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In his notes on elliptic cohomology, Lurie defines the multiplicative group $\mathbb{G}_m$ over a ring spectrum $A$ as $\operatorname{Spec} A[\mathbb{Z}]$. What is the value $\mathbb{G}_m(B)$ of the represented functor at an $A$-algebra $B$? If this is too hard to say in general: Are there at least any specific examples, other than Eilenberg-MacLane spectra, where one does know the answer?

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It is slightly complicated.

One has a number of adjunctions: $$ \begin{eqnarray*} \mathbb{G}_m(B) &=& Hom_{A-alg}(A[\mathbb{Z}],B) \\ &\simeq& Hom_{E_\infty-rings}(\mathbb{S}[\mathbb{Z}],B)\\ &\simeq& Hom_{E_\infty-spaces}(\mathbb{Z},GL_1(B))\\ &\simeq& Hom_{spectra}(H\mathbb{Z},gl_1(B)). \end{eqnarray*} $$ (Note these adjunctions are weak equivalences of spaces, and the last two adjunctions require a fair amount of theory to make rigorous.)

The problem is that it is usually quite difficult to compute the maps out of the Eilenberg-Mac Lane spectrum $H\mathbb{Z}$ unless the target is also an Eilenberg-Mac Lane space. In the case where the algebra $B$ comes from a simplicial commutative ring, this is true and so one at least knows that the set of homotopy classes of maps $[H\mathbb{Z}, gl_1(B)]$ surjects onto $\pi_0(B)^\times$. Even for complex K-theory, the calculation is somewhat involved (but doable), but the only method that I can immediately think of involves the Bousfield-Kuhn functor.

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I forgot an important special case of this. If $\pi_0(B)$ is a $\mathbb{Q}$-algebra, then $\mathbb{G}_m(B)$ is isomorphic to $\pi_0(B)^\times$ after passing to homotopy classes of maps. – Tyler Lawson Aug 13 '10 at 14:24
Tyler, can't we say that if $\pi_0(B)$ is a $\mathbb{Q}$-algebra, then $G_m(B)\simeq GL_1(B)$? – Charles Rezk Aug 13 '10 at 17:52
@Charles: Well, yes. I guess I should have decided once and for all whether I was talking about spaces of maps or homotopy classes of maps – Tyler Lawson Aug 13 '10 at 18:05
Tyler, Margolis shows it is not hard to compute maps out of $HF_p$: the Adams spectral sequence collapses to the 0-line because the Steenrod algebra is injective in the category of bounded below modules. From there, HZ is a Bockstein away. See… – Robert Bruner Apr 11 '15 at 18:11
Bob, that's a lovely result. (I suppose that things like this are predecessors to the Sullivan conjecture.) From that point I guess one needs tools for computing the cohomology of $gl_1(B)$, which may pick up some nontrivial Steenrod algebra summands where $B$ had none. – Tyler Lawson Apr 11 '15 at 20:06

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