The Spectrum of Units of Algebraic $K$-theory

Let $F$ be the cofiber of the map $Y\to\prod_{p}Y_{p}$. By the long exact sequence associated with the fiber sequence

$$\mathrm{Map}(X,F)\rightarrow\mathrm{Map}(X,Y)\rightarrow\prod_{p}\mathrm{Map}(X,Y_{p})$$

it would suffice to show that $\pi_{0}\mathrm{Map}(X,F)=0$. By the arithmetic fracture square of $Y$, we see that $F$ is a rational spectrum, and since both $Y$ and $\prod_{p}Y_{p}$ are concentrated in degrees $[m+2,\infty)$, $F$ is concentrated in degrees $[m+1,\infty)$. Since there are no non-trivial maps between rational spectra concentrated in disjoint homotopical degrees, we deduce that $[X,F]=0$ and the result follows. ∎

First, if $X$ splits then $X\simeq X[0,m]\oplus X[m+1,\infty)$. Taking $p$ completion commutes with the truncations of $X$ since $X$ has finitely generated homotopy. Hence, by $p$-completing the splitting above we obtain a splitting

$$X_{p}\simeq X_{p}[0,m]\oplus X_{p}[m+1,\infty)$$

as desired.

Conversely, note that $X$ splits if and only if the edge map $\alpha\colon X[0,m]\to\Sigma X[m+1,\infty)$ of the cofiber sequence

$$X[m+1,\infty)\rightarrow X\rightarrow X[0,m]$$

is null homotopic. By 2.1.1, this is equivalent to the vanishing of each of the compositions

$$X[0,m]\xrightarrow{\alpha}\Sigma X[m+1,\infty)\rightarrow\Sigma X[m+1,\infty)_{p}\simeq\Sigma X_{p}[m+1,\infty).$$

The above composition fits as the diagonal dashed arrow in the commutative square

in which the vertical maps are induced from the $p$-completion $X\to X_{p}$. By our assumption that $X_{p}$ is split at $m$, the bottom horizontal map labeled $\alpha_{p}$ is $0$, and hence also the diagonal dashed arrow, yielding the result. ∎

Note that since $\mathfrak{gl}_{1}^{\pm 1}R$ has finitely generated homotopy groups, $\pi_{i}\mathfrak{gl}_{1}^{\pm 1}R\to\pi_{i}(\mathfrak{gl}_{1}^{\pm 1}R)_{p}$ is a (classical) $p$-completion. In fact, since $\mathfrak{gl}_{1}^{\pm 1}R_{p}$ has finitely generated $\pi_{0}$ and $p$-complete connected cover, the same hold for it. It would therefore suffices to show that the map $\pi_{i}\mathfrak{gl}_{1}^{\pm 1}R\to\pi_{i}(\mathfrak{gl}_{1}^{\pm 1}R_{p})$ induces an isomorphism on (classical) $p$-completions. For $i=0$ this map is already an isomorphism by definition (both groups are $\{\pm 1\}$), and for $i>0$ this map can be identified with the map $\pi_{i}R\to\pi_{i}R_{p}$, which is a $p$-completion since $R$ has finitely generated homotopy groups. ∎

By the “if” direction of 2.1.2, it would suffice to show that $(\mathfrak{gl}_{1}^{\pm 1}R)_{p}$ splits at $m$ for every prime $p$. By 2.1.3, this spectrum is isomorphic to $(\mathfrak{gl}_{1}^{\pm 1}R_{p})_{p}$. By the ”only if’ direction of 2.1.2, this follows from the assumption that $\mathfrak{gl}_{1}^{\pm 1}R_{p}$ splits at $m$. ∎

For the case of $K(\mathbb{F}_{\ell})$ and $p\neq\ell$ this essentially follows from Quillen’s calculation in [10], but see Theorem 5 of [6] and the discussion preceding it.

Now we turn to $K(\mathbb{Z}[1/p])$. We learned this argument from John Rognes. Dwyer and Mitchell ([8] 1.4, cf. also [7] 8.8) show, using the work of Thomason [14], that the Lichtenbaum-Quillen conjecture implies that $K(\mathbb{Z}[1/p])\rightarrow L_{T(1)}K(\mathbb{Z}[1/p])$ is an equivalence on 1-connected covers. Since Voevodsky proved the Lichtenbaum-Quillen conjecture in [16], it remains to consider $\pi_{0}$. This follows from a calculation in etale cohomology (cf. [12] Section 2).

