 for the invitation. And I learned a lot in last week's workshop. It's a pity that we will be leaving early now. Right now I'm heavily drugged. So excuse me if I sometimes I sound incoherent, but it's a short talk, so it's okay, I mean. So last week the students have already learned what is HK multiplicity. And experts here already know, so I don't have to get into the detail. But my setup will be, R is a Noetherian ring. I mean, an Noetherian commutative ring. And I is an ideal in R, length of R mod I is finite. So either my R is a local ring or R is a graded ring. And in that case, in fact a standard graded ring, in that case I is a homogenous ideal. So that will be my setup. So we all know that once you have ideal of finite length, I can think of a length function and take its limit. So this is a notion introduced, you can say well-defined by Monsky. So this was nothing but HK multiplicity of R with respect to I. And this is a length of, you take R and I, kufrovenius. So q strength, so here always characteristic of R is positive, say p. And q is p to the power, say s. So limit s tends to infinity. And normalize it, qd minus one. Qd and d is the dimension of the ring. So the thing we know already that the limit does exist and in fact it's a positive real constant. And so this belongs to R greater than or equal to one. So in the last week's workshop, the last theorem Smirnau has proved was that, what does this invariant tell? The last theorem he proved was the theorem resolved by Watanabe and Yoshida, which characterizes that the regularity of the ring in terms of this. So it says the EHK of Rm. So this is theorem is one if and only if you assume of course R is a formally unmixed in this case here, if R is regular local ring. So this is a well-known theorem. It's kind of folklore one can say now. So the next question should be asked that what next? What can you say about the next bound? So this was again a conjecture of Watanabe and Yoshida. So let me define this ring which I'm repeatedly going to use, which is coming a quadratic hyper surface. So it is RPDKX0XD is a D dimensional quadratic hyper surface given by X0 square, X1 square plus XD square. So after that the ring is not regular, what is the least number it takes? In usual multiplicity we don't even have to ask that because they are all integer values and here we don't know what is this HK multiplicities are going to take, they are just real numbers. So that was a conjecture. This is a conjecture now also. Conjecture of Watanabe Yoshida, it is as follows. So it says let A be, so excuse me, let A be a formally unmixed, formally unmixed means if I take a completion it said all minimal components or associate components are there, there are no embedded primes and say are of same dimension. Characteristic is not two, that's your setup. Then the conjecture can be put broadly into two parts. One says what is the HK multiplicity of, I'm sorry, of dimension D. So suppose I take any ring of dimension D and characteristic is P, which is not two. Then what is the least bound you can have and I should say one more thing, and not regular. So perhaps I should rewrite this as the formally unmixed non-regular ring of dimension D and characteristic positive of AB positive, not two, so the conjecture is about the lower bound. So it says EHK of, here when I don't write anything I mean I'm trying to look at the multiplicity with respect to the maximal ideal. So this is always greater or equal to EHK RPD. So here if you look at this, I mean the statement this one says if I take any ring then its lower bound is greater or equal to one, which is the multiplicity of a polynomial ring. So the next best class of rings comes is a quadratic hyper surface. So part of that conjecture if ring is not regular then it is greater or equal to the, so this is the best behavior ring, vis-a-vis the HK multiplicity. Okay, and the second part of the conjecture says EHK of RPD is greater or equal to one plus MD. So this is what is one plus MD, I'll just define in a moment. So this is if I take sec of X plus 10 of X and I take my norm of X to be less than pi by two. So it is a convergent power series. So it can be written as one plus MD X D, D from, or D is the back notation. No, yeah D is okay, D is from zero to infinity. So this one, this one, okay. This conjecture is by Monsky, okay. So perhaps I'll change that actually, yeah. So what is the purpose of, mainly I'll be concentrating on this actually. So what is the purpose of this one? So this statement is giving you a characteristic free bound for a D dimensional ring. That's what is doing it. So I would, in that regard, I would like to mention whatever I know, case you can correct me, that for a dimension D equal to four, this was proved by themselves by the author of Watanabe Yoshida. Then five, I think, let me, then D less than equal to six, it was done by Yoshida and D less than equal to the party was by Auberbach and Inescu. So this, in general, this thing is still open question also. Then here, once I talk of characteristic free lower bound, I should mention the work of Seligbas Dao and Heunek Yanzang, which have already given a characteristic free lower bound. They have given, but now there is something I would like to mention as a result of Gessel and Monsky, which says if I take RPD, Quadric Hypersurface, and as P tends to infinity, it is going to be one plus MD. So combining this result, it just suggests that this is going to be