So welcome to Paris Peking Tokyo Seminar. So it's my great pleasure to introduce the speaker today, Yixing Liu from Yale University. So he will tell us on the very some block of conjecture for ranking self ag motives. So please start. Okay, thank you very much for the introduction and the invitation. It's probably one of the largest seminar I have ever given more than 100 people. So I just realized that two years ago, Yi Chaotian also give a talk in the seminar on a similar, on the same topic, but that time was during the middle of our project. So the results at that time was not the final one. So today I have many two girls, one is to state the main results. I mean, our final results obtained from this long-term project. And then I will discuss the main strategy. And some of it, there might be a little bit overlap with each house talk, especially on the arithmetic geometry of certain shimmer variety. Then I will emphasize a new technique. I mean, in the study of Selma group using defamation theory of Galois plantation. So I will start from the main results. Okay, so this talk is based on the joint work with Yi Chaotian and Liang Xiao, who is now in picking and Wei Zhang and the Xing Wenzhu. And we started the liking 2017, I think, and finished at the end of last year. So some notation. So we will fix a CMS tension of number fields like F over F plus, we see the complex conjugation. And then we denote by gamma F, the absolute Galois group of the imaginary field F. So here's our first theorem. So our first data theorem that is more elementary to state. So let N be an integer that is at least the two.  So there is no CM multiplication. And some additional conditions. So we suppose that first, they are not geometrically isogenous to each other. So we consider the M minus one symmetry power of the first one and N's symmetry power of the second one. We assume they are both modular. And we also assume another technical condition is that we assume the F plus is not Q, if N is at least the three. Okay, so I will explain why we need this, I mean, little bit later, but let me finish the statement first. Then the statement is the following. So suppose the central critical value, which are writing a classical way, okay? In a classical way, N is the center of this function. So in particular, if you allow yourself to take N to be one, which I didn't, but if you allow, then this is just simply the central value of the L function of the base change of the second in the curve A prime, okay? So in general, we consider a more, I mean, more complicated L function like this. If the central value does not vanish, then the building some lower color conjecture, which is the generalization of BSD, sorry, predicts that the Selma group should vanish, okay? Then the conclusion is that the blockado Selma group, which I will recall a little bit later, of this galore plantation, okay? So this twist N is somehow echoes this N here, okay? So this is the correct twist of this galore plantation, vanishes, so this Selma group vanishes for all, but finally many rational prime L. So conjecturally, it should vanish for all L, but due to some technical complication, we only prove for all, but finally many L. By the way, this all, but finally many is effective, okay? It's not L, I mean, like some large enough, but there's a way to say how large enough. Oops, sorry. So before continuing, let me give the, recall the definition of blockado Selma group. So in this case, actually, the definition is quite easy. So for galore plantation of F, so gamma F to GLV on a finite dimensional vector space V over a finite tension of QL, I mean, here it's just QL, okay? We define this H1F, this subscript F stands for finite, okay? To be the subspace of all the first cohomology of this galore plantation consisting of those classes, whose localization belongs to the local version of this finite part for every non-archimedean place V of F. So what is this local version? So when L does not divide V, it is just that unremifed subgroup of this H1 of this local galore cohomology, which everybody knows. But when L divides V, you need a little bit, so L, I mean, actually this should be shi-addigo-Hodges theory, but since we consider L, it's the L-addigo-Hodges theory.  When V is a crystalline at this place V, then we know that if you ignore this F, then H1 parameterizes extension of this V by the trivial representation. And this H1F denotes the subspace of H1 consisting of those extensions that are still crystalline. Okay, so that's one interpretation. You can just define this in that way when V is crystalline. But in general, you need a little bit more on the L-addigo-Hodges theory. So that's the definition of block-hadocellular group. And there's a close relation of this fancier definition with the classical definition of the Selma group for an Idb curve, as this is actually the original motivation for block-hadocellular to give this more general definition. Any questions so far? Good, so. Yeah, it's fine. Yeah, so this is our first theorem which is stated in terms of symmetric power of Idb curves. So you may ask, I mean, so when we say this is a unconceivable motive, do we only obtain like things must be like symmetric power? The answer is no. We have another set of theorem which applies to more abstract automorphic representation. Now I will try to state. So since we are going to do automorphic representation, let me give the definition of those automorphic representation we are gonna consider. We call this representation relevant because they are relevant to our work. So there are three of them. So first, we require pi as the irreducible hospital automorphic representation. Second, we require pi to be conjugate self-dual. So if you compose pi with the conjugation of F, it should as equivalent to the contragradient of pi. So the third condition is pi has certain homological weight that is a minimal. So more precisely it says the following. So for every Archimedean place tau of F, this pi tau is the principal series of this N characters. So what are these N characters? These are just, this ARG means the argument to character, just the argument of a complex number which you bifomulate the z over square root of z bar. Okay, so this is the generalization of the weight two case in the classical for the classical modular form. So for those who are familiar with this Archimedean theory, this representation, so this is the admissible representation of GLNC. And this representation is exactly the standard base change of the trivial character of the definite unitary group of rank N, okay? So before we will only consider this kind of automatic representation. So note that for a coefficient field E of this kind of automatic representation and every finite place lambda of E we may attach a Galois representation for the row pi lambda. I mean, actually this is a compatible system of Galois representation from gamma F to GL and E lambda. This is achieved by a series of work I mean, for many years when Harris Taylor, Subushin and the finally, but shouldn't be Harris. And so this is some kind of common knowledge now. So you may attach Galois representation for this kind of automatic representation, okay? So now I will introduce an elementary but very important notation in this talk. So in what follows, we will consider an integer N which is at least two and we will denote by N zero and N one, the unique even and all the numbers in N and N plus one. So sometimes N zero is N plus one and sometimes N one is N plus one depending on the parity of N. So this is a, and we also denote by R alpha which is the floor of half of N alpha if you write in formulas this way, okay?  Because for later argument, the most important thing is not to which one is larger. It is which one is even and which one is all then. So, oops. Now we also give another technical definition. You don't need to memorize this definition, okay? So it's a technical one. We will say that a special, we will call a prime P of F plus is a special inert prime if it is of degree one over Q and inert in F and whose underlying rational prime P is and ramified in F, F is the CM field. So actually the only thing important is this word inert, okay? And this degree one is totally just technical. It's not necessary. And this underlying prime P is ramified in F which simply means P is large enough, okay? So the only thing keeping your mind is the word inert and this is already reflected in this terminology, okay? So the important thing is this P is inert in F. Now I can state the theorem in the abstract version. So we consider two relevant to the plantations I call this pi zero pi one of GLN zero AF and the GLN one AF respectively, okay? And I let E be a number field that is a coefficient field for both pi zero and the pi one. Again, there are some conditions. So first suppose that there exists a, I mean, you can ignore this very, doesn't matter. This is a special inert prime P of F plus such that the even rank one, the even rank one is Steinberg and that all the rank one at P is ramified who the Satake parameter contains one exactly one. So let me explain a little bit about this condition. So since pi one is conjugate self-dual. So it's Satake parameter is given by N1 complex numbers, invertible complex numbers that appear in pair. So it's alpha, alpha inverse, beta, beta inverse, gamma, gamma inverse. And since N1 is all the rank, so one must appear in the Satake parameter. So this condition is a genericity condition, which means that it exactly contains one once, okay? It should not contain three one or five one, just exactly once, okay? And the second condition is easier, which means that for each of them, pi zero and the pi one, it is super-cospital somewhere, okay? So for alpha is zero and one, there exists a non-archimedian place W alpha of F such that pi alpha at W alpha is a super-cospital. Condition C is same as before. We require F plus not to be Q, if N is at least the three. So the conclusion is same as before. So if the central critical value, which I write in the automorphic way, which means half is the center, is not zero, then for all but finitely many primes lambda of E, the block-cuddle-selmo group, H1F with the corrected twist vanishes, okay? So let me explain a little bit why we needed these three conditions, okay? So Y, A, B, and C. So A is needed really due to our method, okay? So the main method is so-called arithmetic level raising. So A is needed for the existence of so-called a level raising prime, which we will go with more details later. So namely, we needed those primes at which the Frobenius acts on this color representation via a special way modulo of various lambda, okay? So I'm vague now, but I will be more precise later, okay? So I mean, as already you can see, this condition must specify some particular elements in this color representation, right? And how about the B? Okay, B is easier. I mean, B is needed so that these two color representation are absolutely irreducible modulo, all but finitely many lambda, okay? So due to our technical reason, we need residual absoluteness for these two color representation.  