 from IIT Mumbai, and the titan is a generation of two, region of happen to competitive, please. I want to thank the organizers for giving me opportunity to talk here. So my title is on a generalization of two results of Apple to committee to rings. So I'll just state Apple's results. So let gamma be a finite dimensional algebra over a field K. So Apple proved that it's bounded derived category DB mod gamma has also written triangles. I'll define later what that is. If and only if gamma has finite global dimension. You also proved that the bounded, if you look at the bounded homotopic category of projective models over gamma, then it has right AR triangles. If and only if gamma is a Gaussian algebra. Okay, so there is a notion of Gaussian algebra of finite dimensional algebras. But for example, if it is commutative, every Artenian Gaussian ring is a Gaussian algebra. So there is Oslander's philosophy. What he said that that study of representation theory of Artenian algebras have natural analogs in Korn-Mokolem modules over Korn-Mokolem local rings, which are free on the punctured set spectrum. So this was a big philosophy of Oslander and Reiten and they literally did a lot of representation theory of maximal Korn-Mokolem modules. And this was what he did. And so we study natural analogs of Apple's results in the context of commutative no theory on rings. So note that I mean, what I want to say is that, I mean, they work when the module is free on the punctured spectrum. Then it is, the analog goes through very well. So our generalization is the following. So let AM be a commutative no theory on local ring of dimension D. Let DBF mod A be bounded derived category of complexes of finitely generated AM modules with finite length homology. So I want the homology should have finite length. Then it can be shown that DBF mod A is a homophonically Smith triangle category. Our generalization of first of Apple's results is the following. Let AM be a no theory on local ring. The following conditions are equal, are equivalent. A is regular if and only if the DBF mod A has oscillatory triangles. So as Oslander's theory philosophy gives us results which can be proved, but the proofs are different. You have to give different proofs. The same proof doesn't generalize. So let KBF prod A be subcategory of KB prod A with finite length homology. By Os, Apple's result of AM is a zero dimensional commutative Goodenstein ring. Then KB prod A has oscillatory triangles. For higher dimensional Goodenstein rings, we prove the following extension of one direction of Abel result. So let AM be a complete Goodenstein local ring. Then KBF prod A has AR triangles. So we believe the converse of this theorem to is also true. That is more precisely, let AM be a complete no theory on local ring. If KBF prod A has oscillatory triangles, then A is Goodenstein. Okay, so the converse is all the true. That is what we conjecture. And we prove conjecture three under the following cases. So you assume that KB prod A has AR triangles. If A is Cohen-McAuley, then A is Goodenstein. And if dimension of A is less than or equal to one, then A is Goodenstein. So Cohen-McAuley case, we have complete conjecture, but otherwise we can do only for dimension one. So let me just define what oscillatory triangles are. So let C be a Kuhl-Smith triangle category with shift function omega, gamma. The triangle NEM, gamma N in C is a right AR triangle ending at M. If M and N are indecomposable, that is the first condition. H is non-zero. And if D is indecomposable, then for every non-isomorphism T from D to M, H composite T is zero, okay? So duly we have what is called left AR triangle. So it is from M, gamma, some inverse M to N to E to M is left AR triangle starting at N. If M and N are indecomposable, W is non-zero. And if D is indecomposable, then for every non-isomorphism T from N to D, we have T composite W is zero. So we say C has AR triangles. If for any indecomposable M in C, there exists an AR triangle ending at M and a left AR triangle ending, starting at M. So you should have both. Then we say it has AR triangles. So if you study Ossler-Riden theory, if you want to prove, I mean, Ossler-Riden triangles naturally occur in representation theory and they are, so this is the triangle version of that. So let us define what is a Seier functor. So let A M be an orthogonal local ring and E B injective hull of K and you just take the Matlis dual of that and let C be a homfinite A triangle and you get a Krull-Smith category. By a right Seier functor on C, we mean a relative functor F from C to C such that we have an isomorphism. It is CD from Homs CD to Homs CDFC dual for any CD in C, which is natural in C and D. Now whenever you have a right Seier functor, you can prove that F will be fully faithful. Okay, now if F is an equivalence, then we say F is a F to be a Seier functor. So there is a very nice result to Reitun and Wanderberg, which characterizes when a triangle category has A R triangles and the result is that if you have a homfinite A linear triangulated Krull-Smith category, then the following are equivalent. C has A R triangles, if and only if C has a Seier functor. Okay, now this result actually came after Hapel's result, okay, and for me, I mean it is easier to prove that triangle category has a Seier functor and then I mean one can, by this result it has A R triangles. So I'll just give a short proof of that if A is a North Indian local ring, then the bounded derived category, Funneled Commercy is a homfinite triangulated A Krull-Smith category. Now Dbfa is a thick subcategory of Dba, so it is