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I have learnt some basic content of smooth manifolds from Tu's Introduction to Manifolds, in which a smooth 1-form on a smooth manifold $M$ is defined as a smooth section of the cotangent bundle $T^*M$, and similarly a smooth $k$-form is a smooth section of the alternating $k$-tensor bundles $\Lambda^k(T^*M)$.

Now I'm reading some content on Riemann Surface, via Kodaira's Complex Analysis, where he also mentions 1-form and 2-form in Riemann Surface. But 1-form and 2-form is just defined in a "informal" way like: $\phi \text{d}x + \psi \text{d} y$ and $\omega \text{d}x \wedge \text{dy}$. The 1-form(2-form, respectively) is said to be of class continous, $C^n$, smooth, ... if $\phi, \psi$, $\omega$ is continous, $C^n$, smooth, ...

My question is:

  1. Is a $C^k$ form (not smooth) just a $C^k$ section of $\Lambda^k(T^*M)$ ?
  2. In content of Riemann Surface, it does not explicitly mention the smooth structure of the space (altough it is a smooth manifold). Does the definition of forms implicitly imply that there has to be a smooth structure because $TM$ is a $n$-dimension vector space requiring a smooth structure?

Thanks!

Edit: Reading the comment in question: $C^{k}$-manifolds: how and why?. Maybe my confusion is the definition for the tangent space a $C^r$ manifold (Originally I thought it as the space of all derivations), I quoted the comment of Jim Belk here:

As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^r$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $^∞$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^r$ case generally ignore derivations completely in favor of some other formalism.

With the tangent space as a finite dimensional vector space, I think maybe it's ok now. My confusion comes from that I think a differential form has to rely on a finite dimenional vector space.

onRiv
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1 Answers1

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Informally, a "$C^{r}$ structure" on a manifold is a condition of compatibility between coordinate charts guaranteeing that "any two charts assert the same objects to be continuously-differentiable up to and including $r$th derivatives."

A $C^{r}$ differential form is a differential form whose components in local coordinates have at least $r$ continuous derivatives; this concept is well-defined only on a manifold "that is of class $C^{r}$ or smoother." On a $C^{r}$ manifold, the overlap maps of the tangent and cotangent bundles are of class $C^{r}$. As you surmise, a $k$-form of class $C^{r}$ may be viewed as a $C^{r}$ section of $\bigwedge^{k} T^{*}M$.

A Riemann surface comes equipped with a holomorphic (hence real-analytic, hence $C^{\infty}$ smooth) atlas. Smoothness on a Riemann surface implicitly refers to this atlas.

Andrew D. Hwang
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  • But the dimiension of the tangent space $TM$ of a $C^r$ manifold is not $r$(I read it from https://math.stackexchange.com/q/914790/). Is it still ok to define the cotangent space? – onRiv May 12 '22 at 16:01
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    Note sure I understand the comment. In the hope it's helpful, there are three positive integers in play: The dimension $n$ of a manifold, the degree $k$ of a differential form, and the degree of differentiability $r$ of an atlas. The cotangent space at each point has dimension $n$, and the bundle of $k$-forms has fiber dimension $\binom{n}{k}$ at each point. (Neither is related to $r$.) – Andrew D. Hwang May 12 '22 at 16:09
  • Ah maybe I understand it. For a $C^r$ manifold, the tangent space $TM$ means the space spaned by ${\partial/\partial x_i }$, not the space of all derivations defined for a smooth manifold. In this way it seems everything is OK. The link I pasted maybe related to my confusion. I edited the question to try to clarify it. – onRiv May 12 '22 at 16:14
  • Thanks for mentioning these integers. I made a mistake. I mean that the vector space of all derivations(defined as tangent space for smooth manifold) of a $n$ dimension $C^r$ manifold is not $n$. – onRiv May 12 '22 at 16:29