So I mean, conjecturally, conjecturally B should, this sentence should automatically be satisfied even without this condition. But our current knowledge doesn't give that, okay? So we put a condition to make sure this is known under condition B. I mean, B is not really needed. It is always should be true that it is absolutely irreducible modulo, all but finitely lambda. Okay, for those experts, let me just give another important remark. You may feel that this condition might be used for the gangloss-proscillic conjecture, but actually no, okay? We have already removed this condition when we use this gangloss-proscillic conjecture, which I will say to later, okay? So this is not for GDP conjecture. This is only to ensure that the color representation is absolutely irreducible, residual absolutely irreducible for all but finitely lambda, okay? So C, so how about C? You see this condition C also in the previous theorem for NP curves. So C is needed because two ingredients we need are not available where F is Q and N is at least the three at this moment. So the first is the cohomology of the stable part of the unitary Schmure variety, which is not compact when F is Q. And somehow I cannot find the literature that is responsible for this cohomology at this moment. And the second, we need a result by Karani Schultzer on this vanishing of non-middle degree cohomology that this recent work. And some of you may know that recently Karani Schultzer extended their result to non-compact Schmure variety as well. However, they still assume F is not Q, okay? Even for their recent result. So when F is Q, there were two difficulties for all these kinds of results. First is that the involving Schmure varieties are not compact anymore. That's the first difficulty. The second is difficulties that the appearance of so-called a Cuspid subgroup. That is something complicated during this stabilization of trace formula. And it only appears in the unitary group case when F is Q, okay? So I mean, C is still some technical issue. So as long as these two things are clear, the one F is Q, then our theorems can automatically extend by removing C. In fact, when we write our paper, we just assume these two things and go through the whole process without assuming C, okay? And the questions for the main results. Okay, good.   We also obtain results. So you can see this is, all these results are in the rank zero case. So L value, the order of vanishing of L value is zero implies the Selma group has rank zero. So you may also ask, can we obtain something in the rank one case as Colivagan did? The answer is partially yes. We do have a partial result toward the rank one case, but I'm not gonna state because first it's technical, second, it's just a partial result, okay? So I'm not gonna state that, okay? So now I'm gonna explain some main steps. So, yeah. So first step is how to use this condition. How to use this condition. This is really some analytical condition, right? I mean, how to really use this analytic condition to obtain some arithmetic information. So this is the GGP conjecture, which is our now state first. So step one is to use Gangl's Prasad to relate this L value to certain period integral on definite Shimura set. So let's start from this condition. The central value does not vanish. So there we can obtain the following conclusion. The conclusion is quite long, okay? So, but let's do this slowly. So first we can find a unique pair of totally positive definite Hermitian space, VN and the VN plus one over F in which VN has dimension N and the VN plus one is simply VN direct sum with a unit line where E has norm one. Okay, that's the first thing. The second thing is that for alpha equals zero and one, we can find actually unique irreducible subrepentation pi alpha of this adelic unitary group contained in the, in this space of automorphic functions on UV alpha, such that, I mean, this is hospital automatically because it's definitely unitary group, okay? Such that the base change of pi alpha is the bigger pi alpha. Okay? So there's that we can find the OE value of the functions F zero in pi zero. I mean, by construction, this representation automatically realized on the space of automorphic functions. We can find the functions F zero in the first representation of F one in the pi one such that the diagonal integration is non-zero. So this is a period integral. I mean, I'm writing this in a fancy way as an integral, but in fact it's just a finite sum because as long as you fix an open complex subgroup, this double quotient is a finite set. Okay, so this is in fact a finite sum but I'm just writing it in a fancy way. Okay, that's the whole, this is the content of the Genghis-Prasad conjecture at least one direction. This is the content of the conjecture and this is known now, okay? Recently known now. So let me explain a little bit of history. So this result, which I mean this one follows from a serious work by originally the beginning by Jackie Wallace who proposed a relative trace formula approach to this conjecture. And then Zhiyue Yun established the fundamental lemma toward this trace formula. And by Wei Zhang who solved this smooth matching problem at non-archimedian places. And then by Hang Xue who solved the smooth matching for archimedian places. And I mean, there's a, and also by Vaspu Jia and the local aspect of this conjecture. And on Buzha Plessis on very important refinement and Buzha Plessis and Zhiyue and Zhiyue and Zhiyue are very important refinement on Wei Zhang's work. And up to this point, we know this conjecture but we have to assume that both Pi zero and the Pi one are super-cospital somewhere. And this is one of the most stubborn condition in all of this the trace formula stuff. Okay, it's because at some point we want to use simple trace formula on the spectral side. So we have to assume the automorphic representation is super-cospital somewhere. But until very recently, by our recent joint work with Buzha Plessis myself and Wei Zhang and Xing Wenzhu, we discovered a new technique that can remove this condition. There's a restriction that the representation is super-cospital somewhere. So we now have this full version of this theorem. So this is now unconditionally known. Okay, so that's the content of the step one, this GDP. So we have a start from the non-vanish of the L function and we obtain some period sum or period integral is non-vanishing. So the next question will naturally be how to use this condition, okay? How to use this condition, okay? So let me put a little bit more condition because apparently we don't want to write a such long expression. So I will put a S, V and R for this quotient. So this S stands for set, okay? Because this is a Schmura set, so S is for set. So we also fix the choices of F0 and F1 as above. And we also fix an open-compact subgroup of this UNVF that fixes F alpha for alpha equals zero and one and we'll carry them implicitly in the notation. So we will not write any complex subgroup in the following discussion, but please remember there are fixed the open-compact subgroups. And we also fix the finite set sigma plus of primes F plus all the side of which everything is N-RMA file, okay? So basically sigma plus will contain all the ramification that appears up to this moment. And the question so far, now we will go to step two. Okay, great. So step two, this is the point where arithmetic geometry enters. So far there's no geometry, there's only a set. So this set will be related to some quite beautiful arithmetic geometry via some better reduction of some integral model of Schumer varieties. So I guess this part, each should have explained some of them two years ago, but I guess you should probably forget when they are. So let me recollect. Oh, it's two years ago. Yeah, yeah, I understand. I mean, so, okay. So I will just hopefully you can remember, you can recall some of them. So let me fix a special inner prime P of F plus, whose underlying rational prime P is called prime to sigma plus, okay? So if you forget what the special inner mean doesn't matter, this only means basically means P is inert in F and the underlying rational prime P is sufficiently large, okay? I mean, nothing else is important. So for alpha equals zero one, we will construct a strictly semi-stable quasi-projective scheme, I would denote by the bolder face which means the integral model. The bolder face MP, P states for the sub prime P and the V and alpha states for the corresponding Hermitian space over spec ZP square. I mean, in general, it should be FP, but since we assume P has a degree one over Q, so FP is just the ZP square, okay? Over spec ZP square of relative dimension N alpha minus one. Of course, I mean, satisfy some condition. Okay, so what's the property of this integral scheme? So I will summarize them as follows. So the generic fiber of this scheme is essentially just a Schmurr variety, but not of our original Hermitian space. Our original Hermitian space is totally definite. So the corresponding Schmurr variety is just a set. It has dimension zero, okay? So this Schmurr variety is attached to a somehow slightly different Hermitian space. We call this a nearby Hermitian space. This Hermitian space is unique up to isomorphism satisfying the following condition. So the signature, instead of being totally definite, this one actually has signature N alpha minus one one at some fixed Archimedean place tau of F plus, okay? So we switch the signature at one Archimedean place. And we also require it is same as our original Hermitian space for all other places other than this Archimedean place