 Okay good afternoon everybody. Good afternoon. It's my pleasure to welcome all of you to this day of celebration of mathematics, in particular the legacy of Srinivas Ramanujan. It's a Ramanujan day event and I'm very pleased that we are joined here by her excellency Nina Malhotra, Indian Ambassador to Italy from Rome. His excellency Vishal Sharma, Permanent Representative of India to UNESCO. He will be joining us later from Paris. Professor Chandrasekhar, Secretary of the Department of Science and Technology, who is joining us from New Delhi. Professor Helge Holden, Secretary General of the International Mathematical Union, joining us from Norway. And Professor Goba Chow, who is a member of the ICTP Scientific Council and also a very eminent mathematician fields medalist joining us from Chicago, who will be introducing today's prize winner. So let me say a few words about today. As I said we are here to celebrate the legacy of the great Indian mathematician Srinivas Ramanujan, who counts among the really stellar figures in the history of mathematics with very far reaching contributions to many areas, in particular the number theory. And this joint prize by the Department of Science and Technology, ICTP and the International Mathematical Union is dedicated to young mathematicians from developing countries who have made outstanding contributions to their field. And I'm very pleased that this year the prize is awarded to a fellow Indian scientist, mathematician, Professor Nina Gupta. She is from the Indian Statistical Institute in Kolkata for her outstanding work in affine algebraic geometry and commutative algebra, in particular for her solution of the Zariski cancellation problem for affine spaces. Okay, hopefully we will understand what these words mean during her talk. There is a second part to the Ramanujan Day, which is on this occasion we would like to announce the appointment of Professor Don Zagir who is in the audience to the ICTP Srinivas Ramanujan International Chair. And he is one of the very eminent number theorists in the world and we are very happy that he is joining ICTP on this chair. But let me say a few words about the idea behind this chair, the idea behind this chair, it's a new initiative at ICTP. First of all to celebrate the unity of science, emphasizing that science is really a common heritage of humankind. It has sprung all over the world in different parts of the world. So to celebrate in particular also contributions from non-western or non-male, non-white mathematicians apart from what are normally celebrated. And the idea is to really to bring distinguished mathematicians and scientists to ICTP on these chairs so that they can interact with young scientists coming from all over the world which is really the fundamental mission of ICTP. So that the next Ramanujan is noticed at ICTP and so this is the purpose of these chairs and I think it's particularly appropriate that the first chair is given to Srinivas Ramanujan because let me say a few words and I will say more about it later. But as I'm sure most of you know that he really is counted among the all-time greats in mathematics along with people like Euler and Gauss in terms of the depth and originality of his contributions. And his collaboration with Hardy really counts among the romantic chapters in the history of mathematics where a kind of a self-taught genius working by himself is noticed by a distinguished mathematician in Cambridge and then the collaboration that flourished and that's sort of the idea that really fits well with the mission of ICTP that and therefore I think it's particularly appropriate that we start our international chairs program with the Srinivas Ramanujan International Chair. And in fact it's also very appropriate that Don is going to be the first occupant of the chair because he's perhaps someone who comes close in the style of his doing mathematics to Ramanujan I would say and I had a good fortune to collaborate with him and this is a very wonderful example of also not only of unity of science coming from different parts of the world but the unity of science in terms of content because it turns out that some of these beautiful ideas that Ramanujan created and the structures that he created a hundred years ago about partitions and mock theta functions which will be talked about they come to have applications in areas of physics a hundred years later in particular in the physics of quantum black holes something that nobody would have anticipated even Ramanujan himself would not have anticipated because black holes were not even known at that time. So it's a particularly appropriate that we celebrate this day dedicated to Ramanujan and with this one very eminent mathematician going to occupy the chair named after him and one very accomplished young mathematician coming from India receiving a prize named after him. So I welcome you all and I give the floor to Professor Claudio Arezzo who will be the moderator. Yes on top of welcoming everybody also on my behalf but now I would like to to give the floor to her excellency Madame Nina Malhotra the ambassador of India to Rome for her welcome message. If I don't see her now maybe we lost temporarily the connection okay so in that case in that case let me ask Professor Vajne representing the secretary of the department of science and technology of India who is one of the three partners of this of today's prize to give a welcome message. Thank you Dr. Arezzo. First of all apology on behalf of Professor Chandrasekhar secretary department of science and technology who would have loved to join but he's held up somewhere and he will try to join the function as soon as he could come out of it. So I would like to convey my personal and also Dr. Chandrasekhar's end of the department of science and technology congratulations to Dr. Nina Gupta a mathematician at Indian statistical institution Kolkata for being awarded the prestigious DST, ICTP IMU Ramanujan Prize for the year 2021. Dr. Gupta this award is recognition of your work and contribution in the field of in affine algebraic geometry and commutative algebra in particular for your solution of Jaraiski cancellation problem for the affine spaces. The award also recognizes your contribution in advancement of mathematics in India. DST, ICTP IMU Ramanujan Prize is given for outstanding contribution by young mathematicians from the developing countries. It's a proud moment for India that this year you are claiming this prize as an Indian. It is also great honor that the award has been instituted in the name of Ramanujan the great Indian mathematician who was recognized among the best mathematicians. I'm sure your contribution will further contribute to enlarge the legacy of great Ramanujan. It is noteworthy the award has been bestowed to a female researcher second year in continuation which will also encourage female researchers around the world to take mathematics as their career choice which is one of the underrepresented area within STEM when we are talking about gender equality. I'm also sure that recognition will motivate you further to expand your research with more note for the outcome in future. It will also inspire researchers and young mathematicians not only in India but in an entire developing world to undertake research in mathematics. I wish you all the best for your future India. Congratulations once again. Thank you. Thank you. Thank you very much and while our staff I think is trying to reconnect with the Embassy of India in Rome I'm very very happy to have with us the Secretary General of the International Mathematical Union. My probably our panelists see it or see it just a little bit of this room. This room is filled with very young faces and students and so on. The International Mathematical Union is of course the largest and by far the most important organization of mathematicians all over the world so we are very happy to have Professor Helge Holden with us and I leave him the floor for his message. Thank you very much. I hope you can hear me. I would first like to thank the ICTP for the opportunity to say a few words on this great occasion. I apologize for not being able to attend the event in person. I've had the pleasure of doing so on previous occasions and I've thoroughly enjoyed it. Where I'm sitting now we are enjoying maybe not the right word but we are having a strong snowstorm so it's clear that I would like to to be in interest in Italy right now. So let me start by saying a few words about the IMU the International Mathematical Union. The IMU was founded in 1920 and indeed we celebrated our centennial only last year not because we forgot about it or we couldn't do the computation but due to the pandemic. The early IMU had its focus to provide a framework on the International Congress of Mathematicians the ICMs that indeed had been taking place in 1987. Furthermore the ICMs served as one of the few meeting places to discuss recent developments in mathematics. The IMU was dissolved before World War II and was reorganized in the aftermath of the war in 1952. We now have about 90 countries worldwide as members. From its reconstruction support for mathematicians in developing countries has always been a core activity of the IMU. The activities and organization have changed over time and currently is organized in a commission for developing countries the CDC with Dipendra Prasad an Indian mathematician as its president. Let me mention some of the ongoing activities of the CDC. We have what is called the breakout graduate fellowships which are thanks to the generous donation by the recipients of the breakthrough prizes in mathematics and thanks to these donations we can offer scholarships to talented students that want to pursue a doctoral degree in a developing country. We hope that this program will help educate the next generation of mathematicians and hopefully result in extended benefits to the home countries all those chosen for the fellowships. We also have what is called the grade graduate research assistantships in developing countries. Here the program offers modest support for emerging research groups working in a developing country thus making it possible for them to fund their most talented students to study full-time as graduate research assistants and pursue a master or PhD graduate degree in mathematics thereby fostering the growth of a mathematics community. He's assumed that the emerging research group has an ongoing collaboration with an international mathematician. Furthermore we have the volunteer lecture program it's a program to foster research and international cooperation between mathematicians in developing countries and the international mathematical community by offering financial assistance to universities in developing countries to host a volunteer lecture for a three to four week intensive course. We are currently also discussing how to extend this to include virtual lectures. We also have what is called the Aval visiting scholar program this program gives an annual grant to support mathematicians professionally based in a developing country to visit an international research collaborator for a period of one month that can be extended to a longer stay. We also support the organization and hosting our conferences. Let me return to the ICM the International Congress of Mathematicians. It remains a flagship event of the IMU and indeed for international mathematics starting in 2014 it has become a tradition to offer a generous program of support for mathematicians from developing countries to attend the ICM. In 2014 the program supported that 1,000 mathematicians from developing countries and this number was similar in 2018 and will be similar this year when we meet in St. Petersburg Russia on July 6 to 14. Let me now turn to the Raman Ujjan Prize. The IMU is a proud supporter of the Raman Ujjan Prize that we're here to celebrate today. It is my experience over a long life visiting universities in many countries that mathematical talent is ubiquitous and talent does not know any boundaries. Mathematical talent is also insensitive to gender, religion, cultural background, geography etc. However our ability to nurture this talent is not equally distributed globally indeed Raman Ujjan himself is a prime example of that. As a Norwegian I can add the example of Nilsenrik Abel at the time of Abel in the early 1800s nor we had no university of its own and no solid education could be offered to him. Thanks to generous support he could travel to the big centres of mathematics in those days that meant Paris and Berlin where his talent was discovered. Fortunately the legacy of Raman Ujjan and Abel will live on forever. The Raman Ujjan Prize is a small step to stimulate and support young and talented mathematicians in developing countries. We hope that by recognising mathematical talent and accomplishments by young mathematicians we encourage young mathematicians to pursue their interest in mathematics. I know that the Raman Ujjan Prize has meant a lot to several of the previous recipients. I have been on the prize committee earlier and I know that the competition to receive the Raman Ujjan Prize is fierce. I'm thus very pleased to congratulate Professor Nina Gupta on receiving the 2021 Raman Ujjan Prize. Congratulations from the IMU. Thank you very much. Thank you Professor Holden. Thank you very much for this message and for pointing all these opportunities that IMU offers also to the people that of course are in this room and our connections. So I see I'm happy to see that the connection with the Embassy in Rome has been re-established so I'm very honoured to introduce you Her Excellency Madame Nina Malhotra, the Ambassador of India in Rome and I leave her the floor for a message. Thank you and I apologise for this technical glitch. I think there was some problem and somehow I'm happy that we are able to reconnect. So let me just begin honorable Secretary-General of the International Mathematical Union Professor Holden, Director of ICTP Mr. Arshish Davulkar, distinguished guests, ladies and gentlemen. I'm delighted to join you all for today for this year's Raman Ujjan Prize Award ceremony. Unfortunately I was unable to join in person but I hope to make up my absence for my absence later this year. This award which is instituted to encourage high level of research in the field of mathematical sciences