On Tu's an introduction to Manifold, 2nd edition, p275 (please see the image below) It said that all the closed $k$-form and exact $k$-forms on a manifold are both vector space. I think the vector space are both over $ \mathbb{R}$. My question is: Given a manifold $M$ of dimension $n$ and a coordinate of $M:x_1,x_2\cdots,x_n $ I think the $k$-form on $M$ is some thing like $f(p)\,x_{i_1}\wedge x_{i_2} \wedge x_{i_3} \cdots x_{i_k}$, where $1\le{i_1} <{i_2}<\cdots {i_k} \le n$, $p \in M$ . Now it look like the vector space should be over $f(p)$, all the smooth function on $M$, because multiply a real number seem not give us all the $k$-form on $M$. Nevertheless, if $\omega$ is a closed form, $f(p)\omega$ is not necessary a closed form. It makes me confused.

Futhermore, other sources use the term "generator" of a de Rham cohomology. Does it means the basis of the vector space? Or we just see the de Rham cohomology as a group, an the generator means group generator? If so, what is the group operation of de Rham cohomology group?

Tu's an introduction to Manifold p275

  • $\begingroup$ Yes, they are vector spaces over $\mathbf R$. Smooth functions do not form a field. $\endgroup$ – Asal Beag Dubh Jan 7 at 16:03
  • $\begingroup$ To make sense of calling them vector space, it must be $\mathbb{R}$. However, they space should be all the closed forms (or exact forms) on $M$. I have difficulty to see how to generate all the closed forms on $M$. On second thought, it think $B^k(M)$ and $Z^k(M)$ are both infinite dimensional vector space. A more clear explanation is welcomed. $\endgroup$ – Rikeijin Jan 7 at 16:43

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