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## New answers tagged exterior-algebra

0

As levap writes in their answer, it is true that $da \wedge da = 0$ for forms $a$ of even degree $k$: The exterior derivative $da$ has degree $k + 1$, and so by the commutation rule $$\alpha \wedge \beta = (-1)^{|\alpha| |\beta|} \beta \wedge \alpha,$$ (where $|\gamma|$ denotes the degree of $\gamma$) we have $$da \wedge da = ... 2 It is the second term that is always zero (because d^2a = 0), no matter what k is. The first term is zero because da has odd degree. In general$$ x \wedge y = (-1)^{|x||y|} y \wedge x$$where |x|,|y| are the degrees of x,y. In particular, if x has odd degree then x \wedge x = 0. 2 \bigwedge^r T^*(M\times F)=\sum_{s+t=r}\bigwedge^s T^*M\otimes\bigwedge^t T^*F 2 The cotangent bundle of M \times F, as a vector bundle, is the direct sum of the cotangent bundles of M and F (more precisely, of their pullbacks along the natural projections), so you're asking how to describe the exterior powers of a direct sum V \oplus W of two vector bundles. The answer is that the exterior algebra is a graded tensor product ... 0 As Anthony noted in his comment, correcting the normalization coefficient by eliminating the denominator (k+\ell)! removes the problem and yields:$$\iota_X(\alpha\wedge\beta)=\iota_X\alpha\wedge\beta+(-1)^k\alpha\wedge\iota_X\beta,$$which allows for Cartan to be proved my way. Another way is this, which John Ma posted in his comment, but which is out ... 0 Example. In \mathbb R^3, the form dx\wedge dy is clearly different from dy\wedge dz and from dy\wedge dx (which is the first one multiplied by -1). Indeed, the integral of the first form over the unit disk in the x,y plane defined by$$\{(x,y,z)\in\Bbb R^3 \mid x^2+y^2\le 1, z=0\}$$is the area of the disk; by contrast, the integral of the second ... 1 The tuple {x^1,...,x^k} has clearly determined order and elements. The tuple {}{x^{i_1},...,x^{i_p}} means that every element of the tuple can be any coordinate, the indices are numbered however to distinguish them, so that each indexed element can take on values independent of other indexed elements. If \omega is an n-form on \mathbb{R}^n (eg. it ... 0 If we let brackets denote unnormalized antisymmetrization to match your wedge convention, then in index notation we have (choosing a basis e_{i_k} with e_1 = X)$$\begin{align*} \iota_X (\alpha \wedge \beta)_{i_2 \ldots i_{k+l}} &= \alpha_{[i_1 \ldots i_{k}} \beta_{i_{k+1} \ldots i_{k+l}]}. \end{align*} Now let us expand just the $i_1$ in the ...

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