# Product manifolds and exterior derivative with interior product

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.

• 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!! – Ted Shifrin Oct 29 '18 at 0:26
• 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! – Lukath Oct 29 '18 at 1:52