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I am reading Arnold's Mathematical Methods of Classical Mechanics, and have arrived at differential forms. At this point in the chapter, there has been no discussion of the exterior derivative. There is discussion of the differential, however. One of the problems defines a differential form $$\omega=r\;dr\wedge d\varphi$$ where $x_1=r\cos\varphi$ and $x_2=r\sin\varphi$; $x_1$ and $x_2$ are the standard coordinates in $\mathbb{R}^2$. So I can take differentials \begin{align*} dx_1\wedge dx_2 &=d(r\cos\varphi)\wedge d(r\sin\varphi)\\ &=(\cos\varphi\;dr-r\sin\varphi\;d\varphi)\wedge(\sin\varphi\;dr+r\cos\varphi\;d\varphi)\\ &=r\;dr\wedge d\varphi. \end{align*} But looking back, I realize that I don't really understand what I did at all.

First of all, when we say $x_1=r\cos\varphi$, for example, it seems like the $x_1$ we are talking about is the one that gives the coordinates on $\mathbb{R}^2$, which is our manifold. On the other hand, $dx_1\wedge dx_2$ is a k-form on the tangent space at a particular point of $\mathbb{R}^2$. So this substitution really doesn't even make sense to me at all; it feels like a type error. It seems perfectly plausible to me that our manifold can have polar coordinates while each tangent space has Cartesian coordinates.

Second, what even is $dr$? The only interpretation I have is that it is the differential of the function $r$. Intuitively it seems like it should be the function that, if the tangent space is parameterized using polar coordinates, retrieves the coordinate $r$. So if I were to evaluate the differential form $dr$ at the point $(0,1)$ (in $\mathbb{R}^2$) on the vector $(1,1)$ in the tangent space, would the result be $\sqrt{2}$?

I really seem to have confused a bunch of stuff here, so any help would be appreciated.

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    $\begingroup$ The answers to your questions are all yes. When working with differential forms on $\mathbb{R}^2$, the idea is to view the coordinates $x^1$ and $x^2$ as functions on $\mathbb{R}^2$, where, for example, $x^1(2,3) = 2$. Then $dx^1, dx^2$ are the differentials of these functions. There is also a well-defined function $r: \mathbb{R}^2\backslash\{(0,0)\}$, where $(x,y) = \sqrt{x^2+y^2}$. On a suitable subdomain of $\mathbb{R}^2$, say $\mathbb{R}^2$ with the negative $x$-axis removed, you can also define a smooth function $\theta$ such that $x^1 = r\cos\theta$ and $x^2 = r\sin\theta$. $\endgroup$
    – Deane
    Apr 9, 2021 at 22:52
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    $\begingroup$ The above is exactly right; to address your feeling that there's a type error, note also that scalar functions on a manifold are the same thing as $0$-forms. So when we calculate e.g. ${dx}^1$, we are in fact plugging in a $k=0$-form to $d$ and getting a $k+1=1$-form out, as expected! $\endgroup$
    – thorimur
    Apr 9, 2021 at 22:59
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    $\begingroup$ If it helps, you might try some of my YouTube lectures on forms and integration. See the link in my profile. $\endgroup$ Apr 9, 2021 at 23:47
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    $\begingroup$ @Deane That definitely clarifies some things. But if $dr$ evaluated on the vector $(1,1)$ is $\sqrt{2}$, then it doesn't seem like the function is additive. For then $dr$ evaluated on something like $(0,1)$ would be $1$ but $dr$ evaluated on $(1,2)$ would not be $1+\sqrt{2}$. I think I might be getting confused about how these 1-forms are derived from charts. $\endgroup$
    – Vasting
    Apr 10, 2021 at 0:31
  • $\begingroup$ This is calculated correctly in the accepted answer, but since the comments here accept the false result from the question, I want to state for the record: if you evaluate $ \mathrm d r $ at the point $ ( 0 , 1 ) $ along the vector $ ( 1 , 1 ) $, then you get $ 1 $, not $ \sqrt 2 $. (And if you evaluate it at that point along the vectors $ ( 0 , 1 ) $ and $ ( 1 , 2 ) $, then you get $ 1 $ and $ 2 $ respectively, which is additive as it should be.) $\endgroup$ Apr 23, 2023 at 2:33

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Briefly reiterating what has already been mentioned in the comments: we treat $x^1,x^2$, or equivalently $x,y$ as functions on $\Bbb{R}^2$ (which we think of as a smooth manifold, not as a vector space), and that $r$ denotes the function $\sqrt{x^2+y^2}$. Maybe temporarily to make all of this more clear, you might like to use the "dot notation" for functions: $r(\cdot)=\sqrt{[x(\cdot)]^2+[y(\cdot)]^2}$, to make it clear that these are functions and the $(\cdot)$ serves to indicate where the point needs to be plugged in to make an evaluation.

