here for degree reasons, the remainder happens to be 0. Okay, so there we have two different factorizations. So of course this begs the question whether what one can do with regularizing infrared singularities. And I have no idea. But I've attempted to think that the thing, the study we did with Dirk for ultraviolet, some of it at least can carry through verbatim in this formula in the interest. You said the masses play no role in it. Why non-infinite singularities? It's crucial. For now, in five minutes I'll answer that question. Because we haven't considered, phi doesn't occur on its own in the phi minitacle. It occurs in this combination. Well, let me write it. It occurs in this combination, psi g q m equals phi q plus sum e in e g. m e squared alpha e psi g. So of course as you rightly say, we now need to consider masses. And we want to understand factorization formally for psi. So that's what I'm going to do now. So another definition. We take a subgraph defined by a subset of edges. And it is called, again I've invented this phrase. It should be taken with a pinch of salt. Mass momentum spanning. And I will abbreviate this by mm. It's mass momentum spanning. If, first of all, the edges of gamma contains all the massive edges. And two, if gamma is momentum spanning. So basically all the vertices contain all the momentum and the edges contain all the non-zero masses. Sort of obvious meaning. And again if we have generic kinematics, again it's a very weak condition on the external momentum, then gamma is mm if and only if the psi polynomial and the quotient vanishes. Here the direction gamma to be connected or not? No, absolutely not. So for example the masses could be on one component? Absolutely, absolutely. That's absolutely right and that's a very good remark. It's an excellent remark. Yep. And it's for that reason that I'm going to state a slightly simplified version of the next proposition to make sure I don't have very cluttered notation. But it works, I mean it works fine. I'll just write it here and then we'll take a break. So the proposition. So let's take gamma contain eg, but I'm going to suppose connected just for simplicity. It really makes no difference. Then we have two factorization formulae for psi, which is the thing we're most interested in. It's psi gamma psi g mod gamma qm plus a remainder term. This is the ultraviolet factorization. And psi g qm, oh sorry this is for any graph, but let's just put gamma not mass momentum spanning. And the infrared factorization occurs when gamma is mass momentum spanning IR gamma g. So when gamma is mass momentum spanning. Of course I wear the degree of r psi uv gamma g in the alpha e in the sub graph is strictly greater than h gamma and the degree as before I'll divide it in the same thing is bigger than h gamma plus one. So this proposition follows from the previous propositions. You just have to take the definition up here. Definition of psi and you plug in the previous factorizations, factorization formulae and you get this coming out. So you notice that in, to answer a question, that in this second formula, so the first formulae uses the two ultraviolet factorizations for psi and for phi. The second formulae uses the infrared factorization for phi but the ultraviolet factorization for psi. So the factorization of psi is both infrared and ultraviolet in some sense. It appears in both. The way the Schringer parameters appear which is that 1 over phi squared per centimeter takes all of the exponential formula in the sub graph. In general the uv is alpha going to zero and the infrared is alpha going to infinity. Here we're in projective space so it's meaningless. So from an algebraic jump to a point of view this is why I said I don't understand what uv or IRR means because we have a picture like this and we have some bad points. And that's why I'm not, so the renormalization only cares about ultraviolet but from the mathematical point of view the infrared is in there as well and the theory works fine without any problem covering all cases. So I'll explain that. Exactly. So mathematically we don't see the difference. So these letters uv and IR are artificial just for physics intuition. So we'll stop here and then I'll continue in five minutes. Okay so let's continue. I think this is number five. Now we want to determine the bad loci which are the bits we're going to need to blow up. So let's take i, any subset strictly contained in the set of internal edges and sort of here we're working with generic momenta which is again a precise but weak condition that I don't particularly care to write down. So let li contained in projective and g-1 space be the locus where alpha e vanishes for all e and i. So it's a coordinate hyperplane, coordinate linear subspace. Then we have two hypersurfaces what's often called the graph hypersurface x psi g equals v psi g containing projective and g-1 space. And we have a family of psi hypersurfaces which I want to call x psi g and this is v the vanishing locus of the polynomial qm. Of course it's a family but we're going to think of it fiber by fiber. So it's a fiber over some point in some appropriate space of kinematics that