# On a question motivating Lurie's treatment of formal moduli problems

Lurie, in his ICM 2010 proceedings paper Moduli Problems for Ring Spectra (pdf), says that one motivating problem (for him, I presume, and possibly others) for thinking about formal moduli problems is this:

...the initial object in the $\infty$-category $CAlg(Sp)$ is given by the sphere spectrum $S$, rather than the discrete ring $\mathbb{Z} \simeq \pi_0(S)$. Therefore it makes sense to ask if the algebraic group $G_\mathbb{C}$ can be defined over the sphere spectrum; it was this question which originally motivated the theory described in this paper.

Before this excerpt, he discusses how any reductive algebraic group over $\mathbb{C}$ descends to a split reductive group scheme over $\mathbb{Z}$, such that the original arises via base change. Thus the natural question whether in the category of derived group schemes over $E_\infty$ ring spectra the reductive group descends down $Spec(\pi_0)\colon Spec(\mathbb{Z}) \to Spec(S)$.

Has this been achieved/is this known?

Yes, this has been achieved in some sense. There is a (unpublished and possibly not yet written) work of Gaitsgory and Lurie where they propose an answer to this question. Given a split reductive group scheme $G$ over $Spec(\mathbb{Z})$ they use a version of the geometric Satake correspondence to construct a stable infinity category which is linear over the sphere spectrum and base changes to the category of representations of $G$. The idea is to look at Whittaker sheaves of spectra over the affine Grassmanian for the complex Langlands dual group. This almost gives you what you want. A small caveat is that the convolution product only gives a braided monoidal structure on this category of spectral representations and they expect that the braiding is unavoidable. So there is no cheap Tannaka duality answer but you get as close to it as you can.
• Ah, so via some version of Tannaka you get something, which isn't quite a group (some sort of derived quantum group?) such that its base change to $\mathbb{Z}$ is a group. That sounds reasonable. – David Roberts Apr 3 '14 at 22:06