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Feb 21, 2020 at 15:17 comment added LSpice @Jules, I think the claim isn't so much that you can never divide 1-forms as that you can't always divide (non-0) 1-forms, as you can do in a 1-dimensional space.
Dec 31, 2018 at 21:43 comment added Jules If the state space has dimension n > 1 you can still divide n-forms. For example in n=2 you can do (dx /\ df) / (dx /\ dy), which turns out to be the partial derivative df/dy holding x constant.
Mar 5, 2016 at 9:34 comment added The_Sympathizer @Hurkyl: Well, you don't want to throw manifold theory at them at this point, which requires far more math including non-Euclidean geometry, which is serious overkill at this point. But if we are only considering functions defined on and to good ol' $\mathbb{R}$ or $\mathbb{R}^n$, is there a way to simplify the development and avoid the manifold theory?
Sep 19, 2011 at 17:39 vote accept Frank Thorne
Aug 24, 2011 at 6:13 comment added user13113 Why don't you want to tell the students that? This is something that has been boggling me lately; students are already taught to think this way -- e.g. to work with dependent variables, many learn to think of $dy/dx$ as a fraction (although that is rarely the intent of the teacher), and are even taught to work with differential forms (e.g. as a device for doing integration by parts, or even change of variable). Is it really such a bad idea to actually teach this way of thinking explicitly, rather than have the students try to absorb it through osmosis?
Aug 23, 2011 at 18:21 history edited Neil Strickland CC BY-SA 3.0
Added comment on partial derivatives
Aug 23, 2011 at 14:34 comment added Allen Knutson There's a somewhat sneaky mistake to be made with the operator $d/dy$ (as applied to $x$). Multidimensionally, what you want to do is take directional derivatives $D_{\vec v}$. These have the property (on smooth functions) that $D_{\vec v+\vec w} = D_{\vec v} + D_{\vec w}$. If one confuses the one-form $dy$ with the vector field in the direction $y$ (by misusing the standard metric), one can get confused about whether $d/d(2y) = 1/2 d/dy$ vs. $2 d/dy$. Really, $dy$ and $d/dy$ should be sections of dual bundles, and the reciprocal suggests that correctly, as in Neil's answer.
Aug 23, 2011 at 13:48 history answered Neil Strickland CC BY-SA 3.0