I don't care to write down. It's not really difficult. And so the remark is that subsets, i contained, in fact any subset contained in the set of edges of g as we remarked earlier this is equivalent to looking at certain families of subgraphs, edge subgraphs, gamma and g defined by a set of edges. So now let v i for example of the first semantic polynomial be the order of vanishing of psi g along the divisor l i. And so we know exactly what these are from these factorization formulae. So factorization formulae, sorry? And i is not a divisor. No? No, did I say divisor? Oh sorry, yeah, thank you. Another comment from a non-native English speaker is factorization is written in s but I think the z is an endangered letter so I take it upon myself to put z in words. Otherwise there's no use for letter z in English if you never use it. So let's compute the valuation along a subgraph of psi g where we apply the factorization formula for the first polynomial and it's psi gamma psi g mod gamma plus the remainder term. But by definition of the remainder term the remainder term vanishes to higher order than psi. So this is, recall that this is of degree hg and this thing, it was that little remark I made earlier, is non-zero. So that means that the valuation is just the valuation psi gamma psi g mod gamma and what's the order of vanishing of psi gamma? Well it's, when all the variables go to zero it's the degree so this is just h gamma. And likewise the valuation of the second of the psi polynomial is h gamma if gamma is not mass momentum spanning but it goes up by one and exactly one when gamma is mass momentum spanning. So these are sort of an ultraviolet sub-divergence if h gamma is positive. Well in some dimension and these are sort of the infrared ones and they have one degree. I and gamma are interchanged. So I, a subset, I and a subset gamma is I think of a subset as a subgraph. So now I need a definition. So right I'm a little bit embarrassed because I needed a word to define the bad loci and I thought about it and the adjectives I could think of were worse than the one that I'm going to propose. So I'm going to try this out and we can see if you think it's ridiculous we can abandon and think of a different word. But the motivation is there's a word in English called a moat, which is a speck of dust or a particle. This leads to notions of irreducibility and there is no word in English, there's no adjective relating to moat in English but the letters M-O-T-I-C stands for members of the inner circle by a coincidence. So this lends to notions of interconnectedness and of course the motive of the graph will be related to its motic subgraphs. So I will define a motic. Again, so this word does not exist but let's try it out. A motic subgraph gamma contained in G it's defined by a subset of edges is a subgraph such that for all strict edge subgraphs oh sorry this should be gamma a motic subgraph big gamma is a subgraph such that for all strict subgraphs defined by a strict subset of edges which is mass momentum spanning then the number of loops of the subgraph is strictly less than the number of loops of the motic graph. So in this sense it's irreducible so it can't be made smaller without changing something and if the small such gamma then it's empty. So if there is no for all subgraphs which is mass momentum if there's no such graph there is no such graph then there's no such graph which is mass You want to say that instead of which? Any gamma that is mass momentum. What difference does that make? It's the same thing. Because it's including the definition of gamma. Anyway. So if there's no graph that is mass momentum spanning let me think. It should be true because Oh yeah, yeah. But yeah that leads me to an important point. So remark. If there are two cases so if it's not quite an intrinsic notion sorry mass momentum spanning in gamma sorry I apologise I forgot the key point. Mass momentum spanning as a subgraph of gamma it's a relative notion then there's a quality hold. Slightly subtle notion. But so if gamma is mass momentum spanning in G apologise this is slightly technical then we think of it as a Feynman subgraph so it inherits all the masses and momentum from the big graph and then this condition is non-trivial but if gamma is not is not mass momentum spanning we think of it we give it no masses or momentum so it is considered it is considered with no kinematics and in which case every subgraph is mass momentum spanning so for physicists if we forget the condition about mass and momentum so if gamma had no masses and no momentum then this condition is equivalent to being one particle irreducible so I can erase this now so in this case so if gamma not gamma has no masses or momentum then this is gamma emotic if and only if it's one particle irreducible so that means that it's not it doesn't have an edge that you can cut so here's an edge a graph that's not one particle irreducible there's an edge that you can cut and the corresponding subgraph has the same number of loops, it has two loops so this is a slightly more subtle notion related to the existence of masses so this is a physics terminology it's called one particle