Proof of the equivalence of spectra $(\mathbb{S}^{-1} \otimes \mathbb{S}^{-1})_{h \Sigma_2} \cong \Sigma^{-1} \mathbb{RP}_{-1}^{\infty}$

Question

$\DeclareMathOperator{\colim}{colim}$$\DeclareMathOperator{\Th}{Th}$I am trying to give a hands-on proof of the equivalence of spectra in the title. I am using the definitions $\mathbb{RP}^{\infty}_{-1} := \colim_k \mathbb{RP}_{-1}^{k}$ and $\mathbb{RP}_{-1}^{k} := \Th(- \mathbb{L}, \mathbb{RP}^{k+1})$ where $\mathbb{L}$ denotes the canonical line bundle on $\mathbb{RP}^{k+1}$. Moreover, if $E$ is a vector bundle over a compact Hausdorff base $B$, then I use the definition $\Th(-E,B) := \Sigma^{-n} \mathrm{Th}(E', B)$, where $E'$ is a vector bundle over $B$ such that $E \oplus E'$ is trivial of rank $n$. Such a complementary bundle exists by Proposition 1.4 in Hatcher's book on K-theory for example.

In the case of the canonical line bundle on $\mathbb{RP}^{k}$, I think we can find explicit complements as follows. I read (see Adams' paper on vector fields on spheres for example) that the virtual bundle $\mathbb{L} -1$ generates $\widetilde{\text{KO}}(\mathbb{RP}^k) \cong \mathbb{Z}/2^{f_k} \mathbb{Z}$, where $f_k$ is the number of integers $i$ such that $0 < i ≤ k$ and $i ≡ 0, 1,2$ or $4\mod 8$. So $2^{f_k}(\mathbb{L} -1) = 0$ and I think the definition of $\widetilde{\text{KO}}$ now implies that $2^{f_k} \mathbb{L} + n_k \mathbb{1} \cong (2^{f_k}+n_k) \mathbb{1}$ for some $n_k \geq 0$.

My question is if we can give a proof of the equivalence $(\mathbb{S}^{-1} \otimes \mathbb{S}^{-1})_{h \Sigma_2} \cong \Sigma^{-1} \mathbb{RP}_{-1}^{\infty}$ in this setting. The fact that Thom spaces of universal bundles can be identified with representation spheres will probably be helpful (applied to the universal $\Sigma_2$-bundle $S^{\infty} \to \mathbb{RP}^{\infty}$). Somehow I keep running into trouble when trying to write down a proof.

Hopefully my question is clear enough. If you can't answer my question but know a different proof of this equivalence (perhaps in a different setting) I would also be interested in that. Thanks in advance!

Answer 1

This is probably not quite what you were looking for, but since you said you'd also be happy to see any proof, here is one using a modern perspective on Thom spectra of virtual bundles:

The flip action of $C_2$ on $(\mathbb{S}^n)^{\otimes 2}$ is a functor $BC_2\to Sp$ which, for nonnegative $n$, factors through the symmetric-monoidal functor $\mathrm{Vect}^{\cong} \to \mathrm{Sp}$ given by taking a vector space $V$ to the suspension spectrum on its one-point compactification $S^V$. Here $\mathrm{Vect}^{\cong}$ is the topological groupoid of real vector spaces with isomorphisms (with the usual topology on $\mathrm{GL}_n$).

The lift to vector spaces is given by the flip action on $\mathbb{R}^n\oplus\mathbb{R}^n$.

The definition of Thom spaces via fibrewise 1-point compactification tells you that the Thom spectrum of a vector bundle, encoded as functor $X\to\mathrm{Vect}^{\cong}$, is obtained by postcomposing with the functor to $\mathrm{Sp} $, and taking the colimit over $X$. For $X=BC_2$, this colimit has another name: it's the homotopy orbits, giving you the equivalence $$ \mathrm{Th}(BC_2,\mathbb{R}^n\oplus\mathbb{R}^n) \simeq ((\mathbb{S}^n)^{\otimes 2})_{hC_2} $$ for nonnegative $n$.

What happens for negative $n$? The crucial fact is that the symmetric-monoidal functor $\mathrm{Vect}^{\cong} \to \mathrm{Sp}$ factors through the group completion of the left hand side, since it takes values in invertible objects on the right. This group completion can be identified with $BO\times\mathbb{Z}$, but we don't need that here. Even if we just call it $G$, we get that we still have Thom spectra for maps $X\to G$ (as colimit), and that for any such map $X\to G$ which arises as formal difference of an actual vector bundle and a trivial vector bundle, the Thom spectrum is a desuspension. Now $G$ contains a point/object "$\mathbb{R}^{-1}$", and from the symmetric-monoidal structure a map $BC_2\to G$ encoding "the flip action on $(\mathbb{R}^{-1})^{\oplus 2}$". This has Thom spectrum given by $(\mathbb{S}^{-1})^{\otimes 2}_{hC_2}$ as before, but the map $BC_2\to G$ is also the negative of the actual vector bundle we get for $\mathbb{R}^1$. In the $n=1$ case we are dealing with the vector bundle corresponding to the regular representation of $C_2$, which splits into a trivial and a sign representation. So the vector bundle splits as $1\oplus L$. The Thom spectrum we found for $n=-1$ is thus $\mathrm{Th}(BC_2,-1-L)$. The equivalence $$ \mathrm{Th}(BC_2,-1-L)\simeq (\mathbb{S}^{-1})^{\otimes 2}_{hC_2} $$ is a shift of the one we were looking to prove.

By definition as a colimit the left hand side can also be written a colimit over skeleta of $BC_2$, as $$ \mathrm{colim}_k \mathrm{Th}(\mathbb{R}P^k,-1-L), $$ and on each of those a complement to $L$ allows you to express the Thom spectrum as an actual desuspension, relating it to the hands-on construction. The advantage is that you don't have to produce compatible maps for those by hand, they come out of the direct description of the Thom spectrum.