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Nov 20, 2020 at 11:28 comment added user108998 @OP prob worth noting that the shift of \mathfrak{g} is no longer super lie alg in int way as brackets are forced to vanish by the parity condition. Nonetheless it's endowed with some extra structure, there is an odd square zero vector field on its symmetric algebra
Nov 20, 2020 at 11:11 history edited YCor CC BY-SA 4.0
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Nov 20, 2020 at 11:05 answer added Ivan Burbano timeline score: 2
S Aug 9, 2020 at 3:43 history suggested RobPratt
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Aug 9, 2020 at 3:38 review Suggested edits
S Aug 9, 2020 at 3:43
Aug 9, 2020 at 2:09 history edited Eggon Viana CC BY-SA 4.0
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Aug 9, 2020 at 1:54 comment added S. Carnahan Suppose $\mathfrak{g}$ is identified with the tangent space of a Lie group $G$ at identity. In this case, $\Pi \mathfrak{g}$ is the fiber of $\Pi TG$ over the identity element.
Aug 9, 2020 at 1:19 comment added LSpice Where do you see this notation?
Aug 9, 2020 at 1:19 history edited LSpice CC BY-SA 4.0
Proofreading
Aug 9, 2020 at 0:17 history asked Eggon Viana CC BY-SA 4.0