irreducible this is when you consider ultraviolet divergences this is a very standard notion in physics because you it just comes up all the time but because we're considering infrared stuff as well this condition has the definition has this extra condition of mass momentum spanning and the point is that gamma emotic subgraph in G implies that from the previous formula here for the valuations implies that the valuation along gamma of the graph polynomial psi is strictly bigger than zero because h gamma is strictly bigger than h lower gamma and and that implies the coordinate hyperplane, sorry, the coordinate subspace L gamma is contained in the graph hypersurface and so we get a bad locus so that's something we need to blow up absolutely, so the emotic subgraphs are precisely the minimal set of subgraphs that we need to blow up in order to get a good compactification and because I was worried I wouldn't have enough time, I actually prepared some pictures on some handout so if you'd like to pass them around it saves you having to so now some properties of emotic graphs oh yeah, sorry, of course what's on that sheet is on this board here so ignore this bottom thing, that's not yet these are the sets of emotic subgraphs in the graph G so let's give them names this one is mass momentum spanning because it contains the massive edge there's only one of them and it meets all the external momentum this one is not mm this is mm this is mm and this is mm so when the variable alpha one goes to zero you can see in this polynomial alpha one divides every monomial so it vanishes along alpha one equals naught and that's an infrared sub divergence that's an infrared sub divergence this one when alpha three and alpha four go to zero again you can check that the whole thing vanishes this is an ultraviolet sub divergence and to go back to the question earlier these graphs here are both infrared and ultraviolet I don't know what to call them we don't really see the difference in this geometric picture so they're sort of mixed they're both infrared and ultraviolet whatever that means and you can check that they all have this minimality condition so if you cut an edge here in this graph well you have to look at if you cut an edge, the edge is massive so so you can't cut that edge and it's motic in these graphs if you cut this edge then the number of loops goes down and you can check the definition on that motic means either UV or infrared problematic it means problematic at all in any dimension of space time sufficiently large dimension of space time and it's minimally problematic so it's not just the problematic sub-graphs it's the minimally problematic sub-graphs for some arbitrarily high dimension that's where you have to do your geomagnetic job your basting of geomagnetic and that's why I avoid the use of the word divergent because divergent refers to a specific well for me it refers to a specific dimension of space time so theorem you take two edge sub-graphs property Q for quotients if gamma 1 is contained in gamma 2 and gamma 2 motic in G then gamma 2 quotient a quotient of a motic graph is motic extensions if gamma 1 contained in gamma 2 and both gamma 1 and the quotient gamma 2 mod gamma 1 a motic then this implies that the big graph gamma 2 is motic unions if gamma 1 and gamma 2 are motic then gamma 1 union gamma 2 is also motic and then contraction if E is an edge in G and the edge contraction is motic then it comes from a motic graph then either and then at least one of gamma or gamma union E is motic I have a suggestion why don't you call them problem motic very good problem motic I like that very much problem motic deserves a citation I wish I had thought of that but now I can now I will now blow ups linear blow ups I have forgotten what number this is 6 maybe linear blow ups and projective space okay so s a finite set and p s is a projective space of dimension mod s minus 1 with coordinates that's exactly what you've been doing all along r for s for s and s now let's choose any subset b be a subset of the power set sorry an element of the power set so the front notation is 2 to the s this is a set of subsets of s with the property that it's closed under unions which is so if i and j are in b implies i union j is in b okay so now we define the iterated blow up so call it p b it's a blow up of projective space p s and what we do is we blow up following an absolutely standard procedure along all the l i where i is in our set so b stands for bad low chi or low side bad or blow up other things we want to blow up and you do it in the following way first of all first of all we blow up the points so the linear subspaces which are in the set to be blown up such that the dimension of that space is zero so we blow up first all the points and then after that we then blow up in the new space i is one element or component one element yeah so the number of elements is number of elements of s-1 yeah so this is s-i equals one and then we blow up the strict transforms sorry oh well then we skip this step we move straight on to the dimension one so we blow up so this could be an empty set then blow up all strict transforms of the l i i