Differential Forms | The exterior derivative.

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  • เผยแพร่เมื่อ 7 พ.ย. 2024

ความคิดเห็น • 43

  • @willgriffin9793
    @willgriffin9793 3 ปีที่แล้ว +5

    Really great video. I watched this after reading chapter 2.4 of Guillemin & Haine, feeling a little shaky on my exterior algebra, and this really blew it all open for me.

  • @rafaelgomezgonzalez400
    @rafaelgomezgonzalez400 4 ปีที่แล้ว +12

    I love this topic, thanks a lot for the effort and clarity in your explanations

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

      th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐

  • @WillHobkirk
    @WillHobkirk 4 หลายเดือนก่อน

    This series is awesome and exactly what I needed. I'm watching the whole thing back to back.

  • @yulflip
    @yulflip 3 ปีที่แล้ว +7

    Typo: Should we not see omega: (T_p R^n)^m -> R on the first board?

  • @dmitryempollon4384
    @dmitryempollon4384 4 ปีที่แล้ว +12

    Dear Michael, thank you for your truly great work (I am a math educator as well)! What's behind the "limited access" videos in the differential forms playlist? Are they placeholders for the future videos?

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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  • @scottstirling804
    @scottstirling804 4 ปีที่แล้ว +7

    So this may be a pedantic thing but if you instead wrote the two form, that you calculated the exterior derivative of at the end, as w=Fdy^dz+Gdz^dx+Hdx^dy when you calculate the exterior derivative you get dw=( F_x+G_y+H_z)dx^dy^dz and you can more easily see that the exterior derivative on two forms is not so secretly the divergence of a vector field. I haven't checked if a similar issue appears when you calculate the exterior derivative of the 1-form. Someone feel free to correct me if I am wrong.

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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    • @Zxv975
      @Zxv975 3 ปีที่แล้ว +1

      This is a good observation. That errant minus sign bothered me a tad, so it's good that it's so easy to explain where it comes from and how to fix it.
      The exterior derivative of the 1 form does have the same issue. The curl of a vector would have the second term (the dx^dz term with dy omitted) as the negative of what's shown here.

    • @shuewingtam6210
      @shuewingtam6210 2 ปีที่แล้ว +1

      The minus sign of G term in three form is due to the fact that he used dx^dz instead of dz^dx in the two form expression.

  • @golden_smaug
    @golden_smaug หลายเดือนก่อน

    I love you, Michael Penn

  • @goodplacetostop2973
    @goodplacetostop2973 4 ปีที่แล้ว +20

    16:24

    • @leonardosaads
      @leonardosaads 4 ปีที่แล้ว +4

      Dude, did you create the channel to signal the moments he says that? lol kkkkkk

    • @stenarsk6877
      @stenarsk6877 4 ปีที่แล้ว +1

      @@leonardosaads lmao

  • @Matias-sc2or
    @Matias-sc2or 3 ปีที่แล้ว +2

    This proposition can be proved using mathematical induction, applying Leibniz rule (hope it is mentioned in one of previous videos. Assume this proposition is true for all forms with order less than or equal to m. Let A be a m+1 form. Split it into a wedge product of two forms A = A1^A2 and calculate dd(A) = dd(A1^A2), using Leibniz rule, and the result is 0.

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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  • @byronwatkins2565
    @byronwatkins2565 3 ปีที่แล้ว +2

    ??? Now I know that a 0-form is a function (possibly of several parameters). "On R^n" leads me to suspect that f() is a vector with n components. I still have no idea what a 1-form is or why it is different from a 0-form, so you might imagine how much I got out of the rest. The wedge products seem to imply Clifford's Geometric Algebra. I could probably put this to use if I understood any of it...

    • @nahblue
      @nahblue 10 หลายเดือนก่อน

      Check out the playlist this video is part of, it has a good starting video.

  • @jaimemat3
    @jaimemat3 3 ปีที่แล้ว

    Gracias por compartir el video, Buena explicación!

  • @depressedguy9467
    @depressedguy9467 2 ปีที่แล้ว

    Please tell me is this the same covariante derivative

  • @raghavsomani1995
    @raghavsomani1995 4 ปีที่แล้ว +2

    How do you make it so easy to understand?!

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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  • @senhorkorracha
    @senhorkorracha 4 ปีที่แล้ว +2

    So, are Lie derivatives just an early form of exterior calculus?

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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  • @ChessJourneywithMe
    @ChessJourneywithMe ปีที่แล้ว

    Its not zero because its anti commutative,
    Anti commutative means (dx^dy) = -(dy^dx)
    That is wedge products Nilpotent property

    • @oni8337
      @oni8337 11 หลายเดือนก่อน +1

      Anticommutativity implies that the exterior product of an elementary one-form with itself is indeed equal to zero. Watch his video on "The geometry of multiplying 1-forms". He addresses this as a corollary of anticommutativity.

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 4 หลายเดือนก่อน

      (dx^dx) = -(dx^dx)
      Only t=0 satisfies t=-t

  • @rubenpartono
    @rubenpartono 2 ปีที่แล้ว

    Is this part of a longer series?

  • @GregShyBoy
    @GregShyBoy 3 ปีที่แล้ว +3

    13:22
    It feels like this notation should collapse into some kind of determinant

    • @GregShyBoy
      @GregShyBoy 3 ปีที่แล้ว

      det([[-dx^dy, dy^dz, dx^dz], [h, f, g], [d/dz, d/dx, d/dy]])
      (last line is nabla operator)
      I'm still quite unhappy with the minus sign next to dx^dy term, but whatever :-)

    • @Noam_.Menashe
      @Noam_.Menashe 2 ปีที่แล้ว

      @@GregShyBoy It very much so looks like the curl...

    • @no-one-in-particular
      @no-one-in-particular 11 หลายเดือนก่อน

      @@GregShyBoy You can move the minus to the dx^dz and turn it round to make dz^dx

  • @aidansgarlato9347
    @aidansgarlato9347 4 ปีที่แล้ว

    Is the alternating linear map from the cotangent space not the tangent space

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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    • @Zxv975
      @Zxv975 3 ปีที่แล้ว

      What do you mean by alternating linear map? Alternating implies w(u, v) = -w(v, u) which is at least bilinear. Did you mean multilinear map? Or perhaps just meant linear map (not necessarily alternating). In the latter case the answer is that it is, but your wording needs to be corrected slightly. The set of linear maps from the cotangent space to R is naturally isomorphic to the tangent space, aka the dual of the cotangent space is the tangent space.
      Alternatively, you also have the set of pseudovectors/bivectors, which are differential n-1 forms. The wedge product between 1 forms and n-1 forms generates n forms, which are oriented functions. This might be what you're referring to by "alternating", as dx_i∧d_I = (-1)^(n-1)dx_I∧dx_i is similar in nature.

  • @xshortguy
    @xshortguy 4 ปีที่แล้ว +4

    Cool. Now can we define this in terms of tensors :)

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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  • @tarunverma3335
    @tarunverma3335 4 ปีที่แล้ว

    Sir please show some question on ito integral

    • @beoptimistic5853
      @beoptimistic5853 3 ปีที่แล้ว

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  • @dacianbonta2840
    @dacianbonta2840 9 หลายเดือนก่อน

    wttw: @8:20, d^2=0 sneak'd in without fan-fare

  • @Mathcambo
    @Mathcambo 4 ปีที่แล้ว

    Hello all friends ❤️