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Apr 3, 2022 at 13:22 vote accept curious math guy
Apr 3, 2022 at 11:10 comment added Marc Hoyois @Z.M You can see this by doing factorization homology in $E_{n-1}$-algebras. An $E_n$-algebra is an $E_1$-algebra in $E_{n-1}$-algebras, so you can integrate it over the circle to get an $E_{n-1}$-algebra (with $S^1$-action).
Apr 3, 2022 at 9:49 comment added Fernando Muro @z.m. It shifts n by -1.
Apr 3, 2022 at 9:42 comment added Z. M What happens if the input is $E_n$ for $n\ge2$?
Apr 3, 2022 at 9:33 history edited Marc Hoyois CC BY-SA 4.0
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Apr 3, 2022 at 6:02 history answered Marc Hoyois CC BY-SA 4.0