the best lower bound. So this, so you can ask them why does it imply this? So EHK-RPD in general need not be monotonic increasing or decreasing. So there are examples. And now, talking of this conjecture, I would also recall a question, I can't by Yoshida, or I think was it a conjecture, something. He says if I fix D, D is a dimension, then the HK-multipacity, EHK-RPD, the HK-multipacity of this Quadric Hypersurface is a decreasing function of P. So that is a, I don't know why he got this because why he thought of this one, because you know the famous examples of Hans and Monsky. So this one says if I take RP, now RP, I'm just noting RP to suggest the underlying characteristic of the RSP. If I take a Fermat-Quadric for example of characteristic P of X4 plus Y, Y4 plus Z4, then what? Then EHK of RP is three plus one upon P square. If P is congruent to plus or minus three mod eight, and the next possibility is P is congruent to plus or one minus mod eight, right? So these are two kind of prime numbers. In that case, this is the three. So this is oscillating as a P varies. So this is a background. Now the main, one of the main theorem of this talk today is the following. So it says there exist polynomials. There exist what? There exist polynomials. FT and GT in rational number QT. So this is a characteristic free polynomial. I meant to say such that EHK of RPD, the Cordic surface is greater than or not. Equal to one plus MD plus FT upon GT, P equal to one or P. I mean there's nothing secretary. If P is greater than two to the, if P is greater than two to the power integral value of D mod two into D minus three, something like that. These, okay? Such that, so you have this formulation of the EHK multiplicity such that, one, the polynomial FTGT is a positive, takes a positive value for all positive rationals. And second thing is a strictly, decreasing, no, increasing function, function of T in a neighborhood of zero. So what is the implication of this theorem on these two results? So this one is saying, since there's a non-negative quantity, this EHK RPD is always greater than or to one plus MD. In fact, P greater than D minus one will do. We don't require such a higher bound. That was a part two of the conjecture, this one. And the second one is saying, this is a increasing function of T. So that means in a neighborhood of zero, that means this is, as P varies, this is a strictly decreasing function of P. As P varies for large enough P onwards. That's what it says. So which kind of says, relates with the conjecture or question of Wattanaway. Here it is saying for all P, but this one is saying for P large enough. This is going to be. But on the other hand, this is saying that this function is going to be strictly decreasing function of P. That implies just not a decreasing function that gives you for free. So this is a thing. So what is happening? So techniques, I suppose people have employed. I suppose some volume function and I didn't quite understand, I must admit. So here what we do, we apply the technique of something called which we call HK density function. So let me quickly recall. So this is this formula, this whole setup what I'm going to work right now is going to work for any graded standard graded ring. In fact, graded domain. So let me recall quickly the notion. I would cannot spend too much time now. So here R of I is a setup is this. It's a graded pair. Or I should strictly say standard graded pair. What do I mean by this? So I mean by this, your R is a standard graded ring. Choose over K, which is a perfect fit. Okay. And I is a graded ideal homogenous ideal. Says that R mod I is a finite length. So the length function makes sense. So now just look at the definition of HK multiplicity. So this is giving you a limiting function of the length function. Now my R is graded, I is graded to Froben's power is graded. So quotient is a graded ring R mod IQ. So I use that. So I define the function as follows. Define using such a graded pair, you define a function F of R of I from R positive to R positive as follows. I'll just define F of R I of X is limit of FS of X, S tends to infinity. Now I'm defining FSX, which is nothing but you take the length of R mod I of Q. So Q is PS now at XQ upon Q to the power D minus one. So instead of taking the length function like this, I take the XQ graded piece of the length of RQ function at each S and go model of QD minus one. So you are constructing a sequence of step functions. Now using the length function and then you take point wise limit of that function and call that to be a HK density function. Is it okay? Are there any questions right now? Okay, so here, so question there, I mean there would be obvious questions here. I mean, does it make sense I mean? So that was the whole point. So here the thing is one is that F of S, the sequence of functions S it converges uniformly, formally to F of RI which is HK density function. And the second, this function is continuous F of RI is a continuous compactly supported function. Having these two information, the following obvious I mean uniform convergence allows me to interchange the limit with the integral. So in particular, you have a continuous compactly supported function. So it is integrable, take its integration, is going to be your HK multiplicity. Yeah, so Q is equal to PS. So at PS level, you take this function. Okay, so this is the thing. So what is happening? I mean why one is suddenly looking at this thing? So