Manjul Bhargava: What is the Birch-Swinnerton-Dyer Conjecture, and what is known about it?
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- čas přidán 31. 05. 2024
- Abstract:
The Birch and Swinnerton-Dyer Conjecture has become one of the central problems of number theory and represents an important next frontier. The purpose of this lecture is to explain the problem in elementary terms, and to describe the implications of Andrew Wiles' groundbreaking work to the problem. It will also summarize what is known to date towards the conjecture - including several recent advances - and, finally, what remains to be done!
Professor Manjul Bhargava is a Canadian-American mathematician. He is the R. Brandon Fradd Professor of Mathematics at Princeton University, the Stieltjes Professor of Number Theory at Leiden University, and also holds Adjunct Professorships at the Tata Institute of Fundamental Research, the Indian Institute of Technology Bombay, and the University of Hyderabad. He is known primarily for his contributions to number theory.
Bhargava was awarded the Fields Medal in 2014. According to the International Mathematical Union citation, he was awarded the prize "for developing powerful new methods in the geometry of numbers, which he applied to count rings of small rank and to bound the average rank of elliptic curves".
This lecture was held at The University of Oslo, May 25, 2016 and was part of the Abel Prize Lectures in connection with the Abel Prize Week celebrations.
Program for the Abel Lecture 2016
1. "Fermat's Last Theorem: abelian and non-abelian approaches" by Abel Laureate Sir Andrew Wiles, University of Oxford
2. "Andrew Wiles' marvelous proof" by professor Henri Darmon, McGill University
3. "What is the Birch-Swinnerton-Dyer Conjecture, and what is known about it?" by professor Manjul Bhargava, Princeton University
4. "From Fermat's Last Theorem to Homer's Last Theorem" - a popular lecture by Simon Singh, author of Fermat's Last Theorem among other achievements. This lecture will never be published because the presentation contained material protected by intellectual property. - Věda a technologie
Prizes for Academic work are a bit of a mystery to students who have their skills in agriculture or physical activity, so the idea of an approach to Mathematical Conjecture from a Condensation Chemistry POV is more effective than turning the pages of recorded history.
Very good lecture! I never really understood the BSD conjecture before, but now I feel like I have a pretty good handle on what it says. Hope I can prove it someday, or at least some key results leading up to its proof.
there are some really weird videos in your favs
@@mashmax98 💀 dude is obsessed with feet
Clear, interesting lecture that is pleasure to hear.
Wonderful lecture !
Another stupid question: Why didn't you start with a more general 2-variable cubic (which includes y^3) ? Excellent talk !
12:05 that solution there is actually a special case of a more general theorem. Use the fact that:
(m² - n²)² + (2mn)² = (m² + n²)² for all m, n. Now divide both sides by (m² + n²)².
The case in the video is m=s and n=1.
s=m/n in your equal m and n are integers, in this it is rational. So effectively they both cover all solutions.
Sorry for the very dumb question, but why is Eq. (1) in his video y = s(x + 1) ? I thought the slope is y = mx + b....so how did we determine that the slope (s) is equal to the y-intercept (b)? I feel like I missed something very easy but I can't find where I went wrong. Is it because the unit circle has a y-intercept at {0,1}? Why then cant we use y = s(x - 1), since {0,-1} is also a y-intercept?
@@kidzbop38isstraightfire92 No bad questions. You are right that the equation of a line is given by y = mx+b, and slope defined as rise over run. Here the run from the point (-1,0) to the y-axis is 1 and the rise is how much you went up after one step.. which is the slope. The y intercept IS the slope here. So we can rewrite the equation as y=sx+s and factor out the m to get y=s(x+1)
To your point about the negative values, they still work here as plugging in negative slopes will make this work as well.
@@hvok99 yep of course, that makes sense now. I knew I was missing something simple. Thanks for explaining it well bro!
Gracias por compartir
Nice ❤️
अद्भुत अति अद्भुत ।डा राहुल ।
His silence speaks more than his words. How is he at computers ?
Excuse me, in the description in the title of third lecture (this lecture) is "Birch--Swinnerton-Dyer". I believe that is a type and it should be "Birch-Swinnerton-Dyer", with one "-" removed.
Sir Peter Swinnerton-Dyer is English baronet.
@@Darrida Thank you for pointing it to me.
10:24 sorry for the stupid question, but why is the slope equation y = s(x + 1) ? I thought y = mx + b....how did we determine that the slope (s) is the y-intercept (b)?
This a specific linear equation for a line with slope s passing through the point (-1,0). We have s= (y - y0)/ (x - x0), so simply plug in P's coordinates to get y = s(x + 1)
Well, in Europe and America usually only the y-intercept form is only taught, which is y=mx+c, where c is the y-intercept.
there's actually another form you can derive using the x-intercept instead, which is y=m(x-d) where d is the x-intercept.
the guy said m=s and d=-1 hence, y=s(x+1)
Starting from (-1,0) the slope is just rise/run. So when you go an unit length in x (from -1 to 0) you go up (or down) a length equal to the slope. So your coordinates after a unit step in the x direction will be (0,s) giving you the y intercept as s.
Thanks everyone for the help! After reading your replies, I realize how elementary this was 🤦♂️
THE RIGHT STATEMENT IS BEABLE PRIZE
No idea what I’m doing here, don’t even know my fractions
La démonstration de π(1-ap/P)? Soit la serie Fp(k)=1/2^k+1/3^k + 1/5^k''''''''' vers l infini la série des premiers quelques soit k de N π(1-1/p^k)=1-Fp(k)+la somme de 1 vers l infini de 1/p^n×fp(P+n) et la somme de 1/p^n×fp(P+n)=1/2fp(k)^2 - 1/2 Fp(2k) d ou π(1-1/p^k)=1-fp(k) +1/2fp(k)^2 -1/2fp(2k) quelques que soit k dans N il y a cette égalité ' on peut vérifier π(1-1/p^k)=e^lnπ(1-1/p^k)=e^-fp(k)-@ le symbole @ le Epsilons donc π(1-1/p^k)=e^-fp(k)-@=1-fp(k)+1/2fp(k)^2 -1/2fp(2k) quelques soit k dans N pour le produit π(1-ap/p)=1/Le(1)=e^lnπ(1-ap/p)=e^-Lp(1)-@ d ou Lp(1)= lnLE(1)/e^@ d ou π(1-ap/p)=1/2(Lp(1)-1)^2 +1/2(1-Lp(2))=t(ln(Le(1))^2 l égalité π(1-1/p^k) =1-fp(k)+1/2fp(k)^2+1/2fp(2k)=e^-fp(k)-@ est valable quelque soit k appartient à N
i am not kidding. this problem is going to be solved in upcoming 2 years.project is on progress. believe me
guess not
I think the mic used in the seminar is too close to the mouth, lots of lip smacking and breathing distracts from the talking. Maybe run an EQ over the audio and filter the high frequencies
Not a helpful comment. Do you really think the people who organized this lecture are reading these CZcams comments?
I love lipsmacking
I really admire you, but I couldn't finish the video because every time you stop talking your lips made annoying sound. Please have a bottled water next time you present something.