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Sep 26, 2022 at 12:19 comment added Neil Strickland Yes, the category $\mathcal{S}_G$ of $G$-spectra is enriched over the category $\mathcal{S}$ of spectra, so for $E\in\mathcal{S}_G$ and $M\in\text{Ab}$ you can define the Moore spectrum $SM\in\mathcal{S}$ and then $EM=E\wedge SM\in\mathcal{S}_G$.
Sep 25, 2022 at 23:57 history asked user340793 CC BY-SA 4.0