also reflects the long-standing association between ICTP Italy and IMU Germany and DST India. Every year on December 22nd, the anniversary of Srinivas Ramachandran's birthday, the Government of India observes National Mathematics Day to honour India's great scientist and his contribution in the field of mathematics and to embrace his vision and passion for the subject that he loved. I feel proud in stating that this award not only highlights and perpetuates the legacy of Srinivas Raman Ujjan but also showcases India's historic contribution to the field of mathematics. I'm intrigued by the fact that the wealth of ideas he created a century ago are being researched upon and finding applications in diverse contexts. I would also like to take this opportunity to congratulate this year's winner Professor Nina Gupta, my name is Seik and heartiest congratulations to you, a mathematician at the premier in Indian Statistical Institute in Kolkata. Her selection stands testimony to the fact that India has an abundance of talent and a vast pool of researchers who have the potential to become world-class mathematicians, scientists, technocrats and entrepreneurs. I wish Professor Gupta many more prestigious awards and honours. I once again extend my gratitude for the invitation and look forward to a visit to ICTP in Trieste in the near future. I wish all the best for the success of this event. Thank you. Of course we are also looking forward to have a real I noticed that also his excellency the permanent representative of India to UNESCO Vishal Sharma has managed to connect now so I would be very happy to leave him the floor. Thank you. Am I audible? Yes, we can hear you well. Thank you. Excellencies, colleagues, ladies and gentlemen, good afternoon. The United Nations Educational, Scientific and Cultural Organization and ICTP have forged a strong and enduring relationship over the past 50 years. That relationship dates back to the 1960s when UNESCO first lent its support to ICTP and today ICTP is a UNESCO category one institute. UNESCO's broad mandate in education, science and culture presents an excellent counterpoint to the center's focus on the basic sciences particularly physics and mathematics. India is the chair of UNESCO's FA commission in the present biennium. I feel therefore honoured to participate in this award ceremony instituted in the name of the great mathematician Ramanujan. Mathematics in India has a long and rich history. Precursors to the decimal system can be traced back to Vedic Samhitas. Since I'm joining this ceremony from France, I'm reminded of the French mathematician Laplace who while referring to the decimal notation said that it is India that gave us the ingenious method of expressing all numbers by means of 10 symbols. Ramanujan was one such spark of genius in the long tradition of Indian mathematicians. Despite his economic background he created a name for himself by following his passion for mathematics. His methodology of inductive reasoning and reliance on intuition were very different from those of his colleagues at Trinity. His contribution is extraordinary. Through his methods he brought the eastern and western thoughts together. His partnership with Mr. Hardy made him adapt to the western discipline and rigor while Hardy appreciated his intuition. A young man from Tamil Nadu in free independent India could have been easily forgotten by the world but the universe has something else in mind. Let me cite an analogy from physics wherein the seemingly infinite is sought to be understood through calculus which is the magic of the infinite decimal. I thank all the people behind this prize for keeping the memory of Ramanujan alive. The award for the year 2021 is special for two reasons. As an Indian it makes me extremely proud that the prize is being awarded to an Indian mathematician. The second reason is that it is being awarded to an Indian lady mathematician. Recognizing the contributions of women in the field of mathematics would encourage more women to join the SDEM disciplines. I once again congratulate the awardee. Thank you so much. Thank you very much and really I really wish I don't want to thank you all, DST, the ambassador in Rome and I am new for this partnership which is I think building a beautiful experience for everybody. So now we approach the mathematical side of the day so we would like to understand a bit more about Professor Gupta's work and we asked one of the greatest experts in the fields at the crossroad of many fields because as you will see also Professor Gupta's work is about mixing algebra, geometry, and other parts of mathematics. So I'm very happy to have Professor Gobao Chau with us to introduce us to the work of today's winner, this year's winner. Let me just say two words about Professor Gobao Chau quickly. This is not a comprehensive introduction but he's the Francis and Rose UN Distinguished Service Professor at the University of Chicago. He's also the scientific director of the Vietnamese Institute for Advanced Studies in Mathematics which is an institution with whom we collaborate closely and of course he's a 2010 Fields Medalist and our friendship with ICTP also is now a member of, he has been a member for the past two years of the ICTP scientific course. So Professor Gobao Chau, thank you very much for being with us and I leave you the floor for mathematical introduction to the winner's work. Thank you. Thank you Professor Aridzo. Could you hear me well? Very well. Okay. It's my great honor and pleasure to say some words to introduce Professor Nina Gupta's work today. Professor Nina Gupta had done a astonishing couple of works interestingly, algebraic geometries in particular her stunning solution to the Zavisky consummation problems. So let me say a few words to situate her work with some kind of proper historical context and significance. Algebraic geometries as the independent phoenix study has after was born in Italy as you all know with illustrious name is Castell Nuovo and Rick Press and Severi. On that period we inherit a large body of extreme body of work on curves on surfaces and so on. So after this floating period that it had been realized that the works done by Italian mathematician was based on a part on in geometric intuition. It lacked some proper mathematical foundations. So there's a second date. The foundation of algebraic geometries has been done in America led by Oskar Zavisky and Andrei. So instead of thinking about how curves move, how surface moves, and what we need to sell. Zavisky and Bay have put algebraic geometries on the foundation of the Angebras which was born in Germany also in the end of the 19th century. So for example the simplest object in algebraic geometries would be a five-space vector space leaving two dimensions, three dimensions. And when you think about it from the algebraic geometry perspective, you think about polynomial in two variables and three variables. So there's a lot of this period development in one side bought a firm foundation for algebraic geometries that we are on work on. And also we had developed a great deal on knowledge about Angebras, I mean about polynomials, which seemed very simple. We learned on in high school maybe, but it had been seemed very intricate. So one of the problems that had been asked by Zavisky in that period about the cancellation program. So in order of cancellation numbers, for example, if x plus 2 would be 10, then x would be 8. So what if you replace 2 by a five-space dimension and 10 by a five-space dimension, put the other vector x should be a five-space dimension 8. So that was roughly what the question of Zavisky's cancellation program is. And that is one of the longest, long-standing problem in algebraic geometries. And that when it was studying that Freninda Huttak gave a good solution to that problem in characteristic B. And so forth, Freninda Huttak came in, I would say that in the long and glorious tradition with Indian mathematics, in practical algebra geometries, when you talk about Indian algebra geometries, of course, we all know about the school studying vector random connections, one of which one of the illustrious representatives is Alit Nahasimhan, who was the former head of ICTP, who passed away last year, was our dear friend of harvest. But there's also the another branch, started by Abhiyanka, who was a student of post-causality, who has been working on a lot of singularities from very close to the risk cancellation problem. So Freninda Huttak really was already such a herself in this branch of algebra geometries, Indian algebra geometries, and some kind of the proof that is in the school, its strivings and forfeit admires. So that's what I had to say about the history around these problems, the algebra geometries, Indian mathematics, and it was a great pleasure to me to say these few words about Freninda Huttak, who cannot wait to hear her talk on her work. Thank you very much. Thank you Professor Chow. I think we were all the mathematicians were trying to steal a good idea from your blackboard also while you were, but the blackboard was not full. So thank you very much. So now I would like to ask the ICTP director to present the prize to Professor Gupta, which unfortunately I promise next year, hopefully there will be another fifth wave or a sixth wave, we will recover everything, but unfortunately we have to do it virtually. So Professor Gupta, this is for you. And in this moment we imagine that you are receiving from the hands of the ICTP director a statue. Congratulations. Congratulations. So of course the prize comes with a statue, a certificate which now I don't know if it's movable or will be, it's movable, yes. And the money prize of course funded by the Department of Science and Technology of India, which of course we are again very grateful. Okay, so also my congratulations Professor Gupta, and now we really would like to hear from you. So let me introduce Professor Gupta's talk. She's giving the Ramanujan prize lectures on GA actions and their applications. I guess we now have to fix also the technology. You know how to share the screen and we actually would like to move on the next the other side to enjoy your talk. Thank you. Oh sorry, just one comment. If somebody wants to ask some questions to Professor Gupta, please use the Q&A section of the Zoom link. Okay, I'm not ready. You can start. Thank you. Thank you everyone for being here and presenting me this award. It is indeed a matter of great honor for me to having received this prize. More so because it is named after Srinivas Ramanujan, one of the greatest mathematician the world has ever seen. So I know this prize brings a lot of responsibilities on me and I hope and wish to continue at least the legacy of Ramanujan and this prize which is honored to me is my future work. I take this opportunity to thank Professor Hatish Dabholkar, Director ICTP, Professor Chandrasekharan, the Secretary General of Department of Science and Technology from India, Professor Holden, the General Secretary of International Mathematical Union. It's a wonderful thing that a mathematical union is there which of course the ICM every four years and we all know the pride and prestige of being going to ICM and being there. I also take this opportunity to thank his Excellency Nina Madham and Vishal Sir for being here. Finally, I would like to thank Professor Chow for a wonderful introduction and he had in fact said whatever I wanted to say about Zariski and Abhyankar and this Italian School of Geometry and the work. This is very wonderfully explained. Thank you for being here and I hope to have meet you in person and hopefully we will meet sometime. Unlike Professor Ramanujan, who was primarily self-taught, I was probably taught and guided by several people. I can't even, on this occasion, I would like to say thanks to many people, especially to my teachers from schools and colleges who brought up my foundation stones. I would like to thank the teachers from Indian Statistical Institute who introduced me to this world of higher mathematics and its sophistications. I also thank my colleagues and my students who were discussing this with them. They had been quite enriching for me. A special thanks goes to Professor S.M. Hartford-Ecker who is my grand supervisor and who empowered me with several techniques which I could use in my future research. I thank the people of the Authorities of Tartar Institute of Fundamental Research for facilitating my visit there so that I can interact with professors from TIFR like Professor S.M. Hartford-Ecker, Professor Ravirao and Professor Gulja. I also thank many mathematicians from abroad who were kind enough to visit ISI like Professor Onoda, Professor T.S. Anuma, Professor S. Kuroda, Professor David Wright and Professor Avinash Sathe who had been paying visits to ISI regularly. Their visits had a lot of impact on my research. The moment they visit I am excited with new problems and new interactions so I really very thankful to them. Finally I would like to thank my family and for having faith in me and supporting me to pursue my research, my passion in mathematics. Lastly I would like to thank my supervisor Professor Amartya Kumar Datta who introduced me to this world of affine algebraic geometry and commutative algebra. This extensive work would not have been possible without his continuous support and guidance. I really owe a lot to him. Now it's time to pay my tribute to GA actions and their applications. I will try to make this talk with just a simple talk and with a quick introduction we will discuss a few applications which this action has been playing. GA is this additive group K plus and it has its two equivalent formulations, exponential maps which we can talk in arbitrary characteristic and local important derivations in characteristic zero. We will discuss in briefly the three major breakthroughs which was achieved using local important derivations and exponential maps in the area of affine algebraic geometry. Namely the solution to the Zaris case cancellation problem for the affine space affine plane in characteristic zero and the non-triviality of the Russell chorus three-fold which led to the solution of the linearization conjecture. It was a major problem and one of the biggest result in the 1990s and finally my solution to the Zaris case cancellation problem in positive characteristic