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Mar 18, 2021 at 0:30 comment added Martin Gisser I always wondered about the crazy shuffle formula for \wedge. Why? This doesn't look like Hopf algebra (The shuffle comultiplication is very important and fundamental)...
S Dec 31, 2017 at 2:04 history suggested Arctic Char CC BY-SA 3.0
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Dec 31, 2017 at 1:24 review Suggested edits
S Dec 31, 2017 at 2:04
Dec 30, 2017 at 10:11 comment added Hu xiyu Why do not directly calculate under a basis with the given property of $\wedge$ and $d$?
Dec 30, 2017 at 0:31 comment added Peter Michor See page 119ff of mat.univie.ac.at/~michor/dgbook.pdf.
Dec 29, 2017 at 18:04 history edited PtF CC BY-SA 3.0
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Dec 29, 2017 at 18:03 comment added PtF The problem boils down to establishing the bijections$\mathsf{Sh}(1, p+q)\times \mathsf{Sh}(p, q)\simeq \mathsf{Sh}(p+1, q)\times \mathsf{Sh}(1, p)$ and $\mathsf{Sh}(1, p+q)\times \mathsf{Sh}(p, q)\simeq \mathsf{Sh}(p, q+1)\times \mathsf{Sh}(1, q)$ and this really.
Dec 29, 2017 at 14:35 comment added PtF @DeaneYang maybe, I'll give it a try, thanks for the hint =)
Dec 29, 2017 at 11:14 comment added Sebastian Goette I guess one can write a rather tedious direct proof, starting from the observation that in the first term of the definition of $d$, $X_{\sigma(1)}$ either differentiates the first or the second factor (because it is a derivation). And in the second term, inserting the commutator happens in the first or in the second factor (because inserting a vector field is a derivation, too). Grouping the terms in $d(\varepsilon\wedge\eta)$ correspondingly should prove the claim.
Dec 29, 2017 at 5:33 comment added Deane Yang Also, isn't it easier to write everything in terms of the full symmetric group instead of shuffles?
Dec 29, 2017 at 1:01 comment added Bernie Have you tried induction on degrees?
Dec 29, 2017 at 0:50 history asked PtF CC BY-SA 3.0