and our chosen special inner prime P. So by Hauser principle, this will force that V prime and V are different at the P, okay? Otherwise you only change the Hauser environment by one place, it's not gonna work. So we change the Hauser environment at two places. So you get this another Hermitian space. Remember, because we have already assumed this P is co-prime to Sigma plus, which means that the original Hermitian space should be unremifed at the P, which means it should admit a self-dual lattice at the P. Now, since V prime is different from V at P, this will imply that the V prime does not admit a self-dual lattice, okay? And at the level structure at the P, we will require it is a stabilizer of an almost a self-dual lattice. I mean, if this one does not admit a self-dual lattice, then it admit an almost self-dual lattice. And we require that level at P should be the stabilizer of an almost a self-dual lattice, in which case it is a special maximal subgroup, but not hyperspecial. Oh, sorry, sorry, it is special, okay, special maximum. Okay, this is the property of its generic fiber, okay? So how about special fiber? The special fiber, which I will denote by this Roman N, I mean, this is just the base changes from Cp square to Fp square. So this is a scheme defined over Fp square. And since our original scheme is strictly semi stable, this scheme is a normal crossing divisor of the integral scheme. So how does this normal crossing cross, okay? So this is a, what's the strata of this normal crossing divisor? So I mean, actually it's quite simple. It's not very, I mean, the strata only has two layers. So it is a union of, so the top dimension strata only have two kinds of them. The first one, we call this MP-circ and the second one, we call MP-bullet, okay? So in which the MP-circ is very easy. This is a PNF minus one vibration over this Schumerer set, okay? So this Schumerer set has a canonical structure. I mean, originally it's a set. A set can always be regarded as a scheme over Fp bar. And there is a canonical Fp square structure on this set. So in fact, you can regard this set as a zero dimensional scheme over Fp square.  And this scheme is smooth. And this is smooth vibration over smooth scheme. So this is smooth. I mean, of course it should be smooth. Otherwise it's not strictly semi stable. So this is a smooth strata. I mean, and there's another one. And by the way, this one is not geometrically irreducible, okay? I just use a... I mean, by the way, both of them are not geometrically irreducible. I just put them as because due to the nature of their property, okay? And the second one, this bullet one is much more mysterious, okay? We only know it's smooth. We don't know much about the structure, okay? However, inside this scheme, you can consider so-called a basic locus, okay? So this basic locus, you can have a very nice description. And this basic locus, which I didn't define, but it doesn't really matter. I mean, the point is that inside this smooth scheme, there are some very important and meaningful cycles, okay? So this basic locus is a delinuistic variety vibration of dimension R alpha. So what is R alpha? If you remember, this is the flow of half of N alpha over, also over this set, but not the exact same, but I will cheat a little bit. You can ignore this essentially, okay? Also over this set.  So there is a... So this smooth scheme, which has dimension N alpha minus one as well, contains a bunch of a family of very special cycles of dimension flow of half N alpha, parameterized by this set. And each of them is actually a delinuistic variety, okay? And what is the intersection? The intersection should also be smooth and off dimension N alpha minus two, right? I mean, this is the property of strictly semi stable scheme. So this intersection, you can characterize either as a divisor here or as a divisor here, but since this guy is so mysterious, we don't know it's the characterization of itself. So we can characterize this intersection as a divisor in this one. I mean, this is just a projected space or its divisor should be easy. So in fact, this is a Fermat hypersurface in this vibration, okay? So I think this gives a pretty clear, clean characterization of the structure of this special fiber of this integral scheme. And then the definition of this integral scheme actually use some modular interpretation of some abelian scheme with additional structure, okay? So this basic locus in black one meet the white one. Say that again, the basic locus. Yeah, basic locus in this black one meet the circle one. You mean this one? Yeah, yeah. So this basic locus, so you have basic locus in there. This meets another component. In N zero. Oh no, I mean the dimension is half of it, right? Yeah, yeah, yeah. So for example, when, so for example, let me give an example. Suppose we consider V four, right? V four, okay? So the dimension of the, this scheme is dimension three. So dimension three, we have two parts. The first part is a P three over a set. The second part is a three-fold. It's mysterious, we don't know what that is, okay? And inside this three-fold, we have a family of divisors because it's a fiber of dimension two, okay? A family of divisors, they are certain delinquent surfaces. Yeah. And they are over this set, okay? So that's the one example, okay? Thank you. Any other questions?  So let me continue.  Just interrupt me if you want me to go back, okay? So I try to keep it as simple as possible. So now at this moment, there's no representation, right? We are purely doing geometry at this moment. So how does this representation pi alpha come in? So for simplicity, I will just pretend that this coefficient just to be q, okay, otherwise there's too many notations. So I will just pretend that all this automorphic representation are defined over q. For example, if they correspond to symmetric power of elliptic curve, then they are defined over q. So in the below, you see this t, it always mean heck algebra. That's a common notation. So we have this abstract spherical heck algebra, which I denote by t alpha of this uv alpha away from p and the sigma plus, okay? And this is representation pi alpha because it's conjugate self-dual, it will give rise to a homomorphism pi alpha, sorry, phi alpha from this heck algebra to z. I mean, c is because we assume it's q, otherwise should it be oe, okay? To z, and for, I mean, this is just given by set up, I mean, the set up a parameter, symmetric polynomial of set up a parameter. So in particular for everywhere else, we can take this homomorphism modular L. So we denote by M L for L, the kernel of the composition of this homomorphism with the quotient to map c to f L. Which is a maximal ideal of t alpha, okay? So this is, so let me explain the logic of this notation. This M is a maximal ideal, alpha goes with the parity, okay? So if M zero, this means it's responsible for the representation whose rank has parity even, has even parity. And this L, which means modular L, okay? Okay, now, so our goal is to study the local Galois homology. Let me explain the thing a little bit here. So it looks complicated. So first you have the, so let's start from the she first. So this is the nearby cycle of coefficient, constant coefficient c alpha twisted by alpha, okay? So this is the nearby cycle for the integral model and the both of these MP. So the nearby cycle is a sheaf on the special fiber, sorry, the base change of the special fiber to the FP bar, okay? And this is the first layer, then this is the middle cohomology, middle cohomology of this nearby cycle sheaf.  So this is a C alpha module, and we have a HEC algebra acting on this C alpha module. So we can localize this T alpha module at this maximal ideal. So we get a localized T alpha module, which is here, okay? But it itself receives a Galois action of QP square, because the Galois group of QP square acts on this nearby cycle cohomology. And since the action of the Galois group and the HEC algebra commutes with each other, so this entire localized HEC module will receive a Galois action of QP square. So it makes sense to take the first cohomology, okay? So I hope this explains this long expression. Okay, so I think it's clear, okay? So our goal is to study this one, okay? So study what? Study what about this one, right? I mean, what's the actual goal of this one? So, I mean, at front this point, it is really different when alpha is zero and alpha is one, okay? I mean, when alpha is zero and alpha is one, the point of studying this cohomology is completely different, okay? So let me first start from the alpha equal one case, which means this is all the dimensional, sorry, this is an even dimensional cohomology because N1 is odd. So this is an even dimensional cohomology. So when it's even dimensional, what we need to understand is to ramify the subspace of this H1, which will boils down to the competition of certain theta cycle because it's even dimensional, okay? So theta cycle on this MPV and alpha. So you may realize this is not smooth. So what does it mean by theta cycle? It actually means the theta cycle on components of this, on smooth components of this scheme and that they are interaction with each other, okay? I mean, this study rely heavily on the recent work of Shao and Zhu on this theta cycle on the special fiber of Schmer varieties, okay? So I'm not going to talk about details of this aspect anymore. I mean, rather I will talk about the even case, the even rank case. This is much more, first, it's much more difficult. Second, it's much more interesting, okay? So when alpha equals zero, what we need to understand is the singular quotient, which is H1 quotient out by its ramified subspace. This is called, usually called H1c, okay? The singular quotient, which boils down to the so-called erismatic level raising phenomenon, okay? So that's our, so in what follows, we will stop explaining further steps toward the main series, but to explain it more on this erismatic level raising phenomenon, which employs new ideas from the theory of Galois definition, okay? So any question? So, I mean, so one, so it is better. Okay, so the emphasis now is just that this local Galois convolution, okay? So it's better to keep in mind of this thing from this moment. Good, so let me continue with the erismatic level raising. So before I mentioned that there's a very special condition. I mean, there is so-called a level raising prime, which requires some special element in the modular area of its Galois plantation. So what exactly are those special things? So I will explain this at least partially toward this step, okay? So here's a definition. We say that a special inner prime P is a level raising prime with respect to L, if the underlying rational prime P is a co-priming sigma plus, which we have already assumed, okay? I just repeat. Second, I want L does not divide the P times P square minus one. This is somehow elementary.  