which I could achieve by proving a non-triviality of the three-fold constructed by T. Asanuma. So before that let me just define what is a GA action. Throughout my talk K will always denote an algebraically closed field and V is an affine variety over the field K. GA is the additive group K plus so K is a field so K plus is an additive group so it has an algebraic group structure which you can realize that has closed subgroup of the GLN 2K. So this group action means this group is acting on a variety V in an usual way so these are the two axioms of the group actions which you can see it's the identity axioms and this is the additive action so this is an additive group action on an affine variety. So right from the 19th century the study of group actions on an affine variety was a very central topic of research in fact it led to the Hilbert basis theorem one of the fundamental results from commutative algebra and algebraic geometry which Hilbert discovered and proved to solve a major problem a fundamental problem in invariant theory. So invariant theory is a theory which is related to study of the ring of invariants of a group action. In fact the Hilbert's 14th problem so Hilbert has posed some 23 problem in the first ICM 1900 and this 14th problem actually originated from the study of invariant theory and this one of the early results in the study of GA actions the additive group action was this result of more error which was in fact a there was a probably a gap in the proof which was taken up by Wittzenberg and finally Sheshadri gave a complete argument that which says that if the additive group GA acts on the affine space a and k by linear transformation then the ring of invariant is always finite to genetic. So this was you can see there's a study of GA action was very early in the history it is as early as in 90s 19th century but this theorem of Nagata and Popu and then there were some partial results for finite groups which you can see as a standard exercise on integral extensions and commutative algebra and in fact Amy Neuder had given a constructive proof in characteristic 0 in 1916 and positive characteristic in 1926. So this theorem says that the ring of invariant of an algebraic group action on an affine variety V is finitely generated if and only if the group G is reductive. Now the group additive group GA is far away from being reductive it's an inubotent group and hence in particular the ring of invariant of a GA action did not be finitely generated and explicit examples have been constructed which eventually led to the counter examples to the Hilbert's 14th problem. Therefore this study of GA actions was thought of to be not so interesting in fact pathological because you don't have any control on the ring of invariant which is not finitely generated so it is outside thought of a subject outside algebraic geometry and not many geometers took up this subject. However few people like a notable result by Gabriel on Slice Theorem and Noyes and Rensler Dick Smear had many early landmark results in 1970s. Mianishi was probably the first mathematician who has systematically investigated GA actions in his Piaffer book in 1978. He systematically wrote down the various theorems and results of local important derivations and proved his results so in fact Mianishi had given an algebraic characterization of the affine spaces and in fact some classification of the surfaces so he promoted the study of local important derivations and GA actions. In 90s many mathematicians most notably Makalimannam, Devani and Finstern, Van Dinesen, Kalliman, Degli and Frodenberg, Dachshund Douglas and SM Hartford and many more recent factors not easy for me to write the names of all the people who have been working in this area of GA actions and proving important results and making discoveries. So I will be in fact not talking about all those results but I will just talk about the few breakthroughs which was achieved using GA actions. The first being the solution of the Zariski cancellation problem for the affine plane and the second one was the linearization conjecture which was solved using solved by Russell and Corus and Russell but there was a very important step left behind where Makalimannam helped them. So the conjecture was that every faithful algebraic C star action on C3 is linearizable. So C star is a reductive group. So reductive group was thought of a very good group because their ring of invariant is finitely generated. However to achieve something we want some solution for a reductive group we have to study the action of the unipotent groups which are far away from being reductive and so it was the additive group actions which came for rescue the reductive group actions and finally in 2010 this say 20th century this 20th last decade exponential maps which is algebraic version of GA action in arbitrary characteristic helped me in solving the Zariski cancellation problem in positive characteristic in higher dimension. So thus in these two cases these two problems a certain invariant associated to a GA action was quite useful and those invariants were useful in distinguishing between two rings where all known methods failed. So this is you can take the quality of GA action and the takeaway which we can take from today's talk is that we should not ignore additive group action just because of its pathological nature that the ring of invariant is not finitely generated it has many more things to contribute and it should not remain confined to a specialist. So now I just start from the scratch from the definition so B is a finitely generated k algebra V is an affine variety GA is the additive group k plus so the action of the additive group on the affine variety B is the usual action of the group action. So this is I am writing it in the geometrical language but you can convert it into an algebraic notion and then you can define it for any k algebra B. So the any k algebra B the exponential map of B it is then algebraic way of ring theoretic version of saying the GA action is a k algebra homomorphism phi u from the ring B to the polynomial ring in one variable over B which satisfies this condition that is this map phi u and if you compose it with the evaluation map u equal to 0 then this composition is identity and secondly if I use this suffix phi V for this variable V then this composition phi u and this phi V this overall is phi V plus Q. So this is the two actions of the group actions written in a ring theoretic way. So when k is an algebraically closed field and B is an finitely generated k algebra and V is the maximal spectrum of B then GA action on V is same as exponential map on the ring B. So whether you study GA action or you study exponential maps you are studying the same thing when over an algebraically closed field. Now comes the local important derivations so derivations we are all familiar it is the k linear so if B is an integral look in so for defining local important derivation we don't need Q field of rationals to be contained in B however if you want to use some nice results of local important derivations it is desirable to have Q inside the ring. So if a derivation is a usual derivation which is the linear and which satisfies the Leibniz rule so derivation D is said to be a local important derivation if for every element x in B you can find an integer n which is depending on x such that nth derivative of x is equal to 0. So this is the definition of local important derivations and it looks apparently it looks far away from exponential maps but we will see in the next slide that these two are coincides. So local important derivations are in a way a generalization of the partial derivative so if I I will denote the set of all local important derivations by the set of L and D's on B. Now I just give you a glimpse which is not incomplete proof but how exponential maps and local important derivations are the same thing when Q is contained in the field of rationals are contained in the ring B. So an exponential map IU is usually a homomorphism ring homomorphism from B to B U so suppose an element B goes to this element which is a polynomial in U and because of the first condition the first coordinate is going to be B this is identity and then this exponential map if I look at the coefficient of U and called this is a linear map from B to B defining D of B equal to B1 then one can see that B is actually a derivation and in fact there's a local important derivation because it is a the image is inside a polynomial ring. Conversely any local important derivation you can define a map like this where this coefficient is D B D square B and like this like if I define it like this then that will give rise to an exponential map on the ring B. Moreover an element B is in the kernel of the derivation if and only you can see that phi of B is B that is B is in the ring of invariant of the exponential map. Thus over an algebraically closed field of characteristic 0 any G action on B is same as exponential map on B and which is same as the local important derivations on B and the ring of invariant of a G action is same as the ring of invariant of an exponential map and which is same as studying the kernel of the local important derivations. So what I have seen is that the study of GA actions is same as the study of local important derivations in characteristic 0 or the study of exponential maps in arbitrary characteristic. So I now come to the first vector which was achieved by using local important derivations. So let K be a field. So the Zariski cancellation problem for the affine plane asks whether if you adjoin a variable to the ring B and suppose it becomes a polynomial ring in three variables does this necessarily imply that B is isomorphic to the polynomial ring in two variables. So this was the cancellation problem for the affine plane and in his attempt to solve this problem our Indian mathematician C.P. Ramanujam he proved his famous topological characterization of the affine plane. Soon Mianishi gave the following algebraic characterization. So what was Mianishi's algebraic characterization of the affine plane? Mianishi proved that if K is a field of characteristic 0 algebraically closed field of characteristic 0 and B is a finitely generated K algebra of dimension 2 such that the units of B are trivial B is a unique factorization domain and there exist a non-zero local important derivation on B then B must be a polynomial ring in two variables. Now you see if this condition is satisfied that B1 is equal to K3 then it is quite obvious that the units of B are trivial and B is a UMD. So there exist so the difficult part would be to show the existence of a non-zero local important derivation on B and which will solve the cancellation problem for the affine plane. Mianishi and Suji with an important contribution by Fujita in 1980s they proved that the ring B is a polynomial ring if B is a stably polynomial ring and so this theorem of Fujita Mianishi and Suji is crucially used the Mianishi's characterization which based on local important derivations and their proof is quite beautiful but how it's not so self-contained Makalimannam recently in 2008 have given a very simplified self-contained proof and we will be giving a glimpse of that proof modulo important flame up so to give the proof I define the Makalimannam invariant so the Makalimannam invariant of a ring is the offering R is the intersection of all the kernels of derivations and which is intersection of the ring of exponential maps and we call it the ak invariant and it is also nowadays it's called Makalimann invariant so for example it's easy to see the Makalimann invariant of the polynomial rings Cxyz is C because we have the three partial derivatives which if they vanish then F must be a constant function so I give the proof this proof was due to Krishula and Makalimannam and an important claim is needed here so what they prove that if B is an affine domain over a field K which does not have any non-zero local important derivation then the ring of Makalimann invariant of B1 is B and the proof is quite elementary you can see just so it's a suppose B1 is a stably polynomial ring then we have to prove this so from this one can easily deduce that B is a finitely generated of dimension two the units of B are trivial since this happens and B is a UFD because the K3 is a UFD so retract of the UFD is UFD and enough to show that there exists a non-zero local important derivation on B because that's a result characterization of the affine plane by Nianishi now that follows immediately because the Makalimann invariant of Bx is same as the Makalimann invariant of K3 which is K as we have seen before so from this theorem lemma of Krishula and Makalimannam we can see that B1 is a Makalimann invariant of B1 cannot be B so it must be so B must have a non-zero local important derivation so the contrapositive of this lemma will give the proof now we given means with my Nikhilesh Dasgupta we have given an algebraic characterization of the affine three space which again uses local important derivation so if I define lnd star of B all those lnd local important derivations which whose image has identity of the ring B and ml star of B the intersection of all the kernel of D where D has a slice so D has a slice means D the image of D has identity then we proved this theorem for the affine plane affine this is not affine there two-dimensional K algebras and this is a result for the three-dimensional K algebras so we proved that all these three conditions are equivalent if the ml star of D is BK or mlB of is K and it has a this lnd star of B is non-empty so the similar things we obtained for the three-dimensional space but we have to assume in this case that the ring B is a UFT and K is an algebraically closed field this theorem is does not require that however we cannot extend at least this characterization to the higher dimension now I come to this famous cubic defined by Russell so you can see that this ring is so simple so it's a three-fold which is given by this one equation and it's a linear polynomial in y it's so nice to see this ring so simple and it was a very important question for some time whether A is a polynomial ringing three variables over C or the C field of complex numbers because it was very tempting to believe that A is C3 reason being that this ring is a smooth ring which is a unifap