 will talk to us on bettie cones over fiber products. So thank you for the introduction. And first of all, I would like to thank the organizers of the school for giving me this opportunity to present my work. So today I will be speaking on bettie cones over fiber products. So this is a joint work with Rajiv Kumar and my PhD thesis advisor, Anand Narayan Hariaran. So this is the outline. We'll define what graded bettie numbers are, what bettie tables are, and see some examples. After that, we'll review some known results. And finally, we'll come to some of our recent results that we have obtained. So throughout the talk, I will be assuming that k is an infinite field. R will denote a standard graded k algebra. And all modules will be finitely generated. What are bettie numbers? So suppose we have a graded module, m over standard graded k algebra, r. Then the ij graded bettie number of m, one way to define that is, it is dimension as a k vector space of the jth component of tor i mk. In other words, what we are saying is that if we have a graded free resolution of m, f dot, then the ith free module that appears in the resolution has beta ij many copies of r shifted by degree minus j. So we tabulate these bettie numbers into a table, which we call as bettie table. And so the ijth entry of this bettie table is beta i, i plus j, and not beta ij. So in the previous talk, Professor Deepankar defined regularity. So we can read regularity from the bettie table as the index of last non-zero row. And similarly, index of the last non-zero column gives us a projective dimension of m. And one thing here to notice is that I'm not assuming my rings to be polynomial rings or regular rings. So regularity of modules, so projective dimension of modules need not be finite. So bettie tables are in this case infinite matrices. So let's see some examples. So if my r is a polynomial ring in two variables, and if I take these two ideals, they are generated by regular sequences, of course. So it's graded free resolution will be given by the Kozol complex. The only difference is that we have to take care of the shifts, depending on the degrees of the generators. And these are the bettie tables. For example, we can say that regularity in this case is 2, in this case it is 3, and projective dimensions are 2. Let's see a few more definitions. We say that the module m is pure. If for every i, beta ij is non-zero for at most 1j. So what this says is that if you consider a bettie table, let's say let's consider this bettie table. So here you can see that in column index by 1, there are two non-zero entries. So this is something that is not a pure module. Whereas here, it's a pure module because you take any column, it contains at most one non-zero entry. A pure module is said to be linear if the non-zero entries appear in a single row. So for example, this table is a bettie table of a linear module. And a standard grade k-algebra is said to be Kozol if the residue field has a linear resolution. So for example, this bettie table will look something like this, with this entry being equal to 1. So what we do is we generate a cone of all finitely generated, bettie tables of finitely generated graded R modules, so which we call as bettie cone. So it is all non-negative rational linear combinations of bettie tables of finitely generated graded modules. So the CIs are rational numbers, non-negative. And we have it subset, namely bettie cone of pure R modules in which we consider linear combination of bettie tables of pure R modules. So we can see that this is clearly a subset of this set. And the natural question that one can ask is when does the equality hold? In other words, when can I write bettie table of any module as rational linear combination of bettie table of pure R modules? We have some results like Eisenberg and Schreyer in 2009. They proved that if your ring is a polynomial ring, then the equality holds. In other words, every bettie table can be decomposed into bettie table of pure R modules. And using this, they resolved the multiplicity conjecture of Herzog, Funnich and Srinivasan. So the statement was conjectured by Weisen-Solderberg and they had also provided a proof for D equal to two. So this result is for polynomial ring. So after their work, several people tried to analyze this bettie cone for other classes of rings. For example, this paper studies bettie cone for KXY modulo quadric, convenient stamp studied for rational normal curve, Georgia Tand Sam studied for three non-colinear points in plane and let's see one result from this paper of Anand Narayan and Kumar where they consider Cohen-McCauley standard graded K-algebra with this special Hilbert series of dimension at most one. And they prove that in this case also the bettie cone and pure cone are equal. If you see both the results, we have put some condition on the ring and we are trying to look for this equality. So natural question to ask is what about the converse? So suppose you know that this equality is true, what can you say about the ring? So we started with looking for converse of this result in particular and we started from the case of bettie cone over fiber product. So I'll define what fiber products are and we'll see that converse of this result is true in some case. What is fiber product? So suppose you have two standard graded K-algebras with the same residue field K, let pi one and pi two denote the usual projections then as the name suggests the fiber product is defined as all ordered pairs where a comes from R1, b comes from R2 such that pi one of a is equal to pi two of b. And this fiber product fits into this Cartesian diagram. Let's see an example. So if R1 and R2 are these two polynomial rings then the fiber product is you put together all the variables and go more by all products Xi, Y, J. So you can see that these two are domains but this is not even an integral domain and that is some of the properties as follows. So if R1 and R2 are standard graded with maximal ideals M1 and M2 then M1 direct sum M2 is maximal ideal of the fiber product and we have M1 into M2 equal to zero. We have the short exact sequence of R modules and from this it follows that the depth of the fiber product is minimum of one depth of R1 and depth of R2. So we can see that depth of R can take only two values either it is zero or it is one whereas the dimension is maximum of R1 and R2. So if one of the R1 or R2 has dimension at least two then these are never cohen, these are not cohen Macaulay, the fiber products. So in order to understand the Betty cone one has to understand free resolutions and for that we'll have to understand how the scissor g's look like. So that's where we started and we have this result that if R is fiber product and if we consider a finitely generated R1 module not an R module then we can think of this M as an R module and considering M as an R module its first scissor g module is isomorphic to this isomorphism is a graded isomorphism. So it can be decomposed as some copies of M2 with some appropriate shifts that come from generators of M and first scissor g of M over an R1 module. With some more work we were able to show that if you start with any R module then it's second scissor g module can be decomposed into an R1 module and an R2 module. So we had to analyze first scissor g of module that is a sub module of a free module because first scissor g of M is a sub module of a free module and once we have this decomposition we can use this lemma to get some information about higher scissor g's. What we were also able to show is that if R is fiber product and suppose there exists a pure R module of infinite projective dimension then first of all such a module should have linear resolution and existence of just one module like this forces the ring to be causal. So we know that there is this result of Audemars-Weisenberg and Piwa which says that ring is causal if and only if k has finite regularity if and only if regularity of k is zero or equivalently every module has finite regularity. So in case of fiber product we can add this statement to the list as an equivalent statement that existence of one pure module forces the ring to be causal. And in particular it follows that in case of fiber products suppose we have this equality then it forces the ring to be causal. This is the converse that we have obtained. So suppose R is a fiber product with depth one and if we have the equality of bety cone and pure cone then we do get that the Hilbert series of R should have this form and in particular it forces the ring to be Cohen-McCauley. So just in case you are wondering about depth zero case because we know that depth of R can be zero or one there is a partial converse that was obtained by Rajiv Kumar and Anand Narayanan. Yeah. So what this result says is that it gives some characterization in case of bety in case of fiber products for Cohen-McCauleyness in terms of this equality. So how much time do I have? Yeah, so this is a sketch. So what we do is because depth is one we can pick a non-zero divisor from each RJ then we consider this ideal because of this equality we can decompose this module into pure modules where this M zero has linear resolution then we analyze this M zero to conclude that the ring should have dimension one. So in particular it is Cohen-McCauley and then we write Hilbert series of R in this form and show that this F of z should be a linear polynomial so that it has this form one plus N z. These are some of the references.