Okay, the third one is a really important condition. We want to do more the L-Satake parameter of pi zero P. Recall that we have explained that since pi zero is a conjugate self-dual, it's Satake parameter comes in pairs. So it comes in pairs like L for L for inverse. So we want that the more L-Satake parameter contains the pair P P inverse exactly ones and does not contain a pair negative one negative one, okay? So here's the proposition, the main theorem of the so-called level-raising isomorphism. Okay, suppose L is effectively sufficiently large and that P is a level-raising prime with respect to L, then we have the following canonical isomorphism. So this is the following. So recall that this is the right-hand side, sorry, this is the left-hand side, is the singular part of the local Galois homology we have introduced, but now we take the quotient instead of localization, okay? So this singular part of the local Galois homology is canonical isomorphic to another space which is simply the CL-valued automorphic functions on the Schmura-Sat module, the same, this is again a hacker module, the same hacker ideal, okay? So you may see, this chooses a very different, here it's the singular Galois homology of the nearby cycle homology of some integral model of a different Hermitian space, of a different Hermitian space, but here it's the Hermitian, sorry, CL-valued automorphic form on the totally definite Hermitian space, okay? Although I use the same notation, but here it's a different Schmura-variety, okay? Of FL vector space of finite dimension, okay? Actually, there's a version modular, not just L, but modular L to a power, but for simplicity I will not state that one, okay? But I mean, actually we need to do L to some power more generally, but for the simplicity of this talk, let me just do FL version. And the proof of, I mean, if you have looked at our paper, our paper has more than 200 pages. The proof of the single isomorphism probably took more than 100 pages, okay? So this is the real technical result of the work. So now I will explain, so this is the meaning of the level raising isomorphism. And probably you can already feel the flavor of this isomorphism because remember that we stop at some non-vanishing of certain period integral, right? On this Schmura set. And how can you use that? So here is a very important link. We already linked this Schmura set to certain important arithmetic geometry of some Schmura-variety. And that period integral will also relate to certain non-vanishing, sorry, non-vanishing of certain cohomology class that lives in some kind of this local Galois cohomology, okay? So this is the most important isomorphism that you link this period integral into some arithmetic property of some cohomology class of a Schmura-variety of positive dimension, okay? So, okay, so how we prove this? So the proof of this uses two main ingredients and these two main integrants are somehow technically quite different. So the first part is we studied the geometry under the intersection theory on this special fiber. We can show that the right-hand side, which is the Schmura set side, is canonically a sub-quotient of the left-hand side with the Galois cohomology side, okay? This is through a very subtle study of some intersection theory of certain the Linn-Lustiger cycle on the special fiber of this integral model. So if you already know this guy is a sub-quotient of this guy and they are both finite sets, how can we conclude they are isomorphic? So you only need to compare their cardinality, right? I mean, they're both finite dimensional FL vector spaces and one is a sub-quotient of the other one. So to show they're same, you only need to compare cardinality, okay? So how to compare cardinality? This, we have a new idea, a new technique using Galois deformation. I don't think such kind of application has ever been used before. So this is actually a new application of Galois deformation, okay? So now I will consider more about this Galois deformation theory. So now it will be more like a number theory and less arithmetic geometry. So recall that we have associated the Galois representation associated to this pi naught and we assume it's residual absolutely irreducible, which is the case for L sufficient large by our condition in the theorem. So we denote this row bar pi naught L as the residual representation. So it's a representation with coefficient FL. So we take a special inner prime P of F that is a level raising prime with respect to L. So we need to consider a global deformation problem. We call it all mix for the polarized Galois representation. I mean, sorry. Before the details, let me give a, why this isomorphism is called the level raising, okay? So the reason is the following. So this ideal is responsible for the representation pi naught, okay? So remember this pi naught originally, it is unremifed at the P, okay? So this pi naught should not appear in this cohomology at all because this is more likely is ramified at the P, okay? So if you don't consider this, it shouldn't appear at all. I mean, this whole thing shouldn't just be zero. However, if you take this, got this, set up a parameter modular L, then we will have some extra congruence so that this pi naught will congruent to some other representation that is ramified at P. So in some sense, you raise the level of your original pi naught so that it is now ramified at P. So that's why I call it's a level raising isomorphism, okay? So this is, I mean, this is the significance of this isomorphism because our priority, this space should just be trivial if you don't have conditions like this. Because of this special kind of condition, you will have some non-trivial thing on this left hand side. So now this Galois deformation theory should really see both the original representation pi naught which is un-ramified at P and it should also be available to see the raised one, the new new automorphic representation that appear in this cohomology. And those representation will necessarily be ramified at P, okay? So we need to consider a global Galois deformation problem that encodes both the original pi zero and those that are level raised from pi zero. So that's why we call the mix because it is mixed from both ramified and un-ramified. So the deformation problem are mixed for the polarized Galois representation, classifying deformations raw with the following local restriction. So for V inside the sigma plus, for those the Galois representation could be ramified, we put a no restriction. And the second for V above L, we say that the Rho V is the fontanella fire with the correct horchate weights. Basically it says that it's still un-ramified. Okay, so I will not go to the details. Okay, the key is the following. So when V is P, we want the Rho V to be tamely ramified. So tamely ramified Galois representation were determined by two things. First is a Frobenius generator. The second is a TAM generator. So if you look at our condition here, so the TAM generator, the Frobenius generator should have two special Satake parameters, module L, which should first be P negative n zero and the second one should be P negative n zero plus two. Sorry, you see this power because it is a tated twist, okay? So this is the Frobenius and the TAM generator, you should see some little monogamy here because if this S and S prime remain to be P negative n zero and P negative n zero plus two, then it is possible that you could put a new potent X here, right? Because then you will obtain a little, you will see a small, like a Steinberger representation on this block. However, this phi and the T should subject to a relation because they subject to a relation themselves. So more precisely the relations here, so S should be a lifting of P to the negative n zero and S prime, the product should still remain this number and the restriction, the relation between phi and the T will give this relation. So X times S minus P negative n zero should be zero. So this gives you somehow a semi stable anode, right? So this is a semi stable anode in the local deformation problem. So this local deformation problem has a nodal singularity. Okay, for VLs, we require Roeby to be unremifed. Now, we, this is the mixed version because you see both the unremifed and then remifed and you can quotient out by the corresponding ideal to get the purely unremifed one and then the remifed one. You want this X to be zero, right? I mean, if X zero, of course, this is unremifed because the TX by trivial representation, trivial matrix. So which means you should the modular R are mixed by the ideal generated by X. So you recover R unremifed and then the remifed one, you should require S to, you should remain S to be its original value P to the n negative n zero. So in some sense, you need to modular by this second ideal. And the congruent ideal is the ideal that it is satisfied both condition. So you take congruent, this is so called the congruence, this congruence stands for congruence. It's the tensor product of two quotient algebra over the R mix. Okay, sorry, I have to be a little bit quick. Okay, so how do we, by the way, our original goal is to study two things, right? First thing is the certain Galois cohomology and second is certain automorphic form space. So how are they related to Galois deformation? So here's the key proposition. So there exists an R RAM module, H RAM, such that this Galois cohomology is naturally a module so there's a universal Galois representation realized on R RAM of dimension n zero and the realizing on the space, somehow the coefficient is this H R RAM. So this is a Galois representation. So this Galois representation is some multiplicity space tensor with the universal Galois representation, okay? And you can compute it's H 1 sin just simply by computing H 1 sin on this