precision domain which has trivial units and it is dominated by an affine three space in fact all projective A models are free and if I look at the variety defined by this ring A then this variety is diffeomorphic to R6 so topologically there is no obstruction and in fact the logarithm codara dimension is also minus infinity we call this a famous topological characterization of the plane by CP Ramanujan which says that if these conditions are satisfied by an affine this surface then that surface is necessarily the affine plane so all the known ingredients were not able to distinguish this ring A from the polynomial ring and the reason so this was realized that it is very important to know whether this ring is a polynomial ring or not because if it is not then it would lead to a step towards the solution of the linear resistance conjecture for the affine three space and this conjecture was as old as in 70s and it was proved for two dimensional rings but for three dimensionals this several steps have been proved so it was a very big problem for a long time almost running over two decades where several small steps were proved and however if this ring turns out to be a polynomial ring then it would have led to counter examples to the linearism conjecture as well as the abhankar satis conjecture so so abhankar had proved this epimorphism theorem for the affine plane epimorphism theorem for the affine line and they have along with satay they have made a conjecture that whenever a hyper plane is isomorphic to the affine plane then that hyper plane is actually a coordinate so and in this example you see that suppose this ring is a polynomial ring in three variables then if you put x is equal to a non-zero number non-zero constant then this y becomes a coordinate and therefore a mod x minus lambda is c2 however if you put x is equal to 0 then this is z square plus t2 which is far away from being smooth so it cannot be a polynomial ring so it's quite easy to see that x minus lambda cannot be a coordinate in A so if this happens then that would be a counter example to the abhankar satay conjecture fortunately makhalimanov came up with a very important tool in fact he developed the invariant which we named a k invariant and we now call it the makhaliman invariant he developed this tool to distinguish this ring a from the polynomial ring and this eventually led to the solution of the linearization conjecture by Russell and porous kaliman and makhalimanov so makhalimanov proved that this ring this porous 3-fold is in fact this is the russell cubic porous and russell have listed down some three-folds and which has non-linearizable c star action and they were wanted to know whether they are all polynomial rings or not and eventually kalimanov and makhalimanov they showed that none of this porous 3-fold is a polynomial ring which leads to the solution of the linearization conjecture so this theorem of makhalimanov this the idea is actually quite simple this even anybody could have thought about this idea sorry it's not anybody because it was a good machinery which was developed by him so if I look at small x the image of capital x in A then it's easy to see that x is not an unit or it's quite difficult to show that if b is any local component derivation on the ring a then the b of x mastery is 0 therefore this makhaliman invariant is not equal to c and hence a is not a polynomial ring so very easily using an tool from local component derivations makhalimanov could distinguish this ring a from the polynomial ring. I now come to this asonomous 3-fold and asonoma one of the greatest mathematician who had worked on affine fibrations he constructed this ring a in 1987 this is a ring over a field of positive characteristic where this number p does not divide this integer s this is here and let r be this subring of a generated by x small x is the image of capital x in this ring a so asonoma proved that this ring a is a stable polynomial ring over r he also showed that a is not a polynomial ring over the subring r although all the fiber rings of a over r are polynomial ring so if I look at all the fibers over the ring r then they are polynomial ring so in particular a is an a2 fibration over the ring r and a is since a is not a polynomial ring over r it is a counter example to the a2 fibration problem over a pid in positive characteristic however note that avinash shathe had proved that if b is a pid containing the field of rational numbers then any a2 fibration over b is necessarily a polynomial ring so this was an example given by asonoma to show that this theorem of sathe does not extend to positive characteristic he also used this ring to give counter examples to the linear addition problem in positive characteristic in higher dimension and it was an open problem for some time to know whether this ring a is itself a polynomial ring over k and I showed that a is not and hence this lead to the solution of the zevinsky cancellation problem in positive characteristic that yeah the cancellation problem does not go in my solution I made use of another tool of locomotive another invariant associated to a exponential map so exponential map because it works in arbitrary characteristic I used the dachshund invariant which was defined by dachshund which is the sub ring of a ring b generated by all those elements which lies inside the ring of invariant for some exponential non-trivial exponential map on b so you can see very easily that the dachshund invariant of a polynomial ring is the whole ring if this polynomial ring is n is greater than 1 and for that we just make use of this exponential maps phi xj where i not equal to j if I define it like this then these are exponential maps with the ring of invariants to be this sub ring and if I vary my i say i equal to 1 and 2 then I will get that the dachshund invariant is the whole ring and what I showed is that the dachshund invariant of this ring a is a proper sub ring of a and hence a is not a polynomial ring you know and this was achieved everything was achieved by making study of exponential maps and dachshund invariant now I proved in a more general setting I consider this ring a where f is a general polynomial and if I look at small f to be the projection on the line x is equal to 0 hyperplan x equal to 0 and small x the image of capital x and a then we proved this theorem that a is a polynomial ring in three variables a is a stably polynomial ring f is a coordinate in this z t and g is a coordinate and x and g forms a coordinate here in x and g so this is an equivalence of five statement and in fact in one statement it proves it connects the affine vibration problem the epimorphism problems and with just a this study of this kind of a three-fold has been reduced to the study of this small f z t and with this theorem we can see the non-privility of the russel corus threefold and asanomus threefold immediately means it was an easy outcome to look at from this theorem that none of these three-folds are trivial and it works in arbitrary characteristic for any field so and in fact the when I this the achievement of this five here this five equivalence is an outcome of a study of exponential maps and dachshund invariant I actually have proved the equivalence of 10 statements so this five statements involves the dachshund invariant of the ring t and to achieve an equivalence of these three we have to make use of the study of dachshund invariant and exponential maps and I stop here I sorry for overshooting so these are the invariants my major invariant is fradenberg I really thank professor fradenberg for writing these excellent books which helped me to get introduced to this subject as early in my career and also the lecture notes of Magali Manavan Digle which were quite useful in my studies thank you everyone for paying attention and for this wonderful ceremony they have arranged and I look forward to the talk of professor dawn and see the glimpse of Ramanujan's work on partition functions in this talk thank you everyone shall I stop sharing here Professor Gupta thank you very much for this beautiful talk and for showing us also I guess the fundamental message that we all received is about the unity of mathematics and the beauty of mixing different subjects to obtain deep and fantastic results I think there are I don't see questions in the sense that is there any quick question from the audience on this room or online but if you could write it quickly because we are a bit behind schedule so I mean as for me of course I keep all my questions for your next visit to ICTP which has to be very soon okay so in that case professor Gupta our congratulation my personal and the congratulations from everybody here and on from the panelist and thanks again now we take a little break also to prepare a little bit of technology but the show is definitely not over we we are very happy to have one of the greatest number theorists who can actually help us to read some parts of Ramanujan's work for non-experts or at least not particularly experts and to to convey with us the importance and the fun I think that we can get from Ramanujan's original work and how this can inspire also research today so we take a little break for short coffee and then we reconvene at four with the international chair lecture by professor Zakiya thanks to everybody thanks to DST to the department of science and technology to the ambassadors their excellencies and to the professor Alden for the international mathematical union for this partnership and we will meet again in 15 14 minutes thanks again okay welcome back everybody it's getting to be evening in ICTP so we'll continue with the second part of the Ramanujan day which is talked by our new international Srinivas Ramanujan international chair recipient of the first chair professor Don Zakiya who will take tell us about Ramanujan and partitions but before that let me say tell you a bit about Don he's a mathematician working in you know many topics apart from number theory he's a number theorist in topology algebraic geometry and mathematical physics he's an emeritus scientific member and director of the Max Plunk Institute for Mathematics in Bonn and was the and he has a longer association with ICTP he first visited ICTP more than 10 years ago and has been part of the center as the distinguished staff associate and collaborated on numerous research papers with scientists here both in mathematics and outside of mathematics I'm as I told you I also collaborated with him on something quite far away from number theory on the face of it and in addition to his ICTP affiliation he has also been a distinguished affiliated member of the Trieste based Institute for geometry and physics which is jointly run by ICTP and CISA and he has shared his time for many years between Germany and professorship in some other country for example University of Maryland then University of Utrecht then at College de France he's a recipient of many prizes including the carousel medal the coal prize in number theory the Pri Ali Kartan the Chauviné prize the Form Stout Prize he's the foreign member of the Royal Netherlands Academy of Arts and Sciences in 1997 member of the National Academy of Sciences in 2017 and he became the honorary member of the London Mathematics Society in 2019 recently in 2021 he was awarded the Fudan Jean-Gy Science Award and he shares this prize with Benedict Gross emeritus professor of mathematics at Harvard University and University of California at San Diego for their formulation and proof of the Gross Zagia formula which is one of his very famous contributions to number theory and as I said Don's style of doing mathematics number theory which I have had the privilege to see from during our collaboration comes closest in some respects to the way Ramanujan worked instead of Ramanujan who did his calculations on a slate Don prefers to do them on pari his favorite program but it's as it was said about Ramanujan I think by little would that every number is his personal friend I mean of course not in some trivial way that Ramanujan saw patterns of interesting patterns among numbers with some deep connections and I think the same could be said in some respects about Don's sort of relationship with numbers so Don and Ramanujan are partitions thank you very much so I'm really I hope everything is working technically you can hear me not just people here but people in the rest of the universe I'm extremely happy to be here because as I teach well Professor double carbon he calls me Don as he just said I've been connected with this dude for about 10 well for eight years as a kind of a member and I absolutely love it as a place for what they do I'm extremely happy that now my connection has been solidified and it's a more permanent one through this new international chair and of course extremely proud that it's the chair bearing the name of Ramanujan who together with Euler is one of my two great heroes in mathematics so in any lecture I give I will praise one of them at the expense of all other mathematicians and often both before I start I would also like to say my personal congratulations to Nina Gupta for the prize and also for a really beautiful lecture I certainly won't be able to do as good a job of making complicated mathematics somewhat understandable to non-experts but I'll achieve something by avoiding complicated mathematics as much as I can so this talk is not meant for professional mathematicians it's meant for students even non-mathematicians it's a popular lecture but I'll try to make some part of it fun even if you are a professional mathematician because of course many of the people in this room if you're here or out there on zoom are so I want to talk as my title says about Ramanujan himself a little bit about him as a person some of his work but in particular on partitions but because this Ramanujan day is specifically geared today towards the prize but also towards introducing the new international chair who is me at one point I'll say in three minutes of just some more personal comments it doesn't mean that I'm confusing myself with Ramanujan I'm not that confused also finally I should apologize if you came to this talk because of the poster showing a very young lively looking mathematician and you want your money back I can't help it I didn't ask to get older the picture is old and I look many