Note that for $p=2$ the desired result can also be deduced from the fiber square

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∎

For $(1)$, this follows from 2.2.1 and the fact that $\mathfrak{gl}_{1}$ depends only on the connective cover of a commutative ring spectrum.

For $(2)$, again by 2.2.1 we can identify the map $K(\mathbb{Z})_{p}\rightarrow L_{T(1)}K(\mathbb{Z})$ with the map

$$K(\mathbb{Z})_{p}\rightarrow K(\mathbb{Z}[1/p])_{p}.$$

Taking $p$-completion from the localization sequence

$$K(\mathbb{Z})\rightarrow K(\mathbb{Z}[1/p])\rightarrow\Sigma K(\mathbb{F}_{p})$$

and using that by Quillen’s computation $K(\mathbb{F}_{p})_{p}\simeq\mathbb{Z}_{p}$ we get the map $K(\mathbb{Z})_{p}\to K(\mathbb{Z}[1/p])_{p}$ induces an isomorphism on $\pi_{i}$ for $i\neq 1$ and an injection on $\pi_{1}$ with cokernel $\mathbb{Z}_{p}$. Hence, the same holds for the map $\mathfrak{gl}_{1}K(\mathbb{Z})_{p}\to\mathfrak{gl}_{1}K(\mathbb{Z}[1/p])_{p}$ and the result follows. ∎

By definition $\pi_{0}\mathfrak{gl}_{1}^{\pm 1}R\simeq\mathbb{Z}/2$ and $\pi_{1}\mathfrak{gl}_{1}^{\pm 1}R\simeq\pi_{1}\mathfrak{gl}_{1}R\simeq\pi_{1}R$. Since $\pi_{1}R$ is $p$-local and hence has no $2$-torsion, the map $\eta\colon\pi_{0}\mathfrak{gl}_{1}^{\pm 1}R\to\pi_{1}\mathfrak{gl}_{1}^{\pm 1}R$ is the zero map and we obtain the desired splitting by 2.3.1. ∎

Since $\pi_{1}K(\mathbb{F}_{\ell})\simeq\mathbb{F}_{\ell}^{\times}$ and $\pi_{0}K(\mathbb{F}_{\ell})\simeq\mathbb{Z}$, we see that $\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{F}_{\ell})_{2}$ has the correct homotopy groups. By 2.3.1, it remains to show that the map

$$\eta\colon\pi_{0}\mathfrak{gl}_{1}^{\pm 1}(K(\mathbb{F}_{\ell})_{2})\to\pi_{1}\mathfrak{gl}_{1}^{\pm 1}(K(\mathbb{F}_{\ell})_{2})$$

sends the non-trivial element of the group $\pi_{0}\mathfrak{gl}_{1}^{\pm 1}(K(\mathbb{F}_{\ell})_{2})\simeq\mathbb{Z}/2$ to the element $-1\in\pi_{1}\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{F}_{\ell})_{2}\simeq(\mathbb{F}_{\ell}^{\times})_{2}$.

Now, there is a canonical map of spectra

$$\mathrm{pic}(\mathbb{F}_{\ell})\to\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{F}_{\ell})\to\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{F}_{\ell})_{2}$$

which is given on $\pi_{0}$ and $\pi_{1}$ by the reduction mod $2$ and $2$-completion maps respectively:

$$\mathbb{Z}\twoheadrightarrow\mathbb{Z}/2\quad\text{and}\quad\mathbb{F}_{\ell}^{\times}\twoheadrightarrow(\mathbb{F}_{\ell}^{\times})_{2}.$$

Hence, it would suffice to show that the map $\eta\colon\pi_{0}\mathrm{pic}(\mathbb{F}_{\ell})\to\pi_{1}\mathrm{pic}(\mathbb{F}_{\ell})$ sends the generator $\Sigma\mathbb{F}_{\ell}\in\mathrm{pic}(\mathbb{F}_{\ell})$ to the element $-1\in\pi_{1}\mathrm{pic}(\mathbb{F}_{\ell})\simeq\mathbb{F}_{\ell}^{\times}$. This follows from the facts that on the Picard spectrum of a ring $S$ we have $\eta\cdot L=\dim(L)$ (see, e.g., [5, Proposition 3.20]) and that $\dim(\Sigma S)=-1\in S^{\times}$.