universal Galois representation. In some sense, I'm separating the Galois part and the HEC part, okay? And the computation is quite easy. So the H 1 sin, the left hand side of this isomorphism one we want to consider is actually isomorphic to this module. That's H RAM tensor, this is R RAM module but the tensor with R con and also tensor with FL, okay? Similarly, this CL module I would denote by Hn is naturally an Rn module and this quotient is actually the Hn tensor, R con, tensor FL. So you see, it's very similar. I mean, the shape is very similar. The only difference is that here it's obtained from some module over H RAM and here it's obtained from some module over Rn, okay? So you want to show that these two sides have the same cardinality. You only need to show that these two modules, H RAM over Rn and Hn over Rn, they are free modules of the same rank that you are done. And this is the case actually. So here's the proposition. So suppose this Gallo Apprentice residual episode irreducible, which we have already been supposed. The second is that this is so called a Taylor-wise condition. We want its base change to F zeta L remains absolutely irreducible. And the third one is a little bit technical, okay? So for V in sigma plus, we want every polarized local lifting of this local Gallo Apprentice is a mini mullogram effect, okay? Then both Rm and Rn are finite free modules over Rm and Rn respectively, and of the same rank, okay? Then through this proposition, we obtain the level raising isomorphism, which is the isomorphism between this one and this one. Okay, since I don't have time, I will ignore because my original plan is to explain what this mini mullogram if I mean, but I think I will not explain anymore. So because it's quite some Gallo Appreformation theory. So I think I will stop this moment. Thank you. Thank you very much. Thank you very much for the excellent lecture. So are there any questions? So let me ask one question. So you need to be able to be sufficiently large. And so it's conditioning is basically from this level raising isomorphism. Yes, yes. So the three conditions, okay, so let us go back. So the three conditions, I mean, as I explained, C is purely technical. B shouldn't be there because I mean, this is because we don't know much about the Gallo Appreformation yet. I mean, the only key condition that is special to our approach is A is this condition. So essentially, I think a B and a C should be removable in future. But A is very hard. I mean, if we continue to use this method, A cannot be removed. A can be improved, but not completely removed. Thank you. So are there more questions? So maybe you can show us the last slide, yeah. Okay, so I will be quick. So the point is, what does it mean by a lifting is minimally ramified, okay? So what is a minimally ramified polarized local lifting? So this concept was actually raised by Closier-Harris Taylor like a decade ago when they studied Gallo Appreformation for general rank. So let me give a, so I will just explain the simplified situation. Suppose that we have a Gallo Apprentice, n zero dimensional galvanization with coefficient FL. So suppose it's a terminally ramified. And so the time generator will expire a unipotent element of Jordan type m1 to ms, okay? So they are lifting to whatever a ring with residual field FL is so-called a minimally ramified. If rho v is still terminally ramified and the rho v is conjugate to a unipotent element of the same Jordan type, okay? So in fact, when you raise the, for example, if your original Jordan type is two 11111, then when you lift, it is possible that the Jordan type will be refined. It could be 1111111, okay? So the minimally ramified means that the Jordan type won't be ramified. So this Jordan type will somehow will characterize how ramified a reputation, attain a unipotent matrix could be. So this minimally ramified, which means you cannot ramify, have a further ramification when you lift, okay? So we have extended this notion to any place v plus. I mean, this is quite technical of the ramifying F. I mean, this is a technical work, but the essential idea is this one. And there's a proposition as the following. Suppose L is larger than N zero, then the local deformation problem classifying minimally ramified polarized lifting is a formal smooth over C L of pure relative dimension N zero square, okay? And for experts, you know that this is a very important local ingredient for any, like R equal T theorem, okay? So the point is that how does it, so when does, I mean, in our main theorem, we only see where L is large enough, right? We, in the theorem, we never mentioned the word minimally ramified, but here it comes an additional condition like a minimally ramified. So actually we proved a theorem which we call the automatic minimality. So if you have a compatible system of Galovar plantation coming from automotive form, so it is expected that for L large enough, or, and if it's a place V, for L large enough, or local deformation of this Galovar plantation should always be minimally ramified automatically, okay? So let me say this is a conjecture, so we call the rigidity of automotive plantation. So we fix the finite set of sigma plus of primes of A plus, therefore all but finitely many primes lambda of E, depending on sigma plus, we have, this is a relatively absolutely irreducible, these are all the conjectures, okay? The third one, for any V plus, every polarized local lifting is automatically minimally ramified, okay? So homework verified above conjecture for N equal two. This is not hard, okay? So theorem, suppose pi is super cuspid or somewhere, then the above conjecture holds, okay?  So originally we also need a pi to be a twist of Steinberg and some place not above sigma plus in order to deal with the Taylor-Weier's condition. But the later, Toby G told us an argument to remove this restriction. So now the theorem is very clean. You only need to pi to be super cuspid or somewhere, okay? So this conjecture, the automatic minimality is true. For a large enough, yeah, that's it. Thank you. Thank you very much. So are there more questions? Okay, so thank you very much. Yeah, okay, great. Yeah, thank you very much for the beautiful lecture. Okay.


Welcome to the series, Photographic Chemistry, presented by the Foundation of the American Institute for Conservation of Historic and Artistic Works. This program was made possible by grants from the National Endowment for the Humanities and the Andrew W. Mellon Foundation. Each program in this series is presented as a short video. Depending on your video viewer, you should be able to pause, return to a previous section, or skip ahead to a later section by using a scroll bar or on-screen icons. You will find an outline of the course and short quizzes to test your understanding on the course webpage. The KSP value communicates the magnitude of the solubility of a material within a solvent, in our case, salts and water. We use these values to be able to understand the properties of these photographic materials in our photographic process. And I'll refer to these KSP values quite often as we talk about photographic chemistry. To illustrate in the table shown, I'm comparing the relevant silver halide grains of silver chloride bromine iodine with something we know a lot about, table salt. And as you can see in these KSP values at room temperature, there are orders of magnitudes. When a scientist has orders of magnitudes, these are factors of ten difference between table salt and sodium chloride. In fact, if you just compare the ratios, compare the numbers, sodium chloride is about 700 million times more soluble than silver chloride. 700 million times. Or we would say that that's about eight orders of magnitude difference between the solubility of sodium chloride and silver chloride. And if you compare the values, you can now understand that there are even big differences between the silver halides, the silver chloride bromides and iodines. And this is a key point. These are the values we use to engineer and develop specific silver halide grains for specific applications. The composition of a grain is very much determined by these KSP values when we go to do an emulsification process. Now you'll see KSP values in two different forms, either in the KSP or if we want to communicate rather than always saying 1.77 times 10 to the minus 10 all the time, we do the same trick that we do for pH. The little p means negative log of. So I can take these bigger numbers and convert them into the negative log of values, much like we do with pH. So 1.77 times 10 to the minus 10 is equal to 9.75 pKSP. You just have to be a little bit more comfortable with that number because what this means is, remember in pH, the higher the pH, the less H there is in solution. In other words, a pH of 14 means a sparingly small amount of H plus in solution. It's mostly basic or OH, and we'll get to that. Same is true here with the pKSP. The higher the value, the less of it that there is, the smaller the magnitude of the number. So we see that between sodium chloride and silver chloride, there's an 8 order of magnitude difference. Each click of the number is a factor of 10. So there's 8 factors of 10 difference, or that's 700 million, roughly, right? That's 700 million time difference in the solubility between silver chloride and sodium chloride. And there's even an additional 7 orders of magnitude difference between silver chloride and silver iodine. This is the key point. So get used to understanding these numbers. These are very, very small numbers, and what that means is that these are very, very insoluble salts. What this means is that the concentration of ions, and this goes back to the definition of KSP, this means the concentrations of ions in solution can be no more than these very, very small values in molarity. So let's see how we use these KSP values in the emulsification process. You have completed this unit. Depending on your video viewer, you should be able to scroll back to any point in the video as desired. The short quiz found in the course materials on the website may help you confirm your understanding of the concepts introduced here. Many thanks to the instructor, production editor, coordinator, and the collaborative workshops and photograph conservation committee for their work to make this program possible.


