How much do we know about where the Balmer spectrum functor lands? Can we say it lands in schemes? ringed spaces?

I guess it lands in "spectral spaces."

@JonathanBeardsley It lands in locally ringed spaces, but the sheaf of rings isn't the same sheaf of rings that comes from the fact that they are spectral spaces

@JamesCameron Thanks! Do you know, if your input was the derived category of a commutative ring, say a Noetherian commutative ring, is the Balmer spectrum isomorphic to Spec(R) as a locally ringed space?

@K.J.Moi Interesting. If you map out of the dualizing spectrum D_X rather than tensoring with it, you can also get the additional right adjoint to to q_* for q: X->* with X finitely dominated - this adjoint is given by E |-> F_X(D_X, q^* E) for E a spectrum and F_X the internal hom.

@JonathanBeardsley This is true for the subcategory of compacts: to get that this is true as spaces you use the Hopkins/Neeman classification of thick subcategories, and then there is a comparison map between the Balmer spectrum and the Zariski spectrum of the endomorphism ring of the unit that tells you this is true as ringed spaces. There is a discussion of this stuff in Balmer's "Spectra Spectra Spectra" paper.

So in this paper arxiv.org/pdf/0801.2664.pdf a definition of "operadic module" is given. For the commutative operad, does this give bimodules or left modules? It seems like (3) of Definition 1.1, the "equivariance axiom," indicates that it gives left modules (or, rather, bimodules where the left and right action agree), but I'm not sure I'm understanding correctly. Is it clear to anyone else?

I'm trying to see if this notion is different from Lurie's "operadic modules."

Whoa Clark! :-)

Ah okay... I believe I've resolved my issue. If the operad is Ass, you get bimodules, if it's Comm, you get left modules, which I believe is compatible with Lurie's constructions.