and b such that the dimension of l i equals one so I don't know think of these as the lines the p ones and you keep going so keep blowing up in order with the dimension two and so on what needs to take the transforms oh so when you when you blow up like here so when you blow up you can look at the complement in projective space of all the bad points it's an open subset in that open subset you have some divisor like this this x psi g there's a map that way and look at its inverse image on that open space it's just this part here and take its risky closure so it's exactly this part here the full inverse image also involves these red exceptional devices so the part up here that maps down to here is all the red stuff the strict transform is the red stuff that does not involve the exceptional devices so it's the bit that's strict I don't know the total transform is everything including the exceptional devices absolutely and the first point is that it's well defined the key point is because of this condition star at the case stage the strict transforms of the Li1 Li2 for dim Li1 so there are two things we want to blow up but because we've already blown up their intersection by the condition star they are disjoint so they're far apart from each other and it does not depend on the order so as long as you blow up in this strict order of dimension it doesn't matter in each step what you choose so that's a very well known fact so PB has a stratification I have to accelerate it a little bit so it's code I mentioned one strata so it's the red stuff in that picture code I mentioned one strata are following they are the DI which are the sorry my colours are mixed up I take that back now the stratification are the DI which is the total inverse image of the original coordinate hyperplanes Li so that's what the sides of the triangle we started off with and then the deep we also have new sides so this is the total inverse image of the Li corresponding to everything that was blown up so they're not yeah these are divisors no sorry this is great I've taken one and there's a product structure so this divisor DI is isomorphic to P a space of the same type of P I cross P S minus I where B I B lower B upper I is the set of subsets in B which were contained in I and B lower I is the sets J minus I where J is in B and G contains I so it has this nice recursive structure that every divisor is a product of spaces of the same type I'll rub this out I was going to talk about operads but I'm out of time so I'll just talk about hopfartipus instead and then the DI that often gets neglected in this business is isomorphic to a point where I cross P B subscript I and that's a blow up of P S minus I so it has this nice recursive structure and of course the union of these divisors of all the D's is strict normal crossing ok so let's apply this for Feynman graph definition for example last time about these modular spaces they have this very nice operatic structure but I actually forgot to mention this product structure which can be thought of as particles going in different directions that was an omission last time but in any case it doesn't matter because if we take G, a Feynman graph in general we get the same structure because we can set to be the projective space blown up in all the motic subgraphs and there's this condition of being closed under unions and that was a property that was satisfied by motic subgraphs so actually what I should have done here was to draw these circles in white so so far we've been discussing the whites we've defined this blow up it's got nothing to do with the red line yet so there's this recursive structure in the white the white devices which are just related to projective space and have got nothing to do with graph hypersurfaces for now ok so now let Y Psi G and Y Psi G Q M be contained in sorry no be contained in this compactified space for graph to be the strict transforms so in this picture it's the red line now now that I've changed the colours of respectively the graph hypersurface Psi G and the graph hypersurface so it's the what remains upstairs and the theorem which is absolutely key to everything following so first of all this space Pg is well defined and that was the property that motic subgraphs were closed under unions it has this stratification by devices so what are they they are of two types they are the subgraphs which are motic with at least two edges and they are just the edge devices that's just the edges of this triangle in this case and so again by the so we had this recursive structure in the general setting so D gamma the facet cross 1 D gamma is itself a product of the same structure for the subgraph and quotient graph and that uses the quotients and the extension property of motic subgraphs and the edge divisor is a point cross Pg double slash e and that uses the contraction property of motic subgraphs and now we want to use the factorization formulae so why the graph hypersurface when it intersects with a coordinate hyperplane or this version upstairs then by the contraction deletion formulae that's exactly the same situation but for the contracted graph so that was contraction deletion then