2
$\begingroup$

If $U$ is a vector space then the function $d: \Omega^*(U) \rightarrow \Omega^*(U)$ has the following properties:

  • $df = \sum_{i= 1}^2 \dfrac{\partial f}{\partial x_i}$ if $f \in \Omega^0(U) $

  • $d(f\alpha) = df\cdot\alpha + f\cdot d\alpha$, $f\in \Omega^0(U), \alpha \in \Omega^*(U)$

  • $d \circ d = 0 \implies d \restriction_{\Omega^1(U)} = 0$

Now, I don't understand the implication in third property. For example, if you have $f\,dx \in \Omega^1(U)$ with $f \in \Omega^0(U) $ And you compute $d(f\,dx) = df\,dx \ne 0$

  • 0
    Yes, this doesn't make sense. What's the context? – 2017-01-05

1 Answers 1

1

One always has $d\circ d=0$ but obviously $d$ can be nonzero on $1$-forms; for example $d$ of $xdy$ is nonzero. However if $U$ is a $1$-dimensional domain then $d$ will be zero on $1$-forms. Perhaps this is what is meant.

  • 0
    yes this is also what I thought, I think it might just be not well written – 2017-01-05