I'm James Milan. Welcome to this episode of Talk of the Town. It's great to have you here. Our subject for today is re-precincting. It is a process that is ongoing as we speak, and it is something that is, of course, related to the fact that every 10 years a census is taken, populations change, and where those populations are located in our town change as well, and we need to make sure that our precincts reflect those differences. A big concept that people are concerned about, and my guest today will be speaking to, I'm sure, consistently, is the idea of equity. This idea of re-precincting, either reducing the number of precincts, or I guess in some years you might add them. I don't know. But anyway, changing that configuration is all about preserving equity and making sure that it is a better situation for voters coming out of the new data from the census, and those are the things that inform these decisions. All right, that's enough of an introduction, and I want to apologize first to my guest who is the authority here, and that is our town clerk, Julie Brazil, and then also to the rest of you for, again, a typically lengthy introduction from James, but really wanting to set a little bit of the framework for our conversation. Julie, appreciate your patience, and thanks for joining us. I'm happy to be here. We really do appreciate it. So I was mentioning that this is, the re-precincting is directly connected to census data that's just coming available. Why don't you, though, do a better job than I of explaining what this process is, how unusual it is to be considering the kinds of changes that we are this year, or whether, in fact, this happens every 10 years, and it's par for the course. Sure. So I think, like a lot of processes, this one is specific to its time. We've had the same general map for 50 years. Every 10 years, we do go through the process, and the last two times, from the 2000 census and the 2010 census, we did move lines. And when you move the lines, redraw the precinct boundary that changes who votes where, and you have to re-elect all of your town meeting members. So it's a substantial process. The last two times, we did this and changed the lines. It affected 13 precincts and 15 precincts out of 21. So, yeah, you have to go where the numbers in the data take you. I have introduced the additional wrinkle this year of recommending that we revisit how many precincts we want to have. We've had 21 for 50 years, and I'm sure it solved the problem at that time. The select board always makes the decision to approve the number of precincts in the map. That's their job. But I do think that with the changes coming in elections, with early voting in person and voting by mail, slightly expanding the size of our precincts can make elections more efficient and more flexible going forward. All right. Yes, and I'm sure that we will hear more of that further on in the conversation. But thanks for that basic information that this structure, basic structure of 21 precincts here in Arlington, has been the case for half century now. And so in a sense, this is an innovative idea. Let's put it that way. Some people might say revolutionary, I don't know, but actually looking at changing the number of precincts is a relatively rare thing. Okay. Let me ask you, what are the options that you are considering? You were just mentioning that one of them would be reducing and there are certain reasons that would argue for that. But if you can just lay out what the different options are that you're considering, and what do you see as the respective advantages and disadvantages that pertain to those? Sure. So yes, I am recommending 16 precincts, so a reduction of five. That number was arrived at sort of from a couple different sort of mathematical ways. Well before we started playing with the maps. We currently vote in eight polling locations. So if we had 16 precincts, each location would serve two precincts. Right now, we double up and triple up. And so some precincts actually serve three precincts, and that's a little unwieldy at times in terms of just parking and traffic flow and confusing signage. And then there's a lot of math involved, and this is very confusing for people, so I don't want to make it tedious this afternoon. But we'll use me as a test subject. There you go. If you're still awake at the end of this. So the number of town meeting members is related to the number of precincts. The law says we should be aiming for 240 town meeting members. The number of town meeting members has to be divisible by three, because we normally elect one third every year. So if you do the math of 16 precincts and 15 town meeting members, then you get to a 240 exactly. And I thought, well, okay, so mathematically 16 is a comfortable number, you know, sort of well balanced. So that was sort of how I got to 16.  Right. So what I gather is that, you know, it's math I can follow. So I appreciate that. And that is the fact that, you know, by reducing to 16, with those eight sites that currently exist, wouldn't have to change those at all. But instead they would all be, you know, basically you could house two precincts at each site. And then it also would work out well in terms of, you know, having the right number of a number divisible by three. In this case, it sounds like five town meeting members in a 16 precinct Arlington five town meeting members from each precinct would would be elected each year. Do I have that right? Yes, except for next year, we would elect all of them. So it would be a really big town election. Despite the work involved in the confusion potential, you know, for voters, navigating that process. I'm choosing to see it as an opportunity. We often talk about how sad we are when turnout is low in a local election. And having the opportunity for every voter to choose all at once all of their town meeting members is good. I mean, I think it's democracy. I think that's that's an interesting process. And we'll do everything we can to make that easy for people. But all of the people interested in running for town meeting with further hat in the ring, and we would choose 15 if we went for the 16 precinct model. To be clear, if we move any line, and even if we were to keep a 21 precinct map, the same process happened. So if a line moves at all, then everybody gets reelected. And so, and given the way the 21 precinct map looks from the state, I think the vast majority of precinct boundaries are going to change. And so I don't really think we're saving ourselves any trouble by trying to stay with 21 precincts just to avoid that election outcome. Yeah, obviously, you're talking about, you know, whatever the, you know, whatever the the tumult is that could come, you know, on that particular day, you're talking about it happening once, which is in this first election after this redrawing of the lines. And then after that, things will proceed in a very familiar way, one way or the other. But let me ask you, you were saying that in 2000 and in 2010, after those censuses, that, that, you know, 13 and 15 respectively of the precincts were were affected by those. And is it true then that it is only precincts which are affected where the lines do change that would need to reelect town meeting members from those precincts? Or is it really going to be all the town meeting members from all the precincts no matter what? No, it's only you only reelect everyone if the line moves. And so, and so over the past 50 years, I looked it up. I'm pretty sure three precincts have never gone through this joyous exercise. They've just chugged along, electing their, their four every year. And I kind of want to, that helps me segue into thinking about equity, because that's, that's a very different experience between precincts. And that's one of the things that we considered as a re-precincting working group got started is there a virtue in trying to, you know, save two or three corners of Arlington from going through the same experience. And, and I have a sort of a number of thoughts about that that we can explore. Yeah, I think that equity has such different meanings, or it can be applied in so many different ways in what we're talking about here, equity between precincts, equity within precincts, equity within populations within precincts, etc. So let's kind of just establish what, what the terms are as you are, as you are referencing equity, you know, explain to us what you mean by that. Sure. I want to start actually with sort of the, the official charge that we're given under state law. Everybody gets the rebalancing of the number of residents per precinct. But we're also called on to look at the demographic data. And redrawn precincts and ward boundaries must not result in the dilution of minority group member votes. And the Voting Rights Act prohibits any voting practice, which results in a denial or abridgment of the right to vote on account of race or color or membership in a language minority group. So we really did start when we sat down with that, you know, sort of very clear understanding that we needed to look at the demographic data that we had on residents and be sure that we were being as equitable as possible using as many different ways to think about it as possible. There's pure demographic data when the mapping software lets us look at the lines, sort of with underlying data. So we can look at distribution by race. We can import information from the assessor's database and look at distribution of people who own versus rent. We can estimate household income using American Community Survey data from 2019. So if we do all of that, it allows us to get a pretty good picture of where to put the lines. And is the goal there in terms of equity to maintain some kind of demographic character of precincts? Or is the goal to make them all as close to evenly matched or comparable in the various ways that you were just saying, income and own versus rent and racial and ethnic diversity, etc.? So, well, the goal is to not dilute any minority subgroup. And so what that means is a practical matter is, if all of the subgroups evenly dispersed, then we would strive to have each precinct have the same profile. But since they're concentrated geographically on the map, we need to make sure that if we see a cluster that we want to keep that cluster together rather than accidentally drawing a line that carves just a corner off, leaving them very different from the rest of the precinct that they're now in. Conversely, if we had a very large spread of a characteristic that was too