years old I don't know from when it dates but I'm no longer as youthful as then of course in my heart I'm equally so I want to as I say start by saying something about Ramanujan himself we heard a few words by Atish and a few words by the other people in their opening remarks so the story you use the word Atish one of the most romantic stories maybe except for Galois's duel it's the only other really romantic story in mathematics which is not famous for romance and it starts with a letter an incredible letter that's kind of epic making so if I could have the first slide you can see of course the letter was handwritten and there are facsimiles but here I have a type version it's easier to read and if you can make it here as big as possible and if you're out there apparently you can enlarge either the screen or the blackboard depending which is relevant so Ramanujan wrote introducing himself to Hardy an utterly incredible letter and you have to put yourself in the place of a great mathematician like Professor Hardy at Cambridge you get a letter from a complete unknown full of the most incredible looking and bizarre looking formulas and a lot of them there were a hundred formulas that you've ever seen the letter says I'm an clerk in the shipping in the port authorities it didn't sound very mathematical I have no formal education I've a salary of 20 pounds a year I'm 23 years old which was not completely true well he was 25 and some of the local mathematicians find by results startling could you please look at them and so look at the second slide you can see an extract from on the you don't have to be able to read it it's incomprehensible anywhere at least that's what Hardy thought on the left you see an facsimile of one page of many pages that were something like a hundred formulas very very complex forbidding incredible formulas and on the right from a book where another page has been reproduced at least set in print and what you can see is that they're extremely complicated and you can't see more hardy of course could see much more he spent the night trying to decide he called his friend little wood who is another with whom he had a great collaboration they spent half of the night studying this letter is this man a genius or a madman or a cheat an imposter had to be one of the three and after a night of studying the formulas they couldn't begin to prove most of the things but they decided it had to be a real genius because no madman and no imposter could have invented such a fantastic formulas so this is the famous story and Romani Hardy within three weeks he spent another feverish three weeks with little wood trying to prove all of the formulas about a third of them they succeeded another third with great difficulty they saw how maybe you go and about a third of them they didn't have a clue that never seen anything like it so it's an amazing story and then he wrote to Romani John inviting him so maybe the next slide we can see what the two looked like so they're supposed to stand but certainly my new John and a famous picture of Hardy with this pipe Hardy was very modest and claimed to be not such a great mathematician although he was a rather great one and he had a famous quote that if he had to grade mathematicians for pure talent he would give himself 25 little wood his collaborator 30 Hilbert of whom we heard the Hilbert problems 80 Romani John 100 that's a pretty strong statement later he took it back said I think I was being too romantic but he was talking about pure talent not the importance of the work which is important but certainly not more important than Hilbert's so let me tell just in a few words the high points of his life while already said he was an underpaid unheard of person in the background in a milieu in Madras where nobody did mathematics he was by definition self-taught there was no one to teach him he had a famous book of formulas 19th century book and he went through the whole very silly book he proved all of the formulas and started as a teenager finding his own and and he discovered more and more when he wrote that famous letter he actually as we now know sent to at least two other British mathematicians whose names have not been forgotten the others we don't know and they all threw it away so their names just suppressed but Hardy as I said together with little wood took it seriously and invited him to England and so he came to England in 1914 justice world war one broke out and he stayed in fact until 1918 and only he badly want to leave at the end he was very depressed by being in England but he couldn't leave until after the war was over so at the end of 1918 or even earlier I think early 1919 he was showered with honors while he was in England in that sense they treated him well he was the became a fellow of the Royal Society and naturally the first Indian to become that he became the fellow of Trinity again the first Indian to become that but he was very unhappy he felt the racism at all times he couldn't stand I mean his health suffered from the climate he couldn't stand the food well nobody can stand British food especially then but it wasn't so much the British food it's the fact that he was of course a strict vegetarian and he couldn't really feed himself so he became very depressed actually threw himself under a train in 1918 and only by miracle was saved and then as I say it went back to India and in 1920 in so I'll use my chalk for the first time so he was born in 1887 in 1920 in January he wrote his last letter so his first letter we saw before it was 1913 last letter to Hardy in which he's full of happiness and tells that he's discovered a wonderful new class of functions called mock theta functions and will tell him all about it and three months later in April Hardy was amazed he got the news that he had died unexpectedly and of course very ill so that is completely amazing life in a nutshell I said before that I want to kind of also introduce myself though many people in this room know me personally or through courses but just very briefly Ramanujan had three great loves and all of them I share as I teach suggest so I wanted to just say briefly what they are one of them is formulas so that's very unpopular these days mathematicians are supposed to care about you know categories and topos is in something very abstract and and functurists and so on but but some people still haven't learned that they care about formulas and I'm one of course Ramanujan had a perfectly good excuse that was more than 100 years ago and he was self-educated I've no excuse I learned from very fancy mathematicians but like Ramanujan I love formulas secondly I love asymptotics which was his perhaps his greatest love so much so that I'm currently giving a course on it and I'll say a few more words about each of these three things and finally a theme in all of not at all but in a great amount of Ramanujan's work but he didn't realize it at all until he started working with Hardy he didn't even really understand the definition is multi-reforms I'm not going to explain but there are but later I'll say a few words if you're not a mathematician to give a feeling or some indication but there are special kind of special functions just like the sine and cosine functions so some specific functions with amazing properties and those properties follow naturally from their definition if you understand the definition but Ramanujan didn't understand the definition but he still found many of those amazing properties indirectly and sometimes cast the light on them which wasn't visible maybe even to harder to people who did know so just to say a few words on on each of these and my own connection also with Ramanujan the formulas the most famous I won't talk about it but it's called well actually there are two so there's a plural it's there it's not an identity but two identities and they were they're called the Robles Ramanujan all the brief year is from Ramanujan but they were discovered by Ramanujan and published by him when he came to England in those four years then soon after he was informed that they had been discovered independently and slightly early by Robbs with completely different proof and then as far as I remember the history they got together and wrote a joint paper with the third proof and many people call those two formulas the most beautiful formulas in mathematics it's a pair you have to see them together and three years ago two or three years ago I gave a whole course of lectures here at the ICTP called the Robles Ramanujan identities and the icosahedron showing that these things are connected with satonic solids with client steric icosahedron of course with multiple forms with Aperi's proof of the irrational the of say of two with conformal field theory and mirror symmetry and many other topics so this was really fun and these identities are a great love of mine I won't talk about them today as some topics I already said that too I particularly love this field in all of its incarnations and I'm giving a talk a course right now actually I spent half of yesterday when I wasn't preparing the course analyzing some numbers that that teacher had just given me from a physics problem using one of these asymptotic methods so let me give you one example could we have the next slide this is if it comes no yeah can I make that bigger or at least here it doesn't matter you don't have to be able to read it all you have to be able to see is it's got my name I'm the author and it's called on an approximate identity of Ramanujan Ramanujan and the surprising thing in the middle I'm supposed to have a laser pointer whether I know how this works we'll soon find out no I don't know how it works I do in the middle you see this complicated formula it doesn't matter you aren't supposed to be able to read it there's a formula which has a left hand side and a right hand side as most formulas doing mathematics connected by an equal sign that's kind of typical and both sides are functions of q and x but then the sentence ends these things are equal very nearly and nobody had any idea what that meant sort of they aren't equal this was in his so-called lost notebooks there's a whole story about the last notebooks and Bruce Bant and Archie Cohen asked me if well Bruce Bant asked Archie Cohen to make sense of it he discovered that us that numerically these things were close to extremely high precision but were a little bigger you could see the numbers at the bottom but indeed the left hand side is equal to the right hand side very very nearly but for certain values of the variables and in that paper I proved so this is many years ago that I already have that love then so that's about formulas about asymptotics so this approximate identity and finally for the modular forms well already said that Ramanujan himself didn't really understand deeply at least when he came to England what multiple forms were he learned from Hardy but it never became didn't really internalize that he's always worked with the formal aspects but in that letter that I mentioned the letter of january 1920 where he's over his depression but he's still depressed and very ill but he writes terribly happily that he's discovered this new class of functions that he called mock theta functions in the year 2007 I was invited to a conference in Paris within for me a resistible title it was called black holes I knew what those were from the from the newspaper black rings I'd never heard and modular forms I thought I have to go and there was a very very nice lecture from by Ashok Sen who's also a member of the scientific counsellor or at the ICTP and I told him that the mystery here presented this was in the quantum theory of black holes there was a puzzle that the experts had found and they didn't know what the what the mechanism could be and I said I think it's in the work my doctoral student Sanders Vegas would just finished a little before doctoral thesis in which he discovered the class of functions that explained these mock theta functions that hardly had discovered Ramanujan it discovered but had not defined he just described a few examples and didn't say in detail what their defining property was those Sander's makers had figured it out and so I told Sen should we work on this and he said no no but I have a colleague here in Paris that was Atish and soon later we were joined by third colleague Samir Murti I don't have to write Atish's name because you all have it and I just should say that we're all three connected with ICTP I've been here for 10 years Atish has now become the director but Samir Murti was here before any of us he was supposed to have for I think three years here and normally learned fluid Italian but he learned transgender dialect to some extent for a very unusual reason usually if people teach tango it's because they come from Argentina but Samir loves tango and he was a tango teacher not in the class he gave the class and of course the students weren't physicists they're people from the city and afterwards they'd go and have a drink and of course chat to each other and Trieste in a dialect he picked up quite a few like all Indians he knows many languages and so it's we pick up others a little faster okay so that's a little of the history his life and a little bit of my life and how it mixes but bear in mind these three things complicated formulas or beautiful forms or simple formulas asymptotic statements and modular forms because all will play a role so now I should keep an eye on the time because we're being very strict but I'm not known for being very strict with time now starts the lecture the mathematical part of the lecture which I'll try to keep low key namely partitions so first let me say I actually well I have a lot of boards I guess I don't have to keep a race let me remind you or tell you if you don't know what a partition is so imagine you have some objects so for instance you might have four objects oh they might be you know cups on the table or they might be anything and now you can imagine putting them in groups or maybe it's five objects five people and you put two of them at one table and one of the table by themselves and one at another table by themselves and now we ask in how many ways can you do that well if they're people at a party they have names and you know which people and then it's very easy to count but if you say I don't care which two