the factorization properties said that when you intersect this graph hypersurface with one of these facets you get two irreducible components which have this nice product structure and likewise for the other hypersurface and when we intersect with one of the blown up facets we get sorry one of the blown up facets we get the other graph hypersurface corresponding to the first semantic so this is if gamma is not mm so it's a pure ultraviolet subdivergence or potentially ultraviolet subdivergence and in the case when it is mm we get this and so if gamma is, so this is an infrared thing okay so as Pierre said essentially this defines a sort of upright structure on this geometry I'm in the generic situation I'm always in the generic situation so this is a fight of a family and we blow up the fibres and it will be nicely behaved on some big open set how does this hypersurface no no so that's the point they're absolutely highly singular they're extraordinarily singular and it's good that they should be singular because otherwise I should say these are y psi g and they are very singular and we don't understand the singularities in general and in fact they need to be singular because if they were smooth we wouldn't the odd numbers would be all wrong and we wouldn't get multiplicative values so because we see multiplicative values we should expect these to be very singular and you can also in this setting you can ask what is the discriminant and again it's something fiendishly complicated in the study of Landau singularities in physics and it's a tricky business so there's divisors of two types there are so each the divisors are here there are six divisors there are the ones corresponding to each edge so this is edge two and so I should draw the quotient graph here so I contract the edge two and I get the quotient here here I contract the edge three and I get this graph and here I contract the edge four and then there are other um divisors which come from blowing up and they corresponded to subgraphs so here we get the subgraph was three four and I will write it three four tensor two and it will be a product of the corresponding spaces corresponding graphs this one is two three that's how we label so the D is the set of all these white divisors there are six of them here and they fall into two sets those that corresponded just to edges contracting an edge in the graph and those which we actually blew up and they correspond to the motic subgraphs okay so very quickly the idea of the proof of this is rather simple let's just look for example at the local coordinates for a single blow up um L I where for convenience I is just the set one up to I in P A minus one then the local the coordinates for a single blow up um and on some I find chart are given by just changing variables so as I explained last time we can blow up um see what happens locally by passing to new variables um and what we do is we write um psi G let's say for example equals psi I psi G mod I plus R take the factorization formula and we write it in the new coordinates so write these in the new coordinates so that's how you compute the strict transform and then you take the limit um well you don't need to take the limit you just get rid of the exceptional divisor something will factor off um and actually in the limit you will get um you will get the R will go to zero because it's of higher degree in the first variables and it will become um psi I psi G mod I and so the zero locus um um so that the intersection of the strict transform with this uh corresponding facet will have this equation and the zeros of this equation is a union of two pieces and it's exactly of this type so that's the essential reason why this works it follows from the factorization formulae and for physicists acquainted with the old literature they should recognize these types of ideas in a different language were used by by Hepp um so I think if I understand correctly he blew up everything but there's sufficient work by by Spear where he defined a sort of minimal set of coordinates to understand the singularities of these polynomials and I suspect that it's very similar to the story that I'm telling here so um remark in the case where um G has no masses or momenta um and primitive then this um and log divergent then this is the same space Pg is the same space as defined by so they did this first in the case of um in this particular case um there's another case that um I did with uh Dirk Kreimer for something called single scale graphs in order to understand renormalization in the context of algebraic geometry and actually it's a different it's slightly different from this one it's a different compactification in algebraic interpretation of renormalization but I presume one can renormalize directly in this geometry by jazzing it up a bit okay so um so here's the example of um of this of this situation um we had this is the graph this is the list of emotic subgraphs and so it tells us we should blow up um so in we blow up first the points which are the subvarieties defined by these so we blow up the points first of all the points um L 1 2 3 L 1 2 4 and L 1 3 4 in any order and then secondly we