big to encompass in one precinct, we would want to put the line as evenly as possible so that the two resulting precincts have sort of equal weight in that. So what we're trying to do is prevent minority voices from being overshadowed. Town meeting is a representative thing, and so if a very small section of a precinct has a different opinion about a decision that town meeting is making, their voices might get lost. And so that's our goal, is to make sure that we're not dividing things up in such a way that voices get lost. Well, I probably don't need to tell you this at all, but that sounds like a lot of needles to thread, all kind of intertwined as well with each other and overlapping. Does the 16 precinct model help you do that any better? Actually, it does. We drew both a 21 precinct map from scratch just to go through the exercise, and we drew the 16 precinct map from scratch. And yes, the re-precincting working group felt that the 16 precinct model worked better for a number of ways, in a number of ways. The slightly larger precinct was easier to envision polling locations that were sort of better situated in the sort of pairs of precincts. There is always a question that comes up about transportation. Somebody's always going to be farthest from the polls, and you can't help that. So we definitely want to be sure that we're being thoughtful about where the possible polling locations could be. To be clear, the select board could always change the polling locations. The map is set for 10 years. We can move out around the polling locations to solve problems. In general, from an equity standpoint, though, it would make more sense for us to have the polling locations closer to transportation spines. If you have all of the polling locations in the schools that tend to be sort of in the leafy suburbs, those people probably, on average, have cars and could drive into the center to a polling location on the central spine. You have to think about a lot of difference at angles. We thought about it from the perspective of sort of that question of do you make everybody, if you move to 16 precincts, then obviously everybody goes through electing their town meeting members. That's, of course, equitable because everybody has the same experience. A couple comments raised an interesting consideration for us, and that is, some people are like, well, you shouldn't do that because it's expensive to run. But if it's expensive to run, then that actually argues for making sure that every precinct goes through the same experience. If you were to isolate three or four precincts in a 21 precinct model and say, well, you guys don't have to do it, that's very inequitable because those people don't have the same outreach issues as everyone else. And there's been a lot of thinking and talking about that. There was a comment specific to that where someone said it is expensive to run, and I think the town should look for ways to remedy that. But in spite of that issue, I support the 16 precinct option. So it's nice to know that people are really thinking about it. The same way that we are, that's always nice. And so I mean, I think we have to look at solutions. One solution that I have been exploring would be to address some of that equity issue for campaigning by setting up a system where all candidates for town meeting can submit written statements that we can publish. Volunteers have been doing that on the League of Women Voters website. I would like to be able to do it on the elections page. And I'd like to build it into our process so everyone gets the same form and invitation at the same time. And I think it encourages more participation, which is good. Incumbents should introduce themselves to their precinct every time they run, just because new people move into the precinct and wouldn't know them. So I think the more we build that habit in and can provide that service, that would be great. It raises, of course, campaign finance issues. So I'm talking to a lot of lawyers, but that's a goal. I would like to do that in order to sort of address the equity in campaigning, give everybody a level playing field to campaign from for free. Yeah, you know, we had said at the outset that equity was going to kind of permeate this conversation in a lot of different ways. And that's yet another one, what you just spent a couple of minutes on that hadn't I have to confess initially occurred to me, which is, of course, equity in campaigning, equity for those who wish to represent their fellow precinct members at town meeting. Yeah, you've got to think about that as well. It's a very important consideration. I want to just take a second, though, Julie, and take a step back or step aside. And just ask you, you've referred a couple of times now to the study committee. And I can't remember exactly what the official designation of the group is. But anyway, the group that is kind of been looking at this and really working at it in the way that such work happens here in town, usually by a lot of volunteer energy, etc. But anyway, who is what is the makeup of that study group and how much of a representation from, you know, the Arlington citizenry is on that group. So the repressing working group is is what we're calling ourselves. It starts when the Secretary of State's office, the division that's managing this process, contacts the town clerk and says who should receive our emails when we have the data to communicate with you on the mapping process. So I talked to the town manager. The town clerk is almost always sort of point in this process. We, of course, invited Adam Korowski, our wonderful and very experienced mapping expert, our director of GIS. He's retired in the middle of this process. However, he's staying on on a part-time contract basis to help get us through this process because it's important to the town and important to Adam. Kelly Linema from planning, it's very helpful to have input from the planning department because one of the things that we get to play with as we rebalance the population numbers is if there's an area of town where we expect there to be new residential growth, it's a good idea, not required, but a good idea to have that precinct be on the lower end of the range so that they're not wildly out of balance in 10 years from the rest of the precincts. And then it occurred to us since equity was clearly key to the effort and something we felt really called to do. We invited Jill Harvey, Arlington's director of diversity, equity and inclusion, and she brings more information and more perspective, more resources to the conversation. And so we felt like the four of us could pull together the information. Technically, we're working for the select board since they are the ones making the final decision, but this is such a big process and that community input is so important that we definitely invited every resident to weigh in with their perspectives and their recommendations on places on the map that we should really look at where we might, where we need to look at the underlying demographic data and make sure we're not dividing a group awkwardly. Yes, so okay, so the working group itself is four people from different levels, understandable of town government. You've explained well what contributions they would make to that. So then I imagine you need to go through the same process in terms of soliciting community input as other departments in town have to do around different things. The planning department, you know, comes to mind with all the plans they have. So I would imagine, I believe in fact that there's a forum, a community forum probably coming up in the goal of which I'm sure, at least in large part, is just that, is getting a sense of what people's reaction feelings and feedback are about them. Is that right? Yes, absolutely. So on the elections page, we have a link to the re-precincting process where we're sort of collecting as much information as we can. It announces the dates of things like the forum, which is next Wednesday evening, September 22nd. It's a Zoom forum and it's being hosted by the League of Women Voters of Arlington and volunteers from the Election Modernization Committee. And it's a chance for people to ask some questions and then break out into small discussion groups and really dig in to the perspectives. I think that's what's important is as we're looking at the comments come in online, we have, there's a, you know, for every person who says, I think 16 doesn't make sense. Somebody else is like, I've never thought of 16, but that's brilliant. So, you know, it's great to hear all of those perspectives. And it of course tells us when there are questions and confusions that we can try to clarify. There are videos on the web page. We have a 12-minute video that I prepared explaining what I mean when I say more flexible election. I mean, there's a lot of election reform that I think Arlington really wants and it's easier to do with 16 precincts. So, I wanted to talk about that. And then Kelly Linemum sort of walks everybody through the process, a little more detail looking at the maps and how we created them, both 16 and 21, you know, sort of the trade-offs and pluses in-liners from each.  And then of course, I think the forum will be very helpful for people who want to talk to their neighbors and sort of think about real-world impacts. So, when does this all come down, Julie? As in, you know, what is the timeline for when this decision needs to be made? Sure. The select board has to take a vote to finalize the map and the entire submission packet by October 30th. But sort of any time. I got the data files, the official census data files Monday afternoon. So, hopefully in the next few days, we'll dig in and see how we need to adjust the draft maps. And then we'll present those again. I of course have to do that in coordination with the select board. And so, there was a hearing on Monday and on the question of 16 versus 21 precincts. The select board wanted to just think about that. The sooner they can feel comfortable deciding that question, then we can really fine-tune one map really well. So, keep your eye on the precincting links on the web page. There's sort of it'll be constantly updated. There's a fact sheet. There's information and more.  