triangulated. Now Balmer proved that Dba has split adipotence. This is in 2001 and using this fact, you can show that Dbfa also has split adipotence. And if K is the homotopic category, then if X, Y is bounded complexes in Dbfa and if you take F to be a minimal projective resolution of X, then Horm Dba X, F, X, Y is Horm K, F, Y, this automatically shows you can localize and so Fp localize is zero. So using that it follows that Horm Dba has finite length. So Dbfa is a finite, Horm finite category. And you take any complex and you just look at the homology. It's clear that if C is the sum of all the homology, then S cannot be the direct sum of C plus one non-zero. So X is the finite direct sum. Now, so every complex X can be written as a finite direct sum of intercomposables. So you also have that if you have a finite, Horm finite which split adipotence, then the endomorphism rings of intercomposables are local. And because of that, you have that Dbfa is a Horm finite Cool-Smith standard category. Now, I'll just prove the first, only the forward implication. So let Am be regular and E injective Hall of A mod M. So you can D is the dual with respect to A and E currently E is the dual with respect to E. Now you know that Dfg is sum of K bounded projective A so the point is that, so, oops, sorry. So D is a equivalence from Kbf project to Kbf project, then you go from to Kbf injective A and then you take projective resolution, okay. So you take the composite of all these maps, then the first thing you show is F is dense and then you want to show F is equivalent but that is easier if you can directly show it a safe function. I mean, you can show that F is a safe function. So you have a homomorphism, HormKxy to HormKyfx dual. So the only, my main contribution was to guess this, that this is the function that we want to do. I mean, that requires some manipulation but once you have that this is the right thing then the proof is actually straightforward. So the contribution is to guess the F, okay. So proof of the second theorem, so you assume AM is a complete Goldenstein local ring and you'd look at the category of finite length AM model which has also a finite projective dimension. So as A is Goldenstein, then you know that a module has finite project dimension if and only if the module has finite injective dimension. So if M belongs to S, then the dual with respect to E also belongs to S and for each M belongs to S, you fix a minimal projective resolution PM and IM is the dual of PM dual which is a minimal injective resolution of M and FF is the thick subcategory of PM in M belongs to S and IF is thick subcategory of IM belongs to S in KBFG injective A. So you can show that FF is contained in KBF project and IF is contained in KBF E. So you have three two balances, the dual with respect to A from KBF project to KBF project opposite and then E is taken dual with respect to the injective hull and P is the projective resolution functor and then you can show that D induces an equivalence from FF to F opposite, E induces an equivalence from FF to I opposite and P induces equivalence from PR from IF to F opposite and you look at this, I mean this is the same as in the theorem one, I mean essentially it's the same functor but you have to do it on this subcategory. Now there's a very non-trivial result due to Neiman which shows that he classified thick subcategories of KBF project and it follows that KBF project does not have proper thick subcategories. So FF is KBF project. This is a extremely non-trivial result of Neiman, I mean. And so once you have Neiman result then the rest of the theorem is okay. I mean you can just prove that and eta is a natural isomorphism and we have proved that this is an equivalent. So this is a right cell functor but this is also equivalent which you just proved. So KBF project has also a right-hand triangle. So now the converse part that if you have a North Korean local ring then the BFA has a triangle if and only if A is regular local. Now we all know that a ring is regular local if and only if the project dimension of residue field is finite. Now if ring is not regular local then how to construct a resolution of K and then there is a process. I mean this was independently proved by Gulukshina scholar that one can use the take process. I'll define what that is to yield a minimal resolution of K and the previous result follows theorem following result you take a minimal resolution of K then if A is not regular that does not exist your right-hand triangle in KBF project ending at X. So this is exactly, I mean you basically you actually need some control on X so that is given by Gulukshina scholar and then you show this. My complexes are all co-homological so let us define what take process is. So an associative algebra X over A is called a non-positive D algebra if the following hypothesis are satisfied X is non-positively graded so and then each XI is a finalized A module X has a unit element and then you assume X is strictly skew commutative so that is X, Y is minus one degree X and degree Y, Y, X and X square is zero if X is odd degree of X is odd and you have a skew derivation from DX to X is that DXN is X to the N plus one and D square is zero for XY homogeneous you have DX, Y is DX, Y plus minus one X, X, D, Y now you have all done your Kuzul complex the Kuzul complex is a very simple example of this if you write it homologically so usually you write positively but I am writing index homologically so this is how it will be and then you have takes process of killing cycles so you let X be a non-positive DAG algebra and let rho be