8:02 PM
@TylerLawson oh -- I guess so. I was worried about a $lim^1$ term but I guess I don't have to worry since by assumption all the mapping spaces in the $holim$ are zero.
Anyway, if Saal's conjecture holds, and the dissonant spectra are exactly those in the colimit closure of shifts of $H\mathbb Z$, then the harmonic spectra would be exactly those spectra $X$ such that $F(H\mathbb Z, X) = 0$.
As it is, we know that being harmonic implies this condition.
Which is already striking -- it hadn't occurred to me before that there can be no nonzero map $H\mathbb Z \to \Sigma^\infty Y$ for a space $Y$, for instance.