this function has several advantage. In fact, this comes very naturally. It is first of all is additive function, whatever you mean take components, take other things. It's a multiplicative function with respect to segue product, tensor product that is for free surprisingly, which we don't know about multiplicity as far as segue product is concerned. And the one thought that there may be something because you have continuous functions. So if I take a continuous compactly supported function, take its Fourier transform. So class of such functions which are continuous compactly supported will sit inside embed in this class of holomorphic functions. So you hope to do some kind of complex analysis or analysis to imply. So that's one thing. Then question people ask, is it does it really help you in computing HK multiplicity? So answer is neither difficult or easier than that. But if I compute HK multiplicity of two pair of ideal, I can talk about this segue product which I get it for free. That's one thing. But many times and since I'm taking a length function this invariant of taking a length invariant corresponding to a pair and looking in a spread out fashion rather than adding it up somehow that seems to hold many more invariants of the graded pair. So for example, if your ring is strongly F regular which is same as F regular in graded case. So if I look at the support of the function, I said compactly supported the largest suppose. So that corresponds to a F pure threshold that was a result of Watanabe and myself. So this seems to carry many more familiar invariants of the ring also. And perhaps it will give more invariants of the ring. So here philosophy is also that though I may not be able to compute the function. It's true, but the nature I can figure out how this function varies vis-a-vis the as characteristic changes or whatever changes. The shape of the function may give me the information which I need. So quadric hypersurface is one of such example. So the first result you prove here, this assertion, it is not actually computing the function at all. But the nature of the function gives me this information. So how much time I have? Okay, so theorem. So this is about quadric hypersurface. So it says there would be an interval. I call it IP just to denote that the interval depends on characteristic P inside a unit interval. And this interval I call it difficult range. And you take N naught to be a sum midpoint approximately. Point of the interval zero and D minus one. So that N naught approximation depends on the order and even property of your D of the quadric hypersurface. So here what I get, I want to write now the density function for the quadric hypersurface. So it looks like as follows. F of R or D, I'm sorry, what is it? R P D of X looks like F infinity of O. F infinity of R X, if X minus N naught doesn't belong to this difficult range. This is square of round cricket, never mind. And is equal to F of R infinity X plus, I'm going to define this quickly, say something about these functions of X minus N naught. If X minus N naught belongs to the difficult range. So this will be the, this is how the function would look. Now what are these, but so the property of this is that F infinity you are looking is a continuous function. Infinity shows that it doesn't depend on the characteristics. So this is a characteristic free part of the thing. And second, mu infinity is a function in fact from the interval 01 to interval 02 and it is a continuous function. So it's a bounded continuous function. That is the point here. And third, the very important thing that length of the interval IP tends to zero as P tends to infinity. So suppose I have this information. So I claim that in that case, you prove this is automatic here. So it's very easy. So now just look at it. So I'm giving part of Y, A or Monsky. I don't know what should I say here. So look at this, look at that formulation. So this one is telling me, I look at the EHK of RPT. So now I have to integrate this function. I'll integrate over this and integrate over this. But this is a non-negative part. So it's greater than or equal to integration of F infinity of RPX, RD of X DX, zero to infinity. But now this is saying the IP length of IP tending to zero says that this is same thing as saying limit of P tending to infinity, sorry, integral F of RP DX, zero to infinity. So this is what you get. And this is now equal to limit of, now my density function formula says this is the limit of EHK of RPD, right? So I'm just applying the shape of the function. And but now this is the theorem result what we have proved, it says the limit exist, but actually the result of Gessel-Monsky says this is equal to one plus empty. So you always get it for free. So this is the thing here. Now what? So this, so how are you getting all this thing? So this is the first part of this result. You get this formula for density function of Godric hyper surface. Now you reach the formula, a final form, formula like that. You need to know how this function behaves in the interval IP. So actually I don't know how it behaves but I have something as here. So what here to formulate, to arrive at this, even the first formula and the second one, what we are using is the classification of ACM bundle on Godric hyper surface. This comes via matrix factorization and which is relies on known as periodicity theorem, but that's all over C, but later classification of maximal Kohan-Makalai module was