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Lecture 12: Smooth Surfaces I (Discrete Differential Geometry)

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  • čas přidán 14. 08. 2024
  • Full playlist: • Discrete Differential ...
    For more information see geometry.cs.cmu...

Komentáře • 28

  • @namjookim998
    @namjookim998 Před rokem +5

    I can FEEL you are sincerely interested in students to LEARN and not other things.
    So that I can invest all myself to contemplate EVERY sentences you speak.
    As more I contemplate it, more I realize every sentence is so accurately and carefully spoken to make us actually LEARN.
    Thank you so much.

  • @DiegoAndrade
    @DiegoAndrade Před 2 lety

    The best explanation of a Riemannian Metric ... thanks!

  • @themathguy3149
    @themathguy3149 Před 2 lety +5

    Do you do the animations yourself? Are they from some textbook, i would like very much to know which framework you use to do the diagrams/ilustrations. You're such an exemplary explainer, congratulations!

    • @keenancrane
      @keenancrane  Před 2 lety +2

      Thank you for the kind words.
      There are course notes here: www.cs.cmu.edu/~kmcrane/Projects/DDG/paper.pdf
      I do almost all of the animations and illustrations myself-see keenan.is/questionable#figures for some more info.

  • @amit2.o761
    @amit2.o761 Před 2 lety +2

    i watch your lecture and they are great even though I am only in 10th i can understand all of it even if I'm in class and have no background in maths but my favourite subject is maths and wish to become an mathmatiusn but I'm from India, in india we don't even heard about calculus until 11th i don't know about your system but from 1 to 10 class the things we know about maths is algebra and geometry . but in this lecture i have couple of small questions like
    is g applicable for n dimension and is I always 2x2 or its just in 3d and if j is not generalized for any dimension ?, that what is analog for j in nd ?, and isn't tensor calc applicable for nd and still have g, I and j ?
    sorry for little grammatical error

  • @TiredGradStudent
    @TiredGradStudent Před 2 lety

    For clarity, 17:00, df is a vector-valued 1-form. Is du and dv essentially just renamed dx and dy such that we know we are explicitly talking about velocity fields with normalized basis covectors pointing along the x-axis and y-axis? I noticed that you specify partial/partial x and partial/partial y as being unit vectors in the u and v directions, respectively.
    Also thank you so much for making this playlist public! I'm going through all these lectures on my own time to build upon my limited understanding of how meshes are built and how math is performed on them in order to work on my project for my dissertation. Your lectures have been a tremendous help!

  • @CarlosValero
    @CarlosValero Před rokem

    Amazing explanations!

  • @haejinsong1835
    @haejinsong1835 Před 2 lety

    Thank you so much for sharing these lectures. They've been invaluable in understanding the big picture of Differential Geometry: clear and intuitive explanations! I have a quick question on the notations for the parametrization functions. I believe the phi_i's @1:11:23 are the same objects as the "f"'s on all other slides. Is there a reason for using a different notation on that slide? Riemannian metric reminds me of kernel functions in ML (which often uses phi notation), and I also wonder if there is any meaningful/useful connection between the two.

  • @hanxw7034
    @hanxw7034 Před rokem

    It is beautiful!!

  • @lordnaive
    @lordnaive Před 3 lety +1

    This is the latest 2021 video on this course right?

  • @SVVV97
    @SVVV97 Před 2 lety

    I'm a bit unsure if the cube-folding example for a (non-)embedding actually works:
    If the domain is open then we can easily undo the folding because the "glued edges" don't actually meet - and if it's closed the map won't even be injective so the continuity aspect isn't needed. Or am I missing something here?

    • @porglezomp7235
      @porglezomp7235 Před 2 lety +3

      He mentions that the edges are open and closed in such a way that an open edge always meets a closed edge when you fold it (doesn't explicitly show on the diagram, but you can make up such an assignment yourself pretty easily). In this case where a closed edge meets an open edge, that's where you run into the trouble because you end up changing the neighborhoods.

    • @SVVV97
      @SVVV97 Před 2 lety +1

      @@porglezomp7235 oh okay - I must've missed that / been distracted. Thanks! :D

  • @columbus8myhw
    @columbus8myhw Před 3 lety +2

    Is your definition of regular homotopy really enough? Don't you need the differential of h_t to be continuous in t as well?
    Consider this scenario in the 1D case. Imagine a figure-eight curve whose image is two geometric circles that are tangent to each other: one of them with radius t, the other with radius 1−t, and the curve is arclength-parametrized. This is an immersion for all t, but if we call this a regular homotopy we violate Whitney-Graustein.

    • @columbus8myhw
      @columbus8myhw Před 3 lety +1

      Also you can find the Nelson Max film here: czcams.com/video/WSk8XkPALus/video.html

    • @columbus8myhw
      @columbus8myhw Před 3 lety +1

      Also also a neat way to visualize Riemannian metrics is through Tissot ellipses en.wikipedia.org/wiki/Tissot%27s_indicatrix

    • @keenancrane
      @keenancrane  Před 3 lety

      A regular homotopy is just a continuous path in the space of surface immersions. The time derivative does not need to be nonzero. For instance, it's fine for motion to stop and then start again; that's still a regular homotopy.

    • @columbus8myhw
      @columbus8myhw Před 3 lety

      @@keenancrane No, I mean that the space derivative should be time continuous

    • @Toctave
      @Toctave Před 3 lety +1

      @@columbus8myhw Have you found out more information about your question ? The usual definition of regular homotopy I found in other places online is the one given here (a homotopy through immersions) , but I don't see why contracting a circle like you suggest would violate this definition.
      Someone on MathExchange said that this construction is not a regular homotopy because you have points with tangents in all directions contracting into a point with a single tangent. That makes intuitive sense to me, but as you said, it does not seem to violate the definition.

  • @diramtabaa
    @diramtabaa Před 3 lety

    First comment on the last video...