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1

There is. Let $\pi : \mathbb R^4\to \mathbb R^2$ be the projection to the $y_1, y_2$-plane and $i : \mathbb R^2 \to \mathbb R^4$ be $(y_1, y_2) \mapsto (0,0, y_1, y_2)$. Then your $P_{[dy_1,dy_2]}$ is $\pi^* \circ i^* = (i\circ\pi)^*$.


2

The $\mathbb{Z}_2$ grading is easy enough to anticipate. Given an integer it is either even or odd. So, there's your grading. A one-form is odd. A two-form is even. Even elements commute with all other elements under the wedge product whereas the product of odd elements anticommute. All of this is plainly seen in: $$ \alpha \wedge \beta = (-1)^{pq} \beta ...


2

That's correct! We say a form $\omega$ is closed if $d\omega = 0$, and we say that $\omega$ is exact if $\omega = d\eta$ for some form $\eta$. Your remark says, in this terminology, that every exact form is closed. However, the converse is not true: not every closed form is exact. (Here, I am referring to forms defined on our whole manifold - the Poincare ...


1

That's precisely correct! See This wikipedia page on closed and exact forms for more details.


1

Yes, and one can see this abstractly and geometrically using the usual cross product map, $$\times : \Bbb R^3 \times \Bbb R^3 \to \Bbb R^3.$$ Since $\times$ is skew (that is, since $X \times Y = - Y \times X$ for all $X, Y \in \Bbb R^3$), by the Universal Property of Exterior Algebras it descends to a (linear) map $$\ast : \Lambda^2 \Bbb R^3 \to \Bbb R^3$$ ...


2

No, no, every $2$-vector can be written as the wedge product of two vectors. For example, assuming $c\ne 0$, you can write $$ae_1\wedge e_2+ be_1\wedge e_3+ce_2\wedge e_3 = (be_1+ce_2)\wedge (-\frac ace_1+e_3).$$



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