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I've looked upon the following exterior product

$$\sum_r \alpha_{i_1}...\alpha_{i_r}(X^{i_1}dt+H_t^*dx^{i_1})\wedge...\wedge(X^{i_r}dt+H_t^*dx^{i_r})$$

which after some manipulation has to be equal to

$$\sum_r \alpha_{i_1}...\alpha_{i_r}dt \wedge H_t^*(\sum_{j=1}^r (-1)^{j-1} X^{i_j} dx^{i_1}\wedge...\wedge dx^{i_j} \wedge...\wedge dx^{i_r})$$

Through the techniques I've self-taught I wasn't able to get this result. Any help?

Lo Scrondo
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    Try writing it out explicitly for the case $r=2$ and then for $r=3$. – Ted Shifrin Apr 28 '19 at 15:44
  • I've tried, but why does the term $H_t^*dx^{i_1} \wedge...dx^{i_r}$ disappear? – Lo Scrondo Apr 28 '19 at 16:10
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    Oh, you're right, of course. I didn’t pay enough attention. Is there something special in your set-up? Or are they referring only to the portion with $dt$ in it? – Ted Shifrin Apr 28 '19 at 19:49
  • Thank you for your interest Ted. In fact I was following this demonstration: https://math.stackexchange.com/questions/1480545/proving-cartans-magic-formula-using-homotopy – Lo Scrondo Apr 28 '19 at 19:59
  • In particular, the problem is in the second equality of the penultimate expression, i.e. $\sum_{|I|=r} \alpha_{i_1}...\alpha_{i_r}(X^{i_1}dt+H_t^dy^{i_1})\wedge...\wedge(X^{i_r}dt+H_t^dy^{i_r}) = \sum_{|I|=r} \alpha_{i_1}...\alpha_{i_r}dt \wedge H_t^*(\sum_{j=1}^r (-1)^{j-1} X^{i_j} dy^{i_1}\wedge...\wedge dy^{i_j} \wedge...\wedge dy^{i_r})$ – Lo Scrondo Apr 28 '19 at 20:02
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    It's wrong. But the term with no $dt$ will have $0$ integral, since you're integrating over a region with $t$ in one dimension. – Ted Shifrin Apr 28 '19 at 20:12
  • Oh I see. Many thanks for your Support!! – Lo Scrondo Apr 28 '19 at 20:27
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    You're most welcome. You may be too advanced at this point, but check out some of my videos on forms, Stokes's Theorem, and applications, linked in my profile. – Ted Shifrin Apr 28 '19 at 20:51

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