TITLE: $K$ theory and singular cohomology
QUESTION [9 upvotes]: For cell complexes${}^1$ $X$ we have an isomorphism
$$
K^*(X)\otimes \mathbb{Q}\cong H^{*}(X;\mathbb{Q}),
$$
which is induced by the Chern character. 
What is the analogous statement for $KO(X)$?
${}^1$:Hatcher states finite, but I've seen arbitrary CW-complexes stated as well.
edit: The footnote seems wrong, as by the comments.

REPLY [12 votes]: In the following, $X$ is a finite complex. The Adams operator $\psi^{-1}$ (complex conjugation) acts on $K^0(X)$. After inverting $2$, the group $KO^0(X) \otimes \Bbb Z[1/2]$ maps isomorphically to the subset of $K^0(X) \otimes \Bbb Z[1/2]$ fixed by $\psi^{-1}$.
We can then tensor this with $\Bbb Q$. Then $K^0(X) \otimes \Bbb Q \cong \prod_{n \geq 0} H^{2n}(X;\Bbb Q)$. The operator $\psi^{-1}$ acts on this by fixing the factors with $n$ even and negating the factors with $n$ odd. The fixed set is $KO^0(X) \otimes \Bbb Q \cong \prod_{m \geq 0} H^{4m}(X; \Bbb Q)$.