people at this table I'm just you know the restaurant I'm told that there's a group of 11 people I have to make some tables that can see them I just have to know how to get numbers that add up to four so in this case we had four objects for four people and we've divided this up as a pair that will sit at one table would be in one group and two singletons which will be by themselves so that's one way I could do it but I don't identify which they are so if you think that in the case of four how could you do it well one really stupid way I mean perfectly good way of course is to put all four people at one table or all four objects in one group another way would be three plus one I'm not sure I have to draw the picture each time it's pretty clear but that's why I took a very small number four so three plus one if you have two groups the other way to have two groups would be two plus two you could have two groups of two pairs another way would be one pair and two singletons that's the one I had before and the last way would be one plus one plus one so I could put everybody at their own table or every object in the group all by itself but again I'm not labeling the group so it doesn't matter which one is which group and so if you count this up each of this is called the partition of four so two plus one plus one is a partition of four so p of four or p of n for any n is the number of partitions of four or of course of any other number n but in the case of four I can tell you the answer it's five because we just counted the one two three four five so once you believe that then of course you can make a little table I'll do it here so it's n if you're if you're taking notes I'm sure nobody is but theoretically if you are leave a little space so let's consider the numbers one two three four five six seven eight nine ten eleven twelve thirteen fourteen and so the number of partitions one you can only divide into one group two into two groups two pairs or one doubles and three there are three ways four we just saw is five here it's seven here it's 11 here it's 15 if you don't see the pattern don't be surprised there is no particular pattern except that the numbers grow okay so that's the first few values and now comes the place where if you're not a mathematician you'll see that mathematicians are just as crazy as you were always told they think very abstractly any mathematician if you ask how many participants said of zero elements are they will say well there's one if I have no objects there's only one way to dispose them if I'm bringing a party of zero people to the restaurant the restaurant has a unique thing to do they put out zero tables but they've got to do it I mean that's it so there's one not zero that would be important in a moment but of course only mathematicians would think of that asking how many ways can you divide up zero into non-mpsets and if there's one way namely no non-mpsets the mpset of non-mpsets okay so that's the beginning and now you could just say we continue but you see that to compute p of four it only took me a few seconds at the board and would have taken less if I weren't trying to get the board when you get to 200 it's already a little harder so dot dot dot and again dot dot dot but here the numbers three nine seven two nine nine nine oh two nine three eight eight so in one talk I gave once I said you know of course you could just hire a team of graduate students and let them count partitions for several billion years and maybe you would get this number but obviously that's not a good way so this number was computed and I have to look up the quote to get the exact name the indefatigable no the practiced and enthusiastic computer a computer in those days was a person not a machine and that was the famous major mcnaen by the way there's a movie that some of you may have seen about brahman brahman ujan's life very very nice movie by many people including my friend ken honor if you have a chance to see it you should and to major mcnaen plays a not terrible slapping role in that movie and he was this indefatigable you know experienced and to make computer and he computed this but even he didn't sit and count all the partitions there's a wonderful formula recursive formula that Euler had found and so you can compute this number from its predecessors in actually only about 15 steps but of course you have to have done it 200 times to have all the predecessors so you make a table and it can be done but it's important to be persevering and not make any mistakes this number is absolutely correct so we have this number p of n and if you love asymptotics as brahman ujan did then you will ask well how big is this we see it's very big but what is very big i mean it's certainly bigger than n squared you know n squared if n is 240 000 that would just be this much the other end it's way smaller than two to the end two to the 200 would have 130 digits or something like that and 200 factorial the number we've all heard of one times two times three to 200 would fill the whole board so this number grows not as fast as an exponential and so the the thing that i'll talk about the most is the is the growth of this that's the african topics but at the very end i want to come back to the other thing and i'll mention it now so you know what i'm talking about again when i get there all by that time i'll have removed this table we heard from a tish that brahman ujan one of his amazing facts was that every number as little would said was his personal friend he saw patterns which you can learn to do too and the trick is to look if you compute a table of numbers with a computer most people copy it from the mathematical into their tech file publish it and never actually read it but if you spend a week reading the numbers and thinking what they're telling you you will find messages of course brahman ujan had neither mathematical nothing for him nor tech and had to do things by himself so when he saw this he said oh look the number of parties of four is five that's visible by five now it's true that this also is by five but there's no pattern there but here i take the number of parties of nine it's 30 that's also visible by five when i get to 14 that's why i stopped there not with the board ran out it's also visible by five and you went up i think to 50 he didn't go as far as medj mcman had gone and they always were so what he discovered is the first brahman ujan congruence which is if the number you see that these numbers are spaced in a space of five each one is a multiple of five plus four and then he found that this is divisible by five and then he looked further and he found the same thing if you do multiples of seven then i should have another color i'm sure i do somewhere i can also just make it's too much trouble if you take this number you see parties of five is to this by seven parties of 12 also are and again you see the step size is seven and indeed you find that if you take seven k plus five then this is divisible always by seven and again he checked many examples and then if you go one further try the next prime not five or seven the next prime is 11 then you find that if you take p of 11 k plus six i think there's only one in my table but if you went on to 17 then this is always divisible by 11 but very remarkably there's not a similar thing for the next prime 13 or for any other prime these three are unique these became known as the famous raman raman ujan congruence is there many proofs actually he proved all three but his proofs were a little complicated get in fact more than one i think first but their proofs of this tend to be a little bit complicated this are quite complicated and these are very complicated and a few years ago i found a very simple proof of all three it's one line each and it's certainly the same line so it's a very uniform proof and at the very last at the end of the talk i'll just project that onto the screen so you see it but the intention is not that you read it or study it actually you cannot understand it even if you're a mathematician unless you know the theory of complex multiplication which is a somewhat specialized subject which is behind many of the formulas in raman ujan's first letter but i'll show it to you just visually so you can see that it's only a few lines for what used to be many pages for proof for this last one so i'll end with that well i could do it even now what's the difference but i'd rather bond with the asymptotes so the question is now the asymptotic this number is very big can we understand why is this number so this is you know mathematical notation it's 3.9 well 3.97 so 4.0 times 10 to the 12th so how can we know the size well you see roughly this is less than exponential because we were exponential we tend to the end which is 200 so you could say 12 is of the order of the square root of 200 and that's in fact correct so already raman ujan i think it meant even in this letter i didn't check it certainly he knew that there's an approximate formula which says that it's roughly e to the square root of n but actually the constant isn't exactly one it's exactly e to the power of pi times the square root of two-thirds n and that's still asymptotic equality means the ratio of b of n and the thing on the right should be one then is very large and it's slightly wrong it's off by constant four square to three and by factor of n so if you take let's call this you know rough rough approximation if you do that for 200 then you'll find that the rough approximation is roughly 4.1 times 10 to the 12th so it is telling you minutes within three or four percent it is telling you roughly the size of the number but it's far from giving you a close feeling for the number and so now comes the absolutely amazing story of the most famous collaboration that were several between hardy and raman ujan once again it's their paper there's more than one but their big paper on partitions and so what they did first there are several steps and as i've already hinted they used deeply the fact that it's a multiple form and in fact so deeply that somewhere hardy wrote in our member reading but i couldn't find the reference when i was repairing hardy wrote that it's clear to any mathematician that what we found could not have been found could not have been even guessed if you didn't know the theory of multiple forms and actually he was wrong because after this story i'll tell the generalization of this problem that they suggest in their paper in their paper they got only the analog this rough thing and because that thing has no multiple properties they basically implied that you can't do it but a few months go actually i went i started that in great detail and it is wonderful asymptotics and for those of you are in ictp or even virtually i'm giving as i mentioned of course the asymptotics i'll spend at least a week or two weeks on that problem so there is the generalization that works and where their amazing discoveries have analogs it doesn't work as well but it works pretty well and that's despite the lack of modularity but first let me go back and say what they actually found so this is the original let me call the first hardy ramanujan approximation and so it's a formula just like this it's slightly more complicated but it's actually not that much more complicated the first part is exactly the same except there's a remarkable fact which is obvious if you know about multiple forms and the dedicated eight functions you have to replace n by n minus one twenty fourth you take the same n we have before and similarly the denominator but here too you replace the n by n minus the twenty fourth but that doesn't quite do the trick for you you have to take this and then you multiply by a factor which is very near to one but it's off again by the square root so pretty by a few percent in the case of 200 and here you have exactly the same number again pi times the square root of two thirds and minus 24 now if you're not a professional mathematician but you have to realize about this formula it's not oh my god that's so complicated that's what these guys do this is an extremely simple formula because there are several little bits it's a bit long to write that's like you know very long word in English that doesn't make it more complicated if the question is what is the meaning and the meaning of this is very simple we're combining square root that you're learning elementary school maybe pi that you learn in high school exponential function you're learning in any good high school this is a completely elementary function it's not it's not higher this is not higher mathematics so there's a completely elementary expression sure it's complicated but it's only two terms it doesn't go on and on it's not an infinite series that's it and now let me give those some numerics I have them written down here except that I can't read my own printing in a second I'll show you a slide so you can see the numbers more neatly printed for the moment I'd rather not do that so let me write again our aim for 200 so p of 200 is the number I've already told you so now you know I know it by heart so it's okay and remember I told you if you took that rough approximate you get roughly 4.1 it's a 41 instead of 39.7 it wouldn't be very good but if you take this number t 100 then this one I didn't learn by heart well the beginning I very much know because rather remarkably it starts exactly the same way but the nine has become 89 actually 899 so here we're only up by three in this place 893 1 oh I ran out of digits 397 2998 993 993 185 but of course this number is no longer an integer so it goes on 89575 dot dot dot so it's whatever it is this is already really amazing because before we got 4.1 but suddenly this elementary formula just two terms is giving us the number of partitions all the way up here the error is about 36 000 but now that's where maybe ordinary mortals would have stopped and I think Hardy basically says he would have never gone further only Ramanjan saw that they had to go further and aim to actually get the number the full number and so you look numerical and remember they were doing all these calculates with no computers had calculations you look at the difference and here we're off by about 36 000 but here I made a little table it's in my notes so I can give you a little table here if I put n 196 198 199 200 201 so I'm just taking a couple