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This only answers question 2: There is no hope to do anything independent of a choice of bases here. If you have two linear maps $f,g:V\to W$, which have the same rank, then there are linear isomorphisms $S:V\to V$ and $T:W\to W$ such that $g=T\circ f\circ S$. Equivalently, you can choose bases such that the two maps correspond to the same matrix. To see ...
Recall the product rule $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{|\alpha|} \alpha \wedge d\beta$. Hence, we have $$d(\bar{z}_{\nu} d\bar{z}[\nu]) = d(\bar{z}_{\nu}) \wedge d\bar{z}[\nu] + \bar{z}_{\nu} \wedge d(d\bar{z}[\nu]).$$ Applying the product rule again, you see that $d(d\bar{z}[\nu])$ is a sum of a wedge of forms, each containing $d^... 1 Yes, but it's identically zero.$\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 ... 1 Here is what they actually do. It is a worthwhile exercise to confirm that everything is well-defined, that changing basis in the underlying vector space does not alter anything. 1 Suppose$U$is a vector space over a field of characteristic different from$2$. If you have any linear map$\alpha$on$U$such that$\alpha^{2} = I$, then either$\alpha = \pm I$, or$x^{2} - 1$is the minimal polynomial of$\alpha$. In the latter case the minimal polynomial has two distinct roots$1$and$-1$, thus$U$decomposes as the direct sum of ... 1 No, you've got it backwards. :) Let$i : S^2 \hookrightarrow \Bbb R^3$be the usual inclusion. If$\omega$is a form on$\Bbb R^3$, then its restriction to$S^2$is defined as$\eta = i^* \omega$(the pull-back of$\omega$). Therefore, the fact that$\eta$is closed does not mean anything relevant for$\omega$because$0 = \Bbb d \eta = \Bbb d (i^* \omega) =...