# The K(1)-local Spanier-Whitehead dual of KO

Let $D_1KO$ be the $K(1)$-local Spanier-Whitehead dual of $KO$, i.e. the spectrum $$D_1KO = F(KO,L_{K(1)}S^0).$$

I am interested in what this is. In fact I know that $D_1KO = \Sigma^{-1} KO$. One way to see this is to use the fact that the Gross-Hopkins dual $I_1 = \Sigma^2 P$ for a spectrum $P$ such that $P \wedge KO \simeq \Sigma^4 KO$, along with the equivalence $$I_1KO = D_1KO \wedge I_1.$$

Here is a claimed direct proof, due to Hahn and Mitchell (see Lemma 8.16).

Start with the cofiber sequence $$\Sigma^{-1} KO \to \Sigma^{-1} KO \xrightarrow{\delta} L_{K(1)}S^0,$$ and define a map $$\phi:\Sigma^{-1}KO \to F(KO,\Sigma^{-1}KO) \xrightarrow{\delta_*} F(KO,L_{K(1)}S^0),$$ where the first map is the adjoint to the (desuspension) of the multiplication map. The claim is that $\phi$ induces a weak equivalence on homotopy groups.

How does one show this? Namely, how do we compute $\pi_*F(KO,L_{K(1)}S^0)$?

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If everything is completed at an odd prime $p$, and $k$ is a topological generator of $\mathbb{Z}_p^\times$, then $L_{K(1)}S$ is the fibre of $\psi^k-1\colon KU\to KU$. This means that $F(KU,L_{K(1)}S)$ is the fibre of the corresponding self-map $(\psi^k-1)_*$ of $F(KU,KU)$, which is not too hard to understand. A slight adjustment of this should work for $KO$ at $p=2$, but I forget what is the best way to organise it.
Thanks Neil! I was being a bit lazy with my notation - everything is really at the prime 2. In this case $F(KO,L_{K(1)}S^0)$ is the fibre of the self-map $(\psi^3-1)_\ast$ of $F(KO,KO)$ so I guess this is the map to understand? –  Drew Heard Apr 30 '14 at 9:36