∎

Since $\pi_{1}K(\mathbb{Z})\simeq\mathbb{Z}/2$, by 2.3.1 it suffices to show that the map $\eta\colon\pi_{0}\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{Z})\to\pi_{1}\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{Z})$ is the non-zero map. Here, as in the case of $\mathbb{F}_{\ell}$, one can use the map $\mathrm{pic}(\mathbb{Z})\to\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{Z})$. Indeed, this map is given on $\pi_{0}$ and $\pi_{1}$ by the maps $\mathbb{Z}\twoheadrightarrow\mathbb{Z}/2$ and $\mathbb{Z}/2\simeq\mathbb{Z}/2$ respectively, so it is enough to show that $\eta$ takes the generator of $\pi_{0}\mathrm{pic}(\mathbb{Z})$ to a non-zero class. This follows as above from the fact that

$$\eta\cdot\Sigma\mathbb{Z}=-1\in\pi_{1}\mathrm{pic}(\mathbb{Z}).$$

∎

Recall that $L_{T(1)}K(\mathbb{Z})[0,\infty)\simeq K(\mathbb{Z}[1/2])_{2}$. It follows from 2.2.2 that we have a cofiber sequence

$$\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{Z})\rightarrow\mathfrak{gl}_{1}^{\pm 1}L_{T(1)}K(\mathbb{Z})\rightarrow\Sigma\mathbb{Z}_{2}.$$

Since the map $\pi_{1}L_{T(1)}K(\mathbb{Z})\rightarrow\pi_{1}\Sigma\mathbb{Z}_{2}=\mathbb{Z}_{2}$ can be identified with the $2$-completion of the $2$-adic valuation map $\mathbb{Z}[1/2]^{\times}\to\mathbb{Z}$ it is in particular a surjection. Hence, the above sequence induces a cofiber sequence

$$\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{Z})[0,1]\rightarrow\mathfrak{gl}_{1}^{\pm 1}L_{T(1)}K(\mathbb{Z})[0,1]\rightarrow\Sigma\mathbb{Z}_{2}$$

bewtween the $1$-truncations. The resulting boundary map $\Sigma\mathbb{Z}_{2}\to\Sigma\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{Z})[0,1]$ is trivial on $\pi_{1}$ and hence lifts to a map of spectra

$$\Sigma\mathbb{Z}_{2}\rightarrow\Sigma^{2}\pi_{1}\mathfrak{gl}_{1}^{\pm 1}K(\mathbb{Z})\simeq\Sigma^{2}\mathbb{Z}/2.$$

Since $\pi_{0}\mathrm{Map}_{\text{Sp}}(\Sigma\mathbb{Z}_{2},\Sigma^{2}\mathbb{Z}/2)=0$ the boundary map must vanish and the sequence splits.

∎

This is immediate. The desired map is induced by any choice of splitting $\Sigma\pi_{1}\mathfrak{gl}_{1}R=\mathfrak{sl}_{1}R[0,1]\rightarrow\mathfrak{sl}_{1}R$. ∎

The result after $p$-completion follows immediately from the first claim by taking $p$-completions from the decomposition. The map $\mathrm{pic}(S)\rightarrow\mathfrak{gl}_{1}K(S)$ restricts to a map

$$\Sigma S^{\times}\simeq\mathrm{pic}(S)[1,\infty)\rightarrow\mathfrak{gl}_{1}K(S)[1,\infty)\simeq\mathfrak{sl}_{1}K(S)$$

which is a section of the $1$-truncation map and hence induces the desired splitting. ∎

Since $R$ is $T(1)$-local, we have $\hom(\mathbb{Z},R)=0$. Hence, the result follows by applying $\hom(\mathbb{Z},-)$ to the fiber sequence

$$\mathrm{F}_{\log}R\rightarrow\mathfrak{gl}_{1}R\xrightarrow{\log}R.$$

∎

For an $L_{1}$-local commutative ring spectrum $R$, let $\mathrm{F}_{\text{ahr}}{R}$ be the fiber of the $L_{1}$-localization map $\mathfrak{gl}_{1}R\to L_{1}\mathfrak{gl}_{1}R$. By [1, Theorem 4.11], $\mathrm{F}_{\text{ahr}}(R)$ is $1$-truncated. Consider the commutative diagram where the rows are fiber sequences