So, we encourage people to keep it. Keep an eye on it. There's a Google form if people want to submit information online. You can print just a PDF on regular letter size paper from home if you have a printer. And, you know, just sort of circle a neighborhood and say, I'm really worried about this. And then, you know, drop that off at town hall or send us a photo, email us a photo of that page and we'll include all of that data. The data from people, you know, sort of block by block neighborhood by neighborhood will help us fine-tune. We might think it makes perfect sense to put a line there because the data we get is in census blocks and it, you know, that it's not as small as maybe people are aware of, you know, the situation in their neighborhood. So, we definitely encourage people to take a map and say, worry about this location. And then whatever version of the map we're doing will have that neighborhood in mind. Right. It's like locals, you know, very, very local scouts, the people who know these things the best, kind of sharing that information with you. All right, Julie, I've got to give you your due. You know, before we went on air, I said, well, we have up to half an hour and that should probably be enough. And you're like, I'm not sure. There's a lot to cover here. And we're right because we're right up against the half hour and I'm sure we could keep going. We are going to have to call it here and certainly I think there's plenty of good content for folks to digest here. And I'll just ask you remind people one more time how it is that they can either get more information or provide their own feedback as succinctly as you can and then we'll sign off. So, on the, on the reprecincting page, you know, it's a link on the elections page, then there's social media, then linking it. There are, there's instructions. We can accept written documents that you mail or drop off at Town Hall. You can email us written comments. You can email us a picture of a map. You can use an online Google form and just write your comments say, you know, this neighborhood has a lot of people who share the same interest and keep an eye on that. And then we'll present all of that information to the select board so that they're aware of it when they make their final vote. All right. Well, thank you very much. It has been a most informative half hour for me to digest and I'm sure for the rest of our audience as well. Complicated stuff and glad that you're glad it's your problem and not mine. But really appreciate, you know, how hard it is to kind of, as I mentioned earlier, thread all these needles. We look forward to seeing what the, what the end results of all of this is. I have been speaking on the subject of representing with our town clerk, Julie Brazil. I am James Milan. This is Talk of the Town. We sure do appreciate Julie taking the time to explain this complicated situation to us as best she can. And we appreciate your time that you've given us in tuning in. We'll see you next time. Thanks for joining us.


Hello everybody, welcome to my channel. If you haven't yet, please click like and make sure to subscribe so you get the videos This will be a good one. Let's talk about dental hygienists starting their own business And let's talk about your first year in business and what to expect a very popular Question for dental hygienists who are thinking about starting their business or have already started and maybe your first year didn't go as planned So I left myself some notes. So I didn't forget to mention certain topics So the first year in business really does depend on a number of factors Do you have a mobile business? Do you have a patient in your in your or sorry? Do you have a practice in your own home or do you have a standalone practice? I can talk specifically about a mobile practice and having a practice in my own home because that's what I have done personally I've been in business now for almost about two and a half years So the first year in business, this is when you likely don't have anything so you'll have a lot of expenses My expenses include getting a sterilizer. I needed new instruments. I needed, you know, a dental software I needed an intraoral camera. I needed masks, gloves, bibs Now we need PPE. You might need things like toothpaste, mouthwashes to have on hand for your patients I needed a patient chair for my office and I needed another patient chair to travel with me mobile I needed a saddle stool to sit on one for my office and one for mobile My big expenses probably did include my compressor unit that houses the air, the water, the suction That was about six thousand dollars and of course the sterilizer. That was about ten thousand dollars I needed a new laptop specific for my patients.  So I did get a new laptop I needed one anyway, so it just kind of worked out But all of these things you will need for your first year of practice and talking about dental supplies So that's my second point here Or sorry, I meant to say you can't only buy and bulk. Well, no, you can you can so let me just kind of fix That off a bit here So talking about dental supplies as a dental hygienist owning your own practice You're not a dentist where you're going to see hundreds of patients a day, right? You might see one patient a day, maybe not even that at first, but the thing is you can't buy 10 Bibs, you know, you can't buy 10 masks 10, you know So you have to buy and bulk meaning boxes of 300 gloves per container Bibs come in like hundreds, but the thing is sure you can buy less you can buy less bibs less masks Less anything, but you might be paying $20 for 10 versus buying a thousand for $50, you know Like I'm just picking out numbers here. So obviously you're going to buy in bulk what I did buy in bulk probably the most of was my Pouches for sterilizing they don't come in packs of 10. They don't even come in packs of a hundred I don't think it's like thousands. I still have my same pouches that I'm still using I haven't needed more for two and a half years So you will get a lot of expenses initially But you need that to get started if you're lucky and maybe you know a dental office who is okay to let you use Their sterilizer, maybe you know a dental office that is okay to say here. Well, you can pay me for a hundred a hundred sterilization pouches you can pay me for the Internal and external indicators you don't have to buy thousands and that would definitely help you out if you're able to do that I just didn't have that luxury. So a lot of expenses Working two jobs. So this will really determine. Well, are you going to make money in your first year of business? Probably not because are you going to just quit your job and work full-time in your own business? If you're able to do that then that means you can market you can advertise more you can truly work on your business I didn't have that luxury either. I was still working part-time in a dental office because I needed to make the money I didn't have a choice. You know, I couldn't just have a A slow week with my new patients and make no money, you know, I had to constantly be making money We have to pay the bills. But if you're working another job, you don't have as much time to really focus on your own business So you're not making as much. So that's just kind of how that comes into play The last point is well, you're waiting to spend the money on probably a lot of things You might be thinking do I really need an interoral camera right away? Maybe not so you're waiting to make money to spend money If you don't have an interoral camera, you know that it doesn't really set you behind So that's a bad example, but let's just say you were waiting to purchase Let's just say business cards. I don't know. I just thought of something now Let's say you're waiting to purchase business cards. You don't want to spend a hundred dollars on business cards You're waiting until you make the money. Well, if you had those business cards now If you just spent the money for those business cards, you could hand out three business cards per patient And you could get patients from those business cards. So certain things and that's my last point here So certain things I know you want to wait to spend the money But I suggest for certain things getting them right away because that will actually help you make morning Make more money now and not later. I do have another video of Tips on things that I suggest you do need right away in your dental hygiene practice And then I have another video of things where you don't need right away So definitely check that out on my youtube channel Moving to the next one the second year you guys gets better because you're more established You probably have a rhythm at this point You might not be working at the the other dental office or other job anymore because you want to focus on your own business You can try to start to make money back that way But you won't be making you know hundreds of thousands of dollars like a dentist would okay So don't be thinking that these things take time. You need to build up patients You need to have your patients come back Even if you're amazing and see a hundred patients. Well, are you that amazing where they want to come back to you? Maybe not. So you have to think about all of this Um, so the second year the third year the fourth year things will hopefully get better If things don't get better then you need to look at your business plan and figure out what you're doing wrong I did forget to mention in your first year you likely have a business loan to pay off That business loan might take you several years to pay off So you're truly not making as much money as you could until you pay off that business loan So just kind of things to expect it's still exciting like i'm hoping i'm not discouraging any of you This is extremely exciting Start your business you can do it another common point people ask me is well Is it worth it to start your old business or do you make more money working for somebody else? What I want to say is if you're thinking that you're not meant to be a business owner You're truly not yes, it's about the money But if that's your first thought Then maybe you will be more comfortable working for somebody else because that's completely different Um, if you do well, of course you can make more money doing your own thing because you can set your own hours You can determine what your expenses are you can kind of go that way working for someone else It's a time clock. Well, you have to work more hours to make the same money If you want to make more money, okay fine make more hours, but at least in your own business you have that flexibility. Okay But that's another video for another time having your own business is amazing I love to talk about all of that. So comment below of any questions and thank you guys so much for watching