of numbers on each side of our 200 and here I put p of n minus t of n I'll just give you very approximate numbers I've written down here it's 33 576 here it's minus 34 889 here it's 36 202 here it's minus 37 641 of course with some decimals and here 39 282 so if you look at this you see aha it's again a very regularly smoothly growing function 33 34 36 237.6 it's but with a sign minus one to the end so it alternates and so that tells you that there has to be a correction term and in fact there is a correction term and that correction term is minus one to the end because when n is even it's zero and then it's almost the same thing here I'll just put it very quickly you take the same square root we have before but you put a one half in front so that means that that thing of 100 decibels this is 50 decibels interrupted the square root and the denominator maybe I won't write the whole thing it's hardly any longer it's almost identical to this you've changed this number you know you just change the constants so it's an extremely similar expression except it alternates and so let's call this first one t t1 for the first term because now it's not the hardy-brained approximation it's just the beginning that's t1 here's t2 and so now if I take my printed notes these numbers I did not try to learn by heart uh I can't quite find them then if you do t2 so this was t1 as I told you we were already very close we're only off here by 36202.97848 and similarly there's a t3 and a t4 and a t5 and each of these is a similar something very regular so I'm sorry which is oscillatory but instead of period one n period two the next is period three it depends whether n is 3k 3k plus one the 3k plus two the next is period four but they're all completely explicit functions and so you can write them all now so already we went from you know three times ten to twelve down to 36 000 the next is even more startling we're down to only 87 58391 the next one we're down to just five the next we're down to 1.42421 I have to show you that I really do love numbers and when you add this all up not the p of 200 but when you add up the rest what do you get well the beginning of course we did expect 39721 against 999029387.861 so we're off by only 0.14 but we know that this is an integer so if you have it within an error of 0.14 you just round it off and you've got it if we get up the next slide now I can show you the same table but a little more carefully typed uh next one that was the last one I'm sorry you can show that that's the way to go back excuse me previous one that's the formless I didn't write that's the one I did write in the other side is very simple just so if you see I was applying they look almost the same except the minus one to the end but now the little table on the next slide can you make it bigger you can't it's not very important you can see some numbers and believe that the first few are exactly what they were on my table the first five but here I've gone up to 12 and the errors are even smaller like 0.004 and the final answer the number ends 388.005 five something so this was the absolutely amazing formula that you know completely dumbfounded well even their authors were amazed that this could work and the mathematician of Europe this is one of the most famous uh surprises in the history of mathematics that such a formula could be true that gave the number to that asthmatic position actually a small number of years later Rob Marker found a small variant of the method and then you got a new series his first term essentially the same but where the rest was actually convergent you actually got the number on the nose but that's in a sense a counter it's like a letdown because we have many exact forms of mathematics but this was an approximate identity so as I said what's behind this is the modularity and I don't want to say what modularity is especially is the time running out has it run out when am I supposed to stop five I think and we started late so a little after five then I'm still okay but I do have two more topics so let me let me just say briefly that I said that multiple forms are behind this let me just say in words what a multiple form is and why it's important when you go to high school you maybe in elementary school it's really good when you learn how to graph for instance the function x squared y equals x squared and you get a parabola and this parabola has a symmetry it's the same left and right so it's got a two fold symmetry then a bit later you learn about things like the sine curve and this also is the symmetry first of all you can also reflect it in one point it's even or odd but on top of it you can shift it by 360 degrees or 2 pi and it goes into itself that's an infinite symmetry group slightly non-commutative but very slightly a multireform is a function an explicit function of the same general type so it's given by a formula some kind of a formula usually but it has an infinite symmetry group but one which is highly non-abeled so the different symmetries do not commute with each other it's a much more subtle structure and therefore it's much more productive and so the reason that all of this worked in the first place was a form of Euler which is Euler had the wonderful idea if you have a small table of numbers like one one two three then instead of giving this a table zero one two three gives one one two three you could give it as a polynomial and then you say I can read off the numbers of the table from the coefficient this polynomial and then Euler invented basically power series except they were also invented in Japan at about the same time he said why don't we continue and even if we have an infinite table of numbers we can make a whole power series an infinite degree polynomial and this function if it is nice properties will tell you something about the numbers and we call this a generating function it generates and it gives birth to them and actually the Japanese translation of generating function is the mother function the numbers the coefficients are its children so in this case Euler had proved a formula but that's why he invented generating functions for this very example it's very easy to prove this form of Euler that this infinite sum so what are these numbers this is the sum of the pn times x to the n you see why it's very important to me actually I should have called it q but I called it x you see why it's important to have the one because without that this formula wouldn't work so this was Euler but then it was found based by Riemann but he died before he could publish it and then in his the papers he left behind were analyzed by Weber and Dedekind and Dedekind they both wrote about analysis and Dedekind explained this thing and it's now called the Dedekind data function this thing is whatever it is a modular form so in this case the function that underlines these underlies these numbers the mother function the functions coefficients are these partitions is itself a member of this class of functions multiple forms with their infinite group of symmetry and as I said hardly was convinced that the method would have never worked without that this is the thing that we now know that in Ramanujan's last letter about the mock theta functions those mock theta functions are not ordinary theta functions which are also multiple forms but they have a nearly multiple property that was identified by my student's fakers my mention that's what led to this wonderful collaboration with Atitians and Samir that I already talked about so just to tie things together but in their paper of I think it's 1917 this big paper Hardin Ramanujan actually mentioned a more a different or a more general problem instead of saying for instance p of n p of four for instance five remember because we had the five partitions that I had listed before for three plus one etc there were five of them you could also say how about p2 I write a number not as a sum of positive numbers but as a sum of squares and then there are many fewer for instance if I take 11 well the biggest square in the left is nine what's left is two if you want that as a sum of squares four it's too big you have to take one plus one if it's not that you could take a four and another four but now the remainder is three you can't use four anymore it's that or you could have just one four and seven ones well I can't count the one two three four five six seven or you could of course have 11 ones and this is all there is so we can ask about the number of ways of writing the numbers the sum of squares or if that matter cubes are higher powers and they asked this problem and in their paper in the introduction they gave a very crude formula even cruder than this this formula if you know even simple that is you'll see that this formula can make even weaker by just take the log then the log is roughly pi times the square root of two thirds n that's much weaker than this that in this case would give you not an error of five percent but an error per factor of about a million so it would be a rather poor approximation so oh sorry in the case I'm going to do but in fact I overlooked in their paper that in the introduction they mentioned this but at the very end of the paper they come back to this but I'd overlooked it it's not even a displayed equation and they actually do know the analog of this formula for p2 or also p3 for partitioned into squares so they know this crude approximation something's esoteric but they're they it's implicit what they say you can't expect to go to do better simply because there's no modularity and so I came back to this a year ago for various reasons and studied the both theoretical numerically for actually many weeks or a couple of months very intensively and it turned out that so first of all the numbers grow much more slowly but still they're big next slide piece I no longer have to ask Mr. McMahon or Major McMahon we've got Paris which was already mentioned Paris did this calculation not just this is not p2 of 200 before McMahon did the argument 200 I'm doing p2 for the argument 100 000 and Paris needed two seconds to compute all of p2 a table all the way up to 100 000 so it's a lot faster than Major McMahon was and the final number is the one you see here you don't have to read it but you see it's very long I think 60 did or 59 or 60 digits and so again the question is what is the asymptotics and first you do some experimentation and you find an approximate formula this one which as I said they had actually already found I didn't realize but then I went into more and more detail and something very amusing happened I'll just say it in words there's again the circle method which is what this method is called Fardin Ramanujan could be applied but it's way way trickier and so there's again the term t1 which is a smoothly growing function of F just an elementary function there's a term t2 which is a combination of smoothly growing in period two t3 is period three t4 is period four but it's completely different from here here the big contribution this is huge this is much smaller this is much smaller a few terms do it but when you do this for that huge number that I just showed you p2 of 100 000 then the first term I'll just give a couple of digits I've been written down here the first term I wonder if it's 3.90 I'll just put the first digit it's about four times 10 to the 59 well if you see the number there it starts 3.88 I'm just rounding it off t2 as you'd expect is you know quite a bit smaller it's 10 to the 26 so that's about what you'd expect t3 the third term is 8 times 10 to the 16 so it's still getting smaller no surprise yet the t4 is as it happens first of all minus 8 but now it's 10 to 32 it's way way bigger even than t2 it's the next actually if you keep going it's the next biggest contribution and then I'll do just two more but I ran out of space here I'm too short I should have started higher the t5 is approximately 3 times 10 to the 15 so it's about the same t6 is minus 1 times 10 to the 5th so they have wildly varying sizes there's no pattern it's 60 digits 26 16 32 15 5 this and if you continue it's it's completely erratic and if you continue this analysis of both the American also theoretically you find that the contributions the ordering of the k if I look at you know this tk so if I look at tk then one is the most important as you see here the second most important is four as you can see here 32 the third in this table is two but that's actually true if you go further but it's a completely weird ordering and this was a complete surprise and there are several other surprise which those of you who will come to the course will be able to see so you get this rather bizarre behavior asymptotic behavior but the basic principle remains big but true has been hard to pronounce and that there's one main term which gives almost everything it's after all already giving you 26 an error of only 10 to the 26 which means that 33 digits are already correct more than half and you get more and more precision but it does not give the exact integer as far as I know so now my time is surely up can ask for the very last slide and this will just be visual fun so I mentioned the um remanution congruences that there are three of them five seven and 11 remanution proved all three with difficult complicated proofs this is many proofs some fairly easy this is several some not so difficult and 11 I think the previous proofs roll very long and here's this little it's half a page I wrote about seven years ago I have published it can you enlarge that too maybe Marko if it's possible just so you can see it I don't expect you to read it even if you're a mathematician but you can see there are four displayed lines and the first three are the first line of the two line proof in each case the first line there's one first line for the prime five one first line for the prime seven one for the prime 11 and each two line proof the second line is always the same it's the fourth line on the page it's the same for all of them so it's incredibly short each proof is two lines and it's exactly the same proof each time so very uniform and I thought that was a nice place to end because I think this is a proof and the point of view that remanution would have also enjoyed so thank you very much but have there been questions by the way from the audience Claudia will know I wouldn't know or not yes I was