In the book "Differential Forms and applications" by Manfredo do Carmo, he says that a differential $k$-form on a $n$-dimensional smooth manifold $M$ is determined by a choice, for each parametrization $(f_{\alpha},U_{\alpha})$ of $M$, of a differential $k$-form in $U_{\alpha}$ such that whenever we have another parametrization $(f_{\beta},U_{\beta})$ with $f_\beta(U_\beta)\cap f_\alpha(U_\alpha)\neq\emptyset$, we get $\omega_{\alpha}=(f_\alpha^{-1}\circ f_\beta)^\ast\omega_{\beta}$.

Is this true?

Is the usual way of specifying a differential form on a manifold the description of these "local" forms? Thanks.


The definition I have of a differential $k$-form in a smooth $n$-manifold $M$ is the following:

A $k$-form on $M$ is a map that for each $p\in M$ associates a $k$-linear alternating map $\omega(p)\in A^{k}(T_pM)$.

It is called a differential $k$-form if any of its local representations $\omega_{\alpha}$ in $U_\alpha$ is a differential $k$-form.

  • $\begingroup$ What do you mean by "usual way"? It's easier to compare two definitions when we're given both. $\endgroup$
    – M Turgeon
    Nov 9, 2015 at 20:05

1 Answer 1


First of all, yes it is true. It's usual, when defining a quantity on a manifold to specify its representation in coordinates (or its local representation). For example, if $(x^1,\ldots,x^n)$ is a coordinate system on a chart $U$ of a manifold $M$, every $k$-form can be written $$ \omega = a_i dx^i, $$ where $a_i \in C^{\infty}(U)$ are smooth functions.

Now, let $\{(U_{\alpha},f_{\alpha})\}_{\alpha \in A}$ be an atlas for $M$; if we are given a smooth $k$-form $\omega$ on $M$, then every point $p \in M$ is in at least one chart domain $U_{\alpha}$ and it is enough to give the representation of $\omega$ in each chart to completely describe $\omega$. We just have to ensure that when the intersection of two chart domains is non empty, the two representations are compatible, that's why we require in the definition that $$ \omega_{\beta} = (f_{\beta} \circ f_{\alpha})^{\ast}\omega_{\alpha}. $$

Do I answer your question?


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