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I usually do computations in equivariant homotopy theory, but I would like to learn chromatic homotopy theory where one may use the equivariant techniques, e.g., slice spectral sequences, etc.

For this, I am looking for those papers which are dealt with the above kind of literature.

Any reference will be highly appreciated.

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  • $\begingroup$ Several results used equivariant methods in the context of algebraic K theory, and in particular for work related to the redshift conjecture (this is where chromatic homotopy enters). Not sure if they use the kind of tools you are looking for. For example arxiv.org/abs/2011.08233 and arxiv.org/abs/1606.03328 $\endgroup$
    – Shay Ben Moshe
    Dec 25, 2020 at 16:09
  • $\begingroup$ @ShayBenMoshe: Thank you for the comment. I will go through these articles. $\endgroup$
    – Surojit Ghosh
    Dec 26, 2020 at 2:34

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The canonical answer to this question is of course the celebrated solution by Hill, Hopkins and Ravenel to the Kervaire invariant one problem

Hill, Michael A., Michael J. Hopkins, and Douglas C. Ravenel. "On the nonexistence of elements of Kervaire invariant one." Annals of Mathematics (2016): 1-262 (arXiv:0908.3724).

Beside that amazing paper, equivariant homotopy theory can be used in the computation of Picard groups of some local categories. For example the following paper uses $C_4$-equivariant homotopy theory to study the Picard group of the $K(2)$-local category

Beaudry, Agnes, Irina Bobkova, Michael Hill, and Vesna Stojanoska. "Invertible $ K (2) $-Local $ E $-Modules in $ C_4 $-Spectra." (arXiv:1901.02109) (2019).

In both cases the intuition is that equivariant techniques are useful to run descent arguments along a Galois extension (typically by some small subgroup of the Morava stabilizer group). For example the slice spectral sequence allows one to resolve Borel $G$-spectra (i.e. "spectra with a $G$-action") by pieces that are not Borel anymore but which are "simpler" in some sense.

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  • $\begingroup$ Thank you so much, Denis, for this beautiful answer. $\endgroup$
    – Surojit Ghosh
    Dec 24, 2020 at 15:44

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