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2

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 ...


2

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^...


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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 ...


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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.


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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 ...


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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) =...



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