should blow up the line L 3 4 and this subgraph plays no role so um here's a picture of the um this is a picture of the coordinate hyperplanes in p3 the three bad points are our red dots here we blow them up they're just they're disjoint so it doesn't matter which order we do it um you can think of a blow up as simply taking a piece of cheese and cutting off the corners so after blowing up um the piece of cheese will look like this that's the first step where we've blown up these three red corners and then there remains to blow up um this line which is the red line here and then the second step we blow up that so we cut this edge off the cheese and we get a uh a figure a polytope of this of this shape and you can check that it's a picture of this space pg and it has um facets in one-to-one correspondence with either edges of the graph or motic subgraphs no no no no it has triangles okay so um so we so what happened to the coordinate simplex we had sigma in p and g-1 r was the coordinate simplex and so it's just this the locus where all the projective coordinates are real and positive so it's the tetrahedron on the left in that figure and we define um the Feynman Feynman polytope for want of a better name Feynman Feynman polytope um which is sigma tilde g which is the inverse image of sigma and it's contained in this blown up in the real points of this blown up space and so pi g is the the map from pg to p and g-1 so sigma tilde is is this um polytope and the polytope is a nice way to visualize all the geometry that's going on in the Feynman integral and hence the motive so the facets of sigma tilde g are in one to one correspondence with either the edges e of the original graph when you write pi g inverse of sigma do you mean you take the inside of sigma you take the inverse image and then you take what I mean do you want me to write yeah analytic topology yeah sorry bit sloppy yeah sigma that's the interior and that's yeah um so the facets are the edges of the graph and the motic subgraphs and of course it has this nice structure so sigma g intersected with d gamma is isomorphic to products of Feynman polytopes we have this recursive structure and um so from this we get a hopf algebra so I think that the correct concept here is that of an operad but um I'm slightly out of time so it's quicker to define hopf algebra so let f be the ring generated the uh the vector the free z module generated by disjoint unions of Feynman graphs um so it's it's graded so normally um when you do hopf algorithms of graphs you grade with respect to the loop number um that's possible but I prefer to grade by the number of edges for reasons that will become clear and this is an algebra where the multiplication um is disjoint union of graphs you just put them side by side I should perhaps say that all my graphs are always labeled the edges are labeled and on this we have a absolutely absolutely so we have a co-product following um con and chimera um which to a connected graph associates the linear combination gamma tensor G mod gamma gamma subgraph NG so this is much more general than the con crime hopf algebra because it contains the infrared so my understanding is that the con crime hopf algebra sees the ultraviolet information um and maybe I'm I'm ignorant of the literature but um it seems to me that the infrared case hasn't been considered um this contains um this contains all the infrared sorry gamma is gamma is emotic which is a so in the so in the special case um with no masses and no momenta this is just ultraviolet then we get this is called the core hopf algebra defined by the concrimer and is in some sense the limit of the concrimer hopf algebra is in all for all dimensions um and then on this we have a differential so the fact that this is co-associative follows from the properties of quotients and extensions of the motic graphs are stable under quotients and extensions we also have a a differential for every edge we have a differential f to f which sends g to g contracted e and then we can define d to be the sum following maximum minus one to the i d e i and then the the result the corollary is that f is a differential of the hopf algebra I wouldn't actually need this but it's just enumerate unordering yeah yeah yeah otherwise you have irritating you have to worry whether this is plus or minus because when I play enumeration doesn't work here no so I just take a disjoint union of graphs which have ordered edges and then the Leibniz rule tells me to differentiate a product so I start with the lowest edge in each component there are different ways to get around this I mean that's one way to do it so what that means is that one delta is co-associative which is the key fact for the renormalization group equation no no no it's not so yeah so this is commutative but not absolutely not co-commutative yeah so it functions on some groups so as Doug explained to us the hopf algebra of graphs gives you the theoretical explanation for the first formula and all the recursive structure renormalization and the renormalization group equation so but bizarrely this holds in the infrared case as well and I don't know what that means the differential