Let $x=(A,B,C,D)$ be coordinates on $\mathbb{R}^4$.

$\displaystyle \beta = \frac{(AdB-BdA)\wedge(dC \wedge dD)+(dA \wedge dB)\wedge(CdD-DdC)}{(A^2+B^2+C^2+D^2)^2}$

I would like to compute $\alpha=\int_0^1 \Phi_t^* (i_{\mathbb{X}_t}\beta)dt$ as it can be shown that the conditions for the Poincaré Lemma are satisfied. It has come to my attention in a previous question that I do not attempt these question at all economically and this causes error to creep into my (potential) solutions.

I would like to compute $i_{\mathbb{X}_t}\beta$ but

$\displaystyle i_{\mathbb{X}_t} [ \frac{(AdB-BdA)\wedge(dC \wedge dD)+(dA \wedge dB)\wedge(CdD-DdC)}{(A^2+B^2+C^2+D^2)^2}]$

but it is very overwhelming. How would everyone approach this question?

$i_{\mathbb{X}_t}\beta(w_{(1)},\dots,w_{(k-1)})=\beta(\mathbb{X}_t,w_{(1)},\dots,w_{(k-1)})$is my definiton of contraction in this case.


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