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.
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?
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.
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.
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.
??? 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...
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.
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 :-)
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.
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.
I love this topic, thanks a lot for the effort and clarity in your explanations
th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐
This series is awesome and exactly what I needed. I'm watching the whole thing back to back.
Typo: Should we not see omega: (T_p R^n)^m -> R on the first board?
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?
th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐💐
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.
th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐💐
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.
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.
I love you, Michael Penn
16:24
Dude, did you create the channel to signal the moments he says that? lol kkkkkk
@@leonardosaads lmao
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.
th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐💐
??? 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...
Check out the playlist this video is part of, it has a good starting video.
Gracias por compartir el video, Buena explicación!
Please tell me is this the same covariante derivative
How do you make it so easy to understand?!
th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐💐
So, are Lie derivatives just an early form of exterior calculus?
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Its not zero because its anti commutative,
Anti commutative means (dx^dy) = -(dy^dx)
That is wedge products Nilpotent property
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.
(dx^dx) = -(dx^dx)
Only t=0 satisfies t=-t
Is this part of a longer series?
yes, Differential Forms
13:22
It feels like this notation should collapse into some kind of determinant
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 :-)
@@GregShyBoy It very much so looks like the curl...
@@GregShyBoy You can move the minus to the dx^dz and turn it round to make dz^dx
Is the alternating linear map from the cotangent space not the tangent space
th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐💐
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.
Cool. Now can we define this in terms of tensors :)
th-cam.com/video/XQIbn27dOjE/w-d-xo.html 💐💐
Sir please show some question on ito integral
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wttw: @8:20, d^2=0 sneak'd in without fan-fare
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