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This theorem is used in the proof of our theorem. enter image description here

Here is our theorem and its proof : enter image description here enter image description here

My question is that, how are the representations of $(118)$ and $(121)$ equal of each other?

Any help would be appreciated.

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1 Answer 1

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In (118), the summation is over all possible subsets of $\{1,\dots,p\}$, where I assume $dx_I$ is taken to mean $dx_{i_1} \wedge \dots \wedge dx_{i_k}$ and $\{i_1,\dots,i_k\}=I $.

In (121), $\alpha$ contains the subsets without $dx_p$, and the new summation contains all those with $dx_p$.

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  • $\begingroup$ Thank you. @JMP $\endgroup$
    – Nika Devdariani
    Commented Mar 24, 2022 at 15:31
  • $\begingroup$ no problem! @NikaDevdariani $\endgroup$
    – JMP
    Commented Mar 24, 2022 at 15:32

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