Something from Nothing

Поділитися
Вставка
  • Опубліковано 11 вер 2024

КОМЕНТАРІ • 16

  • @math-life-balance
    @math-life-balance  Рік тому +25

    Isn't it terrific!
    Please share the video if you also enjoyed it :)

  • @altan3850
    @altan3850 Рік тому +7

    This particular piece of visual masterpiece is the closest to being the ultimate mixture of mathematics and art that I've ever seen in my life. The storytelling, the images, the examples, the fact that starting from a fun paradox for primary school 0 = 1 we end up discussing cutting edge research just make watching the video a touching and cosy, yet engaging and satisfying experience ❤❤ Huge congrats to Jeremiah Heller, Ivo and of course Elka! The suspense has more than paid off❤

  • @DMVMedien
    @DMVMedien Рік тому +1

    Great idea! And so nicely drawn und presented! Thank you! 👌

  • @PeakMathLandscape
    @PeakMathLandscape Рік тому

    Incredible!

  • @effy1219
    @effy1219 Рік тому

    please ignore comments trying to correct your content(why dont they create their video!), we want you to do this for fun! devoting time and putting effort into making video is already not easy, we dont want you to spending more efforts or create any mental stress, you dont have to be perfect and we still love your content!!!

  • @RohitSingh-nm9wd
    @RohitSingh-nm9wd Рік тому

    Just amazing.

  • @user-rk2ny4pw1l
    @user-rk2ny4pw1l Рік тому +2

    Shouldn't R+M be isomorphic to M (rather than R as in the video) for a flasque ring?

    • @jeremiahheller1908
      @jeremiahheller1908 Рік тому

      Yes, thank you!! R+M is isomorphic to M. The example of interest here has M=R which is the source of the typo.

  • @ligmamale4389
    @ligmamale4389 Рік тому

    This is great

  • @JakubWaniek
    @JakubWaniek Рік тому

    Around 4:20, are you assuming that your flasque ring R is commutative? Unless P is a right module and M is a left module, I think we need R commutative for the tensor products to make sense. If I'm correct, then we have only shown that a commutative flasque ring must have vanishing K-theory. This is enough to set up the contradiction in argument that follows, but I thought it was worth pointing out anyway

    • @jeremiahheller1908
      @jeremiahheller1908 Рік тому

      We only gave an informal description. In the formal definition M is a bimodule which is finitely generated projective as a right module and the isomorphism between M and R+M is a bimodule isomorphism.

  • @matti1610
    @matti1610 Рік тому

    Wow nice!

  • @TranquilSeaOfMath
    @TranquilSeaOfMath Рік тому +1

    I like this approach. It is something nice visually and the lecture is like a fairy tale. 🐉🧮