## Values of the multiplicative group over a ring spectrum

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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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## 1 Answer

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 2010 at 14:24

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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 2010 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 2010 at 18:05
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