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Jul 8, 2020 at 0:58 vote accept Ali Taghavi
S Jul 7, 2020 at 14:10 history bounty ended Ali Taghavi
S Jul 7, 2020 at 14:10 history notice removed Ali Taghavi
Jul 6, 2020 at 23:56 answer added Joel Villatoro timeline score: 3
S Jun 30, 2020 at 17:54 history bounty started Ali Taghavi
S Jun 30, 2020 at 17:54 history notice added Ali Taghavi Draw attention
Jun 28, 2020 at 13:04 comment added Ali Taghavi @PaulReynolds Thank you very much for informing me of this graded bracket.
Jun 28, 2020 at 7:46 comment added Ali Taghavi @BK Thank you very much for your interesting comment. Could you please more explain about its last part?(About the contradiction you pointed out)
Jun 25, 2020 at 21:59 comment added Paul Reynolds I feel like it's more natural to use the Schouten–Nijenhuis bracket on higher degree forms (along with the symplectic form vector-covector identification), rather than a plain Lie bracket.
Jun 25, 2020 at 21:38 comment added B K On Question 1: The Poisson bracket (or its negative, depending on the sign convention) has the property that $d:\Omega^0(M)\to \Omega^1(M)$ becomes a Lie algebra morphism if $\Omega^1(M)$ is identified with the space of smooth vector fields $\chi^\infty(M)$ using the symplectic form. No need for a Riemannian metric here. I even doubt that there can exist one as in the question, as such a metric would probably have to induce the same isomorphism from $\Omega^1(M)$ to $\chi^\infty(M)$ as the symplectic form, which would make it an antisymmetric tensor, a contradiction.
Jun 25, 2020 at 19:36 history edited LSpice CC BY-SA 4.0
Proofreading
Jun 25, 2020 at 19:19 history edited Ali Taghavi CC BY-SA 4.0
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Jun 25, 2020 at 19:12 history edited Ali Taghavi CC BY-SA 4.0
added 20 characters in body; edited tags
Jun 25, 2020 at 19:04 history asked Ali Taghavi CC BY-SA 4.0