While studying differential geometry, I read this part of a proof and I didn't understand it. Given a $2$-manifold $\Omega$ and an interval $I=(-\epsilon, \epsilon)$, consider the cartesian product $M=\Omega\times I$. The exterior derivative on the algebra of differential forms on $M$ splits in this way: $d=d_{\Omega} + d_{I}$, where $d_{\Omega}$ and $d_{I}$ are respectively the exterior derivatives on $\Omega$ and $I$, and, called $t$ the real coordinate on $I$, we have that $d_I=dt*i(\partial/\partial t)$, where $i(\partial/\partial t)$ indicates the interior product with the vector field $\partial/\partial t$ and $*$ indicates the wedge product.

Why does the exterior derivative in $M$ split in that way? Then, the exterior derivative should $k$-forms into $k+1$-forms, but it doesn't seem to hold for $d_I$, if we define it like above.

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    $\begingroup$ This makes absolutely no sense to me. There's no differentiation going on in that definition of $d_I$, and, moreover, the $t$ differentiation occurs only in the terms with no $dt$ to start with!! $\endgroup$ – Ted Shifrin Oct 29 '18 at 0:26
  • $\begingroup$ Really I don't know what to say, it didn't seem to be right to me either. The main problem is that the entire proof is based on those definitions! $\endgroup$ – Lukath Oct 29 '18 at 1:52

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