structures this equation and of course these differentials edge contractions commute so so the proof actually what's nice about this geometric picture is the proof of this is completely obvious if you have in your mind this polytope because if we think of this as a as a simplicial complex there's a map there's the boundary map the interior the interior stratum is indexed by the graph G and we want to take the boundary of this polytope so the boundary of the polytope is the sum of all the facets the facets being so the facets of two types either they're the edges and that gives us the differential or they are of the blown up with the corresponding subgraphs and then so a fine graph for you has labelled edges so for example if you have various labelling of the same graph you take it many times you mean yeah it's heavy handed but yeah so the remark is that the boundary operator d squared equals zero so if you take the boundary of the boundary you get zero and that's so this statement the fact that you've got a differential graded Hopf algebra is just this fact and in the handout I explained how to see that we can look for example at the two skeleton of this polytope we look at this edge here and this edge can be viewed either in the boundary of this facet and it gives the same edge and that's exactly the co-associativity so once you have this geometric picture this statement is completely trivial so proof look at the code I mentioned two skeleton of sigma tilde g and so next time we want to study the graph motive so mg which will be a relative co-homology pg minus yg relative to d minus d intersection yg so here yg is the union of these in the generic case it will need both both hypersurfaces and so the first thing to do next time is to define a Betty homology class which will be the sigma tilde the Feynman polytope and from that define a matific period and then there will be so then we can then speak about the matific period associated to a Feynman amplitude and the main theorem which says that the conjugates are most of the conjugates of the same type will follow from some spectral sequence argument applied to the geometry of this the inside of the big category that you call h so I only had time to define the category h which is when you think of a period as a number in the classical sense so here what you need to do is you need a good notion of a period function a family of periods and you need a tenacian category of not mixed hot structures but variations of mixed hot structures so the Betty will be the category of finite dimensional of local systems of finite dimensional q vector spaces equipped with an increasing filtration the Duran will be integrable algebraic vector bundles with regular singularities at infinity equipped with an increasing filtration and a decreasing hodge filtration satisfying Griffith's transversality the comparison will be given by the Riemann-Hilbert correspondence and you can set up that this is a tenacian category and you can set up the definition of motific periods in families in exactly the same way I know that this tenacian category goes to h by a fully faithful function so if you specialize at a point no you don't you can't take limits so I can't I don't understand what happens if a momentum goes to zero the homology can jump so I don't know how to do that that's something that someone needs to look at that could be very complicated but over the generic point yes superficially the formula for the motific correction will be the same though you're in a different op-file so they could be different relate so you could have something like the motific logarithm that you view as a function and then you look at the motific logarithm at a point and at this particular rational point it could have some special functional equations that aren't true in general specialization is a tricky business but with generic the ideas that will have a space of we look at all Feynman diagrams with a certain number of incoming momenta and a certain fixed set of masses that's a space of kinematics and on some huge open set in this we have a tenacian category of variations of mixed-hot structure and every graph with that kinematics defines a motivic period on this tenacian category of cosmic Galar group acting on all those Feynman diagrams and it works fine but I won't talk about that because it's it's not enough time


Good morning. I'm Mary Ellen from the channel Mary Ellen after 60 and I'm here today to talk to you about why I'm no longer subscribed to any beauty boxes just to chit chat and to announce two winners of the IT cosmetics today special value giveaway. I've loved makeup ever since I can remember. I've just loved everything about it. The very first subscription box or bag or whatever subscription service that I ever subscribed to was called Cosmetic and I'll type the words up here because I'm not going to put any links for any of these subscription services below. I'm sure I read about this Cosmetic in a magazine and if any of you all got that let me know. It was so cool. At the time that a lot of the brands that they had in their monthly subscription we couldn't find around here. 