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\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{ {}{}{}}{}{ {}{}{}} {{{{{}}{ {}{}}{}{}{{}{}}}}}{}{{{{{}}{ {}{}}{}{}{{}{}}}}}{{}}{}{}{}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@setlinewidth{0.39998pt}\pgfsys@invoke{ }{}{}{}{}{{}}{}{}{{}}\pgfsys@moveto{-3.85065pt}{-7.32637pt}\pgfsys@lineto{19.34937pt}{-7.32637pt}\pgfsys@stroke\pgfsys@invoke{ }{{}{{}}{}{}{{}}{{{}}}}{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{19.54935pt}{-7.32637pt}\pgfsys@invoke{ }\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}.$$

The right vertical arrow is the top row of the chromatic fracture square, so its fiber agrees with that of the bottom row of the chromatic fracture square which is a rational spectrum. Call this fiber $Q$. Note also that the fiber of the map $\mathrm{F}_{\text{ahr}}R\to\mathrm{F}_{\log}R$ agrees with $\Omega Q$.

Since $\mathrm{F}_{\text{ahr}}R$ is 1-truncated, the homotopy groups of $\mathrm{F}_{\log}R$ in degrees $\geq 3$ agree with those of $\Omega Q$, which are rational. But on the other hand, the homotopy groups of $\mathrm{F}_{\log}R$ sit in an long exact sequnce with the homotopy groups of $\mathfrak{gl}_{1}R$ and $R$, which in degrees $\geq 3$ are the homotopy groups of a $T(1)$-local ring, and hence derived $p$-complete. Hence, these groups vanish and $\mathrm{F}_{\log}R$ is 2-truncated. ∎

First, $\Omega\mathbb{G}_{m}R\simeq\hom(\mathbb{Z},\Omega\mathfrak{gl}_{1}R)$ is $p$-complete since $\Omega\mathfrak{gl}_{1}R$ is. Let $\mu_{p}(R):=\hom(\mathbb{Z}/p,\mathfrak{gl}_{1}R)$. Hence, by the long exact sequence of homotopy groups associated with the fiber sequence

$$\mu_{p}R\rightarrow\mathbb{G}_{m}R\xrightarrow{p}\mathbb{G}_{m}R,$$

the claim is equivalent to the fact that $\mu_{p}(R)$ is $1$-truncated, which follows from the vanishing of the $T(1)$-local suspension spectrum of $B^{2}\mathbb{Z}/p$ (see, e.g., [5, Proposition 4.1]). We can also argue using the logarithmic fiber: the truncatedness claim follows immediately from 3.1.2 and 3.1.3. Consider the fracture square for $L_{1}\mathfrak{gl}_{1}R$.

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Let $Q$ be the fiber of bottom row. Since there are no maps from $\mathbb{Z}$ to a $T(1)$-local spectrum, we find that $\mathrm{Map}(\mathbb{Z},L_{1}\mathfrak{gl}_{1}R)=\mathrm{Map}(\mathbb{Z},Q)=Q$ and is therefore rational. As mentioned before, the fiber $\mathrm{F}_{\text{ahr}}$ of $\mathfrak{gl}_{1}R\rightarrow L_{1}\mathfrak{gl}_{1}R$ 1-truncated. Therefore the fiber sequence

$$\mathrm{Map}(\mathbb{Z},\mathrm{F}_{\text{ahr}})\rightarrow\mathbb{G}_{m}R\rightarrow\mathrm{Map}(\mathbb{Z},L_{1}\mathfrak{gl}_{1}R)$$

shows that $\pi_{2}\mathbb{G}_{m}R$ must inject into a rational group, so it is torsion free. ∎

The general formula for the relationship between log and $\theta$ in degree 0 (i.e. between $\log_{0}$ and $\theta_{0}$) is Theorem 1.9 of [11]. In loc. sit. just below, there is a simplified formula: if $\epsilon^{2}=0$ then $\log_{0}(1+\epsilon)=\epsilon-\theta_{0}(\epsilon)$.