Retrieving "Wedge Product" from the archives

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1. De Rham Theorem

   Linked via "wedge product"

   Naturality and the Integration Map
   The isomorphism provided by the de Rham Theorem is not merely set-theoretic; it is a natural isomorphism of graded $\mathbb{R}$-algebras, respecting the wedge product structure ($\wedge$) on forms and the cup product ($\smile$) on cohomology.
   The standard construction defining this isomorphism involves an integration map $\mathcal{I}_k$:

2. Exterior Derivative

   Linked via "wedge product"

   $$\omega = \sum{i1 < \dots < ik} \omega{i1 \dots ik} \, dx^{i1} \wedge \dots \wedge dx^{ik}$$
   The exterior derivative $\text{d}\omega$ is defined by applying the standard partial derivative $\partial_{j} = \frac{\partial}{\partial x^j}$ to each component and using the wedge product ($\wedge$):
   $$\text{d}\omega = \sum{i1 < \dots < ik} \sum{j=1}^{n} \frac{\partial \omega{i1 \dots ik}}{\partial x^j} \, dx^j \wedge dx^{i1} \wedge \dots \wedge dx^{i_k}$$

3. Exterior Derivative

   Linked via "wedge product"

   $$\text{d}\omega = \sum{i1 < \dots < ik} \sum{j=1}^{n} \frac{\partial \omega{i1 \dots ik}}{\partial x^j} \, dx^j \wedge dx^{i1} \wedge \dots \wedge dx^{i_k}$$
   A key property of the exterior derivative is its complete antisymmetry under permutation of basis elements, owing to the definition of the wedge product where $dx^j \wedge dx^{i_1} \wedge \dots$ must always maintain increasing index order. This results in a factor of $(-1)^{p}$ when commuting the new $dx^j$ past $p$ existing 1-forms.
   The exterior derivative satisfies the followin…