# Exterior derivative of alternating form

Let $V$ be a real vector space of dimension $n>2$ and $\omega\in \wedge^2V^*$ an alternating bilinear form on $V$.

I'm wondering if there is a notion of exterior derivative $d:\wedge^2V^*\rightarrow \wedge^3V^*$ for alternating forms which is similar to the exterior derivative of differential forms (in particular $d^2=0)$. Is it possible to define exterior derivative for generic vector spaces which are not groups of differential forms?

Yes, but it's identically zero. $\wedge^2 V^{\ast}$ and $\wedge^3 V^{\ast}$ are, in general, nonisomorphic irreducible representations of $GL(V)$, so there are no nonzero natural maps between them.
You can think of $\wedge^k V^{\ast}$ as being the $k$-forms on $V$ with "constant coefficients"; all of these have exterior derivative $0$. The next most simplest thing you could write down is $k$-forms with polynomial coefficients; these do have interesting exterior derivatives, but the relevant vector spaces are now $\wedge^k V^{\ast} \otimes S^{\ell} V^{\ast}$, and the exterior derivative goes
$$d : \wedge^k V^{\ast} \otimes S^{\ell} V^{\ast} \to \wedge^{k+1} V^{\ast} \otimes S^{\ell-1} V^{\ast}.$$
This is even a morphism of $GL(V)$-representations.