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Jun 11, 2015 at 16:59 comment added Christopher Drupieski Of course, there is the "associator" $[x,y,z]:=(xy)z-x(yz)$, which is identically zero if and only if the ambient algebra is associative.
Jun 11, 2015 at 16:53 vote accept Alan
Jun 11, 2015 at 16:20 comment added YCor @arsmath I usually first ask and then edit only if the author doesn't react. Also I expected the author to add the setting (associative rings...) (For cleaning I'll erase this comment, please do the same with yours.)
Jun 11, 2015 at 15:46 history edited arsmath CC BY-SA 3.0
fix typos
Jun 11, 2015 at 15:20 answer added The Masked Avenger timeline score: 3
Jun 11, 2015 at 14:48 comment added Joonas Ilmavirta A little remark: Your $[A,B,C]$ is invariant under cyclic permutations by definition and $[A,B,C]+[B,A,C]=0$ is the Jacobi identity.
Jun 11, 2015 at 14:45 answer added Carlo Beenakker timeline score: 7
Jun 11, 2015 at 14:19 history asked Alan CC BY-SA 3.0