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I was trying to buy one season of the subscription box for one of my viewers and it really aggravated me that they didn't have e-gift cards for the price of one box. So that's a whole other video but anyway I'm no longer subscribed to that. I do miss it. When I canceled my Ipsy I got an email a couple days later and they said do you want to try the Ipsy Glam plus box and it was like $25 or $27 a month and I got two of those and I've canceled that. So this is the first time in many many years that I don't have a subscription service and I tell you I do kind of miss having something to look forward to but I'll tell you what I miss more than anything is the envelope from Ipsy coming to my mailbox every month but I don't think I'm going to be subscribed to anything. Jim and I are doing very well staying in even though we're not used to being in the home all the time 24-7 together. It's gone a lot better than I kind of thought because Jim somebody that likes to get out and about I thought it would be hard on him but it hasn't been. I tell you something I have learned to kind of kind of enjoy my skincare since I've been home. It's just something that has been normal so I don't dread doing my skincare like I did. I tell you I'm so happy I listened to my mother and her frugal ways. She lived through the depression and there's so many things that I've learned from her and I'm not wasteful. I've never been wasteful and I'm happy that I'm not. The last nine years of my career I worked with a fabulous psychiatrist and anytime he saw a patient I was in the room with him. One of the questions he always asked the patients that were anxious or depressed is how much cable news do you watch in a day and a lot of the time it was 24-7 and he told them he said watch your local news and read your newspaper because it will provide to you all the information you need and you won't get stressed out and I pretty well followed that. I've never been one to have cable news on Jim is but a lot of times he watches from another room. On March 6th I had some lab work done. I have to get routine lab work and my cardiologist from St. Louis wanted my electrolytes done and a panel of some sort. Well it wasn't fasting but a lot of times my blood work isn't fasting and when I saw the results and nobody has said anything to me my non-fasting blood sugar was 117. Well it has never been that high and it's not bad for a non-fasting blood sugar but on March 6th I decided to truly clean up what I was eating and I have been excellent ever since and to tell you the truth I'm really proud of myself because when you're in what do you think of a lot of times oh what sounds good what can I eat next and I'm just determined to the next time I have lab work done to go in and have it drawn non-fasting that my blood sugar is going to be lower and that is really motivating to me. I plan to continue uploading videos on Sunday as long as I have something I want to talk about and usually I find something I want to talk about. I have two sisters that watch me every Sunday one of them comments and one of them doesn't and the one of the one that usually doesn't she said to me she said when she brought up my video last Sunday the reds the box was red and that meant she wasn't subscribed to me she said I never did hit the unsubscribe button she said so something's going on so if you all are wanting to be subscribed make sure that the button beneath my videos is not red. I went to announce the winners of the QVC TSV giveaway. The first place winner receives it celebration foundation it's a powder foundation and the brush and that's what I have on today I have the celebration foundation on today and I did my little random comment picker and the first name that came up was Rose Brennan. Rose since your number one even though you said second you'll get the brush and the powder unless you don't want the brush but it's yours and so I hit the button and another name came up and that this name was Cindy Forster. Cindy you also get an IT cosmetic celebration foundation and if you want it I will give you my brush but that's just optional so congratulations to Cindy and congratulations to Rose. Since I've been an adult any time that I bring new clothing, bedding, brushes, no matter what it is I never use anything unless I wash it because you never know where it's been so even though I wash this brush of course you need to wash your brush and of course you need to wash this brush. The only thing I didn't ever launder was something that had to be dry cleaned and I would hang it outside for half a day and that's just what my mother told me to do and if you're still here this is going to be another giveaway. 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It's red since it gives since this is it since is it had so many different and that just kind of ticked me off like the $60 and they haven't changed anyway. I was trying to buy a subscription one but when I come when I I promise I didn't unsubscribe I promise I didn't unsubscribe but you know what I miss more than anything and I'm not gonna I'm not gonna and I just didn't need to get in the meantime