given by book why over Godric hyper surface in any characteristic was book wise, Eisenberg and Herzog. So that comes into play. So that's, so using that Achingar and has given the classification of this sheaf, so I'll just write it in a, so Q is QD, QD is your, proge of your RPD. So what he does, we know that this, I mean it's a well known that if I take this vector bundle over my Godric hyper surface, projective Godric hyper surface, it is spritz up in terms of the twist of the line bundle, ample line bundle or a spinar bundle. That is well known. What he, what Achingar has done that if I take such a thing, it will split up and it they all the bundles which are O of QD, QI and a spinar bundle of SJ. So he's telling when this bundle exactly will occur, they're saying. So he did that. Now we go through that whole machinery again and we say this bundles ranks occur and in a uniform way, if I look at the like mu and not, which is a rank of a spinar bundle, I give a density function for the rank. So rank density function, you think, is spread out those things. So this is the U of mu and not P. So this is there and then these rank functions, mu I, I'm not writing everything here. So these are the density function, rank density function associated with the ranks of these bundles. So if I look, so my user density function is a positive linear combination of these density functions. That's one thing you didn't get it. Now these functions, we know outside the difficult range, which I denoted IP, they behave like a polynomial. Like what means rationals, there's no characteristic involved. Very good. IP is a problem and it is the where characteristic P behavior shows up. So what you do now, so I need to only know about this actually. So what you do that difficult range is IP, you write, I'm writing this term, is a union of IP of N1 up to NL. So this NL stands for the indexing, L belongs to N. Sorry, I'll just explain. And this Ni belongs to the fixed integers, zero to D minus four. So you divide into infinite number of intervals. Actually the membership of element belonging to this interval depends on the periodic expansion of that element. It's a lot of messy and technical work. So here what you're doing, you are not able to cover the entire IP in terms of this. But approximately, means it covers other than outside a set of measure zero. But I can always ignore my set of measure zero because when I integrate, I don't have to worry about that. So this is there. Now notice this, I said this indexing, this stands for indexing, but the way I'm indexing them, they're independent of the characteristic, this indexing. So though interval depends on the length of the interval depends on the P, which P you're taking, this indexing doesn't. So it gives you a kind of ground to compare various P characteristic changing there. So what we do is on each IP, this IP of N1 up to NL, you write your mu and not on this function. I'm just writing it mu P of X as a polynomial function. H1, I'm sorry, this is a, I can't explain everything. It's a really, at times I myself get confused what I'm doing actually. So the thing is, you have a difficult range and there will be fixed set of polynomials, okay? And for each interval, there will be polynomial depending on the interval, the polynomial depend, but the number of variables depend independent of the polynomial, P equal to one over P. So this is what you express for those intervals and then you integrate, okay? So you integrate. So the thing is coming here. Here some bit of analysis comes in. You have infinite interval on each interval each IP you will integrate this function. It's a nice polynomial function, but when you integrate, you get a power series when you take a sum of infinite interval, but since these are being very well actually this function. So they come like a power series over one matrix actually. So it's something like a integration of mu I, mu and not will correspond to summation of A of T, T equal to one over P where L is equal to zero to infinity power L, L equal to zero to infinity. So this is a matrix ALT with rational polynomials and T equal to P is the answer here, but this infinite, this is a power series given terms of a power of one single matrix. So what you prove that eigenvalue of this matrix is less than one if T is large enough. So here where am? My theorem. So that's where you get in a neighborhood of this thing, this P. So P, if T is less than something, this is a invertible matrix. So you can write as a rational polynomial and that's what amounts are saying P greater than thing than you have this formulation. And it's strictly increasing also involves similar kind of logic. So I don't want to get into that. So that's how you get. So you may ask question that, I'm constructing infinitely many intervals and saying that on each interval there is a polynomial function, but perhaps they may just coincide. I'm making a heavy weather out of nothing, but you take the first example where characteristic P shows, so which is I think D equal to four. So there I was able to compute brutally this F density function. So in fact, this density function, this number, so there will be infinite interval and there, so on which the density function is a polynomial. So point of singularity of this function, that function is in, has infinite number of singular points. So it has a actually limit point. So they genuinely occur the different polynomials. So that's all I have to say.