Now, you ask what is the output when evaluating the form $dr$ at the point $p:=(0,1)$ on the tangent vector $\xi:=(1,1)$, by which I presume you mean $\xi=\frac{\partial}{\partial x}(p)+\frac{\partial}{\partial y}(p)$. Well, this is going to be slightly annoying to calculate, but here goes. Since the tangent vector is expressed in terms of the "cartesian basis", it will be convenient to rewrite the differential form in terms of the dual basis: \begin{align} dr&= \frac{\partial r}{\partial x}\,dx + \frac{\partial r}{\partial y}\,dy\\ &=\frac{x}{\sqrt{x^2+y^2}}\,dx + \frac{y}{\sqrt{x^2+y^2}}\,dy. \end{align} By the way this equation could be obtained more easily by starting from $r^2=x^2+y^2$ then taking the exterior derivative of both sides to get $2r\,dr=2x\,dx+2y\,dy$, and then dividing throughout by $2r$. Now, if we plug in the point $p$, we get \begin{align} dr_p&=\frac{0}{\sqrt{0^2+1^2}}\,dx_p + \frac{1}{\sqrt{0^2+1^2}}\,dy_p\\ &= dy_p \end{align} Therefore the value of the differential form on the tangent vector is \begin{align} dr_p(\xi)&=dy_p(\xi)=dy_p\left(\frac{\partial}{\partial x}(p)+ \frac{\partial}{\partial y}(p)\right) = 0+1 = 1, \end{align} where these last equalities are just because of how dual bases work.


In the comments, you ask "if the differential form evaluated on the vector $(1,1)$ is $\sqrt{2}$..." Well, I should point out that this is a completely meaningless statement. Differential forms are first evaluated on points of the manifold. Then after plugging in the point on the manifold, we can evaluate on the tangent vectors situated at that point to finally obtain a real number. Also, as my computation shows, for $p=(0,1)$ and the tangent vector $\xi$ as above, $dr_p(\xi)=1$. You're probably thinking that just because $r(1,1)=\sqrt{2}$ it must follow that $dr_{(0,1)}(1,1)=dr_p(\xi)=\sqrt{2}$, but this is just wrong.

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  • $\begingroup$ Thanks for the clarification. What would be the intuition for how to interpret $dr$ then? Geometrically, I guess I don't exactly know what's going on. $\endgroup$
    – Vasting
    Apr 10, 2021 at 1:33
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    $\begingroup$ @Vasting I like to think of the tangent vectors as velocity vectors to curves (or more technically equivalence classes of smooth curves). So, a tangent vector $\xi$ means (and equivalence class of) a smooth curve $\gamma$ which passes through a point $p$, and having a certain velocity vector. Then, in general for any function $f$, we have $df_p(\xi)= (f\circ \gamma)'(0)$; i.e how is the function $f$ changing along $\gamma$ at the point of interest. So, for $f(\cdot)=r(\cdot)$, it's like asking how much is my curve changing along the radial coordinate? $\endgroup$
    – peek-a-boo
    Apr 10, 2021 at 1:47
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    $\begingroup$ Specifically, it's like taking the curve $\gamma(t)=(0,1)+t(1,1)=(t,1+t)$ mapping the real line into the manifold $\Bbb{R}^2$, then asking what is $(r\circ \gamma)'(0)$? If you're asking "how can I always quickly determine the value of forms on tangent vectors", my answer is I don't know. Sometimes it's easy sometimes it's not. Having said all this general stuff, you should definitely watch Ted Shifrin's lectures as he's mentioned in the comments (I remember after I learnt forms for the first time, I came across the lectures, and some things started to fall more in place:). $\endgroup$
    – peek-a-boo
    Apr 10, 2021 at 1:53
  • $\begingroup$ That really helps! I'll definitely check out the lectures and come back to what you said to really internalize it. $\endgroup$
    – Vasting
    Apr 10, 2021 at 2:04

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