I was going to ask if there were any questions from the audience here yes in that case well you can never microphone or I can repeat the question but the microphone is much better can there exist any other primes apart from five seven and 11 for which such properties would hold true yes that's known it's not my work there are many deep results behind this some of them are due to ser others to other people but it is known such comments of this sort exist exactly for the three primes five seven eleven the reason it's quite sophisticated but it's known that's a fact not an opinion but certainly you would find from the tables there there aren't any other exams for all the primes up to 500 but in fact it's it's a theorem very surprising yeah I wouldn't know how to prove it don't ask me but I know roughly it has to do with pediatric analysis of multiple forms for these primes yes these oh would you need the microphone hi so I have two questions first small technicality but I see the error there the difference between PNN and TNN it is growing with N so is this some kind of a series that you know that you mean that no no you're at the table to the right one to the right yeah I see the the error growing with N so is this series an asymptotic series which oh no that's absolutely true the board the formula is on the board remember this was the form for the leading term and this is obviously increasing and the second term I didn't write it out completely but it was displayed as the sum so many slides it was a very similar formula with minus one to the end so indeed there's a sign minus one to the end and what's left is completely it's a regular function and it has nothing to do with there with N being an integer here I could make a table I'm in a graph 198 199 and this function is a smooth it's an analytic function it's monotone and a very simple function so is that function but then we specialize to an integer and then we take it with a plus or minus sign but what I mean is so that the p of n is exact and t of n is our approximation well sorry this is I said when I wrote it that's right I didn't have time to write an infinite number of digits and I was even two days but but as magnitude it's growing as a magnitude that difference is growing these numbers without their signs are uniformly that's what I just drew it's a monotone the increasing function of n specialized integers is that what you're asking so these numbers the differences are growing they go to infinity and they grow exactly well I even wrote that the difference t two of n I mean if you still are connected we could go back two slides I think and actually see yeah there it is it's minus one to the n times practically the same expression as t one with playing with the exponents a little and that function is so smoothly a very simple smoothly growing function so there's no mystery in these error terms each term individually is very simple but the amazing thing is that when you add up and if you keep adding them up if you extend this table I went to five then next slide the previous I showed you up to 12 and the error was better but if I went up to a thousand the errors would become huge it's not a convergence series you have to yeah that that was my question so the about the convergence of no but that I said very briefly but I ran over and I said a few years later they showed that they're formed if you stopped at the right place you have to take approximately the fourth root of n terms and if you stop there of course there's some finite difference but they show that difference is actually going to zero like one over the fourth root of n so in particular since it's an integer you're looking for if the error is less than one you're finished but then a few years later I think less than 10 years later I don't remember 10 years later rather much approved the famous formula still used in the multiple properties he made some very very brilliant change in their method and then he had the new series with the t1 star and a t2 star and a t3 star and t1 star to this accuracy would be completely indistinguishable they would differ in the 30th or no the 12th place to the left they differ by an exponentially small amount so the first few terms are absolutely identical but if you keep going then these terms of part of Ramananj would eventually start blowing up but rather much with terms that start out exactly the same they tend to zero fast enough that the series converges and what's more they converge to the exact integer so though each so if that's what you're asking very good question indeed that is what happened that was Radomacher's famous formula so now some books talk right away about the Hardy Brother Ramanuj and Radomacher formula which is true enough but the real breakthrough was seeing that well intending this method and and seeing that something like that is true so you said you're yes I know then I had a very benignly provocative question but this our approximation which is sort of the sum of t1 to you know tm whichever whichever it is so it's it's very beautiful because it's in terms of you know pi and and and rationals and so on but I guess it does get more and more complicated why is that more intellectually satisfying than say guessing the form of the solution and fitting it for you know for some data and you get some you know you get some best fit parameters for the it is just you say slightly provocative if I can rephrase the question even more productive productively it's why are you a mathematician doll why you waste your time just take a computer and compute the first 10 000 values will never need more than that anyway that's of course true enough if you if you take the bottom line approach you're paying your taxes you just want to know how big the number is you don't need any of this you can just compute it major McMahon did this in a few days but Paris can compute this number in a thousandth of a second you can get a huge table that's not the problem but it's like the famous answer about Everest why do we want to know because it's there there is such a formula and this form is utterly amazing that this very simple even though it's a little ugly but still very elementary expression already gives you know all but the last five digits of a 13 digit number if it were a bigger n it would give you know a huge chunk of it and then an equally simple correction gives most of the next and another and this is just a beautiful phenomenon plus the actual proof which goes into the modularity you see a modular function is a function in one version of the upper half plate in the unit disk in the so-called circle method and there's a big contribution from the point here at zero from the origin but then the smaller compute this gives t1 there's a smaller one at minus one then there are two more at t3 prime and t3 double prime and then two more at the points of order four and so on and so in the circle methods when you actually start it it's absolutely beautiful the modularity of the function this infinite symmetry tells you that because you know how the series looks at the beginning the first few terms you also know everything about how it looks near one then near minus one then near the cube roots can be then near i and minus i and it's this information that you put together and so the surprise to me and put hard in which i'm actually didn't expect so we knew this for p2 there is no modularity but nevertheless there's a contribution from q equals one q equals minus one except that now these terms don't go down smooth in size it's much more subtle and so the reason we're doing pure mathematics is to understand the structures behind things we don't actually care about the answer nobody could care less the p of 200 is that number i guess but so there could be for example a different type of approximation where maybe it's impossible because if you had another one that's even close to that good let's say you thought of another one let's call it you for you you you found it so you have 200 but it better be at least as good as this or else you're kind of wasting our time but that means that these two only different the sixth place so there aren't any two elementary functions that are that close to glue to each other i mean the the numerical values of this function pin it down much more than uniquely that there is one is the miracle but there can't possibly be two if instead of pi you use some other you know irrational no it's not i didn't choose pi that's a fact of life that's like saying if instead of rome you decide that the capital of italy is deli but it isn't i mean i didn't i didn't write this down at random this is the result of their computation this is a fact the fact happens to involve a number we know it's called pi but it wasn't the choice that's not how mathematics where it would be nice if we could choose the numbers but but you can't this formula gives this and no other elementary formula will give opposed eight digits of this 13-digit number correctly because if it did then two elementary functions would be the same within an incredibly small error and you can show very easily that that doesn't happen so we do know actually that this if it's an approximation of this type now you might have something completely different you know that but it would basically have to depend on telepathy if it's given by explicit formulas it's got to be this one the formula is given the mathematics itself forces it has solved in mathematics you don't know what you're going to find but once you found it it was there you didn't invent it you just found it and it's always like that thanks for both questions by the way are there any others are you already had a question maybe somebody should if if anyone has one otherwise you're welcome to too I'll send you a bell afterwards anyway so yeah uh given that successive terms in this asymptotic series they're rather erratic as you pointed out how does the method of optimal truncation work in this case oh well since I never mentioned optimal truncation that's not a legitimate question ask it again in my course when I get to optimal truncation and I'll try to explain anyway I don't do optimal truncation there is actually nothing to do with this but what I did say about generating functions is this there was Oilers maybe his greatest single idea because they're universal mathematics if you have a bunch of mathematics then you always make a generating function it doesn't always help almost always helps but you think of a series of numbers as the same as a power series you think the power series is the same as a series of numbers and you use that interplay to go between the properties of the function maybe differential equation modularity and the properties of the numbers but there's no universal rule you you apply this otherwise we wouldn't need mathematicians it's an art each time you have a new function you have to go deeply into its properties find out does it have a differential equation does it have some other property is it multi-nourished of some other symmetries and those will guide you to find out whatever you can about the coefficients so I don't know quite if I'm answering your question but it's not about truncation that's not what you do here at all but you are picking out coefficients yes I don't know how our sketch is maybe you want us to stop at some point yes it's my time to be provocative but it's getting late in Calcutta and New Daily so I think we should at least have our so no I well first of all let's thank Don properly again he's a beautiful lecture as usual he transmits us a lot of energy for doing something good ourselves and so find just few final remarks first of all final congratulations to professor Gupta still with us I saw her a few seconds ago and and of course congratulations to Don Zaghiere but for staying for the new appointment as Ramanujan International Chair so I guess we can just just a couple of pre-announcements so the new the nomination for the new Ramanujan prize for the 2022 Ramanujan prize will be open soon it's not open yet but should be a question of few weeks so please stay tuned for being able to nominate new excellent candidates for this year's prize and well okay I will allow me to add the personal thanks to all the people who have made this possible unfortunately organizing the zoom event like this requires a lot of work and a lot of stress without even the pleasure of going together for dinner with all our people our friends and our partners but I really want to thank our IT team Marco the director's office and especially Shira and Joanna and Kutu and Fabrizio all this all the people in this room know I mean they are institutions of course of this of the ICTP and the PIO of course for the great work that it was required to to make this happen hopefully as I said at least next year we will be able to thank properly sorry I don't see the script is there something going on sorry I don't hear yes no no no and then of course I mean the final the closing remark was actually to thank once again but I will never be tired to thank once again the international mathematical union professor Holden is still with us and the department of science and technology of the government of India for this great partnership that we have together to keep on moving this this idea ahead so thank you Helge and as I said waiting for you here next year okay so final thing for the students actually professor Gupta will be for all the students of ICTP at least for the past six or seven years in mathematics professor Gupta has kindly accepted to meet you only you it's not open to the public to and also you will meet also another eminent mathematician professor Mustafa Fahl that I will thank tomorrow he is not among the hundred people still with us to to have just a free conversation with you about what it means to be a mathematician's experience future worries everything it's just a private session between you and them and professor Gupta and Fahl so I strongly invite you to be part of this of this meeting and for this thanks again to professor Gupta and finally at least for those of us who are here we can actually move to the terrace and have a little refreshment and coffee all together to celebrate this day and thank you it was really a beautiful day for mathematics for Trieste thank you very much thank you goodbye Helge goodbye professor Gupta goodbye