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Feb 8, 2023 at 12:57 comment added Neil Strickland This was asked at mathoverflow.net/questions/431226/… and answered in the comments. The category of $G$-equivariant spectra (like any stable $\infty$-category) is enriched over the category of spectra, so you can just define $EA=E\wedge M(A)$ again.
Feb 7, 2023 at 18:03 comment added anon No, I want to see whether the generalisation of spectrum with coefficients is possible in the equivariant case. I am curious whether there is such a thing as an equivariant spectrum with coefficients.
Feb 7, 2023 at 17:43 comment added user43326 Do you want $G$ go act on $A$? In that case there are obstructions, cf. jstor.org/stable/2154066
S Feb 7, 2023 at 13:39 review First questions
Feb 7, 2023 at 13:44
S Feb 7, 2023 at 13:39 history asked anon CC BY-SA 4.0