less than zero be negative integer and T is a cycle in degree rho plus one if P is odd then one can join X here algebra to X and if rho is even we can divide variable to X and in the DG algebra X, T you send T to this small T the cycle T is killed and by takes process we can construct a DG resolution of A mod I for any ideal I the problem with this process is that if you start with M primary ideal and you take say the Kuzul complex of I then S star of X has finite length that is fine but if T is a cycle of degree minus one and you join a divided variable T then S star of Y need not have finite length okay so note that but each component H I Y has finite length the whole may not be finite length but each H I Y has finite length okay so then so to go into the category of DBA that is you want finite homology so you get good filtration of DG algebras and so let X be a non positive DG algebra over A we assume X I is a finitely generated free A module for all I less than equal to zero and each component has finite length then by good filtration F I we assume F I F dot I is sub complex of F I with F I N is a direct sum and of X N for all N F I contain F I plus one and union of F I is X and F I N is a direct sum and of F I plus one N for all N and then X you can multiply things so what we want is that there is some C depending on X and F so that F I F J is contained in F I plus J plus C with the appropriate degrees but here is the main point that is you want the homology of F star I to finite for all I that is the total homology is finite so you are not leaving that category of finite finitely generated homology so that is the point okay and we say F is a proper good filtration if F I is not X for all I there is a lemma on good filtration so you assume rho is less than zero and H I X is zero for I zero because an I less than rho plus two and suppose HP plus one X is non-zero and you take a cycle and you join a variable which kills it now you assume X has a good filtration then Z will also have a good filtration if X has a good filtration so I did not write that I am sorry about it I should have mentioned here X has a good filtration if the filtration on X is proper then the constructed filtration on Z is also proper and if rho is even then regardless of filtration on X the constructed filtration on Z is proper and then we have two propositions so the point is that so if U is a minimal complex and V is proper sub-complex where VT, VN is a direct cement or UN in the inclusion V to U is not a retraction in KBF project and now we want a lemma to extend a homotopy from V to U so there the condition is that if W is such that HN of W is 0 for N less than or equal to M and VN is UN for all N bigger than M then G from U to W by chain map so the G ratio to V is null homotopic then G is null homotopic so we can extend the whole trick here is that you can extend a homotopy from V to U the condition is that HN of W should be 0 for N less than or equal to M so why is this useful so I will just give the proof so now we prove the results so suppose if possible that there is an AR triangle ending at X so we have AR triangle U to K to X to U and we assume U is a complex is finite homology so H of U1 is 0 for I bigger than or equal to M now you can take 2 cases so A is a complete intersection so let Y1 be the causal complex on a minimal generators of M and Y2 be Y1, T1, TR be DG complex by killing cycles of degree minus 1 in Y1 then take short that Y2 is a minimal resolution of K and you take a proper good filtration on Y2 then we may assume Fij is Y2 for I bigger than or equal to M for I bigger than or equal to I0 so you have an inclusion from Fiy0 to X now I proved the previous result that phi of YI0 is not a retraction so because this is an AR triangle you have G0 phi I0 is 0 as alpha is an AR triangle and by second proposition we can extend the homotopy from G compass phi I to G compass phi I plus 1 keep on doing that you have homotopy on X and it follows that G is null homotopic so just go back see the thing is if you have AR triangle G is non-zero but we have shown G is null homotopic so G is 0 that is not possible so we have a contradiction the case 2 is that A is not a complete intersection then the point is that this state process doesn't end so if Y1 is a causal complex you take YI killing Y minus 1 and keep on doing that then it is known that YI is not except for all I and let F be a good filtration of YM so I am assuming that so we may assume that Fij is Y as M for I bigger than M and we have an inclusion then phi I0 is not a retraction so G comma phi I0 is 0 then as considered before we can extend to a homotopy on YM and then we have done on YM then see the previous result on extending homotopy did not use anything so we can keep on doing that so we have a homotopy on G is null homotopic which is a contradiction that finishes the proof and so thank you any question or remark could you go back to that thing where you said it's not a retraction the composition to my are you assuming U dot is indicable? no no I mean I'm just assuming VN is a direct sum and it's a it's in project I'm not assuming it's indicable so nothing because I'm a bit puzzled then if I take U dot to be V dot direct sum itself take the direct sum of two copies of V dot and take the inclusion as one sum it's a sub-complex it's a direct sum and minimum everything will be satisfied the retraction because I took a direct sum of two copies and I included one copy you must have some condition I'll think about it until you okay thank you as a question or remark so it's not the case we thank you people