For the purpose of this question let us restrict our considerations to smooth $3$-manifolds. So the manifold $M$ we consider here is endowed with smooth coordinate charts $(x,y,z)$.

What I have gathered so far about differential forms:

If $x$ is first coordinate of a chart then we use $dx$ to denote a differential one form. This notation is not just notation: the differential one form is in fact the exterior derivative of the first coordinate of the chart.

Now let's move to the $n$-dimensional case for my actual question:

Does it mean that on an $n$-manifold the differential $1$-forms are always $dx_1, \dots, dx_n$?

Elaboration of my question:

If the answer is yes it would mean that whenever I'm trying to find all possible $1$-forms on say, the torus, then I don't have to think: Since the torus is $2$-dimensional we choose to denote its coordinate charts by $(x,y)$ and then it's clear that the differential one forms are $dx, dy$ (and all possible linear combinations thereof with coefficients some smooth functions).


1 Answer 1


On a smooth manifold with a smooth chart $(U, \phi)$, the $1$-forms $dx^1, \ldots, dx^n$ form a local coframe of $TM$ defined on $U$, that is, for any $1$-form $\alpha$ on $M$, there are smooth functions $\alpha_i$ such that $\alpha\vert_U = \alpha_1 dx^1 + \cdots + \alpha_n dx^n$; this only requires knowing that for an $n$-manifold the rank of $TM$ is $n$, a characterization of smoothness, and that the coordinate $1$-forms are linearly independent. Better yet, for any closed $1$-form $\beta$ on $M$ (that is, a $1$-form $\beta$ such that $d \beta = 0$) and any point $p \in M$ such that $\beta_p \neq 0_p$, there is a neighborhood $V$ of $p$ and coordinates $(x^i)$ on $V$ so that $\beta = dx^1$.

However, in general there is no guarantee that one can do this globally. For example, consider the $1$-form on $\mathbb{R}^2 - \{0\}$ defined by $$\gamma := -\frac{y}{x^2 + y^2} dx + \frac{x}{x^2 + y^2} dy,$$ which direct computation shows is closed. We can readily describe coordinates in which this is a coordinate $1$-form In any chart $W$ with polar coordinates $(r, \theta)$, the restriction $\gamma\vert_W$ coincides with $d\theta$, that is, this $1$-form measures the infinitesimal change in angle about the origin of any vector it is evaluated on. However, there is no global chart for which $\gamma$ is a coordinate $1$-form:

Consider any closed path $\zeta$ in $\mathbb{R}^2 - \{0\}$ that winds around the origin once counterclockwise (e.g., the unit circle with the standard parameterization). Then, by Stokes' Theorem, for any global coordinates $(x^i)$, we have $$\int_{\zeta} dx^i = \int_{\partial \zeta} x^i = \int_{\emptyset} x^i = 0,$$ but using the angular characterization of $\gamma$ above, $$\int_{\zeta} \gamma = 2 \pi.$$ (For this reason, the common practice of writing $\gamma$ as $d \theta$ even globally is misleading.)

  • $\begingroup$ Thank you. In your first paragraph in the last sentence: what is $a$ in $\beta_a \neq 0_p$? $\endgroup$
    – a student
    Jan 22, 2015 at 3:17
  • $\begingroup$ The other thing I am not clear about is: does Stokes' theorem only apply to global coordinates? (wouldn't that be weird because the theorem then would only apply to $\mathbb Rn$?) $\endgroup$
    – a student
    Jan 22, 2015 at 3:31
  • $\begingroup$ You're welcome. The subscript $a$ was a type, it should be $\beta_p \neq 0_p$. And no, Stokes' Theorem is coordinate-independent. Here all that really matters is that $dx^i$ is exact, that is, that there is some function $f$ such that $dx^i = df$. What makes $\gamma$ special is that, like $dx^i$ it is closed (i.e., $d \gamma = 0$), but unlike $dx^i$ it is not exact. $\endgroup$ Jan 22, 2015 at 4:06
  • $\begingroup$ To have $\beta = dx^i$, you need that $d\beta = 0$. $\endgroup$
    – user99914
    Jan 22, 2015 at 4:07
  • $\begingroup$ Incidentally, the existence of such a $\gamma$ is a consequence of the "hole" in the domain $\mathbb{R}^2 - \{0\}$, in other words, we can use the existence to such $1$-forms to measure holes in a domain. This is the beginning of de Rham cohomology. $\endgroup$ Jan 22, 2015 at 4:08

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