Timeline for $C_2$-equivariant Betti realization of MGL

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Jun 19, 2018 at 22:48	comment	added	Denis Nardin		@SeanTilson Late to the party... I don't know if it is obvious but it is true, since $KGL$ is the motivic spectrum represented by the $\mathbb{P}^1$-spectrum that has $BGL_\infty\times\mathbb{Z}$ in each degree. This quite clearly realizes to $BU\times\mathbb{Z}$ with the conjugation action and so the corresponding spectrum is $K\mathbb{R}$
Oct 13, 2017 at 9:33	comment	added	Sean Tilson		Similarly, is it obvious that $KGL$ realizes to $K\mathbb{R}$?
Oct 13, 2017 at 8:12	comment	added	Lennart Meier		Have you tried arguing directly from the definitions? The building blocks $BGL(n)$ of $MGL$ realize to $BGL(n,\mathbb{C})$, which is equivalent to $BU(n)$; Thom spaces are quotients (i.e. colimits) and thus preserves by colimits etc.
Oct 12, 2017 at 9:00	history	edited	Drew Heard	CC BY-SA 3.0	added 3 characters in body
Oct 12, 2017 at 8:51	history	edited	Drew Heard	CC BY-SA 3.0	edited title
Oct 12, 2017 at 8:23	history	edited	Drew Heard	CC BY-SA 3.0	Changed title
Oct 12, 2017 at 7:29	history	asked	Drew Heard	CC BY-SA 3.0