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Let $MGL$ denote the motivic spectrum representing algebraic cobordism. Over $\mathop{Spec}(\mathbb{C})$ there is a Betti realization functor $\mathop{SH}(\mathbb{R}) \to \mathop{SH}$, which takes $MGL$ to $MU$.

Over $\mathop{Spec}(\mathbb{R})$ there is a $C_2$-equivariant Betti realization functor $\mathop{SH}(\mathbb{R}) \to \mathop{SH}(C_2)$. The analog of $MU$ in the $C_2$-equivariant homotopy category is $M\mathbb{R}$, the real bordism spectrum of Araki, Landweber, and Hu-Kriz, and it seems logical that the equivariant Betti realization takes $MGL$ to $M\mathbb{R}.$

Is there a reference that proves that the $C_2$-equivariant Betti realization of $MGL$ is $M\mathbb{R}$?

I have seen it claimed that this is proved in work of Hu and Kriz, but I was unable to find it there (that is not to say that it I didn't miss something!).

Note that it is known (for example, by work of Heller and Ormsby) Betti realization commutes with colimits. Since $MGL$ is a colimit of certain Thom spectra, one can reduce to checking the equivariant Betti realization of these.

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    $\begingroup$ 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. $\endgroup$
    – Lennart Meier
    Commented Oct 13, 2017 at 8:12
  • $\begingroup$ Similarly, is it obvious that $KGL$ realizes to $K\mathbb{R}$? $\endgroup$
    – Sean Tilson
    Commented Oct 13, 2017 at 9:33
  • $\begingroup$ @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 $K\mathbb{R}$ $\endgroup$
    – Denis Nardin
    Commented Jun 19, 2018 at 22:48

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