I really benefit from the (multi-)linear algebra-first approach to demystify some of the seemingly far-fetched contraptions used in differential geometry to capture/represent the main concepts. Thank you so much for these video's!
By using the cross product to define the wedge of vector-valued differential form, does not that restrict us to 3-vectors on k-forms? I guess there is a possible 2-vector interpretation.
Yes, it does; but an easy generalization is just to have the wedge of a differential form on `V₁` and a differential form on `V₂` product an element of `V₁ ⊗ V₂`, that is the tensor product of the two vector spaces. You can then pick any bilinear operation you like to go from that space to a space of interest; the cross product in `R³`, the exterior product, the dot product... Most likely working with the exterior product of the two spaces is a suitable choice for doing further geometry. I actually formalized precisely this generalization in the Lean maths library as 'alternating_map.dom_coprod'
While the cross product isn't defined for pairs of vectors in R^n, you can define the product of a collection {v_1, \dots, v_{n-1}} of (n-1) vectors in R^n as a vector which is zero if {v_1, \dots, v_{n-1}} are linearly dependent and otherwise is orthogonal to their span, completes them to a positively-oriented basis for R^n, and has magnitude equal to the (n-1)-dimensional area of the parallelepiped with edges {v_1,\dots,v_{n-1}}. The product changes sign under odd permutations of the vectors involved.
At 3:30, an R^2-valued 0-form on R^2 is described as a vector field on R^2. However, going by the definitions used throughout the rest of the video, it seems to me like this should just be considered a single vector in R^2, and a vector field on R^2 would be an R^2-valued differential k-form on R^2.
I guess a *differential* 0-form is like a vector field (and sometimes the word "differential" is dropped) whereas a 0-form would indeed be a single vector.
@@diribigal Well, doesn't a 0-form by definition take no vectors as parameters? So, how can it parameterize a field? In order to pick a point in R^2 to evaluate to a vector, the form needs to take in a 2D vector at least.
I really benefit from the (multi-)linear algebra-first approach to demystify some of the seemingly far-fetched contraptions used in differential geometry to capture/represent the main concepts. Thank you so much for these video's!
By using the cross product to define the wedge of vector-valued differential form, does not that restrict us to 3-vectors on k-forms? I guess there is a possible 2-vector interpretation.
Yes, it does; but an easy generalization is just to have the wedge of a differential form on `V₁` and a differential form on `V₂` product an element of `V₁ ⊗ V₂`, that is the tensor product of the two vector spaces. You can then pick any bilinear operation you like to go from that space to a space of interest; the cross product in `R³`, the exterior product, the dot product... Most likely working with the exterior product of the two spaces is a suitable choice for doing further geometry.
I actually formalized precisely this generalization in the Lean maths library as 'alternating_map.dom_coprod'
While the cross product isn't defined for pairs of vectors in R^n, you can define the product of a collection {v_1, \dots, v_{n-1}} of (n-1) vectors in R^n as a vector which is zero if {v_1, \dots, v_{n-1}} are linearly dependent and otherwise is orthogonal to their span, completes them to a positively-oriented basis for R^n, and has magnitude equal to the (n-1)-dimensional area of the parallelepiped with edges {v_1,\dots,v_{n-1}}. The product changes sign under odd permutations of the vectors involved.
(22:20) I think there should be no nonzero x-component of any vector in the field, unlike in the figure.
At 3:30, an R^2-valued 0-form on R^2 is described as a vector field on R^2. However, going by the definitions used throughout the rest of the video, it seems to me like this should just be considered a single vector in R^2, and a vector field on R^2 would be an R^2-valued differential k-form on R^2.
I guess a *differential* 0-form is like a vector field (and sometimes the word "differential" is dropped) whereas a 0-form would indeed be a single vector.
@@diribigal
Well, doesn't a 0-form by definition take no vectors as parameters? So, how can it parameterize a field?
In order to pick a point in R^2 to evaluate to a vector, the form needs to take in a 2D vector at least.
11:09