Homotopy fixed points of complex conjugation on $BU(n)$

Question

Stably it is known that $(\mathbb Z\times BU)^{hC_2}\simeq \mathbb Z\times BO$ holds. The homotopy fixed point spectral sequence for KU with complex conjugation action can be completely calculated and spits out the 'correct homotopy' groups. But from that point of view this seems like a computational fact, using real Bott periodicity. If one uses the full genuine C_2-homotopy type of $KU$ then i think that one can also get the above equivalence without a priori knowledge of real Bott periodicity.

So i'm wondering how much calculational input this really needs and what happens unstably. There are of course models for $BU(n)$ with $BU(n)^{C_2}=BO(n)$, but what about the homotopy fixed points. Is there an equivalence $BU(n)^{hC_2}\simeq BO(n)$ ?

Answer 1

I think the answer is yes, after Bousfield-Kan $2$-completion. For $n=1$, $BO(1) \to BU(1)^{hC_2}$ is an equivalence, since $BO(1) \simeq K(\mathbb{Z}/2, 1)$, while $BU(1) \simeq K(\mathbb{Z}(1), 2)$ where the generator of $C_2$ reverses the sign in $\mathbb{Z}(1)$, and $H^i(C_2; \mathbb{Z}(1)) = (0, \mathbb{Z}/2, 0, \dots)$. The cases $n\ge2$ then follow by induction over $n$ from Carlsson's form of the Sullivan conjecture (Invent. Math. 1991). To see this, use the $C_2$-equivariant homotopy fiber sequence $$ S(\mathbb{C}^n) \to BU(n-1) \to BU(n) $$ where $S(\mathbb{C}^n) = S^{2n-1}$ is the unit sphere in $\mathbb{C}^n$, with $C_2$-fixed points $S(\mathbb{R}^n) = S^{n-1}$. We get a vertical map of homotopy fiber sequences from $$ S(\mathbb{R}^n) \to BO(n-1) \to BO(n) $$ to $$ (S(\mathbb{C}^n))^{hC_2} \to BU(n-1)^{hC_2} \to BU(n)^{hC_2} \,. $$ Carlsson proves that the left hand vertical map is a $2$-complete equivalence. By induction the middle vertical map is a $2$-complete equivalence. Hence the right hand vertical map is also a $2$-complete equivalence. (I have not carefully checked what happens at $\pi_0$.)